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# FRM Practice Questions
1. A share ‘X’ is currently selling at Rs. 75. The risk free rate of interest is
Rs.9% per annum. What should be the fair contract price of a 2 month
futuress contract?
2. A share ‘Y’ is available at Rs.100. The risk free rate of interest if 8.5%
compounded continuously. The share is expected to yield a dividend of
Rs.2.50 in 1 month from now. Determine the value of a 3 month
futuress contract if one contract involves 2000 shares.
3. Consider a 10 month forward contract on a stock with a price of \$50.
We assume that the risk free rate of interest continuously compounded
is 8% per annum. We also assume that dividends of \$0.75 per share are
expected after 3 months, 6 months and 9 months. Calculate the
forward price.
4. A 1 year long forward contract on a non-dividend paying stock is
entered into when the stock price is Rs.40 & the risk free rate of
interest rate is 10% per annum with continuously compounding.
i. What are the forward price &the initial value of the forward
contract?
ii. What will be the forward price 6 months later, if the price of the
stock is Rs.45 & the risk free interest is still 10%?
5. The company ‘B’ rated ‘BBB’ by standard & Poor’s has the opportunity
to borrow either at a fixed rate of 8% or at a floating rate of LIBOR+75
basis points. ‘A’ corporation which is rated ‘AAA’ by standard & Poor’s,
it can borrow either in long term fixed coupon debt market at 7% or at
a floating rate of LIBOR+25 basis points.
i. You are required to arrange a swap between them so as to lower
their cost of borrowing.
ii. Is the swap is contracted with swap dealers & the swap dealer
charges 6 basis points per year, show the benefit of the swap to
both the parties.
6. Company ‘A’ &’B’ have been offered the following rates per annum on
Rs.100 million loan for 7 years. Company A requires a floating rate loan
& B requires fixed rate loan.
Fixed rate Floating rate
Company A 7% LIBOR+25 basis points
Company B 8% LIBOR+75 basis points
i.Design a swap that will appear equally attractive to both companies.
How swaps lower the borrowing costs for both companies.
ii.Design a swap that will net a swap dealer, acting as intermediaries
earn a profit of 6 basis points that will appear equally attractive to
both companies.
7. From the following data, calculate the values of call & put options by
using B & S model. The current price of the share is Rs.125.94, Exercise
price is Rs.125. Time to expiry is 35 days. Standard deviation is 0.83 &
continuously compounded rate of return is 4.56% per annum. There are
no dividends on the stock.
8. An investor wants to earn by writing a call option. The current price of
the stock is Rs.28 & he wants to write a 4 months call option with the
strike price of Rs.30. The investor wants to determine the appropriate
premium to charge for the call option. The stocks standard deviation is
assumed to be 30%. The risk free rate interest rate is assumed to be
10%. Determine the call option premium.
9. A stock is currently selling for 58.875. The risk less interest rate is 8%
per year. Estimate the value of a call option using B&S model with a
strike price of 60 & the time to expiration of 3 months. The standard
deviation of the stocks annual returns is 0.22. Also calculate the put
option price using put-call parity.
10. The current price of a share is Rs.50. It is believed at the end of one
month the price will be (55/-) / (45/-). What will a European call option
with an exercise price of 53/- be valued? The risk free rate of interest is
15% per annum continuously compounded. Calculate the hedge ratio
also.
11. Share ‘X’ is currently available ay Rs. 100. The risk fee rate of interest is
8% per annum compounded continuously. What should be the ideal
contract price of one month futures contract?
12. A forward contract on 200 shares, currently trading at Rs.112 per share,
is due on 45 days, if the annual risk free rate of interest is 9% calculate
the value of the contract price. How would the value be changed if a
dividend of Rs.4 per share is expected to be paid in 25 days before due
date.
13. A certain share index provides a dividend yield of 3.5% per annum. The
current value of the index is 1003. The continuously compounded risk
free rate of return is 8%.
i. Find the value of a one month futures contract on the given
index per unit
ii. Find the value of a one month futures contract on the given
index assuming that each contract has 200 units.
14. A share is currently available at Rs.100. The risk free rate of interest is
9% per annum compounded quarterly. What should be the fair price of
a 45 days futures contract?
15. Using the data given below obtain the value of a futures contract to an
index:
Spot value of index= 1216
Risk free rate of return= 7% p.a.
Time to expiration= 146 days
Contract multiplier= 200
16. A stock index is currently at 820. The continuously compounded risk
free rate of return is 9% p.a. & the dividend yield on the index is 3% p.a.
What should the futures price for a contract with 3 months to
expiration be?
17. The current spot price of a 100rupee share is 802.60. Obtain the fare
price of December futures contract on this share assuming the risk free
rate return to be 9% & the market lot size as 250. The maturity fate is
73 days from today. How would the value of the contract be affected if
a dividend of 8% is expected in 30 days time?
18. The risk free rate of interest is 6% p.a. with continuous compounding,
and the dividend yield n a stock index is 3.2% p.a. The current value of
index is Rs. 4,400. What is the 6 months futures price?
19. The 2 month interest rate in Switzerland & India are, respectively 3% &
6% p.a. with continuous compounding. The spot price of the Swizz franc
is Rs. 33,778 & the 2onth forward rate is Rs. 33,924. What arbitrage
opportunities does this create? | 1,509 | 5,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-39 | latest | en | 0.940994 |
https://www.physicsforums.com/threads/coordinate-representation-of-the-momentum-operator.614132/ | 1,537,678,766,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159006.51/warc/CC-MAIN-20180923035948-20180923060348-00044.warc.gz | 839,758,227 | 13,498 | # Coordinate representation of the momentum operator
1. Jun 15, 2012
### MSLion
The position operator in coordinate representation is:
Xab=aδ(a-b)
this is diagonal as expected
The momentum operator turns out to look like
Pab=-ih∂aδ(a-b)
Now, this is not supposed to be diagonal because it does not commute with X.
However it looks pretty diagonal to me.
What am I missing?
2. Jun 15, 2012
### haael
The momentum operator is not what you wrote here. It is equal to:
$$P_{ab} = e^{i(a-b)}$$
with some scaling factors.
3. Jun 15, 2012
### MSLion
What exactly are you saying?
Anyone familiar with quantum mechanics in dirac notation?
4. Jun 15, 2012
### geoduck
I don't see how it looks diagonal. For something to be diagonal, it needs to be multiplied by a delta function. An example of a diagonal matrix is:
Oab=δ(a-b) f(a,b)
for arbitrary function f(a,b).
The delta function is essentially the identity matrix, and f(a,b) are the diagonal components.
5. Jun 15, 2012
### MSLion
From what I understood on distributions the support of the derivative of a distribution is contained within the support of the original distribution.
Therefore ∂aδ(a-b) should be zero wherever δ(a-b) is.
6. Jun 15, 2012
### fzero
Distributions are supported on functions $f(a)$, not on the domain of the functions themselves. Given a function $f(a)$, we have $\delta[f]=f(0)$ and $\delta'[f]=f'(0)$. These are generally not equal. The same argument can be made for delta functions which are shifted away from the origin.
$\delta'$ is as independent of $\delta$ as an ordinary function is from its derivative. There's no unitary operator that can map one to another (for all test functions $f$).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 458 | 1,781 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-39 | latest | en | 0.905336 |
https://studyrankersonline.com/2069/polynomial-power-divided-another-polynomial-power-remainder | 1,566,515,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317516.88/warc/CC-MAIN-20190822215308-20190823001308-00235.warc.gz | 646,845,350 | 12,016 | # If the polynomial x(power 4)-6x(power 3)+16x(power 2)-15x+10 is divided by another polynomial x(power 2)-2x-k , the remainder comes out to be (x+a) . Find k and a ?
65 views
+1 vote
answered Jun 7, 2016 by Basic (37 points)
PROVING BY REMAINDER THEOREM:
X4-6X3+16X2-15X+10 = (x2-2x-k)(q(x)) +(x+a)
[any number=divisor x quotient + remainder] {q(x) is a function of x which i am considering as the quotient}
X4-6X3+16X2-15X+10-x-a = (x2-2x-k)(q(x))
X4-6X3+16X2-16X+10= (x2-2x-k)(q(x))+a
When x=2,
16-48+64-32+10=(4- 4-k)(q(x))+a
-22 = -k(q(x))+a
• a=k(q(x))-22
When x= -k,
solve it and divide the equations the q(x) will be eliminated and solve the rest coz i am still figuring this out
GOOD LUCK
+1 vote
+1 vote | 294 | 732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-35 | latest | en | 0.748375 |
https://www.dataunitconverter.com/kibibyte-per-second-to-gibibit-per-day | 1,701,541,036,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00667.warc.gz | 809,351,165 | 16,492 | # KiBps to Gibit/Day → CONVERT Kibibytes per Second to Gibibits per Day
info 1 KiBps is equal to 0.6591796875 Gibit/Day
Input Kibibytes per Second (KiBps) - and press Enter.
KiBps
You are converting .
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
## Kibibytes per Second (KiBps) Versus Gibibits per Day (Gibit/Day) - Comparison
Kibibytes per Second and Gibibits per Day are units of digital information used to measure storage capacity and data transfer rate.
Both Kibibytes per Second and Gibibits per Day are the "binary" units. One Kibibyte is equal to 1024 bytes. One Gibibit is equal to 1024^3 bits. There are 131,072 Kibibyte in one Gibibit. Find more details on below table.
Kibibytes per Second (KiBps) Gibibits per Day (Gibit/Day)
Kibibytes per Second (KiBps) is a unit of measurement for data transfer bandwidth. It measures the number of Kibibytes that can be transferred in one Second. Gibibits per Day (Gibit/Day) is a unit of measurement for data transfer bandwidth. It measures the number of Gibibits that can be transferred in one Day.
## Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day) Conversion Formula and Steps
KiBps to Gibit/Day Calculator Tool allows you to easily convert from Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day). This converter uses the below formula and steps to perform the conversion.
The formula of converting the Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day) is represented as follows :
diamond Gibit/Day = KiBps x 8 ÷ 10242 x 60 x 60 x 24
Source Data Unit Target Data Unit
Kibibyte
Equal to 1024 bytes
(Binary Unit)
Gibibit
Equal to 1024^3 bits
(Binary Unit)
The conversion diagram below will help you to visualize the steps involved in calculating Kibibyte to Gibibit in a simplified manner.
÷ 1024
÷ 1024
x 8
Mebibyte [MiB]
Gibibyte [GiB]
x 1024
x 1024
÷ 8
Now let us apply the above formula and see how to manually convert Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day). We can further simplify the formula to ease the calculation.
FORMULA
Gibibits per Day = Kibibytes per Second x 8 ÷ 10242 x 60 x 60 x 24
STEP 1
Gibibits per Day = Kibibytes per Second x 8 ÷ (1024x1024) x 60 x 60 x 24
STEP 2
Gibibits per Day = Kibibytes per Second x 8 ÷ 1048576 x 60 x 60 x 24
STEP 3
Gibibits per Day = Kibibytes per Second x 0.00000762939453125 x 60 x 60 x 24
STEP 4
Gibibits per Day = Kibibytes per Second x 0.00000762939453125 x 86400
STEP 5
Gibibits per Day = Kibibytes per Second x 0.6591796875
Example : If we apply the above Formula and steps, conversion from 1 Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day) will be processed as below.
1. = 1 x 8 ÷ 10242 x 60 x 60 x 24
2. = 1 x 8 ÷ (1024x1024) x 60 x 60 x 24
3. = 1 x 8 ÷ 1048576 x 60 x 60 x 24
4. = 1 x 0.00000762939453125 x 60 x 60 x 24
5. = 1 x 0.00000762939453125 x 86400
6. = 1 x 0.6591796875
7. = 0.6591796875
8. i.e. 1 KiBps is equal to 0.6591796875 Gibit/Day.
Note : Result rounded off to 40 decimal positions.
You can use above formula and steps to convert Kibibytes per Second to Gibibits per Day using any of the programming language such as Java, Python or Powershell.
### Unit Definitions
#### Kibibyte
A Kibibyte (KiB) is a binary unit of digital information that is equal to 1024 bytes (or 8,192 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'kibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'kilobyte' (KB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
arrow_downward
#### Gibibit
A Gibibit (Gib or Gibit) is a binary unit of digital information that is equal to 1,073,741,824 bits and is defined by the International Electro technical Commission(IEC). The prefix 'gibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'gigabit' (Gb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day)
Apply the formula as shown below to convert from 1 Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day).
A B C
1 Kibibytes per Second (KiBps) Gibibits per Day (Gibit/Day)
2 1 =A2 * 0.00000762939453125 * 60 * 60 * 24
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kibibytes per Second (KiBps) to Gibibits per Day (Gibit/Day) Conversion
You can use below code to convert any value in Kibibytes per Second (KiBps) to Kibibytes per Second (KiBps) in Python.
kibibytesperSecond = int(input("Enter Kibibytes per Second: "))
gibibitsperDay = kibibytesperSecond * 8 / (1024*1024) * 60 * 60 * 24
print("{} Kibibytes per Second = {} Gibibits per Day".format(kibibytesperSecond,gibibitsperDay))
The first line of code will prompt the user to enter the Kibibytes per Second (KiBps) as an input. The value of Gibibits per Day (Gibit/Day) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Gibibits(Gibit) are there in a Kibibyte(KiB)?expand_more
There are 0.00000762939453125 Gibibits in a Kibibyte.
#### What is the formula to convert Kibibyte(KiB) to Gibibit(Gibit)?expand_more
Use the formula Gibit = KiB x 8 / 10242 to convert Kibibyte to Gibibit.
#### How many Kibibytes(KiB) are there in a Gibibit(Gibit)?expand_more
There are 131072 Kibibytes in a Gibibit.
#### What is the formula to convert Gibibit(Gibit) to Kibibyte(KiB)?expand_more
Use the formula KiB = Gibit x 10242 / 8 to convert Gibibit to Kibibyte.
#### Which is bigger, Gibibit(Gibit) or Kibibyte(KiB)?expand_more
Gibibit is bigger than Kibibyte. One Gibibit contains 131072 Kibibytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,972 | 6,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.75209 |
https://www.convertunits.com/from/keel/to/decigram | 1,610,822,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00528.warc.gz | 739,172,739 | 12,680 | ## ››Convert keel [coal] to decigram
keel decigram
## ››More information from the unit converter
How many keel in 1 decigram? The answer is 4.6424836208069E-9.
We assume you are converting between keel [coal] and decigram.
You can view more details on each measurement unit:
keel or decigram
The SI base unit for mass is the kilogram.
1 kilogram is equal to 4.6424836208069E-5 keel, or 10000 decigram.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between keel [coal] and decigrams.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of keel to decigram
1 keel to decigram = 215401944.6656 decigram
2 keel to decigram = 430803889.3312 decigram
3 keel to decigram = 646205833.9968 decigram
4 keel to decigram = 861607778.6624 decigram
5 keel to decigram = 1077009723.328 decigram
6 keel to decigram = 1292411667.9936 decigram
7 keel to decigram = 1507813612.6592 decigram
8 keel to decigram = 1723215557.3248 decigram
9 keel to decigram = 1938617501.9904 decigram
10 keel to decigram = 2154019446.656 decigram
## ››Want other units?
You can do the reverse unit conversion from decigram to keel, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Decigram
The SI prefix "deci" represents a factor of 10-1, or in exponential notation, 1E-1.
So 1 decigram = 10-1 grams-force.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 556 | 1,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-04 | latest | en | 0.681526 |
http://catalog.flatworldknowledge.com/bookhub/reader/30?e=wright-ch15_s01 | 1,493,591,702,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125881.93/warc/CC-MAIN-20170423031205-00207-ip-10-145-167-34.ec2.internal.warc.gz | 61,026,719 | 15,383 | Please wait while we create your Custom Book...
# Money and Banking, v. 1.0
by Robert E. Wright and Vincenzo Quadrini
## 15.1 A More Sophisticated Money Multiplier for M1
### Learning Objectives
1. How do the simple money multiplier and the more sophisticated one developed here contrast and compare?
2. What equation helps us to understand how changes in the monetary base affect the money supply?
In Chapter 14 “The Money Supply Process”, you learned that an increase (decrease) in the monetary base (MB, which = C + R) leads to an even greater increase (decrease) in the money supply (MS, such as M1M1 is a measure of the money supply that includes currency in circulation plus checkable deposits. or M2M2 is a measure of the money supply that includes M1 plus time deposits and noninstitutional (retail) money market funds.) due to the multiple deposit creation process. You also learned a simple but unrealistic upper-bound formula for estimating the change that assumed that banks hold no excess reserves and that the public holds no currency.
### Stop and Think Box
You are a research associate for Moody’s subsidiary, High Frequency Economics, in West Chester, Pennsylvania. A client wants you to project changes in M1 given likely increases in the monetary base. Because of a glitch in the Federal Reserve’s computer systems, currency, deposit, and excess reserve figures will not be available for at least one week. A private firm, however, can provide you with good estimates of changes in banking system reserves, and of course the required reserve ratio is well known. What equation can you use to help your client? What are the equation’s assumptions and limitations?
You cannot use the more complex M1 money multiplier this week because of the Fed’s computer glitch, so you should use the simple deposit multiplier from Chapter 14 “The Money Supply Process”: ΔD = (1/rr) × ΔR. The equation provides an upper-bound estimate for changes in deposits. It assumes that the public will hold no more currency and that banks will hold no increased excess reserves.
To get a more realistic estimate, we’ll have to do a little more work. We start with the observation that we can consider the money supply to be a function of the monetary base times some money multiplier (m):
$△MS=m×△MB$
This is basically a less specific version of the formula you learned in Chapter 14 “The Money Supply Process”, except that instead of calculating the change in deposits (ΔD) brought about by the change in reserves (ΔR), we will now calculate the change in the money supply (ΔMS) brought about by the change in the monetary base (ΔMB). Furthermore, instead of using the reciprocal of the required reserve ratio (1/rr) as the multiplier, we will use a more sophisticated one (m1, and later M2) that doesn’t assume away cash and excess reserves.
We can add currency and excess reserves to the equation by algebraically describing their relationship to checkable depositsDeposits that can easily, cheaply, and quickly be drawn upon by check in order to make payments. Also known as transaction deposits. in the form of a ratio:
C/D = currency ratio
ER/D = excess reserves ratio
Recall that required reserves are equal to checkable deposits (D) times the required reserve ratio (rr). Total reserves equal required reserves plus excess reserves:
$R=rrD+ER$
So we can render MB = C + R as MB = C + rrD + ER. Note that we have successfully removed C and ER from the multiple deposit expansion process by separating them from rrD. After further algebraic manipulations of the above equation and the reciprocal of the reserve ratio (1/rr) concept embedded in the simple deposit multiplier, we’re left with a more sophisticated, more realistic money multiplier:
$m 1 =1+(C/D)/[rr+(ER/D)+(C/D)]$
So if
Required reserve ratio (rr) = .2
Currency in circulation = $100 billion Deposits =$400 billion
Excess reserves = $10 billion $m 1 =1.25/(.2+.025+.25)$ $m 1 =1.25/.475=2.6316$ Practice calculating the money multiplier in Exercise 1. ### Exercises 1. Given the following, calculate the M1 money multiplier using the formula m1 = 1 + (C/D)/[rr + (ER/D) + (C/D)]. Currency Deposits Excess Reserves Required Reserve Ratio Answer: m1 100 100 10 .1 1.67 100 100 10 .2 1.54 100 1,000 10 .2 3.55 1,000 100 10 .2 1.07 1,000 100 50 .2 1.02 100 1,000 50 .2 3.14 100 1,000 0 1 1 Once you have m, plug it into the formula ΔMS = m × ΔMB. So if m1 = 2.6316 and the monetary base increases by$100,000, the money supply will increase by $263,160. If m1 = 4.5 and MB decreases by$1 million, the money supply will decrease by \$4.5 million, and so forth. Practice this in Exercise 2.
2. Calculate the change in the money supply given the following:
Change in MB m1 Answer: Change in MS
100 2 200
100 4 400
−100 2 −200
−100 4 −400
1,000 2 2,000
−1,000 2 −2,000
10,000 1 10,000
−10,000 1 −10,000
### Stop and Think Box
Figure 15.1 U.S. MB and M1, 1959–2007
Figure 15.2 U.S. m1, 1959–2007.
Figure 15.3 U.S. currency and checkable deposits, 1959–2007
Figure 15.4 U.S. currency ratio, 1959–2007
In Figure 15.1 “U.S. MB and M1, 1959–2007”, M1 has increased because MB has increased, likely due to net open market purchases by the Fed. Apparently, m1 has changed rather markedly since the early 1990s. In Figure 15.2 “U.S. m”, the M1 money multiplier m has indeed dropped considerably since about 1995. That could be caused by an increase in rr, C/D, or ER/D. Figure 15.3 “U.S. currency and checkable deposits, 1959–2007” shows that m decreased primarily because C/D increased. It also shows that the increase in C/D was due largely to the stagnation in D coupled with the continued growth of C. The stagnation in D is likely due to the advent of sweep accounts. Figure 15.4 “U.S. currency ratio, 1959–2007” isolates C/D for closer study.
### Key Takeaways
• The money multipliers are the same because they equate changes in the money supply to changes in the monetary base times some multiplier.
• The money multipliers differ because the simple multiplier is merely the reciprocal of the required reserve ratio, while the other multipliers account for cash and excess reserve leakages.
• Therefore, m1 and m2 are always smaller than 1/rr (except in the rare case where C and ER both = 0).
• ΔMS = m × ΔMB, where ΔMS = change in the money supply; m = the money multiplier; ΔMB = change in the monetary base. A positive sign means an increase in the MS; a negative sign means a decrease. | 1,662 | 6,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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} | 2.984375 | 3 | CC-MAIN-2017-17 | latest | en | 0.916494 |
https://origin.geeksforgeeks.org/frequency-substring-string/?ref=lbp | 1,675,112,326,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00630.warc.gz | 467,018,692 | 38,610 | # Frequency of a Substring in a String
• Difficulty Level : Medium
• Last Updated : 13 Oct, 2022
Given an input string and a pattern, the task is to find the frequency of occurrences of the string pattern in given string.
Examples:
Input: pattern = “man”, string = “dhimanman”
Output: 2
Input: pattern = “nn”, string = “Banana”
Output: 0
Input: pattern = “aa”, string = “aaaaa”
Output : 4
## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
Approach:
A simple solution is to match characters one by one. And whenever we see a complete match, increment count. For this, we can use Naive pattern searching.
Below is the implementation of the above approach.
## C++
`// Simple C++ program to count occurrences` `// of pat in txt.` `#include ` `using` `namespace` `std;` `int` `countFreq(string& pat, string& txt)` `{` ` ``int` `M = pat.length();` ` ``int` `N = txt.length();` ` ``int` `res = 0;` ` ``/* A loop to slide pat[] one by one */` ` ``for` `(``int` `i = 0; i <= N - M; i++) {` ` ``/* For current index i, check for` ` ``pattern match */` ` ``int` `j;` ` ``for` `(j = 0; j < M; j++)` ` ``if` `(txt[i + j] != pat[j])` ` ``break``;` ` ``// if pat[0...M-1] = txt[i, i+1, ...i+M-1]` ` ``if` `(j == M) {` ` ``res++;` ` ``}` ` ``}` ` ``return` `res;` `}` `/* Driver program to test above function */` `int` `main()` `{` ` ``string txt = ``"dhimanman"``;` ` ``string pat = ``"man"``;` ` ``cout << countFreq(pat, txt);` ` ``return` `0;` `}`
## Java
`// Simple Java program to count occurrences` `// of pat in txt.` `class` `GFG {` ` ``static` `int` `countFreq(String pat, String txt)` ` ``{` ` ``int` `M = pat.length();` ` ``int` `N = txt.length();` ` ``int` `res = ``0``;` ` ``/* A loop to slide pat[] one by one */` ` ``for` `(``int` `i = ``0``; i <= N - M; i++) {` ` ``/* For current index i, check for` ` ``pattern match */` ` ``int` `j;` ` ``for` `(j = ``0``; j < M; j++) {` ` ``if` `(txt.charAt(i + j) != pat.charAt(j)) {` ` ``break``;` ` ``}` ` ``}` ` ``// if pat[0...M-1] = txt[i, i+1, ...i+M-1]` ` ``if` `(j == M) {` ` ``res++;` ` ``j = ``0``;` ` ``}` ` ``}` ` ``return` `res;` ` ``}` ` ``/* Driver program to test above function */` ` ``static` `public` `void` `main(String[] args)` ` ``{` ` ``String txt = ``"dhimanman"``;` ` ``String pat = ``"man"``;` ` ``System.out.println(countFreq(pat, txt));` ` ``}` `}` `// This code is contributed by 29AjayKumar`
## Python3
`# Simple python program to count` `# occurrences of pat in txt.` `def` `countFreq(pat, txt):` ` ``M ``=` `len``(pat)` ` ``N ``=` `len``(txt)` ` ``res ``=` `0` ` ``# A loop to slide pat[] one by one` ` ``for` `i ``in` `range``(N ``-` `M ``+` `1``):` ` ``# For current index i, check` ` ``# for pattern match` ` ``j ``=` `0` ` ``while` `j < M:` ` ``if` `(txt[i ``+` `j] !``=` `pat[j]):` ` ``break` ` ``j ``+``=` `1` ` ``if` `(j ``=``=` `M):` ` ``res ``+``=` `1` ` ``j ``=` `0` ` ``return` `res` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ``txt ``=` `"dhimanman"` ` ``pat ``=` `"man"` ` ``print``(countFreq(pat, txt))` `# This code is contributed` `# by PrinciRaj1992`
## C#
`// Simple C# program to count occurrences` `// of pat in txt.` `using` `System;` `public` `class` `GFG {` ` ``static` `int` `countFreq(String pat, String txt)` ` ``{` ` ``int` `M = pat.Length;` ` ``int` `N = txt.Length;` ` ``int` `res = 0;` ` ``/* A loop to slide pat[] one by one */` ` ``for` `(``int` `i = 0; i <= N - M; i++) {` ` ``/* For current index i, check for` ` ``pattern match */` ` ``int` `j;` ` ``for` `(j = 0; j < M; j++) {` ` ``if` `(txt[i + j] != pat[j]) {` ` ``break``;` ` ``}` ` ``}` ` ``// if pat[0...M-1] = txt[i, i+1, ...i+M-1]` ` ``if` `(j == M) {` ` ``res++;` ` ``j = 0;` ` ``}` ` ``}` ` ``return` `res;` ` ``}` ` ``/* Driver program to test above function */` ` ``static` `public` `void` `Main()` ` ``{` ` ``String txt = ``"dhimanman"``;` ` ``String pat = ``"man"``;` ` ``Console.Write(countFreq(pat, txt));` ` ``}` `}` `// This code is contributed by 29AjayKumar`
## PHP
`
## Javascript
``
Output
`2`
Time Complexity: O(M * N)
Efficient Approach:
An efficient solution is to use KMP algorithm
Below is the implementation of the above approach.
## C++
`// C++ program to count occurrences` `// of pattern in a text.` `#include ` `using` `namespace` `std;` `void` `computeLPSArray(string pat, ``int` `M, ``int` `lps[])` `{` ` ``// Length of the previous longest` ` ``// prefix suffix` ` ``int` `len = 0;` ` ``int` `i = 1;` ` ``lps[0] = 0; ``// lps[0] is always 0` ` ``// The loop calculates lps[i] for` ` ``// i = 1 to M-1` ` ``while` `(i < M) {` ` ``if` `(pat[i] == pat[len]) {` ` ``len++;` ` ``lps[i] = len;` ` ``i++;` ` ``}` ` ``else` `// (pat[i] != pat[len])` ` ``{` ` ``// This is tricky. Consider the example.` ` ``// AAACAAAA and i = 7. The idea is similar` ` ``// to search step.` ` ``if` `(len != 0) {` ` ``len = lps[len - 1];` ` ``// Also, note that we do not` ` ``// increment i here` ` ``}` ` ``else` `// if (len == 0)` ` ``{` ` ``lps[i] = len;` ` ``i++;` ` ``}` ` ``}` ` ``}` `}` `int` `KMPSearch(string pat, string txt)` `{` ` ``int` `M = pat.length();` ` ``int` `N = txt.length();` ` ``// Create lps[] that will hold the longest` ` ``// prefix suffix values for pattern` ` ``int` `lps[M];` ` ``int` `j = 0; ``// index for pat[]` ` ``// Preprocess the pattern (calculate lps[]` ` ``// array)` ` ``computeLPSArray(pat, M, lps);` ` ``int` `i = 0; ``// index for txt[]` ` ``int` `res = 0;` ` ``int` `next_i = 0;` ` ``while` `(i < N) {` ` ``if` `(pat[j] == txt[i]) {` ` ``j++;` ` ``i++;` ` ``}` ` ``if` `(j == M) {` ` ``// When we find pattern first time,` ` ``// we iterate again to check if there` ` ``// exists more pattern` ` ``j = lps[j - 1];` ` ``res++;` ` ``}` ` ``// Mismatch after j matches` ` ``else` `if` `(i < N && pat[j] != txt[i]) {` ` ``// Do not match lps[0..lps[j-1]]` ` ``// characters, they will match anyway` ` ``if` `(j != 0)` ` ``j = lps[j - 1];` ` ``else` ` ``i = i + 1;` ` ``}` ` ``}` ` ``return` `res;` `}` `// Driver code` `int` `main()` `{` ` ``string txt = ``"geeksforgeeks"``;` ` ``string pat = ``"eeks"``;` ` ``int` `ans = KMPSearch(pat, txt);` ` ``cout << ans;` ` ``return` `0;` `}` `// This code is contributed by akhilsaini`
## Java
`// Java program to count occurrences of pattern` `// in a text.` `class` `KMP_String_Matching {` ` ``int` `KMPSearch(String pat, String txt)` ` ``{` ` ``int` `M = pat.length();` ` ``int` `N = txt.length();` ` ``// create lps[] that will hold the longest` ` ``// prefix suffix values for pattern` ` ``int` `lps[] = ``new` `int``[M];` ` ``int` `j = ``0``; ``// index for pat[]` ` ``// Preprocess the pattern (calculate lps[]` ` ``// array)` ` ``computeLPSArray(pat, M, lps);` ` ``int` `i = ``0``; ``// index for txt[]` ` ``int` `res = ``0``;` ` ``int` `next_i = ``0``;` ` ``while` `(i < N) {` ` ``if` `(pat.charAt(j) == txt.charAt(i)) {` ` ``j++;` ` ``i++;` ` ``}` ` ``if` `(j == M) {` ` ``// When we find pattern first time,` ` ``// we iterate again to check if there` ` ``// exists more pattern` ` ``j = lps[j - ``1``];` ` ``res++;` ` ``// We start i to check for more than once` ` ``// appearance of pattern, we will reset i` ` ``// to previous start+1` ` ``if` `(lps[j] != ``0``)` ` ``i = ++next_i;` ` ``j = ``0``;` ` ``}` ` ``// mismatch after j matches` ` ``else` `if` `(i < N` ` ``&& pat.charAt(j) != txt.charAt(i)) {` ` ``// Do not match lps[0..lps[j-1]] characters,` ` ``// they will match anyway` ` ``if` `(j != ``0``)` ` ``j = lps[j - ``1``];` ` ``else` ` ``i = i + ``1``;` ` ``}` ` ``}` ` ``return` `res;` ` ``}` ` ``void` `computeLPSArray(String pat, ``int` `M, ``int` `lps[])` ` ``{` ` ``// length of the previous longest prefix suffix` ` ``int` `len = ``0``;` ` ``int` `i = ``1``;` ` ``lps[``0``] = ``0``; ``// lps[0] is always 0` ` ``// the loop calculates lps[i] for i = 1 to M-1` ` ``while` `(i < M) {` ` ``if` `(pat.charAt(i) == pat.charAt(len)) {` ` ``len++;` ` ``lps[i] = len;` ` ``i++;` ` ``}` ` ``else` `// (pat[i] != pat[len])` ` ``{` ` ``// This is tricky. Consider the example.` ` ``// AAACAAAA and i = 7. The idea is similar` ` ``// to search step.` ` ``if` `(len != ``0``) {` ` ``len = lps[len - ``1``];` ` ``// Also, note that we do not increment` ` ``// i here` ` ``}` ` ``else` `// if (len == 0)` ` ``{` ` ``lps[i] = len;` ` ``i++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Driver program to test above function` ` ``public` `static` `void` `main(String args[])` ` ``{` ` ``String txt = ``"geeksforgeeks"``;` ` ``String pat = ``"eeks"``;` ` ``int` `ans` ` ``= ``new` `KMP_String_Matching().KMPSearch(pat, txt);` ` ``System.out.println(ans);` ` ``}` `}`
## Python3
`# Python3 program to count occurrences of` `# pattern in a text.` `def` `KMPSearch(pat, txt):` ` ``M ``=` `len``(pat)` ` ``N ``=` `len``(txt)` ` ``# Create lps[] that will hold the longest` ` ``# prefix suffix values for pattern` ` ``lps ``=` `[``None``] ``*` `M` ` ``j ``=` `0` `# index for pat[]` ` ``# Preprocess the pattern (calculate lps[]` ` ``# array)` ` ``computeLPSArray(pat, M, lps)` ` ``i ``=` `0` `# index for txt[]` ` ``res ``=` `0` ` ``next_i ``=` `0` ` ``while` `(i < N):` ` ``if` `pat[j] ``=``=` `txt[i]:` ` ``j ``=` `j ``+` `1` ` ``i ``=` `i ``+` `1` ` ``if` `j ``=``=` `M:` ` ``# When we find pattern first time,` ` ``# we iterate again to check if there` ` ``# exists more pattern` ` ``j ``=` `lps[j ``-` `1``]` ` ``res ``=` `res ``+` `1` ` ``# We start i to check for more than once` ` ``# appearance of pattern, we will reset i` ` ``# to previous start+1` ` ``if` `lps[j] !``=` `0``:` ` ``next_i ``=` `next_i ``+` `1` ` ``i ``=` `next_i` ` ``j ``=` `0` ` ``# Mismatch after j matches` ` ``elif` `((i < N) ``and` `(pat[j] !``=` `txt[i])):` ` ``# Do not match lps[0..lps[j-1]]` ` ``# characters, they will match anyway` ` ``if` `(j !``=` `0``):` ` ``j ``=` `lps[j ``-` `1``]` ` ``else``:` ` ``i ``=` `i ``+` `1` ` ``return` `res` `def` `computeLPSArray(pat, M, lps):` ` ``# Length of the previous longest` ` ``# prefix suffix` ` ``len` `=` `0` ` ``i ``=` `1` ` ``lps[``0``] ``=` `0` `# lps[0] is always 0` ` ``# The loop calculates lps[i] for` ` ``# i = 1 to M-1` ` ``while` `(i < M):` ` ``if` `pat[i] ``=``=` `pat[``len``]:` ` ``len` `=` `len` `+` `1` ` ``lps[i] ``=` `len` ` ``i ``=` `i ``+` `1` ` ``else``: ``# (pat[i] != pat[len])` ` ``# This is tricky. Consider the example.` ` ``# AAACAAAA and i = 7. The idea is similar` ` ``# to search step.` ` ``if` `len` `!``=` `0``:` ` ``len` `=` `lps[``len` `-` `1``]` ` ``# Also, note that we do not increment` ` ``# i here` ` ``else``: ``# if (len == 0)` ` ``lps[i] ``=` `len` ` ``i ``=` `i ``+` `1` `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` ` ``txt ``=` `"geeksforgeeks"` ` ``pat ``=` `"eeks"` ` ``ans ``=` `KMPSearch(pat, txt)` ` ``print``(ans)` `# This code is contributed by akhilsaini`
## C#
`// C# program to count occurrences of pattern` `// in a text.` `using` `System;` `public` `class` `KMP_String_Matching {` ` ``int` `KMPSearch(String pat, String txt)` ` ``{` ` ``int` `M = pat.Length;` ` ``int` `N = txt.Length;` ` ``// create lps[] that will hold the longest` ` ``// prefix suffix values for pattern` ` ``int``[] lps = ``new` `int``[M];` ` ``int` `j = 0; ``// index for pat[]` ` ``// Preprocess the pattern (calculate lps[]` ` ``// array)` ` ``computeLPSArray(pat, M, lps);` ` ``int` `i = 0; ``// index for txt[]` ` ``int` `res = 0;` ` ``int` `next_i = 0;` ` ``while` `(i < N) {` ` ``if` `(pat[j] == txt[i]) {` ` ``j++;` ` ``i++;` ` ``}` ` ``if` `(j == M) {` ` ``// When we find pattern first time,` ` ``// we iterate again to check if there` ` ``// exists more pattern` ` ``j = lps[j - 1];` ` ``res++;` ` ``// We start i to check for more than once` ` ``// appearance of pattern, we will reset i` ` ``// to previous start+1` ` ``if` `(lps[j] != 0)` ` ``i = ++next_i;` ` ``j = 0;` ` ``}` ` ``// mismatch after j matches` ` ``else` `if` `(i < N && pat[j] != txt[i]) {` ` ``// Do not match lps[0..lps[j-1]] characters,` ` ``// they will match anyway` ` ``if` `(j != 0)` ` ``j = lps[j - 1];` ` ``else` ` ``i = i + 1;` ` ``}` ` ``}` ` ``return` `res;` ` ``}` ` ``void` `computeLPSArray(String pat, ``int` `M, ``int``[] lps)` ` ``{` ` ``// length of the previous longest prefix suffix` ` ``int` `len = 0;` ` ``int` `i = 1;` ` ``lps[0] = 0; ``// lps[0] is always 0` ` ``// the loop calculates lps[i] for i = 1 to M-1` ` ``while` `(i < M) {` ` ``if` `(pat[i] == pat[len]) {` ` ``len++;` ` ``lps[i] = len;` ` ``i++;` ` ``}` ` ``else` `// (pat[i] != pat[len])` ` ``{` ` ``// This is tricky. Consider the example.` ` ``// AAACAAAA and i = 7. The idea is similar` ` ``// to search step.` ` ``if` `(len != 0) {` ` ``len = lps[len - 1];` ` ``// Also, note that we do not increment` ` ``// i here` ` ``}` ` ``else` `// if (len == 0)` ` ``{` ` ``lps[i] = len;` ` ``i++;` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `Main(String[] args)` ` ``{` ` ``String txt = ``"geeksforgeeks"``;` ` ``String pat = ``"eeks"``;` ` ``int` `ans` ` ``= ``new` `KMP_String_Matching().KMPSearch(pat, txt);` ` ``Console.WriteLine(ans);` ` ``}` `}` `// This code is contributed by Princi Singh`
## Javascript
``
Output
`2`
Time Complexity: O(M + N)
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# The mean of set S is 20. What is the median of set S? 1. In
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The mean of set S is 20. What is the median of set S? 1. In [#permalink]
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16 Feb 2008, 10:08
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The mean of set S is 20. What is the median of set S?
1. In set S there are as many numbers bigger than 20 as there are numbers smaller than 20
2. All numbers in set S are even integers
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16 Feb 2008, 19:37
marcodonzelli wrote:
The mean of set S is 20. What is the median of set S?
1. In set S there are as many numbers bigger than 20 as there are numbers smaller than 20
2. All numbers in set S are even integers
A. Median is 20 as it is in the middle.
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16 Feb 2008, 19:52
(1) It is possible that this could imply the median is 20 (and therefore known): i.e. S={17,18,19,21,22,23} median=20
But it is also possible that it could be something else: i.e. S= {2,5,12,21,22,23} median =(12+21)/2=33/2
Not suff.
(2) All numbers in S are even...definitely not suff. by itself.
(1) U (2): there are an equal number of even integers less than 20 and even integers greater than 20 in S.
ex. 1 S={16,18,22,24) median =20
ex. 2 S={2,4,22,24} median = (4+22)/2 = 13
Not suff.
I think it's (E).
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16 Feb 2008, 21:00
jankynoname wrote:
(1) It is possible that this could imply the median is 20 (and therefore known): i.e. S={17,18,19,21,22,23} median=20 But it is also possible that it could be something else: i.e. S= {2,5,12,21,22,23} median =(12+21)/2=33/2 Not suff.
(2) All numbers in S are even...definitely not suff. by itself. (1) U (2): there are an equal number of even integers less than 20 and even integers greater than 20 in S. ex. 1 S={16,18,22,24) median =20 ex. 2 S={2,4,22,24} median = (4+22)/2 = 13 Not suff.I think it's (E).
yup. you are correct.
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Re: tricky statistics [#permalink] 16 Feb 2008, 21:00
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Nokia Nokia. My first cellular phone was Nokia. Those days Nokia created best cellular phones. Nokia, once a dominant force in the mobile phone industry, has experienced a dramatic decline. From a market capitalization of $550 billion in 2000, it plummeted to just$18 billion. This decline prompts a critical question: Why did Nokia fail while companies like Apple soared to unprecedented heights ? While Nokia excelled at producing high-quality cellular phones, it missed the boat on the smartphone revolution. Smart phones slowly gain more and more customers, and, of course, who bought smart phone did not buy cellular phone. Nokia did not lost it's domain of cellular phones. Nokia just missed this moment when new market of smartphones was created. Those two markets could not coexist. Such situation defined by " Struggle of Informations " model as "antagonistic informations" where only one information will survive and other will totally disappear. The critical ques
### "Life", "Death" and "The Winner Takes It All"
I want to talk about how one simple differential equation became very useful mathematical model that reflect how different ecosystems (both in nature and in technology) evolve.
"Life"
One of most known ecosystems used for mathematical modeling is population of rabbits. We assume that next generation of rabbits will have all already living rabbits plus new born rabbits. Number of new born rabbits is proportional to number of existing rabbits. We'll take simplest case when number of additional rabbits equal to number of existing rabbits (when 2 rabbits bring 2 new rabbits). This means that change of number of rabbits equal to number of rabbits, or as differential equation: $$\large{}\color{blue}\frac{d x}{d t} = x$$ General solution for this differential equation is: $$\Large{}\color{blue} x = e^t$$(where X is population number, T is time). This equation is known as "exponential function".
Many times this function used to represent population growth of living elements of system. Living objects are such that can reproduce themself (like rabbits or virus). Living object may be not only "life" elements, but also "non real" elements like piece of information. In case someone reveal some information number of people that "know" this information also will grow exponentially.
Exponential growth is very fast growth, it is often used to express "blowup" growth. While exponential growth used by many to express population growth in reality it's wrong representation. No population grow accordingly exponential function because that growth cannot be infinite. There is another function that more close to describe population growth.
"Death"
Any population has growth limit. When population is small it grows fast because there is almost no death. When population closer to resource limits more species start to die. Original differential equation get "death" element. We assume that death rate proportional to number of species and proportional to relation of number of species to maximum number environment allow. $$\large{}\color{blue}\frac{d x}{d t} = x-x*\frac{x}{L}$$ or $$\large{}\color{blue}\frac{d x}{d t} = x-\frac{x^2}{L}$$ (L is population maximum limited by environment) When current population significantly smaller than maximum population growth very close to exponential, but when x became close to maximum L death rate became very close to birth rate and makes growth almost 0. General solution to this function is $$\large{}\color{blue}x = \frac{L}{1+e^{-t}}$$ This function is known as "logistic function".
This function was introduced by Belgian mathematician Pierre Verhulst. It is used in many domains, from ecology and health to machine learning. This function has "S" shape and is one of "Sigmoid" functions. Logistic function is interesting for TRIZ (Theory of Inventive Problem Solving), because one of TRIZ laws of system evolution is that useful parameters in technical systems grow with S-curve. It's very interesting to apply logistic function on "S-curve" law of TRIZ.
Direct analogy of "logistic function" and "S-curve" is wrong because number of elements of some species in ecological system and number representing some system's parameter (e.g. car speed or screen size of TV) are very different numbers in their nature. But, if we consider that there is correlation between useful parameter and number of subsystems in system and number of relations between subsystems, we can find that such analogy may be possible. What we can say is that "Complexity" of system grows accordingly to "logistic function". Many interesting questions come from it. Is it possible to express "complexity" of technical system mathematically? Why complexity has limit in technical system? (I have some ideas on these issues, but I'll represent them in another blog).
"The Winner Takes It All"
There is one important aspect in population growth that should be considered. Competition. Ecological model where we have only one type of species (e.g. rabbits) is too simplified. In real world resources may be useful for many different species, which leads to direct competition between species. We can express competition in differential equation by adding part that reduce growth rate. Number of differential equations will be equal to number of species in system (one differential equation for each species). Let's see how it looks like for 2 species. $$\large{}\color{blue}\frac{d x_{1}}{d t} = x_{1}-a_{1}x_{1}^2-b_{12}x_{1}x_{2}$$ $$\large{}\color{blue}\frac{d x_{2}}{d t} = x_{2}-a_{2}x_{2}^2-b_{21}x_{2}x_{1}$$ ("a" is coefficient of internal competition in some of species, "b" is competition coefficient between species). Now we build equation where growth of species "1" depends also on species "2". General form for many information types. $$\large{}\color{blue}\frac{d x_{i}}{d t} = x_{i}-a_{i}x_{i}^2-\sum_{j \neq i}^n b_{ij}x_{i}x_{j}$$ This mathematical model was represented by Russian scientist Dmitrii Chernavskii and is called "Struggle of Informations". By "Information" in this model we mean some entity that can reproduce itself. It may be some living species (like virus or rabbit). It may be technological standard, rumors, ... anything that can be replicated. Those informations may coexist and do not affect each other (coefficient "b" is very low), or informations may not coexist ("b" is very high). Informations with high "b" coefficient are called "antagonistic". Accordingly to "Struggle of Information" model when in system there are 2 antagonistic "informations" it leads to "pure" state where only one of two informations will survive, another information will disappear with time. This effect is known as "The winner takes it all". It does not matter that some information has very little advantage over another information, it still will lead to situation that another information will disappear totally.
I'll give an example from technological domain. When mechanical clocks invented part of them had 12 hours and part of them had 24 hours
Those two types of clock are different in way they represent time, in our words those are two different informations. Lets think, are those informations antagonistic each to other? Lets imagine in some town two men decide to meet tomorrow at 11 o'clock. In case those two men has different types of clock they will not meet. With time all men in this town will have only one type of clock. Clock with 24 hours totally disappeared. It's not that 12 hours clock has very big advantage over 24 hours clock, but because those two types are antagonistic only one will survive.
Another good example of "Struggle of Informations" is relations between programming languages. Programming languages may be antagonistic in case they try to solve similar problems, e.g. many scripting languages are antagonistic each to other. We can see that last years one scripting language usage significantly grow, while other scripting languages used much less than before. Programming languages may be not antagonistic (like "C" and "JavaScript") because they are solve different problems and programmers do not need to select one of them.
Professor Chernavskii used "Struggle of Informations" model to explain why all living forms on our planet use same genetic code. This model can be very useful for understanding process how one of many technologies became standard, while other disappear. This model also used to explain some historical processes where countries were united because languages are antagonistic. This model is also useful in economics, it can be good tool to understand how different currencies behave locally and globally. | 1,866 | 8,714 | {"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": 8, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-33 | latest | en | 0.934151 |
https://ai-solutions.com/_freeflyeruniversityguide/plane_change_maneuver.htm | 1,696,303,833,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00519.warc.gz | 102,622,591 | 4,598 | Plane Change Maneuver
# Plane Change Maneuver
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Plane change maneuvers are known as one of the more costly maneuvers for spacecraft to perform. However, the math for them is relatively simple. In fact, you only need to have a basic understanding of trigonometry and vector math to calculate the Δv needed in a plane change.
Say we wanted to make an inclination change. If we burned normal to the orbital plane, we would be adding Δv in the y direction of the VNB attitude system. However, this would make our resultant vector larger than what we started with, thus changing the shape of the orbit. See the diagrams below:
VNB Attitude System
Burn in the Normal Direction
Our resultant velocity is the hypotenuse in this triangle. Because of that, it will definitely be larger than our original velocity. With this, we are not only changing the inclination of the orbit, but we are changing the eccentricity. How can we burn in such a way that the resultant velocity vector has the same magnitude as our original velocity and not change the eccentricity of the orbit? The trick is to tilt the Δv vector backwards. Doing this, we will get an isosceles triangle like the diagram below:
Inclination Change Burn
With this, we can break down the Δv vector into two components: the normal component, and the negative velocity component. Using trigonometry, we can calculate these values quite easily. The formulas for the the Δv are as follows:
Now that we have worked out the math, let's try writing a Mission Plan that calculates and performs a plane change.
## Modeling Plane Change Maneuvers
Problem: We have a spacecraft in a circular, equatorial orbit with a SMA of 7200 km. How much Δv would we need to change the inclination to 30º without changing the shape of the orbit?
We can build our own calculator for this in FreeFlyer. But first, let's add in the necessary objects in the Object Browser.
Open a new Mission Plan and save it as "CircularPlaneChange.MissionPlan"
Create a new Spacecraft with the following orbital elements:
oA: 7200 km
oE: 0
oI: 0 deg
oRAAN: 0 deg
oW: 0 deg
oTA: 0 deg
Create a new ImpulsiveBurn object
Open the ImpulsiveBurn editor
Change the attitude system to "VNB"
Click "Ok" to close the editor
Create a new ViewWindow object
Open the ViewWindow editor
Check "Spacecraft1" in the "Available Objects" section
Check "Show Name"
Change the history mode to "Unlimited"
Go into "Viewpoints" on the left-hand side
Change the reference frame to "Inertial"
Click "Ok" to close the editor
### Building the Mission Sequence
Drag and drop a FreeForm script editor into the Mission Sequence
Name this "Calculate Plane Change"
In this script, we will assign an inclination angle, calculate the Δv required, and assign it to the ImpulsiveBurn object To do this, we write:
// Inclination to change to in degrees Variable theta = 30; theta = rad(theta); // Calculate the plane change Variable norm = Spacecraft1.VMag * sin(theta); Variable negVel = Spacecraft1.VMag * ( 1 - cos(theta)); // Assigns the values to the burn ImpulsiveBurn1.BurnDirection = { -negVel, norm, 0};
Remember, we are using the VNB attitude system. This is why we need to put a negative sign in front of the "negVel" variable, as VNB's x-component is the positive direction of the velocity vector.
Now, let's propagate the Spacecraft for a day so we can visualize its original orbit.
In the Mission Sequence, drag and drop a "While...End" loop
Inside that loop, drag and drop a "Step" command
Drag and drop an "Update" command after the "Step" command inside the loop
Drag and drop a FreeForm script editor below the While loop
Rename this to "Perform Plane Change"
In this FreeForm, we need to change the Spacecraft object's color, maneuver the Spacecraft, and report the total Δv used for the plane change. To do this, we write:
// Change the spacecraft tail color Spacecraft1.Color = ColorTools.Lime; Maneuver Spacecraft1 using ImpulsiveBurn1; // Reports the total Delta V used for the plane change Report ImpulsiveBurn1.BurnDirection.Norm();
Now, we need to propagate the Spacecraft for one day to visualize the new orbit.
In the Mission Sequence, drag and drop another "While...End" loop
Inside the loop, drag and drop a "Step" command
Drag and drop an "Update" command after the "Step" command inside the loop
Your Mission Sequence should look something like this:
Mission Sequence Example
Save and run your Mission Plan, then try to answer these questions:
Try solving this problem by hand. How do your calculations compare to FreeFlyer's calculations?
Change Spacecraft1's SMA to 30,000 km and run the Mission Plan again. How does the amount of Δv needed change for slower Spacecraft?
If you were to use a bi-elliptic transfer and wanted to perform a plane change as well, where in the transfer would be the most efficient position to perform a plane change? | 1,150 | 4,940 | {"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-2023-40 | latest | en | 0.9171 |
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# Australian embryologists have found evidence that suggests
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Re: Australian embryologists have found evidence that suggests [#permalink]
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16 Jul 2014, 02:06
SidKaria wrote:
Australian embryologists have found evidence that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved as a kind of snorkel.
A. that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved
B. that has suggested the elephant descended from an aquatic animal, its trunk originally evolving
C. suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving
D. to suggest that the elephant had descended from an aquatic animal and its trunk originally evolved
E. to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
[Reveal] Spoiler: OE
a) incorrect- parallel clauses start with the same word."that" is missing in the 2nd clause.
can u please explain me how" is descended" correct ? why do we need "is" ??
and why" descended from an aquatic animal " is incorrect. dont understand why if we remove "is" we are only left with a modifier.(as per one of the earlier explanation) descended from an aquatic animal is a modifier. so is descended here not a verb ??
eg.humans evolved from apes. here evolved is a verb.
b)incorrect- has-incorrect verb. simnple past tense should be used. only a comma between the two clauses distorts the meaning. it seems that aquatic animal's trunk evoved.
c)incorrect- with after aquatic animals modifies the animals. distorts meaning.
had is incorrect verb.
(according to one of the explanations to this question "had descended" changes the intended meaning ) the meaning of descend here becomes"to pass from higher to lower") please explain.
d) same errors. has-incorrect verb. and no "that"
e)correct.
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Re: Australian embryologists have found evidence that suggests [#permalink]
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01 Aug 2014, 00:41
"its" in B is ambiguous. the pattern in B do not allow us to infer that "its" refers to "elephant".
b is wong
by contrast, "its" in E grammatically refers to the subject of the previous clause, the elephant. this is no more ambiguous.
E is better without regarding the idioms "be descended from"
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Re: Australian embryologists have found evidence that suggests [#permalink]
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23 Aug 2014, 12:21
Zatarra wrote:
TommyWallach wrote:
Hey All,
(E) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
ANSWER: "descended" and "evolved" are parallel, and everything else is clear.
Hope that helps!
-t
Tommy
Isnt it appropriate to say "elephant has descended" instead of "elephant is descended"
Both the usage are different.
Consider the following example:
A- Elephant is descended from ... : In this descended is a predicate adjective ... similar to the one in sentence " The laundry is washed" . Hence it shows a state of being of the elephant.
Also the sentence can be rewritten as Descended from an aquatic animal, the elephant has a huge trunk.
B- Elephant has descended from an aquatic animal ...: In this sentence "Descend" is used in the capacity of an action verb. Hence it conveys a different mean as in the "elephant is descending from ...."
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Re: Australian embryologists have found evidence that suggests [#permalink]
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11 Sep 2015, 10:47
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Re: Australian embryologists have found evidence that suggests [#permalink]
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27 Dec 2015, 23:12
TommyWallach wrote:
(E) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
ANSWER: "descended" and "evolved" are parallel, and everything else is clear.
-t
Hi,
In my pre-thinking step before looking at the answer choices, I thought that we should have 'was descended' instead of 'is descended' since 'evolved' is in past and the action of descending took place in past. But to my surprise, I found the answer choice in present tense 'is descended'.
Can you please let me know where I have gone wrong in my thinking process?
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Re: Australian embryologists have found evidence that suggests [#permalink]
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21 Jan 2016, 21:03
Amardeep Sharma wrote:
Australian embryologists have found evidence that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved as a kind of snorkel.
(a)that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved
(b) that has suggested the elephant descended from an aquatic animal, its trunk originally evolving
(c)suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving
(d) to suggest that the elephant had descended from an aquatic animal and its trunk originally evolving
(e) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
Amar
A That x that is nearly always wrong.
B evidence that hsa suggested, trunk evolving wrong
C had needs another event that happened after that
D same as C
E correct. Also to note, universal true statements should be in present tense.
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Re: Australian embryologists have found evidence that suggests [#permalink]
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23 Jan 2016, 02:41
Hi Experts,
As the OA is E. I have 1 question to ask.
We use infinitive to express intention.
Eg-
Jack excercises daily to get slim.
But in E, there is no intention-
Australian embryologists FOUND evidence.
If we SHOW evidence to do BLAH BLAh BLAH...!!!--> Then I understand because here is the intention.
So, how does FOUND + Intention correct..??
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Re: Australian embryologists have found evidence that suggests [#permalink]
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30 Jan 2016, 22:03
Deleted the wrong post
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Last edited by AryamaDuttaSaikia on 02 Feb 2016, 23:36, edited 1 time in total.
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Re: Australian embryologists have found evidence that suggests [#permalink]
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31 Jan 2016, 15:00
The answer requires a past perfect tense."The number" refers to the number of applications in the 1970 s which rose to 3000 in the 1990 s. So it refers to a double past action. The action started in the past in the 1970 s and was carried forward to the 90 s. "E" is out because "the number of applications" is redundant out here.
I think you posted on wrong thread.
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Re: Australian embryologists have found evidence that suggests [#permalink]
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01 Feb 2016, 04:54
Amardeep Sharma wrote:
Australian embryologists have found evidence that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved as a kind of snorkel.
(a)that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved
(b) that has suggested the elephant descended from an aquatic animal, its trunk originally evolving
(c)suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving
(d) to suggest that the elephant had descended from an aquatic animal and its trunk originally evolving
(e) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
Amar
we discuss again and again this hard question.
a.
we need the second "that" before "its" . if we dont have "that", the ending clause is parallel to main clause "embryologiy have found". no sense.
b,
structured as in b, "its" is ambiguous, refering to elephant or animal.
c
"its" refers to animal. no sense.
d
we need the second "that" before "its"
a is correct.
though I can analyze above thing, I possibly can not do so in the test room. this is a hard question
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Re: Australian embryologists have found evidence that suggests [#permalink]
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11 Jul 2016, 06:48
TommyWallach wrote:
Hey All,
I got asked to take this one on by private message, so here I am! The answer is certainly E.
Australian embryologists have found evidence that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved as a kind of snorkel.
(A) that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved
PROBLEM: There's no reason to use a comma with a list of two things.
(B) that has suggested the elephant descended from an aquatic animal, its trunk originally evolving
PROBLEM: You have to say "that" after the verb "to suggest".
(C) suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving
PROBLEM: The use of the past perfect tense here is incorrect, because the action isn't complete. I know it may seem like the descent has ended, but in the present, the elephant is STILL descended from an aquatic animal. That's an eternal truth. Also, using the prepositional phrase "with its trunk originally evolving" ends up modifying "an aquatic animal", when we really want to be referring to the elephant.
(D) to suggest that the elephant has descended from an aquatic animal and its trunk originally evolved
PROBLEM: We don't want to change tense in parallel unless there's a significant reason ("yesterday I went to the store but today I will stay at home"). No good reason from present perfect "has descended" to past "evolved".
(E) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
ANSWER: "descended" and "evolved" are parallel, and everything else is clear.
Hope that helps!
-t
Amardeep Sharma wrote:
Australian embryologists have found evidence that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved as a kind of snorkel.
(a)that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved
(b) that has suggested the elephant descended from an aquatic animal, its trunk originally evolving
(c)suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving
(d) to suggest that the elephant had descended from an aquatic animal and its trunk originally evolving
(e) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
Amar
Since people are having doubts on A and E, let me focus on that... with a point everyone else has missed :
A)
evidence :
that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved
Due to the missing that, we can interpret it as :
researchers have found that:
evidence that suggests that the elephant is descended from an aquatic animal
and researchers have found its trunk originally evolved ....
No!!!!!
researchers have found evidence pointing to each of the things.... we need a second that to clarify that the intended meaning is :
researchers have found evidence that its trunk originally evolved
More than 'its' or 'a comma issue', the second that is required here to clarify the meaning
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Re: Australian embryologists have found evidence that suggests [#permalink]
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11 Jul 2016, 07:07
That is missing in A, which should be present.
So E is perfect
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Re: Australian embryologists have found evidence that suggests [#permalink]
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31 Jul 2016, 03:35
In order to eliminate answer choice A is it also possible to to say, that because of the combination " , and ..." a second IC is introduced? With the second IC the logical connection between the evidence and the trunk is missing?
Kind regards.
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Re: Australian embryologists have found evidence that suggests [#permalink]
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26 Feb 2017, 09:35
(a)that suggests that the elephant is descended from an aquatic animal,and its trunk originally evolved ...to what its referring is not clear
(b) that has suggested the elephant descended from an aquatic animal, its trunk originally evolving ....awkward
(c)suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving ..change meaning
(d) to suggest that the elephant had descended from an aquatic animal and its trunk originally evolving ..change meaning
(e) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved..correct
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Re: Australian embryologists have found evidence that suggests [#permalink]
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27 Feb 2017, 08:25
GMAT TIGER wrote:
Amardeep Sharma wrote:
Australian embryologists have found evidence that suggests that the elephant is descended from an aquatic animal, and its trunk originally evolved as a kind of snorkel.
(a)..
(b) that has suggested the elephant descended from an aquatic animal, its trunk originally evolving
(c)suggesting that the elephant had descended from an aquatic animal with its trunk originally evolving
(d) to suggest that the elephant had descended from an aquatic animal and its trunk originally evolving
(e) to suggest that the elephant is descended from an aquatic animal and that its trunk originally evolved
Amar
It is an example of how tricky can gmat questions be.
Most of the people are intimidited by A but the problem is with "its", which refers to elephant's.
If the setence were not properly paralled without "that x ......... and that y ......", "its" is never be a clear referant for "elephant". That construction is only in E, which is tghe best and OA.
excellent posting for us to know.
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,343 | 18,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-22 | latest | en | 0.948176 |
https://www.physicsforums.com/threads/subspace-induced-relative-topology-definition.378606/ | 1,521,570,616,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647519.62/warc/CC-MAIN-20180320170119-20180320190119-00097.warc.gz | 839,771,219 | 14,616 | # Subspace/induced/relative Topology Definition
1. Feb 15, 2010
### filter54321
I'm having trouble understanding the definition of a Subspace/Induced/Relative Topology. The definitions I'm finding either don't define symbols well (at all).
If I understand correctly the definition is:
Given:
-topological space (A,$$\tau$$)
-$$\tau$$={0,A,u1,u2,...un}
-subset B$$\subset$$A
The subspace topology on B will be the intersection of B and every part of the topology of A
OR
$$\tau$$B={0,B,B$$\bigcap$$u1,B$$\bigcap$$u2,...B$$\bigcap$$un}
...I apologize in advance for my LATEX work.
2. Feb 15, 2010
### Hurkyl
Staff Emeritus
That looks correct idea for the subspace topology... but there is a serious problem in your understanding of topology: you seem to be assuming there are only finitely many open sets.
3. Feb 15, 2010
### filter54321
The u's are finite for illustrative purposes. I wanted to avoid using the form given by Wikipedia because it has insufficient textual explanation.
They have this definition, but don't specify exactly what a U is:
$$\tau$$B={B$$\bigcap$$U|U$$\in$$$$\tau$$}
It seems that me that you need to say how U fits into the first topology.
4. Feb 15, 2010
### Hurkyl
Staff Emeritus
Yes they do: they say a U is an element of $\tau$.
This is standard set-builder syntax for replacement: on the right of | you introduce a variable and its domain, and on the left of the | you have a function of that variable indicating what should go into the set you're building. | 409 | 1,510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-13 | longest | en | 0.891718 |
https://physics.stackexchange.com/questions/696986/what-is-the-fundamental-definition-of-force/697000 | 1,660,328,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00219.warc.gz | 414,878,618 | 89,755 | # What is the fundamental definition of force?
As I pick up more physics I see that the definitions of force commonly provided in books and classrooms are misleading.
• "A force is a push or pull." This seems to be a "correct" definition but it doesn't provide enough information.
• "A force is the influence of one body on another." This is not sufficient because as other people have pointed out to me, force is more so the relationship between two bodies as opposed to how one acts on another. This is more evident with forces such as electricity and gravity.
• "$$\vec{F} = m \cdot \vec{a}$$." My understanding is that this is not a mathematical definition, but rather a scientific observation. Rigorous application of the scientific method led us to conclude that the relationship between force and acceleration is proportional, and the constant of proportionality is the mass of the given object. It's not a definition in the sense that we define velocity as displacement over time.
Can someone please provide an intuitive, natural definition which describes the inherent behavior between objects/bodies in the physical world? I understand that there are many different kinds of forces but since we call them all "forces" there must be a good way of defining all of them in a singular manner.
• Just because different things are referred to by the same word doesn't mean there's one definition that works for them all at once. In a car crash, the guy with the right-wing bumper sticker in the right lane might have the right of way but didn't signal right. In this case, what is the true, natural definition of "right"? Feb 28 at 20:48
• It is quite rare for anything to have a single and universally accepted definition in any hard field. You always have a wide range of different definitions with overlapping regimes of validity and different degrees of usefulness and precision. Feb 28 at 20:52
• Rather than looking for a definition, understand that all of physics is models (conceptual + formal descriptions) of observed phenomena. It has been observed that objects interact, sometimes over distance, causing each other to move; and that there's directionality to these influences. These interactions have then been termed "forces", have been modeled as vector quantities, and have then been related to other quantities and concepts within some theoretical framework. All that together is a complete description of what a force is. So, any concise def. has to be to some extent a summary. Feb 28 at 21:14
• Related: Are Newton's "laws" of motion laws or definitions of force and mass? and links therein. Mar 1 at 8:43
• I would say that the most basic definition of a force is the exchange of bosons between two fermions. I don't know how much that helps with the intuition about forces, but at least is an explanation for any force in physics, which is what I believe you are looking for. Mar 4 at 16:11
(Look at the section Some Further Clarification for a bit of meta-commentary on what we are trying to do when we are defining something. I think it has some important information.)
In Newtonian mechanics, a force is a mathematical vector we prescribe onto a model of a physical system by declaring a force law.
In other words, it's an intermediate mathematical gadget we invoke to do calculations in our models. It is invoked between the inputs (initial conditions) and outputs (predictions) of data but it is never measured directly (time, position, velocity, etc are what are ultimately recorded directly).
This is similar to how the wavefunction is invoked as a mathematical gadget to do calculations for models of quantum systems; the wavefunction is also invoked between inputs and outputs but is never directly measured. Consider the example below.
Example 1. Suppose I want to model a binary star system. I model the two stars as point objects with masses $$m_{A}$$ and $$m_{B}$$, and then I appeal to Newton's law of universal gravitation to declare the force law as $$\vec{F}_{\text{A on B}} = -\frac{Gm_{A}m_{B}}{r^{2}}\hat{r}$$ where $$\vec{r}$$ is the vector from star $$A$$ to star $$B$$. This is something I put into my model manually, because this law was very successful for Newton to make astronomical predictions.
Another example is given below.
Example 2. Suppose I want to model a harmonic oscillator put into a fluid with some drag. Then I postulate two force laws: the spring force $$\vec{F} = -k(\vec{x} - \vec{x}_{0}),$$ and the linear drag force $$\vec{F} = -b\vec{v}$$ where $$b, k$$ are some positive constants.
One important point to understand is that neither Newton's first nor second law are used to define what is a force. It's the force law specific to the situation that defines the force, and then Newton's laws relate it to motion.
Some forces are "more fundamental" in the sense that we can derive other forces from the more fundamental ones. For example, the spring and drag forces come from more elementary forces that act on the molecules of the substances. As far as we can tell the fundamental forces can be written in terms of fields, which are yet another slew of mathematical gadgets that we invoke. To define a field, we ascribe a vector (or tensor, etc) to every point in spacetime. The most well-known examples are the electric and magnetic fields.
Given a system with electric field $$\vec{E} = \vec{E}(x, y, z, t)$$ and magnetic field $$\vec{B} = \vec{B}(x, y, z, t)$$, the Lorentz force law states that the force on a particle of electric charge $$q$$ and velocity $$\vec{v}$$ is $$\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}.$$
Non-relativistic gravitation can also be put into a field-theoretic form described here. The force law for that is $$\vec{F} = m\vec{g}$$ where $$\vec{g}$$ is the "gravitational field" and $$m$$ is the "gravitational charge" in analogy with $$\vec{F} = q\vec{E}$$ for electric fields.
## Some Further Clarification
I thought about this question some more, and I realized there are a few more points that need to be mentioned.
A lot of the other answers to this questions either rely on vague intuition or they define force in terms of other things and inevitably it shifts the burden on asking what those other things are (e.g. you can say force is a change in momentum per time, but then it leaves open the question of what is momentum). I think I can give an account for why this is the case.
Let me give a related example. What are lines and points in Euclidean geometry? For a long time, lines and points were considered primitive notions that don't have any explicit definition. They were primitive things that were characterized by axioms of Euclidean geometry (the axioms told us how we could treat these concepts but there was no explicit definition in the form of "a line is blah-blah-blah" or "a point is such-and-such"). However, around the 19th and 20th century, set theory began to be developed and people made a reformulation of geometry in terms of real analysis, which was itself founded on set theory. In this new formulation, the notion of a set was the primitive (not explicitly defined) notion, and everything else was defined in terms of sets. In particular, points and lines now had concrete definitions: a point on the plane is an ordered pair of real numbers $$(x, y)$$ and a line was a set of points $$(x, y)$$ such that $$ax+by = c$$ for some real constants $$a, b, c$$. Now lines and points could be explicitly defined in terms of other things.
Now to define force, we have two options:
1. Option 1 is accept the notion of a force as a primitive concept with no explicit definition, and build axioms around how you want to characterize it.
2. Option 2 is to start in a different theory (that has its own various primitive notions) and give an explicit definition of force in terms of the elements of that theory.
I think you can see pretty clearly how these options map on to the scenario involving points and lines in Euclidean geometry. Both options are perfectly tenable.
If we start with Newtonian mechanics, then mathematically speaking force is going to have to be a primitive notion. If we start with some other formalism like Lagrangian mechanics, then the Lagrangian $$\mathcal{L}(q, \dot{q}, t)$$ will be the primitive notion, and force will be defined as $$F_{i} = \frac{\partial\mathcal{L}}{\partial q_{i}}.$$ For $$\mathcal{L} = T-U$$, force ends up being defined as the negative gradient of potential energy: $$\vec{F} = -\nabla U$$.
The above options are the only two ways you can define anything rigorously, and force just happens to be a primitive concept in Newtonian mechanics, because it starts with force.
Although force itself is primitive, it is supposed to be the mathematical concretization of the intuitive (but vague) notion of pushes and pulls (and more generally influences between bodies). The desired characterization that justifies force as the concretization of the notion of pushes and pulls is done through the axioms of Newtonian mechanics. You need to actually do and solve problems with Newtonian mechanics to understand exactly what this means.
## Regarding Newton's Laws of Motion
As I've said, what exactly is the force in a given scenario is specified by the relevant force law. If you come across a new scenario that no one else has analyzed, you will have to guess the force law and empirically test whether or not your guess leads to correct predictions.
Of course, like I've said before, the force law can come from other theories such as electromagnetism where force is defined by the electric and magnetic fields.
Newton's first and second laws are not definitions of force so much as they are axiomatic characterizations of force. There is a subtle difference, because at no point do we say "a force is defined as blah-blah-blah" in either of the laws. The role of Newton's first and second laws are to relate force to the motion of objects, and in the process of doing this they elucidate what it means for a force to be "a push or a pull" or to be "an influence of one body on another."
Newton's third law is different from the other two laws, because unlike the first two laws the third law gives a constraint on what the possible force laws (which are the things that specify what the force is in a given scenario) there can be. In many cases, we actually ignore this law (for example when we consider a spring attached to a wall, we simplify our scenario by ignoring the fact that the motion of the spring imparts some momentum to the Earth). What the law truly means is that any time we have a force without an opposite force, the system we are analyzing is not truly a closed/isolated system.
• Best answer so far IMO. "Time, position, velocity, etc are what are ultimately recorded directly" not sure about this statement: can we directly measure velocity? Are those 3 things the only things that can be "directly" measured? Feb 28 at 22:26
• @Quillo I'm not 100% sure. I've seen people argue that time and position might be the only "directly" measurable things. In any case, as far as I can tell, every measure of force relies on a measure of something else. Maybe the feeling of pressure (force divided by area) seems like it doesn't involve position, but actually that occurs because our nerves send actual electrical signals which have to be carried by something... so even that involves changes in position. Feb 28 at 22:46
• What is the force law for the normal force acting on a body from the ground? Sometimes forces exist without an explicit definition of their magnitude, but as enfocers of constraints. Mar 1 at 13:30
• @Quillo One minor addendum is that I think we have to be careful in what we mean. For example, sometimes when people say "gravity" they may refer to the observable phenomena that things fall down. This is obviously not a mathematical gadget. Other times, when people say "gravity" they may refer to the force vector in Newtonian mechanics or to a term in a Lagrangian. Clearly, these two have different meanings. The same applies to the other forces/interactions. Mar 2 at 16:11
• @MaximalIdeal I totally agree. I was just replying to a previous comment of ACuriousMind by noticing that the definition of "force" as "gadget" also applies perfectly to the classical concept as well as to "fundamental interactions". Mar 2 at 16:18
"A force is a push or pull." This seems to be a "correct" definition but it doesn't provide enough information.
That is the most commonly cited qualitative definition. It's broader than using Newton's 2nd law since, as discussed below, Newton's 2nd law only addresses the influence of a net force. A force (push or pull) does not require that there is an influence.
Insofar as whether or not it provides enough information it depends on what kind of information you are looking for.
"$$\mathbf F = m \mathbf a$$." My understanding is that this is not a mathematical definition, but rather a scientific observation.
Newton's 2nd law provides information on what a force does. But if you are looking for a better mathematical definition of the effect of a force, I think you are better off defining the effect of a net force as the change in momentum of an object, or
$$F_{net}=\frac{dp}{dt}$$
where, for the case of constant mass,
$$\frac{dp}{dt}=m\frac{dv}{dt}=ma$$
The reason I believe this is a better mathematical description of the effect of a force is that conservation of momentum is one of the fundamental laws of physics.
The emphasis is on net force, because though "pushing or pulling" is a force, there may be no effect unless there is a net force. I can push and pull on a wall all day, but if it doesn't move (cause a change in momentum) my force has no effect (at least, macroscopically) on the wall.
"A force is the influence of one body on another." This is not sufficient because as other people have pointed out to me, force is more so the relationship between two bodies as opposed to how one acts on another. This is more evident with forces such as electricity and gravity.
I have a few issues of what you have been told here. For one thing, the influence may due to contact between bodies, or the influence may be due to a field between the two bodies. But the main reason not to define force as "the influence of one body on another", in my view, is as I said above, a force does not necessarily influence a body (read rigid body) unless it is a net force.
I'm actually more concerned with being accurate than being precise. Would it be fair to say that this definition applies to all forces in physics? "A force is a push or pull resulting from an object's interaction with another object."
I would say the "push or pull" definition applies at least to two of the four fundamental forces, i.e., the gravitational and electromagnetic force. I'm not so sure in the case of the other two, the strong and weak forces. As far as your original statement
I understand that there are many different kinds of forces but since we call them all "forces" there must be a good way of defining all of them in a singular manner.
That, of course is the Holy Grail. Gravity still has not been combined with the other three.
Hope this helps.
• I'm actually more concerned with being accurate than being precise. Would it be fair to say that this definition applies to all forces in physics? "A force is a push or pull resulting from an object's interaction with another object." Feb 28 at 21:36
• @EthanDandelion I have revised my answer to respond to your follow up question. Feb 28 at 21:49
I think that force has at least $$4$$ layers of meaning.
The primary meaning is an intensive quantity, something that we feel with our muscles, mainly when pushing or pulling. As such it is not measurable, because even if we can say that force A is bigger that B, it is not possible to precise how much.
In order to measure the force and treat it as an extensive quantity, we use the Hooke's law to make load cells, strain gages, and other devices. That is the second layer.
The discovery that net force is proportional to acceleration leads to the third level. As that is a more universal property than elasticity (which can have a short range, depending on the material), this discovery is 'promoted' to the standard way to measure the net force. And if a load cell is not perfectly linear with acceleration for a given mass, the Newton's second law prevails, and it is a way to fine tuning the degree of non-linearity of the load cell.
The fourth level derives from the observation that sometimes acceleration of particles is a function of the position in a system of coordinates, and for such cases, the (conservative) force can be defined as minus the gradient of a scalar function. In that way for example , even the force between $$2$$ protons of a $$H_2$$ molecule can be known as a function of their momentarily distance.
I happen to like Aristotle's definition although he didn't use the term force. Essentially, force is that which causes change. More precisely, he wrote in his Physics:
... anything which can cause change must cause something to be changed and it must be something that can be changed. Similarly, what can be changed must be changed by something and it must be something that has the ability to cause change ... when something changes, it inevitably does so in respect of substance, quantity, quality or place ... the upshot is that there are as many kinds of change as there are categories of being.
$$200^b26$$
His categories of being are four:
• actual existence and their change is 'coming to be' and 'passing away'.
• the number of things and their change is an increase or decrease in number. He means here integral number, like for example the number of atoms.
• quality, these are continuous things such as length or mass and their change is what he terms alteration - their continuous increase or decrease
• place, this is position and change here is just change of position, that is motion.
Thus force is that which can cause things to come to be, like particles coming into existence; or to pass away, like particles annihilating; and such forces obviously changes the number of particles, either their increase or decrease; more, forces are what causes change in volume, say pressure.
For Aristotle, the world is a network of forces inhering in matter and acting on matter and thus causing change in substance, number, quality and place.
It's worth seeing how Aristotle's definition of force stacks up against the classical definition, that of Newton. This is usually expressed symbolically as $$F=ma$$. But this is not what Newton wrote in his Principia, what he actually wrote was:
The alteration of motion is ever proportional to the motive force impressed and is made in the direction of the right line in which that force is impressed.
Obviously, Newton's definition is much narrower than that of Aristotle's. He focuses only on motion. Thus his term 'motive force', a force that causes change in motion. The crucial question is whether Newton's law a specialisation of Aristotle's? Well we see a 'motive force is impressed' and this causes an 'alteration in motion'. Alteration is obviously change and change is what we are looking for when characterising a force according to Aristotle. And Aristotle does specify change in place as one of the kinds of changes possible. However, here Newton is not talking about change in spatial position but change in velocity. But equally soundly, we argue that velocities constitute a space, the space of velocities. So, yes, it does fit. Of course this is really an extension of Aristotle's definition as did not concieve of a space of velocities; however, he left his theory open to extension because although he characterised four main categories of being, he recognised that there were other more specialised senses.
Whilst Arostotle's law is broader, we see that Newton's law is quantitative, it says the change of motion is 'proportional' to the motive force as well as specifying the direction of change. Aristotle's definition is qualitative and as he said himself, one can become more precise as this law is specialised to more specific domains, as is here by Newton.
It's also worth noting that Newton says 'impressed' and this means that the force should act by contact. In fact, Newton felt philosophically that all forces should act by contact and this is why he understood his theory of gravity to be incomplete since it had forces acting at a distance. Newton doesn't say why forces should act by contact but it's likely the original source was Aristotle. In fact, he wrote:
Everything that cause change is changed ... as long it is capable of changing ... For to act on something changeable, in so far as it is changeable, is precisely to change it, and it takes contact to do this, so the agent of change is acted upon at the same time.
$$202^a3$$
Thus forces take 'contact' to act on and cause change. In Newton's language, they need to be 'impressed' upon. But more, we also see that the preceding passage is a qualitative statement of Newton's third law:
To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each another are always equal and directed to contrary parts.
Again, Newton provided the crucial quantifying information: these two forces are equal and acting in opposite directions.
Finally, I'd like to add a few words about what Aristotle meant by change. Change is familiar and ever since the discovery of calculus it has been straight-forward to model. Aristotle thought otherwise. He struggled to characterise change and his struggle had nothing to do with the lack of a suitable calculus. What he was stuck on was ontology. It is easier to say what something is or not. But change, which ontological category does that lie in? Aristotle himself said:
Also, the process of change does seem to be an actuality but an incomplete one and the reason for this is that the potential of which it is the actuality is incomplete. This makes it hard to grasp what change is. For it has to be assigned either to privation or to potentiality or to simple actuality. But none of these seem to be possible.
$$201^b31$$
Eventually, Aristotle assigns it to a 'special kind of actuality'. That Aristotle struggled with this question is to understand that Newtonian mechanics is not really about change, it is more akin to geometry. It is motion tackled by geometrical means. And the same is true for it's classical completion, GR. However, the interpretational question that Aristotle found elusive reared it's ontological head in QM. The quantum wave describes change at the fundamental level, in a sense, it is our unit of change, but what is the ontological status of a quantum wave? Like Aristotle, we can certainly say it is not actual and only when it is 'complete' (to use Aristotle's language) or 'measured' (to use QM language) is it actual in the sense an observable yields up a measurable value.
One that doesn't seem to have been mentioned so far, from Lagrangian mechanics:
Forces are potential energy gradients. Wherever a change in the configuration of system would increase the potential energy in the system, there is force opposing that change.
The sum of $$-F\cdot d$$ over all the moving parts determines how much potential energy is increased by any small change, and motion against or with a force is in fact the mechanism by which energy is transformed from kinetic to potential form and vice-versa.
In short, potential energy wants out, and force is the expression of that.
I'm a bit out on a limb here but I think all the existing answers are missing the point, so I'm going to add my 2 cents.
The definition of a physical quantity is how you measure it.
Any definition in terms of mathematical or common language reference to other quantities is begging the question, "okay, but how do we measure those?" Force is mass times acceleration. Okay, acceleration is what an accelerometer measures, but mass is the ratio of acceleration to force, and we're right back where we started.
Force is what a calibrated scale measures.
A scale is a machine that measures the difflection of a known spring element and applies Hooke's law to relate force as a linear function of displacement.
Ultimately we could calibrate a scale using only times and distances, starting with the a priori definitions of $$c$$ and $$h$$ as basis constants, by looking at the work done on a beam of light in an interaction, as energy per quantum if light is a function whose only inputs are distance, time, and fundamental constants.
There may be other machines that will output the same value as a calibrated scale. If so, force could be equally well defined as what one of those machines measures.
Force is the rate of change of momentum. The total momentum of an isolated system is constant, but its parts can mutially exchange momentum. A change in momentum of one part is then accompanied by an equal but opposite change of momentum of some other part. The rates of change are opposite, which constitutes Newton's third law.
Okay. One more. I think force is that, in a freely falling frame, what causes free particles to have velocities, in the mean, that change over time. Kinetic energies change. As do potential energies.
Free particles can collect in stable structures. Forces can balance. No change in kinetic energy will be seen although forces are at work.
What causes force? Charge. Charged particles want to be with each other or they don't.
Curvature of spacetime causes tidal forces that drives particles away from each other. The falling of a stone on Earth is caused not by the curvature of spacetime but by electric charges keeping up the surface of the Earth.
If there weren't charges in particles (only mass), all particles would have constant velocities, but tidal effects would cause relative accelerations. Which can be called forces. If no particles had charge all mass in the universe would collapse to black holes, though if they didn't have charge from the start, it remains to be seen if they could collect in lumps, and it could even be asked if they could have mass or exist at all.
I would argue that Newton's first law defines what is meant by a "force". Many mistakenly believe that Newton's first law is redundant and can be derived from Newton's second law $$F=ma$$ but this is wrong. If that were the case then why would there even be a "Newton's first law"?
Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.
What is a force? A force is something that disturbs a body's state of rest or its uniform motion in a straight line.
Edit: The mathematics of Newton's first law $$F=ma=0$$ can be derived from Newton's second law. But that would be to completely miss the point of Newton's first law.
Second Edit: The definition of Newton's first law above was taken from the top part of "Newton's laws of motion" wikipedia page. However further down the wording "net force" is used instead of "force" which is more accurate.
• "Something that disturb's a body state of uniform motion." Is it fair to say that this definition applies to absolutely any force in physics? Thank you. Feb 28 at 21:30
• I would say yes this definition applies to all of the four forces; electromagnetic force, gravitional force, strong force and weak force. I am as @Bob D a bit unsure if it applies to strong and weak force but I would guess it does.
– ludz
Feb 28 at 22:06
• @EthanDandelion No, this is obviously a definition of force that only works in classical mechanics -- as would be the case with any definition of any concept in classical mechanics.
– ACat
Feb 28 at 23:04
• @BobD After sitting down and thinking about it for a couple of days I think this definition may actually be viable, if we use my previous definition: "Something that disturb's a body state of uniform motion." This is TECHNICALLY correct because if the body is already at rest, adding a singular force will disturb that state. However, my thinking now is that this definition simply isn't descriptive enough. To address this disturbance of uniform motion we effectively have to ignore all of the forces keeping it in equilibrium. Mar 5 at 0:01
• @EthanDandelion If the object is sitting at rest next to a wall and we then add a singular force then it will not "disturb the state" since the wall will immediately counter this. It still needs to be net force.
– ludz
Mar 5 at 9:56
I really don't think one can do better than to say a force is the measure of a specific kind of interaction. It is interaction, not "force", that is the most fundamental concept; and I don't think there's a better definition of this than its usual meaning: "what is happening amongst a number of things when they are having an influence on each other by their mutual presence". This is like your definition 2), but note that it is not a definition of force.
One can consider that the most basic object in classical mechanics of a single point-like object is its kinematic state:
$$(X, \mathbf{p})$$
where $$X$$ is the position, a point, and $$\mathbf{p}$$ is the momentum, a vector. To describe the system over time, we assign a state to it at each point in time. "Influence", then, shows up as a change therein, and we can translate "interaction" into agents of such change. We like to represent these interactions as having a summatory effect: if we let the state be denoted by $$\mathbf{с}$$, then
$$\frac{d\mathbf{с}}{dt} = \text{(sum of all 1-body interactions)} + \text{(sum of all 2-body interactions)} + \cdots$$
Force, then, is the measure of effect each interaction individually contributes to the change in the momentum:
$$\frac{d\mathbf{p}}{dt} = \text{(sum of all 1-body forces)} + \text{(sum of all 2-body forces)} + \cdots$$
where there is a 1:1 correspondence between interactions and forces.
When it comes to the "inherent behavior" of objects, though, there isn't really so much a way we can access it as that we have to invent a description language suitable for it. The above is one of many languages we can invent for such a purpose. (*)
But I also submit this does not make the question meaningless, for while those languages are invention, it is far from nonsense to ask what the meaning is that we have invented or created, or can invent or create, for a word or term therein that is suitable and appropriate to how it behaves. A language where nobody knows what the terms mean is quite a useless one, regardless of its fundamental ontological status. Although, of course, some theorists of language might argue that language is inherently empty of meaning, but if we wanna go down that rabbit hole then we are both a) really far from physics now and well into the realms of philosophy, especially non-analytic philosophy, and b) I am not sure if it's useful to actually trying to get one's head around the theory in question.
(*) Why choose a sum, for example? Actually, we don't have to. However we find empirically that, for some reason, things "factor more nicely" with sums; e.g. gravitation could be many different interactions together. Of course - maybe it really is. We don't know, and maybe we need to not be so dogmatic as to just avoid ever entertaining such ideas!
A NET force is an interaction that causes mass to accelerate in the direction of the net force. Since there are only two known macroscopic fundamental forces, that interaction must necessarily be either electromagnetic or gravitational for forces applied to macroscopic objects.
• Hmm… Pauli exclusion causes deceleration upon contact, detectable as a net force, but is not one of those two fundamental interactions. Mar 1 at 0:23
I guess one way to think about it in a generalized, abstract way, is in the style of abstract mathematical definitions (e.g. a topological space is any object of the form [set + some subsets] that satisfies such and such axioms).
So, in that spirit (but without attempting that level of rigor), perhaps one could go about it this way.
Within the context of Newtonian dynamics, you could say that there are the following building blocks*:
1. A notion/concept that there can be these directed influences of different magnitudes between objects that act either on contact or over distance, that are capable of changing the state of motion of objects they act upon, and can be represented as vectors**. Qualitatively, these influences "push" or "pull". These are then termed "forces".
$$\hspace{9pt}\large \&$$
1. The idea that, given a description (model) that explains the dynamics of a system, some elements of the model (vectors, vector-valued functions, ...) can be designated as forces iff these model elements exhibit roles and behaviors expressed by Newton's laws (law of inertia, $$\textbf F=m\textbf a$$ or equivalents, action-reaction).
Inline footnotes:
* These are probably incomplete (additional assumptions could potentially be included, like flat spacetime).
** Vector here being an abstract mathematical object (so, distinct from how one chooses to represent it (e.g., from a particular coordinate representation).
Then, when you are able to successfully map this framework onto an actual physical situation (and you may not be able to) and onto the accompanying contextualized mathematical description, you call "force" whatever ends up being mapped to the concept of the force as defined above.
Further, if within a wider theoretical conceptualization there's an explanation for the origin of the force, then it's considered a real force (an actual physical thing), if there isn't, then it's a fictitious force. (And then perhaps you could build on top of that to define inertial frames, etc. I know this seems backwards, but, if you start with inertial frames, then you have to bolt on fictitious forces, and explain the sense in which these are "forces", why use the term, and so on. I think this works, but I could be missing something.)
In other words, if it's a vector-like influence, and if it is behaving in accordance to Newton's laws, it can be called a force in some sense compatible with $$1\ \&\ 2$$ (if it looks like a duck, and quacks like a duck, ...) — unless there exists within a larger theoretical context some overriding reason that modifies the meaning, or alters the fundamental description of the system (e.g. when people invoke GR and say that gravity is not really a force in the above sense, but that you can treat it as such under certain circumstances).
Also, in a larger theoretical context this conceptualization can be related (perhaps with caveats/constraints) to other qualitatively different ideas (e.g. negative gradient of the potential), equivalences can be established between different theoretical descriptions, etc.
And then there are uses of the term outside of this context, where the meaning may be more loose, and Newtonian mechanics might not even apply (e.g. strong interaction being called strong force).
An ontological answer to round up all the other attempts:
What is the fundamental definition of force?
Since you know of $$F=ma$$ and do not accept that as your answer, I guess you want to know what force "really" is. Objectively, in the universe, without any human concepts or "theories".
Herein lies the crux. This kind of answer cannot be answered. This is the age old problem of Plato's cave. We do not know the answer to questions of that kind for any physical phenomenon.
We do not know what an electron actually is. We do not know what a photon is. We don't know what electromagnetic fields really are. All we know about anything is our models, which describe some aspects of the behaviour of physical objects in relationship to other objects at varying levels of detail and correctness.
Yes, we have become quite refined, and our knowledge is substantial. We've split the atom and watched the big bang, but we still are not a single bit of knowledge closer to knowing what all that stuff actually is. It would still be quite possible that I am just a brain in suspension, with all my nerve cells connected to some computer feeding me totally wrong information. Or I could still be a simulation in a computer, just like you and everyone else, in an actual universe which behaves totally different than what we are fed.
(Seeing all the other answers, my own best guess at a helpful answer would be to simple re-iterate Newton's First and Second Law - for all intents and purposes, everything that makes objects move is a force and vice versa, by definition - fully accepting that this is only useful for the electromagnetic force and maybe not so much for the other three fundamental forces).
• Quoting the answer physics.stackexchange.com/a/697000/226902 , "In Newtonian mechanics, a force is a mathematical vector we prescribe onto a model of a physical system by declaring a force law.". This is "really" what a force is, because this is exactly the description of the concept of "force" that is used ever yday by "real" humans to make calculations on "real" paper or running simulations on "real" computers. Nature would probably work the way it does with no regard for humans inventing and using a formal "force" concept. Mar 2 at 10:51
• True about "reality", but the question also asks about a definition. That is part of a human model, not something separate from it. Hence, we can ask whether the term force is given definition. Apr 26 at 2:43
A comprehensive and exhaustive definition of force is yet to evolve through discussions, studies, in depth analysis etc. F=ma helps to calculate the magnitude of force in cases where motion is involved.
In the gravitational field for example, every cubic centimeter space in the gravitational field is filled with 'Force'. The moment an object with mass and volume is brought into that space, it experiences force in the form of 'gravitational pull'. It will experience force irrespective of spot 'A' or spot 'B' in the gravitational field. The object is only a medium for the gravitational force to act. Irrespective of whether object is there or not, the 'causal gravitation pull' is present through out the gravitational field through out the time (Is it a wave?)
In an IC engine, when fuel burns inside the cylinder, high pressure is developed which exerts the force/pressure on the piston which in turn causes the shaft to rotate.Here, force is derived by burning of fuel, and that force is obtained during the combustion stroke only and not through out the time.This is a case force being developed by spending energy over a period of time, as opposed to 'gravitational force' which is present all the time without burning any fuel.
In the case of magnetism, both attraction and repulsion forces are there. Only magnetic materials experience magnetic force. Between magnets, there is attractive and repulsive force. To explain away all these, we need much more insight.
This discussion is inconclusive. We can rest assured that a comprehensive out look will emerge as the discussions, exchange of ideas proceed.
As you probably know, one of the observations about the universe is that there exist frames of references where particles move in straight lines when they seem to be far away in isolation from all the other particles.
But in general, the goal of physics is to predict the trajectories of interacting particles that are not in isolation (non-straight lines in general).
Then it follows that the equations of physics should be able to tell us the acceleration of a particle, if we provide the current state of the system as input (positions and velocities of all the particles in the system).
Acceleration is the quantity we care about in the end, as the goal of physics is to predict trajectories.
Now, how do we actually make inferences about how acceleration behaves? Obviously: by making observations about acceleration.
Observation 1: In an isolated system of two particles (say, billiard balls), their accelerations at any point of time during the collision is observed to have the property:
$$-\frac{a_1(t)}{a_2(t)}=c$$
The minus sign indicates the opposite directions of $$a_1$$ and $$a_2$$. $$c$$ is always the same positive constant. This observation allows you to say "Particle 2 is $$c$$ times as massive as Particle 1". We're defining relative mass right now. In any interaction involving the two particles, particle 1 "suffers" $$c$$ times as much change in velocity as particle 2, in the same amount of time.
We've defined mass as the constant measuring measuring resistance to change in the state of motion. We can further choose a reference mass, call it 1 unit, and relative to it, ascribe mass values ($$m_i$$) to all the particles of the universe. Then the equation would become:
$$-\frac{a_2}{a_1}=\frac{m_1}{m_2}$$
or $$m_1a_1+m_2a_2=0$$
or $$m_1v_1+m_2v_2=constant$$
Moving on, so far we've only observed a relationship between two accelerations $$a_1$$ and $$a_2$$ and that happen in pair. There's still no formula to measure either $$a_1$$ or $$a_2$$, given the initial state of the particles (positions, velocities).
Enter gravitation. The acceleration formula that has been observed to explain the motion of planets is $$\frac{Gm}{r^2}$$, $$m$$ being the mass of the sun and $$r$$ being the distance between Sun and Earth.
For the first time, we actually are able to calculate the acceleration in a system of two interacting bodies.
But, hold on, this acceleration formula coupled with the previous observation is enough to calculate the trajectory of the system. Then, who cares about Force?? Why introduce a middle man called Force??
I mean..you could introduce a quantity with the formula $$F_1=m_1a_1=\frac{Gm_1m_2}{r^2}$$. Then the calculation of acceleration ($$a_1=\frac{F_1}{m_1})$$ would go one step down the road for no reason. Why introduce Force when acceleration itself has a much simpler formula?
For one, you can't deny Force has a convenient property. From the first observation, $$m_1a_1=-m_2a_2$$. Unlike acceleration, this guy occurs in equal an opposite pairs.
For another, gravity isnt't the only interaction that we've observed:
Enter electromagnetism: Now we're in a realm where the observed laws are best described in terms of the quantity $$ma$$. The observed laws are $$F=k\frac{q_1q_2}{r^2}$$ and $$F=qvB$$. $$m$$ no longer simply cancels out in the division. In fact, it sticks around and acceleration ends up having an uglier formula.
This, combined with the "equal and opposite property", now makes Force the natural quantity to work with. Acceleration becomes the derived quantity by $$a=\frac{F}{m}$$
Another reason Force is a convenient quantity: Work energy theorem
With just gravity, you could simply write down the work energy theorem as:
$$\frac{1}{2}(v_2^2-v_1^2)=\int \frac{GM}{x^2}dx$$
No force involved in the above formula. You could forcefully involve it by multiplying both sides by $$m$$.
This again changes in electromagnetism:
$$\frac{1}{2}m(v_2^2-v_1^2)=\int k \frac{q_1q_2}{x^2}dx$$
This is a fundamental equation in which the RHS is naturally an integral of Force. You can no longer cancel out the mass | 9,594 | 43,344 | {"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": 80, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-33 | latest | en | 0.964038 |
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### Machine Learning Notes 4
Gradient descent algorithm for multiple linear regression
(i.e. for linear regression with multiple variables)
Terminology (notation used):
$n$ - this is the number of features (the number of input variables)
$X_1, X_2, \dots, X_n$ - these are the features
$X_j$ - the $j$-th feature ($j = 1,2,\dots,n$)
$Y$ - the target (output) variable
$m$ - the size of the training set (the number of training examples)
$X_j^{(i)}$ - the $i$-th training value from the $j$-th feature ($i = 1,2,\dots,m$)
Let's assume that we have the following features and training examples.
# $X_1$ $X_2$ $X_3$ ... $X_n$ $Y$ 1 2100 5 1 ... 45 460 2 1400 3 2 ... 40 235 3 1530 3 2 ... 30 315 ... ... ... ... ... ... ... m 850 2 1 ... 35 175
Here we have that $X_1^{(3)} = 1530$, $X_2^{(2)} = 3$, $X_n^{(3)} = 30$, $X_n^{(m)} = 35$
$\overrightarrow{X}^{(i)}$ - this is a row vector of all $i$-th training values (basically the $i$-th row from the table above without the Y values), so this is the $i$-th training example
$\overrightarrow{X}^{(i)} = (X_1^{(i)}, X_2^{(i)}, \dots, X_n^{(i)})$, for $i=1,2,\dots,m$
Note: The training examples are the rows from the table above.
The linear regression model for a single feature was defined as follows.
$f_{w,b} = w \cdot x + b \tag{1}$
The linear regression model for multiple features is defined as follows.
$f_{w,b} = \sum_{j=1}^n w_j \cdot x_j + b \tag{2}$
We can rewrite this model in vector notation. Let us denote
$\overrightarrow {w} = (w_1, w_2, \dots, w_n)$ - this is a row vector of $n$ scalars (these are the weights assigned to the features)
$b$ - this is a scalar (a number), this value $b$ is the bias
$\overrightarrow {x} = (x_1, x_2, \dots, x_n)$ - this is a row vector of $n$ scalars (the input variables to our model)
Then for the multiple linear regression model we obtain
$f_{\overrightarrow{w},b} = \overrightarrow{w} \cdot \overrightarrow{x} + b \tag{3}$
Here $\overrightarrow{w}, \overrightarrow{b}$ are the parameters of the multiple linear regression model.
In this notation $\overrightarrow{w}\cdot \overrightarrow{x}$ is the dot product of the vectors $\overrightarrow{w}$ and $\overrightarrow{x}$ i.e.
$\overrightarrow{w} \cdot \overrightarrow{x} = \sum_{j=1}^n w_j \cdot x_j \tag{4}$
Note: This is called multiple linear regression, and not multivariate linear regression. The term multivariate linear regression in machine learning is used for something else.
For example in the table above $Y$ might be the price of a house (in 1000s of US dollars) which we are trying to estimate (i.e. build a model for), $X_1$ might be the size of the house in square feet, $X_2$ might be the number of bedrooms, $X_3$ might be the number of floors, and $X_n$ might be the age of the house in years (let us say n=4). Then one possible linear model for the price (Y) of the house could be as follows:
$f_{w,b}(x) = 0.1 \cdot x_1 + 4 \cdot x_2 + 10 \cdot x_3 + (-2) \cdot x_4 + 80 \tag{5}$
Here we have $b=80$, and this is called the base price.
We denote
$\hat{y}^{(j)} = f_{\overrightarrow{w}, b} (x^{(j)}) = \sum_{i=1}^n w_i \cdot x_i^{(j)} + b\tag{6}$
This value is the predicted value (by the linear regression model) which
corresponds to the $j$-th training example (here $j=1,2,\dots,m$).
We also have
$y^{(j)}$ - the actual value corresponding to the $j$-th training example ($j=1,2,\dots,m$).
The least mean squares (LMS) cost function is defined as follows
$$J(\overrightarrow{w},b) = J(w_1, w_2, \dots, w_n, b) = \frac{1}{2m}\sum_{i=1}^m (\hat{y}^{(i)} - y^{(i)})^2 \tag{7}$$
$$J(\overrightarrow{w},b) = \frac{1}{2m}\sum_{i=1}^m (f_{\overrightarrow{w},b}(\overrightarrow{x}^{(i)}) - y^{(i)})^2 \tag{8}$$
$$J(\overrightarrow{w},b) = \frac{1}{2m} \sum_{i=1}^m (w_1 x_1^{(i)} + w_2 x_2^{(i)} + \dots + w_n x_n^{(i)} + b - y ^ {(i)})^2 \tag{9}$$
So we get that
$$J(\overrightarrow{w},b) = \frac{1}{2m}\sum_{i=1}^m [ (\sum_{j=1}^n w_j \cdot x_j^{(i)}) + b - y^{(i)}]^2 \tag{10}$$
The gradient descent algorithm essentially varies the $w_j$ and $b$ parameters of the model to find the optimal i.e. the minimal value for $J$ (since $J$ is a function of the parameters $b$ and $w_j$, $j=1,2,\dots,n$).
So we compute
$$\frac{\partial{J}}{\partial{w_j}} = \frac{1}{m}\sum_{i=1}^m (f_{\overrightarrow{w},b}(\overrightarrow{x}^{(i)}) - y^{(i)}) \cdot x_j^{(i)} \tag{11}$$
(here $j=1,2,\dots,n$)
and
$$\frac{\partial{J}}{\partial{b}} = \frac{1}{m}\sum_{i=1}^m (f_{\overrightarrow{w},b}(\overrightarrow{x}^{(i)}) - y^{(i)}) \tag{12}$$
Now that we have formulas for these partial derivatives, the gradient descent algorithm for multiple linear regression goes as follows.
1. Pick a small positive number $\alpha \gt 0$.
2. Repeat these four actions until convergence
2.1.
$$temporary\_w_{j} = w_{j} - \alpha \cdot \frac{\partial{J}}{\partial{w_j}}$$ for $j=1,2,\dots,n$
(use the partial derivative expression from (11))
2.2.
$$temporary\_b = b - \alpha \cdot \frac{\partial{J}}{\partial{b}}$$
(use the partial derivative expression from (12))
2.3.
$$w_j = temporary\_w_j$$ for $j=1,2,\dots,n$
2.4.
$$b = temporary\_b$$
• The temporary variables here are used to underline the fact that the algorithm is doing a simultaneous update of the (n+1) parameters $w_1, w_2, \dots, w_n, b$. That means we should not change any of the parameters before we have computed the new values for all the parameters.
• Convergence means that the values of the (n+1) parameters $w_1, w_2, \dots, w_n, b$ have settled i.e. it means they all change by very little after certain number of iterations (say after K iterations/steps). To detect convergence we can use some sort of threshold e.g. we can assume the algorithm has converged if all parameters change only by less then 0.05% after certain number of iterations/steps.
This completes our description of the gradient descent algorithm for multiple linear regression.
### Generating a minimal set algebra from a given set of sets
This is a Python program which generates the minimal set algebra which contains all the elements/sets of the set C. In this program Omega is the universal set. Then C is a subset of the powerset of Omega. The program generates the minimal set algebra C3 which contains all the elements of C. A set algebra is also known as a field of sets. Of course here Omega and C are both finite sets.
import itertools as it
Empty = frozenset({})
def main():
# Omega ---> the universal set
Omega = frozenset({1, 2, 3, 4, 5, 6})
# C ---> a set of subsets of Omega
# C = {frozenset({1}), frozenset({1,3})}
# C = {frozenset({1,2}), frozenset({2,3,4}), frozenset({5})}
C = {frozenset({1, 2}), frozenset({2, 3, 5})}
# Now we generate the sets C1, C2, C3
# C3 is the minimal boolean algebra
# which contains all the elements of C
print("=" * 60)
C1 = gen_C1(Omega, C)
C2 = gen_C2(Omega, C1)
C3 = gen_C3(Omega, C2)
print("The set of sets C is given as input.")
print("Size: ", len(C), "---> C = ", make_lst(C))
print("Added all complements to C ---> result: C1")
print("Size: ", len(C1), "---> C1 = ", make_lst(C1))
print("Added all intersections to C1 ---> result: C2")
print("Size: ", len(C2), "---> C2 = ", make_lst(C2))
print("Added all unions of disjoint sets to C2 ---> result: C3")
print("Size: ", len(C3), "---> C3 = ", make_lst(C3))
print("Checking if C3 is indeed a set algebra ---> ", is_set_algebra(Omega, C3))
print("=" * 60)
def make_lst(C):
P = []
for A in C:
P.append(str(set(A)) if not Empty == A else "{}")
return P
def powerset(Omega):
n = len(Omega)
ps = set({})
for k in range(0, n + 1):
z = it.combinations(Omega, k)
for t in z:
return ps
def gen_C1(Omega, C):
ps = powerset(Omega)
C1 = frozenset({})
C1 = C1 | {Empty}
C1 = C1 | {Omega}
for A in C:
C1 = C1 | {A}
C1 = C1 | {Omega - A}
return C1
def gen_C2(Omega, C1):
P = Empty
k = 0
while (True):
k = k + 1
sz1 = len(P)
Z = Empty
S = C1 if k == 1 else P
for A in S:
for B in S:
Z = Z | {A & B}
P = P | Z
sz2 = len(P)
if sz2 == sz1:
break
return P
def gen_C3(Omega, C2):
P = Empty
k = 0
while True:
k = k + 1
sz1 = len(P)
Z = Empty
S = C2 if k == 1 else P
for A in S:
for B in S:
# print("A=",A)
# print("B=",B)
if A & B == Empty or A & B == A or A & B == B:
Z = Z | {A | B}
P = P | Z
sz2 = len(P)
if sz2 == sz1:
break
return P
def is_set_algebra(Omega, U):
if Omega not in U:
return False
if Empty not in U:
return False
for A in U:
B = Omega - A
if B not in U:
return False
for A in U:
for B in U:
if not (A | B in U):
return False
if not (A & B in U):
return False
return True
if __name__ == "__main__":
main()
### Equation of a plane passing through 3 points
Equation of a plane passing through 3 points
Let's assume we're given 3 distinct points in the 3-dimensional space and these points are not collinear. On a side note, this implies the given 3 points are also distinct.
Let's also assume that their coordinates in some 3-dimensional coordinate system $S = \{O,\overrightarrow{e_1},\overrightarrow{e_2},\overrightarrow{e_3}\}$ are as follows.
$P_1 = (x_1, y_1, z_1)$
$P_2 = (x_2, y_2, z_2)$
$P_3 = (x_3, y_3, z_3)$
The coordinate system $S$ does not need to be orthogonal.
Then the equation of the plane passing through these points is as follows.
$$\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix} = 0$$
Note: This is a determinant in the expression above.
### Machine Learning Notes 3
for univariate linear regression
(i.e. for linear regression with one variable)
Let $J(w,b)$ be any cost function of two variables. If we want to be specific, we can assume that $J(w,b)$ is the usual mean squared error cost function from the univariate linear regression model.
See here: ML Notes 2
1) The algorithm first sets two arbitrary values into $w$ and $b$ (e.g. $w := 0,\ b := 0$)
2) Then a loop begins. At each step of that loop the algorithm applies these two updates simultaneously
2.1) $w := w - \alpha \cdot \frac{\partial J}{\partial w}(w,b)$
2.2) $b := b - \alpha \cdot \frac{\partial J}{\partial b}(w,b)$
Here $\alpha$ is some small positive number e.g. $\alpha=0.01$
This number $\alpha$ is called the learning rate. It controls the size of the step which the algorithm takes downhill in each iteration of the loop.
3) The loop repeats step 2) until the values of $(w,b)$ eventually settle at some point $(w_0, b_0)$. Settle means that after a number of steps, the values of $w$ and $b$ no longer change much with each additional step of the algorithm. At that point, the algorithm exits the loop. The point $(w_0, b_0)$ is the point where $J(w,b)$ has a local minimum. We also say that the algorithm converges to the point $(w_0, b_0)$ after a number of steps.
Note 1: A correct implementation ensures that the updates in step 2) are done simultaneously. What does that mean? That means in the two right hand sides of the assignments 2.1) and 2.2), we should make sure we are using "the old values" of $w$ and $b$. Usually this is implemented in the following way.
$tmp\_w := w - \alpha \cdot \frac{\partial J}{\partial w}(w,b)$
$tmp\_b := b - \alpha \cdot \frac{\partial J}{\partial b}(w,b)$
$w := tmp\_w$
$b := tmp\_b$
Note 2: Choosing a good value for $\alpha$ is very important. If $\alpha$ is too small, the gradient descent algorithm will work correctly but might be too slow. If $\alpha$ is too big, the gradient descent algorithm may overshoot (as they call it), i.e. it may miss the local minimum, i.e. the algorithm may diverge instead of converge (this is usually detected by the condition that after several iterations of the algorithm, the cost $J(w,b)$ actually gets bigger instead of getting smaller).
This algorithm is also known as batch gradient descent which refers to the fact that at each step of the algorithm all the training examples are used (i.e. at each step the entire training set is used). There are also other versions of the gradient descent algorithm which at each step use only subsets of the entire training set.
### Real Analysis - Basic Differentiation Rules
$C' = 0$, where $C$ is a constant
$(C \cdot x)' = C$, where $C$ is a constant
$(x^n)' = n \cdot x^{n-1}$, $x \in \mathbb{R}$, $n \in \mathbb{Z}$
$(x^a)' = a \cdot x^{a-1}$, $x \in \mathbb{R}$, $x \gt 0$, $a \in \mathbb{R}$
$(\sin{x})' = \cos{x}$
$(\cos{x})' = -\sin{x}$
$(\tan{x})' = (tg\ {x})' = 1 / \cos^2{x}$
$(\cot{x})' = (cotg\ {x})' = -1 / \sin^2{x}$
$(\arcsin{x})' = 1 / \sqrt{1-x^2}$
$(\arccos{x})' = -1 / \sqrt{1-x^2}$
$(arctan\ {x})' = (arctg\ {x})' = 1 / (1 + x^2)$
$(arccot\ {x})' = (arccotg\ {x})' = -1 / (1 + x^2)$
$(e^x)' = e^x$
$(a^x)' = a^x \cdot \ln{a}\ \ \$ ($a \gt 0$ is a constant)
$(\ln{x})' = 1 / x$
$(\log_{a}x)' = 1 / (x \cdot \ln{a})\ \ \$ ($a \gt 0$ is a constant)
$$(u + v)' = u' + v'$$
$$(u - v)' = u' - v'$$
$$(uv)' = u' \cdot v + v' \cdot u$$
$$\left(\frac{u}{v}\right)' = \frac{u'v - v'u}{v^2}$$
$$(f(g(x)))' = f'(g(x)) \cdot g'(x)$$
### Machine Learning Notes 2
Supervised learning terminology/notation
• Training set is the dataset used to train the model
• $x$ - the input variable, the input feature
• $y$ - the output variable, the target variable
• $m$ - the total number of the training examples (training data points)
• $(x^{(i)}, y^{(i)})$ - the $i$-th training example (for $i=1,2,\dots, m$); here the expression $(i)$ is just a superscript, it does not denote exponentiation
• $f$ - hypothesis function, prediction function; also $f$ is called the model
• $\widehat{y} = f(x)$, where $x$ is the input variable and $\widehat{y}$ is the prediction for the target variable $y$
• $\widehat{y}^{(i)} = f(x^{(i)})$, where $x^{(i)}$ is the $i$-th training example, and $\widehat{y}^{(i)}$ is the prediction value corresponding to $x^{(i)}$
• Cost function is a function which estimates how well a given model predicts the values for the target variable $y$; this is a function which measures how well the model fits the training data. Cost function is a general concept which is applicable to any supervised learning model.
Linear regression with one variable (univariate linear regression)
When we have only one input variable we call this model/algorithm univariate linear regression. In a univariate linear regression model, we have $f(x) = wx + b$, where $x$ is the input variable, and $w, b$ are numbers which are called parameters of the model. So $f(x)$ is a linear function of $x$. It's also written sometimes as $f_{w,\ b}(x) = wx + b$. By varying the parameters $w$ and $b$ we get different linear models. The parameter $w$ is called the slope, and $b$ is called the y-intercept because $b=f(0)$ and so $b$ is the point where the graph of $f(x)$ intercepts the $y$ axis.
When we have more than one input variable (more than one feature) the model is called multivariate linear regression. In this case we try to predict the values of the target variable $y$ based on several input variables $x_1, x_2, \dots, x_k$. Here we have $k \in \mathbb{N}, k \ge 2$.
In a univariate linear regression model the most commonly used cost function is the mean squared error cost function.
$$J(w,b) = \frac{1}{m} \cdot \sum_{i=1}^m (\widehat{y}^{(i)} - y^{(i)})^2$$
In the last formula
$\widehat{y}^{(i)} = f_{w,b}(x^{(i)}) = wx^{(i)} + b$
for $i=1,2,\dots,m$.
So for the cost function we also get the following expression
$$J(w,b) = \frac{1}{m} \cdot \sum_{i=1}^m (f_{w,b}(x^{(i)}) - y^{(i)})^2$$
Note that $J(w,b) \ge 0$
In practice an additional division by 2 is performed in the formulas given above. This is done just for practical reasons, to make further computations simpler. In this way we get the final version of our cost function.
$$J(w,b) = \frac{1}{2m} \cdot \sum_{i=1}^m (\widehat{y}^{(i)} - y^{(i)})^2$$
$$J(w,b) = \frac{1}{2m} \cdot \sum_{i=1}^m (f_{w,b}(x^{(i)}) - y^{(i)})^2$$
Then the task is to find values of $w,b$ such that the value of $J(w,b)$ is as small as possible.
The smaller the value of $J(w,b)$, the better the model.
All this info about univariate linear regression can be summarized with the following picture
(pictures credits go to Andrew Ng's Coursera ML Course).
### An application of the AM-GM inequality
I found this problem and its solution in a FB group dedicated to maths. Here they are.
Problem
If $a,b,c \gt 0$ prove that
$$\frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} \ge 2(a+b+c)$$
Solution:
By the AM-GM inequality we have that
$$\frac{a^2}{c} + c + \frac{b^2}{c} + c \ge 2\sqrt{a^2} + 2\sqrt{b^2} = 2(a+b)$$
$$\frac{b^2}{a} + a + \frac{c^2}{a} + a \ge 2\sqrt{b^2} + 2\sqrt{c^2} = 2(b+c)$$
$$\frac{c^2}{b} + b + \frac{a^2}{b} + b \ge 2\sqrt{c^2} + 2\sqrt{a^2} = 2(c+a)$$
When we sum up these 3 inequalities we get the desired inequality.
Note: This solution is due to Ghimisi Dumitrel, Professor at Colegiul National Ionita Asan, Caracal.
### Machine Learning Notes 1
1) Machine learning (ML) related definitions
Definition A (Athur Samuel, 1959): Machine learning is the field of study which gives computers the ability to learn without being explicitly programmed.
Definition B (Tom Mitchell, 1998): Well-posed machine learning problem: A computer program is said to learn from experience E with respect to some task T and some performance measure P, if its performance on T, as measured by P, improves with experience E.
2) Types of ML algorithms
In general any ML algorithm can be classified into one of the following two categories
• Supervised learning (the "right answers" are given in the initial data set; i.e. the data set is labeled)
• Regression: trying to predict the values of some quantity which we view as a continuous function
• Classification: trying to predict the values of some quantity whose possible values form a small finite discrete set (that set could be e.g. {Yes, No}; {0,1,2,3}; {Cat, Dog, Other}; etc.); trying to classify a set of data points into a small finite number of classes
• Unsupervised learning (the "right answers" are not given in the initial data set; i.e. the data set is unlabeled); in unsupervised learning the task is to automatically find some structure in the unlabeled data set that is given
• Clustering: a data set is given, the task is to break it down into several separate clusters (i.e. into several separate disjoint data sets). Examples: 1) Google news (news articles from various sources are grouped together if they are about the same story); 2) Market segmentation: find market segments/clusters in a dataset which contains data about all customers of a given company.
• Non-clustering:
• Independent component analysis (used in signal processing). Example: separating different voices (or say different audio tracks) out of one (or a few) given chaotic audio recordings (see e.g. (1) cocktail party effect, (2) cocktail party problem and algorithm). The so-called "cocktail party algorithm" allows you to find structure in a chaotic environment (identifying individual voices and music from a mesh of sounds at a cocktail party).
• Anomaly detection (detect unusual data points in a given dataset). Anomaly detection (also referred to as outlier detection and sometimes as novelty detection) is generally understood to be the identification of rare items, events or observations which deviate significantly from the majority of the data and do not conform to a well defined notion of normal behavior (Source: Wikipedia).
• Dimensionality reduction (compress data points using fewer numbers). Dimensionality reduction, or dimension reduction, is the transformation of data from a higher-dimensional space into a lower-dimensional space so that the lower-dimensional representation retains some meaningful properties of the original data, ideally close to its intrinsic dimension (Source: Wikipedia).
3) GNU Octave
This is a powerful framework for quick prototyping of solutions/algorithms for ML problems (see here: GNU Octave)
### NumPy operators and their corresponding NumPy universal functions (ufuncs)
NumPy operators and their corresponding NumPy universal functions (ufuncs)
This table was compiled from the 3 tables shown on pages 53, 72, 75 of the book Python Data Science Handbook by Jake VanderPlas. I put it here just for my own reference.
Arithmetic Operator Equivalent ufunc Description + np.add Addition (e.g., 1 + 1 = 2) - np.subtract Subtraction (e.g., 3 - 2 = 1) - np.negative Unary negation (e.g., -2) * np.multiply Multiplication (e.g., 2 * 3 = 6) / np.divide Division (e.g., 3 / 2 = 1.5) // np.floor_divide Floor division (e.g., 3 // 2 = 1) ** np.power Exponentiation (e.g., 2 ** 3 = 8) % np.mod Modulus/remainder (e.g., 9 % 4 = 1) Comparison Operator Equivalent ufunc == np.equal != np.not_equal < np.less <= np.less_equal > np.greater >= np.greater_equal Logical Operator Equivalent ufunc See also & np.bitwise_and np.logical_and | np.bitwise_or np.logical_or ^ np.bitwise_xor np.logical_xor ~ np.bitwise_not np.logical_not
These rules are listed in the book Python Data Science Handbook by Jake VanderPlas (see page 65).
Broadcasting in NumPy is simply a set of rules for applying binary universal functions (like addition, subtraction, multiplication, etc.) on NumPy arrays of different sizes.
Here are the NumPy broadcasting rules.
• Rule 1: If the two arrays differ in their number of dimensions, the shape of the
one with fewer dimensions is | 6,634 | 21,398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.718501 |
https://www.cfd-online.com/Wiki/Scalar_dissipation | 1,498,411,444,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320545.67/warc/CC-MAIN-20170625170634-20170625190634-00543.warc.gz | 844,084,874 | 10,463 | # Scalar dissipation
Jump to: navigation, search
Scalar dissipation is a very important quantity in non-premixed combustion modelling. It often provides the connection between the mixing field and the combustion modelling. It is specially important in flamelet and RANS models.
In a laminar flow the scalar dissipation rate is defined (units are 1/s) as
$\chi \equiv 2 D \left( \frac{\partial Z}{\partial x_j} \right)^2$
where $D$ is the diffusion coefficient of the scalar.
In turbulent flows, the scalar dissipation is seen as a scalar energy dissipation} and its role is to destroy (dissipate) scalar variance (scalar energy) analogous to the dissipation of the turbulent energy $\epsilon$. This term is known as the turbulent scalar dissipation and is written as
$\chi_t \equiv 2 D \left( \frac{\partial \widetilde{Z''}}{\partial x_j} \right) ^2$
Opposite to the kinetic energy dissipation, most of the scalar dissipation occur at the finest scales.
In Conditional Moment Closure (CMC) and Flamelet based on conserved scalar models, the quantity of interest is the " main scalar dissipation rate", $\widetilde{\chi}$. From Favre Averaging the laminar dissipation rate
$\widetilde{\chi} = 2D \widetilde{\left(\frac{\partial Z}{\partial x_j}\right) ^2} \approx 2 D \left( \frac{\partial \tilde{Z}}{\partial x_j} \right) ^2 + \chi_t$
Under RANS assumptions gradient of the scalar fluctuations are much larger than gradients of the means, and therefore the mean scalar dissipation rate is approximately the turbulent dissipation rate $\widetilde{\chi} \approx \chi_t$ | 394 | 1,578 | {"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": 7, "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-2017-26 | longest | en | 0.868631 |
https://viindoo.com/blog/business-management-3/fixed-asset-ratios-1277 | 1,722,724,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00446.warc.gz | 482,335,692 | 20,813 | # Fixed Asset Ratios: Explanation, Types With Examples
Fixed Asset Ratios are important in financial analysis because they provide insight into a company's use of its fixed assets, which can help investors and analysts evaluate a company's efficiency, profitability, and risk. By analyzing these ratios, investors and analysts can gain a better understanding of a company's financial position and make informed investment decisions. In this article, let's find out about this concept with Viindoo
## Explanation of Fixed Asset Ratios
Fixed asset ratios are financial ratios used to evaluate a company's utilization and management of its fixed assets. Fixed assets are assets that a company owns and uses for long-term operations and are not easily converted into cash. Examples of fixed assets include property, plant, and equipment (PPE), land, buildings, and machinery.
By analyzing these ratios, analysts can determine the level of efficiency and effectiveness of a company's fixed assets management. Moreover, these ratios are helpful in making informed decisions regarding investment opportunities, creditworthiness, and overall financial health.
These ratios are important because fixed assets represent a significant investment for many companies. They are also an essential component of the company's operational and production processes. Therefore, a company's ability to manage and utilize these assets efficiently and effectively can have a significant impact on its profitability and financial performance.
Furthermore, fixed asset ratios can be used to compare a company's performance with that of its competitors in the same industry. This comparison can provide insight into a company's strengths and weaknesses relative to its peers, which can be useful for investors and analysts when making investment decisions. Overall, fixed asset ratios provide a valuable tool for assessing a company's financial health, and they are an essential component of financial analysis.
Explanation of Fixed Asset Ratios
## Types of Fixed Asset Ratios
There are several types of fixed asset ratios used in financial analysis. Below are the four main types of fixed asset ratios:
Property, Plant, and Equipment (PPE) to Total Assets Ratio:
This ratio measures the proportion of a company's total assets that are represented by its property, plant, and equipment. A high PPE to total assets ratio indicates that a company has invested heavily in fixed assets.
Formula:
PPE to Total Assets Ratio = Property, Plant, and Equipment / Total Assets
Fixed Asset Turnover Ratio:
This ratio measures how efficiently a company is using its fixed assets to generate revenue. A high fixed asset turnover ratio indicates that a company is generating a significant amount of revenue relative to the amount invested in fixed assets.
Formula:
Fixed Asset Turnover Ratio = Sales / Net Fixed Assets
Depreciation to Fixed Assets Ratio:
This ratio measures the level of depreciation on a company's fixed assets relative to the total value of those assets. A high depreciation to fixed assets ratio indicates that a company's fixed assets are aging and losing value more quickly than they are being replaced.
Formula:
Depreciation to Fixed Assets Ratio = Depreciation / Net Fixed Assets
Net Fixed Asset Ratio:
This ratio measures the proportion of a company's total assets that are represented by its net fixed assets. Net fixed assets are the value of a company's fixed assets minus any accumulated depreciation.
Formula:
Net Fixed Asset Ratio = Net Fixed Assets / Total Assets
These fixed asset ratios can be used together to get a more comprehensive picture of a company's fixed asset utilization and management. They are valuable tools for investors and analysts in evaluating a company's financial health and identifying potential risks and opportunities.
Types of Fixed Asset Ratios
## Property, Plant, and Equipment (PPE) to Total Assets Ratio
### Definition and Explanation
The Property, Plant, and Equipment (PPE) to Total Assets ratio measures the percentage of a company's total assets that are tied up in property, plant, and equipment. This ratio is also known as the fixed assets ratio or the capital asset ratio. It is used to evaluate a company's capital expenditure and investment in long-term assets.
### Formula
PPE to Total Assets Ratio = (Property, Plant, and Equipment / Total Assets) x 100
Example and Interpretation
Let's assume that Company ABC has a property, plant, and equipment value of \$2,000,000 and total assets of \$10,000,000. Using the formula above, we can calculate the PPE to Total Assets ratio as follows:
PPE to Total Assets Ratio = (\$2,000,000 / \$10,000,000) x 100
PPE to Total Assets Ratio = 20%
This means that 20% of Company ABC's total assets are tied up in property, plant, and equipment.
Interpretation of this ratio can vary depending on the industry in which the company operates. For example, a manufacturing company will typically have a higher PPE to Total Assets ratio than a service-based company, as the former requires more significant investment in property, plant, and equipment to produce goods. A higher PPE to Total Assets ratio can indicate that a company has invested heavily in fixed assets, which could be a good thing if these assets are being utilized efficiently to generate revenue. However, if the assets are not being utilized efficiently, this could negatively impact the company's profitability and financial performance. A low PPE to Total Assets ratio may indicate that a company has a more asset-light business model or may be leasing its assets instead of owning them outright. In such cases, the company may have less risk and less capital tied up in long-term assets.
Property, Plant, and Equipment (PPE) to Total Assets Ratio
## Fixed Asset Turnover Ratio
### Definition and Explanation
The Fixed Asset Turnover ratio measures how efficiently a company is using its fixed assets to generate revenue. It calculates the amount of sales generated per dollar of net fixed assets. This ratio is used to evaluate a company's ability to generate revenue from its investment in fixed assets.
### Formula
Fixed Asset Turnover Ratio = Sales / Net Fixed Assets
### Example and Interpretation
Let's assume that Company XYZ had sales of \$1,000,000 and net fixed assets of \$500,000 during a given year. Using the formula above, we can calculate the Fixed Asset Turnover ratio as follows:
Fixed Asset Turnover Ratio = \$1,000,000 / \$500,000
Fixed Asset Turnover Ratio = 2
This means that for every dollar invested in fixed assets, Company XYZ generated \$2 in sales during the year.
A high fixed asset turnover ratio is generally desirable, as it indicates that a company is generating significant revenue from its investment in fixed assets. A low ratio, on the other hand, may indicate that a company's fixed assets are not being utilized efficiently to generate revenue. It could be due to factors such as outdated equipment, poor production processes, or low demand for the company's products or services.
Interpretation of the fixed asset turnover ratio should be done within the context of the company's industry and its competitors. Comparing a company's ratio to the industry average or its competitors' ratios can provide additional insights into the company's efficiency in utilizing its fixed assets.
Fixed Asset Turnover Ratio
## Depreciation to Fixed Assets Ratio
### Definition and Explanation
The Depreciation to Fixed Assets ratio measures the extent to which a company's fixed assets have been depreciated during a specific period relative to the value of its fixed assets. It is an efficiency ratio that evaluates how much of a company's fixed assets' value has been used up or consumed over time.
### Formula
Depreciation to Fixed Assets Ratio = Depreciation Expense / Net Fixed Assets
### Example and Interpretation
Let's assume that Company XYZ incurred \$50,000 in depreciation expenses during a given year and had net fixed assets of \$500,000 at the beginning of the year. Using the formula above, we can calculate the Depreciation to Fixed Assets ratio as follows:
Depreciation to Fixed Assets Ratio = \$50,000 / \$500,000
Depreciation to Fixed Assets Ratio = 0.10 or 10%
This means that for every dollar of net fixed assets, Company XYZ has depreciated 10 cents during the year.
A higher ratio indicates that a company's fixed assets are being depreciated more rapidly than the industry average, which could indicate that the assets are being used heavily or that the company is using accelerated depreciation methods. However, a higher ratio could also mean that a company's fixed assets are being underutilized or are in poor condition, resulting in higher depreciation expenses.
Conversely, a lower Depreciation to Fixed Assets ratio may indicate that the company is using straight-line depreciation methods, which spread out the depreciation of assets evenly over their useful lives. A lower ratio could also indicate that a company's fixed assets are being underutilized, which could result in lower depreciation expenses.
It is important to note that the interpretation of the Depreciation to Fixed Assets ratio should be done in conjunction with other financial ratios and metrics to obtain a comprehensive understanding of the company's financial position and performance.
Depreciation to Fixed Assets Ratio
## Net Fixed Asset Ratio
### Definition and Explanation
The Net Fixed Asset ratio measures the proportion of a company's total assets that are invested in fixed assets net of accumulated depreciation. It indicates the extent to which a company's operations rely on its fixed assets to generate revenue.
### Formula
Net Fixed Asset Ratio = Net Fixed Assets / Total Assets
### Example and Interpretation
Let's assume that Company XYZ has net fixed assets of \$500,000 and total assets of \$1,000,000 during a given year. Using the formula above, we can calculate the Net Fixed Asset ratio as follows:
Net Fixed Asset Ratio = \$500,000 / \$1,000,000
Net Fixed Asset Ratio = 0.50 or 50%
This means that 50% of Company XYZ's total assets are invested in fixed assets net of accumulated depreciation.
A higher Net Fixed Asset ratio indicates that a company has invested a larger proportion of its assets in fixed assets, indicating that its operations rely more heavily on these assets to generate revenue. This could be a positive sign if the company's fixed assets are being utilized efficiently and generating significant revenue. However, a high ratio could also indicate that the company has invested too much in fixed assets, which could lead to underutilization of these assets and lower profitability.
On the other hand, a lower Net Fixed Asset ratio may indicate that the company is relying more on current assets, such as inventory or accounts receivable, to generate revenue. This could be a positive sign if the company's operations are more reliant on current assets or if the company is in a growth phase that requires higher levels of investment in working capital.
As with other financial ratios, the Net Fixed Asset ratio should be interpreted in conjunction with other financial ratios and metrics to obtain a comprehensive understanding of a company's financial position and performance.
Net Fixed Asset Ratio
## Problems with the Fixed Asset Ratio
While the Fixed Asset Ratio can be a useful tool for evaluating a company's asset structure and investment decisions, there are some potential problems and limitations to consider:
1. Fixed assets may not accurately reflect a company's true value: The Fixed Asset Ratio is based solely on a company's investments in fixed assets, which may not accurately reflect its true value. For example, a company may have significant intangible assets, such as patents or trademarks, that are not reflected in the Fixed Asset Ratio.
2. A higher Net Fixed Asset ratio indicates that a company has invested a larger proportion of its assets in fixed assets, indicating that its operations rely more heavily on these assets to generate revenue. This could be a positive sign if the company's fixed assets are being utilized efficiently and generating significant revenue. However, a high ratio could also indicate that the company has invested too much in fixed assets, which could lead to underutilization of these assets and lower profitability.
3. Different industries have different asset structures: The Fixed Asset Ratio may not be as useful for comparing companies in different industries, as some industries may require more or less investment in fixed assets. For example, a service-based company may have a lower Fixed Asset Ratio than a manufacturing company, even if both companies are equally profitable.
4. Does not consider changes in asset utilization: The Fixed Asset Ratio is based on historical data and does not take into account changes in a company's asset utilization over time. For example, a company may have recently increased its investments in fixed assets, but this improvement may not be reflected in the Fixed Asset Ratio until the next reporting period.
5. May not reflect the impact of depreciation: The Fixed Asset Ratio does not take into account the impact of depreciation on a company's fixed assets, which can impact their true value and efficiency. Depreciation reduces the book value of fixed assets over time, and may not reflect their true value or usefulness in generating revenue.
Using fixed asset software with Fixed Asset Ratios features can help companies streamline their fixed asset management processes and gain valuable insights into their financial performance. These software solutions can also help ensure that companies are accurately tracking their fixed assets and calculating depreciation, which can help them comply with regulatory requirements and avoid costly mistakes.
Problems with the Fixed Asset Ratio
There are many accounting software system solutions available that include Fixed Asset Ratios as part of their features. These software programs can help companies manage and track their fixed assets, calculate depreciation, and generate financial reports. Try Viindoo accounting software for 15 Days free trial.
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FirstOrderSAGCostFunction Class Referenceabstract
## Detailed Description
The class is about a stochastic cost function for stochastic average minimizers.
The cost function must be written as a finite sample-specific sum of cost. For example, least squares cost function,
$f(w)=\frac{ \sum_i^n{ (y_i-w^T x_i)^2 } }{2}$
where $$n$$ is the sample size, $$(y_i,x_i)$$ is the i-th sample, $$y_i$$ is the label and $$x_i$$ is the features
A stochastic average minimizer uses average sample gradients ( get_average_gradient() ) to reduce variance related to stochastic gradients.
Well known stochastic average methods are: SVRG, Johnson, Rie, and Tong Zhang. "Accelerating stochastic gradient descent using predictive variance reduction." Advances in Neural Information Processing Systems. 2013.
SAG, Schmidt, Mark, Nicolas Le Roux, and Francis Bach. "Minimizing finite sums with the stochastic average gradient." arXiv preprint arXiv:1309.2388 (2013).
SAGA, Defazio, Aaron, Francis Bach, and Simon Lacoste-Julien. "SAGA: A fast incremental gradient method with support for non-strongly convex composite objectives." Advances in Neural Information Processing Systems. 2014.
SDCA, Shalev-Shwartz, Shai, and Tong Zhang. "Stochastic dual coordinate ascent methods for regularized loss." The Journal of Machine Learning Research 14.1 (2013): 567-599.
Definition at line 70 of file FirstOrderSAGCostFunction.h.
Inheritance diagram for FirstOrderSAGCostFunction:
[legend]
## Public Types
typedef rxcpp::subjects::subject< ObservedValueSGSubject
typedef rxcpp::observable< ObservedValue, rxcpp::dynamic_observable< ObservedValue > > SGObservable
typedef rxcpp::subscriber< ObservedValue, rxcpp::observer< ObservedValue, void, void, void, void > > SGSubscriber
## Public Member Functions
virtual ~FirstOrderSAGCostFunction ()
virtual int32_t get_sample_size ()=0
virtual float64_t get_cost ()=0
virtual void begin_sample ()=0
virtual bool next_sample ()=0
virtual SGVector< float64_tobtain_variable_reference ()=0
int32_t ref ()
int32_t ref_count ()
int32_t unref ()
virtual CSGObjectshallow_copy () const
virtual CSGObjectdeep_copy () const
virtual const char * get_name () const =0
virtual bool is_generic (EPrimitiveType *generic) const
template<class T >
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
template<>
void set_generic ()
void unset_generic ()
virtual void print_serializable (const char *prefix="")
virtual bool save_serializable (CSerializableFile *file, const char *prefix="")
virtual bool load_serializable (CSerializableFile *file, const char *prefix="")
void set_global_io (SGIO *io)
SGIOget_global_io ()
void set_global_parallel (Parallel *parallel)
Parallelget_global_parallel ()
void set_global_version (Version *version)
Versionget_global_version ()
SGStringList< char > get_modelsel_names ()
void print_modsel_params ()
char * get_modsel_param_descr (const char *param_name)
index_t get_modsel_param_index (const char *param_name)
void build_gradient_parameter_dictionary (CMap< TParameter *, CSGObject * > *dict)
bool has (const std::string &name) const
template<typename T >
bool has (const Tag< T > &tag) const
template<typename T , typename U = void>
bool has (const std::string &name) const
template<typename T >
void set (const Tag< T > &_tag, const T &value)
template<typename T , typename U = void>
void set (const std::string &name, const T &value)
template<typename T >
get (const Tag< T > &_tag) const
template<typename T , typename U = void>
get (const std::string &name) const
SGObservableget_parameters_observable ()
void subscribe_to_parameters (ParameterObserverInterface *obs)
void list_observable_parameters ()
virtual void update_parameter_hash ()
virtual bool parameter_hash_changed ()
virtual bool equals (CSGObject *other, float64_t accuracy=0.0, bool tolerant=false)
virtual CSGObjectclone ()
## Public Attributes
SGIOio
Parallelparallel
Versionversion
Parameterm_parameters
Parameterm_model_selection_parameters
uint32_t m_hash
## Protected Member Functions
virtual void load_serializable_pre () throw (ShogunException)
virtual void load_serializable_post () throw (ShogunException)
virtual void save_serializable_pre () throw (ShogunException)
virtual void save_serializable_post () throw (ShogunException)
template<typename T >
void register_param (Tag< T > &_tag, const T &value)
template<typename T >
void register_param (const std::string &name, const T &value)
bool clone_parameters (CSGObject *other)
void observe (const ObservedValue value)
void register_observable_param (const std::string &name, const SG_OBS_VALUE_TYPE type, const std::string &description)
## Member Typedef Documentation
inherited
Definition at line 130 of file SGObject.h.
inherited
Definition at line 127 of file SGObject.h.
typedef rxcpp::subscriber< ObservedValue, rxcpp::observer > SGSubscriber
inherited
Definition at line 133 of file SGObject.h.
## Constructor & Destructor Documentation
virtual ~FirstOrderSAGCostFunction ( )
virtual
Definition at line 74 of file FirstOrderSAGCostFunction.h.
## Member Function Documentation
virtual void begin_sample ( )
pure virtualinherited
Initialize to generate a sample sequence
void build_gradient_parameter_dictionary ( CMap< TParameter *, CSGObject * > * dict )
inherited
Builds a dictionary of all parameters in SGObject as well of those of SGObjects that are parameters of this object. Dictionary maps parameters to the objects that own them.
Parameters
dict dictionary of parameters to be built.
Definition at line 635 of file SGObject.cpp.
CSGObject * clone ( )
virtualinherited
Creates a clone of the current object. This is done via recursively traversing all parameters, which corresponds to a deep copy. Calling equals on the cloned object always returns true although none of the memory of both objects overlaps.
Returns
an identical copy of the given object, which is disjoint in memory. NULL if the clone fails. Note that the returned object is SG_REF'ed
Definition at line 734 of file SGObject.cpp.
bool clone_parameters ( CSGObject * other )
protectedinherited
Definition at line 759 of file SGObject.cpp.
CSGObject * deep_copy ( ) const
virtualinherited
A deep copy. All the instance variables will also be copied.
Definition at line 232 of file SGObject.cpp.
bool equals ( CSGObject * other, float64_t accuracy = 0.0, bool tolerant = false )
virtualinherited
Recursively compares the current SGObject to another one. Compares all registered numerical parameters, recursion upon complex (SGObject) parameters. Does not compare pointers!
May be overwritten but please do with care! Should not be necessary in most cases.
Parameters
other object to compare with accuracy accuracy to use for comparison (optional) tolerant allows linient check on float equality (within accuracy)
Returns
true if all parameters were equal, false if not
Definition at line 656 of file SGObject.cpp.
T get ( const Tag< T > & _tag ) const
inherited
Getter for a class parameter, identified by a Tag. Throws an exception if the class does not have such a parameter.
Parameters
_tag name and type information of parameter
Returns
value of the parameter identified by the input tag
Definition at line 381 of file SGObject.h.
T get ( const std::string & name ) const
inherited
Getter for a class parameter, identified by a name. Throws an exception if the class does not have such a parameter.
Parameters
name name of the parameter
Returns
value of the parameter corresponding to the input name and type
Definition at line 404 of file SGObject.h.
pure virtual
Get the AVERAGE gradient value wrt target variables
Note that the average gradient is the mean of sample gradient from get_gradient() if samples are generated (uniformly) at random.
WARNING This method returns $$\frac{\sum_i^n{ \frac{\partial f_i(w) }{\partial w} }}{n}$$
For least squares, that is the value of $$\frac{\frac{\partial f(w) }{\partial w}}{n}$$ given $$w$$ is known where $$f(w)=\frac{ \sum_i^n{ (y_i-w^t x_i)^2 } }{2}$$
Returns
virtual float64_t get_cost ( )
pure virtual
Get the cost given current target variables
For least squares cost function, that is the value of $$f(w)$$.
Returns
cost
Implements FirstOrderStochasticCostFunction.
SGIO * get_global_io ( )
inherited
get the io object
Returns
io object
Definition at line 269 of file SGObject.cpp.
Parallel * get_global_parallel ( )
inherited
get the parallel object
Returns
parallel object
Definition at line 311 of file SGObject.cpp.
Version * get_global_version ( )
inherited
get the version object
Returns
version object
Definition at line 324 of file SGObject.cpp.
pure virtual
Get the SAMPLE gradient value wrt target variables
WARNING This method does return $$\frac{\partial f_i(w) }{\partial w}$$ instead of $$\sum_i^n{ \frac{\partial f_i(w) }{\partial w} }$$
For least squares cost function, that is the value of $$\frac{\partial f_i(w) }{\partial w}$$ given $$w$$ is known where the index $$i$$ is obtained by next_sample()
Returns
Implements FirstOrderStochasticCostFunction.
SGStringList< char > get_modelsel_names ( )
inherited
Returns
vector of names of all parameters which are registered for model selection
Definition at line 536 of file SGObject.cpp.
char * get_modsel_param_descr ( const char * param_name )
inherited
Returns description of a given parameter string, if it exists. SG_ERROR otherwise
Parameters
param_name name of the parameter
Returns
description of the parameter
Definition at line 560 of file SGObject.cpp.
index_t get_modsel_param_index ( const char * param_name )
inherited
Returns index of model selection parameter with provided index
Parameters
param_name name of model selection parameter
Returns
index of model selection parameter with provided name, -1 if there is no such
Definition at line 573 of file SGObject.cpp.
SGObservable* get_parameters_observable ( )
inherited
Get parameters observable
Returns
RxCpp observable
Definition at line 415 of file SGObject.h.
virtual int32_t get_sample_size ( )
pure virtual
Get the sample size
Returns
the sample size
bool has ( const std::string & name ) const
inherited
Checks if object has a class parameter identified by a name.
Parameters
name name of the parameter
Returns
true if the parameter exists with the input name
Definition at line 304 of file SGObject.h.
bool has ( const Tag< T > & tag ) const
inherited
Checks if object has a class parameter identified by a Tag.
Parameters
tag tag of the parameter containing name and type information
Returns
true if the parameter exists with the input tag
Definition at line 315 of file SGObject.h.
bool has ( const std::string & name ) const
inherited
Checks if a type exists for a class parameter identified by a name.
Parameters
name name of the parameter
Returns
true if the parameter exists with the input name and type
Definition at line 326 of file SGObject.h.
bool is_generic ( EPrimitiveType * generic ) const
virtualinherited
If the SGSerializable is a class template then TRUE will be returned and GENERIC is set to the type of the generic.
Parameters
generic set to the type of the generic if returning TRUE
Returns
TRUE if a class template.
Definition at line 330 of file SGObject.cpp.
void list_observable_parameters ( )
inherited
Print to stdout a list of observable parameters
Definition at line 878 of file SGObject.cpp.
bool load_serializable ( CSerializableFile * file, const char * prefix = "" )
virtualinherited
Load this object from file. If it will fail (returning FALSE) then this object will contain inconsistent data and should not be used!
Parameters
file where to load from prefix prefix for members
Returns
TRUE if done, otherwise FALSE
Definition at line 403 of file SGObject.cpp.
void load_serializable_post ( ) throw ( ShogunException )
protectedvirtualinherited
Can (optionally) be overridden to post-initialize some member variables which are not PARAMETER::ADD'ed. Make sure that at first the overridden method BASE_CLASS::LOAD_SERIALIZABLE_POST is called.
Exceptions
ShogunException will be thrown if an error occurs.
Definition at line 460 of file SGObject.cpp.
void load_serializable_pre ( ) throw ( ShogunException )
protectedvirtualinherited
Can (optionally) be overridden to pre-initialize some member variables which are not PARAMETER::ADD'ed. Make sure that at first the overridden method BASE_CLASS::LOAD_SERIALIZABLE_PRE is called.
Exceptions
ShogunException will be thrown if an error occurs.
Definition at line 455 of file SGObject.cpp.
virtual bool next_sample ( )
pure virtualinherited
Get next sample
Returns
false if reach the end of the sample sequence
void observe ( const ObservedValue value )
protectedinherited
Observe a parameter value and emit them to observer.
Parameters
value Observed parameter's value
Definition at line 828 of file SGObject.cpp.
virtual SGVector obtain_variable_reference ( )
pure virtualinherited
Obtain a reference of target variables Minimizers will modify target variables in place.
This method will be called by FirstOrderMinimizer::minimize()
For least squares, that is $$w$$
Returns
reference of variables
bool parameter_hash_changed ( )
virtualinherited
Returns
whether parameter combination has changed since last update
Definition at line 296 of file SGObject.cpp.
void print_modsel_params ( )
inherited
prints all parameter registered for model selection and their type
Definition at line 512 of file SGObject.cpp.
void print_serializable ( const char * prefix = "" )
virtualinherited
prints registered parameters out
Parameters
prefix prefix for members
Definition at line 342 of file SGObject.cpp.
int32_t ref ( )
inherited
increase reference counter
Returns
reference count
Definition at line 186 of file SGObject.cpp.
int32_t ref_count ( )
inherited
display reference counter
Returns
reference count
Definition at line 193 of file SGObject.cpp.
void register_observable_param ( const std::string & name, const SG_OBS_VALUE_TYPE type, const std::string & description )
protectedinherited
Register which params this object can emit.
Parameters
name the param name type the param type description a user oriented description
Definition at line 871 of file SGObject.cpp.
void register_param ( Tag< T > & _tag, const T & value )
protectedinherited
Registers a class parameter which is identified by a tag. This enables the parameter to be modified by set() and retrieved by get(). Parameters can be registered in the constructor of the class.
Parameters
_tag name and type information of parameter value value of the parameter
Definition at line 472 of file SGObject.h.
void register_param ( const std::string & name, const T & value )
protectedinherited
Registers a class parameter which is identified by a name. This enables the parameter to be modified by set() and retrieved by get(). Parameters can be registered in the constructor of the class.
Parameters
name name of the parameter value value of the parameter along with type information
Definition at line 485 of file SGObject.h.
bool save_serializable ( CSerializableFile * file, const char * prefix = "" )
virtualinherited
Save this object to file.
Parameters
file where to save the object; will be closed during returning if PREFIX is an empty string. prefix prefix for members
Returns
TRUE if done, otherwise FALSE
Definition at line 348 of file SGObject.cpp.
void save_serializable_post ( ) throw ( ShogunException )
protectedvirtualinherited
Can (optionally) be overridden to post-initialize some member variables which are not PARAMETER::ADD'ed. Make sure that at first the overridden method BASE_CLASS::SAVE_SERIALIZABLE_POST is called.
Exceptions
ShogunException will be thrown if an error occurs.
Reimplemented in CKernel.
Definition at line 470 of file SGObject.cpp.
void save_serializable_pre ( ) throw ( ShogunException )
protectedvirtualinherited
Can (optionally) be overridden to pre-initialize some member variables which are not PARAMETER::ADD'ed. Make sure that at first the overridden method BASE_CLASS::SAVE_SERIALIZABLE_PRE is called.
Exceptions
ShogunException will be thrown if an error occurs.
Definition at line 465 of file SGObject.cpp.
void set ( const Tag< T > & _tag, const T & value )
inherited
Setter for a class parameter, identified by a Tag. Throws an exception if the class does not have such a parameter.
Parameters
_tag name and type information of parameter value value of the parameter
Definition at line 342 of file SGObject.h.
void set ( const std::string & name, const T & value )
inherited
Setter for a class parameter, identified by a name. Throws an exception if the class does not have such a parameter.
Parameters
name name of the parameter value value of the parameter along with type information
Definition at line 368 of file SGObject.h.
void set_generic ( )
inherited
Definition at line 73 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 78 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 83 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 88 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 93 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 98 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 103 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 108 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 113 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 118 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 123 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 128 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 133 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 138 of file SGObject.cpp.
void set_generic ( )
inherited
Definition at line 143 of file SGObject.cpp.
void set_generic ( )
inherited
set generic type to T
void set_global_io ( SGIO * io )
inherited
set the io object
Parameters
io io object to use
Definition at line 262 of file SGObject.cpp.
void set_global_parallel ( Parallel * parallel )
inherited
set the parallel object
Parameters
parallel parallel object to use
Definition at line 275 of file SGObject.cpp.
void set_global_version ( Version * version )
inherited
set the version object
Parameters
version version object to use
Definition at line 317 of file SGObject.cpp.
CSGObject * shallow_copy ( ) const
virtualinherited
A shallow copy. All the SGObject instance variables will be simply assigned and SG_REF-ed.
Reimplemented in CGaussianKernel.
Definition at line 226 of file SGObject.cpp.
void subscribe_to_parameters ( ParameterObserverInterface * obs )
inherited
Subscribe a parameter observer to watch over params
Definition at line 811 of file SGObject.cpp.
int32_t unref ( )
inherited
decrement reference counter and deallocate object if refcount is zero before or after decrementing it
Returns
reference count
Definition at line 200 of file SGObject.cpp.
void unset_generic ( )
inherited
unset generic type
this has to be called in classes specializing a template class
Definition at line 337 of file SGObject.cpp.
void update_parameter_hash ( )
virtualinherited
Updates the hash of current parameter combination
Definition at line 282 of file SGObject.cpp.
## Member Data Documentation
SGIO* io
inherited
io
Definition at line 600 of file SGObject.h.
inherited
parameters wrt which we can compute gradients
Definition at line 615 of file SGObject.h.
uint32_t m_hash
inherited
Hash of parameter values
Definition at line 618 of file SGObject.h.
Parameter* m_model_selection_parameters
inherited
model selection parameters
Definition at line 612 of file SGObject.h.
Parameter* m_parameters
inherited
parameters
Definition at line 609 of file SGObject.h.
Parallel* parallel
inherited
parallel
Definition at line 603 of file SGObject.h.
Version* version
inherited
version
Definition at line 606 of file SGObject.h.
The documentation for this class was generated from the following file:
SHOGUN Machine Learning Toolbox - Documentation | 4,766 | 20,868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-30 | latest | en | 0.623895 |
https://quant.stackexchange.com/questions/17171/black-model-delta-strike-relationship-regardless-of-expiry?noredirect=1 | 1,631,807,363,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053657.29/warc/CC-MAIN-20210916145123-20210916175123-00563.warc.gz | 512,999,627 | 38,153 | Black model: Delta - strike relationship regardless of expiry?
While wandering through some QuantLib experimental classes for FX trading, I've found this Black Delta Calculator.
By reading its .cpp, it seems that no use of options time to expiry is made at all.
Usual Black, Scholes & Merton $d_{1}$ argument of cumulative Normal distribution has time to expiry as argument: on the contrary, Black Delta Calculator makes a weird use of $d_{1}$ and $d_{2}$, expressing them like at line 222 or 250:
d1_ = std::log(forward_/strike)/stdDev_ + 0.5*stdDev_; // .cpp line # 222
d2_ = std::log(forward_/strike)/stdDev_ - 0.5*stdDev_; // .cpp line # 250
This is quite different than what usual Itö correction produces over classic GBM process (squared variance and of course time adjustment for annual basis).
By reading that code it seems that going from Delta to strike and back can be made regardless of expiry date, while common sense says that there are a lot of options whose Delta can match any strike if you can search all over implied volatility surface without an expiry boundary.
Questions
• It must be I am missing some important relations which allow for such a simplification: could you show me which one?
• Is this possible relation viable just for FX options or can it be extended to any use of Black model (e.g. interest rates)?
• Time to expiry is already included in the inputs (the two discounts $e^{-rt}$ and $e^{-qt}$ and the standard deviation $\sigma \sqrt{t}$). Mar 31 '15 at 8:59
• You should give it an answer, Luigi... something like: «As far as I know, there's no way to go straight from Delta to strike without knowing $T$... and you're a lazy ass who doesn't read my codes from the beginning» ...Ahahah, always thank you for your help :) Mar 31 '15 at 9:39
• Ok. I'll skip the lazy ass part, though. :) Mar 31 '15 at 9:42
The time to expiry is required, but it's included in the inputs: the two discounts $e^{-rT}$ and $e^{-qT}$ and the standard deviation $\sigma\sqrt{T}$. You might argue it could be documented more clearly, and I might agree with you. | 523 | 2,087 | {"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} | 2.78125 | 3 | CC-MAIN-2021-39 | latest | en | 0.905223 |
https://www.jiskha.com/questions/1393233/The-function-f-x-2x-3-10-2x-2-202-275x-0-87-is-increasing-on-the-open-interval | 1,569,082,725,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00174.warc.gz | 895,013,637 | 5,453 | # Math (Calculus)
The function f(x)=-2x^3+10.2x^2+202.275x+0.87
is increasing on the open interval (?,?).
It is decreasing on the open interval (-oo,?) and the open interval (?,+oo)
The function has a local maximum at ?
I used derivative
-6x^2+20.4x+202.275 and find the root but it doesn't work with this problem.
1. 👍 0
2. 👎 0
3. 👁 66
1. Yes, it does work,
let's set
-6x^2+20.4x+202.275 = 0
6x^2 - 20.4x - 202.275 = 0
I used the formula and got
x = 7.75 or x = -4.35
remember that the function is increasing when the derivative is postive, and decreasing when ....
so the function is increasing for
-4.35 < x < 7.75
and decreasing for
x < -4.35 OR x > 7.75
You should have a general idea what a function like
f(x) = -2x^3 + ... looks like
so x = 7.75 will produce a local maximum.
Plug x = 7.75 into f(x) to find that local maximum
1. 👍 0
2. 👎 0
posted by Reiny
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More Similar Questions | 1,101 | 3,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-39 | latest | en | 0.874112 |
https://ebs.sakarya.edu.tr/DersDetay/DersinDetayliBilgileri/21174/29775?Disaridan= | 1,582,442,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145747.6/warc/CC-MAIN-20200223062700-20200223092700-00129.warc.gz | 346,435,045 | 23,551 | Ders Bilgileri
#### Ders Tanımı
Ders Kodu Yarıyıl T+U Saat Kredi AKTS
INTEGRATED CIRCUIT DESIGN ELE 427 7 3 + 0 3 5
Dersin Dili Türkçe Dersin Seviyesi Lisans Dersin Türü SECMELI Dersin Koordinatörü Doç.Dr. HALİL İBRAHİM ESKİKURT Dersi Verenler Dersin Yardımcıları Dersin Kategorisi Dersin Amacı Giving students an understanding of the integrated circuit production techniques, analog and digital integrated circuit design and implementation methods. Dersin İçeriği Integrated circuit design methods. Investigation of different architectures used for integrated circuit design. Implementing a digital design into a programmable device.
Dersin Öğrenme Çıktıları Öğretim Yöntemleri Ölçme Yöntemleri 1 - Explains integrated circuit fabrication techniques. 1 - 2 - A - 2 - Explains analog and digital integrated circuit design and implementation methods. 1 - 2 - A - 3 - Explains architectures and structures used in VLSI design. 1 - 2 - A - 4 - Designs an example of digital integrated circuit. 9 - 13 - 14 - 16 - C - D - 5 - Simulates the designed circuit in a computer with a virtual programmable architecture chosen. 9 - 13 - 14 - 15 - 16 - C - D - 6 - Analyzes the simulation results. 9 - 13 - 14 - 16 - C - D - 7 - Transfers the circuit implemented in a computer to the FPGA platform and tests the design as real-time. 13 - 14 - 15 - 16 - C - D -
Öğretim Yöntemleri: 1:Lecture 2:Question-Answer 9:Simulation 13:Lab / Workshop 14:Self Study 16:Project Based Learning 15:Problem Solving Ölçme Yöntemleri: A:Testing C:Homework D:Project / Design
#### Ders Akışı
Hafta Konular ÖnHazırlık
1 The methods of integrated circuit design; application approaches
2 The concept of ASIC and the design architecture. Physical structure, design of the devices and usage
3 The synthesis of the design. The test methods of the integrated circuits
4 Basic design process. Design with either schematic or hardware definition languages
5 VLSI technology; device miniaturization. The development of VLSI design
6 The steps of CMOS manifacturing : Forming layers, oxidation, diffusion, masking, etching, metallization and protocols
7 CMOS technology; basic CMOS contents, inverters, switches, sequential circuits, multiplexers, decoders, Flip/Flops
8 The historic developments of programmable logic devices. PLA, PAL and PROM.
9 Midterm exam
10 The advanced PLD structures. SPLD, EPLD and CPLD
11 The field programmable gate array (FPGA) architectures.
12 FPGA programming technologies. SRAM, EPROM and Antifuse structures.
13 The main blocks of an FPGA. CLBs, interconnection lines and switch matrix.
14 Examples about integrated circuit design with VHDL programming languages and CAD softwares.
Ders Notu
Ders Kaynakları
#### Dersin Program Çıktılarına Katkısı
No Program Öğrenme Çıktıları KatkıDüzeyi
1 2 3 4 5
1 To have latest knowledge and skills intended for research and practice in electronic technology X
2 To utilize equipments and instruments used in electronic technology X
3 To develop curriculum related to electronic technology and have a skill to transfer those accumulation by oral and written way X
4 To have knowledge and skills for planning, designing and managing procedures independently or in cooperation X
5 To have an open mind to ethic auditing and positive criticism, and have a constructive and interpreting attitude against scientific and social problems X
6 To disseminate and realize the environmental awareness X
7 To cooperate with social organizations and the society
8 To contribute to the education of people who work under his/her responsibility and to manage some activities for their vocational careers and social rights
9 To appropriate self learning and life- long learning principles X
10 To congregate in national or international scale to see individual applications on the premises and to perform some activities and mobility for professional advancement in electronic technology
#### Değerlendirme Sistemi
YARIYIL İÇİ ÇALIŞMALARI SIRA KATKI YÜZDESİ
AraSinav 1 70
KisaSinav 1 10
Odev 1 10
KisaSinav 2 10
Toplam 100
Yıliçinin Başarıya Oranı 50
Finalin Başarıya Oranı 50
Toplam 100
#### AKTS - İş Yükü
Etkinlik Sayısı Süresi(Saat) Toplam İş yükü(Saat)
Course Duration (Including the exam week: 16x Total course hours) 14 3 42
Hours for off-the-classroom study (Pre-study, practice) 14 3 42
Mid-terms 1 5 5
Assignment 1 10 10
Project / Design 3 7 21
Performance Task (Laboratory) 1 15 15
Toplam İş Yükü 135
Toplam İş Yükü /25(s) 5.4
Dersin AKTS Kredisi 5.4
; ; | 1,184 | 4,494 | {"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-2020-10 | latest | en | 0.663845 |
http://cn.metamath.org/mpeuni/axgroth6.html | 1,660,149,659,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00669.warc.gz | 10,755,282 | 7,800 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > axgroth6 Structured version Visualization version GIF version
Theorem axgroth6 9851
Description: The Tarski-Grothendieck axiom using abbreviations. This version is called Tarski's axiom: given a set 𝑥, there exists a set 𝑦 containing 𝑥, the subsets of the members of 𝑦, the power sets of the members of 𝑦, and the subsets of 𝑦 of cardinality less than that of 𝑦. (Contributed by NM, 21-Jun-2009.)
Assertion
Ref Expression
axgroth6 𝑦(𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ 𝒫 𝑧𝑦) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦))
Distinct variable group: 𝑥,𝑦,𝑧
Proof of Theorem axgroth6
Dummy variables 𝑤 𝑣 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 axgroth5 9847 . 2 𝑦(𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ ∃𝑤𝑦 𝒫 𝑧𝑤) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦))
2 biid 251 . . . 4 (𝑥𝑦𝑥𝑦)
3 pweq 4298 . . . . . . . . 9 (𝑧 = 𝑣 → 𝒫 𝑧 = 𝒫 𝑣)
43sseq1d 3779 . . . . . . . 8 (𝑧 = 𝑣 → (𝒫 𝑧𝑦 ↔ 𝒫 𝑣𝑦))
54cbvralv 3319 . . . . . . 7 (∀𝑧𝑦 𝒫 𝑧𝑦 ↔ ∀𝑣𝑦 𝒫 𝑣𝑦)
6 ssid 3771 . . . . . . . . . 10 𝒫 𝑧 ⊆ 𝒫 𝑧
7 sseq2 3774 . . . . . . . . . . 11 (𝑤 = 𝒫 𝑧 → (𝒫 𝑧𝑤 ↔ 𝒫 𝑧 ⊆ 𝒫 𝑧))
87rspcev 3458 . . . . . . . . . 10 ((𝒫 𝑧𝑦 ∧ 𝒫 𝑧 ⊆ 𝒫 𝑧) → ∃𝑤𝑦 𝒫 𝑧𝑤)
96, 8mpan2 663 . . . . . . . . 9 (𝒫 𝑧𝑦 → ∃𝑤𝑦 𝒫 𝑧𝑤)
10 pweq 4298 . . . . . . . . . . . . 13 (𝑣 = 𝑤 → 𝒫 𝑣 = 𝒫 𝑤)
1110sseq1d 3779 . . . . . . . . . . . 12 (𝑣 = 𝑤 → (𝒫 𝑣𝑦 ↔ 𝒫 𝑤𝑦))
1211rspccv 3455 . . . . . . . . . . 11 (∀𝑣𝑦 𝒫 𝑣𝑦 → (𝑤𝑦 → 𝒫 𝑤𝑦))
13 pwss 4312 . . . . . . . . . . . 12 (𝒫 𝑤𝑦 ↔ ∀𝑣(𝑣𝑤𝑣𝑦))
14 vpwex 4977 . . . . . . . . . . . . 13 𝒫 𝑧 ∈ V
15 sseq1 3773 . . . . . . . . . . . . . 14 (𝑣 = 𝒫 𝑧 → (𝑣𝑤 ↔ 𝒫 𝑧𝑤))
16 eleq1 2837 . . . . . . . . . . . . . 14 (𝑣 = 𝒫 𝑧 → (𝑣𝑦 ↔ 𝒫 𝑧𝑦))
1715, 16imbi12d 333 . . . . . . . . . . . . 13 (𝑣 = 𝒫 𝑧 → ((𝑣𝑤𝑣𝑦) ↔ (𝒫 𝑧𝑤 → 𝒫 𝑧𝑦)))
1814, 17spcv 3448 . . . . . . . . . . . 12 (∀𝑣(𝑣𝑤𝑣𝑦) → (𝒫 𝑧𝑤 → 𝒫 𝑧𝑦))
1913, 18sylbi 207 . . . . . . . . . . 11 (𝒫 𝑤𝑦 → (𝒫 𝑧𝑤 → 𝒫 𝑧𝑦))
2012, 19syl6 35 . . . . . . . . . 10 (∀𝑣𝑦 𝒫 𝑣𝑦 → (𝑤𝑦 → (𝒫 𝑧𝑤 → 𝒫 𝑧𝑦)))
2120rexlimdv 3177 . . . . . . . . 9 (∀𝑣𝑦 𝒫 𝑣𝑦 → (∃𝑤𝑦 𝒫 𝑧𝑤 → 𝒫 𝑧𝑦))
229, 21impbid2 216 . . . . . . . 8 (∀𝑣𝑦 𝒫 𝑣𝑦 → (𝒫 𝑧𝑦 ↔ ∃𝑤𝑦 𝒫 𝑧𝑤))
2322ralbidv 3134 . . . . . . 7 (∀𝑣𝑦 𝒫 𝑣𝑦 → (∀𝑧𝑦 𝒫 𝑧𝑦 ↔ ∀𝑧𝑦𝑤𝑦 𝒫 𝑧𝑤))
245, 23sylbi 207 . . . . . 6 (∀𝑧𝑦 𝒫 𝑧𝑦 → (∀𝑧𝑦 𝒫 𝑧𝑦 ↔ ∀𝑧𝑦𝑤𝑦 𝒫 𝑧𝑤))
2524pm5.32i 556 . . . . 5 ((∀𝑧𝑦 𝒫 𝑧𝑦 ∧ ∀𝑧𝑦 𝒫 𝑧𝑦) ↔ (∀𝑧𝑦 𝒫 𝑧𝑦 ∧ ∀𝑧𝑦𝑤𝑦 𝒫 𝑧𝑤))
26 r19.26 3211 . . . . 5 (∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ 𝒫 𝑧𝑦) ↔ (∀𝑧𝑦 𝒫 𝑧𝑦 ∧ ∀𝑧𝑦 𝒫 𝑧𝑦))
27 r19.26 3211 . . . . 5 (∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ ∃𝑤𝑦 𝒫 𝑧𝑤) ↔ (∀𝑧𝑦 𝒫 𝑧𝑦 ∧ ∀𝑧𝑦𝑤𝑦 𝒫 𝑧𝑤))
2825, 26, 273bitr4i 292 . . . 4 (∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ 𝒫 𝑧𝑦) ↔ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ ∃𝑤𝑦 𝒫 𝑧𝑤))
29 selpw 4302 . . . . . 6 (𝑧 ∈ 𝒫 𝑦𝑧𝑦)
30 impexp 437 . . . . . . . . 9 (((𝑧𝑦𝑧𝑦) → (¬ 𝑧𝑦𝑧𝑦)) ↔ (𝑧𝑦 → (𝑧𝑦 → (¬ 𝑧𝑦𝑧𝑦))))
31 vex 3352 . . . . . . . . . . . 12 𝑦 ∈ V
32 ssdomg 8154 . . . . . . . . . . . 12 (𝑦 ∈ V → (𝑧𝑦𝑧𝑦))
3331, 32ax-mp 5 . . . . . . . . . . 11 (𝑧𝑦𝑧𝑦)
3433pm4.71i 541 . . . . . . . . . 10 (𝑧𝑦 ↔ (𝑧𝑦𝑧𝑦))
3534imbi1i 338 . . . . . . . . 9 ((𝑧𝑦 → (¬ 𝑧𝑦𝑧𝑦)) ↔ ((𝑧𝑦𝑧𝑦) → (¬ 𝑧𝑦𝑧𝑦)))
36 brsdom 8131 . . . . . . . . . . . 12 (𝑧𝑦 ↔ (𝑧𝑦 ∧ ¬ 𝑧𝑦))
3736imbi1i 338 . . . . . . . . . . 11 ((𝑧𝑦𝑧𝑦) ↔ ((𝑧𝑦 ∧ ¬ 𝑧𝑦) → 𝑧𝑦))
38 impexp 437 . . . . . . . . . . 11 (((𝑧𝑦 ∧ ¬ 𝑧𝑦) → 𝑧𝑦) ↔ (𝑧𝑦 → (¬ 𝑧𝑦𝑧𝑦)))
3937, 38bitri 264 . . . . . . . . . 10 ((𝑧𝑦𝑧𝑦) ↔ (𝑧𝑦 → (¬ 𝑧𝑦𝑧𝑦)))
4039imbi2i 325 . . . . . . . . 9 ((𝑧𝑦 → (𝑧𝑦𝑧𝑦)) ↔ (𝑧𝑦 → (𝑧𝑦 → (¬ 𝑧𝑦𝑧𝑦))))
4130, 35, 403bitr4ri 293 . . . . . . . 8 ((𝑧𝑦 → (𝑧𝑦𝑧𝑦)) ↔ (𝑧𝑦 → (¬ 𝑧𝑦𝑧𝑦)))
4241pm5.74ri 261 . . . . . . 7 (𝑧𝑦 → ((𝑧𝑦𝑧𝑦) ↔ (¬ 𝑧𝑦𝑧𝑦)))
43 pm4.64 828 . . . . . . 7 ((¬ 𝑧𝑦𝑧𝑦) ↔ (𝑧𝑦𝑧𝑦))
4442, 43syl6bb 276 . . . . . 6 (𝑧𝑦 → ((𝑧𝑦𝑧𝑦) ↔ (𝑧𝑦𝑧𝑦)))
4529, 44sylbi 207 . . . . 5 (𝑧 ∈ 𝒫 𝑦 → ((𝑧𝑦𝑧𝑦) ↔ (𝑧𝑦𝑧𝑦)))
4645ralbiia 3127 . . . 4 (∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦) ↔ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦))
472, 28, 463anbi123i 1157 . . 3 ((𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ 𝒫 𝑧𝑦) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦)) ↔ (𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ ∃𝑤𝑦 𝒫 𝑧𝑤) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦)))
4847exbii 1923 . 2 (∃𝑦(𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ 𝒫 𝑧𝑦) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦)) ↔ ∃𝑦(𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ ∃𝑤𝑦 𝒫 𝑧𝑤) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦)))
491, 48mpbir 221 1 𝑦(𝑥𝑦 ∧ ∀𝑧𝑦 (𝒫 𝑧𝑦 ∧ 𝒫 𝑧𝑦) ∧ ∀𝑧 ∈ 𝒫 𝑦(𝑧𝑦𝑧𝑦))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 382 ∨ wo 826 ∧ w3a 1070 ∀wal 1628 = wceq 1630 ∃wex 1851 ∈ wcel 2144 ∀wral 3060 ∃wrex 3061 Vcvv 3349 ⊆ wss 3721 𝒫 cpw 4295 class class class wbr 4784 ≈ cen 8105 ≼ cdom 8106 ≺ csdm 8107 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1869 ax-4 1884 ax-5 1990 ax-6 2056 ax-7 2092 ax-8 2146 ax-9 2153 ax-10 2173 ax-11 2189 ax-12 2202 ax-13 2407 ax-ext 2750 ax-sep 4912 ax-nul 4920 ax-pow 4971 ax-pr 5034 ax-un 7095 ax-groth 9846 This theorem depends on definitions: df-bi 197 df-an 383 df-or 827 df-3an 1072 df-tru 1633 df-ex 1852 df-nf 1857 df-sb 2049 df-eu 2621 df-mo 2622 df-clab 2757 df-cleq 2763 df-clel 2766 df-nfc 2901 df-ral 3065 df-rex 3066 df-rab 3069 df-v 3351 df-dif 3724 df-un 3726 df-in 3728 df-ss 3735 df-nul 4062 df-if 4224 df-pw 4297 df-sn 4315 df-pr 4317 df-op 4321 df-uni 4573 df-br 4785 df-opab 4845 df-id 5157 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-dom 8110 df-sdom 8111 This theorem is referenced by: grothomex 9852 grothac 9853
Copyright terms: Public domain W3C validator | 4,052 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | latest | en | 0.259935 |
https://www.mathworks.com/matlabcentral/cody/problems/308-matrix-with-different-incremental-runs/solutions/260039 | 1,508,261,796,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822145.14/warc/CC-MAIN-20171017163022-20171017183022-00773.warc.gz | 983,234,739 | 11,464 | Cody
# Problem 308. Matrix with different incremental runs
Solution 260039
Submitted on 12 Jun 2013 by Suman Saha
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
%% a = [3 2 4]; y_correct = [1 1 1;2 2 2;3 0 3;0 0 4]; assert(isequal(matInc(a),y_correct))
ans = []
2 Pass
%% a = [0 1 1 2 0 3]; y_correct = [0 1 1 1 0 1;0 0 0 2 0 2;0 0 0 0 0 3]; assert(isequal(matInc(a),y_correct))
ans = [] | 198 | 516 | {"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-2017-43 | latest | en | 0.631173 |
https://support.buildsoft.eu/knowledge-base/what-is-the-difference-between-a-1st-and-a-2nd-order-calculation/ | 1,719,153,000,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00037.warc.gz | 490,860,331 | 13,705 | 1. Home
2. Diamonds
3. Calculation core & results
4. What is the difference between a 1st and a 2nd order calculation
# What is the difference between a 1st and a 2nd order calculation
We perfer to explain the difference using an example:
• A column loaded with a bending moment M and a point load P needs to be calculated (Figure A = intitial state).
• Using the geometry and supports the reactions M0 and deformation δ0 can be calculated (Figure B = iteration 1).
• You could stop the calculation but because of the deformation δ0, the load P has an eccentricity (Figure C= iteration 2).
Resulting in an additional moment ΔM1. Increasing the total moment M1 = M0 + ΔM1.
Resulting in additional deformation Δδ1. Increasing the total deformation δ1 = δ0 + Δδ1.
• Again you could stop the calculation but because of the additional deformation Δδ1, the load P has an additional eccentricity (Figure D = iteration 3).
Resulting in an additional moment ΔM2. Increasing the total moment M2 = M1 + ΔM2.
Resulting in additional deformation Δδ2. Increasing the total deformation δ2 = δ1 + Δδ2.
• You could continue the calculation and do an additional iteration, but it will almost not affect the final result. Δδ2 has become quite small during iteration 3, that the affect of Δδ2 will be neglectable. You stop the calculations
Conclusions:
• The calculation in ‘Iteration 1’ is called a first order calculation.
Equilibrium expressed as function of the initial geometry.
If the boundary conditions are adequately defined, a solution will always be found.
Because of the simplicity, this type of calculation is used in hand calculations.
• The calculation from ‘Iteration 1’ to ‘Iteration 3) is called a second order calculation.
Equilibrium expressed as function of the a priori unknown deformation state.
Results in larger internal forces and additional deformations relative to a 1st order calculation. The additional effects are called ‘second order effects’.
Convergence
is no certainty. For some structures the deformations and internal forces will keep increasing until it collapses.
Because of its complexity, this type of calculation is done by software. | 504 | 2,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.891283 |
https://www.freestatistics.org/blog/index.php?v=date/2011/Nov/26/t1322342301bhgt99944gbggvx.htm/ | 1,675,046,373,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00517.warc.gz | 795,182,861 | 62,084 | ## Free Statistics
of Irreproducible Research!
Author's title
Author*The author of this computation has been verified*
R Software Modulerwasp_decompose.wasp
Title produced by softwareClassical Decomposition
Date of computationSat, 26 Nov 2011 16:18:13 -0500
Cite this page as followsStatistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?v=date/2011/Nov/26/t1322342301bhgt99944gbggvx.htm/, Retrieved Mon, 30 Jan 2023 02:39:32 +0000
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Dataseries X:
235.1
280.7
264.6
240.7
201.4
240.8
241.1
223.8
206.1
174.7
203.3
220.5
299.5
347.4
338.3
327.7
351.6
396.6
438.8
395.6
363.5
378.8
357
369
464.8
479.1
431.3
366.5
326.3
355.1
331.6
261.3
249
205.5
235.6
240.9
264.9
253.8
232.3
193.8
177
213.2
207.2
180.6
188.6
175.4
199
179.6
225.8
234
200.2
183.6
178.2
203.2
208.5
191.8
172.8
148
159.4
154.5
213.2
196.4
182.8
176.4
153.6
173.2
171
151.2
161.9
157.2
201.7
236.4
356.1
398.3
403.7
384.6
365.8
368.1
367.9
347
343.3
292.9
311.5
300.9
366.9
356.9
329.7
316.2
269
289.3
266.2
253.6
233.8
228.4
253.6
260.1
306.6
309.2
309.5
271
279.9
317.9
298.4
246.7
227.3
209.1
259.9
266
320.6
308.5
282.2
262.7
263.5
313.1
284.3
252.6
250.3
246.5
312.7
333.2
446.4
511.6
515.5
506.4
483.2
522.3
509.8
460.7
405.8
375
378.5
406.8
467.8
469.8
429.8
355.8
332.7
378
360.5
334.7
319.5
323.1
363.6
352.1
411.9
388.6
416.4
360.7
338
417.2
388.4
371.1
331.5
353.7
396.7
447
533.5
565.4
542.3
488.7
467.1
531.3
496.1
444
403.4
386.3
394.1
404.1
462.1
448.1
432.3
386.3
395.2
421.9
382.9
384.2
345.5
323.4
372.6
376
462.7
487
444.2
399.3
394.9
455.4
414
375.5
347
339.4
385.8
378.8
451.8
446.1
422.5
383.1
352.8
445.3
367.5
355.1
326.2
319.8
331.8
340.9
394.1
417.2
369.9
349.2
321.4
405.7
342.9
316.5
284.2
270.9
288.8
278.8
324.4
310.9
299
273
279.3
359.2
305
282.1
250.3
246.5
257.9
266.5
315.9
318.4
295.4
266.4
245.8
362.8
324.9
294.2
289.5
295.2
290.3
272
307.4
328.7
292.9
249.1
230.4
361.5
321.7
277.2
260.7
251
257.6
241.8
287.5
292.3
274.7
254.2
230
339
318.2
287
295.8
284
271
262.7
340.6
379.4
373.3
355.2
338.4
466.9
451
422
429.2
425.9
460.7
463.6
541.4
544.2
517.5
469.4
439.4
549
533
506.1
484
457
481.5
469.5
544.7
541.2
521.5
469.7
434.4
542.6
517.3
485.7
465.8
447
426.6
411.6
467.5
484.5
451.2
417.4
379.9
484.7
455
420.8
416.5
376.3
405.6
405.8
500.8
514
475.5
430.1
414.4
538
526
488.5
520.2
504.4
568.5
610.6
818
830.9
835.9
782
762.3
856.9
820.9
769.6
752.2
724.4
723.1
719.5
817.4
803.3
752.5
689
630.4
765.5
757.7
732.2
702.6
683.3
709.5
702.2
784.8
810.9
755.6
656.8
615.1
745.3
694.1
675.7
643.7
622.1
634.6
588
689.7
673.9
647.9
568.8
545.7
632.6
643.8
593.1
579.7
546
562.9
572.5
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 6 seconds R Server 'Gertrude Mary Cox' @ cox.wessa.net
\begin{tabular}{lllllllll}
\hline
Summary of computational transaction \tabularnewline
Raw Input & view raw input (R code) \tabularnewline
Raw Output & view raw output of R engine \tabularnewline
Computing time & 6 seconds \tabularnewline
R Server & 'Gertrude Mary Cox' @ cox.wessa.net \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=147450&T=0
[TABLE]
[ROW][C]Summary of computational transaction[/C][/ROW]
[ROW][C]Raw Input[/C][C]view raw input (R code) [/C][/ROW]
[ROW][C]Raw Output[/C][C]view raw output of R engine [/C][/ROW]
[ROW][C]Computing time[/C][C]6 seconds[/C][/ROW]
[ROW][C]R Server[/C][C]'Gertrude Mary Cox' @ cox.wessa.net[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=147450&T=0
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=147450&T=0
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 6 seconds R Server 'Gertrude Mary Cox' @ cox.wessa.net
Classical Decomposition by Moving Averages t Observations Fit Trend Seasonal Random 1 235.1 NA NA 1.12478036463889 NA 2 280.7 NA NA 1.14434096703054 NA 3 264.6 NA NA 1.08034073104924 NA 4 240.7 NA NA 0.979030802194402 NA 5 201.4 NA NA 0.925484201792753 NA 6 240.8 NA NA 1.11361515899907 NA 7 241.1 241.129916299572 230.416666666667 1.04649511594751 0.999875932857974 8 223.8 225.596343447582 235.879166666667 0.956406395001318 0.992037355658652 9 206.1 219.614781810963 241.729166666667 0.908515860288394 0.938461420039584 10 174.7 214.609970790079 248.425 0.863882341914377 0.814034871524599 11 203.3 239.412379081157 258.308333333333 0.926847291342352 0.8491624400553 12 220.5 252.154933827686 271.058333333333 0.930260769801161 0.874462365867018 13 299.5 321.448168459236 285.7875 1.12478036463889 0.931720972110565 14 347.4 344.656426920148 301.183333333333 1.14434096703054 1.00796031312797 15 338.3 340.199296207404 314.9 1.08034073104924 0.994417107182238 16 327.7 323.04345106907 329.9625 0.979030802194402 1.01441462105336 17 351.6 319.172507909102 344.870833333333 0.925484201792753 1.10159863800091 18 396.6 398.075658773705 357.4625 1.11361515899907 0.996293019326399 19 438.8 387.765684025401 370.5375 1.04649511594751 1.1316112231614 20 395.6 366.219963725942 382.9125 0.956406395001318 1.0802251083615 21 363.5 356.38805909463 392.275 0.908515860288394 1.01995560940913 22 378.8 343.623599535475 397.766666666667 0.863882341914377 1.10236898895209 23 357 369.190309187656 398.329166666667 0.926847291342352 0.966980961080807 24 369 367.960771408308 395.545833333333 0.930260769801161 1.0028242918062 25 464.8 437.933234972151 389.35 1.12478036463889 1.06134899770637 26 479.1 434.034224532596 379.2875 1.14434096703054 1.1038300044563 27 431.3 398.560202782626 368.920833333333 1.08034073104924 1.08214517402589 28 366.5 349.444648368246 356.929166666667 0.979030802194402 1.04880701911274 29 326.3 318.968130147872 344.65 0.925484201792753 1.02298621448083 30 355.1 372.230506958602 334.254166666667 1.11361515899907 0.95397876681691 31 331.6 335.493252983823 320.5875 1.04649511594751 0.988395435827109 32 261.3 289.667601859378 302.870833333333 0.956406395001318 0.902068434035127 33 249 259.101152388748 285.191666666667 0.908515860288394 0.96101463735062 34 205.5 232.992667124066 269.704166666667 0.863882341914377 0.882002006915411 35 235.6 237.539375179903 256.2875 0.926847291342352 0.991835563352669 36 240.9 227.127043033494 244.154166666667 0.930260769801161 1.06063988146262 37 264.9 262.139437148798 233.058333333333 1.12478036463889 1.01053089485973 38 253.8 256.918851360444 224.5125 1.14434096703054 0.987860558522938 39 232.3 236.198495165064 218.633333333333 1.08034073104924 0.983494834874625 40 193.8 210.357005736495 214.8625 0.979030802194402 0.921290923121263 41 177 196.279774463546 212.083333333333 0.925484201792753 0.901774013566909 42 213.2 231.636593134969 208.004166666667 1.11361515899907 0.920407251352439 43 207.2 213.297506611685 203.820833333333 1.04649511594751 0.971413136944045 44 180.6 192.588367740099 201.366666666667 0.956406395001318 0.937751340432579 45 188.6 180.980144852199 199.204166666667 0.908515860288394 1.04210326582523 46 175.4 170.566369391478 197.441666666667 0.863882341914377 1.02833870842046 47 199 182.650706213866 197.066666666667 0.926847291342352 1.08951125415847 48 179.6 182.982293419888 196.7 0.930260769801161 0.981515733808588 49 225.8 220.836564842287 196.3375 1.12478036463889 1.02247560389855 50 234 225.273055534687 196.858333333333 1.14434096703054 1.03873940647096 51 200.2 212.467010439683 196.666666666667 1.08034073104924 0.942263928812773 52 183.6 190.780468987616 194.866666666667 0.979030802194402 0.962362662039153 53 178.2 177.762378059343 192.075 0.925484201792753 1.0024618366689 54 203.2 210.895510798611 189.379166666667 1.11361515899907 0.963510314802481 55 208.5 196.540503567575 187.808333333333 1.04649511594751 1.0608500345493 56 191.8 177.620607658328 185.716666666667 0.956406395001318 1.07982965787927 57 172.8 166.644521673399 183.425 0.908515860288394 1.03693777788066 58 148 157.572139165182 182.4 0.863882341914377 0.939252337272975 59 159.4 167.828873279816 181.075 0.926847291342352 0.949776977494432 60 154.5 166.330625640448 178.8 0.930260769801161 0.928872836286797 61 213.2 197.947284421886 175.9875 1.12478036463889 1.07705443205579 62 196.4 197.665829705075 172.733333333333 1.14434096703054 0.993596112656579 63 182.8 184.292624457861 170.5875 1.08034073104924 0.991900791134467 64 176.4 166.941068954182 170.516666666667 0.979030802194402 1.05666029997935 65 153.6 159.796415992041 172.6625 0.925484201792753 0.961223060269701 66 173.2 198.042535838497 177.8375 1.11361515899907 0.874559595324733 67 171 195.90824610169 187.204166666667 1.04649511594751 0.872857592279391 68 151.2 192.783634045745 201.570833333333 0.956406395001318 0.784298940874423 69 161.9 199.135320126962 219.1875 0.908515860288394 0.813014988485105 70 157.2 204.797707189835 237.066666666667 0.863882341914377 0.767586718411281 71 201.7 235.959872920907 254.583333333333 0.926847291342352 0.854806359671202 72 236.4 252.608435952964 271.545833333333 0.930260769801161 0.935835729745849 73 356.1 323.791460885567 287.870833333333 1.12478036463889 1.09978193688638 74 398.3 348.146666869591 304.233333333333 1.14434096703054 1.14405805915469 75 403.7 345.655016899203 319.95 1.08034073104924 1.16792750072458 76 384.6 326.176349636092 333.1625 0.979030802194402 1.17911675824777 77 365.8 317.803562527283 343.391666666667 0.925484201792753 1.15102548596697 78 368.1 390.493795566186 350.654166666667 1.11361515899907 0.942652621320866 79 367.9 370.241251229597 353.791666666667 1.04649511594751 0.993676417142009 80 347 337.149194344548 352.516666666667 0.956406395001318 1.02921794214755 81 343.3 315.898535587777 347.708333333333 0.908515860288394 1.08674134674679 82 292.9 295.253387407786 341.775 0.863882341914377 0.992029261955474 83 311.5 310.393434143126 334.891666666667 0.926847291342352 1.00356504273336 84 300.9 304.730171667615 327.575 0.930260769801161 0.987430940472173 85 366.9 359.990642287528 320.054166666667 1.12478036463889 1.0191931592126 86 356.9 356.948556141001 311.925 1.14434096703054 0.9998639687984 87 329.7 327.851901935454 303.470833333333 1.08034073104924 1.00563699052418 88 316.2 290.009320085027 296.220833333333 0.979030802194402 1.09030978696579 89 269 269.427732062741 291.120833333333 0.925484201792753 0.998412442329279 90 289.3 319.616830759058 287.008333333333 1.11361515899907 0.905146325720524 91 266.2 295.94445839364 282.795833333333 1.04649511594751 0.899493105716221 92 253.6 266.163914702221 278.295833333333 0.956406395001318 0.952796325842001 93 233.8 250.265835647443 275.466666666667 0.908515860288394 0.934206618315099 94 228.4 235.61670973763 272.741666666667 0.863882341914377 0.969370976508132 95 253.6 251.465255732322 271.3125 0.926847291342352 1.00848922154857 96 260.1 253.922429290309 272.958333333333 0.930260769801161 1.02432857438769 97 306.6 309.867617288308 275.491666666667 1.12478036463889 0.989454795835385 98 309.2 316.462726344933 276.545833333333 1.14434096703054 0.977050294583454 99 309.5 298.160537510451 275.9875 1.08034073104924 1.03803139940728 100 271 269.147805408268 274.9125 0.979030802194402 1.00688170051739 101 279.9 253.925871682712 274.370833333333 0.925484201792753 1.10229020046348 102 317.9 306.109606893032 274.879166666667 1.11361515899907 1.03851689996482 103 298.4 288.527424259362 275.708333333333 1.04649511594751 1.03421711390514 104 246.7 264.219221699052 276.2625 0.956406395001318 0.933694370960618 105 227.3 249.928927682586 275.095833333333 0.908515860288394 0.909458549306765 106 209.1 236.369007277048 273.6125 0.863882341914377 0.884633744537051 107 259.9 252.643124165069 272.583333333333 0.926847291342352 1.02872382083982 108 266 252.751851154975 271.7 0.930260769801161 1.05241563527423 109 320.6 304.717060535232 270.9125 1.12478036463889 1.05212356484691 110 308.5 309.625289066926 270.570833333333 1.14434096703054 0.99636564225643 111 282.2 293.609602180906 271.775 1.08034073104924 0.961140228057406 112 262.7 268.539990451906 274.291666666667 0.979030802194402 0.978252809043158 113 263.5 257.330882308475 278.05 0.925484201792753 1.02397348361838 114 313.1 315.208770754687 283.05 1.11361515899907 0.99330992361146 115 284.3 304.626007459688 291.091666666667 1.04649511594751 0.933275534714883 116 252.6 291.508684169756 304.795833333333 0.956406395001318 0.866526500640722 117 250.3 293.431695459395 322.979166666667 0.908515860288394 0.853009418795511 118 246.5 296.185660435102 342.854166666667 0.863882341914377 0.832248258196859 119 312.7 335.669332150774 362.1625 0.926847291342352 0.931571549883332 120 333.2 353.530101216768 380.033333333333 0.930260769801161 0.942494002217077 121 446.4 447.82661559612 398.145833333333 1.12478036463889 0.996814357283742 122 511.6 476.289014740199 416.2125 1.14434096703054 1.07413772765484 123 515.5 466.018478597226 431.3625 1.08034073104924 1.10617931192711 124 506.4 433.90237223755 443.195833333333 0.979030802194402 1.16708281033034 125 483.2 417.663307900721 451.291666666667 0.925484201792753 1.15691273535299 126 522.3 509.033489178475 457.1 1.11361515899907 1.02606215721275 127 509.8 482.495294000233 461.058333333333 1.04649511594751 1.05659061619729 128 460.7 440.146193032898 460.208333333333 0.956406395001318 1.04669768202577 129 405.8 413.280079362439 454.895833333333 0.908515860288394 0.981900701882417 130 375 384.470836268994 445.05 0.863882341914377 0.97536656782371 131 378.5 400.865315369281 432.504166666667 0.926847291342352 0.944207407047232 132 406.8 390.914955903152 420.220833333333 0.930260769801161 1.04063554964314 133 467.8 458.896329018108 407.9875 1.12478036463889 1.01940235826454 134 469.8 453.750265777059 396.516666666667 1.14434096703054 1.03537129437369 135 429.8 418.8165914898 387.670833333333 1.08034073104924 1.02622486485344 136 355.8 373.904101243069 381.9125 0.979030802194402 0.95158089685863 137 332.7 350.878054188852 379.129166666667 0.925484201792753 0.948192672719657 138 378 418.974503257588 376.229166666667 1.11361515899907 0.902202871680723 139 360.5 388.899387067677 371.620833333333 1.04649511594751 0.926974976016778 140 334.7 349.957069984274 365.908333333333 0.956406395001318 0.95640302399103 141 319.5 328.852457562389 361.966666666667 0.908515860288394 0.971560323338575 142 323.1 312.390653365513 361.6125 0.863882341914377 1.03428190478528 143 363.6 335.553476239357 362.0375 0.926847291342352 1.08358287351086 144 352.1 338.514141957561 363.891666666667 0.930260769801161 1.04013379755385 145 411.9 412.442899958522 366.6875 1.12478036463889 0.99868369668001 146 388.6 422.68140852218 369.366666666667 1.14434096703054 0.91936856498766 147 416.4 401.220541832835 371.383333333333 1.08034073104924 1.03783320290089 148 360.7 365.333502428859 373.158333333333 0.979030802194402 0.987317061265791 149 338 347.808531586239 375.8125 0.925484201792753 0.971799048339886 150 417.2 424.449777789333 381.145833333333 1.11361515899907 0.982919586323988 151 388.4 408.307511072187 390.166666666667 1.04649511594751 0.951243828407879 152 371.1 385.049214627531 402.6 0.956406395001318 0.963772904611624 153 331.5 377.227141639995 415.2125 0.908515860288394 0.87878088135123 154 353.7 367.833902167626 425.791666666667 0.863882341914377 0.961575314063398 155 396.7 404.57270453465 436.504166666667 0.926847291342352 0.980540692818845 156 447 415.489344572066 446.6375 0.930260769801161 1.07583986410142 157 533.5 512.763935314605 455.879166666667 1.12478036463889 1.04043978770206 158 565.4 530.292372209315 463.404166666667 1.14434096703054 1.06620428584409 159 542.3 507.152451931925 469.4375 1.08034073104924 1.06930371318168 160 488.7 463.856635489689 473.791666666667 0.979030802194402 1.05355828204135 161 467.1 439.643557693299 475.041666666667 0.925484201792753 1.06245159704093 162 531.3 526.902372417247 473.145833333333 1.11361515899907 1.00834619051453 163 496.1 490.160870724548 468.383333333333 1.04649511594751 1.01211669398798 164 444 440.445070031336 460.520833333333 0.956406395001318 1.00807122206728 165 403.4 409.78607878308 451.05 0.908515860288394 0.984416067031743 166 386.3 382.008771594538 442.2 0.863882341914377 1.01123332427041 167 394.1 403.120643778214 434.9375 0.926847291342352 0.977622967423179 168 404.1 397.577948666853 427.383333333333 0.930260769801161 1.01640445944001 169 462.1 470.280043625224 418.108333333333 1.12478036463889 0.982606015849266 170 448.1 470.209703352849 410.9 1.14434096703054 0.952979057651948 171 432.3 438.613835386277 405.995833333333 1.08034073104924 0.985605024563996 172 386.3 392.554638024873 400.9625 0.979030802194402 0.984066834475978 173 395.2 367.829839818356 397.445833333333 0.925484201792753 1.07440984177673 174 421.9 440.30023355242 395.379166666667 1.11361515899907 0.958209802879359 175 382.9 412.56325787704 394.233333333333 1.04649511594751 0.92810009783789 176 384.2 378.621366647793 395.879166666667 0.956406395001318 1.01473406902943 177 345.5 361.58552691203 397.995833333333 0.908515860288394 0.955513908287753 178 323.4 344.7178505019 399.033333333333 0.863882341914377 0.93815855352178 179 372.6 370.333420846978 399.5625 0.926847291342352 1.00612037430443 180 376 372.984179565235 400.945833333333 0.930260769801161 1.00808565242172 181 462.7 454.003534431929 403.6375 1.12478036463889 1.01915506137844 182 487 462.966978649018 404.570833333333 1.14434096703054 1.05191087584932 183 444.2 436.750247625217 404.270833333333 1.08034073104924 1.01705723675096 184 399.3 396.507474888733 405 0.979030802194402 1.00704280571773 185 394.9 375.947107504913 406.216666666667 0.925484201792753 1.0504137207515 186 455.4 453.111447944071 406.883333333333 1.11361515899907 1.00505074869839 187 414 425.448228992144 406.545833333333 1.04649511594751 0.973091369967001 188 375.5 386.758791058595 404.3875 0.956406395001318 0.9708893726041 189 347 365.022745250121 401.779166666667 0.908515860288394 0.950625692550279 190 339.4 345.725713234134 400.2 0.863882341914377 0.981703087181572 191 385.8 368.67281944999 397.770833333333 0.926847291342352 1.04645631477678 192 378.8 368.007284446798 395.595833333333 0.930260769801161 1.0293274508667 193 451.8 442.305818639684 393.2375 1.12478036463889 1.02146519661332 194 446.1 446.807930577074 390.45 1.14434096703054 0.998415581889605 195 422.5 419.964453516539 388.733333333333 1.08034073104924 1.00603752641975 196 383.1 378.933871989343 387.05 0.979030802194402 1.01099434048686 197 352.8 355.370508751721 383.983333333333 0.925484201792753 0.9927666795966 198 445.3 423.345442756659 380.154166666667 1.11361515899907 1.05185967539979 199 367.5 393.660939845238 376.170833333333 1.04649511594751 0.933544486644971 200 355.1 356.321157537679 372.5625 0.956406395001318 0.996572873903651 201 326.2 335.393771756466 369.166666666667 0.908515860288394 0.972588126164903 202 319.8 315.802988616075 365.5625 0.863882341914377 1.01265666104504 203 331.8 336.298815936144 362.841666666667 0.926847291342352 0.98662256385405 204 340.9 334.785346705274 359.883333333333 0.930260769801161 1.01826439942758 205 394.1 401.780919418716 357.208333333333 1.12478036463889 0.980882816859924 206 417.2 405.754698384854 354.575 1.14434096703054 1.0282074407535 207 369.9 379.433670423342 351.216666666667 1.08034073104924 0.974873947236403 208 349.2 340.143855747399 347.429166666667 0.979030802194402 1.02662445344692 209 321.4 317.99637173599 343.6 0.925484201792753 1.01070335565601 210 405.7 377.761462248297 339.220833333333 1.11361515899907 1.0739581469889 211 342.9 349.2459429659 333.729166666667 1.04649511594751 0.981829587161391 212 316.5 312.167062301784 326.395833333333 0.956406395001318 1.01388018859602 213 284.2 289.827915880251 319.0125 0.908515860288394 0.980581870924481 214 270.9 270.294386745977 312.883333333333 0.863882341914377 1.00224056911175 215 288.8 285.426485232591 307.954166666667 0.926847291342352 1.01181920719326 216 278.8 283.043467471626 304.2625 0.930260769801161 0.985007718038746 217 324.4 338.273008080293 300.745833333333 1.12478036463889 0.958988722869075 218 310.9 340.708450583893 297.733333333333 1.14434096703054 0.91251038671683 219 299 318.578977327281 294.8875 1.08034073104924 0.938542783043818 220 273 286.325716691771 292.458333333333 0.979030802194402 0.953459588451441 221 279.3 268.533097334342 290.154166666667 0.925484201792753 1.04009525370444 222 359.2 321.115571160544 288.354166666667 1.11361515899907 1.11860038023636 223 305 300.85426464596 287.4875 1.04649511594751 1.0137798789687 224 282.1 274.915033216483 287.445833333333 0.956406395001318 1.02613522694432 225 250.3 261.296732384445 287.608333333333 0.908515860288394 0.957914772664416 226 246.5 248.092610558777 287.183333333333 0.863882341914377 0.993580580432484 227 257.9 264.626487269383 285.5125 0.926847291342352 0.974581201833603 228 266.5 264.442128162143 284.266666666667 0.930260769801161 1.00778193645679 229 315.9 320.83891242839 285.245833333333 1.12478036463889 0.984606254923981 230 318.4 327.94428071414 286.579166666667 1.14434096703054 0.970896639229824 231 295.4 311.912374732765 288.716666666667 1.08034073104924 0.947060854039817 232 266.4 286.248210086597 292.379166666667 0.979030802194402 0.930660841230788 233 245.8 273.719665048555 295.758333333333 0.925484201792753 0.897999052996056 234 362.8 331.119547338886 297.3375 1.11361515899907 1.09567678174158 235 324.9 311.03142964855 297.2125 1.04649511594751 1.04458896763945 236 294.2 284.327666153954 297.2875 0.956406395001318 1.03472167861674 237 289.5 270.38567647008 297.6125 0.908515860288394 1.07069281102261 238 295.2 256.389480550913 296.7875 0.863882341914377 1.15137329100123 239 290.3 273.813861044814 295.425 0.926847291342352 1.06020929288341 240 272 274.174981466188 294.729166666667 0.930260769801161 0.992067177484405 241 307.4 331.294683234679 294.541666666667 1.12478036463889 0.927874836380177 242 328.7 336.092942016869 293.7 1.14434096703054 0.978003280960008 243 292.9 315.234422480741 291.791666666667 1.08034073104924 0.929149798093177 244 249.1 282.695144133634 288.75 0.979030802194402 0.881161226746248 245 230.4 264.268157637747 285.545833333333 0.925484201792753 0.871841700716087 246 361.5 315.069568859812 282.925 1.11361515899907 1.1473656478733 247 321.7 293.895072124909 280.8375 1.04649511594751 1.09460835009603 248 277.2 266.351210954575 278.491666666667 0.956406395001318 1.04073114218833 249 260.7 250.947222542659 276.216666666667 0.908515860288394 1.03886385893625 250 251 238.147165097488 275.670833333333 0.863882341914377 1.05397013605957 251 257.6 255.686272771643 275.866666666667 0.926847291342352 1.00748466942559 252 241.8 255.740313877962 274.9125 0.930260769801161 0.94549035438889 253 287.5 307.997669932096 273.829166666667 1.12478036463889 0.933448620125552 254 292.3 313.654322888346 274.091666666667 1.14434096703054 0.931917651599059 255 274.7 298.133528992175 275.9625 1.08034073104924 0.921399216413563 256 254.2 272.953787651799 278.8 0.979030802194402 0.931293176719999 257 230 259.814264916619 280.733333333333 0.925484201792753 0.885247775266739 258 339 314.220437301075 282.1625 1.11361515899907 1.07886044240713 259 318.2 298.508371427711 285.245833333333 1.04649511594751 1.06596675489571 260 287 278.397946504946 291.0875 0.956406395001318 1.03089840856603 261 295.8 271.487251950679 298.825 0.908515860288394 1.08955392149955 262 284 265.334262299485 307.141666666667 0.863882341914377 1.07034801136782 263 271 292.760164425337 315.866666666667 0.926847291342352 0.925672386241309 264 262.7 302.997560983861 325.7125 0.930260769801161 0.867003678666552 265 340.6 378.572951228333 336.575 1.12478036463889 0.899694494535004 266 379.4 397.92549893542 347.733333333333 1.14434096703054 0.953444805660905 267 373.3 387.752294052421 358.916666666667 1.08034073104924 0.962728024375099 268 355.2 362.620771247779 370.3875 0.979030802194402 0.979535724822811 269 338.4 355.57488651295 384.204166666667 0.925484201792753 0.951698257766794 270 466.9 445.979670863315 400.479166666667 1.11361515899907 1.04690870571788 271 451 436.615203958567 417.216666666667 1.04649511594751 1.03294616383262 272 422 413.59794551832 432.45 0.956406395001318 1.02031454598052 273 429.2 404.584825482929 445.325 0.908515860288394 1.06084057771492 274 425.9 394.009537127631 456.091666666667 0.863882341914377 1.08093830191232 275 460.7 431.038056566189 465.058333333333 0.926847291342352 1.06881513820406 276 463.6 439.722637625386 472.6875 0.930260769801161 1.05430096231469 277 541.4 539.360304353462 479.525 1.12478036463889 1.00378169403657 278 544.2 556.659895324643 486.445833333333 1.14434096703054 0.977616682233994 279 517.5 531.779719180135 492.233333333333 1.08034073104924 0.973147303920972 280 469.4 485.415709613012 495.8125 0.979030802194402 0.967006198407175 281 439.4 460.867995387746 497.975 0.925484201792753 0.953418341905724 282 549 555.791405666948 499.0875 1.11361515899907 0.987780657279508 283 533 522.693787641567 499.470833333333 1.04649511594751 1.01971749540957 284 506.1 477.709054196575 499.483333333333 0.956406395001318 1.05943145844529 285 484 453.82638511056 499.525 0.908515860288394 1.06648713225893 286 457 431.685605764372 499.704166666667 0.863882341914377 1.0586408115017 287 481.5 462.967945752933 499.508333333333 0.926847291342352 1.04002880635057 288 469.5 464.231132823106 499.033333333333 0.930260769801161 1.01134966357136 289 544.7 560.267159381187 498.1125 1.12478036463889 0.972214756620072 290 541.2 568.289260402091 496.608333333333 1.14434096703054 0.952331915646401 291 521.5 534.768661869371 495 1.08034073104924 0.975188033975311 292 469.7 483.46988589365 493.825 0.979030802194402 0.971518627539339 293 434.4 454.524572421292 491.120833333333 0.925484201792753 0.955723906599623 294 542.6 541.68561365296 486.420833333333 1.11361515899907 1.00168803882546 295 517.3 503.14613095493 480.791666666667 1.04649511594751 1.0281307321557 296 485.7 454.496273984564 475.2125 0.956406395001318 1.06865562558274 297 465.8 426.930530163273 469.920833333333 0.908515860288394 1.09104401557289 298 447 401.543311051077 464.8125 0.863882341914377 1.11320494626081 299 426.6 426.685736160593 460.3625 0.926847291342352 0.99979906485423 300 411.6 423.900452365685 455.679166666667 0.930260769801161 0.970982686390073 301 467.5 506.905704248778 450.670833333333 1.12478036463889 0.922262259196361 302 484.5 509.656090103864 445.370833333333 1.14434096703054 0.950641048753607 303 451.2 476.011630359431 440.6125 1.08034073104924 0.947875999708881 304 417.4 426.478055320909 435.6125 0.979030802194402 0.978713898153381 305 379.9 399.616365965762 431.791666666667 0.925484201792753 0.950661765520755 306 484.7 479.606208601924 430.675 1.11361515899907 1.01062077868617 307 455 451.898393047718 431.820833333333 1.04649511594751 1.00686350515957 308 420.8 415.498803228385 434.4375 0.956406395001318 1.01275863307048 309 416.5 396.729948774186 436.679166666667 0.908515860288394 1.04983251525855 310 376.3 378.57123977567 438.220833333333 0.863882341914377 0.99400049571379 311 405.6 407.986592057761 440.1875 0.926847291342352 0.99415031742655 312 405.8 412.892366589704 443.845833333333 0.930260769801161 0.982822722908917 313 500.8 505.054503231976 449.025 1.12478036463889 0.991576150287245 314 514 520.451039892852 454.804166666667 1.14434096703054 0.987604905364047 315 475.5 499.058899288482 461.945833333333 1.08034073104924 0.952793348997343 316 430.1 461.715005609889 471.604166666667 0.979030802194402 0.931527012928401 317 414.4 447.683701696374 483.729166666667 0.925484201792753 0.925653532683334 318 538 555.749645098486 499.05 1.11361515899907 0.968061796790999 319 526 545.014656385464 520.8 1.04649511594751 0.965111660461447 320 488.5 523.36550447795 547.220833333333 0.956406395001318 0.933382112157491 321 520.2 522.797880837454 575.441666666667 0.908515860288394 0.995030812226529 322 504.4 522.75320264118 605.120833333333 0.863882341914377 0.964891266952644 323 568.5 587.879927579884 634.279166666667 0.926847291342352 0.967034207717101 324 610.6 615.890770906481 662.0625 0.930260769801161 0.991409562934846 325 818 773.441157989372 687.6375 1.12478036463889 1.05761115962132 326 830.9 814.355944925196 711.6375 1.14434096703054 1.02031550844309 327 835.9 791.90776153794 733.016666666667 1.08034073104924 1.05555222539633 328 782 736.084308629861 751.85 0.979030802194402 1.06237830481077 329 762.3 710.270563034197 767.458333333333 0.925484201792753 1.07325298227698 330 856.9 866.879800333338 778.4375 1.11361515899907 0.988487676919567 331 820.9 819.353351031104 782.95 1.04649511594751 1.0018876458697 332 769.6 747.694609452155 781.775 0.956406395001318 1.02929724284611 333 752.2 706.053100823126 777.15 0.908515860288394 1.06535896396896 334 724.4 665.016626805688 769.8 0.863882341914377 1.08929607291107 335 723.1 704.801713382722 760.429166666667 0.926847291342352 1.02596231857816 336 719.5 698.742120716897 751.125 0.930260769801161 1.02970749675403 337 817.4 837.605191207168 744.683333333333 1.12478036463889 0.97587742838837 338 803.3 847.374949911389 740.491666666667 1.14434096703054 0.947986484712584 339 752.5 796.06707335248 736.866666666667 1.08034073104924 0.94527210732507 340 689 717.715243203688 733.0875 0.979030802194402 0.959990757510581 341 630.4 676.351567038492 730.808333333333 0.925484201792753 0.932059642827918 342 765.5 812.405458805634 729.520833333333 1.11361515899907 0.942263486418971 343 757.7 761.264151303384 727.441666666666 1.04649511594751 0.995318114878677 344 732.2 694.733605328957 726.4 0.956406395001318 1.05392915267616 345 702.6 660.350967567868 726.845833333333 0.908515860288394 1.06397966309906 346 683.3 626.861823371136 725.633333333333 0.863882341914377 1.09003288208772 347 709.5 670.716904243607 723.654166666667 0.926847291342352 1.05782334620018 348 702.2 671.811071431153 722.175 0.930260769801161 1.04523433724322 349 784.8 808.360901726557 718.683333333333 1.12478036463889 0.970853486757915 350 810.9 816.692307732883 713.679166666666 1.14434096703054 0.992907600967907 351 755.6 765.822034302814 708.870833333333 1.08034073104924 0.986652206589851 352 656.8 689.107147304566 703.866666666667 0.979030802194402 0.953117381773016 353 615.1 646.169213507526 698.195833333333 0.925484201792753 0.951917836910123 354 745.3 768.747104509708 690.316666666667 1.11361515899907 0.96949958657124 355 694.1 713.286710633507 681.595833333333 1.04649511594751 0.973100984011792 356 675.7 642.633366961261 671.925 0.956406395001318 1.05145489596206 357 643.7 601.191443132089 661.729166666667 0.908515860288394 1.07070718878907 358 622.1 564.611901616689 653.575 0.863882341914377 1.1018187859992 359 634.6 599.685644953357 647.016666666667 0.926847291342352 1.05822109523625 360 588 594.835868816648 639.429166666667 0.930260769801161 0.988507974762438 361 689.7 711.578237934234 632.6375 1.12478036463889 0.969253924912392 362 673.9 717.616220424852 627.1 1.14434096703054 0.939081337377003 363 647.9 670.88259114215 620.991666666667 1.08034073104924 0.965742752240712 364 568.8 602.254877264895 615.154166666667 0.979030802194402 0.944450632900096 365 545.7 563.616022707613 608.995833333333 0.925484201792753 0.96821236092341 366 632.6 674.140856689574 605.3625 1.11361515899907 0.938379559290377 367 643.8 NA NA 1.04649511594751 NA 368 593.1 NA NA 0.956406395001318 NA 369 579.7 NA NA 0.908515860288394 NA 370 546 NA NA 0.863882341914377 NA 371 562.9 NA NA 0.926847291342352 NA 372 572.5 NA NA 0.930260769801161 NA
\begin{tabular}{lllllllll}
\hline
Classical Decomposition by Moving Averages \tabularnewline
t & Observations & Fit & Trend & Seasonal & Random \tabularnewline
1 & 235.1 & NA & NA & 1.12478036463889 & NA \tabularnewline
2 & 280.7 & NA & NA & 1.14434096703054 & NA \tabularnewline
3 & 264.6 & NA & NA & 1.08034073104924 & NA \tabularnewline
4 & 240.7 & NA & NA & 0.979030802194402 & NA \tabularnewline
5 & 201.4 & NA & NA & 0.925484201792753 & NA \tabularnewline
6 & 240.8 & NA & NA & 1.11361515899907 & NA \tabularnewline
7 & 241.1 & 241.129916299572 & 230.416666666667 & 1.04649511594751 & 0.999875932857974 \tabularnewline
8 & 223.8 & 225.596343447582 & 235.879166666667 & 0.956406395001318 & 0.992037355658652 \tabularnewline
9 & 206.1 & 219.614781810963 & 241.729166666667 & 0.908515860288394 & 0.938461420039584 \tabularnewline
10 & 174.7 & 214.609970790079 & 248.425 & 0.863882341914377 & 0.814034871524599 \tabularnewline
11 & 203.3 & 239.412379081157 & 258.308333333333 & 0.926847291342352 & 0.8491624400553 \tabularnewline
12 & 220.5 & 252.154933827686 & 271.058333333333 & 0.930260769801161 & 0.874462365867018 \tabularnewline
13 & 299.5 & 321.448168459236 & 285.7875 & 1.12478036463889 & 0.931720972110565 \tabularnewline
14 & 347.4 & 344.656426920148 & 301.183333333333 & 1.14434096703054 & 1.00796031312797 \tabularnewline
15 & 338.3 & 340.199296207404 & 314.9 & 1.08034073104924 & 0.994417107182238 \tabularnewline
16 & 327.7 & 323.04345106907 & 329.9625 & 0.979030802194402 & 1.01441462105336 \tabularnewline
17 & 351.6 & 319.172507909102 & 344.870833333333 & 0.925484201792753 & 1.10159863800091 \tabularnewline
18 & 396.6 & 398.075658773705 & 357.4625 & 1.11361515899907 & 0.996293019326399 \tabularnewline
19 & 438.8 & 387.765684025401 & 370.5375 & 1.04649511594751 & 1.1316112231614 \tabularnewline
20 & 395.6 & 366.219963725942 & 382.9125 & 0.956406395001318 & 1.0802251083615 \tabularnewline
21 & 363.5 & 356.38805909463 & 392.275 & 0.908515860288394 & 1.01995560940913 \tabularnewline
22 & 378.8 & 343.623599535475 & 397.766666666667 & 0.863882341914377 & 1.10236898895209 \tabularnewline
23 & 357 & 369.190309187656 & 398.329166666667 & 0.926847291342352 & 0.966980961080807 \tabularnewline
24 & 369 & 367.960771408308 & 395.545833333333 & 0.930260769801161 & 1.0028242918062 \tabularnewline
25 & 464.8 & 437.933234972151 & 389.35 & 1.12478036463889 & 1.06134899770637 \tabularnewline
26 & 479.1 & 434.034224532596 & 379.2875 & 1.14434096703054 & 1.1038300044563 \tabularnewline
27 & 431.3 & 398.560202782626 & 368.920833333333 & 1.08034073104924 & 1.08214517402589 \tabularnewline
28 & 366.5 & 349.444648368246 & 356.929166666667 & 0.979030802194402 & 1.04880701911274 \tabularnewline
29 & 326.3 & 318.968130147872 & 344.65 & 0.925484201792753 & 1.02298621448083 \tabularnewline
30 & 355.1 & 372.230506958602 & 334.254166666667 & 1.11361515899907 & 0.95397876681691 \tabularnewline
31 & 331.6 & 335.493252983823 & 320.5875 & 1.04649511594751 & 0.988395435827109 \tabularnewline
32 & 261.3 & 289.667601859378 & 302.870833333333 & 0.956406395001318 & 0.902068434035127 \tabularnewline
33 & 249 & 259.101152388748 & 285.191666666667 & 0.908515860288394 & 0.96101463735062 \tabularnewline
34 & 205.5 & 232.992667124066 & 269.704166666667 & 0.863882341914377 & 0.882002006915411 \tabularnewline
35 & 235.6 & 237.539375179903 & 256.2875 & 0.926847291342352 & 0.991835563352669 \tabularnewline
36 & 240.9 & 227.127043033494 & 244.154166666667 & 0.930260769801161 & 1.06063988146262 \tabularnewline
37 & 264.9 & 262.139437148798 & 233.058333333333 & 1.12478036463889 & 1.01053089485973 \tabularnewline
38 & 253.8 & 256.918851360444 & 224.5125 & 1.14434096703054 & 0.987860558522938 \tabularnewline
39 & 232.3 & 236.198495165064 & 218.633333333333 & 1.08034073104924 & 0.983494834874625 \tabularnewline
40 & 193.8 & 210.357005736495 & 214.8625 & 0.979030802194402 & 0.921290923121263 \tabularnewline
41 & 177 & 196.279774463546 & 212.083333333333 & 0.925484201792753 & 0.901774013566909 \tabularnewline
42 & 213.2 & 231.636593134969 & 208.004166666667 & 1.11361515899907 & 0.920407251352439 \tabularnewline
43 & 207.2 & 213.297506611685 & 203.820833333333 & 1.04649511594751 & 0.971413136944045 \tabularnewline
44 & 180.6 & 192.588367740099 & 201.366666666667 & 0.956406395001318 & 0.937751340432579 \tabularnewline
45 & 188.6 & 180.980144852199 & 199.204166666667 & 0.908515860288394 & 1.04210326582523 \tabularnewline
46 & 175.4 & 170.566369391478 & 197.441666666667 & 0.863882341914377 & 1.02833870842046 \tabularnewline
47 & 199 & 182.650706213866 & 197.066666666667 & 0.926847291342352 & 1.08951125415847 \tabularnewline
48 & 179.6 & 182.982293419888 & 196.7 & 0.930260769801161 & 0.981515733808588 \tabularnewline
49 & 225.8 & 220.836564842287 & 196.3375 & 1.12478036463889 & 1.02247560389855 \tabularnewline
50 & 234 & 225.273055534687 & 196.858333333333 & 1.14434096703054 & 1.03873940647096 \tabularnewline
51 & 200.2 & 212.467010439683 & 196.666666666667 & 1.08034073104924 & 0.942263928812773 \tabularnewline
52 & 183.6 & 190.780468987616 & 194.866666666667 & 0.979030802194402 & 0.962362662039153 \tabularnewline
53 & 178.2 & 177.762378059343 & 192.075 & 0.925484201792753 & 1.0024618366689 \tabularnewline
54 & 203.2 & 210.895510798611 & 189.379166666667 & 1.11361515899907 & 0.963510314802481 \tabularnewline
55 & 208.5 & 196.540503567575 & 187.808333333333 & 1.04649511594751 & 1.0608500345493 \tabularnewline
56 & 191.8 & 177.620607658328 & 185.716666666667 & 0.956406395001318 & 1.07982965787927 \tabularnewline
57 & 172.8 & 166.644521673399 & 183.425 & 0.908515860288394 & 1.03693777788066 \tabularnewline
58 & 148 & 157.572139165182 & 182.4 & 0.863882341914377 & 0.939252337272975 \tabularnewline
59 & 159.4 & 167.828873279816 & 181.075 & 0.926847291342352 & 0.949776977494432 \tabularnewline
60 & 154.5 & 166.330625640448 & 178.8 & 0.930260769801161 & 0.928872836286797 \tabularnewline
61 & 213.2 & 197.947284421886 & 175.9875 & 1.12478036463889 & 1.07705443205579 \tabularnewline
62 & 196.4 & 197.665829705075 & 172.733333333333 & 1.14434096703054 & 0.993596112656579 \tabularnewline
63 & 182.8 & 184.292624457861 & 170.5875 & 1.08034073104924 & 0.991900791134467 \tabularnewline
64 & 176.4 & 166.941068954182 & 170.516666666667 & 0.979030802194402 & 1.05666029997935 \tabularnewline
65 & 153.6 & 159.796415992041 & 172.6625 & 0.925484201792753 & 0.961223060269701 \tabularnewline
66 & 173.2 & 198.042535838497 & 177.8375 & 1.11361515899907 & 0.874559595324733 \tabularnewline
67 & 171 & 195.90824610169 & 187.204166666667 & 1.04649511594751 & 0.872857592279391 \tabularnewline
68 & 151.2 & 192.783634045745 & 201.570833333333 & 0.956406395001318 & 0.784298940874423 \tabularnewline
69 & 161.9 & 199.135320126962 & 219.1875 & 0.908515860288394 & 0.813014988485105 \tabularnewline
70 & 157.2 & 204.797707189835 & 237.066666666667 & 0.863882341914377 & 0.767586718411281 \tabularnewline
71 & 201.7 & 235.959872920907 & 254.583333333333 & 0.926847291342352 & 0.854806359671202 \tabularnewline
72 & 236.4 & 252.608435952964 & 271.545833333333 & 0.930260769801161 & 0.935835729745849 \tabularnewline
73 & 356.1 & 323.791460885567 & 287.870833333333 & 1.12478036463889 & 1.09978193688638 \tabularnewline
74 & 398.3 & 348.146666869591 & 304.233333333333 & 1.14434096703054 & 1.14405805915469 \tabularnewline
75 & 403.7 & 345.655016899203 & 319.95 & 1.08034073104924 & 1.16792750072458 \tabularnewline
76 & 384.6 & 326.176349636092 & 333.1625 & 0.979030802194402 & 1.17911675824777 \tabularnewline
77 & 365.8 & 317.803562527283 & 343.391666666667 & 0.925484201792753 & 1.15102548596697 \tabularnewline
78 & 368.1 & 390.493795566186 & 350.654166666667 & 1.11361515899907 & 0.942652621320866 \tabularnewline
79 & 367.9 & 370.241251229597 & 353.791666666667 & 1.04649511594751 & 0.993676417142009 \tabularnewline
80 & 347 & 337.149194344548 & 352.516666666667 & 0.956406395001318 & 1.02921794214755 \tabularnewline
81 & 343.3 & 315.898535587777 & 347.708333333333 & 0.908515860288394 & 1.08674134674679 \tabularnewline
82 & 292.9 & 295.253387407786 & 341.775 & 0.863882341914377 & 0.992029261955474 \tabularnewline
83 & 311.5 & 310.393434143126 & 334.891666666667 & 0.926847291342352 & 1.00356504273336 \tabularnewline
84 & 300.9 & 304.730171667615 & 327.575 & 0.930260769801161 & 0.987430940472173 \tabularnewline
85 & 366.9 & 359.990642287528 & 320.054166666667 & 1.12478036463889 & 1.0191931592126 \tabularnewline
86 & 356.9 & 356.948556141001 & 311.925 & 1.14434096703054 & 0.9998639687984 \tabularnewline
87 & 329.7 & 327.851901935454 & 303.470833333333 & 1.08034073104924 & 1.00563699052418 \tabularnewline
88 & 316.2 & 290.009320085027 & 296.220833333333 & 0.979030802194402 & 1.09030978696579 \tabularnewline
89 & 269 & 269.427732062741 & 291.120833333333 & 0.925484201792753 & 0.998412442329279 \tabularnewline
90 & 289.3 & 319.616830759058 & 287.008333333333 & 1.11361515899907 & 0.905146325720524 \tabularnewline
91 & 266.2 & 295.94445839364 & 282.795833333333 & 1.04649511594751 & 0.899493105716221 \tabularnewline
92 & 253.6 & 266.163914702221 & 278.295833333333 & 0.956406395001318 & 0.952796325842001 \tabularnewline
93 & 233.8 & 250.265835647443 & 275.466666666667 & 0.908515860288394 & 0.934206618315099 \tabularnewline
94 & 228.4 & 235.61670973763 & 272.741666666667 & 0.863882341914377 & 0.969370976508132 \tabularnewline
95 & 253.6 & 251.465255732322 & 271.3125 & 0.926847291342352 & 1.00848922154857 \tabularnewline
96 & 260.1 & 253.922429290309 & 272.958333333333 & 0.930260769801161 & 1.02432857438769 \tabularnewline
97 & 306.6 & 309.867617288308 & 275.491666666667 & 1.12478036463889 & 0.989454795835385 \tabularnewline
98 & 309.2 & 316.462726344933 & 276.545833333333 & 1.14434096703054 & 0.977050294583454 \tabularnewline
99 & 309.5 & 298.160537510451 & 275.9875 & 1.08034073104924 & 1.03803139940728 \tabularnewline
100 & 271 & 269.147805408268 & 274.9125 & 0.979030802194402 & 1.00688170051739 \tabularnewline
101 & 279.9 & 253.925871682712 & 274.370833333333 & 0.925484201792753 & 1.10229020046348 \tabularnewline
102 & 317.9 & 306.109606893032 & 274.879166666667 & 1.11361515899907 & 1.03851689996482 \tabularnewline
103 & 298.4 & 288.527424259362 & 275.708333333333 & 1.04649511594751 & 1.03421711390514 \tabularnewline
104 & 246.7 & 264.219221699052 & 276.2625 & 0.956406395001318 & 0.933694370960618 \tabularnewline
105 & 227.3 & 249.928927682586 & 275.095833333333 & 0.908515860288394 & 0.909458549306765 \tabularnewline
106 & 209.1 & 236.369007277048 & 273.6125 & 0.863882341914377 & 0.884633744537051 \tabularnewline
107 & 259.9 & 252.643124165069 & 272.583333333333 & 0.926847291342352 & 1.02872382083982 \tabularnewline
108 & 266 & 252.751851154975 & 271.7 & 0.930260769801161 & 1.05241563527423 \tabularnewline
109 & 320.6 & 304.717060535232 & 270.9125 & 1.12478036463889 & 1.05212356484691 \tabularnewline
110 & 308.5 & 309.625289066926 & 270.570833333333 & 1.14434096703054 & 0.99636564225643 \tabularnewline
111 & 282.2 & 293.609602180906 & 271.775 & 1.08034073104924 & 0.961140228057406 \tabularnewline
112 & 262.7 & 268.539990451906 & 274.291666666667 & 0.979030802194402 & 0.978252809043158 \tabularnewline
113 & 263.5 & 257.330882308475 & 278.05 & 0.925484201792753 & 1.02397348361838 \tabularnewline
114 & 313.1 & 315.208770754687 & 283.05 & 1.11361515899907 & 0.99330992361146 \tabularnewline
115 & 284.3 & 304.626007459688 & 291.091666666667 & 1.04649511594751 & 0.933275534714883 \tabularnewline
116 & 252.6 & 291.508684169756 & 304.795833333333 & 0.956406395001318 & 0.866526500640722 \tabularnewline
117 & 250.3 & 293.431695459395 & 322.979166666667 & 0.908515860288394 & 0.853009418795511 \tabularnewline
118 & 246.5 & 296.185660435102 & 342.854166666667 & 0.863882341914377 & 0.832248258196859 \tabularnewline
119 & 312.7 & 335.669332150774 & 362.1625 & 0.926847291342352 & 0.931571549883332 \tabularnewline
120 & 333.2 & 353.530101216768 & 380.033333333333 & 0.930260769801161 & 0.942494002217077 \tabularnewline
121 & 446.4 & 447.82661559612 & 398.145833333333 & 1.12478036463889 & 0.996814357283742 \tabularnewline
122 & 511.6 & 476.289014740199 & 416.2125 & 1.14434096703054 & 1.07413772765484 \tabularnewline
123 & 515.5 & 466.018478597226 & 431.3625 & 1.08034073104924 & 1.10617931192711 \tabularnewline
124 & 506.4 & 433.90237223755 & 443.195833333333 & 0.979030802194402 & 1.16708281033034 \tabularnewline
125 & 483.2 & 417.663307900721 & 451.291666666667 & 0.925484201792753 & 1.15691273535299 \tabularnewline
126 & 522.3 & 509.033489178475 & 457.1 & 1.11361515899907 & 1.02606215721275 \tabularnewline
127 & 509.8 & 482.495294000233 & 461.058333333333 & 1.04649511594751 & 1.05659061619729 \tabularnewline
128 & 460.7 & 440.146193032898 & 460.208333333333 & 0.956406395001318 & 1.04669768202577 \tabularnewline
129 & 405.8 & 413.280079362439 & 454.895833333333 & 0.908515860288394 & 0.981900701882417 \tabularnewline
130 & 375 & 384.470836268994 & 445.05 & 0.863882341914377 & 0.97536656782371 \tabularnewline
131 & 378.5 & 400.865315369281 & 432.504166666667 & 0.926847291342352 & 0.944207407047232 \tabularnewline
132 & 406.8 & 390.914955903152 & 420.220833333333 & 0.930260769801161 & 1.04063554964314 \tabularnewline
133 & 467.8 & 458.896329018108 & 407.9875 & 1.12478036463889 & 1.01940235826454 \tabularnewline
134 & 469.8 & 453.750265777059 & 396.516666666667 & 1.14434096703054 & 1.03537129437369 \tabularnewline
135 & 429.8 & 418.8165914898 & 387.670833333333 & 1.08034073104924 & 1.02622486485344 \tabularnewline
136 & 355.8 & 373.904101243069 & 381.9125 & 0.979030802194402 & 0.95158089685863 \tabularnewline
137 & 332.7 & 350.878054188852 & 379.129166666667 & 0.925484201792753 & 0.948192672719657 \tabularnewline
138 & 378 & 418.974503257588 & 376.229166666667 & 1.11361515899907 & 0.902202871680723 \tabularnewline
139 & 360.5 & 388.899387067677 & 371.620833333333 & 1.04649511594751 & 0.926974976016778 \tabularnewline
140 & 334.7 & 349.957069984274 & 365.908333333333 & 0.956406395001318 & 0.95640302399103 \tabularnewline
141 & 319.5 & 328.852457562389 & 361.966666666667 & 0.908515860288394 & 0.971560323338575 \tabularnewline
142 & 323.1 & 312.390653365513 & 361.6125 & 0.863882341914377 & 1.03428190478528 \tabularnewline
143 & 363.6 & 335.553476239357 & 362.0375 & 0.926847291342352 & 1.08358287351086 \tabularnewline
144 & 352.1 & 338.514141957561 & 363.891666666667 & 0.930260769801161 & 1.04013379755385 \tabularnewline
145 & 411.9 & 412.442899958522 & 366.6875 & 1.12478036463889 & 0.99868369668001 \tabularnewline
146 & 388.6 & 422.68140852218 & 369.366666666667 & 1.14434096703054 & 0.91936856498766 \tabularnewline
147 & 416.4 & 401.220541832835 & 371.383333333333 & 1.08034073104924 & 1.03783320290089 \tabularnewline
148 & 360.7 & 365.333502428859 & 373.158333333333 & 0.979030802194402 & 0.987317061265791 \tabularnewline
149 & 338 & 347.808531586239 & 375.8125 & 0.925484201792753 & 0.971799048339886 \tabularnewline
150 & 417.2 & 424.449777789333 & 381.145833333333 & 1.11361515899907 & 0.982919586323988 \tabularnewline
151 & 388.4 & 408.307511072187 & 390.166666666667 & 1.04649511594751 & 0.951243828407879 \tabularnewline
152 & 371.1 & 385.049214627531 & 402.6 & 0.956406395001318 & 0.963772904611624 \tabularnewline
153 & 331.5 & 377.227141639995 & 415.2125 & 0.908515860288394 & 0.87878088135123 \tabularnewline
154 & 353.7 & 367.833902167626 & 425.791666666667 & 0.863882341914377 & 0.961575314063398 \tabularnewline
155 & 396.7 & 404.57270453465 & 436.504166666667 & 0.926847291342352 & 0.980540692818845 \tabularnewline
156 & 447 & 415.489344572066 & 446.6375 & 0.930260769801161 & 1.07583986410142 \tabularnewline
157 & 533.5 & 512.763935314605 & 455.879166666667 & 1.12478036463889 & 1.04043978770206 \tabularnewline
158 & 565.4 & 530.292372209315 & 463.404166666667 & 1.14434096703054 & 1.06620428584409 \tabularnewline
159 & 542.3 & 507.152451931925 & 469.4375 & 1.08034073104924 & 1.06930371318168 \tabularnewline
160 & 488.7 & 463.856635489689 & 473.791666666667 & 0.979030802194402 & 1.05355828204135 \tabularnewline
161 & 467.1 & 439.643557693299 & 475.041666666667 & 0.925484201792753 & 1.06245159704093 \tabularnewline
162 & 531.3 & 526.902372417247 & 473.145833333333 & 1.11361515899907 & 1.00834619051453 \tabularnewline
163 & 496.1 & 490.160870724548 & 468.383333333333 & 1.04649511594751 & 1.01211669398798 \tabularnewline
164 & 444 & 440.445070031336 & 460.520833333333 & 0.956406395001318 & 1.00807122206728 \tabularnewline
165 & 403.4 & 409.78607878308 & 451.05 & 0.908515860288394 & 0.984416067031743 \tabularnewline
166 & 386.3 & 382.008771594538 & 442.2 & 0.863882341914377 & 1.01123332427041 \tabularnewline
167 & 394.1 & 403.120643778214 & 434.9375 & 0.926847291342352 & 0.977622967423179 \tabularnewline
168 & 404.1 & 397.577948666853 & 427.383333333333 & 0.930260769801161 & 1.01640445944001 \tabularnewline
169 & 462.1 & 470.280043625224 & 418.108333333333 & 1.12478036463889 & 0.982606015849266 \tabularnewline
170 & 448.1 & 470.209703352849 & 410.9 & 1.14434096703054 & 0.952979057651948 \tabularnewline
171 & 432.3 & 438.613835386277 & 405.995833333333 & 1.08034073104924 & 0.985605024563996 \tabularnewline
172 & 386.3 & 392.554638024873 & 400.9625 & 0.979030802194402 & 0.984066834475978 \tabularnewline
173 & 395.2 & 367.829839818356 & 397.445833333333 & 0.925484201792753 & 1.07440984177673 \tabularnewline
174 & 421.9 & 440.30023355242 & 395.379166666667 & 1.11361515899907 & 0.958209802879359 \tabularnewline
175 & 382.9 & 412.56325787704 & 394.233333333333 & 1.04649511594751 & 0.92810009783789 \tabularnewline
176 & 384.2 & 378.621366647793 & 395.879166666667 & 0.956406395001318 & 1.01473406902943 \tabularnewline
177 & 345.5 & 361.58552691203 & 397.995833333333 & 0.908515860288394 & 0.955513908287753 \tabularnewline
178 & 323.4 & 344.7178505019 & 399.033333333333 & 0.863882341914377 & 0.93815855352178 \tabularnewline
179 & 372.6 & 370.333420846978 & 399.5625 & 0.926847291342352 & 1.00612037430443 \tabularnewline
180 & 376 & 372.984179565235 & 400.945833333333 & 0.930260769801161 & 1.00808565242172 \tabularnewline
181 & 462.7 & 454.003534431929 & 403.6375 & 1.12478036463889 & 1.01915506137844 \tabularnewline
182 & 487 & 462.966978649018 & 404.570833333333 & 1.14434096703054 & 1.05191087584932 \tabularnewline
183 & 444.2 & 436.750247625217 & 404.270833333333 & 1.08034073104924 & 1.01705723675096 \tabularnewline
184 & 399.3 & 396.507474888733 & 405 & 0.979030802194402 & 1.00704280571773 \tabularnewline
185 & 394.9 & 375.947107504913 & 406.216666666667 & 0.925484201792753 & 1.0504137207515 \tabularnewline
186 & 455.4 & 453.111447944071 & 406.883333333333 & 1.11361515899907 & 1.00505074869839 \tabularnewline
187 & 414 & 425.448228992144 & 406.545833333333 & 1.04649511594751 & 0.973091369967001 \tabularnewline
188 & 375.5 & 386.758791058595 & 404.3875 & 0.956406395001318 & 0.9708893726041 \tabularnewline
189 & 347 & 365.022745250121 & 401.779166666667 & 0.908515860288394 & 0.950625692550279 \tabularnewline
190 & 339.4 & 345.725713234134 & 400.2 & 0.863882341914377 & 0.981703087181572 \tabularnewline
191 & 385.8 & 368.67281944999 & 397.770833333333 & 0.926847291342352 & 1.04645631477678 \tabularnewline
192 & 378.8 & 368.007284446798 & 395.595833333333 & 0.930260769801161 & 1.0293274508667 \tabularnewline
193 & 451.8 & 442.305818639684 & 393.2375 & 1.12478036463889 & 1.02146519661332 \tabularnewline
194 & 446.1 & 446.807930577074 & 390.45 & 1.14434096703054 & 0.998415581889605 \tabularnewline
195 & 422.5 & 419.964453516539 & 388.733333333333 & 1.08034073104924 & 1.00603752641975 \tabularnewline
196 & 383.1 & 378.933871989343 & 387.05 & 0.979030802194402 & 1.01099434048686 \tabularnewline
197 & 352.8 & 355.370508751721 & 383.983333333333 & 0.925484201792753 & 0.9927666795966 \tabularnewline
198 & 445.3 & 423.345442756659 & 380.154166666667 & 1.11361515899907 & 1.05185967539979 \tabularnewline
199 & 367.5 & 393.660939845238 & 376.170833333333 & 1.04649511594751 & 0.933544486644971 \tabularnewline
200 & 355.1 & 356.321157537679 & 372.5625 & 0.956406395001318 & 0.996572873903651 \tabularnewline
201 & 326.2 & 335.393771756466 & 369.166666666667 & 0.908515860288394 & 0.972588126164903 \tabularnewline
202 & 319.8 & 315.802988616075 & 365.5625 & 0.863882341914377 & 1.01265666104504 \tabularnewline
203 & 331.8 & 336.298815936144 & 362.841666666667 & 0.926847291342352 & 0.98662256385405 \tabularnewline
204 & 340.9 & 334.785346705274 & 359.883333333333 & 0.930260769801161 & 1.01826439942758 \tabularnewline
205 & 394.1 & 401.780919418716 & 357.208333333333 & 1.12478036463889 & 0.980882816859924 \tabularnewline
206 & 417.2 & 405.754698384854 & 354.575 & 1.14434096703054 & 1.0282074407535 \tabularnewline
207 & 369.9 & 379.433670423342 & 351.216666666667 & 1.08034073104924 & 0.974873947236403 \tabularnewline
208 & 349.2 & 340.143855747399 & 347.429166666667 & 0.979030802194402 & 1.02662445344692 \tabularnewline
209 & 321.4 & 317.99637173599 & 343.6 & 0.925484201792753 & 1.01070335565601 \tabularnewline
210 & 405.7 & 377.761462248297 & 339.220833333333 & 1.11361515899907 & 1.0739581469889 \tabularnewline
211 & 342.9 & 349.2459429659 & 333.729166666667 & 1.04649511594751 & 0.981829587161391 \tabularnewline
212 & 316.5 & 312.167062301784 & 326.395833333333 & 0.956406395001318 & 1.01388018859602 \tabularnewline
213 & 284.2 & 289.827915880251 & 319.0125 & 0.908515860288394 & 0.980581870924481 \tabularnewline
214 & 270.9 & 270.294386745977 & 312.883333333333 & 0.863882341914377 & 1.00224056911175 \tabularnewline
215 & 288.8 & 285.426485232591 & 307.954166666667 & 0.926847291342352 & 1.01181920719326 \tabularnewline
216 & 278.8 & 283.043467471626 & 304.2625 & 0.930260769801161 & 0.985007718038746 \tabularnewline
217 & 324.4 & 338.273008080293 & 300.745833333333 & 1.12478036463889 & 0.958988722869075 \tabularnewline
218 & 310.9 & 340.708450583893 & 297.733333333333 & 1.14434096703054 & 0.91251038671683 \tabularnewline
219 & 299 & 318.578977327281 & 294.8875 & 1.08034073104924 & 0.938542783043818 \tabularnewline
220 & 273 & 286.325716691771 & 292.458333333333 & 0.979030802194402 & 0.953459588451441 \tabularnewline
221 & 279.3 & 268.533097334342 & 290.154166666667 & 0.925484201792753 & 1.04009525370444 \tabularnewline
222 & 359.2 & 321.115571160544 & 288.354166666667 & 1.11361515899907 & 1.11860038023636 \tabularnewline
223 & 305 & 300.85426464596 & 287.4875 & 1.04649511594751 & 1.0137798789687 \tabularnewline
224 & 282.1 & 274.915033216483 & 287.445833333333 & 0.956406395001318 & 1.02613522694432 \tabularnewline
225 & 250.3 & 261.296732384445 & 287.608333333333 & 0.908515860288394 & 0.957914772664416 \tabularnewline
226 & 246.5 & 248.092610558777 & 287.183333333333 & 0.863882341914377 & 0.993580580432484 \tabularnewline
227 & 257.9 & 264.626487269383 & 285.5125 & 0.926847291342352 & 0.974581201833603 \tabularnewline
228 & 266.5 & 264.442128162143 & 284.266666666667 & 0.930260769801161 & 1.00778193645679 \tabularnewline
229 & 315.9 & 320.83891242839 & 285.245833333333 & 1.12478036463889 & 0.984606254923981 \tabularnewline
230 & 318.4 & 327.94428071414 & 286.579166666667 & 1.14434096703054 & 0.970896639229824 \tabularnewline
231 & 295.4 & 311.912374732765 & 288.716666666667 & 1.08034073104924 & 0.947060854039817 \tabularnewline
232 & 266.4 & 286.248210086597 & 292.379166666667 & 0.979030802194402 & 0.930660841230788 \tabularnewline
233 & 245.8 & 273.719665048555 & 295.758333333333 & 0.925484201792753 & 0.897999052996056 \tabularnewline
234 & 362.8 & 331.119547338886 & 297.3375 & 1.11361515899907 & 1.09567678174158 \tabularnewline
235 & 324.9 & 311.03142964855 & 297.2125 & 1.04649511594751 & 1.04458896763945 \tabularnewline
236 & 294.2 & 284.327666153954 & 297.2875 & 0.956406395001318 & 1.03472167861674 \tabularnewline
237 & 289.5 & 270.38567647008 & 297.6125 & 0.908515860288394 & 1.07069281102261 \tabularnewline
238 & 295.2 & 256.389480550913 & 296.7875 & 0.863882341914377 & 1.15137329100123 \tabularnewline
239 & 290.3 & 273.813861044814 & 295.425 & 0.926847291342352 & 1.06020929288341 \tabularnewline
240 & 272 & 274.174981466188 & 294.729166666667 & 0.930260769801161 & 0.992067177484405 \tabularnewline
241 & 307.4 & 331.294683234679 & 294.541666666667 & 1.12478036463889 & 0.927874836380177 \tabularnewline
242 & 328.7 & 336.092942016869 & 293.7 & 1.14434096703054 & 0.978003280960008 \tabularnewline
243 & 292.9 & 315.234422480741 & 291.791666666667 & 1.08034073104924 & 0.929149798093177 \tabularnewline
244 & 249.1 & 282.695144133634 & 288.75 & 0.979030802194402 & 0.881161226746248 \tabularnewline
245 & 230.4 & 264.268157637747 & 285.545833333333 & 0.925484201792753 & 0.871841700716087 \tabularnewline
246 & 361.5 & 315.069568859812 & 282.925 & 1.11361515899907 & 1.1473656478733 \tabularnewline
247 & 321.7 & 293.895072124909 & 280.8375 & 1.04649511594751 & 1.09460835009603 \tabularnewline
248 & 277.2 & 266.351210954575 & 278.491666666667 & 0.956406395001318 & 1.04073114218833 \tabularnewline
249 & 260.7 & 250.947222542659 & 276.216666666667 & 0.908515860288394 & 1.03886385893625 \tabularnewline
250 & 251 & 238.147165097488 & 275.670833333333 & 0.863882341914377 & 1.05397013605957 \tabularnewline
251 & 257.6 & 255.686272771643 & 275.866666666667 & 0.926847291342352 & 1.00748466942559 \tabularnewline
252 & 241.8 & 255.740313877962 & 274.9125 & 0.930260769801161 & 0.94549035438889 \tabularnewline
253 & 287.5 & 307.997669932096 & 273.829166666667 & 1.12478036463889 & 0.933448620125552 \tabularnewline
254 & 292.3 & 313.654322888346 & 274.091666666667 & 1.14434096703054 & 0.931917651599059 \tabularnewline
255 & 274.7 & 298.133528992175 & 275.9625 & 1.08034073104924 & 0.921399216413563 \tabularnewline
256 & 254.2 & 272.953787651799 & 278.8 & 0.979030802194402 & 0.931293176719999 \tabularnewline
257 & 230 & 259.814264916619 & 280.733333333333 & 0.925484201792753 & 0.885247775266739 \tabularnewline
258 & 339 & 314.220437301075 & 282.1625 & 1.11361515899907 & 1.07886044240713 \tabularnewline
259 & 318.2 & 298.508371427711 & 285.245833333333 & 1.04649511594751 & 1.06596675489571 \tabularnewline
260 & 287 & 278.397946504946 & 291.0875 & 0.956406395001318 & 1.03089840856603 \tabularnewline
261 & 295.8 & 271.487251950679 & 298.825 & 0.908515860288394 & 1.08955392149955 \tabularnewline
262 & 284 & 265.334262299485 & 307.141666666667 & 0.863882341914377 & 1.07034801136782 \tabularnewline
263 & 271 & 292.760164425337 & 315.866666666667 & 0.926847291342352 & 0.925672386241309 \tabularnewline
264 & 262.7 & 302.997560983861 & 325.7125 & 0.930260769801161 & 0.867003678666552 \tabularnewline
265 & 340.6 & 378.572951228333 & 336.575 & 1.12478036463889 & 0.899694494535004 \tabularnewline
266 & 379.4 & 397.92549893542 & 347.733333333333 & 1.14434096703054 & 0.953444805660905 \tabularnewline
267 & 373.3 & 387.752294052421 & 358.916666666667 & 1.08034073104924 & 0.962728024375099 \tabularnewline
268 & 355.2 & 362.620771247779 & 370.3875 & 0.979030802194402 & 0.979535724822811 \tabularnewline
269 & 338.4 & 355.57488651295 & 384.204166666667 & 0.925484201792753 & 0.951698257766794 \tabularnewline
270 & 466.9 & 445.979670863315 & 400.479166666667 & 1.11361515899907 & 1.04690870571788 \tabularnewline
271 & 451 & 436.615203958567 & 417.216666666667 & 1.04649511594751 & 1.03294616383262 \tabularnewline
272 & 422 & 413.59794551832 & 432.45 & 0.956406395001318 & 1.02031454598052 \tabularnewline
273 & 429.2 & 404.584825482929 & 445.325 & 0.908515860288394 & 1.06084057771492 \tabularnewline
274 & 425.9 & 394.009537127631 & 456.091666666667 & 0.863882341914377 & 1.08093830191232 \tabularnewline
275 & 460.7 & 431.038056566189 & 465.058333333333 & 0.926847291342352 & 1.06881513820406 \tabularnewline
276 & 463.6 & 439.722637625386 & 472.6875 & 0.930260769801161 & 1.05430096231469 \tabularnewline
277 & 541.4 & 539.360304353462 & 479.525 & 1.12478036463889 & 1.00378169403657 \tabularnewline
278 & 544.2 & 556.659895324643 & 486.445833333333 & 1.14434096703054 & 0.977616682233994 \tabularnewline
279 & 517.5 & 531.779719180135 & 492.233333333333 & 1.08034073104924 & 0.973147303920972 \tabularnewline
280 & 469.4 & 485.415709613012 & 495.8125 & 0.979030802194402 & 0.967006198407175 \tabularnewline
281 & 439.4 & 460.867995387746 & 497.975 & 0.925484201792753 & 0.953418341905724 \tabularnewline
282 & 549 & 555.791405666948 & 499.0875 & 1.11361515899907 & 0.987780657279508 \tabularnewline
283 & 533 & 522.693787641567 & 499.470833333333 & 1.04649511594751 & 1.01971749540957 \tabularnewline
284 & 506.1 & 477.709054196575 & 499.483333333333 & 0.956406395001318 & 1.05943145844529 \tabularnewline
285 & 484 & 453.82638511056 & 499.525 & 0.908515860288394 & 1.06648713225893 \tabularnewline
286 & 457 & 431.685605764372 & 499.704166666667 & 0.863882341914377 & 1.0586408115017 \tabularnewline
287 & 481.5 & 462.967945752933 & 499.508333333333 & 0.926847291342352 & 1.04002880635057 \tabularnewline
288 & 469.5 & 464.231132823106 & 499.033333333333 & 0.930260769801161 & 1.01134966357136 \tabularnewline
289 & 544.7 & 560.267159381187 & 498.1125 & 1.12478036463889 & 0.972214756620072 \tabularnewline
290 & 541.2 & 568.289260402091 & 496.608333333333 & 1.14434096703054 & 0.952331915646401 \tabularnewline
291 & 521.5 & 534.768661869371 & 495 & 1.08034073104924 & 0.975188033975311 \tabularnewline
292 & 469.7 & 483.46988589365 & 493.825 & 0.979030802194402 & 0.971518627539339 \tabularnewline
293 & 434.4 & 454.524572421292 & 491.120833333333 & 0.925484201792753 & 0.955723906599623 \tabularnewline
294 & 542.6 & 541.68561365296 & 486.420833333333 & 1.11361515899907 & 1.00168803882546 \tabularnewline
295 & 517.3 & 503.14613095493 & 480.791666666667 & 1.04649511594751 & 1.0281307321557 \tabularnewline
296 & 485.7 & 454.496273984564 & 475.2125 & 0.956406395001318 & 1.06865562558274 \tabularnewline
297 & 465.8 & 426.930530163273 & 469.920833333333 & 0.908515860288394 & 1.09104401557289 \tabularnewline
298 & 447 & 401.543311051077 & 464.8125 & 0.863882341914377 & 1.11320494626081 \tabularnewline
299 & 426.6 & 426.685736160593 & 460.3625 & 0.926847291342352 & 0.99979906485423 \tabularnewline
300 & 411.6 & 423.900452365685 & 455.679166666667 & 0.930260769801161 & 0.970982686390073 \tabularnewline
301 & 467.5 & 506.905704248778 & 450.670833333333 & 1.12478036463889 & 0.922262259196361 \tabularnewline
302 & 484.5 & 509.656090103864 & 445.370833333333 & 1.14434096703054 & 0.950641048753607 \tabularnewline
303 & 451.2 & 476.011630359431 & 440.6125 & 1.08034073104924 & 0.947875999708881 \tabularnewline
304 & 417.4 & 426.478055320909 & 435.6125 & 0.979030802194402 & 0.978713898153381 \tabularnewline
305 & 379.9 & 399.616365965762 & 431.791666666667 & 0.925484201792753 & 0.950661765520755 \tabularnewline
306 & 484.7 & 479.606208601924 & 430.675 & 1.11361515899907 & 1.01062077868617 \tabularnewline
307 & 455 & 451.898393047718 & 431.820833333333 & 1.04649511594751 & 1.00686350515957 \tabularnewline
308 & 420.8 & 415.498803228385 & 434.4375 & 0.956406395001318 & 1.01275863307048 \tabularnewline
309 & 416.5 & 396.729948774186 & 436.679166666667 & 0.908515860288394 & 1.04983251525855 \tabularnewline
310 & 376.3 & 378.57123977567 & 438.220833333333 & 0.863882341914377 & 0.99400049571379 \tabularnewline
311 & 405.6 & 407.986592057761 & 440.1875 & 0.926847291342352 & 0.99415031742655 \tabularnewline
312 & 405.8 & 412.892366589704 & 443.845833333333 & 0.930260769801161 & 0.982822722908917 \tabularnewline
313 & 500.8 & 505.054503231976 & 449.025 & 1.12478036463889 & 0.991576150287245 \tabularnewline
314 & 514 & 520.451039892852 & 454.804166666667 & 1.14434096703054 & 0.987604905364047 \tabularnewline
315 & 475.5 & 499.058899288482 & 461.945833333333 & 1.08034073104924 & 0.952793348997343 \tabularnewline
316 & 430.1 & 461.715005609889 & 471.604166666667 & 0.979030802194402 & 0.931527012928401 \tabularnewline
317 & 414.4 & 447.683701696374 & 483.729166666667 & 0.925484201792753 & 0.925653532683334 \tabularnewline
318 & 538 & 555.749645098486 & 499.05 & 1.11361515899907 & 0.968061796790999 \tabularnewline
319 & 526 & 545.014656385464 & 520.8 & 1.04649511594751 & 0.965111660461447 \tabularnewline
320 & 488.5 & 523.36550447795 & 547.220833333333 & 0.956406395001318 & 0.933382112157491 \tabularnewline
321 & 520.2 & 522.797880837454 & 575.441666666667 & 0.908515860288394 & 0.995030812226529 \tabularnewline
322 & 504.4 & 522.75320264118 & 605.120833333333 & 0.863882341914377 & 0.964891266952644 \tabularnewline
323 & 568.5 & 587.879927579884 & 634.279166666667 & 0.926847291342352 & 0.967034207717101 \tabularnewline
324 & 610.6 & 615.890770906481 & 662.0625 & 0.930260769801161 & 0.991409562934846 \tabularnewline
325 & 818 & 773.441157989372 & 687.6375 & 1.12478036463889 & 1.05761115962132 \tabularnewline
326 & 830.9 & 814.355944925196 & 711.6375 & 1.14434096703054 & 1.02031550844309 \tabularnewline
327 & 835.9 & 791.90776153794 & 733.016666666667 & 1.08034073104924 & 1.05555222539633 \tabularnewline
328 & 782 & 736.084308629861 & 751.85 & 0.979030802194402 & 1.06237830481077 \tabularnewline
329 & 762.3 & 710.270563034197 & 767.458333333333 & 0.925484201792753 & 1.07325298227698 \tabularnewline
330 & 856.9 & 866.879800333338 & 778.4375 & 1.11361515899907 & 0.988487676919567 \tabularnewline
331 & 820.9 & 819.353351031104 & 782.95 & 1.04649511594751 & 1.0018876458697 \tabularnewline
332 & 769.6 & 747.694609452155 & 781.775 & 0.956406395001318 & 1.02929724284611 \tabularnewline
333 & 752.2 & 706.053100823126 & 777.15 & 0.908515860288394 & 1.06535896396896 \tabularnewline
334 & 724.4 & 665.016626805688 & 769.8 & 0.863882341914377 & 1.08929607291107 \tabularnewline
335 & 723.1 & 704.801713382722 & 760.429166666667 & 0.926847291342352 & 1.02596231857816 \tabularnewline
336 & 719.5 & 698.742120716897 & 751.125 & 0.930260769801161 & 1.02970749675403 \tabularnewline
337 & 817.4 & 837.605191207168 & 744.683333333333 & 1.12478036463889 & 0.97587742838837 \tabularnewline
338 & 803.3 & 847.374949911389 & 740.491666666667 & 1.14434096703054 & 0.947986484712584 \tabularnewline
339 & 752.5 & 796.06707335248 & 736.866666666667 & 1.08034073104924 & 0.94527210732507 \tabularnewline
340 & 689 & 717.715243203688 & 733.0875 & 0.979030802194402 & 0.959990757510581 \tabularnewline
341 & 630.4 & 676.351567038492 & 730.808333333333 & 0.925484201792753 & 0.932059642827918 \tabularnewline
342 & 765.5 & 812.405458805634 & 729.520833333333 & 1.11361515899907 & 0.942263486418971 \tabularnewline
343 & 757.7 & 761.264151303384 & 727.441666666666 & 1.04649511594751 & 0.995318114878677 \tabularnewline
344 & 732.2 & 694.733605328957 & 726.4 & 0.956406395001318 & 1.05392915267616 \tabularnewline
345 & 702.6 & 660.350967567868 & 726.845833333333 & 0.908515860288394 & 1.06397966309906 \tabularnewline
346 & 683.3 & 626.861823371136 & 725.633333333333 & 0.863882341914377 & 1.09003288208772 \tabularnewline
347 & 709.5 & 670.716904243607 & 723.654166666667 & 0.926847291342352 & 1.05782334620018 \tabularnewline
348 & 702.2 & 671.811071431153 & 722.175 & 0.930260769801161 & 1.04523433724322 \tabularnewline
349 & 784.8 & 808.360901726557 & 718.683333333333 & 1.12478036463889 & 0.970853486757915 \tabularnewline
350 & 810.9 & 816.692307732883 & 713.679166666666 & 1.14434096703054 & 0.992907600967907 \tabularnewline
351 & 755.6 & 765.822034302814 & 708.870833333333 & 1.08034073104924 & 0.986652206589851 \tabularnewline
352 & 656.8 & 689.107147304566 & 703.866666666667 & 0.979030802194402 & 0.953117381773016 \tabularnewline
353 & 615.1 & 646.169213507526 & 698.195833333333 & 0.925484201792753 & 0.951917836910123 \tabularnewline
354 & 745.3 & 768.747104509708 & 690.316666666667 & 1.11361515899907 & 0.96949958657124 \tabularnewline
355 & 694.1 & 713.286710633507 & 681.595833333333 & 1.04649511594751 & 0.973100984011792 \tabularnewline
356 & 675.7 & 642.633366961261 & 671.925 & 0.956406395001318 & 1.05145489596206 \tabularnewline
357 & 643.7 & 601.191443132089 & 661.729166666667 & 0.908515860288394 & 1.07070718878907 \tabularnewline
358 & 622.1 & 564.611901616689 & 653.575 & 0.863882341914377 & 1.1018187859992 \tabularnewline
359 & 634.6 & 599.685644953357 & 647.016666666667 & 0.926847291342352 & 1.05822109523625 \tabularnewline
360 & 588 & 594.835868816648 & 639.429166666667 & 0.930260769801161 & 0.988507974762438 \tabularnewline
361 & 689.7 & 711.578237934234 & 632.6375 & 1.12478036463889 & 0.969253924912392 \tabularnewline
362 & 673.9 & 717.616220424852 & 627.1 & 1.14434096703054 & 0.939081337377003 \tabularnewline
363 & 647.9 & 670.88259114215 & 620.991666666667 & 1.08034073104924 & 0.965742752240712 \tabularnewline
364 & 568.8 & 602.254877264895 & 615.154166666667 & 0.979030802194402 & 0.944450632900096 \tabularnewline
365 & 545.7 & 563.616022707613 & 608.995833333333 & 0.925484201792753 & 0.96821236092341 \tabularnewline
366 & 632.6 & 674.140856689574 & 605.3625 & 1.11361515899907 & 0.938379559290377 \tabularnewline
367 & 643.8 & NA & NA & 1.04649511594751 & NA \tabularnewline
368 & 593.1 & NA & NA & 0.956406395001318 & NA \tabularnewline
369 & 579.7 & NA & NA & 0.908515860288394 & NA \tabularnewline
370 & 546 & NA & NA & 0.863882341914377 & NA \tabularnewline
371 & 562.9 & NA & NA & 0.926847291342352 & NA \tabularnewline
372 & 572.5 & NA & NA & 0.930260769801161 & NA \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=147450&T=1
[TABLE]
[ROW][C]Classical Decomposition by Moving Averages[/C][/ROW]
[ROW][C]t[/C][C]Observations[/C][C]Fit[/C][C]Trend[/C][C]Seasonal[/C][C]Random[/C][/ROW]
[ROW][C]1[/C][C]235.1[/C][C]NA[/C][C]NA[/C][C]1.12478036463889[/C][C]NA[/C][/ROW]
[ROW][C]2[/C][C]280.7[/C][C]NA[/C][C]NA[/C][C]1.14434096703054[/C][C]NA[/C][/ROW]
[ROW][C]3[/C][C]264.6[/C][C]NA[/C][C]NA[/C][C]1.08034073104924[/C][C]NA[/C][/ROW]
[ROW][C]4[/C][C]240.7[/C][C]NA[/C][C]NA[/C][C]0.979030802194402[/C][C]NA[/C][/ROW]
[ROW][C]5[/C][C]201.4[/C][C]NA[/C][C]NA[/C][C]0.925484201792753[/C][C]NA[/C][/ROW]
[ROW][C]6[/C][C]240.8[/C][C]NA[/C][C]NA[/C][C]1.11361515899907[/C][C]NA[/C][/ROW]
[ROW][C]7[/C][C]241.1[/C][C]241.129916299572[/C][C]230.416666666667[/C][C]1.04649511594751[/C][C]0.999875932857974[/C][/ROW]
[ROW][C]8[/C][C]223.8[/C][C]225.596343447582[/C][C]235.879166666667[/C][C]0.956406395001318[/C][C]0.992037355658652[/C][/ROW]
[ROW][C]9[/C][C]206.1[/C][C]219.614781810963[/C][C]241.729166666667[/C][C]0.908515860288394[/C][C]0.938461420039584[/C][/ROW]
[ROW][C]10[/C][C]174.7[/C][C]214.609970790079[/C][C]248.425[/C][C]0.863882341914377[/C][C]0.814034871524599[/C][/ROW]
[ROW][C]11[/C][C]203.3[/C][C]239.412379081157[/C][C]258.308333333333[/C][C]0.926847291342352[/C][C]0.8491624400553[/C][/ROW]
[ROW][C]12[/C][C]220.5[/C][C]252.154933827686[/C][C]271.058333333333[/C][C]0.930260769801161[/C][C]0.874462365867018[/C][/ROW]
[ROW][C]13[/C][C]299.5[/C][C]321.448168459236[/C][C]285.7875[/C][C]1.12478036463889[/C][C]0.931720972110565[/C][/ROW]
[ROW][C]14[/C][C]347.4[/C][C]344.656426920148[/C][C]301.183333333333[/C][C]1.14434096703054[/C][C]1.00796031312797[/C][/ROW]
[ROW][C]15[/C][C]338.3[/C][C]340.199296207404[/C][C]314.9[/C][C]1.08034073104924[/C][C]0.994417107182238[/C][/ROW]
[ROW][C]16[/C][C]327.7[/C][C]323.04345106907[/C][C]329.9625[/C][C]0.979030802194402[/C][C]1.01441462105336[/C][/ROW]
[ROW][C]17[/C][C]351.6[/C][C]319.172507909102[/C][C]344.870833333333[/C][C]0.925484201792753[/C][C]1.10159863800091[/C][/ROW]
[ROW][C]18[/C][C]396.6[/C][C]398.075658773705[/C][C]357.4625[/C][C]1.11361515899907[/C][C]0.996293019326399[/C][/ROW]
[ROW][C]19[/C][C]438.8[/C][C]387.765684025401[/C][C]370.5375[/C][C]1.04649511594751[/C][C]1.1316112231614[/C][/ROW]
[ROW][C]20[/C][C]395.6[/C][C]366.219963725942[/C][C]382.9125[/C][C]0.956406395001318[/C][C]1.0802251083615[/C][/ROW]
[ROW][C]21[/C][C]363.5[/C][C]356.38805909463[/C][C]392.275[/C][C]0.908515860288394[/C][C]1.01995560940913[/C][/ROW]
[ROW][C]22[/C][C]378.8[/C][C]343.623599535475[/C][C]397.766666666667[/C][C]0.863882341914377[/C][C]1.10236898895209[/C][/ROW]
[ROW][C]23[/C][C]357[/C][C]369.190309187656[/C][C]398.329166666667[/C][C]0.926847291342352[/C][C]0.966980961080807[/C][/ROW]
[ROW][C]24[/C][C]369[/C][C]367.960771408308[/C][C]395.545833333333[/C][C]0.930260769801161[/C][C]1.0028242918062[/C][/ROW]
[ROW][C]25[/C][C]464.8[/C][C]437.933234972151[/C][C]389.35[/C][C]1.12478036463889[/C][C]1.06134899770637[/C][/ROW]
[ROW][C]26[/C][C]479.1[/C][C]434.034224532596[/C][C]379.2875[/C][C]1.14434096703054[/C][C]1.1038300044563[/C][/ROW]
[ROW][C]27[/C][C]431.3[/C][C]398.560202782626[/C][C]368.920833333333[/C][C]1.08034073104924[/C][C]1.08214517402589[/C][/ROW]
[ROW][C]28[/C][C]366.5[/C][C]349.444648368246[/C][C]356.929166666667[/C][C]0.979030802194402[/C][C]1.04880701911274[/C][/ROW]
[ROW][C]29[/C][C]326.3[/C][C]318.968130147872[/C][C]344.65[/C][C]0.925484201792753[/C][C]1.02298621448083[/C][/ROW]
[ROW][C]30[/C][C]355.1[/C][C]372.230506958602[/C][C]334.254166666667[/C][C]1.11361515899907[/C][C]0.95397876681691[/C][/ROW]
[ROW][C]31[/C][C]331.6[/C][C]335.493252983823[/C][C]320.5875[/C][C]1.04649511594751[/C][C]0.988395435827109[/C][/ROW]
[ROW][C]32[/C][C]261.3[/C][C]289.667601859378[/C][C]302.870833333333[/C][C]0.956406395001318[/C][C]0.902068434035127[/C][/ROW]
[ROW][C]33[/C][C]249[/C][C]259.101152388748[/C][C]285.191666666667[/C][C]0.908515860288394[/C][C]0.96101463735062[/C][/ROW]
[ROW][C]34[/C][C]205.5[/C][C]232.992667124066[/C][C]269.704166666667[/C][C]0.863882341914377[/C][C]0.882002006915411[/C][/ROW]
[ROW][C]35[/C][C]235.6[/C][C]237.539375179903[/C][C]256.2875[/C][C]0.926847291342352[/C][C]0.991835563352669[/C][/ROW]
[ROW][C]36[/C][C]240.9[/C][C]227.127043033494[/C][C]244.154166666667[/C][C]0.930260769801161[/C][C]1.06063988146262[/C][/ROW]
[ROW][C]37[/C][C]264.9[/C][C]262.139437148798[/C][C]233.058333333333[/C][C]1.12478036463889[/C][C]1.01053089485973[/C][/ROW]
[ROW][C]38[/C][C]253.8[/C][C]256.918851360444[/C][C]224.5125[/C][C]1.14434096703054[/C][C]0.987860558522938[/C][/ROW]
[ROW][C]39[/C][C]232.3[/C][C]236.198495165064[/C][C]218.633333333333[/C][C]1.08034073104924[/C][C]0.983494834874625[/C][/ROW]
[ROW][C]40[/C][C]193.8[/C][C]210.357005736495[/C][C]214.8625[/C][C]0.979030802194402[/C][C]0.921290923121263[/C][/ROW]
[ROW][C]41[/C][C]177[/C][C]196.279774463546[/C][C]212.083333333333[/C][C]0.925484201792753[/C][C]0.901774013566909[/C][/ROW]
[ROW][C]42[/C][C]213.2[/C][C]231.636593134969[/C][C]208.004166666667[/C][C]1.11361515899907[/C][C]0.920407251352439[/C][/ROW]
[ROW][C]43[/C][C]207.2[/C][C]213.297506611685[/C][C]203.820833333333[/C][C]1.04649511594751[/C][C]0.971413136944045[/C][/ROW]
[ROW][C]44[/C][C]180.6[/C][C]192.588367740099[/C][C]201.366666666667[/C][C]0.956406395001318[/C][C]0.937751340432579[/C][/ROW]
[ROW][C]45[/C][C]188.6[/C][C]180.980144852199[/C][C]199.204166666667[/C][C]0.908515860288394[/C][C]1.04210326582523[/C][/ROW]
[ROW][C]46[/C][C]175.4[/C][C]170.566369391478[/C][C]197.441666666667[/C][C]0.863882341914377[/C][C]1.02833870842046[/C][/ROW]
[ROW][C]47[/C][C]199[/C][C]182.650706213866[/C][C]197.066666666667[/C][C]0.926847291342352[/C][C]1.08951125415847[/C][/ROW]
[ROW][C]48[/C][C]179.6[/C][C]182.982293419888[/C][C]196.7[/C][C]0.930260769801161[/C][C]0.981515733808588[/C][/ROW]
[ROW][C]49[/C][C]225.8[/C][C]220.836564842287[/C][C]196.3375[/C][C]1.12478036463889[/C][C]1.02247560389855[/C][/ROW]
[ROW][C]50[/C][C]234[/C][C]225.273055534687[/C][C]196.858333333333[/C][C]1.14434096703054[/C][C]1.03873940647096[/C][/ROW]
[ROW][C]51[/C][C]200.2[/C][C]212.467010439683[/C][C]196.666666666667[/C][C]1.08034073104924[/C][C]0.942263928812773[/C][/ROW]
[ROW][C]52[/C][C]183.6[/C][C]190.780468987616[/C][C]194.866666666667[/C][C]0.979030802194402[/C][C]0.962362662039153[/C][/ROW]
[ROW][C]53[/C][C]178.2[/C][C]177.762378059343[/C][C]192.075[/C][C]0.925484201792753[/C][C]1.0024618366689[/C][/ROW]
[ROW][C]54[/C][C]203.2[/C][C]210.895510798611[/C][C]189.379166666667[/C][C]1.11361515899907[/C][C]0.963510314802481[/C][/ROW]
[ROW][C]55[/C][C]208.5[/C][C]196.540503567575[/C][C]187.808333333333[/C][C]1.04649511594751[/C][C]1.0608500345493[/C][/ROW]
[ROW][C]56[/C][C]191.8[/C][C]177.620607658328[/C][C]185.716666666667[/C][C]0.956406395001318[/C][C]1.07982965787927[/C][/ROW]
[ROW][C]57[/C][C]172.8[/C][C]166.644521673399[/C][C]183.425[/C][C]0.908515860288394[/C][C]1.03693777788066[/C][/ROW]
[ROW][C]58[/C][C]148[/C][C]157.572139165182[/C][C]182.4[/C][C]0.863882341914377[/C][C]0.939252337272975[/C][/ROW]
[ROW][C]59[/C][C]159.4[/C][C]167.828873279816[/C][C]181.075[/C][C]0.926847291342352[/C][C]0.949776977494432[/C][/ROW]
[ROW][C]60[/C][C]154.5[/C][C]166.330625640448[/C][C]178.8[/C][C]0.930260769801161[/C][C]0.928872836286797[/C][/ROW]
[ROW][C]61[/C][C]213.2[/C][C]197.947284421886[/C][C]175.9875[/C][C]1.12478036463889[/C][C]1.07705443205579[/C][/ROW]
[ROW][C]62[/C][C]196.4[/C][C]197.665829705075[/C][C]172.733333333333[/C][C]1.14434096703054[/C][C]0.993596112656579[/C][/ROW]
[ROW][C]63[/C][C]182.8[/C][C]184.292624457861[/C][C]170.5875[/C][C]1.08034073104924[/C][C]0.991900791134467[/C][/ROW]
[ROW][C]64[/C][C]176.4[/C][C]166.941068954182[/C][C]170.516666666667[/C][C]0.979030802194402[/C][C]1.05666029997935[/C][/ROW]
[ROW][C]65[/C][C]153.6[/C][C]159.796415992041[/C][C]172.6625[/C][C]0.925484201792753[/C][C]0.961223060269701[/C][/ROW]
[ROW][C]66[/C][C]173.2[/C][C]198.042535838497[/C][C]177.8375[/C][C]1.11361515899907[/C][C]0.874559595324733[/C][/ROW]
[ROW][C]67[/C][C]171[/C][C]195.90824610169[/C][C]187.204166666667[/C][C]1.04649511594751[/C][C]0.872857592279391[/C][/ROW]
[ROW][C]68[/C][C]151.2[/C][C]192.783634045745[/C][C]201.570833333333[/C][C]0.956406395001318[/C][C]0.784298940874423[/C][/ROW]
[ROW][C]69[/C][C]161.9[/C][C]199.135320126962[/C][C]219.1875[/C][C]0.908515860288394[/C][C]0.813014988485105[/C][/ROW]
[ROW][C]70[/C][C]157.2[/C][C]204.797707189835[/C][C]237.066666666667[/C][C]0.863882341914377[/C][C]0.767586718411281[/C][/ROW]
[ROW][C]71[/C][C]201.7[/C][C]235.959872920907[/C][C]254.583333333333[/C][C]0.926847291342352[/C][C]0.854806359671202[/C][/ROW]
[ROW][C]72[/C][C]236.4[/C][C]252.608435952964[/C][C]271.545833333333[/C][C]0.930260769801161[/C][C]0.935835729745849[/C][/ROW]
[ROW][C]73[/C][C]356.1[/C][C]323.791460885567[/C][C]287.870833333333[/C][C]1.12478036463889[/C][C]1.09978193688638[/C][/ROW]
[ROW][C]74[/C][C]398.3[/C][C]348.146666869591[/C][C]304.233333333333[/C][C]1.14434096703054[/C][C]1.14405805915469[/C][/ROW]
[ROW][C]75[/C][C]403.7[/C][C]345.655016899203[/C][C]319.95[/C][C]1.08034073104924[/C][C]1.16792750072458[/C][/ROW]
[ROW][C]76[/C][C]384.6[/C][C]326.176349636092[/C][C]333.1625[/C][C]0.979030802194402[/C][C]1.17911675824777[/C][/ROW]
[ROW][C]77[/C][C]365.8[/C][C]317.803562527283[/C][C]343.391666666667[/C][C]0.925484201792753[/C][C]1.15102548596697[/C][/ROW]
[ROW][C]78[/C][C]368.1[/C][C]390.493795566186[/C][C]350.654166666667[/C][C]1.11361515899907[/C][C]0.942652621320866[/C][/ROW]
[ROW][C]79[/C][C]367.9[/C][C]370.241251229597[/C][C]353.791666666667[/C][C]1.04649511594751[/C][C]0.993676417142009[/C][/ROW]
[ROW][C]80[/C][C]347[/C][C]337.149194344548[/C][C]352.516666666667[/C][C]0.956406395001318[/C][C]1.02921794214755[/C][/ROW]
[ROW][C]81[/C][C]343.3[/C][C]315.898535587777[/C][C]347.708333333333[/C][C]0.908515860288394[/C][C]1.08674134674679[/C][/ROW]
[ROW][C]82[/C][C]292.9[/C][C]295.253387407786[/C][C]341.775[/C][C]0.863882341914377[/C][C]0.992029261955474[/C][/ROW]
[ROW][C]83[/C][C]311.5[/C][C]310.393434143126[/C][C]334.891666666667[/C][C]0.926847291342352[/C][C]1.00356504273336[/C][/ROW]
[ROW][C]84[/C][C]300.9[/C][C]304.730171667615[/C][C]327.575[/C][C]0.930260769801161[/C][C]0.987430940472173[/C][/ROW]
[ROW][C]85[/C][C]366.9[/C][C]359.990642287528[/C][C]320.054166666667[/C][C]1.12478036463889[/C][C]1.0191931592126[/C][/ROW]
[ROW][C]86[/C][C]356.9[/C][C]356.948556141001[/C][C]311.925[/C][C]1.14434096703054[/C][C]0.9998639687984[/C][/ROW]
[ROW][C]87[/C][C]329.7[/C][C]327.851901935454[/C][C]303.470833333333[/C][C]1.08034073104924[/C][C]1.00563699052418[/C][/ROW]
[ROW][C]88[/C][C]316.2[/C][C]290.009320085027[/C][C]296.220833333333[/C][C]0.979030802194402[/C][C]1.09030978696579[/C][/ROW]
[ROW][C]89[/C][C]269[/C][C]269.427732062741[/C][C]291.120833333333[/C][C]0.925484201792753[/C][C]0.998412442329279[/C][/ROW]
[ROW][C]90[/C][C]289.3[/C][C]319.616830759058[/C][C]287.008333333333[/C][C]1.11361515899907[/C][C]0.905146325720524[/C][/ROW]
[ROW][C]91[/C][C]266.2[/C][C]295.94445839364[/C][C]282.795833333333[/C][C]1.04649511594751[/C][C]0.899493105716221[/C][/ROW]
[ROW][C]92[/C][C]253.6[/C][C]266.163914702221[/C][C]278.295833333333[/C][C]0.956406395001318[/C][C]0.952796325842001[/C][/ROW]
[ROW][C]93[/C][C]233.8[/C][C]250.265835647443[/C][C]275.466666666667[/C][C]0.908515860288394[/C][C]0.934206618315099[/C][/ROW]
[ROW][C]94[/C][C]228.4[/C][C]235.61670973763[/C][C]272.741666666667[/C][C]0.863882341914377[/C][C]0.969370976508132[/C][/ROW]
[ROW][C]95[/C][C]253.6[/C][C]251.465255732322[/C][C]271.3125[/C][C]0.926847291342352[/C][C]1.00848922154857[/C][/ROW]
[ROW][C]96[/C][C]260.1[/C][C]253.922429290309[/C][C]272.958333333333[/C][C]0.930260769801161[/C][C]1.02432857438769[/C][/ROW]
[ROW][C]97[/C][C]306.6[/C][C]309.867617288308[/C][C]275.491666666667[/C][C]1.12478036463889[/C][C]0.989454795835385[/C][/ROW]
[ROW][C]98[/C][C]309.2[/C][C]316.462726344933[/C][C]276.545833333333[/C][C]1.14434096703054[/C][C]0.977050294583454[/C][/ROW]
[ROW][C]99[/C][C]309.5[/C][C]298.160537510451[/C][C]275.9875[/C][C]1.08034073104924[/C][C]1.03803139940728[/C][/ROW]
[ROW][C]100[/C][C]271[/C][C]269.147805408268[/C][C]274.9125[/C][C]0.979030802194402[/C][C]1.00688170051739[/C][/ROW]
[ROW][C]101[/C][C]279.9[/C][C]253.925871682712[/C][C]274.370833333333[/C][C]0.925484201792753[/C][C]1.10229020046348[/C][/ROW]
[ROW][C]102[/C][C]317.9[/C][C]306.109606893032[/C][C]274.879166666667[/C][C]1.11361515899907[/C][C]1.03851689996482[/C][/ROW]
[ROW][C]103[/C][C]298.4[/C][C]288.527424259362[/C][C]275.708333333333[/C][C]1.04649511594751[/C][C]1.03421711390514[/C][/ROW]
[ROW][C]104[/C][C]246.7[/C][C]264.219221699052[/C][C]276.2625[/C][C]0.956406395001318[/C][C]0.933694370960618[/C][/ROW]
[ROW][C]105[/C][C]227.3[/C][C]249.928927682586[/C][C]275.095833333333[/C][C]0.908515860288394[/C][C]0.909458549306765[/C][/ROW]
[ROW][C]106[/C][C]209.1[/C][C]236.369007277048[/C][C]273.6125[/C][C]0.863882341914377[/C][C]0.884633744537051[/C][/ROW]
[ROW][C]107[/C][C]259.9[/C][C]252.643124165069[/C][C]272.583333333333[/C][C]0.926847291342352[/C][C]1.02872382083982[/C][/ROW]
[ROW][C]108[/C][C]266[/C][C]252.751851154975[/C][C]271.7[/C][C]0.930260769801161[/C][C]1.05241563527423[/C][/ROW]
[ROW][C]109[/C][C]320.6[/C][C]304.717060535232[/C][C]270.9125[/C][C]1.12478036463889[/C][C]1.05212356484691[/C][/ROW]
[ROW][C]110[/C][C]308.5[/C][C]309.625289066926[/C][C]270.570833333333[/C][C]1.14434096703054[/C][C]0.99636564225643[/C][/ROW]
[ROW][C]111[/C][C]282.2[/C][C]293.609602180906[/C][C]271.775[/C][C]1.08034073104924[/C][C]0.961140228057406[/C][/ROW]
[ROW][C]112[/C][C]262.7[/C][C]268.539990451906[/C][C]274.291666666667[/C][C]0.979030802194402[/C][C]0.978252809043158[/C][/ROW]
[ROW][C]113[/C][C]263.5[/C][C]257.330882308475[/C][C]278.05[/C][C]0.925484201792753[/C][C]1.02397348361838[/C][/ROW]
[ROW][C]114[/C][C]313.1[/C][C]315.208770754687[/C][C]283.05[/C][C]1.11361515899907[/C][C]0.99330992361146[/C][/ROW]
[ROW][C]115[/C][C]284.3[/C][C]304.626007459688[/C][C]291.091666666667[/C][C]1.04649511594751[/C][C]0.933275534714883[/C][/ROW]
[ROW][C]116[/C][C]252.6[/C][C]291.508684169756[/C][C]304.795833333333[/C][C]0.956406395001318[/C][C]0.866526500640722[/C][/ROW]
[ROW][C]117[/C][C]250.3[/C][C]293.431695459395[/C][C]322.979166666667[/C][C]0.908515860288394[/C][C]0.853009418795511[/C][/ROW]
[ROW][C]118[/C][C]246.5[/C][C]296.185660435102[/C][C]342.854166666667[/C][C]0.863882341914377[/C][C]0.832248258196859[/C][/ROW]
[ROW][C]119[/C][C]312.7[/C][C]335.669332150774[/C][C]362.1625[/C][C]0.926847291342352[/C][C]0.931571549883332[/C][/ROW]
[ROW][C]120[/C][C]333.2[/C][C]353.530101216768[/C][C]380.033333333333[/C][C]0.930260769801161[/C][C]0.942494002217077[/C][/ROW]
[ROW][C]121[/C][C]446.4[/C][C]447.82661559612[/C][C]398.145833333333[/C][C]1.12478036463889[/C][C]0.996814357283742[/C][/ROW]
[ROW][C]122[/C][C]511.6[/C][C]476.289014740199[/C][C]416.2125[/C][C]1.14434096703054[/C][C]1.07413772765484[/C][/ROW]
[ROW][C]123[/C][C]515.5[/C][C]466.018478597226[/C][C]431.3625[/C][C]1.08034073104924[/C][C]1.10617931192711[/C][/ROW]
[ROW][C]124[/C][C]506.4[/C][C]433.90237223755[/C][C]443.195833333333[/C][C]0.979030802194402[/C][C]1.16708281033034[/C][/ROW]
[ROW][C]125[/C][C]483.2[/C][C]417.663307900721[/C][C]451.291666666667[/C][C]0.925484201792753[/C][C]1.15691273535299[/C][/ROW]
[ROW][C]126[/C][C]522.3[/C][C]509.033489178475[/C][C]457.1[/C][C]1.11361515899907[/C][C]1.02606215721275[/C][/ROW]
[ROW][C]127[/C][C]509.8[/C][C]482.495294000233[/C][C]461.058333333333[/C][C]1.04649511594751[/C][C]1.05659061619729[/C][/ROW]
[ROW][C]128[/C][C]460.7[/C][C]440.146193032898[/C][C]460.208333333333[/C][C]0.956406395001318[/C][C]1.04669768202577[/C][/ROW]
[ROW][C]129[/C][C]405.8[/C][C]413.280079362439[/C][C]454.895833333333[/C][C]0.908515860288394[/C][C]0.981900701882417[/C][/ROW]
[ROW][C]130[/C][C]375[/C][C]384.470836268994[/C][C]445.05[/C][C]0.863882341914377[/C][C]0.97536656782371[/C][/ROW]
[ROW][C]131[/C][C]378.5[/C][C]400.865315369281[/C][C]432.504166666667[/C][C]0.926847291342352[/C][C]0.944207407047232[/C][/ROW]
[ROW][C]132[/C][C]406.8[/C][C]390.914955903152[/C][C]420.220833333333[/C][C]0.930260769801161[/C][C]1.04063554964314[/C][/ROW]
[ROW][C]133[/C][C]467.8[/C][C]458.896329018108[/C][C]407.9875[/C][C]1.12478036463889[/C][C]1.01940235826454[/C][/ROW]
[ROW][C]134[/C][C]469.8[/C][C]453.750265777059[/C][C]396.516666666667[/C][C]1.14434096703054[/C][C]1.03537129437369[/C][/ROW]
[ROW][C]135[/C][C]429.8[/C][C]418.8165914898[/C][C]387.670833333333[/C][C]1.08034073104924[/C][C]1.02622486485344[/C][/ROW]
[ROW][C]136[/C][C]355.8[/C][C]373.904101243069[/C][C]381.9125[/C][C]0.979030802194402[/C][C]0.95158089685863[/C][/ROW]
[ROW][C]137[/C][C]332.7[/C][C]350.878054188852[/C][C]379.129166666667[/C][C]0.925484201792753[/C][C]0.948192672719657[/C][/ROW]
[ROW][C]138[/C][C]378[/C][C]418.974503257588[/C][C]376.229166666667[/C][C]1.11361515899907[/C][C]0.902202871680723[/C][/ROW]
[ROW][C]139[/C][C]360.5[/C][C]388.899387067677[/C][C]371.620833333333[/C][C]1.04649511594751[/C][C]0.926974976016778[/C][/ROW]
[ROW][C]140[/C][C]334.7[/C][C]349.957069984274[/C][C]365.908333333333[/C][C]0.956406395001318[/C][C]0.95640302399103[/C][/ROW]
[ROW][C]141[/C][C]319.5[/C][C]328.852457562389[/C][C]361.966666666667[/C][C]0.908515860288394[/C][C]0.971560323338575[/C][/ROW]
[ROW][C]142[/C][C]323.1[/C][C]312.390653365513[/C][C]361.6125[/C][C]0.863882341914377[/C][C]1.03428190478528[/C][/ROW]
[ROW][C]143[/C][C]363.6[/C][C]335.553476239357[/C][C]362.0375[/C][C]0.926847291342352[/C][C]1.08358287351086[/C][/ROW]
[ROW][C]144[/C][C]352.1[/C][C]338.514141957561[/C][C]363.891666666667[/C][C]0.930260769801161[/C][C]1.04013379755385[/C][/ROW]
[ROW][C]145[/C][C]411.9[/C][C]412.442899958522[/C][C]366.6875[/C][C]1.12478036463889[/C][C]0.99868369668001[/C][/ROW]
[ROW][C]146[/C][C]388.6[/C][C]422.68140852218[/C][C]369.366666666667[/C][C]1.14434096703054[/C][C]0.91936856498766[/C][/ROW]
[ROW][C]147[/C][C]416.4[/C][C]401.220541832835[/C][C]371.383333333333[/C][C]1.08034073104924[/C][C]1.03783320290089[/C][/ROW]
[ROW][C]148[/C][C]360.7[/C][C]365.333502428859[/C][C]373.158333333333[/C][C]0.979030802194402[/C][C]0.987317061265791[/C][/ROW]
[ROW][C]149[/C][C]338[/C][C]347.808531586239[/C][C]375.8125[/C][C]0.925484201792753[/C][C]0.971799048339886[/C][/ROW]
[ROW][C]150[/C][C]417.2[/C][C]424.449777789333[/C][C]381.145833333333[/C][C]1.11361515899907[/C][C]0.982919586323988[/C][/ROW]
[ROW][C]151[/C][C]388.4[/C][C]408.307511072187[/C][C]390.166666666667[/C][C]1.04649511594751[/C][C]0.951243828407879[/C][/ROW]
[ROW][C]152[/C][C]371.1[/C][C]385.049214627531[/C][C]402.6[/C][C]0.956406395001318[/C][C]0.963772904611624[/C][/ROW]
[ROW][C]153[/C][C]331.5[/C][C]377.227141639995[/C][C]415.2125[/C][C]0.908515860288394[/C][C]0.87878088135123[/C][/ROW]
[ROW][C]154[/C][C]353.7[/C][C]367.833902167626[/C][C]425.791666666667[/C][C]0.863882341914377[/C][C]0.961575314063398[/C][/ROW]
[ROW][C]155[/C][C]396.7[/C][C]404.57270453465[/C][C]436.504166666667[/C][C]0.926847291342352[/C][C]0.980540692818845[/C][/ROW]
[ROW][C]156[/C][C]447[/C][C]415.489344572066[/C][C]446.6375[/C][C]0.930260769801161[/C][C]1.07583986410142[/C][/ROW]
[ROW][C]157[/C][C]533.5[/C][C]512.763935314605[/C][C]455.879166666667[/C][C]1.12478036463889[/C][C]1.04043978770206[/C][/ROW]
[ROW][C]158[/C][C]565.4[/C][C]530.292372209315[/C][C]463.404166666667[/C][C]1.14434096703054[/C][C]1.06620428584409[/C][/ROW]
[ROW][C]159[/C][C]542.3[/C][C]507.152451931925[/C][C]469.4375[/C][C]1.08034073104924[/C][C]1.06930371318168[/C][/ROW]
[ROW][C]160[/C][C]488.7[/C][C]463.856635489689[/C][C]473.791666666667[/C][C]0.979030802194402[/C][C]1.05355828204135[/C][/ROW]
[ROW][C]161[/C][C]467.1[/C][C]439.643557693299[/C][C]475.041666666667[/C][C]0.925484201792753[/C][C]1.06245159704093[/C][/ROW]
[ROW][C]162[/C][C]531.3[/C][C]526.902372417247[/C][C]473.145833333333[/C][C]1.11361515899907[/C][C]1.00834619051453[/C][/ROW]
[ROW][C]163[/C][C]496.1[/C][C]490.160870724548[/C][C]468.383333333333[/C][C]1.04649511594751[/C][C]1.01211669398798[/C][/ROW]
[ROW][C]164[/C][C]444[/C][C]440.445070031336[/C][C]460.520833333333[/C][C]0.956406395001318[/C][C]1.00807122206728[/C][/ROW]
[ROW][C]165[/C][C]403.4[/C][C]409.78607878308[/C][C]451.05[/C][C]0.908515860288394[/C][C]0.984416067031743[/C][/ROW]
[ROW][C]166[/C][C]386.3[/C][C]382.008771594538[/C][C]442.2[/C][C]0.863882341914377[/C][C]1.01123332427041[/C][/ROW]
[ROW][C]167[/C][C]394.1[/C][C]403.120643778214[/C][C]434.9375[/C][C]0.926847291342352[/C][C]0.977622967423179[/C][/ROW]
[ROW][C]168[/C][C]404.1[/C][C]397.577948666853[/C][C]427.383333333333[/C][C]0.930260769801161[/C][C]1.01640445944001[/C][/ROW]
[ROW][C]169[/C][C]462.1[/C][C]470.280043625224[/C][C]418.108333333333[/C][C]1.12478036463889[/C][C]0.982606015849266[/C][/ROW]
[ROW][C]170[/C][C]448.1[/C][C]470.209703352849[/C][C]410.9[/C][C]1.14434096703054[/C][C]0.952979057651948[/C][/ROW]
[ROW][C]171[/C][C]432.3[/C][C]438.613835386277[/C][C]405.995833333333[/C][C]1.08034073104924[/C][C]0.985605024563996[/C][/ROW]
[ROW][C]172[/C][C]386.3[/C][C]392.554638024873[/C][C]400.9625[/C][C]0.979030802194402[/C][C]0.984066834475978[/C][/ROW]
[ROW][C]173[/C][C]395.2[/C][C]367.829839818356[/C][C]397.445833333333[/C][C]0.925484201792753[/C][C]1.07440984177673[/C][/ROW]
[ROW][C]174[/C][C]421.9[/C][C]440.30023355242[/C][C]395.379166666667[/C][C]1.11361515899907[/C][C]0.958209802879359[/C][/ROW]
[ROW][C]175[/C][C]382.9[/C][C]412.56325787704[/C][C]394.233333333333[/C][C]1.04649511594751[/C][C]0.92810009783789[/C][/ROW]
[ROW][C]176[/C][C]384.2[/C][C]378.621366647793[/C][C]395.879166666667[/C][C]0.956406395001318[/C][C]1.01473406902943[/C][/ROW]
[ROW][C]177[/C][C]345.5[/C][C]361.58552691203[/C][C]397.995833333333[/C][C]0.908515860288394[/C][C]0.955513908287753[/C][/ROW]
[ROW][C]178[/C][C]323.4[/C][C]344.7178505019[/C][C]399.033333333333[/C][C]0.863882341914377[/C][C]0.93815855352178[/C][/ROW]
[ROW][C]179[/C][C]372.6[/C][C]370.333420846978[/C][C]399.5625[/C][C]0.926847291342352[/C][C]1.00612037430443[/C][/ROW]
[ROW][C]180[/C][C]376[/C][C]372.984179565235[/C][C]400.945833333333[/C][C]0.930260769801161[/C][C]1.00808565242172[/C][/ROW]
[ROW][C]181[/C][C]462.7[/C][C]454.003534431929[/C][C]403.6375[/C][C]1.12478036463889[/C][C]1.01915506137844[/C][/ROW]
[ROW][C]182[/C][C]487[/C][C]462.966978649018[/C][C]404.570833333333[/C][C]1.14434096703054[/C][C]1.05191087584932[/C][/ROW]
[ROW][C]183[/C][C]444.2[/C][C]436.750247625217[/C][C]404.270833333333[/C][C]1.08034073104924[/C][C]1.01705723675096[/C][/ROW]
[ROW][C]184[/C][C]399.3[/C][C]396.507474888733[/C][C]405[/C][C]0.979030802194402[/C][C]1.00704280571773[/C][/ROW]
[ROW][C]185[/C][C]394.9[/C][C]375.947107504913[/C][C]406.216666666667[/C][C]0.925484201792753[/C][C]1.0504137207515[/C][/ROW]
[ROW][C]186[/C][C]455.4[/C][C]453.111447944071[/C][C]406.883333333333[/C][C]1.11361515899907[/C][C]1.00505074869839[/C][/ROW]
[ROW][C]187[/C][C]414[/C][C]425.448228992144[/C][C]406.545833333333[/C][C]1.04649511594751[/C][C]0.973091369967001[/C][/ROW]
[ROW][C]188[/C][C]375.5[/C][C]386.758791058595[/C][C]404.3875[/C][C]0.956406395001318[/C][C]0.9708893726041[/C][/ROW]
[ROW][C]189[/C][C]347[/C][C]365.022745250121[/C][C]401.779166666667[/C][C]0.908515860288394[/C][C]0.950625692550279[/C][/ROW]
[ROW][C]190[/C][C]339.4[/C][C]345.725713234134[/C][C]400.2[/C][C]0.863882341914377[/C][C]0.981703087181572[/C][/ROW]
[ROW][C]191[/C][C]385.8[/C][C]368.67281944999[/C][C]397.770833333333[/C][C]0.926847291342352[/C][C]1.04645631477678[/C][/ROW]
[ROW][C]192[/C][C]378.8[/C][C]368.007284446798[/C][C]395.595833333333[/C][C]0.930260769801161[/C][C]1.0293274508667[/C][/ROW]
[ROW][C]193[/C][C]451.8[/C][C]442.305818639684[/C][C]393.2375[/C][C]1.12478036463889[/C][C]1.02146519661332[/C][/ROW]
[ROW][C]194[/C][C]446.1[/C][C]446.807930577074[/C][C]390.45[/C][C]1.14434096703054[/C][C]0.998415581889605[/C][/ROW]
[ROW][C]195[/C][C]422.5[/C][C]419.964453516539[/C][C]388.733333333333[/C][C]1.08034073104924[/C][C]1.00603752641975[/C][/ROW]
[ROW][C]196[/C][C]383.1[/C][C]378.933871989343[/C][C]387.05[/C][C]0.979030802194402[/C][C]1.01099434048686[/C][/ROW]
[ROW][C]197[/C][C]352.8[/C][C]355.370508751721[/C][C]383.983333333333[/C][C]0.925484201792753[/C][C]0.9927666795966[/C][/ROW]
[ROW][C]198[/C][C]445.3[/C][C]423.345442756659[/C][C]380.154166666667[/C][C]1.11361515899907[/C][C]1.05185967539979[/C][/ROW]
[ROW][C]199[/C][C]367.5[/C][C]393.660939845238[/C][C]376.170833333333[/C][C]1.04649511594751[/C][C]0.933544486644971[/C][/ROW]
[ROW][C]200[/C][C]355.1[/C][C]356.321157537679[/C][C]372.5625[/C][C]0.956406395001318[/C][C]0.996572873903651[/C][/ROW]
[ROW][C]201[/C][C]326.2[/C][C]335.393771756466[/C][C]369.166666666667[/C][C]0.908515860288394[/C][C]0.972588126164903[/C][/ROW]
[ROW][C]202[/C][C]319.8[/C][C]315.802988616075[/C][C]365.5625[/C][C]0.863882341914377[/C][C]1.01265666104504[/C][/ROW]
[ROW][C]203[/C][C]331.8[/C][C]336.298815936144[/C][C]362.841666666667[/C][C]0.926847291342352[/C][C]0.98662256385405[/C][/ROW]
[ROW][C]204[/C][C]340.9[/C][C]334.785346705274[/C][C]359.883333333333[/C][C]0.930260769801161[/C][C]1.01826439942758[/C][/ROW]
[ROW][C]205[/C][C]394.1[/C][C]401.780919418716[/C][C]357.208333333333[/C][C]1.12478036463889[/C][C]0.980882816859924[/C][/ROW]
[ROW][C]206[/C][C]417.2[/C][C]405.754698384854[/C][C]354.575[/C][C]1.14434096703054[/C][C]1.0282074407535[/C][/ROW]
[ROW][C]207[/C][C]369.9[/C][C]379.433670423342[/C][C]351.216666666667[/C][C]1.08034073104924[/C][C]0.974873947236403[/C][/ROW]
[ROW][C]208[/C][C]349.2[/C][C]340.143855747399[/C][C]347.429166666667[/C][C]0.979030802194402[/C][C]1.02662445344692[/C][/ROW]
[ROW][C]209[/C][C]321.4[/C][C]317.99637173599[/C][C]343.6[/C][C]0.925484201792753[/C][C]1.01070335565601[/C][/ROW]
[ROW][C]210[/C][C]405.7[/C][C]377.761462248297[/C][C]339.220833333333[/C][C]1.11361515899907[/C][C]1.0739581469889[/C][/ROW]
[ROW][C]211[/C][C]342.9[/C][C]349.2459429659[/C][C]333.729166666667[/C][C]1.04649511594751[/C][C]0.981829587161391[/C][/ROW]
[ROW][C]212[/C][C]316.5[/C][C]312.167062301784[/C][C]326.395833333333[/C][C]0.956406395001318[/C][C]1.01388018859602[/C][/ROW]
[ROW][C]213[/C][C]284.2[/C][C]289.827915880251[/C][C]319.0125[/C][C]0.908515860288394[/C][C]0.980581870924481[/C][/ROW]
[ROW][C]214[/C][C]270.9[/C][C]270.294386745977[/C][C]312.883333333333[/C][C]0.863882341914377[/C][C]1.00224056911175[/C][/ROW]
[ROW][C]215[/C][C]288.8[/C][C]285.426485232591[/C][C]307.954166666667[/C][C]0.926847291342352[/C][C]1.01181920719326[/C][/ROW]
[ROW][C]216[/C][C]278.8[/C][C]283.043467471626[/C][C]304.2625[/C][C]0.930260769801161[/C][C]0.985007718038746[/C][/ROW]
[ROW][C]217[/C][C]324.4[/C][C]338.273008080293[/C][C]300.745833333333[/C][C]1.12478036463889[/C][C]0.958988722869075[/C][/ROW]
[ROW][C]218[/C][C]310.9[/C][C]340.708450583893[/C][C]297.733333333333[/C][C]1.14434096703054[/C][C]0.91251038671683[/C][/ROW]
[ROW][C]219[/C][C]299[/C][C]318.578977327281[/C][C]294.8875[/C][C]1.08034073104924[/C][C]0.938542783043818[/C][/ROW]
[ROW][C]220[/C][C]273[/C][C]286.325716691771[/C][C]292.458333333333[/C][C]0.979030802194402[/C][C]0.953459588451441[/C][/ROW]
[ROW][C]221[/C][C]279.3[/C][C]268.533097334342[/C][C]290.154166666667[/C][C]0.925484201792753[/C][C]1.04009525370444[/C][/ROW]
[ROW][C]222[/C][C]359.2[/C][C]321.115571160544[/C][C]288.354166666667[/C][C]1.11361515899907[/C][C]1.11860038023636[/C][/ROW]
[ROW][C]223[/C][C]305[/C][C]300.85426464596[/C][C]287.4875[/C][C]1.04649511594751[/C][C]1.0137798789687[/C][/ROW]
[ROW][C]224[/C][C]282.1[/C][C]274.915033216483[/C][C]287.445833333333[/C][C]0.956406395001318[/C][C]1.02613522694432[/C][/ROW]
[ROW][C]225[/C][C]250.3[/C][C]261.296732384445[/C][C]287.608333333333[/C][C]0.908515860288394[/C][C]0.957914772664416[/C][/ROW]
[ROW][C]226[/C][C]246.5[/C][C]248.092610558777[/C][C]287.183333333333[/C][C]0.863882341914377[/C][C]0.993580580432484[/C][/ROW]
[ROW][C]227[/C][C]257.9[/C][C]264.626487269383[/C][C]285.5125[/C][C]0.926847291342352[/C][C]0.974581201833603[/C][/ROW]
[ROW][C]228[/C][C]266.5[/C][C]264.442128162143[/C][C]284.266666666667[/C][C]0.930260769801161[/C][C]1.00778193645679[/C][/ROW]
[ROW][C]229[/C][C]315.9[/C][C]320.83891242839[/C][C]285.245833333333[/C][C]1.12478036463889[/C][C]0.984606254923981[/C][/ROW]
[ROW][C]230[/C][C]318.4[/C][C]327.94428071414[/C][C]286.579166666667[/C][C]1.14434096703054[/C][C]0.970896639229824[/C][/ROW]
[ROW][C]231[/C][C]295.4[/C][C]311.912374732765[/C][C]288.716666666667[/C][C]1.08034073104924[/C][C]0.947060854039817[/C][/ROW]
[ROW][C]232[/C][C]266.4[/C][C]286.248210086597[/C][C]292.379166666667[/C][C]0.979030802194402[/C][C]0.930660841230788[/C][/ROW]
[ROW][C]233[/C][C]245.8[/C][C]273.719665048555[/C][C]295.758333333333[/C][C]0.925484201792753[/C][C]0.897999052996056[/C][/ROW]
[ROW][C]234[/C][C]362.8[/C][C]331.119547338886[/C][C]297.3375[/C][C]1.11361515899907[/C][C]1.09567678174158[/C][/ROW]
[ROW][C]235[/C][C]324.9[/C][C]311.03142964855[/C][C]297.2125[/C][C]1.04649511594751[/C][C]1.04458896763945[/C][/ROW]
[ROW][C]236[/C][C]294.2[/C][C]284.327666153954[/C][C]297.2875[/C][C]0.956406395001318[/C][C]1.03472167861674[/C][/ROW]
[ROW][C]237[/C][C]289.5[/C][C]270.38567647008[/C][C]297.6125[/C][C]0.908515860288394[/C][C]1.07069281102261[/C][/ROW]
[ROW][C]238[/C][C]295.2[/C][C]256.389480550913[/C][C]296.7875[/C][C]0.863882341914377[/C][C]1.15137329100123[/C][/ROW]
[ROW][C]239[/C][C]290.3[/C][C]273.813861044814[/C][C]295.425[/C][C]0.926847291342352[/C][C]1.06020929288341[/C][/ROW]
[ROW][C]240[/C][C]272[/C][C]274.174981466188[/C][C]294.729166666667[/C][C]0.930260769801161[/C][C]0.992067177484405[/C][/ROW]
[ROW][C]241[/C][C]307.4[/C][C]331.294683234679[/C][C]294.541666666667[/C][C]1.12478036463889[/C][C]0.927874836380177[/C][/ROW]
[ROW][C]242[/C][C]328.7[/C][C]336.092942016869[/C][C]293.7[/C][C]1.14434096703054[/C][C]0.978003280960008[/C][/ROW]
[ROW][C]243[/C][C]292.9[/C][C]315.234422480741[/C][C]291.791666666667[/C][C]1.08034073104924[/C][C]0.929149798093177[/C][/ROW]
[ROW][C]244[/C][C]249.1[/C][C]282.695144133634[/C][C]288.75[/C][C]0.979030802194402[/C][C]0.881161226746248[/C][/ROW]
[ROW][C]245[/C][C]230.4[/C][C]264.268157637747[/C][C]285.545833333333[/C][C]0.925484201792753[/C][C]0.871841700716087[/C][/ROW]
[ROW][C]246[/C][C]361.5[/C][C]315.069568859812[/C][C]282.925[/C][C]1.11361515899907[/C][C]1.1473656478733[/C][/ROW]
[ROW][C]247[/C][C]321.7[/C][C]293.895072124909[/C][C]280.8375[/C][C]1.04649511594751[/C][C]1.09460835009603[/C][/ROW]
[ROW][C]248[/C][C]277.2[/C][C]266.351210954575[/C][C]278.491666666667[/C][C]0.956406395001318[/C][C]1.04073114218833[/C][/ROW]
[ROW][C]249[/C][C]260.7[/C][C]250.947222542659[/C][C]276.216666666667[/C][C]0.908515860288394[/C][C]1.03886385893625[/C][/ROW]
[ROW][C]250[/C][C]251[/C][C]238.147165097488[/C][C]275.670833333333[/C][C]0.863882341914377[/C][C]1.05397013605957[/C][/ROW]
[ROW][C]251[/C][C]257.6[/C][C]255.686272771643[/C][C]275.866666666667[/C][C]0.926847291342352[/C][C]1.00748466942559[/C][/ROW]
[ROW][C]252[/C][C]241.8[/C][C]255.740313877962[/C][C]274.9125[/C][C]0.930260769801161[/C][C]0.94549035438889[/C][/ROW]
[ROW][C]253[/C][C]287.5[/C][C]307.997669932096[/C][C]273.829166666667[/C][C]1.12478036463889[/C][C]0.933448620125552[/C][/ROW]
[ROW][C]254[/C][C]292.3[/C][C]313.654322888346[/C][C]274.091666666667[/C][C]1.14434096703054[/C][C]0.931917651599059[/C][/ROW]
[ROW][C]255[/C][C]274.7[/C][C]298.133528992175[/C][C]275.9625[/C][C]1.08034073104924[/C][C]0.921399216413563[/C][/ROW]
[ROW][C]256[/C][C]254.2[/C][C]272.953787651799[/C][C]278.8[/C][C]0.979030802194402[/C][C]0.931293176719999[/C][/ROW]
[ROW][C]257[/C][C]230[/C][C]259.814264916619[/C][C]280.733333333333[/C][C]0.925484201792753[/C][C]0.885247775266739[/C][/ROW]
[ROW][C]258[/C][C]339[/C][C]314.220437301075[/C][C]282.1625[/C][C]1.11361515899907[/C][C]1.07886044240713[/C][/ROW]
[ROW][C]259[/C][C]318.2[/C][C]298.508371427711[/C][C]285.245833333333[/C][C]1.04649511594751[/C][C]1.06596675489571[/C][/ROW]
[ROW][C]260[/C][C]287[/C][C]278.397946504946[/C][C]291.0875[/C][C]0.956406395001318[/C][C]1.03089840856603[/C][/ROW]
[ROW][C]261[/C][C]295.8[/C][C]271.487251950679[/C][C]298.825[/C][C]0.908515860288394[/C][C]1.08955392149955[/C][/ROW]
[ROW][C]262[/C][C]284[/C][C]265.334262299485[/C][C]307.141666666667[/C][C]0.863882341914377[/C][C]1.07034801136782[/C][/ROW]
[ROW][C]263[/C][C]271[/C][C]292.760164425337[/C][C]315.866666666667[/C][C]0.926847291342352[/C][C]0.925672386241309[/C][/ROW]
[ROW][C]264[/C][C]262.7[/C][C]302.997560983861[/C][C]325.7125[/C][C]0.930260769801161[/C][C]0.867003678666552[/C][/ROW]
[ROW][C]265[/C][C]340.6[/C][C]378.572951228333[/C][C]336.575[/C][C]1.12478036463889[/C][C]0.899694494535004[/C][/ROW]
[ROW][C]266[/C][C]379.4[/C][C]397.92549893542[/C][C]347.733333333333[/C][C]1.14434096703054[/C][C]0.953444805660905[/C][/ROW]
[ROW][C]267[/C][C]373.3[/C][C]387.752294052421[/C][C]358.916666666667[/C][C]1.08034073104924[/C][C]0.962728024375099[/C][/ROW]
[ROW][C]268[/C][C]355.2[/C][C]362.620771247779[/C][C]370.3875[/C][C]0.979030802194402[/C][C]0.979535724822811[/C][/ROW]
[ROW][C]269[/C][C]338.4[/C][C]355.57488651295[/C][C]384.204166666667[/C][C]0.925484201792753[/C][C]0.951698257766794[/C][/ROW]
[ROW][C]270[/C][C]466.9[/C][C]445.979670863315[/C][C]400.479166666667[/C][C]1.11361515899907[/C][C]1.04690870571788[/C][/ROW]
[ROW][C]271[/C][C]451[/C][C]436.615203958567[/C][C]417.216666666667[/C][C]1.04649511594751[/C][C]1.03294616383262[/C][/ROW]
[ROW][C]272[/C][C]422[/C][C]413.59794551832[/C][C]432.45[/C][C]0.956406395001318[/C][C]1.02031454598052[/C][/ROW]
[ROW][C]273[/C][C]429.2[/C][C]404.584825482929[/C][C]445.325[/C][C]0.908515860288394[/C][C]1.06084057771492[/C][/ROW]
[ROW][C]274[/C][C]425.9[/C][C]394.009537127631[/C][C]456.091666666667[/C][C]0.863882341914377[/C][C]1.08093830191232[/C][/ROW]
[ROW][C]275[/C][C]460.7[/C][C]431.038056566189[/C][C]465.058333333333[/C][C]0.926847291342352[/C][C]1.06881513820406[/C][/ROW]
[ROW][C]276[/C][C]463.6[/C][C]439.722637625386[/C][C]472.6875[/C][C]0.930260769801161[/C][C]1.05430096231469[/C][/ROW]
[ROW][C]277[/C][C]541.4[/C][C]539.360304353462[/C][C]479.525[/C][C]1.12478036463889[/C][C]1.00378169403657[/C][/ROW]
[ROW][C]278[/C][C]544.2[/C][C]556.659895324643[/C][C]486.445833333333[/C][C]1.14434096703054[/C][C]0.977616682233994[/C][/ROW]
[ROW][C]279[/C][C]517.5[/C][C]531.779719180135[/C][C]492.233333333333[/C][C]1.08034073104924[/C][C]0.973147303920972[/C][/ROW]
[ROW][C]280[/C][C]469.4[/C][C]485.415709613012[/C][C]495.8125[/C][C]0.979030802194402[/C][C]0.967006198407175[/C][/ROW]
[ROW][C]281[/C][C]439.4[/C][C]460.867995387746[/C][C]497.975[/C][C]0.925484201792753[/C][C]0.953418341905724[/C][/ROW]
[ROW][C]282[/C][C]549[/C][C]555.791405666948[/C][C]499.0875[/C][C]1.11361515899907[/C][C]0.987780657279508[/C][/ROW]
[ROW][C]283[/C][C]533[/C][C]522.693787641567[/C][C]499.470833333333[/C][C]1.04649511594751[/C][C]1.01971749540957[/C][/ROW]
[ROW][C]284[/C][C]506.1[/C][C]477.709054196575[/C][C]499.483333333333[/C][C]0.956406395001318[/C][C]1.05943145844529[/C][/ROW]
[ROW][C]285[/C][C]484[/C][C]453.82638511056[/C][C]499.525[/C][C]0.908515860288394[/C][C]1.06648713225893[/C][/ROW]
[ROW][C]286[/C][C]457[/C][C]431.685605764372[/C][C]499.704166666667[/C][C]0.863882341914377[/C][C]1.0586408115017[/C][/ROW]
[ROW][C]287[/C][C]481.5[/C][C]462.967945752933[/C][C]499.508333333333[/C][C]0.926847291342352[/C][C]1.04002880635057[/C][/ROW]
[ROW][C]288[/C][C]469.5[/C][C]464.231132823106[/C][C]499.033333333333[/C][C]0.930260769801161[/C][C]1.01134966357136[/C][/ROW]
[ROW][C]289[/C][C]544.7[/C][C]560.267159381187[/C][C]498.1125[/C][C]1.12478036463889[/C][C]0.972214756620072[/C][/ROW]
[ROW][C]290[/C][C]541.2[/C][C]568.289260402091[/C][C]496.608333333333[/C][C]1.14434096703054[/C][C]0.952331915646401[/C][/ROW]
[ROW][C]291[/C][C]521.5[/C][C]534.768661869371[/C][C]495[/C][C]1.08034073104924[/C][C]0.975188033975311[/C][/ROW]
[ROW][C]292[/C][C]469.7[/C][C]483.46988589365[/C][C]493.825[/C][C]0.979030802194402[/C][C]0.971518627539339[/C][/ROW]
[ROW][C]293[/C][C]434.4[/C][C]454.524572421292[/C][C]491.120833333333[/C][C]0.925484201792753[/C][C]0.955723906599623[/C][/ROW]
[ROW][C]294[/C][C]542.6[/C][C]541.68561365296[/C][C]486.420833333333[/C][C]1.11361515899907[/C][C]1.00168803882546[/C][/ROW]
[ROW][C]295[/C][C]517.3[/C][C]503.14613095493[/C][C]480.791666666667[/C][C]1.04649511594751[/C][C]1.0281307321557[/C][/ROW]
[ROW][C]296[/C][C]485.7[/C][C]454.496273984564[/C][C]475.2125[/C][C]0.956406395001318[/C][C]1.06865562558274[/C][/ROW]
[ROW][C]297[/C][C]465.8[/C][C]426.930530163273[/C][C]469.920833333333[/C][C]0.908515860288394[/C][C]1.09104401557289[/C][/ROW]
[ROW][C]298[/C][C]447[/C][C]401.543311051077[/C][C]464.8125[/C][C]0.863882341914377[/C][C]1.11320494626081[/C][/ROW]
[ROW][C]299[/C][C]426.6[/C][C]426.685736160593[/C][C]460.3625[/C][C]0.926847291342352[/C][C]0.99979906485423[/C][/ROW]
[ROW][C]300[/C][C]411.6[/C][C]423.900452365685[/C][C]455.679166666667[/C][C]0.930260769801161[/C][C]0.970982686390073[/C][/ROW]
[ROW][C]301[/C][C]467.5[/C][C]506.905704248778[/C][C]450.670833333333[/C][C]1.12478036463889[/C][C]0.922262259196361[/C][/ROW]
[ROW][C]302[/C][C]484.5[/C][C]509.656090103864[/C][C]445.370833333333[/C][C]1.14434096703054[/C][C]0.950641048753607[/C][/ROW]
[ROW][C]303[/C][C]451.2[/C][C]476.011630359431[/C][C]440.6125[/C][C]1.08034073104924[/C][C]0.947875999708881[/C][/ROW]
[ROW][C]304[/C][C]417.4[/C][C]426.478055320909[/C][C]435.6125[/C][C]0.979030802194402[/C][C]0.978713898153381[/C][/ROW]
[ROW][C]305[/C][C]379.9[/C][C]399.616365965762[/C][C]431.791666666667[/C][C]0.925484201792753[/C][C]0.950661765520755[/C][/ROW]
[ROW][C]306[/C][C]484.7[/C][C]479.606208601924[/C][C]430.675[/C][C]1.11361515899907[/C][C]1.01062077868617[/C][/ROW]
[ROW][C]307[/C][C]455[/C][C]451.898393047718[/C][C]431.820833333333[/C][C]1.04649511594751[/C][C]1.00686350515957[/C][/ROW]
[ROW][C]308[/C][C]420.8[/C][C]415.498803228385[/C][C]434.4375[/C][C]0.956406395001318[/C][C]1.01275863307048[/C][/ROW]
[ROW][C]309[/C][C]416.5[/C][C]396.729948774186[/C][C]436.679166666667[/C][C]0.908515860288394[/C][C]1.04983251525855[/C][/ROW]
[ROW][C]310[/C][C]376.3[/C][C]378.57123977567[/C][C]438.220833333333[/C][C]0.863882341914377[/C][C]0.99400049571379[/C][/ROW]
[ROW][C]311[/C][C]405.6[/C][C]407.986592057761[/C][C]440.1875[/C][C]0.926847291342352[/C][C]0.99415031742655[/C][/ROW]
[ROW][C]312[/C][C]405.8[/C][C]412.892366589704[/C][C]443.845833333333[/C][C]0.930260769801161[/C][C]0.982822722908917[/C][/ROW]
[ROW][C]313[/C][C]500.8[/C][C]505.054503231976[/C][C]449.025[/C][C]1.12478036463889[/C][C]0.991576150287245[/C][/ROW]
[ROW][C]314[/C][C]514[/C][C]520.451039892852[/C][C]454.804166666667[/C][C]1.14434096703054[/C][C]0.987604905364047[/C][/ROW]
[ROW][C]315[/C][C]475.5[/C][C]499.058899288482[/C][C]461.945833333333[/C][C]1.08034073104924[/C][C]0.952793348997343[/C][/ROW]
[ROW][C]316[/C][C]430.1[/C][C]461.715005609889[/C][C]471.604166666667[/C][C]0.979030802194402[/C][C]0.931527012928401[/C][/ROW]
[ROW][C]317[/C][C]414.4[/C][C]447.683701696374[/C][C]483.729166666667[/C][C]0.925484201792753[/C][C]0.925653532683334[/C][/ROW]
[ROW][C]318[/C][C]538[/C][C]555.749645098486[/C][C]499.05[/C][C]1.11361515899907[/C][C]0.968061796790999[/C][/ROW]
[ROW][C]319[/C][C]526[/C][C]545.014656385464[/C][C]520.8[/C][C]1.04649511594751[/C][C]0.965111660461447[/C][/ROW]
[ROW][C]320[/C][C]488.5[/C][C]523.36550447795[/C][C]547.220833333333[/C][C]0.956406395001318[/C][C]0.933382112157491[/C][/ROW]
[ROW][C]321[/C][C]520.2[/C][C]522.797880837454[/C][C]575.441666666667[/C][C]0.908515860288394[/C][C]0.995030812226529[/C][/ROW]
[ROW][C]322[/C][C]504.4[/C][C]522.75320264118[/C][C]605.120833333333[/C][C]0.863882341914377[/C][C]0.964891266952644[/C][/ROW]
[ROW][C]323[/C][C]568.5[/C][C]587.879927579884[/C][C]634.279166666667[/C][C]0.926847291342352[/C][C]0.967034207717101[/C][/ROW]
[ROW][C]324[/C][C]610.6[/C][C]615.890770906481[/C][C]662.0625[/C][C]0.930260769801161[/C][C]0.991409562934846[/C][/ROW]
[ROW][C]325[/C][C]818[/C][C]773.441157989372[/C][C]687.6375[/C][C]1.12478036463889[/C][C]1.05761115962132[/C][/ROW]
[ROW][C]326[/C][C]830.9[/C][C]814.355944925196[/C][C]711.6375[/C][C]1.14434096703054[/C][C]1.02031550844309[/C][/ROW]
[ROW][C]327[/C][C]835.9[/C][C]791.90776153794[/C][C]733.016666666667[/C][C]1.08034073104924[/C][C]1.05555222539633[/C][/ROW]
[ROW][C]328[/C][C]782[/C][C]736.084308629861[/C][C]751.85[/C][C]0.979030802194402[/C][C]1.06237830481077[/C][/ROW]
[ROW][C]329[/C][C]762.3[/C][C]710.270563034197[/C][C]767.458333333333[/C][C]0.925484201792753[/C][C]1.07325298227698[/C][/ROW]
[ROW][C]330[/C][C]856.9[/C][C]866.879800333338[/C][C]778.4375[/C][C]1.11361515899907[/C][C]0.988487676919567[/C][/ROW]
[ROW][C]331[/C][C]820.9[/C][C]819.353351031104[/C][C]782.95[/C][C]1.04649511594751[/C][C]1.0018876458697[/C][/ROW]
[ROW][C]332[/C][C]769.6[/C][C]747.694609452155[/C][C]781.775[/C][C]0.956406395001318[/C][C]1.02929724284611[/C][/ROW]
[ROW][C]333[/C][C]752.2[/C][C]706.053100823126[/C][C]777.15[/C][C]0.908515860288394[/C][C]1.06535896396896[/C][/ROW]
[ROW][C]334[/C][C]724.4[/C][C]665.016626805688[/C][C]769.8[/C][C]0.863882341914377[/C][C]1.08929607291107[/C][/ROW]
[ROW][C]335[/C][C]723.1[/C][C]704.801713382722[/C][C]760.429166666667[/C][C]0.926847291342352[/C][C]1.02596231857816[/C][/ROW]
[ROW][C]336[/C][C]719.5[/C][C]698.742120716897[/C][C]751.125[/C][C]0.930260769801161[/C][C]1.02970749675403[/C][/ROW]
[ROW][C]337[/C][C]817.4[/C][C]837.605191207168[/C][C]744.683333333333[/C][C]1.12478036463889[/C][C]0.97587742838837[/C][/ROW]
[ROW][C]338[/C][C]803.3[/C][C]847.374949911389[/C][C]740.491666666667[/C][C]1.14434096703054[/C][C]0.947986484712584[/C][/ROW]
[ROW][C]339[/C][C]752.5[/C][C]796.06707335248[/C][C]736.866666666667[/C][C]1.08034073104924[/C][C]0.94527210732507[/C][/ROW]
[ROW][C]340[/C][C]689[/C][C]717.715243203688[/C][C]733.0875[/C][C]0.979030802194402[/C][C]0.959990757510581[/C][/ROW]
[ROW][C]341[/C][C]630.4[/C][C]676.351567038492[/C][C]730.808333333333[/C][C]0.925484201792753[/C][C]0.932059642827918[/C][/ROW]
[ROW][C]342[/C][C]765.5[/C][C]812.405458805634[/C][C]729.520833333333[/C][C]1.11361515899907[/C][C]0.942263486418971[/C][/ROW]
[ROW][C]343[/C][C]757.7[/C][C]761.264151303384[/C][C]727.441666666666[/C][C]1.04649511594751[/C][C]0.995318114878677[/C][/ROW]
[ROW][C]344[/C][C]732.2[/C][C]694.733605328957[/C][C]726.4[/C][C]0.956406395001318[/C][C]1.05392915267616[/C][/ROW]
[ROW][C]345[/C][C]702.6[/C][C]660.350967567868[/C][C]726.845833333333[/C][C]0.908515860288394[/C][C]1.06397966309906[/C][/ROW]
[ROW][C]346[/C][C]683.3[/C][C]626.861823371136[/C][C]725.633333333333[/C][C]0.863882341914377[/C][C]1.09003288208772[/C][/ROW]
[ROW][C]347[/C][C]709.5[/C][C]670.716904243607[/C][C]723.654166666667[/C][C]0.926847291342352[/C][C]1.05782334620018[/C][/ROW]
[ROW][C]348[/C][C]702.2[/C][C]671.811071431153[/C][C]722.175[/C][C]0.930260769801161[/C][C]1.04523433724322[/C][/ROW]
[ROW][C]349[/C][C]784.8[/C][C]808.360901726557[/C][C]718.683333333333[/C][C]1.12478036463889[/C][C]0.970853486757915[/C][/ROW]
[ROW][C]350[/C][C]810.9[/C][C]816.692307732883[/C][C]713.679166666666[/C][C]1.14434096703054[/C][C]0.992907600967907[/C][/ROW]
[ROW][C]351[/C][C]755.6[/C][C]765.822034302814[/C][C]708.870833333333[/C][C]1.08034073104924[/C][C]0.986652206589851[/C][/ROW]
[ROW][C]352[/C][C]656.8[/C][C]689.107147304566[/C][C]703.866666666667[/C][C]0.979030802194402[/C][C]0.953117381773016[/C][/ROW]
[ROW][C]353[/C][C]615.1[/C][C]646.169213507526[/C][C]698.195833333333[/C][C]0.925484201792753[/C][C]0.951917836910123[/C][/ROW]
[ROW][C]354[/C][C]745.3[/C][C]768.747104509708[/C][C]690.316666666667[/C][C]1.11361515899907[/C][C]0.96949958657124[/C][/ROW]
[ROW][C]355[/C][C]694.1[/C][C]713.286710633507[/C][C]681.595833333333[/C][C]1.04649511594751[/C][C]0.973100984011792[/C][/ROW]
[ROW][C]356[/C][C]675.7[/C][C]642.633366961261[/C][C]671.925[/C][C]0.956406395001318[/C][C]1.05145489596206[/C][/ROW]
[ROW][C]357[/C][C]643.7[/C][C]601.191443132089[/C][C]661.729166666667[/C][C]0.908515860288394[/C][C]1.07070718878907[/C][/ROW]
[ROW][C]358[/C][C]622.1[/C][C]564.611901616689[/C][C]653.575[/C][C]0.863882341914377[/C][C]1.1018187859992[/C][/ROW]
[ROW][C]359[/C][C]634.6[/C][C]599.685644953357[/C][C]647.016666666667[/C][C]0.926847291342352[/C][C]1.05822109523625[/C][/ROW]
[ROW][C]360[/C][C]588[/C][C]594.835868816648[/C][C]639.429166666667[/C][C]0.930260769801161[/C][C]0.988507974762438[/C][/ROW]
[ROW][C]361[/C][C]689.7[/C][C]711.578237934234[/C][C]632.6375[/C][C]1.12478036463889[/C][C]0.969253924912392[/C][/ROW]
[ROW][C]362[/C][C]673.9[/C][C]717.616220424852[/C][C]627.1[/C][C]1.14434096703054[/C][C]0.939081337377003[/C][/ROW]
[ROW][C]363[/C][C]647.9[/C][C]670.88259114215[/C][C]620.991666666667[/C][C]1.08034073104924[/C][C]0.965742752240712[/C][/ROW]
[ROW][C]364[/C][C]568.8[/C][C]602.254877264895[/C][C]615.154166666667[/C][C]0.979030802194402[/C][C]0.944450632900096[/C][/ROW]
[ROW][C]365[/C][C]545.7[/C][C]563.616022707613[/C][C]608.995833333333[/C][C]0.925484201792753[/C][C]0.96821236092341[/C][/ROW]
[ROW][C]366[/C][C]632.6[/C][C]674.140856689574[/C][C]605.3625[/C][C]1.11361515899907[/C][C]0.938379559290377[/C][/ROW]
[ROW][C]367[/C][C]643.8[/C][C]NA[/C][C]NA[/C][C]1.04649511594751[/C][C]NA[/C][/ROW]
[ROW][C]368[/C][C]593.1[/C][C]NA[/C][C]NA[/C][C]0.956406395001318[/C][C]NA[/C][/ROW]
[ROW][C]369[/C][C]579.7[/C][C]NA[/C][C]NA[/C][C]0.908515860288394[/C][C]NA[/C][/ROW]
[ROW][C]370[/C][C]546[/C][C]NA[/C][C]NA[/C][C]0.863882341914377[/C][C]NA[/C][/ROW]
[ROW][C]371[/C][C]562.9[/C][C]NA[/C][C]NA[/C][C]0.926847291342352[/C][C]NA[/C][/ROW]
[ROW][C]372[/C][C]572.5[/C][C]NA[/C][C]NA[/C][C]0.930260769801161[/C][C]NA[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=147450&T=1
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=147450&T=1
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Classical Decomposition by Moving Averages t Observations Fit Trend Seasonal Random 1 235.1 NA NA 1.12478036463889 NA 2 280.7 NA NA 1.14434096703054 NA 3 264.6 NA NA 1.08034073104924 NA 4 240.7 NA NA 0.979030802194402 NA 5 201.4 NA NA 0.925484201792753 NA 6 240.8 NA NA 1.11361515899907 NA 7 241.1 241.129916299572 230.416666666667 1.04649511594751 0.999875932857974 8 223.8 225.596343447582 235.879166666667 0.956406395001318 0.992037355658652 9 206.1 219.614781810963 241.729166666667 0.908515860288394 0.938461420039584 10 174.7 214.609970790079 248.425 0.863882341914377 0.814034871524599 11 203.3 239.412379081157 258.308333333333 0.926847291342352 0.8491624400553 12 220.5 252.154933827686 271.058333333333 0.930260769801161 0.874462365867018 13 299.5 321.448168459236 285.7875 1.12478036463889 0.931720972110565 14 347.4 344.656426920148 301.183333333333 1.14434096703054 1.00796031312797 15 338.3 340.199296207404 314.9 1.08034073104924 0.994417107182238 16 327.7 323.04345106907 329.9625 0.979030802194402 1.01441462105336 17 351.6 319.172507909102 344.870833333333 0.925484201792753 1.10159863800091 18 396.6 398.075658773705 357.4625 1.11361515899907 0.996293019326399 19 438.8 387.765684025401 370.5375 1.04649511594751 1.1316112231614 20 395.6 366.219963725942 382.9125 0.956406395001318 1.0802251083615 21 363.5 356.38805909463 392.275 0.908515860288394 1.01995560940913 22 378.8 343.623599535475 397.766666666667 0.863882341914377 1.10236898895209 23 357 369.190309187656 398.329166666667 0.926847291342352 0.966980961080807 24 369 367.960771408308 395.545833333333 0.930260769801161 1.0028242918062 25 464.8 437.933234972151 389.35 1.12478036463889 1.06134899770637 26 479.1 434.034224532596 379.2875 1.14434096703054 1.1038300044563 27 431.3 398.560202782626 368.920833333333 1.08034073104924 1.08214517402589 28 366.5 349.444648368246 356.929166666667 0.979030802194402 1.04880701911274 29 326.3 318.968130147872 344.65 0.925484201792753 1.02298621448083 30 355.1 372.230506958602 334.254166666667 1.11361515899907 0.95397876681691 31 331.6 335.493252983823 320.5875 1.04649511594751 0.988395435827109 32 261.3 289.667601859378 302.870833333333 0.956406395001318 0.902068434035127 33 249 259.101152388748 285.191666666667 0.908515860288394 0.96101463735062 34 205.5 232.992667124066 269.704166666667 0.863882341914377 0.882002006915411 35 235.6 237.539375179903 256.2875 0.926847291342352 0.991835563352669 36 240.9 227.127043033494 244.154166666667 0.930260769801161 1.06063988146262 37 264.9 262.139437148798 233.058333333333 1.12478036463889 1.01053089485973 38 253.8 256.918851360444 224.5125 1.14434096703054 0.987860558522938 39 232.3 236.198495165064 218.633333333333 1.08034073104924 0.983494834874625 40 193.8 210.357005736495 214.8625 0.979030802194402 0.921290923121263 41 177 196.279774463546 212.083333333333 0.925484201792753 0.901774013566909 42 213.2 231.636593134969 208.004166666667 1.11361515899907 0.920407251352439 43 207.2 213.297506611685 203.820833333333 1.04649511594751 0.971413136944045 44 180.6 192.588367740099 201.366666666667 0.956406395001318 0.937751340432579 45 188.6 180.980144852199 199.204166666667 0.908515860288394 1.04210326582523 46 175.4 170.566369391478 197.441666666667 0.863882341914377 1.02833870842046 47 199 182.650706213866 197.066666666667 0.926847291342352 1.08951125415847 48 179.6 182.982293419888 196.7 0.930260769801161 0.981515733808588 49 225.8 220.836564842287 196.3375 1.12478036463889 1.02247560389855 50 234 225.273055534687 196.858333333333 1.14434096703054 1.03873940647096 51 200.2 212.467010439683 196.666666666667 1.08034073104924 0.942263928812773 52 183.6 190.780468987616 194.866666666667 0.979030802194402 0.962362662039153 53 178.2 177.762378059343 192.075 0.925484201792753 1.0024618366689 54 203.2 210.895510798611 189.379166666667 1.11361515899907 0.963510314802481 55 208.5 196.540503567575 187.808333333333 1.04649511594751 1.0608500345493 56 191.8 177.620607658328 185.716666666667 0.956406395001318 1.07982965787927 57 172.8 166.644521673399 183.425 0.908515860288394 1.03693777788066 58 148 157.572139165182 182.4 0.863882341914377 0.939252337272975 59 159.4 167.828873279816 181.075 0.926847291342352 0.949776977494432 60 154.5 166.330625640448 178.8 0.930260769801161 0.928872836286797 61 213.2 197.947284421886 175.9875 1.12478036463889 1.07705443205579 62 196.4 197.665829705075 172.733333333333 1.14434096703054 0.993596112656579 63 182.8 184.292624457861 170.5875 1.08034073104924 0.991900791134467 64 176.4 166.941068954182 170.516666666667 0.979030802194402 1.05666029997935 65 153.6 159.796415992041 172.6625 0.925484201792753 0.961223060269701 66 173.2 198.042535838497 177.8375 1.11361515899907 0.874559595324733 67 171 195.90824610169 187.204166666667 1.04649511594751 0.872857592279391 68 151.2 192.783634045745 201.570833333333 0.956406395001318 0.784298940874423 69 161.9 199.135320126962 219.1875 0.908515860288394 0.813014988485105 70 157.2 204.797707189835 237.066666666667 0.863882341914377 0.767586718411281 71 201.7 235.959872920907 254.583333333333 0.926847291342352 0.854806359671202 72 236.4 252.608435952964 271.545833333333 0.930260769801161 0.935835729745849 73 356.1 323.791460885567 287.870833333333 1.12478036463889 1.09978193688638 74 398.3 348.146666869591 304.233333333333 1.14434096703054 1.14405805915469 75 403.7 345.655016899203 319.95 1.08034073104924 1.16792750072458 76 384.6 326.176349636092 333.1625 0.979030802194402 1.17911675824777 77 365.8 317.803562527283 343.391666666667 0.925484201792753 1.15102548596697 78 368.1 390.493795566186 350.654166666667 1.11361515899907 0.942652621320866 79 367.9 370.241251229597 353.791666666667 1.04649511594751 0.993676417142009 80 347 337.149194344548 352.516666666667 0.956406395001318 1.02921794214755 81 343.3 315.898535587777 347.708333333333 0.908515860288394 1.08674134674679 82 292.9 295.253387407786 341.775 0.863882341914377 0.992029261955474 83 311.5 310.393434143126 334.891666666667 0.926847291342352 1.00356504273336 84 300.9 304.730171667615 327.575 0.930260769801161 0.987430940472173 85 366.9 359.990642287528 320.054166666667 1.12478036463889 1.0191931592126 86 356.9 356.948556141001 311.925 1.14434096703054 0.9998639687984 87 329.7 327.851901935454 303.470833333333 1.08034073104924 1.00563699052418 88 316.2 290.009320085027 296.220833333333 0.979030802194402 1.09030978696579 89 269 269.427732062741 291.120833333333 0.925484201792753 0.998412442329279 90 289.3 319.616830759058 287.008333333333 1.11361515899907 0.905146325720524 91 266.2 295.94445839364 282.795833333333 1.04649511594751 0.899493105716221 92 253.6 266.163914702221 278.295833333333 0.956406395001318 0.952796325842001 93 233.8 250.265835647443 275.466666666667 0.908515860288394 0.934206618315099 94 228.4 235.61670973763 272.741666666667 0.863882341914377 0.969370976508132 95 253.6 251.465255732322 271.3125 0.926847291342352 1.00848922154857 96 260.1 253.922429290309 272.958333333333 0.930260769801161 1.02432857438769 97 306.6 309.867617288308 275.491666666667 1.12478036463889 0.989454795835385 98 309.2 316.462726344933 276.545833333333 1.14434096703054 0.977050294583454 99 309.5 298.160537510451 275.9875 1.08034073104924 1.03803139940728 100 271 269.147805408268 274.9125 0.979030802194402 1.00688170051739 101 279.9 253.925871682712 274.370833333333 0.925484201792753 1.10229020046348 102 317.9 306.109606893032 274.879166666667 1.11361515899907 1.03851689996482 103 298.4 288.527424259362 275.708333333333 1.04649511594751 1.03421711390514 104 246.7 264.219221699052 276.2625 0.956406395001318 0.933694370960618 105 227.3 249.928927682586 275.095833333333 0.908515860288394 0.909458549306765 106 209.1 236.369007277048 273.6125 0.863882341914377 0.884633744537051 107 259.9 252.643124165069 272.583333333333 0.926847291342352 1.02872382083982 108 266 252.751851154975 271.7 0.930260769801161 1.05241563527423 109 320.6 304.717060535232 270.9125 1.12478036463889 1.05212356484691 110 308.5 309.625289066926 270.570833333333 1.14434096703054 0.99636564225643 111 282.2 293.609602180906 271.775 1.08034073104924 0.961140228057406 112 262.7 268.539990451906 274.291666666667 0.979030802194402 0.978252809043158 113 263.5 257.330882308475 278.05 0.925484201792753 1.02397348361838 114 313.1 315.208770754687 283.05 1.11361515899907 0.99330992361146 115 284.3 304.626007459688 291.091666666667 1.04649511594751 0.933275534714883 116 252.6 291.508684169756 304.795833333333 0.956406395001318 0.866526500640722 117 250.3 293.431695459395 322.979166666667 0.908515860288394 0.853009418795511 118 246.5 296.185660435102 342.854166666667 0.863882341914377 0.832248258196859 119 312.7 335.669332150774 362.1625 0.926847291342352 0.931571549883332 120 333.2 353.530101216768 380.033333333333 0.930260769801161 0.942494002217077 121 446.4 447.82661559612 398.145833333333 1.12478036463889 0.996814357283742 122 511.6 476.289014740199 416.2125 1.14434096703054 1.07413772765484 123 515.5 466.018478597226 431.3625 1.08034073104924 1.10617931192711 124 506.4 433.90237223755 443.195833333333 0.979030802194402 1.16708281033034 125 483.2 417.663307900721 451.291666666667 0.925484201792753 1.15691273535299 126 522.3 509.033489178475 457.1 1.11361515899907 1.02606215721275 127 509.8 482.495294000233 461.058333333333 1.04649511594751 1.05659061619729 128 460.7 440.146193032898 460.208333333333 0.956406395001318 1.04669768202577 129 405.8 413.280079362439 454.895833333333 0.908515860288394 0.981900701882417 130 375 384.470836268994 445.05 0.863882341914377 0.97536656782371 131 378.5 400.865315369281 432.504166666667 0.926847291342352 0.944207407047232 132 406.8 390.914955903152 420.220833333333 0.930260769801161 1.04063554964314 133 467.8 458.896329018108 407.9875 1.12478036463889 1.01940235826454 134 469.8 453.750265777059 396.516666666667 1.14434096703054 1.03537129437369 135 429.8 418.8165914898 387.670833333333 1.08034073104924 1.02622486485344 136 355.8 373.904101243069 381.9125 0.979030802194402 0.95158089685863 137 332.7 350.878054188852 379.129166666667 0.925484201792753 0.948192672719657 138 378 418.974503257588 376.229166666667 1.11361515899907 0.902202871680723 139 360.5 388.899387067677 371.620833333333 1.04649511594751 0.926974976016778 140 334.7 349.957069984274 365.908333333333 0.956406395001318 0.95640302399103 141 319.5 328.852457562389 361.966666666667 0.908515860288394 0.971560323338575 142 323.1 312.390653365513 361.6125 0.863882341914377 1.03428190478528 143 363.6 335.553476239357 362.0375 0.926847291342352 1.08358287351086 144 352.1 338.514141957561 363.891666666667 0.930260769801161 1.04013379755385 145 411.9 412.442899958522 366.6875 1.12478036463889 0.99868369668001 146 388.6 422.68140852218 369.366666666667 1.14434096703054 0.91936856498766 147 416.4 401.220541832835 371.383333333333 1.08034073104924 1.03783320290089 148 360.7 365.333502428859 373.158333333333 0.979030802194402 0.987317061265791 149 338 347.808531586239 375.8125 0.925484201792753 0.971799048339886 150 417.2 424.449777789333 381.145833333333 1.11361515899907 0.982919586323988 151 388.4 408.307511072187 390.166666666667 1.04649511594751 0.951243828407879 152 371.1 385.049214627531 402.6 0.956406395001318 0.963772904611624 153 331.5 377.227141639995 415.2125 0.908515860288394 0.87878088135123 154 353.7 367.833902167626 425.791666666667 0.863882341914377 0.961575314063398 155 396.7 404.57270453465 436.504166666667 0.926847291342352 0.980540692818845 156 447 415.489344572066 446.6375 0.930260769801161 1.07583986410142 157 533.5 512.763935314605 455.879166666667 1.12478036463889 1.04043978770206 158 565.4 530.292372209315 463.404166666667 1.14434096703054 1.06620428584409 159 542.3 507.152451931925 469.4375 1.08034073104924 1.06930371318168 160 488.7 463.856635489689 473.791666666667 0.979030802194402 1.05355828204135 161 467.1 439.643557693299 475.041666666667 0.925484201792753 1.06245159704093 162 531.3 526.902372417247 473.145833333333 1.11361515899907 1.00834619051453 163 496.1 490.160870724548 468.383333333333 1.04649511594751 1.01211669398798 164 444 440.445070031336 460.520833333333 0.956406395001318 1.00807122206728 165 403.4 409.78607878308 451.05 0.908515860288394 0.984416067031743 166 386.3 382.008771594538 442.2 0.863882341914377 1.01123332427041 167 394.1 403.120643778214 434.9375 0.926847291342352 0.977622967423179 168 404.1 397.577948666853 427.383333333333 0.930260769801161 1.01640445944001 169 462.1 470.280043625224 418.108333333333 1.12478036463889 0.982606015849266 170 448.1 470.209703352849 410.9 1.14434096703054 0.952979057651948 171 432.3 438.613835386277 405.995833333333 1.08034073104924 0.985605024563996 172 386.3 392.554638024873 400.9625 0.979030802194402 0.984066834475978 173 395.2 367.829839818356 397.445833333333 0.925484201792753 1.07440984177673 174 421.9 440.30023355242 395.379166666667 1.11361515899907 0.958209802879359 175 382.9 412.56325787704 394.233333333333 1.04649511594751 0.92810009783789 176 384.2 378.621366647793 395.879166666667 0.956406395001318 1.01473406902943 177 345.5 361.58552691203 397.995833333333 0.908515860288394 0.955513908287753 178 323.4 344.7178505019 399.033333333333 0.863882341914377 0.93815855352178 179 372.6 370.333420846978 399.5625 0.926847291342352 1.00612037430443 180 376 372.984179565235 400.945833333333 0.930260769801161 1.00808565242172 181 462.7 454.003534431929 403.6375 1.12478036463889 1.01915506137844 182 487 462.966978649018 404.570833333333 1.14434096703054 1.05191087584932 183 444.2 436.750247625217 404.270833333333 1.08034073104924 1.01705723675096 184 399.3 396.507474888733 405 0.979030802194402 1.00704280571773 185 394.9 375.947107504913 406.216666666667 0.925484201792753 1.0504137207515 186 455.4 453.111447944071 406.883333333333 1.11361515899907 1.00505074869839 187 414 425.448228992144 406.545833333333 1.04649511594751 0.973091369967001 188 375.5 386.758791058595 404.3875 0.956406395001318 0.9708893726041 189 347 365.022745250121 401.779166666667 0.908515860288394 0.950625692550279 190 339.4 345.725713234134 400.2 0.863882341914377 0.981703087181572 191 385.8 368.67281944999 397.770833333333 0.926847291342352 1.04645631477678 192 378.8 368.007284446798 395.595833333333 0.930260769801161 1.0293274508667 193 451.8 442.305818639684 393.2375 1.12478036463889 1.02146519661332 194 446.1 446.807930577074 390.45 1.14434096703054 0.998415581889605 195 422.5 419.964453516539 388.733333333333 1.08034073104924 1.00603752641975 196 383.1 378.933871989343 387.05 0.979030802194402 1.01099434048686 197 352.8 355.370508751721 383.983333333333 0.925484201792753 0.9927666795966 198 445.3 423.345442756659 380.154166666667 1.11361515899907 1.05185967539979 199 367.5 393.660939845238 376.170833333333 1.04649511594751 0.933544486644971 200 355.1 356.321157537679 372.5625 0.956406395001318 0.996572873903651 201 326.2 335.393771756466 369.166666666667 0.908515860288394 0.972588126164903 202 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par2 <- as.numeric(par2)x <- ts(x,freq=par2)m <- decompose(x,type=par1)m$figurebitmap(file='test1.png')plot(m)dev.off()mylagmax <- length(x)/2bitmap(file='test2.png')op <- par(mfrow = c(2,2))acf(as.numeric(x),lag.max = mylagmax,main='Observed')acf(as.numeric(m$trend),na.action=na.pass,lag.max = mylagmax,main='Trend')acf(as.numeric(m$seasonal),na.action=na.pass,lag.max = mylagmax,main='Seasonal')acf(as.numeric(m$random),na.action=na.pass,lag.max = mylagmax,main='Random')par(op)dev.off()bitmap(file='test3.png')op <- par(mfrow = c(2,2))spectrum(as.numeric(x),main='Observed')spectrum(as.numeric(m$trend[!is.na(m$trend)]),main='Trend')spectrum(as.numeric(m$seasonal[!is.na(m$seasonal)]),main='Seasonal')spectrum(as.numeric(m$random[!is.na(m$random)]),main='Random')par(op)dev.off()bitmap(file='test4.png')op <- par(mfrow = c(2,2))cpgram(as.numeric(x),main='Observed')cpgram(as.numeric(m$trend[!is.na(m$trend)]),main='Trend')cpgram(as.numeric(m$seasonal[!is.na(m$seasonal)]),main='Seasonal')cpgram(as.numeric(m$random[!is.na(m$random)]),main='Random')par(op)dev.off()load(file='createtable')a<-table.start()a<-table.row.start(a)a<-table.element(a,'Classical Decomposition by Moving Averages',6,TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'t',header=TRUE)a<-table.element(a,'Observations',header=TRUE)a<-table.element(a,'Fit',header=TRUE)a<-table.element(a,'Trend',header=TRUE)a<-table.element(a,'Seasonal',header=TRUE)a<-table.element(a,'Random',header=TRUE)a<-table.row.end(a)for (i in 1:length(m$trend)) {a<-table.row.start(a)a<-table.element(a,i,header=TRUE)a<-table.element(a,x[i])if (par1 == 'additive') a<-table.element(a,m$trend[i]+m$seasonal[i]) else a<-table.element(a,m$trend[i]*m$seasonal[i])a<-table.element(a,m$trend[i])a<-table.element(a,m$seasonal[i])a<-table.element(a,m$random[i])a<-table.row.end(a)}a<-table.end(a)table.save(a,file='mytable.tab') | 68,181 | 144,732 | {"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} | 2.671875 | 3 | CC-MAIN-2023-06 | latest | en | 0.506745 |
https://questioncove.com/updates/4df9717e0b8b370c28be11d7 | 1,632,650,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00362.warc.gz | 515,845,468 | 5,492 | Mathematics
OpenStudy (anonymous):
Find the exact solutions to each equation for the interval 0
OpenStudy (anonymous):
Divide each side by sin x
OpenStudy (anonymous):
so its cos x=0?
OpenStudy (anonymous):
no cosx=1
OpenStudy (anonymous):
which means that x=0 or 2*pi right?
OpenStudy (anonymous):
wow.... 0<x<2pi... no = if that is the case => no solution
OpenStudy (anonymous):
the problem states that x is greater than or equal to 0. so i would assume its 0. i just couldnt add in the equal to symbol
OpenStudy (anonymous):
next time type: >= or <= everybody will understand :) if <= then 2 solutions: 0 and 2pi
OpenStudy (anonymous):
okay thanks for the tip and answer
OpenStudy (anonymous):
welcome!
myininaya (myininaya):
sinxcosx=sinx sinxcosx-sinx=0 sinx(cosx-1)=0 sinx=0 cosx-1=0 sinx=0 cosx=1 x=0 x=0 x=pi x=2pi x=2pi
myininaya (myininaya):
you guys missed a solution
OpenStudy (anonymous):
isnt it just 0 and 2pi?
myininaya (myininaya):
and pi
OpenStudy (anonymous):
oh, alright thanks, would have missed that
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Join our real-time social learning platform and learn together with your friends! | 386 | 1,443 | {"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} | 3.125 | 3 | CC-MAIN-2021-39 | latest | en | 0.904584 |
http://betterlesson.com/lesson/resource/2684099/sinusoidal-project-docx | 1,477,373,337,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719908.93/warc/CC-MAIN-20161020183839-00391-ip-10-171-6-4.ec2.internal.warc.gz | 29,504,068 | 21,729 | ## Sinusoidal Project.docx - Section 2: Introduction of the Project
Sinusoidal Project.docx
# Sinusodial Project Day 1 of 3
Unit 8: Graphing Trigonometric Functions
Lesson 11 of 13
## Big Idea: Students collect data, analyze the data, and construct models using their knowledge of trigonometric functions.
Print Lesson
Standards:
40 minutes
### Katharine Sparks
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Environment: Urban | 288 | 1,251 | {"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-2016-44 | longest | en | 0.753918 |
https://homework.cpm.org/category/CC/textbook/cc2/chapter/3/lesson/3.2.3/problem/3-62 | 1,725,869,548,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00677.warc.gz | 281,296,362 | 16,104 | ### Home > CC2 > Chapter 3 > Lesson 3.2.3 > Problem3-62
3-62.
Simplify the following fraction expressions. Show all of your work.
1. $\frac{5}{6}+\frac{1}{4}$
Draw diagrams to represent each fraction.
Now change the diagram so that each whole is divided into equal portions.
1. $\frac{7}{12}-\frac{2}{5}$
Follow the steps in part (a) to solve the problem.
1. $\frac{3}{8}\cdot\frac{2}{3}$
Draw a rectangle. A rectangle divided into 24 equal sections arranged in 3 rows and 8 columns.
Shade in three eighths. The same rectangle with three sections shaded in each row.
Shade in two thirds in the other direction. The same rectangle with 2 rows lightly shaded, and 6 sections, in those 2 rows, are shaded dark.
1. $\frac{4}{5}\cdot\frac{7}{8}$
$\frac{28}{40}$ | 231 | 769 | {"found_math": true, "script_math_tex": 5, "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} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.795741 |
https://casinoonlinewithbonus.com/blackjack-hit-chart/ | 1,726,644,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00721.warc.gz | 142,228,478 | 39,223 | Select Page
# Blackjack hit chart
## Understanding the Blackjack Hit Chart: A Comprehensive Guide
The Blackjack Hit Chart is an essential tool for any serious blackjack player. It provides a detailed strategy on when to hit or stand based on the player’s hand and the dealer’s upcard. By understanding and utilizing this chart, players can greatly improve their chances of winning and minimize their losses.
The chart is based on statistical probabilities and mathematical calculations, taking into account the player’s hand value and the dealer’s upcard. It helps players make informed decisions on whether to take another card (hit) or to keep their current hand (stand).
The chart takes into consideration the player’s hand value, the dealer’s upcard, and the basic principles of blackjack strategy.
The Blackjack Hit Chart is a visual representation of the optimal moves to make in any given situation. It consists of a grid with the player’s hand value on one axis and the dealer’s upcard on the other axis. The cells in the grid indicate whether the player should hit, stand, double down, or split their hand, depending on the specific combination of cards. The chart is color-coded for easier interpretation, with green cells indicating a hit and red cells indicating a stand.
In order to make the most of the Blackjack Hit Chart, players should start by getting familiar with the fundamental game rules.
They must understand the value of each card and the objective of the game, which is to reach a hand value as close to 21 as possible without exceeding it. Once players have a solid understanding of the rules, they can begin to study the chart and learn the optimal moves for each possible hand.
It is worth mentioning that the Blackjack Hit Chart does not provide a foolproof tactic for winning every hand. Blackjack is ultimately a game of chance, and the outcome of each hand depends on several factors, including luck. However, by following the chart’s recommendations, players can greatly increase their odds of winning in the long run.
It is important to note that the strategy for hitting in Blackjack is not universally applicable, as it does not conform to a one-size-fits-all approach.
It is based on the assumption that the player is utilizing basic blackjack strategy, such as when to double down or split pairs. The chart does not account for card counting or other advanced techniques that some players may employ.
To sum up, the Blackjack Hit Chart is a crucial resource for any dedicated blackjack player. By understanding and utilizing this chart, players can make more informed decisions on when to hit or stand, ultimately improving their chances of winning. However, it is important to remember that the chart is not foolproof and that luck still plays a significant role in the outcome of each hand. Nonetheless, by following the chart’s recommendations and employing basic blackjack strategy, players can enhance their overall blackjack experience.
## Mastering the Blackjack Hit Chart: Strategies for Success
Mastering the Blackjack Hit Chart is essential for players who want to improve their chances of success at the blackjack table. While the chart provides a solid foundation, there are strategies and techniques that can further enhance a player’s understanding and application of the hit chart.
One key strategy to master is knowing when to deviate from the hit chart. While the chart provides general guidelines, there may be specific situations where it is beneficial to make a different decision based on the cards in play and the current game dynamics. Experienced players understand the importance of adapting their strategy based on the specific circumstances, and this flexibility can lead to better outcomes.
Another strategy for mastering the Blackjack Hit Chart is understanding the concept of card counting.
While card counting is not explicitly addressed in the chart, it is a technique that can be used in conjunction with the chart to further optimize gameplay. Card counting involves keeping track of the cards that have been played to gain an advantage over the casino. By tracking the values of the cards that have already been dealt, players can adjust their decisions based on the likelihood of certain cards appearing in subsequent rounds.
Furthermore, it is essential to regularly train with the Blackjack Hit Chart. By familiarizing oneself with the chart and repeatedly applying it in simulated or real gameplay, players can develop a deeper understanding of its principles and recommendations.
Practice allows players to internalize the strategies outlined in the chart and make quicker and more accurate decisions at the table.
Furthermore, studying and analyzing the outcomes of various hands can help players refine their understanding of the hit chart. By reviewing past games and analyzing the decisions made, players can identify patterns, mistakes, and areas for improvement. This analysis can help players make better-informed decisions in the future and fine-tune their strategy.
Finally, players should prioritize a disciplined approach when utilizing the Blackjack Hit Chart. While it provides valuable guidance, it is crucial to resist the urge to deviate from its recommendations impulsively.
Emotional decision-making can lead to suboptimal outcomes and negate the advantages offered by the hit chart. Players must trust in the statistical probabilities and make decisions based on logic and strategy rather than emotions.
In conclusion, becoming proficient in the strategy of the Blackjack Hit Chart requires more than just memorizing its suggestions. Players must also be willing to deviate when necessary, understand the concept of card counting, practice using the chart, analyze past games, and maintain discipline in their decision-making. By incorporating these strategies, players can elevate their understanding and utilization of the hit chart, leading to improved success at the blackjack table.
## Utilizing the Blackjack Hit Chart: Tips and Tricks for Winning
Utilizing the Blackjack Hit Chart effectively is crucial for increasing the chances of winning and maximizing profits in blackjack. While the chart provides valuable guidance, there are additional tips and tricks that players can employ to enhance their gameplay and overall success.
One tip for utilizing the Blackjack Hit Chart is to understand the concept of soft hands and hard hands. A soft hand is a hand that contains an Ace, which can be counted as either 1 or 11. Knowing how to play soft hands is essential as it can significantly impact the player’s decisions.
The hit chart provides specific recommendations for soft hands, and players should familiarize themselves with these guidelines to make optimal choices.
Another tip is to pay attention to the dealer’s upcard. The dealer’s upcard plays a crucial role in determining the player’s strategy. Depending on the value of the dealer’s upcard, players may need to adjust their decisions based on the recommendations of the hit chart. For example, if the dealer’s upcard is a 2, it is generally recommended to hit until reaching a hand value of at least 13.
Proper timing plays a crucial role in effectively employing the Blackjack Hit Chart. Players should consider the flow of the game and the cards that have already been played.
If many low-value cards have already been dealt, the chances of receiving high-value cards increase. In such cases, players may want to be more conservative and follow the chart’s recommendations more closely. On the other hand, if many high-value cards have been played, hitting more aggressively might be advantageous.
Additionally, it is crucial to manage bankroll effectively when utilizing the hit chart. Players should set a budget and stick to it to avoid overspending.
It is also recommended to set win and loss limits to maintain discipline and prevent significant financial losses. By managing their bankroll wisely, players can ensure they can keep playing and utilize the hit chart to its full potential.
Lastly, players should practice patience and not get discouraged by short-term losses. Blackjack is a game of probabilities, and even with the best strategies, losses are inevitable at times. It is essential to trust in the hit chart and the principles of basic blackjack strategy, as they provide the best statistical advantage over the long run.
In conclusion, effectively using the Blackjack Hit Chart entails more than just adhering to its suggestions; it also involves incorporating additional strategies and techniques. Understanding soft and hard hands, considering the dealer’s upcard, timing decisions based on the game flow, managing bankroll effectively, and practicing patience are all vital aspects of utilizing the hit chart to its full potential. By incorporating these strategies, players can enhance their chances of winning and make the most out of their blackjack experience.
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People play slot games for various reasons, as these games offer a unique combination of entertainment, simplicity, and the chance to win prizes. | 2,485 | 12,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.942958 |
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###### The average of 9 observations was 9, that of the 1st of 5 being 10 and that of the last 5 being 8. What was the 5th observation? A. 9 B. 10 C. 11 D. 12
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###### Find the breadth and perimeter of the rectangle if its area is 96 cm22 and the length is 12 cm. A. 20 cm B. 40 cm C. 38 cm D. 24 cm
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###### How many times are the hands of a clock at right angles in a day? A. 24 B. 34 C. 54 D. 44
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###### 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. Find the diameter of each sphere. A. 2 cm B. 4 cm C. 6 cm D. 8 cm
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###### A sum of Rs. 1360 has been divided among A, B and C such that A gets of what B gets and B gets of what C gets. B’s share is: A. Rs. 140 B. Rs. 240 C. Rs. 340 D. Rs. 440
517 views Changed status to publish | 318 | 1,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-06 | latest | en | 0.916953 |
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# Find the point on the line $y = 2x + 3$ that is closest to the origin.
## $$\left(-\frac{6}{5}, \frac{3}{5}\right)$$
Derivatives
Differentiation
Volume
### Discussion
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##### Heather Z.
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Michael J.
Idaho State University
Lectures
Join Bootcamp
### Video Transcript
Okay, let's use the distance formula. Shoe expose three. The minus zero doesn't affect the problem. Plus X minus zero squared is just plus X squared. This gives us the square root five X squared plus 12 ax plus nine. So now we have five x squared plus 12 ax plus nine equals. Why therefore wide prime is equivalent to 10 Expose 12. We're just taking the derivative. So that's equal to zero solve for acts and we get access equivalent to night of six divided by five. Which means that we have why equals two times negative six over five plus three, which is three divided by five. So we have negative 6/5 Karmah 3/5 or negative 1.2 comma 0.6
#### Topics
Derivatives
Differentiation
Volume
##### Heather Z.
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Michael J.
Idaho State University
Lectures
Join Bootcamp | 343 | 1,362 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-39 | longest | en | 0.859777 |
https://www.in2013dollars.com/uk/inflation/1844?amount=1 | 1,618,153,945,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00119.warc.gz | 936,163,684 | 17,638 | # £1 in 1844 is worth £127.89 today
£
## Value of £1 from 1844 to 2020
£1 in 1844 is equivalent in purchasing power to about £127.89 today, an increase of £126.89 over 176 years. The pound had an average inflation rate of 2.79% per year between 1844 and today, producing a cumulative price increase of 12,689.25%.
This means that today's prices are 127.89 times higher than average prices since 1844, according to the Office for National Statistics composite price index. A pound today only buys 0.78% of what it could buy back then.
The 1844 inflation rate was 0.00%. The current year-over-year inflation rate (2019 to 2020) is now 1.50%1. If this number holds, £1 today will be equivalent in buying power to £1.01 next year.
Cumulative price change 12,689.25% Average inflation rate 2.79% Converted amount (£1 base) £127.89 Price difference (£1 base) £126.89 CPI in 1844 8.900 CPI in 2020 1,138.243 Inflation in 1844 0.00% Inflation in 2020 1.50% £1 in 1844 £127.89 in 2020
GBP Inflation since 1750
Annual Rate, the Office for National Statistics CPI
## Buying power of £1 in 1844
This chart shows a calculation of buying power equivalence for £1 in 1844 (price index tracking began in 1750).
For example, if you started with £1, you would need to end with £127.89 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When £1 is equivalent to £127.89 over time, that means that the "real value" of a single U.K. pound decreases over time. In other words, a pound will pay for fewer items at the store.
This effect explains how inflation erodes the value of a pound over time. By calculating the value in 1844 dollars, the chart below shows how £1 is worth less over 176 years.
According to the Office for National Statistics, each of these GBP amounts below is equal in terms of what it could buy at the time:
Pound inflation: 1844-2020
Year Pound Value Inflation Rate
1844 £1.00 -
1845 £1.04 4.49%
1846 £1.09 4.30%
1847 £1.22 12.37%
1848 £1.07 -12.84%
1849 £1.00 -6.32%
1850 £0.94 -5.62%
1851 £0.91 -3.57%
1852 £0.91 0.00%
1853 £1.00 9.88%
1854 £1.15 14.61%
1855 £1.18 2.94%
1856 £1.18 0.00%
1857 £1.12 -4.76%
1858 £1.02 -9.00%
1859 £1.01 -1.10%
1860 £1.04 3.33%
1861 £1.07 2.15%
1862 £1.04 -2.11%
1863 £1.01 -3.23%
1864 £1.00 -1.11%
1865 £1.01 1.12%
1866 £1.07 5.56%
1867 £1.13 6.32%
1868 £1.12 -0.99%
1869 £1.07 -5.00%
1870 £1.07 0.00%
1871 £1.08 1.05%
1872 £1.12 4.17%
1873 £1.17 4.00%
1874 £1.12 -3.85%
1875 £1.10 -2.00%
1876 £1.10 0.00%
1877 £1.09 -1.02%
1878 £1.07 -2.06%
1879 £1.02 -4.21%
1880 £1.06 3.30%
1881 £1.04 -1.06%
1882 £1.06 1.08%
1883 £1.04 -1.06%
1884 £1.02 -2.15%
1885 £0.99 -3.30%
1886 £0.98 -1.14%
1887 £0.97 -1.15%
1888 £0.98 1.16%
1889 £0.99 1.15%
1890 £0.99 0.00%
1891 £1.00 1.14%
1892 £1.00 0.00%
1893 £0.99 -1.12%
1894 £0.98 -1.14%
1895 £0.97 -1.15%
1896 £0.96 -1.16%
1897 £0.98 2.35%
1898 £0.98 0.00%
1899 £0.99 1.15%
1900 £1.03 4.55%
1901 £1.03 0.00%
1902 £1.03 0.00%
1903 £1.04 1.09%
1904 £1.04 0.00%
1905 £1.04 0.00%
1906 £1.04 0.00%
1907 £1.06 1.08%
1908 £1.06 0.00%
1909 £1.07 1.06%
1910 £1.08 1.05%
1911 £1.08 0.00%
1912 £1.11 3.13%
1913 £1.10 -1.01%
1914 £1.10 0.00%
1915 £1.24 12.24%
1916 £1.46 18.18%
1917 £1.83 25.38%
1918 £2.24 22.09%
1919 £2.46 10.05%
1920 £2.84 15.53%
1921 £2.60 -8.70%
1922 £2.24 -13.85%
1923 £2.10 -6.03%
1924 £2.09 -0.53%
1925 £2.09 0.00%
1926 £2.08 -0.54%
1927 £2.02 -2.70%
1928 £2.02 0.00%
1929 £2.00 -1.11%
1930 £1.94 -2.81%
1931 £1.87 -4.05%
1932 £1.82 -2.41%
1933 £1.78 -2.47%
1934 £1.78 0.00%
1935 £1.79 0.63%
1936 £1.80 0.63%
1937 £1.87 3.75%
1938 £1.89 1.20%
1939 £1.94 2.98%
1940 £2.27 16.76%
1941 £2.52 10.89%
1942 £2.70 7.14%
1943 £2.79 3.33%
1944 £2.87 2.82%
1945 £2.94 2.75%
1946 £3.03 3.05%
1947 £3.25 7.04%
1948 £3.49 7.61%
1949 £3.60 2.89%
1950 £3.71 3.13%
1951 £4.04 9.09%
1952 £4.42 9.17%
1953 £4.55 3.05%
1954 £4.64 1.98%
1955 £4.84 4.36%
1956 £5.09 5.10%
1957 £5.27 3.53%
1958 £5.44 3.20%
1959 £5.46 0.41%
1960 £5.52 1.03%
1961 £5.71 3.46%
1962 £5.96 4.33%
1963 £6.07 1.89%
1964 £6.27 3.33%
1965 £6.56 4.66%
1966 £6.82 3.94%
1967 £7.00 2.64%
1968 £7.33 4.65%
1969 £7.72 5.37%
1970 £8.21 6.40%
1971 £8.99 9.44%
1972 £9.63 7.13%
1973 £10.51 9.10%
1974 £12.19 16.04%
1975 £15.15 24.24%
1976 £17.65 16.54%
1977 £20.45 15.85%
1978 £22.15 8.30%
1979 £25.11 13.39%
1980 £29.63 17.99%
1981 £33.15 11.87%
1982 £36.00 8.61%
1983 £37.65 4.59%
1984 £39.53 4.98%
1985 £41.93 6.08%
1986 £43.36 3.40%
1987 £45.17 4.17%
1988 £47.38 4.90%
1989 £51.07 7.78%
1990 £55.90 9.46%
1991 £59.18 5.87%
1992 £61.39 3.74%
1993 £62.37 1.59%
1994 £63.88 2.41%
1995 £66.09 3.47%
1996 £67.69 2.41%
1997 £69.81 3.14%
1998 £72.20 3.43%
1999 £73.31 1.54%
2000 £75.48 2.96%
2001 £76.82 1.77%
2002 £78.10 1.67%
2003 £80.36 2.89%
2004 £82.75 2.98%
2005 £85.09 2.82%
2006 £87.81 3.20%
2007 £91.57 4.29%
2008 £95.22 3.99%
2009 £94.72 -0.53%
2010 £99.09 4.61%
2011 £104.25 5.20%
2012 £107.60 3.21%
2013 £110.87 3.04%
2014 £113.48 2.36%
2015 £114.61 0.99%
2016 £116.60 1.74%
2017 £120.78 3.58%
2018 £123.77 2.48%
2019 £126.00 1.80%
2020 £127.89 1.50%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
Click to show 170 more rows
This conversion table shows various other 1844 amounts in today's pounds, based on the 12,689.25% change in prices:
Conversion: 1844 pounds today
Initial value Equivalent value
£1 pound in 1844 £127.89 pounds today
£5 pounds in 1844 £639.46 pounds today
£10 pounds in 1844 £1,278.93 pounds today
£50 pounds in 1844 £6,394.63 pounds today
£100 pounds in 1844 £12,789.25 pounds today
£500 pounds in 1844 £63,946.26 pounds today
£1,000 pounds in 1844 £127,892.51 pounds today
£5,000 pounds in 1844 £639,462.56 pounds today
£10,000 pounds in 1844 £1,278,925.12 pounds today
£50,000 pounds in 1844 £6,394,625.62 pounds today
£100,000 pounds in 1844 £12,789,251.24 pounds today
£500,000 pounds in 1844 £63,946,256.22 pounds today
£1,000,000 pounds in 1844 £127,892,512.44 pounds today
## How to Calculate Inflation Rate for £1 since 1844
Our calculations use the following inflation rate formula to calculate the change in value between 1844 and today:
CPI today CPI in 1844
×
1844 GBP value
=
Today's value
Then plug in historical CPI values. The U.K. CPI was 8.9 in the year 1844 and 1138.2433607545 in 2020:
1138.24336075458.9
×
£1
=
£127.89
£1 in 1844 has the same "purchasing power" or "buying power" as £127.89 in 2020.
To get the total inflation rate for the 176 years between 1844 and 2020, we use the following formula:
CPI in 2020 - CPI in 1844CPI in 1844
×
100
=
Cumulative inflation rate (176 years)
Plugging in the values to this equation, we get:
1138.2433607545 - 8.98.9
×
100
=
12,689%
Politics and news often influence economic performance. Here's what was happening at the time:
• The World's first telegraph message containing "What hath God wrought", is sent by Samuel Morse.
• Discovery of Lake Tahoe in the US by a European, called John C. Fremont.
• Haiti gains its Independence from the Dominican Republic, now celebrated as their National Day.
## Data Source & Citation
Raw data for these calculations comes from the composite price index published by the UK Office for National Statistics (ONS). A composite index is created by combining price data from several different published sources, both official and unofficial. The Consumer Price Index, normally used to compute inflation, has only been tracked since 1988. All inflation calculations after 1988 use the Office for National Statistics' Consumer Price Index, except for 2017, which is based on The Bank of England's forecast.
You may use the following MLA citation for this page: “£1 in 1844 → 2020 | UK Inflation Calculator.” Official Inflation Data, Alioth Finance, 11 Apr. 2021, https://www.officialdata.org/uk/inflation/1844?amount=1.
Special thanks to QuickChart for their chart image API, which is used for chart downloads.
in2013dollars.com is a reference website maintained by the Official Data Foundation. | 3,289 | 7,982 | {"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-2021-17 | longest | en | 0.935361 |
https://www.mersenneforum.org/showthread.php?s=cfb5ec838a83959e5049bbaab5faac59&t=22644&goto=nextoldest | 1,600,565,718,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00348.warc.gz | 977,412,642 | 11,111 | mersenneforum.org Searching for generalized repunit PRP
Register FAQ Search Today's Posts Mark Forums Read
2016-12-06, 14:52 #1
sweety439
Nov 2016
1000101101112 Posts
Searching for generalized repunit PRP
Is there a project to search the (probable) primes of the form (b^n-1)/(b-1) (generalized repunit base b) or (b^n+1)/(b+1) (generalized repunit base -b)? There are links to this, http://www.fermatquotient.com/PrimSerien/GenRepu.txt (base b for 2<=b<=152)
http://www.fermatquotient.com/PrimSerien/GenRepuP.txt (base -b for 2<=b<=152)
These are text files I searched the repunit PRP for b up to +-257 and n up to 10000.
Attached Files
repunit base 2 to 257.txt (6.2 KB, 130 views) repunit base -2 to -257.txt (6.3 KB, 145 views)
2016-12-06, 14:58 #2 sweety439 Nov 2016 23·97 Posts Some bases have algebra factors. These bases are numbers of the form k^r with integer k != 1, 0, -1 and integer r > 1, and numbers of the form -4k^4 with integer k > 0. The list of all such bases are ..., -1024, -1000, -729, -512, -343, -243, -216, -125, -64, -32, -27, -8, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, ... and ..., -5184, -2500, -1024, -324, -64, -4.
2016-12-06, 15:03 #3 sweety439 Nov 2016 8B716 Posts Also, the link is the searching for primes of the form (a+1)^n-a^n: http://www.fermatquotient.com/PrimSerien/PrimPot.txt. More generally, is there a project to search for primes of the forms (a^n-b^n)/(a-b) with a>1, -a
2016-12-06, 15:38 #4 paulunderwood Sep 2002 Database er0rr 2×1,697 Posts Some are cofactors... (a^p-1)/n (a^p+1)/n (a^p-b^p)/n (a^p+b^p)/n
2016-12-07, 01:35 #5 carpetpool "Sam" Nov 2016 311 Posts I don't think there is a project (just yet) but I am starting my own (I don't know if you'd find it interesting or not) for find PRP factors of numbers of the form (b^n-1)/(b-1) or (a+1)^n-a^n. As far as I know for both projects, All prime bases b < 10000 (for (b^n-1)/(b-1)) are tested to 20k digits. PRP factors found and tested. FYI All bases a < 100 (for (a+1)^n-a^n) are tested to 20k digits. PRP factors found and tested. Hope this helps. Check out here. Last fiddled with by carpetpool on 2016-12-07 at 02:08
2016-12-07, 11:26 #6 sweety439 Nov 2016 42678 Posts Before, I used factordb.com to find all repunit PRPs with 152<=b<=257 and -257<=b<=-152 and n<=10000. Since in http://www.fermatquotient.com/, it was already searched for 2<=b<=151 and -151<=b<=-2.
2016-12-07, 11:35 #7
sweety439
Nov 2016
23·97 Posts
These are files that I found the smallest odd prime p such that (b^p-1)/(b-1) or (b^p+1)/(b+1) is (probable) prime for 2<=b<=1025, and the smallest prime p such that (b+1)^p-b^p is (probable) prime for 1<=b<=1024. (not all (probable) primes are founded by me)
Attached Files
least odd prime p such that (n^p-1)(n-1) is prime.txt (8.7 KB, 159 views) least odd prime p such that (n^p+1)(n+1) is prime.txt (8.1 KB, 79 views) least prime p such that (n+1)^p-n^p is prime.txt (9.9 KB, 94 views)
2016-12-07, 11:46 #8
sweety439
Nov 2016
23·97 Posts
Also, I have searched the smallest odd prime p such that (a^p-b^p)/(a-b) is prime for 1<a<=50, -a<b<a, and a and b are coprime. (for (a^p+b^p)/(a+b), note that since p is odd, it equals (a^p-(-b)^p)/(a-(-b)))
All of the "NA" terms were searched to at least p=10000. (0 if no possible prime)
Attached Files
least odd prime p such that (a^p-b^p)(a-b) is prime.txt (17.5 KB, 92 views)
Last fiddled with by sweety439 on 2016-12-07 at 11:48
2016-12-07, 16:03 #9 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22×2,281 Posts 'Those who don't know history are doomed to repeat it.' (George Santayana) Would you please stop reposting in every possible forum thread trivial results that are known for a century? Sadly you forgot to post in http://mersenneforum.org/forumdisplay.php?f=24 that for 1
2016-12-07, 18:12 #10 sweety439 Nov 2016 1000101101112 Posts No, it is (a^p-b^p)/(a-b), not 2^a-1. This a is the base, not the exponent. The file in that post lists the smallest odd prime p such that (a^p-b^p)/(a-b) is prime. (for 1
2016-12-10, 00:46 #11 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22·2,281 Posts (49^16747+46^16747)/95 is a PRP ... Too small for PRPtop - the limit was raised to 30000 digits. (46^45281+11^45281)/57 is a PRP Btw, I wonder if you know how to sieve them properly. Or is it "up to the first 10000 primes"? (Hint: most of the small primes will not divide any of these candidates.) Last fiddled with by Batalov on 2016-12-10 at 22:40
Similar Threads Thread Thread Starter Forum Replies Last Post T.Rex Wagstaff PRP Search 190 2020-07-13 21:44 Batalov And now for something completely different 10 2019-09-12 13:31 Bob Underwood Math 11 2017-01-25 11:19 sweety439 And now for something completely different 1 2016-12-07 15:58 jasong Sierpinski/Riesel Base 5 2 2006-06-07 20:37
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Sun Sep 20 01:35:18 UTC 2020 up 9 days, 22:46, 0 users, load averages: 1.41, 1.31, 1.25 | 1,924 | 5,119 | {"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-2020-40 | latest | en | 0.774426 |
http://washingtoncountyrepublicans.com/algebra-2-polynomials/ | 1,550,914,090,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249495888.71/warc/CC-MAIN-20190223082039-20190223104039-00557.warc.gz | 288,023,361 | 10,499 | # 40algebra 2 Polynomials
40algebra 2 Polynomials
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What to do How to do it Factor general polynomials ï Given a special binomial via slideplayer.com | 704 | 3,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-09 | latest | en | 0.856809 |
https://www.airmilescalculator.com/distance/hsv-to-blf/ | 1,606,903,284,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00632.warc.gz | 543,177,575 | 26,447 | # Distance between Huntsville, AL (HSV) and Bluefield, WV (BLF)
Flight distance from Huntsville to Bluefield (Huntsville International Airport – Mercer County Airport (West Virginia)) is 362 miles / 582 kilometers / 314 nautical miles. Estimated flight time is 1 hour 11 minutes.
Driving distance from Huntsville (HSV) to Bluefield (BLF) is 432 miles / 696 kilometers and travel time by car is about 8 hours 6 minutes.
## Map of flight path and driving directions from Huntsville to Bluefield.
Shortest flight path between Huntsville International Airport (HSV) and Mercer County Airport (West Virginia) (BLF).
## How far is Bluefield from Huntsville?
There are several ways to calculate distances between Huntsville and Bluefield. Here are two common methods:
Vincenty's formula (applied above)
• 361.802 miles
• 582.263 kilometers
• 314.397 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 361.389 miles
• 581.600 kilometers
• 314.039 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Huntsville International Airport
City: Huntsville, AL
Country: United States
IATA Code: HSV
ICAO Code: KHSV
Coordinates: 34°38′13″N, 86°46′30″W
B Mercer County Airport (West Virginia)
City: Bluefield, WV
Country: United States
IATA Code: BLF
ICAO Code: KBLF
Coordinates: 37°17′44″N, 81°12′27″W
## Time difference and current local times
The time difference between Huntsville and Bluefield is 1 hour. Bluefield is 1 hour ahead of Huntsville.
CST
EST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 78 kg (173 pounds).
## Frequent Flyer Miles Calculator
Huntsville (HSV) → Bluefield (BLF).
Distance:
362
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
362
Round trip? | 509 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-50 | latest | en | 0.802888 |
https://www.jwfsanctuary.club/tutorial/creatinganewflame/ | 1,627,526,264,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00024.warc.gz | 851,303,502 | 60,717 | # Start from Scratch – Creating a New Flame
Home » Tutorial » Start from Scratch - Creating a New Flame
A flame in JWildfire is the set of variations and transforms you’ve used to create your design.
We’re going to go through some steps here to make a very simple fractal.
## 1 New From Scratch
After running JWildfire you need to click New From Scratch to begin our fractal (your screen might be differently coloured to this)
Without getting technical, a transform is the first instruction, it will contain one or more variations. A variation is what alters the method of creating the fractal. By default when you click on Add it will put a Linear3D variation, so let’s do that.
You’ll see on your screen, a bunch of dots, and a triangle. The triangle is our control for the variation we just added. Don’t worry about that just now, let’s just add another transform so we can start to play.
Click Add again, and it will place another transform in there, again, by default it will be a Linear3D variation.
## 4 Moving the triangles (affines)
Click and hold your mouse button on a triangle and move it about the screen
## 5) Controlling the triangles
The triangle’s position, size, shape and rotation can all be controlled, here are the buttons that do that. Click one then move the mouse, or in the case of the triangle shape, click a point on the triangle and move it about.
## 6) Changing the Variation
This is where you’ll see the differences. So far we have two Linear3D variations on our transforms, let’s change the second one.
Click the NonLinear tab, then click the dropdown after Linear 3D, this will give you a list of variations to choose from. We’ll choose Hypertile1
## 7) A sort of interesting design, once you’ve done that you’ll get something like this.
Play with the controls, and try different variations. Don’t worry if some of them don’t give you anything, you’ll probably need to add another transform again so you have three. Try changing the variations on each transform. | 452 | 2,012 | {"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-2021-31 | longest | en | 0.882533 |
https://www.bartleby.com/solution-answer/chapter-7-problem-74bpe-accounting-27th-edition/9781337272094/perpetual-inventory-using-weighted-average-beginning-inventory-purchases-and-sales-for-wcs12-are/aa52224f-98de-11e8-ada4-0ee91056875a | 1,603,549,624,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00254.warc.gz | 624,122,811 | 69,481 | # Perpetual inventory using weighted average Beginning inventory, purchases, and sales for WCS12 are as follows: Oct. 1 Inventory 300 units at $8 13 Sale 175 units 22 Purchase 375 units at$10 29 Sale 280 units Assuming a perpetual inventory system and using the weighted average method, determine (a) the weighted average unit cost after the October 22 purchase, (b) the cost of the merchandise sold on October 29, and (c) the inventory on October 31.
### Accounting
27th Edition
WARREN + 5 others
Publisher: Cengage Learning,
ISBN: 9781337272094
### Accounting
27th Edition
WARREN + 5 others
Publisher: Cengage Learning,
ISBN: 9781337272094
#### Solutions
Chapter
Section
Chapter 7, Problem 7.4BPE
Textbook Problem
## Perpetual inventory using weighted averageBeginning inventory, purchases, and sales for WCS12 are as follows: Oct. 1 Inventory 300 units at $8 13 Sale 175 units 22 Purchase 375 units at$10 29 Sale 280 units Assuming a perpetual inventory system and using the weighted average method, determine (a) the weighted average unit cost after the October 22 purchase, (b) the cost of the merchandise sold on October 29, and (c) the inventory on October 31.
Expert Solution
(a)
To determine
Perpetual Inventory System:
Perpetual Inventory System refers to the inventory system that maintains the detailed records of every inventory transactions related to purchases, and sales on a continuous basis. It shows the exact on-hand-inventory at any point of time.
To determine: the weighted average unit cost after October 22 purchases.
### Explanation of Solution
Calculate the weighted average unit cost as follows:
Weighted average method Quantity Unit cost Total cost of inventory in hand 125 units $8$1,000 375 units $10$3,750 500 units \$4,750
Expert Solution
(b)&(c)
To determine
cost of merchandise sold for each sale and inventory balance after each sale of WCS12 as on October 31.
### Want to see the full answer?
Check out a sample textbook solution.See solution
### Want to see this answer and more?
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
See solution | 505 | 2,172 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-45 | latest | en | 0.856969 |
https://www.coursehero.com/file/5569781/ech203264-exam1-sp08/ | 1,496,153,020,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615105.83/warc/CC-MAIN-20170530124450-20170530144450-00146.warc.gz | 1,082,420,524 | 31,456 | ech%203264_exam1_sp08
# ech%203264_exam1_sp08 - ECH 3264 Exam#1 Elementary...
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1 ECH 3264: Elementary Transport Phenomena Exam #1 Feb 1, 2008 Name: Closed book and notes. No calculators. You are allowed one formula sheet (front and back). Start each problem on a new page. DO NOT use the back of a page. Write clearly your assumptions and any parameter definitions. Show your work to get maximum credit. It is helpful to the grader (and to you) if you box intermediate and final answers. Useful information: Relevant volume and surface areas: rL SA r SA L r V r V cylinder sphere cylinder sphere π 2 4 3 4 2 2 3 = = = = 1. (15 pts) Qualitative Questions Consider each of the steady-state temperature profiles given below. Are the profiles physically achievable? If so, which material (left or right) has the larger thermal conductivity? If not, give a reason that the profile cannot be achieved. Note: the slab of material on the left and right have the same dimensions (i.e. length, cross sectional area). There is no heat generation.
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## This note was uploaded on 08/30/2009 for the course ECH 3264 taught by Professor Asthagiri during the Spring '09 term at University of Florida.
### Page1 / 2
ech%203264_exam1_sp08 - ECH 3264 Exam#1 Elementary...
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from Quasimodo
Related concepts
Parents numberWeight: 0.73, real numberWeight: 0.63, problemWeight: 0.62, objectWeight: 0.55 Siblings irrational numberWeight: 0.82, integerWeight: 0.60, complex numberWeight: 0.38, imaginary numberWeight: 0.38, social security numberWeight: 0.37
Related properties
Similarity Property is real numbers 1.00 set of real numbers 0.93 is numbers 0.92 be different from numbers 0.87 be different than numbers 0.86 is real 0.85 set of numbers 0.83 be greater than numbers 0.82 be greater than negative numbers 0.81 be deeper than numbers 0.79
Clauses
Plausibility inference from child typicality
0.60
Rule weight: 0.66
Evidence weight: 0.98
Similarity weight: 0.92
Evidence: 0.33
Plausible(number, is numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
0.57
Rule weight: 0.66
Evidence weight: 0.99
Similarity weight: 0.87
Evidence: 0.48
Plausible(number, be different from numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
0.57
Rule weight: 0.66
Evidence weight: 0.99
Similarity weight: 0.86
Evidence: 0.52
Plausible(number, be different than numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
0.56
Rule weight: 0.66
Evidence weight: 0.98
Similarity weight: 0.85
Evidence: 0.36
Plausible(number, is real)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
0.54
Rule weight: 0.66
Evidence weight: 0.98
Similarity weight: 0.83
Evidence: 0.40
Plausible(number, set of numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
0.54
Rule weight: 0.66
Evidence weight: 0.98
Similarity weight: 0.82
Evidence: 0.41
Plausible(number, be greater than numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
0.52
Rule weight: 0.66
Evidence weight: 0.99
Similarity weight: 0.79
Evidence: 0.51
Plausible(number, be deeper than numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
Plausibility inheritance from parent to child
0.06
Rule weight: 0.09
Evidence weight: 0.73
Similarity weight: 0.92
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.33
¬ Plausible(number, is numbers)
0.06
Rule weight: 0.09
Evidence weight: 0.70
Similarity weight: 0.85
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.36
¬ Plausible(number, is real)
0.05
Rule weight: 0.09
Evidence weight: 0.67
Similarity weight: 0.83
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.40
¬ Plausible(number, set of numbers)
0.05
Rule weight: 0.09
Evidence weight: 0.66
Similarity weight: 0.82
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.41
¬ Plausible(number, be greater than numbers)
0.05
Rule weight: 0.09
Evidence weight: 0.60
Similarity weight: 0.87
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.48
¬ Plausible(number, be different from numbers)
0.05
Rule weight: 0.09
Evidence weight: 0.57
Similarity weight: 0.86
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.52
¬ Plausible(number, be different than numbers)
0.04
Rule weight: 0.09
Evidence weight: 0.58
Similarity weight: 0.79
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.51
¬ Plausible(number, be deeper than numbers)
Remarkability exclusitivity betweem a parent and a child
0.36
Rule weight: 0.58
Evidence weight: 0.73
Similarity weight: 0.85
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.29
¬ Remarkable(number, is real)
0.11
Rule weight: 0.58
Evidence weight: 0.22
Similarity weight: 0.87
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.85
¬ Remarkable(number, be different from numbers)
0.08
Rule weight: 0.58
Evidence weight: 0.17
Similarity weight: 0.79
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.90
¬ Remarkable(number, be deeper than numbers)
0.08
Rule weight: 0.58
Evidence weight: 0.15
Similarity weight: 0.86
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.92
¬ Remarkable(number, be different than numbers)
0.05
Rule weight: 0.58
Evidence weight: 0.10
Similarity weight: 0.83
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.98
¬ Remarkable(number, set of numbers)
0.04
Rule weight: 0.58
Evidence weight: 0.08
Similarity weight: 0.92
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 1.00
¬ Remarkable(number, is numbers)
0.04
Rule weight: 0.58
Evidence weight: 0.09
Similarity weight: 0.82
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.99
¬ Remarkable(number, be greater than numbers)
Remarkability exclusitivity between siblings
0.13
Rule weight: 0.13
Evidence weight: 0.95
Similarity weight: 1.00
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.05
¬ Remarkable(irrational number, is real numbers)
0.11
Rule weight: 0.13
Evidence weight: 0.96
Similarity weight: 0.85
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.04
¬ Remarkable(irrational number, is real)
0.10
Rule weight: 0.13
Evidence weight: 0.73
Similarity weight: 1.00
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.30
¬ Remarkable(integer, is real numbers)
0.09
Rule weight: 0.13
Evidence weight: 0.82
Similarity weight: 0.87
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.20
¬ Remarkable(irrational number, be different from numbers)
0.06
Rule weight: 0.13
Evidence weight: 0.54
Similarity weight: 0.87
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.51
¬ Remarkable(integer, be different from numbers)
0.03
Rule weight: 0.13
Evidence weight: 0.24
Similarity weight: 0.92
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.83
¬ Remarkable(integer, is numbers)
0.01
Rule weight: 0.13
Evidence weight: 0.12
Similarity weight: 0.92
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.96
¬ Remarkable(irrational number, is numbers)
Remarkability from parent implausibility
0.38
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.92
Evidence: 0.33
Plausible(number, is numbers)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
0.36
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.87
Evidence: 0.48
Plausible(number, be different from numbers)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
0.36
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.86
Evidence: 0.52
Plausible(number, be different than numbers)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
0.35
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.85
Evidence: 0.36
Plausible(number, is real)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
0.35
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.83
Evidence: 0.40
Plausible(number, set of numbers)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
0.34
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.82
Evidence: 0.41
Plausible(number, be greater than numbers)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
0.33
Rule weight: 0.42
Evidence weight: 0.99
Similarity weight: 0.79
Evidence: 0.51
Plausible(number, be deeper than numbers)
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.17
¬ Plausible(rational number, is real numbers)
Remarkability from sibling implausibility
0.59
Rule weight: 0.60
Evidence weight: 0.98
Similarity weight: 1.00
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.05
Remarkable(irrational number, is real numbers)
Evidence: 0.03
¬ Plausible(irrational number, is real numbers)
0.55
Rule weight: 0.60
Evidence weight: 0.99
Similarity weight: 0.92
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.96
Remarkable(irrational number, is numbers)
Evidence: 0.16
¬ Plausible(irrational number, is numbers)
0.53
Rule weight: 0.60
Evidence weight: 0.96
Similarity weight: 0.92
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.83
Remarkable(integer, is numbers)
Evidence: 0.28
¬ Plausible(integer, is numbers)
0.50
Rule weight: 0.60
Evidence weight: 0.97
Similarity weight: 0.87
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.20
Remarkable(irrational number, be different from numbers)
Evidence: 0.05
¬ Plausible(irrational number, be different from numbers)
0.50
Rule weight: 0.60
Evidence weight: 0.97
Similarity weight: 0.85
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.04
Remarkable(irrational number, is real)
Evidence: 0.03
¬ Plausible(irrational number, is real)
0.44
Rule weight: 0.60
Evidence weight: 0.84
Similarity weight: 0.87
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.51
Remarkable(integer, be different from numbers)
Evidence: 0.39
¬ Plausible(integer, be different from numbers)
0.41
Rule weight: 0.60
Evidence weight: 0.68
Similarity weight: 1.00
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.30
Remarkable(integer, is real numbers)
Evidence: 0.54
¬ Plausible(integer, is real numbers)
Salient implies Plausible
0.19
Rule weight: 0.28
Evidence weight: 0.69
Similarity weight: 1.00
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.38
¬ Salient(rational number, is real numbers)
Similarity expansion
0.78
Rule weight: 0.85
Evidence weight: 1.00
Similarity weight: 0.92
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.00
¬ Typical(rational number, is numbers)
0.74
Rule weight: 0.85
Evidence weight: 0.93
Similarity weight: 0.93
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.07
¬ Typical(rational number, set of real numbers)
0.74
Rule weight: 0.85
Evidence weight: 1.00
Similarity weight: 0.87
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.03
¬ Remarkable(rational number, be different from numbers)
0.74
Rule weight: 0.85
Evidence weight: 0.93
Similarity weight: 0.93
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.87
¬ Remarkable(rational number, set of real numbers)
0.73
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.87
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.01
¬ Plausible(rational number, be different from numbers)
0.73
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.87
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.02
¬ Salient(rational number, be different from numbers)
0.73
Rule weight: 0.85
Evidence weight: 0.99
Similarity weight: 0.86
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.09
¬ Remarkable(rational number, be different than numbers)
0.72
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.92
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.99
¬ Remarkable(rational number, is numbers)
0.72
Rule weight: 0.85
Evidence weight: 0.97
Similarity weight: 0.86
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.03
¬ Plausible(rational number, be different than numbers)
0.71
Rule weight: 0.85
Evidence weight: 0.97
Similarity weight: 0.86
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.05
¬ Salient(rational number, be different than numbers)
0.69
Rule weight: 0.85
Evidence weight: 0.98
Similarity weight: 0.83
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.02
¬ Typical(rational number, set of numbers)
0.68
Rule weight: 0.85
Evidence weight: 0.94
Similarity weight: 0.85
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.69
¬ Remarkable(rational number, is real)
0.68
Rule weight: 0.85
Evidence weight: 0.87
Similarity weight: 0.92
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.16
¬ Plausible(rational number, is numbers)
0.68
Rule weight: 0.85
Evidence weight: 0.97
Similarity weight: 0.82
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.03
¬ Typical(rational number, be greater than numbers)
0.66
Rule weight: 0.85
Evidence weight: 0.97
Similarity weight: 0.81
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.41
¬ Remarkable(rational number, be greater than negative numbers)
0.65
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.83
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.97
¬ Remarkable(rational number, set of numbers)
0.65
Rule weight: 0.85
Evidence weight: 0.92
Similarity weight: 0.82
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.96
¬ Remarkable(rational number, be greater than numbers)
0.64
Rule weight: 0.85
Evidence weight: 0.81
Similarity weight: 0.93
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.23
¬ Plausible(rational number, set of real numbers)
0.64
Rule weight: 0.85
Evidence weight: 0.87
Similarity weight: 0.87
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.14
¬ Typical(rational number, be different from numbers)
0.63
Rule weight: 0.85
Evidence weight: 0.86
Similarity weight: 0.86
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.15
¬ Typical(rational number, be different than numbers)
0.63
Rule weight: 0.85
Evidence weight: 0.93
Similarity weight: 0.79
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.07
¬ Typical(rational number, be deeper than numbers)
0.63
Rule weight: 0.85
Evidence weight: 0.93
Similarity weight: 0.79
Evidence: 0.92
Remarkable(rational number, is real numbers)
Evidence: 0.87
¬ Remarkable(rational number, be deeper than numbers)
0.60
Rule weight: 0.85
Evidence weight: 0.76
Similarity weight: 0.93
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.39
¬ Salient(rational number, set of real numbers)
0.59
Rule weight: 0.85
Evidence weight: 0.83
Similarity weight: 0.83
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.21
¬ Plausible(rational number, set of numbers)
0.58
Rule weight: 0.85
Evidence weight: 0.80
Similarity weight: 0.85
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.20
¬ Typical(rational number, is real)
0.57
Rule weight: 0.85
Evidence weight: 0.82
Similarity weight: 0.82
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.22
¬ Plausible(rational number, be greater than numbers)
0.57
Rule weight: 0.85
Evidence weight: 0.73
Similarity weight: 0.92
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.43
¬ Salient(rational number, is numbers)
0.57
Rule weight: 0.85
Evidence weight: 0.83
Similarity weight: 0.81
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.28
¬ Salient(rational number, be greater than negative numbers)
0.57
Rule weight: 0.85
Evidence weight: 0.78
Similarity weight: 0.85
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.35
¬ Salient(rational number, is real)
0.56
Rule weight: 0.85
Evidence weight: 0.77
Similarity weight: 0.85
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.27
¬ Plausible(rational number, is real)
0.55
Rule weight: 0.85
Evidence weight: 0.82
Similarity weight: 0.79
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.21
¬ Plausible(rational number, be deeper than numbers)
0.54
Rule weight: 0.85
Evidence weight: 0.78
Similarity weight: 0.81
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.26
¬ Plausible(rational number, be greater than negative numbers)
0.51
Rule weight: 0.85
Evidence weight: 0.76
Similarity weight: 0.79
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.38
¬ Salient(rational number, be deeper than numbers)
0.50
Rule weight: 0.85
Evidence weight: 0.71
Similarity weight: 0.83
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.47
¬ Salient(rational number, set of numbers)
0.50
Rule weight: 0.85
Evidence weight: 0.71
Similarity weight: 0.82
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.47
¬ Salient(rational number, be greater than numbers)
0.47
Rule weight: 0.85
Evidence weight: 0.68
Similarity weight: 0.81
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.32
¬ Typical(rational number, be greater than negative numbers)
Typical and Remarkable implies Salient
0.14
Rule weight: 0.14
Evidence weight: 0.99
Similarity weight: 1.00
Evidence: 0.38
Salient(rational number, is real numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Typical implies Plausible
0.47
Rule weight: 0.48
Evidence weight: 0.98
Similarity weight: 1.00
Evidence: 0.17
Plausible(rational number, is real numbers)
Evidence: 0.03
¬ Typical(rational number, is real numbers)
Typicality and Rermarkability incompatibility between a parent and a child
0.43
Rule weight: 0.51
Evidence weight: 0.93
Similarity weight: 0.92
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.08
¬ Typical(number, is numbers)
0.35
Rule weight: 0.51
Evidence weight: 0.83
Similarity weight: 0.82
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.19
¬ Typical(number, be greater than numbers)
0.35
Rule weight: 0.51
Evidence weight: 0.82
Similarity weight: 0.83
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.20
¬ Typical(number, set of numbers)
0.27
Rule weight: 0.51
Evidence weight: 0.62
Similarity weight: 0.86
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.41
¬ Typical(number, be different than numbers)
0.27
Rule weight: 0.51
Evidence weight: 0.60
Similarity weight: 0.87
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.44
¬ Typical(number, be different from numbers)
0.25
Rule weight: 0.51
Evidence weight: 0.61
Similarity weight: 0.79
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.42
¬ Typical(number, be deeper than numbers)
0.20
Rule weight: 0.51
Evidence weight: 0.46
Similarity weight: 0.85
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.59
¬ Typical(number, is real)
Typicality and Rermarkability incompatibility between siblings
0.12
Rule weight: 0.14
Evidence weight: 0.98
Similarity weight: 0.92
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.02
¬ Typical(irrational number, is numbers)
0.11
Rule weight: 0.14
Evidence weight: 0.82
Similarity weight: 1.00
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.19
¬ Typical(irrational number, is real numbers)
0.11
Rule weight: 0.14
Evidence weight: 0.88
Similarity weight: 0.92
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.13
¬ Typical(integer, is numbers)
0.11
Rule weight: 0.14
Evidence weight: 0.89
Similarity weight: 0.87
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.12
¬ Typical(irrational number, be different from numbers)
0.09
Rule weight: 0.14
Evidence weight: 0.79
Similarity weight: 0.85
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.23
¬ Typical(irrational number, is real)
0.07
Rule weight: 0.14
Evidence weight: 0.60
Similarity weight: 0.87
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.44
¬ Typical(integer, be different from numbers)
0.05
Rule weight: 0.14
Evidence weight: 0.33
Similarity weight: 1.00
Evidence: 0.92
¬ Remarkable(rational number, is real numbers)
Evidence: 0.73
¬ Typical(integer, is real numbers)
Typicality inheritance from parent to child
0.41
Rule weight: 0.48
Evidence weight: 0.92
Similarity weight: 0.92
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.08
¬ Typical(number, is numbers)
0.33
Rule weight: 0.48
Evidence weight: 0.82
Similarity weight: 0.82
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.19
¬ Typical(number, be greater than numbers)
0.33
Rule weight: 0.48
Evidence weight: 0.81
Similarity weight: 0.83
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.20
¬ Typical(number, set of numbers)
0.25
Rule weight: 0.48
Evidence weight: 0.60
Similarity weight: 0.86
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.41
¬ Typical(number, be different than numbers)
0.24
Rule weight: 0.48
Evidence weight: 0.57
Similarity weight: 0.87
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.44
¬ Typical(number, be different from numbers)
0.23
Rule weight: 0.48
Evidence weight: 0.59
Similarity weight: 0.79
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.42
¬ Typical(number, be deeper than numbers)
0.18
Rule weight: 0.48
Evidence weight: 0.43
Similarity weight: 0.85
Evidence: 0.03
Typical(rational number, is real numbers)
Evidence: 0.59
¬ Typical(number, is real) | 7,012 | 21,220 | {"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-2022-49 | latest | en | 0.766878 |
http://www.jiskha.com/display.cgi?id=1310596600 | 1,498,294,823,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320243.11/warc/CC-MAIN-20170624082900-20170624102900-00157.warc.gz | 559,788,288 | 3,779 | # chemistry
posted by .
calculate the molar mass of an unknown whose mass was 0.846 g, and occupied a volume of 354 cm3 at a pressure of 752 torr and a temperature of 100 degrees celcius.
I came up with .0114 moles, I want to know if this is correct.
• chemistry -
yes and no. Yes, n = number of moles = 0.0114 but that isn't the molar mass which is the question.
n = grams/molar mass will get that for you if you solve for molar mass. | 124 | 440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-26 | latest | en | 0.961907 |
https://tuitionphysics.com/2015-oct/principle-of-calorimetry/ | 1,708,774,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474533.12/warc/CC-MAIN-20240224112548-20240224142548-00148.warc.gz | 594,966,471 | 20,608 | The principle of calorimetry is to make a quantifiable measurement of the amount heat energy transferred in a system and its relation to temperature. The principle of calorimetry departs from the classical view point that heat is composed of a fluid called “caloric” which flows from a relatively hotter body to the colder body. The idea that heat energy is a measurement of change in temperature of the body was evolved at a much later time after a series of experiments were conducted using instruments called calorimeters.
It is through such experiments that we began to understand that heat energy transferred (both absorbed/released) is directly proportional to the mass of the body and the change in temperature in the body. The proportionality constant is called the specific heat of the substance measured in calories. For example, one calorie is defined as the amount of energy required to increase the temperature of one gram of water by one degree centigrade. In almost all the edible items we purchase, the amount of energy contained in the item is always mentioned. It is the amount of energy generated in the body that is often stored in a chemical form called ATP. The specific heat for water is 4.18 Jg-1K-1.
Another interesting aspect of heat transfer taught in physics tuition classes is that every exchange of heat need not necessarily cause a change in temperature of the substance. When water at 100 oC is provided further energy, one does not find a change in the temperature of the water but the water at 100 oC is converted into steam at 100 oC. This process is called phase transition. The water keeps absorbing energy while it is transformed into steam at the same temperature. This energy is called Latent heat of vaporization. This heat energy is proportional to the mass of the substance and is independent of the temperature. The proportionality constant is called the specific latent heat of the substance. In the case of water if energy is being lost, water at 0 oC is converted into ice at 0 oC. The energy required for this transition to occur is called the Latent heat of fusion and the proportionality constant is called the specific latent heat of ice.
Similarly, the measurements using the principle of calorimetry explain a lot of important phenomenon in thermodynamics. One classical phenomenon that can be explained using this theory is how the clay pots used as an everyday house item works. The clay pots use the heat energy from its surroundings to cool the contents inside. Clay pots are usually porous, the heat from outside evaporates small quantities of water through the pores. This small reduction in the mass of its contents causes the clay pot as a system to loose energy and hence cooling the water inside. It is an indigenous invention made by our ancestors before any phenomenon of heat transfer was remotely understood. A consolidated effort in understanding the phenomenon of heat transfer has also been applied in the field of architecture where open brick and mud structures are used to naturally cool the house. | 600 | 3,074 | {"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} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.960895 |
https://puzzling.stackexchange.com/questions/50228/what-am-i-i-am-a-group | 1,719,065,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00086.warc.gz | 409,637,836 | 44,163 | # What am I? I am a group
What am I? I am a group.
You can't blame anyone for I being in me.
But I might not anyway.
It depends on how you see the sixth in the third.
But I wasn't invoking Godwin's Law!
You are not in me.
She is not, either.
But the second is, if you cut it.
But not the sixtieth of a minute!
Now, what am I, specifically?
Hint:
If you see an ordinal number, read aloud which part(s) of the riddle you think is/are relevant.
Hint 2:
For example, the fourth line: the sixth (beep) in the third (beep). And watch out for red herrings.
Hint 3 (quite a big hint):
The title counts.
• You're a group. Commented Mar 22, 2017 at 2:29
• A bounty for each puzzle you post. Not sure if you have enough budget. Commented Jan 19, 2018 at 14:26
• @ibrahimmahrir I'm restoring my old questions. As you can see, I still have 222. Commented Jan 19, 2018 at 14:27
• Do we require any knowledge of set theory to solve this? Commented Jan 26, 2018 at 11:12
• @Albino What was correct was that the word "I" is 'interesting'. It's not to be interpreted as "eye". Commented Jan 31, 2018 at 13:26
Could it be the word
ITS?
What am I? I am a group.
Not sure about this. A quick web search threw up various organisations called ITS Group, but none seemed particularly famous or relevant.
You can't blame anyone for I being in me.
The letter I is contained in the word.
But it might not be the case anyway.
It depends on how you see the third.
Except for its third letter, the word is "it".
But I wasn't invoking Godwin's Law!
The third letter gives SS, a Nazi organisation.
You will never be in me
She will never, either.
Wrong pronouns! Not yours or hers, but its.
But the second is.
Except for its second letter, the word is "is".
But not the sixtieth of a minute!
Indeed, not tha kind of second.
Now, what am I, specifically?
ITS.
• I have fixed the question Commented Mar 22, 2017 at 6:07
• It's quite far away Commented Jan 28, 2018 at 12:57
Without being sure, I think about
Vowels
You can't blame anyone for I being in me.
Well, the letter "I" is a vowel!
But I might not anyway.
But it does not appear in the word "vowel".
It depends on how you see the sixth in the third.
The sixth vowel ("Y") is sometimes pronounced like the third ("I"), but sometimes as a consonant.
But I wasn't invoking Godwin's Law!
You are not in me.
The letter "U" (pronounced "You") is not in the word "vowel"
She is not, either.
But the second is, if you cut it.
The second vowel ("E") is in the word "vowel". Not sure about the "cut it" part, maybe because it is not pronounced "ee" like in the alphabet (the "ee" pronounciation like in "see")? It is possible that the second refers to the letter "B", which is almost an "O" when cut?
But not the sixtieth of a minute!
Already said by others: not that kind of second.
• Almost correct~ Commented Dec 21, 2021 at 6:25
• Do you mean that I need to work on my interpretations or to find another answer somewhat similar? Commented Dec 21, 2021 at 6:29
• Another answer. Commented Dec 21, 2021 at 6:58
My thought is:
CODEINS
You can't blame anyone for I being in me.
My deduction is there's the letter 'I' in the word after the answers the author gave in the comments. => "It depends on how you see the sixth in the third" was an hint, the third eye (see Hint 2) refers to that specific 'i' in the riddle.
You are not in me. She is not, either. But the second is
I think that the word has to be formed with the letters of "second" but doesn't contain "you" or even "she"
By combining all the letters altogether we have
we obtain: codeins or secondi
By deduction: "the second is, if you cut it." and "What am I? I am a group."
• Still quite far away Commented Feb 1, 2018 at 22:26
Group 17 in the periodic system of chemical elements, i. e. the halogens (including hydrogen for that purpose).
What am I? I am a group.
Whell, this is a group in the periodic table.
You can't blame anyone for I being in me.
"I" (iodine) is a halogen, and this is due to the atom structure, so nobody (Mendeleev, Courtois etc.) can be blamed for it.
But I might not anyway.
the word "halogens" contains no I's.
It depends on how you see the sixth in the third.
Well, it depends on how you see the "A" (6th word in the riddle) in the whole group (3rd word in the riddle is "I" which refers to it). If "A" is in the group (so this must be the word "halogens", which contains an A), so "I" is not. On the other side, "A" is not in the group itself, but "I" is.
But I wasn't invoking Godwin's Law!
Unsure. Probably a reference to the word "halogen" being invited by a German.
You are not in me.
"U" (uranium) is not a halogen.
She is not, either.
"S" (sulfur) and "He" (helium) are also not halogens.
But the second is, if you cut it.
The 2nd element (helium) is not a halogen, but can be "cut" (not physically of course, and it requires a LOT of energy) into two hydrogen atoms.
But not the sixtieth of a minute!
As pointed by others, not that kind of second.
Now, what am I, specifically?
Hydrogen and the halogens.
• Maybe Hydrogen and Halogens make "HH" which could be a reference to "Heil Hitler" (but not if we don't invoke Godwin's law)? Commented Dec 21, 2021 at 8:32
you are a TEAM.
What am I? I am a group.
Group = team.
*You can't blame anyone for I being in me.
*But I might not anyway.
It depends on how you see the sixth in the third.
There's a common phrase 'there's no "I" in "team"', which has sometimes been updated to the following:
"Sixth in third" refers to the words in the title: the sixth word is "a" and the third word is "I", so "seeing the sixth in the third" probably refers to the above memetic image.
But I wasn't invoking Godwin's Law!
You are not in me.
She is not, either.
But the second is, if you cut it.
But not the sixtieth of a minute!
"You" and "she" are also not in "team", but the second word of the title ("am") is in "team": it can be seen if you cut the word "team" in half.
Now, what am I, specifically?
The word TEAM, specifically.
• nope, not the correct answer Commented Dec 20, 2021 at 9:32 | 1,725 | 6,128 | {"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-2024-26 | latest | en | 0.955539 |
http://phys.org/news/2013-10-eyes-math-translucent-images.html | 1,397,772,039,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00412-ip-10-147-4-33.ec2.internal.warc.gz | 189,080,502 | 15,939 | # Matching eyes to math for translucent images
##### Oct 01, 2013 by Bill Steele
(Phys.org) —Whether it's a rare jade figurine or an ice sculpture, how light passes through a translucent surface is key to its appearance, and humans are sensitive to subtle differences in the result. So Cornell researchers, with colleagues at Harvard and MIT, are using the techniques of perceptual psychology to find out how to create realistic computer graphics of translucent materials. They have discovered that a previously overlooked bit of math is the key.
Their results were reported at the 2013 SIGGRAPH conference, July 21-25 in Anaheim, Calif., and were published in the August issue of ACM Transactions in Graphics.
Artists could use to guide their work on real-world objects, said Kavita Bala, associate professor of . "You could work with clay, scan it in and see if the real sculpture would look as gorgeous as you hoped," she said.
To create a computer-graphic image, an artist builds a "model" – a set of coordinates that describes the shape of the object. Then the computer calculates how light rays will bounce off that object and arrive at the eye. It's no coincidence that one of the first successful computer-animated movies was about toys; smooth plastic surfaces reflect light in a simple way.
When the surface is translucent, some light is reflected and some passes through, bouncing around for a while and then coming back out. Just how light scatters inside is what makes jade look different from ice or human skin from wax. Instead of studying the physics of light, Bala and her collaborators measured human responses to images.
Subjects were shown three images in a row and asked to choose which of those on the ends most resembled the one in the middle. This avoids widely varying subjective descriptions and generates numerical data that can be plotted against the numbers that were fed into the computer to generate the images.
In pilot studies, the researchers learned that one crucial factor in making translucent images different from one another is the "phase function," which determines the possible range of angles at which light going into the surface will be scattered. Previously, they noted, researchers have instead ignored the phase function and used simple models of the angular scattering that miss important visual features in objects with fine details. For their tests, the researchers used images of a delicate statue.
Two different parts of the function control the diffusiveness or "milkiness" and the sharpness or "glassiness" of a computer-generated image. As Bala explains it, if you lay a block of translucent material on a page with text, glassiness is how sharp the text appears, and milkiness is how much light is held back – as if you were looking at the text through milk.
But how much indeed? By asking subjects to compare and performing computer analysis of the results, the researchers were able to establish a scale showing how much a change in the values fed into the phase function would change the appearance of the image.
"If we can determine how similar or dissimilar different materials appear visually, we can build an intuitive interface for designers who edit and visualize materials," Bala explained. The goal is to build software that will allow graphic artists to "turn a knob" and watch how the image changes.
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Suddenly there was a word for chili peppers. Information about archaeological remains of ancient chili peppers in Mexico along with a study of the appearance of words for chili peppers in ancient dialects ... | 1,842 | 9,523 | {"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-2014-15 | longest | en | 0.962958 |
https://socratic.org/questions/57e949f77c014948fa890cf3#314675 | 1,716,928,539,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00288.warc.gz | 460,353,411 | 6,549 | # Question #90cf3
Sep 26, 2016
To find the roots of equations like ${e}^{x} = {x}^{3}$, I recommend that you use a recursive numerical analysis method, called Newton's Method
#### Explanation:
Let's do an example.
To use Newton's method, you write the equation in the form $f \left(x\right) = 0$:
${e}^{x} - {x}^{3} = 0$
Compute $f ' \left(x\right)$:
${e}^{x} - 3 {x}^{2}$
Because the method requires that we do the same computation many times, until it converges, I recommend that you use an Excel spreadsheet; the rest of my answer will contain instructions on how to do this.
Enter a good guess for x into cell A1. For this equation, I will enter 2.
Enter the following into cell A2:
=A1-(EXP(A1) - A1^3)/(EXP(A1) - 3*A1^2)
${x}_{2} = {x}_{1} - \frac{{e}^{{x}_{1}} - {x}_{1}^{3}}{{e}^{{x}_{1}} - 3 {x}_{1}^{2}}$
Copy the contents of cell A2 into A3 through A10. After only 3 or 4 recursions, you can see that the method has converged on
$x = 1.857184$
Sep 26, 2016
We can use the Intermediate Value Theorem to see that each pair has at least one point of intersection.
#### Explanation:
$f \left(x\right) = {e}^{x} - {x}^{2}$ is continuous on the entire real line.
At $x = 0$, we have $f \left(0\right) = 1$.
At $x = - 1$, we have $f \left(- 1\right) = \frac{1}{e} - 1$ which is negative.
$f$ is continuous on $\left[- 1 , 0\right]$, so there is at least one $c$ in $\left(- 1 , 0\right)$ with $f \left(c\right) = 0$.
$g \left(x\right) = {e}^{x} - {x}^{3}$ is continuous on the entire real line.
At $x = 0$, we have $g \left(0\right) = 1$.
At $x = 2$, we have $g \left(2\right) = {e}^{2} - 8$ which is negative.
(Note that ${e}^{2} \approx {2.7}^{2} < 7.3 < 8$.)
$g$ is continuous on $\left[0 , 2\right]$, so there is at least one $c$ in $\left(0 , 2\right)$ with $g \left(c\right) = 0$. | 657 | 1,816 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 28, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-22 | latest | en | 0.845117 |
https://www.fact-archive.com/encyclopedia/Electromotive_force | 1,696,305,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00715.warc.gz | 823,760,579 | 4,556 | Search
# Electromotive force
An electromotive force (emf) is the "force", measured in volts, that is produced by interaction between a current and a magnetic field, at least one of which is changing. Since the word "force" now has a very specific meaning in physics, and an emf is not a force in this sense, the expansion of the acronym is considered obsolete; or at best, an embarrassing historical artifact. (The term is attributed to Alessandro Volta.)
The emf describes the electrical effect of a changing magnetic field. In the presence of a magnetic field, the electric potential and hence the potential difference (commonly known as voltage) is undefined (see the former) — hence the need for distinct concepts of emf and potential difference. Technically, the emf is an effective potential difference included in a circuit to make Kirchhoff's voltage law valid: it is exactly the amount from Faraday's law of induction by which the line integral of the electric field around the circuit is not zero. The emf is then given by L di/dt, where i is the current and L is the inductance of the circuit.
Given this emf and the resistance of the circuit, the instantaneous current can be computed with Ohm's Law, for example, or more generally by solving the differential equations that arise out of Kirchhoff's laws.
Emf is often used as a synonym for any potential difference, irrespective of the source of the potential difference. (E.g., a battery, charged capacitor, or electret might be the source.) This usage is considered obsolete.
In certain cases, it is possible to make a mathematical analogy between electrical circuits and one-dimensional mechanical systems, in which the emf plays the role of the "force" in the equations. The origin of the term "electromotive force", however, did not employ such a strict analogy; it simply referred to the strength with which positive and negative charges could be separated (i.e. moved, hence "electromotive"), and was also called "electromotive power" (although it is not a power in the modern sense). (c.f. Oxford English Dictionary, "electromotive force".) Maxwell's 1865 explication of what are now called Maxwell's equations used the term "electromotive force" for what is now called the electric field. | 487 | 2,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-40 | latest | en | 0.963808 |
https://online.stat.psu.edu/stat462/node/134/ | 1,709,555,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00511.warc.gz | 425,675,502 | 7,992 | # 5.5 - Three Types of MLR Parameter Tests
Let's investigate an example that highlights the differences between the three hypotheses that we learn how to test in the remainder of this lesson.
An Example: Heart attacks in rabbits. When heart muscle is deprived of oxygen, the tissue dies and leads to a heart attack ("myocardial infarction"). Apparently, cooling the heart reduces the size of the heart attack. It is not known, however, whether cooling is only effective if it takes place before the blood flow to the heart becomes restricted. Some researchers (Hale, et al, 1997) hypothesized that cooling the heart would be effective in reducing the size of the heart attack even if it takes place after the blood flow becomes restricted.
To investigate their hypothesis, the researchers conducted an experiment on 32 anesthetized rabbits that were subjected to a heart attack. The researchers established three experimental groups:
• Rabbits whose hearts were cooled to 6º C within 5 minutes of the blocked artery ("early cooling")
• Rabbits whose hearts were cooled to 6º C within 25 minutes of the blocked artery ("late cooling")
• Rabbits whose hearts were not cooled at all ("no cooling")
At the end of the experiment, the researchers measured the size of the infarcted (i.e., damaged) area (in grams) in each of the 32 rabbits. But, as you can imagine, there is great variability in the size of hearts. The size of a rabbit's infarcted area may be large only because it has a larger heart. Therefore, in order to adjust for differences in heart sizes, the researchers also measured the size of the region at risk for infarction (in grams) in each of the 32 rabbits.
With their measurements in hand (coolhearts.txt), the researchers' primary research question was:
Does the mean size of the infarcted area differ among the three treatment groups — no cooling, early cooling, and late cooling — when controlling for the size of the region at risk for infarction?
A regression model that the researchers might use in answering their research question is:
$y_i=(\beta_0+\beta_1x_{i1}+\beta_2x_{i2}+\beta_3x_{i3})+\epsilon_i$
where:
• yi is the size of the infarcted area (in grams) of rabbit i
• xi1 is the size of the region at risk (in grams) of rabbit i
• xi2 = 1 if early cooling of rabbit i, 0 if not
• xi3 = 1 if late cooling of rabbit i, 0 if not
and the independent error terms εi follow a normal distribution with mean 0 and equal variance σ2.
The predictors x2and x3 are "indicator variables" that translate the categorical information on the experimental group to which a rabbit belongs into a usable form. We'll learn more about such variables in Lesson 8, but for now observe that for "early cooling" rabbits x2 = 1 and x3 = 0, for "late cooling" rabbits x2 = 0 and x3 = 1, and for "no cooling" rabbits x2 = 0 and x3 = 0. The model can therefore be simplified for each of the three experimental groups:
$y_i=(\beta_0+\beta_1x_{i1}+\beta_2)+\epsilon_i \text{ for "early cooling" rabbits}$
$y_i=(\beta_0+\beta_1x_{i1}+\beta_3)+\epsilon_i \text{ for "late cooling" rabbits}$
$y_i=(\beta_0+\beta_1x_{i1})+\epsilon_i \text{ for "no cooling" rabbits}$
Thus, $\beta_2$ represents the difference in mean size of the infarcted area — controlling for the size of the region at risk —between "early cooling" and "no cooling" rabbits. Similarly, $\beta_3$ represents the difference in mean size of the infarcted area — controlling for the size of the region at risk —between "late cooling" and "no cooling" rabbits.
Fitting the above model to the researchers' data, statistical software reports:
A plot of the data adorned with the estimated regression equation looks like:
The plot suggests that, as we'd expect, as the size of the area at risk increases, the size of the infarcted area also tends to increase. The plot also suggests that for this sample of 32 rabbits with a given size of area at risk, 1.0 gram say, the average size of the infarcted area differs for the three experimental groups. But, the researchers aren't just interested in this sample. They want to be able to answer their research question for the whole population of rabbits.
How could the researchers use the above regression model to answer their research question? Note that the estimated slope coefficients b2 and b3 are -0.2435 and -0.0657, respectively. If the estimated coefficients b2 and b3 were instead both 0, then the average size of the infarcted area would be the same for the three groups of rabbits in this sample. It can be shown that the mean size of the infarcted area would be the same for the whole population of rabbits — controlling for the size of the region at risk — if the two slopes β2 and β3 simultaneously equal 0. That is, the researchers's question reduces to testing the hypothesis H0 : β2 = β3 = 0.
I'm hoping this example clearly illustrates the need for being able to "translate" a research question into a statistical procedure. Often, the procedure involves four steps, namely:
• formulating a multiple regression model
• determining how the model helps answer the research question
• checking the model
• and performing a hypothesis test (or calculating a confidence interval)
We next learn how to perform three different hypothesis tests for slope parameters in order to answer various research questions. Let's take a look at the different research questions — and the hypotheses we need to test in order to answer the questions — for our heart attacks in rabbits example.
### A research question
Consider the research question: "Is a regression model containing at least one predictor useful in predicting the size of the infarct?" Are you convinced that testing the following hypotheses helps answer the research question?
• H0 : β1 = β2 = β3 = 0
• HA : At least one βi ≠ 0 (for i = 1, 2, 3)
In this case, the researchers are interested in testing that all three slope parameters are zero. We'll soon see that the null hypothesis is tested using the analysis of variance F-test.
### Another research question
Consider the research question: "Is the size of the infarct significantly (linearly) related to the area of the region at risk?" Are you convinced that testing the following hypotheses helps answer the research question?
• H0 : β1 = 0
• HA : β1 ≠ 0
In this case, the researchers are interested in testing that just one of the three slope parameters is zero. You already know how to do this, don't you? Wouldn't this just involve performing a t-test for β1? We'll soon learn how to think about the t-test for a single slope parameter in the multiple regression framework.
### A final research question
Consider the researcher's primary research question: "Is the size of the infarct area significantly (linearly) related to the type of treatment after controlling for the size of the region at risk for infarction?" Are you convinced that testing the following hypotheses helps answer the research question?
• H0 : β2 = β3 = 0
• HA : At least one βi ≠ 0 (for i = 2, 3)
In this case, the researchers are interested in testing whether a subset (more than one, but not all) of the slope parameters are simultaneously zero. We will learn a general linear F-test for testing such a hypothesis. | 1,701 | 7,257 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-10 | latest | en | 0.951161 |
https://solvedlib.com/how-do-these-two-equations-explain-how-a-simple,432252 | 1,701,456,486,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00261.warc.gz | 627,034,220 | 17,370 | # How do these two equations explain how a simple pendulum is defined as a point mass...
###### Question:
How do these two equations explain how a simple pendulum is defined as a point mass at the end of the length of negligable mass?
17 27 I Ir = 3 m, L? T= 21 mg Lim How do these two equations explain why a simple pendulum is defined as a point mass @ the end of a length of negligible mass?
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38. LO.2, 3, 4, 5, 6, 7 Evan is single and has AGI of $277,300 in 2020. His potential itemized deductions before any limitations for the year total$52,300 and consist of the following: Medical expenses (before the AGI limitation) \$29,000 Interest on home mortgage 8,700 State income taxes 9,500 Real... | 2,296 | 8,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.896432 |
https://mailman.anu.edu.au/pipermail/nauty/2004-October/000225.html | 1,675,663,499,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00552.warc.gz | 409,438,319 | 2,722 | # [Nauty] Isomorphism test of labeled graphs
Brendan McKay bdm at cs.anu.edu.au
Thu Oct 7 15:04:02 EST 2004
```Hi. Your mail was delayed (and might have been lost forever) because it
had the sender address akerlund at cs.helsinki.fi but you are subscribed to
the list as arto.akerlund at cs.helsinki.fi . It has to be a perfect match.
(This is a Good Thing because it keeps out tons of spam messages.)
At
http://cs.anu.edu.au/mailman/listinfo/nauty-list
you can unsubscribe/subscribe to fix the problem. It is also ok to
subscribe with two different addresses. You can mark one as
no-delivery so you don't get two copies of everything.
The answer to your question is that the two coloured graphs you
describe are non-isomorphic and nauty finds them non-isomorphic.
Generally speaking, if a graph is adorned with things attached to
the edges or vertices, such as colours, labels, tags, weights,
or whatever they are called, the word "isomorphism" usually implies
that vertices/edges are only mapped to vertices/edges with equal
colours on the vertices.
Brendan.
* Arto J Akerlund, tkol <akerlund at cs.helsinki.fi> [041007 14:39]:
> Someone has asked something similar before and mr. McKay replied as
> follows.
>
> >Sorry, I somehow missed your message before. In order for two labelled
> >graphs to be isomorphic they must have the same number of vertices of
> >each colour. This is not tested by nauty!
>
> >*If* the graphs have the same number of vertices of each colour, then
> >you can test the graphs for isomorphism by finding the canonical graphs
> >and comparing those.
>
> >Incidentally, the values you put in ptn[] are examined for zero or
> >non-zero only, so there is no point in using values other than 0 and 1
> >(but it doesn't hurt either).
>
> But what if the case was like this.
>
> G1:
>
> A- - - - B
> | |
> | |
> B- - - - A
>
>
> G2:
>
> A- - - - A
> | |
> | |
> B- - - - B
>
> Data structures in nauty would thus be:
>
> G1:
> 0 : 1 3;
> 1 : 0 2;
> 2 : 1 3;
> 3 : 0 2;
> lab[0]=0; lab[1]=2; lab[2]=1; lab[3]=3;
> ptn[0]=1; ptn[1]=0; ptn[2]=1; ptn[3]=0;
>
> G2 :
> 0 : 1 3;
> 1 : 0 2;
> 2 : 1 3;
> 3 : 0 2;
> lab[0]=0; lab[1]=1; lab[2]=2; lab[3]=3;
> ptn[0]=1; ptn[1]=0; ptn[2]=1; ptn[3]=0;
>
>
>
> Now there's an equal number of verticles of both colour and if G1 and G2
> were unlabeled, they would be isomorphic. Is there a way to test
> isomorphims for labeled graphs so that graphs G1 and G2 wouldn't be
> isomorphic.
>
> I'm not exactly sure what is the mathematical definition of 'isomorphism
> of labeled graphs', but in my application G1 and G2 shouldn't be found
> identical anyway.
>
> _______________________________________________
> This is the nauty mailing list
> Post messages to nauty-list at cs.anu.edu.au
> nauty page: http://cs.anu.edu.au/~bdm/nauty/
> list page: http://cs.anu.edu.au/mailman/listinfo/nauty-list
``` | 949 | 2,931 | {"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} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.916522 |
https://bridgitmendlermusic.com/what-is-moment-of-inertia-of-rectangle/ | 1,721,410,162,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00049.warc.gz | 121,597,284 | 8,818 | ## What is moment of inertia of rectangle?
When we take a situation when the axis passes through the centroid, the moment of inertia of a rectangle is given as: I = bh3 / 12. Here, b is used to denote the rectangle width (the dimension parallel to the axis) and h is said to be the height (dimension perpendicular to the axis).
How do you find IXY of a rectangle?
1-introduce a strip of width dy and breadth=b. 2- estimate the Ixy=∫h*dy*x/2*y from y=0 to y=h. 3-the value of integration will be Ixy=Ab*h/4.
What is the centroid of a rectangle?
Centroid of rectangle lies at intersection of two diagonals. Diagonals intersect at width (b/2) from reference x-axis and at height (h/2) from reference y-axis.
### What is the dimensional formula of inertia?
Therefore, the moment of inertia is dimensionally represented as M1 L2 T0.
How to figure the moment of inertia?
1) Segment the beam section into parts When calculating the area moment of inertia, we must calculate the moment of inertia of smaller segments. 2) Calculate the Neutral Axis (NA) The Neutral Axis (NA) or the horizontal XX axis is located at the centroid or center of mass. 3) Calculate Moment of Inertia
How do you determine the moment of inertia?
Basically, for any rotating object, the moment of inertia can be calculated by taking the distance of each particle from the axis of rotation ( r in the equation), squaring that value (that’s the r2 term), and multiplying it times the mass of that particle. You do this for all of the particles that make up…
## What is the equation for the moment of inertia?
A hollow cylinder with rotating on an axis that goes through the center of the cylinder, with mass M, internal radius R 1, and external radius R 2, has a moment of inertia determined by the formula: I = (1/2) M ( R 1 2 + R 2 2 )
What is the significance of calculating the moment of inertia?
The moment of inertia calculation identifies the force it would take to slow, speed up or stop an object’s rotation. The International System of Units (SI unit) of moment of inertia is one kilogram per meter squared (kg-m 2). In equations, it is usually represented by the variable I or IP (as in the equation shown).
What is moment of inertia of rectangle? When we take a situation when the axis passes through the centroid, the moment of inertia of a rectangle is given as: I = bh3 / 12. Here, b is used to denote the rectangle width (the dimension parallel to the axis) and h is said to be… | 584 | 2,475 | {"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} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.883104 |
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Pile group analysis using rigid plate assumption | 561 | 2,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.795068 |
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# Business Project management. Risk in a project - Essay Example
## Extract of sample Business Project management. Risk in a project
From the above equation, it can also be inferred that the overall impact of all risks associated with a project may be calculated by adding up all the ‘R’ values of all the risks associated with the project. As far as the probability (P) is concerned, being a variable, it varies with different stages of time in the life cycle of a project. For example, the probability of an accident occurring during a building construction project varies at different stages of the lifecycle. This might directly be related to various other factors in turn which themselves vary with the project cycle. For instance, the implementation stage of the project usually is more susceptible to this kind of risk in terms of the probability of occurring (Hillson and Hulett 2004) and thus it might be at different levels. Determination of probability is more a qualitative exercise than a quantitative one. There is no clear cut way or model of determining the probability though the probability might be estimated based on past data (Mind Tools, 2012). The way to do it is to arrive at mathematical models of probability distributions as close to the real probability as possible. It is also important to note that as many factors as can affect the probability of a given event need to be taken into account in order arrive at a realistic estimate of the probability of an event. A simple way to arrive at a probability level for an event is to assign a score on a scale of 1 to 10, wherein 1 represents the least probability and 10 represent the highest probability. The scores should be greater than zero because if its zero, it is not likely to occur and thus not a risk and if its 10 then it becomes a certainty and then too it is not a risk. The costs associated with an event also vary with various stages in the project lifecycle which means a snapshot sum total of all the factors that affect project cost associated with a certain event (Rochester 2012). Looking at the event based project cost in the above example, it might be seen that the cost of an accident might be more when the project is in progress, for instance the construction work is in full swing and the building as well as people working with project are more vulnerable to any mishap. The cost might be less when for example, the work is off on account of a holiday. On a longer time scale, in the above example, the costs associated with an accident just prior to the completion of the building project might be substantially high, since a lot of money and resources have been invested by then. As seen above, the difficulties in arriving at realistic values of ‘P’ and ‘C’ make it important to consider a qualitative mode of assessing the project risk (Wordpress 2009). The project risk matrix as shown below Consequences IV III I II As seen in the figure above, the project risk as expressed in terms of probability is classified as low and high on the vertical axis. The Consequences or costs associated are represented as low or high on the horizontal axis. This represents four typical situations represented by four quadrants, for the project risk situations. The four combinations are as follows: Quadrant 1: Low probability-Low Cost Risk impact Quadrant 2: Low probability- High Cost Risk impact Quadrant 3: High probability-High Cost Risk imp ...Show more
## Summary
Business Project management Answer 1. Risk in a project is a function of the following: The possibility of an event occurring, with which a particular project risk is associated The costs of the event or consequences and their seriousness over the project life cycle The above two factors are variables and the relationship of these variables with the project risk is denoted as per the following equation given by Pinto (2007)…
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Let us find you another Essay on topic Business Project management. Risk in a project for FREE! | 1,629 | 8,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-13 | latest | en | 0.955287 |
http://www.ck12.org/physical-science/Kinetic-Energy-in-Physical-Science/rwa/Into-the-Wild-Blue-Yonder/r7/ | 1,493,001,082,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118950.30/warc/CC-MAIN-20170423031158-00584-ip-10-145-167-34.ec2.internal.warc.gz | 479,432,423 | 29,534 | # Kinetic Energy
## The energy of moving matter. KE = 1/2 mass X velocity 2
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Into the Wild Blue Yonder
### Into the Wild Blue Yonder
Credit: Robert Lopez
Source: CK-12 Foundation
You’ve probably made paper airplanes. You fold a sheet of paper to resemble an airplane and then just fling it into the air. It seems to fly on its own. Of course, all if its energy actually comes from you. But if you’ve folded the paper correctly, after you let go of the plane it may glide on the air for a while before falling back to the ground.
Amazing But True!
• Did you know that some real airplanes are like paper airplanes? They are called gliders. They don’t have engines, so like paper airplanes, their energy must come from somewhere else.
• Credit: ajmexico
Source: https://www.flickr.com/photos/ajmexico/5479171584
In order to get them going, gliders use tow planes to pull them into the air [Figure2]
• Do you know how real gliders work? Where do they get their energy? And without an engine, how do they stay aloft? Watch this video to find out.
Can You Apply It?
1. How does a glider such as a sailplane gain potential energy?
2. How does a glider convert the potential energy it gains to kinetic energy?
3. A glider applies Bernoulli’s law to generate lift. Explain how.
4. Gliders are constantly descending because of gravity. But some gliders can stay aloft for hours. Explain why.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes | 394 | 1,647 | {"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-2017-17 | latest | en | 0.904893 |
https://www.helpteaching.com/questions/CCSS.Math.Content.5.G.B.3 | 1,627,586,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00625.warc.gz | 832,375,975 | 11,072 | Tweet
# Common Core Standard 5.G.B.3 Questions
Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category. For example, all rectangles have four right angles and squares are rectangles, so all squares have four right angles.
You can create printable tests and worksheets from these questions on Common Core standard 5.G.B.3! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.
Previous Next
Which of the following are parallelograms? Select all correct answers.
Which best explains why the shape shown is not a parallelogram?
1. It has more than four sides.
2. It is a two-dimensional shape.
3. It has angles of the same measure.
4. It has opposite sides that are parallel.
The table shows the side lengths of quadrilateral ABCD.
SideLength
AB4
BC8
CD6
DA6
1. Yes. It is a parallelogram because two of the sides measure the same length.
2. No. It is not a parallelogram because the opposite sides are not the same length.
3. Maybe. It could be a parallelogram, but the table needs to include angle measures in order to determine.
Lucy says the shape shown is not a parallelogram because all four sides are not the same length. Is Lucy's reasoning correct and why?
1. Yes, Lucy is correct. The shape is not a parallelogram because all four sides are not the same length.
2. No Lucy is incorrect. The shape is a parallelogram because all four sides are the same length.
3. No, Lucy is incorrect. The shape is not a parallelogram because it has four sides, but the opposite sides are not parallel.
4. No, Lucy is incorrect. The shape is a parallelogram because it has four sides and the opposite sides are parallel.
John's teacher asked him to draw a quadrilateral with all congruent sides. What could he draw?
1. a rectangle or a rhombus
2. a rectangle or a square
3. a square or a rhombus
4. a trapezoid or a kite
Which of the following is a parallelogram?
1. Rectangle
2. Rhombus
3. Square
4. All of the above
Which is sufficient to prove that a quadrilateral is a rhombus?
1. The diagonals bisect each other.
2. The diagonals are perpendicular.
3. All four sides are congruent.
4. A pair of opposite sides are congruent and parallel.
5. None of the above.
1. All rectangles are always squares.
2. All squares are always rhombuses.
3. All trapezoids are always parallelograms.
4. All parallelograms are always rhombuses.
5. All rectangles are always parallelograms. | 612 | 2,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-31 | latest | en | 0.903623 |
https://www.mathcelebrity.com/binary.php?num=15&check1=5&bchoice=5&pl=Convert | 1,708,675,613,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00071.warc.gz | 920,519,923 | 19,207 | <-- Enter Number or Notation that will be converted
Convert Decimal to:
Base:
Convert to Decimal from:
Convert 15 from decimal to hexadecimal
(base 16) notation:
##### Power Test
Raise our base of 16 to a power
Start at 0 and increasing by 1 until it is >= 15
160 = 1
161 = 16 <--- Stop: This is greater than 15
Since 16 is greater than 15, we use 1 power less as our starting point which equals 0
Work backwards from a power of 0
##### 160 = 1
The highest coefficient less than 15 we can multiply this by to stay under 15 is 15
Multiplying this coefficient by our original value, we get: 15 * 1 = 15
Add our new value to our running total, we get:
0 + 15 = 15
Hexadecimal (10 - 15) are represented by an (A-F) where 15 translates to the letter F
This = 15, so we assign our outside coefficient of F for this digit.
Our new sum becomes 15
Our hexadecimal notation is now equal to F | 247 | 898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-10 | latest | en | 0.899793 |
http://forum.allaboutcircuits.com/threads/star-connection.103244/ | 1,484,847,798,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280723.5/warc/CC-MAIN-20170116095120-00328-ip-10-171-10-70.ec2.internal.warc.gz | 99,019,985 | 14,272 | # star connection
Discussion in 'Homework Help' started by segecious, Nov 7, 2014.
1. ### segecious Thread Starter New Member
Nov 6, 2014
7
0
I have a problem solving the following assignment question.
A star-connected balanced three phase load 30 ohms resistance per phase is supplied by 415 v, three-phase generator of efficiency 90%. calculate the power input to the generator.
2. ### segecious Thread Starter New Member
Nov 6, 2014
7
0
i need solution to that question.
Apr 26, 2005
3,399
1,218
Apr 26, 2005
3,399
1,218
5. ### WBahn Moderator
Mar 31, 2012
18,085
4,917
What power is delivered to a balanced, start-connected load if each phase has a load resistance of R and the three-phase power is 415 V? What assumptions do you need to make, or what additional information do you need, in order to answer that part of the question?
6. ### segecious Thread Starter New Member
Nov 6, 2014
7
0
Vph=415/√3=239.6
V=IR
Iph=v/R = 239.6/30=7.99
VL =√3Vph
Vph=239.6/√3=138.3
IL=138.3/30=4.6A
p=3*I2R
=3*4.62*30=1917w
this is what i got please assist me
7. ### WBahn Moderator
Mar 31, 2012
18,085
4,917
Why are you dividing the voltage by sqrt(3) twice?
Nov 6, 2014
7
0
thanks | 408 | 1,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-04 | latest | en | 0.879205 |
http://www.convertit.com/Go/Beverageonline/Measurement/Converter.ASP?From=Spanish+quintal&To=mass | 1,590,581,564,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00482.warc.gz | 162,907,050 | 3,663 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
(Help)
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Spanish quintal = 45.994266318 mass (mass) ``` Related Measurements: Try converting from "Spanish quintal" to as (Roman as), carat troy (troy carat), catty (Chinese catty), denarius (Roman denarius), dinar (Arabian dinar), earth mass, electron mass (electron rest mass), grain (avoirdupois grain), kin (Japanese kin), long hundredweight (avoirdupois long hundredweight), long ton (avoirdupois long ton), mina (Greek mina), mite (English mite), obol (Greek obol), oz ap (apothecary ounce), pfund (German pfund), scruple (apothecary scruple), tael (Chinese tael), UK quintal (British quintal), wey mass, or any combination of units which equate to "mass" and represent mass. Sample Conversions: Spanish quintal = 5,495.13 Babylonian shekel, .2028 cotton bale (US), .45994266 doppelzentner, 7.69E-24 earth mass, 45,994.27 gram, 4,690.11 hyl, 12.26 kwan (Japanese kwan), 93.87 livre (French livre), 3.62 long quarter, .04526786 long ton (avoirdupois long ton), 107.15 mina (Greek mina), 2.75E+28 neutron mass (neutron rest mass), 63,375 obol (Greek obol), 29,575 pennyweight troy (troy pennyweight), .7605 picul (Chinese picul), 80,935.01 Roman obolus, 1.13 Roman talent, 40,490.59 scrupulum (Roman scrupulum), 3.15 slug, 7.24 stone.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 549 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-24 | latest | en | 0.63985 |
https://oeis.org/A269170 | 1,627,277,718,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00397.warc.gz | 437,603,568 | 4,168 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A269170 a(n) = n OR floor(n/2), where OR is bitwise-OR (A003986). 1
0, 1, 3, 3, 6, 7, 7, 7, 12, 13, 15, 15, 14, 15, 15, 15, 24, 25, 27, 27, 30, 31, 31, 31, 28, 29, 31, 31, 30, 31, 31, 31, 48, 49, 51, 51, 54, 55, 55, 55, 60, 61, 63, 63, 62, 63, 63, 63, 56, 57, 59, 59, 62, 63, 63, 63, 60, 61, 63, 63, 62, 63, 63, 63, 96, 97, 99, 99, 102, 103, 103, 103, 108, 109, 111, 111, 110, 111 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Fibbinary numbers (A003714) give all integers n >= 0 for which a(n) = A003188(n) and also for which a(n) = A032766(n). LINKS Antti Karttunen, Table of n, a(n) for n = 0..8191 FORMULA a(n) = A003986(n,(n-A000035(n))/2). Other identities and observations. For all n >= 0: a(2n) = A163617(n). A003188(n) <= a(n) <= A032766(n). PROG (Scheme) (define (A269170 n) (A003986bi n (/ (- n (A000035 n)) 2))) ;; Here A003986bi implements dyadic bitwise-OR operation (see A003986). (PARI) a(n) = bitor(n, n\2); \\ Michel Marcus, Feb 29 2016 CROSSREFS Cf. A000035, A003714, A003986. Cf. A163617 (even bisection). Cf. also A003188, A048735, A032766. Sequence in context: A008867 A185958 A273062 * A003879 A333885 A270060 Adjacent sequences: A269167 A269168 A269169 * A269171 A269172 A269173 KEYWORD nonn,base AUTHOR Antti Karttunen, Feb 22 2016 STATUS approved
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Last modified July 26 01:33 EDT 2021. Contains 346294 sequences. (Running on oeis4.) | 720 | 1,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.633258 |
http://openstudy.com/updates/4fbfe2aae4b0964abc826a09 | 1,448,735,177,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453656.76/warc/CC-MAIN-20151124205413-00344-ip-10-71-132-137.ec2.internal.warc.gz | 174,097,326 | 9,828 | Eyad 3 years ago Three Numbers forming A.s ,Their sum=15 ..If the numbers (1,1,4) are added to them respectively ,then the resulting numbers form A Geometric sequence ..Find The Three Numbers ?
1. experimentX
let the three numbers be a-d,a,a+d rest is simple
2. experimentX
add them ... get a=5 add those numbers to AS ... use the property of GS .. and solve for d, you will have your number | 102 | 395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2015-48 | longest | en | 0.79989 |
https://www.curriki.org/oer/Unit-3-Project--Classroom-Olympics-and-Statistical-Data?mrid=56562-56827 | 1,560,635,141,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997335.70/warc/CC-MAIN-20190615202724-20190615224724-00282.warc.gz | 728,583,801 | 22,160 | #### Type:
Graphic Organizer/Worksheet
#### Description:
In this project we apply what students learned in the first four lessons (3.1-3.4). Students will summarize, represent, and interpret data on single and multiple variables. The activities are designed to give students experience interpreting linear models. Students will calculate the mean, median, mode and range of data sets.
#### Subjects:
• Mathematics > General
#### Keywords:
algebra linear equations variables inequalities exponents data mean median mode range
English
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#### Collections:
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Update Standards?
#### CCSS.Math.Content.HSS-ID.A.2: Common Core State Standards for Mathematics
Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.
#### CCSS.Math.Content.HSS-ID.A.3: Common Core State Standards for Mathematics
Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
#### CCSS.Math.Content.HSS-ID.A.4: Common Core State Standards for Mathematics
Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.
#### MA.9-12.CCSS.Math.Content.HSS-ID.A.2: Mathematics
Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.
#### MA.9-12.CCSS.Math.Content.HSS-ID.A.3: Mathematics
Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
#### MA.9-12.CCSS.Math.Content.HSS-ID.A.4: Mathematics
Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.
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https://kr.mathworks.com/matlabcentral/cody/problems/42836-juggler-sequence-revisited/solutions/883578 | 1,606,970,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141718314.68/warc/CC-MAIN-20201203031111-20201203061111-00544.warc.gz | 297,648,305 | 17,077 | Cody
# Problem 42836. Juggler sequence revisited
Solution 883578
Submitted on 28 Apr 2016 by LY Cao
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 = 3; l_correct = 6; h_correct = 36; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
2 Pass
x = 5; l_correct = 5; h_correct = 36; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
3 Pass
x = 10; l_correct = 7; h_correct = 36; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
4 Pass
x = 77; l_correct = 19; h_correct = 2322378; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
5 Pass
x = 99; l_correct = 11; h_correct = 37754276; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
6 Pass
x = 117; l_correct = 11; h_correct = 44992; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
7 Pass
x = 1; l_correct = 0; h_correct = 1; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
8 Pass
x = 2; l_correct = 1; h_correct = 2; [l h] = juggler(x); assert(isequal([l h],[l_correct h_correct]))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 440 | 1,312 | {"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-2020-50 | latest | en | 0.661118 |
http://6bdmath.blogspot.com/2007/03/finding-area-of-parallelogram.html?showComment=1175135460000 | 1,529,639,382,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00172.warc.gz | 4,027,329 | 11,544 | ## Wednesday, March 21, 2007
### Finding the Area of A Parallelogram
Hi, I'm Jessie, and this is my partner, Firoze. Firoze: That's FUR-OS-A. Sorry, she's very cautious when it comes to her name pronounciation. Okay, so, today we'll be showing you how to find the area of a parallelogram. Okay, so the first step is to measure the base or the width of the parallelogram.
Next, measure the height or the length of the parallelogram. Remember, the side isn't the height. Imagine a dotted line from the top corners to the bottom, the line is straight, measure this imaginary line, not the side.
Next you multiply the base by the height. As you can see, we've already started the equation.
Your answer should be 28 cm2. Noticted the 2 after the cm. This 2 stands for square. So your answer should be 28cm (squared).
Another way of finding the area of a parallelogram, is to put it on gragh paper. Count up al, the full squares that are in the parallelogram.
Notice some squares aren't full. Try putting the half squares together, and counting them as one.
So that's basically it. It's very simple, we hope we made it clear to you. That's all for now. Firoze and Jessie signing off. We'll see you next time on 6BD' s math blog. Chouw for now!.
Evan said...
Good Job! I Understood It Well Fir-Os-A! Great Pictures!
Evan`o
Evan said...
This comment has been removed by the author.
taihyun said...
great job you guys and I am impressed with your work and I like those pictures that you put up there and try making more longer and explain more better.
gagan said...
Good job fir-os-a and jessie you did a really good job on the post. It made lots of sense and now I learned how to nfind the area of a parallagram before I didn't really now how. I also liked the pictures they were cool.
Ruby said...
This comment has been removed by the author.
Ruby said...
Hey Firoze and Jessie! Great post it makes so much sense and sort of told me how to find the area of a parellelagram because i sort of didnt but sort of did. Also I liked the pictures.
karissa♥ said...
Hi gals! You did a very good job explaining how to find the area of a parallelogram. You did that by taking us step by step. For example, when you said you have to measure the imaganary dotted line, not just the side. You also did a great job on choosing the right picture with the right information. Over all you did a fabulous job!!!
karissa
Dillani☺ said...
hey jessie and firoze
you guys did a really great job on the post. but when you said you dont museur the diangle bla bla bla you should have out the line on the pic. you guys still did a very good job and keep it up!
p.s. fir-os-a everyone knows how to saye ur name lol
lara said...
heyy guys
you did a great job and it looked like you took your time great job and great pics.
later lara:P:PP
Taylor said...
Hey Fir-os-a, and Jessie!
You girls did awesome!
You took us through really good steps! Your both really good at this. Good Job!
Nice pictures too!
Taylor
jagan said...
Good job!I really understood and i thought it was well written out. it was also very long and good!
JaCk said...
hey guys
AWESOME! i understood your blog very well, and it was very clear
Awesome pics
Jack
ilhan said...
Good job! I impressed with your work and it was easy to understand | 835 | 3,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-26 | latest | en | 0.961098 |
http://slideplayer.com/slide/4084394/ | 1,521,956,280,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651820.82/warc/CC-MAIN-20180325044627-20180325064627-00434.warc.gz | 295,758,741 | 33,305 | # Circuits Practice Questions.
## Presentation on theme: "Circuits Practice Questions."— Presentation transcript:
Circuits Practice Questions
Review Formula’s Charging Discharging
Question A wire made of brass and another wire made of silver have the same length, but the diameter of the brass wire is 4 times the diameter of the silver wire. The resistivity of brass is 5 times greater than the resistivity of . Silver. If RB denotes the resistance of the brass wire and RS denotes the resistance of the silver wire, which of the following is true? RB=5/16 RS RB=4/5 RS RB=5/4 RS RB=5/2 RS RB=16/5 RS Let ρs denote the resistivity of silver and let As denote the cross-sectional area of the silver wire.
Question For a ohmic conductor, doubling the voltage without changing the resistance will cause the current to? Decrease by a factor of 4 Decrease by a factor of 2 Remain unchanged Increase by a factor of 2 Increase by a factor of 4 Therefore doubling the voltage, doubles the current.
Question If a 60 watt light bulb operates at a voltage of 120V, what is the resistance of the bulb? 30Ω 240Ω 720Ω 7200Ω
Question A battery whose emf is 40V has an internal resistance of 5Ω. If this battery is connect to a 15Ω resistor R, what will the voltage drop across R be? 10V 30V 40V 50V 70V
Question Three resistors are connected to a 10-V battery as shown below. What is the current through the 2.0 Ω resistor? 4.0Ω 2.0Ω ε=10V 0.25A 0.50A 1.0A 2.0A 4.0A Since all resistors are in series, the amount of current that passes through any one of them is the same. So we need to simply the circuit to determine that current.
Question Determine the equivalent resistance between points a and b?
0.167Ω 0.25 Ω 0.333 Ω 1.5 Ω 2 Ω 12Ω
Question Three identical light bulbs are connected to a source of emf, as shown in the diagram above. What will happen if the middle bulb burns out? All the bulbs will go out The light intensity of the other two bulbs will decrease (but they won’t go out). The light intensity of the other two bulbs will increase. The light intensity of the other two bulbs will remain the same. More current will be drawn from the source emf. If each bulb has a resistance of R, then each individual bulb will draw ε/R. This will be unchanged if any individual bulb goes out. (less current will be drawn from the battery, but the same amount of current will pass through each bulb)
Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected. a) Is Higher b) Is Lower c) Is The Same d) Don’t know Bulbs in parallel are like resistors in parallel. Therefore since the total resistance of parallel resistors is lower, and the voltage ins the same, then the current must increase (double).
Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected. a) Is Higher b) Is Lower c) Is The Same d) Don’t know Since the total resistance goes up by two, the current must go down by two. Therefore lower.
Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the power output from the battery (compared to when only one bulb was connected) a) Is four times higher b) Is twice as high c) Is the same d) Is half as much e) Is one quarter as much f) Don’t know Since the current goes up by two. Therefore the Power goes up by two
Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the light from the first bulb (compared to when only one bulb was connected) a) is four times as bright b) is twice as bright c) is the same d) is half as bright e) is one quarter as bright Since the voltage goes down by a factor of two, the power goes down by 4. Therefore dimmer.
Question What is the voltage drop across the 12 ohm resistor in the portion of the circuit shown? 24V 36V 48V 72V 144V Since the top branch has 4Ω of resistance and the bottom branch has 12Ω of resistance, therefore 3 times as much current will flow through the 4Ω resistor than the 12Ω resistor. This breaks down the 12A into 9A up and 3A down. So form V=IR we have V=(3A)(12Ω)=36V 12A 12Ω We could also have produced a RT and found 3Ω value, therefore a voltage drop of (3Ω)(12A)=36V on both the upper and lower branch
Question What is the current through the 8Ω resistor in the circuit shown? 0.5A 1.0A 1.25A 1.5A 3.0A Since points a and b are grounded, they are all at the same potential (call it zero). Travelling from b to a across the battery , the potential increases by 24V, therefore it must decrease by 24V across the 8Ω resistor as we reach point a. Therefore I=V/R =(24V)/(8Ω)=3A. 24V a b
Question How much energy is dissipated as heat in 20 seconds by a 100 Ω resistor that carries a current of 0.5A? 50 J 100 J 250 J 500 J 1000J
Question What is the time constant for the circuit shown? 0.01 s
200Ω 50Ω 200 uF 50V
Question A 100Ω, 120Ω, and 150Ω resistor are connected to a 9-V battery as in the circuit shown below. Which of the three resistors dissipates the most power? We can also answer this question faster by noticing that the voltage drops across the 100Ω and then across the parallel resistors. Since the 100Ω has a higher value than the parallel combination, it will have a larger voltage drop than the combination so via P=IV, it dissipates the most power. The 100Ω resistor The 120Ω resistor The 150Ω resistor Both the 120Ω and the 150Ω All dissipate the same power 100Ω 120Ω 9V 150Ω Therefore the 100Ω resistor dissipates the most amount of power.
Question A 1.0 F capacitor is connected to a 12 V power supply until it is fully charged. The capacitor is then disconnected from the power supply, and then is used to power a toy car. The average drag force on the car is 2 N. about how far will the car go? 36m 72m 144m 24m 12m This energy is the Work used in moving the car a fixed distance. First we find the energy stored in the capacitor
Question Three capacitors are connected to a 9 V power supply as shown. How much charge is stored by this system of capacitors? 3 uC 30 uC 2.7 uC 27 uC 10 uC Simplify the circuit to find equivalent capacitance. C1=2uF C2=4uF C3=6uF 9V To determine the charge , we use Q=CV
Question What is the resistance of an ideal ammeter and an ideal voltmeter? Measures current Measures Voltage An ammeter is placed in series with other circuit components. In order for the ammeter not to itself resist current and change the total current in the circuit, you want it to have zero resistance. Ideal Ammeter Ideal Voltmeter zero infinite infinite zero A voltmeter is placed in parallel with other circuit components. If it had a low resistance, the current will flow through it instead of the other circuit element. So you want it to have infinite resistance to it won’t affect the circuit element being measured.
Question A light bulb is rated at 100 W in Canada, where the standard wall outlet voltage is 120V. If this bulb was plugged in in england, where standard wall outlet voltage is 240V, which of the following would be true? The bulb would be ¼ as bright. The bulb would be ½ as bright. The bulb’s brightness would the same. The bulb would be trice as bright. The bulb would be 4 times as bright Your first instinct is to say that because brightness depends on power, the bulb is exactly as bright. But that is not correct. The power of the bulb can change. The resistance of a light bulb is a property of the bulb itself, and so will not change no matter what the bulb is hooked to.
Question A current of 6.4A flows in a segment of copper wire. The number of electrons crossing a cross-sectional area of the wire every second is about? 6.4 4 x 1019 4 x 10-19 6.4 x 1019 6.4 x 10-19 Since each electron carries a charge of 1.6 x C. The we calculate that
Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors What is the resistance of the toaster and blender at their rated voltages? Determine the current that is achieved in the blender when it is working properly. Create a circuit that will make both devices work simultaneously. What power must the battery supply to run your circuit?
Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors What is the resistance of the toaster and blender at their rated voltages? Since we know the Power requirements , let’s us a power formula:
Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors Determine the current that is achieved in the blender when it is working properly. This is a job for Ohm’s Law:
Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors Create a circuit that will make both devices work simultaneously. Required to give a current of 2A for blender 60Ω blender Toaster ε=120V
Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors What power must the battery supply to run your circuit? From Ohm’s Law, we can determine current in toaster Since the blender had a current of 2A, this gives the circuit a total current of 12A Using P=IV, we have:
Question Given the following circuit 10Ω a 10Ω 20Ω 40Ω ε=120V 100Ω b
At what rate does the battery deliver energy to the circuit? Determine the current through the 20 Ω resistor. i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? Determine the energy dissipated by the 100 Ω resistor in 10 s Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is Ωm, determine its radius.
Question Given the following circuit
At what rate does the battery deliver energy to the circuit? 10Ω 20Ω 100Ω 40Ω ε=120V Recall: Rate is Power RT We need to write this as a simple circuit with a RT so that we can determine the current, I by using V=IR.
Question Given the following circuit
b) Determine the current through the 20 Ω resistor. We can determine the voltage drop . Recall: from Question a) we have a 2A current 10Ω 2A 10Ω 120V-(2A)(10Ω)-(2A)(10Ω)-(2A)(10Ω) =60V 20Ω 40Ω The ratio of the resistance of left side to right side is 40:120 or 1:3 ε=120V 100Ω Then we use V=IR to find each individual current. 2A 10Ω Therefore ¼ of 2 A goes through the right and ¾ of 2A goes through the left. Thus (1/4)(2A)=0.5A passes through the 20Ω resistor. OR
Question Given the following circuit
c) i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? 10Ω 20Ω 100Ω 40Ω ε=120V a b i) ii) Point a is at a higher potential. Since current flows from a high potential to a low potential.
Question Given the following circuit
d) Determine the energy dissipated by the 100 Ω resistor in 10 s 10Ω 20Ω 100Ω 40Ω ε=120V a b Recall: Energy equals Power multiplied by Time
Question Given the following circuit
e) Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is Ωm, determine its radius. 20Ω 100Ω 10Ω 40Ω ε=120V a b
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. 10Ω R C r ε S b a Determine the current through r at time t=0. Compute the time required for the charge on the capacitor to reach one-half its final value. When the capacitor is fully charged, which plate is positive? Determine the electric potential energy stored in the capacitor when the current r is zero. When the current through r is zero, the switch S is moved to b [t=0]. Determine the current through R as a function of time. Find the power dissipated in R as a function of time.
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. Determine the current through r at time t=0. 10Ω R C r ε S b a
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. Compute the time required for the charge on the capacitor to reach one-half its final value. 10Ω R C r ε S b a Therefore we need
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. When the capacitor is fully charged, which plate is positive? 10Ω R C r ε S b a + Because the top plate is connected to the positive end of the battery, the top is positive.
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. Determine the electric potential energy stored in the capacitor when the current r is zero. 10Ω R C r ε S b a When the current through r is zero, the capacitor is fully charged, with voltage across the plates matching the emf of the battery.
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. When the current through r is zero, the switch S is moved to b [t=0]. Determine the current through R as a function of time. 10Ω R C r ε S b a The current established by the discharging capacitor decreases exponentially.
Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. When the current through r is zero, the switch S is moved to b [t=0]. Find the power dissipated in R as a function of time. 10Ω R C r ε S b a
Question 200Ω ε= 9V 300Ω 400Ω 500Ω Simplify the above circuit so that it consists of one equivalent resistor and the battery. What is the total current through this circuit? Find the voltage across each resistor. Find the current through each resistor. The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same.
Question 200Ω ε= 9V 300Ω 400Ω 500Ω REQ Simplify the above circuit so that it consists of one equivalent resistor and the battery.
Question 200Ω ε= 9V 300Ω 400Ω 500Ω REQ =342.2Ω b) What is the total current through this circuit?
Question 200Ω ε= 9V 300Ω 400Ω 500Ω Find the voltage across each resistor. Find the current through each resistor. Let’s use a VIR chart V I R 200Ω 300Ω 400Ω 500Ω Total 9V 0.0263A 342.2Ω 3.16V 0.0158A 3.16V 0.0105A 5.84V 0.0146A 5.84V 0.0117A
Question 200Ω e) The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same. 300Ω ε= 9V 400Ω 500Ω By removing a resistor from a parallel set, we actually increase the resistance of the total circuit. Therefore by Ohm’s law if the voltage remains the same and the resistance increases, the total current must decrease Now through the 200Ω set, the total resistance remains the same, yet the current decreases, therefore the voltage across each resistor decreases as well as the current . | 4,185 | 16,413 | {"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} | 4.25 | 4 | CC-MAIN-2018-13 | latest | en | 0.889163 |
https://docs.pybinding.site/en/v0.9.5/_api/pybinding.StructureMap.html | 1,670,586,204,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00106.warc.gz | 252,132,552 | 6,235 | # StructureMap¶
class StructureMap(data, sites, hoppings, boundaries=())
A subclass of SpatialMap that also includes hoppings between sites
Attributes
boundaries Boundary hoppings between different translation units (only for infinite systems) data 1D array of values for each site, i.e. hoppings Sparse matrix of hopping IDs num_sites Total number of lattice sites positions Lattice site positions. spatial_map Just the SpatialMap subset without hoppings sub 1D array of sublattices IDs, short for .sublattices sublattices 1D array of sublattices IDs x 1D array of coordinates, short for .positions.x xyz Return a new array with shape=(N, 3). y 1D array of coordinates, short for .positions.y z 1D array of coordinates, short for .positions.z
Methods
__getitem__(idx) Same rules as numpy indexing clipped(v_min, v_max) Clip (limit) the values in the data array, see clip() cropped(**limits) Return a copy which retains only the sites within the given limits plot([cmap, site_radius, num_periods]) Plot the spatial structure with a colormap of data at the lattice sites plot_contour(**kwargs) Contour plot of the xy plane plot_contourf([num_levels]) Filled contour plot of the xy plane plot_pcolor(**kwargs) Color plot of the xy plane with_data(data) Return a copy of this object with different data mapped to the sites
__getitem__(idx)
Same rules as numpy indexing
clipped(v_min, v_max)
Clip (limit) the values in the data array, see clip()
cropped(**limits)
Return a copy which retains only the sites within the given limits
Parameters: **limits Attribute names and corresponding limits. See example.
Examples
Leave only the data where -10 <= x < 10 and 2 <= y < 4:
new = original.cropped(x=[-10, 10], y=[2, 4])
plot(cmap='YlGnBu', site_radius=(0.03, 0.05), num_periods=1, **kwargs)
Plot the spatial structure with a colormap of data at the lattice sites
Both the site size and color are used to display the data.
Parameters: cmap : str Matplotlib colormap to be used for the data. site_radius : Tuple[float, float] Min and max radius of lattice sites. This range will be used to visually represent the magnitude of the data. num_periods : int Number of times to repeat periodic boundaries. **kwargs Additional plot arguments as specified in structure_plot_properties().
plot_contour(**kwargs)
Contour plot of the xy plane
Parameters: **kwargs Forwarded to tricontour().
plot_contourf(num_levels=50, **kwargs)
Filled contour plot of the xy plane
Parameters: num_levels : int Number of contour levels. **kwargs Forwarded to tricontourf().
plot_pcolor(**kwargs)
Color plot of the xy plane
Parameters: **kwargs Forwarded to tripcolor().
with_data(data) → pybinding.results.StructureMap
Return a copy of this object with different data mapped to the sites
boundaries
Boundary hoppings between different translation units (only for infinite systems)
data
1D array of values for each site, i.e. maps directly to x, y, z site coordinates
hoppings
Sparse matrix of hopping IDs
num_sites
Total number of lattice sites
positions
Lattice site positions. Named tuple with x, y, z fields, each a 1D array.
spatial_map
Just the SpatialMap subset without hoppings
sub
1D array of sublattices IDs, short for .sublattices
sublattices
1D array of sublattices IDs
x
1D array of coordinates, short for .positions.x
xyz
Return a new array with shape=(N, 3). Convenient, but slow for big systems.
y
1D array of coordinates, short for .positions.y
z
1D array of coordinates, short for .positions.z | 862 | 3,528 | {"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} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.618419 |
https://www.gamedev.net/forums/topic/629475-collision-with-filloval/ | 1,512,959,740,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512054.0/warc/CC-MAIN-20171211014442-20171211034442-00365.warc.gz | 771,526,215 | 29,026 | # collision with fillOval
This topic is 1946 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I have 4 ovals being drawn to the screen and i need a way to check if a movable ball collides with the 4 that were drawn on the screen.
They 4 are drawn at random points on the screen but itd be near impossible to hit the exact center of the oval.
How would i detect a collision with the edges of the oval?
say i have the main ball moving right with starting coordinates 100, 100 and it comes into contact with a ball at 200, 100. Do i use a formula to measure this? Edited by burnt_casadilla
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If you have a small enough circle or accuracy isn't an issue you can assign bounding rectangles. Otherwise I would check the bounds against circle.x + radius, circle.x - radius, circle.y + radius, and circle.y - radius.
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i changed the balls to rectangles to make it a little easier
the main rectangle has the coords 100, 100 and is 10, 10 high and wide, but the others are created by a random point so i dont have a set point to test the bounds againsst Edited by burnt_casadilla
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You don't need a set point to test against. You should be keeping track of the x and y position of the object so you just need to check if one side is inside another. If the left side of one rectangle is inside another rectangle its (position - width) will be less than the other rectangle's right edge and greater than its left. This can easily be applied to the other sides.
Rectangle to Rectangle you can use [url=http://docs.oracle.com/javase/6/docs/api/java/awt/Rectangle.html#intersects(java.awt.Rectangle)]Rectangle.intersects()[/url].
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but the others are created by a random point so i dont have a set point to test the bounds againsst
[/quote]
hope you understand there is no such thing as random.
If its drawn at some point, that point had to exist, you just need to grab it at that time.
EDIT:: oh java... well i got low knowledge of java. But it should be still same Edited by BaneTrapper
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ok so i have random balls created every 3 seconds at random points x and y. I know these are stored in the array list, but how would i call the points from the list and then set the boundaries within the public Rectangle getBounds() method?
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If you want to make a ball at some random position I would do something like the below.
[code]ArrayList<Ball> balls = new ArrayList<Ball>();
private Ball makeBall()
{
Random rand = new Random();
Ball newBall = new Ball();
newBall.x = rand.nextFloat() * FRUSTUM_WIDTH;
newBall.y = rand.nextFloat() * FRUSTUM_HEIGHT;
return newBall;
}
//Option 2: add the ball to the list inside the method
private void newBall()
{
Random rand = new Random();
balls.add(new Ball(rand.nextFloat() * FRUSTUM_WIDTH, rand.nextFloat() * FRUSTUM_HEIGHT,
}
[/code]
This would have a ball class which stores an x and y position and a radius. The list will store all the balls and be looped through to update and render the balls. Option 1 has a blank constructor and the fields (x, y, radius) must be initialized manually. Option 2 has a constructor public Ball(float xPosition, float yPosition, float radius).
Edit: Then you would call this
[code]for(int i = 0; i < balls.size(); i++)
{
float x = balls.get(i).getX();
float y = balls.get(i).getY();
//check collisions here
}[/code] Edited by David.M
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this is what i have for creating balls
[source lang="java"]class Ball
{
int x;
int y;
int width;
int height;
public Ball(int x, int y, int width, int height)
{
this.x = x;
this.y = y;
this.width = width;
this.height = height;
}//end ball
public void paint(Graphics g)
{
g.setColor(color[getRandomColor()]);
g.fillOval(x, y, width, height);
} //end paint
} //ball class[/source]
[source lang="java"] public void paint(Graphics g)
{
g.drawString("time: " + t, 20, 20);
for (Ball ball : BALLS)
{
ball.paint(g);
}
updateGame();
}[/source]
and this works fine for what i want. but im wondering how to grab these points from the arraylist and the use those points in the getBounds() method for collision against another ball
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Any special reason for having both a width and height? If it's a circle those are the same so you can have just one field, radius, to cover width and height.
[code]ArrayList<Ball> balls = new ArrayList<Ball>();
ballPositionX = balls.get(0).getX(); // gets the first ball's x position
ballPositionY = balls.get(0).getY(); //gets the first ball's y position
Then, circle collision detection is easy. Take two balls. Grab the x and y positions (centers) of both of them. If those positions are within (<) ball1.radius + ball2.radius then they are colliding. Edited by David.M
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well its really an oval lol. so id have to put that in a for loop in order to get each coordinates of each ball getting created every 3 seconds?
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Yes. Loop through balls.size().
Here's a picture of what you're doing for collision detection.
[IMG]http://img4.imageshack.us/img4/4663/20120812163527.jpg[/IMG]
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i currently have this
[source lang="java"] public Rectangle getFixedBallBounds()
{
return new Rectangle (250, 250, 20, 20);
}
public Rectangle getMainBallBounds()
{
return new Rectangle(mainBall.xpos, mainBall.ypos, 10, 10);
}
public boolean checkCollision()
{
if(mainBallRectangle.intersects(fixedBallRectangle))
{
mainBall.xspeed = 0;
mainBall.yspeed = 0;
mainBall.xpos = 100;
mainBall.ypos = 100;
}
return true;
}[/source]
but it does nothing
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Rectangle constructs a rectangle whose upper-left is at the specified x and y coordinates and width and height extend from there. Make sure those are in the proper positions.
Any special reason you're using a boolean function that always returns true?
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Not really i thought that this method returned a boolean value for some reason
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This topic is 1946 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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https://confidenceconnections.com/qa/quick-answer-how-do-you-keep-a-cell-constant-in-a-formula.html | 1,606,874,284,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00359.warc.gz | 217,932,128 | 8,574 | # Quick Answer: How Do You Keep A Cell Constant In A Formula?
## How do you create an absolute cell reference in Excel?
Create an Absolute Reference Click a cell where you want to enter a formula.
Type = (an equal sign) to begin the formula.
Select a cell, and then type an arithmetic operator (+, -, *, or /).
Select another cell, and then press the F4 key to make that cell reference absolute..
## How do you copy an exact formula in Excel without changing the cell reference?
Press F2 (or double-click the cell) to enter the editing mode. Select the formula in the cell using the mouse, and press Ctrl + C to copy it. Select the destination cell, and press Ctl+V. This will paste the formula exactly, without changing the cell references, because the formula was copied as text.
## How do you create a constant in Excel?
Name an array constantClick Formulas > Define Name.In the Name box, enter a name for your constant.In the Refers to box, enter your constant. … Click OK.In your worksheet, select the cells that will contain your constant.In the formula bar, enter an equal sign and the name of the constant, such as =Quarter1.More items…
## How do you do f4 in Excel on a Mac?
For example, the shortcut for Edit Cell in Windows is F2, and on a Mac, it’s Control + U. The shortcut to toggle absolute and relative references is F4 in Windows, while on a Mac, its Command T.
## How do you change a cell reference from relative to absolute?
For example, if you type a relative reference and then press F4, the reference changes to absolute. When you press F4 again, the reference changes to mixed with the row fixed.
## How do you reference a cell in a formula?
Use cell references in a formulaClick the cell in which you want to enter the formula.In the formula bar. , type = (equal sign).Do one of the following, select the cell that contains the value you want or type its cell reference. … Press Enter.
## How do you use the same cell in a formula?
Just select all the cells at the same time, then enter the formula normally as you would for the first cell. Then, when you’re done, instead of pressing Enter, press Control + Enter. Excel will add the same formula to all cells in the selection, adjusting references as needed.
## How do you keep a cell fixed in Excel?
Freeze columns and rowsSelect the cell below the rows and to the right of the columns you want to keep visible when you scroll.Select View > Freeze Panes > Freeze Panes.
## How do you fix a cell in Mac Excel?
If you are using a PC or Windows based laptop, you can make an Excel cell reference absolute (or fixed) by pressing the F4 function key on the keyboard after the cell reference. The equivalent if you are using a Mac is to press ⌘T.
## What is the f4 key on a Mac?
If you are using Mac, the F4 key is usually used for system features, like opening up the launchpad application or changing the brightness of the screen.
## What is absolute cell reference?
In contrast, the definition of absolute cell reference is one that does not change when it’s moved, copied or filled. This way, the reference points back to the same cell, no matter where it appears in the workbook. It’s indicated by a dollar sign in the column or row coordinate.
## What is the constant in a formula?
A constant is a value that doesn’t change (or rarely changes). Because a constant doesn’t change, you could just enter the value right into the formula. For instance, if you want to determine 10% commission on sales, you could use the formula =Sales*.
## How do you do an absolute cell reference on a Mac?
If you’re running MAC, use the shortcut: ⌘ + T to toggle absolute and relative references. You can’t select a cell and press F4 and have it change all references to absolute. You need to have your marker placed inside the reference in the formula before it works when you hit the shortcut.
## Why does my f4 key not work in Excel?
The problem isn’t in Excel, it’s in the computer BIOS settings. The function keys are not in function mode, but are in multimedia mode by default! You can change this so that you don’t have to press the combination of Fn+F4 each time you want to lock the cell.
## How do you reference a value instead of formula in a formula in Excel?
To display the calculated value rather than the formula, you must change the format of the cell containing the formula and re-enter the formula. To do this, follow these steps: Select the cell with the formula, and then click Cells on the Format menu. Click the Number tab. | 1,008 | 4,536 | {"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-2020-50 | latest | en | 0.803984 |
https://artofproblemsolving.com/wiki/index.php?title=Heine-Borel_Theorem&oldid=105228 | 1,713,697,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00034.warc.gz | 103,842,683 | 10,799 | # Heine-Borel Theorem
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The Heine-Borel theorem is an important theorem in elementary topology.
## Statement
Let $E$ be any subset of $\mathbb R^n$. Then $E$ is compact if and only if $E$ is closed and bounded.
This statement does not hold if $\mathbb R^n$ is replaced by an arbitrary metric space $X$. However, a modified version of the theorem does hold:
Let $X$ be any metric space, and let $E$ be a subset of $X$. Then $E$ is compact if and only if $E$ is closed and totally bounded.
In $\mathbb R^n$ the totally bounded sets are precisely the bounded sets, so this new formulation does indeed imply the original theorem. | 180 | 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": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.917023 |
https://www.archivemore.com/what-is-a-line-ray-or-segment-that-divides-a-segment-into-two-congruent-parts-at-a-90-degree-angle/ | 1,716,410,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00693.warc.gz | 569,827,421 | 8,687 | Line Bisector
## What do you call a ray from the vertex of an angle to a point in its interior which divides the angle into two congruent parts *?
Angle bisector is a ray from the vertex of an angle to a point in its interior which divides the angle into two congruent parts.
## Are all rays congruent?
Rays and lines cannot be congruent because they do not have both end points defined, and so have no definite length.
## Why can a ray be bisected?
A line of reflection must exist so that when the figure is folded along this line, each point on one side of the line maps to a corresponding point on the other side of the line. A ray cannot be bisected. Accordingly, can a line bisect a ray? In general, ‘to bisect’ something means to cut it into two equal parts.
## Can a angle bisector be a ray?
The angle bisector is a ray or line segment that bisects the angle, creating two congruent angles. To construct an angle bisector you need a compass and straightedge.
## Can u bisect a ray?
The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides. So whenever you see a triangle with one of its angles bisected, consider using the theorem. How about an angle-bisector problem?
## What do you call the ray that bisect an angle?
An angle bisector is a line or ray that divides an angle into two congruent angles .
## How many angle Bisectors can an angle have?
An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle.
## Does angle bisector bisect opposite side?
The angle bisector theorem is commonly used when the angle bisectors and side lengths are known. It can be used in a calculation or in a proof. An immediate consequence of the theorem is that the angle bisector of the vertex angle of an isosceles triangle will also bisect the opposite side.
## Do angle Bisectors form right angles?
Angle bisector. The angle bisector of an angle of a triangle is a straight line that divides the angle into two congruent angles. The three angle bisectors of the angles of a triangle meet in a single point, called the incenter . For an equilateral triangle the incenter and the circumcenter will be the same.
## Does the angle bisector go through the midpoint?
To bisect a segment or an angle means to divide it into two congruent parts. A bisector of a line segment will pass through the midpoint of the line segment. Any point on the angle bisector of an angle will be equidistant from the rays that create the angle.
## Will drawing an angle bisector always result in two acute angles explain?
No. If the angle bisector is bisecting an obtuse angle it would result in two acute angles. But if the angle bisector is bisecting a reflex angle it would result in two obtuse angles. So it will not always result in two acute angles.
## What is the first step in constructing congruent angles?
Answer use a straightedge and draw an arc across the first arc from a leg use a compass and draw an arc across both the legs of the given angle use a compass and join points to make the new leg of the congruent angle use a straightedge to measure the width between the points where the first arc cuts both legs of given …
## What do you mean by angle bisector of an angle?
The (interior) bisector of an angle, also called the internal angle bisector (Kimberling 1998, pp. 11-12), is the line or line segment that divides the angle into two equal parts. The angle bisectors meet at the incenter. , which has trilinear coordinates 1:1:1.
## How do you know angles are congruent?
Angles are congruent if they have the same angle measure in degrees. They can be at any orientation on the plane. You could say “the measure of angle A is equal to the measure of angle B”. But in geometry, the correct way to say it is “angles A and B are congruent”.
## How do you know if two angles are vertical?
The angles opposite each other when two lines cross. They are always equal. In this example a° and b° are vertical angles.
## What is the vertical angle of a triangle?
Note: In an isosceles triangle, the vertical angle is the angle other than the two equal angles (also known as the base angles). | 980 | 4,344 | {"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} | 4.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.899367 |
http://box2d.org/forum/viewtopic.php?f=2&t=3201&sid=f9bf97a34881dad44713ba06db0ab485&start=10 | 1,519,190,871,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813431.5/warc/CC-MAIN-20180221044156-20180221064156-00236.warc.gz | 50,159,757 | 6,502 | ## Big Changes - Read Before Updating
Erin Catto
Posts: 2948
Joined: Thu Sep 06, 2007 12:34 am
### Re: Big Changes - Read Before Updating
Yeah, you get the shape type off the fixture and cast the shape as needed.
The polygon radius is the skin radius. This means that polygons are held apart according to the radius. This allows me to do continuous physics without a secondary shrunken polygon.
dc443
Posts: 341
Joined: Fri May 16, 2008 10:09 am
### Re: Big Changes - Read Before Updating
I understand the concept of a "skin" of a certain thickness being wrapped around a polygon (with a thickness that i guess is what radius represents) but could you tell me a bit more about what things are now changed due to this? Before, if I create a 1x1 box it would create a core shape which is a (1-2*b2_toislop)x(1-2*b2_toislop) box. If I create a 1x1 box now, is it just basically equivalent to having a rounded out (1+2*b2_toislop)x(1+2*b2_toislop) box with the 1x1 unshrunken box geometry now treated as the core shape?
is the b2_toislop constant still there and being used in that way?
With the core shapes there were certain limitations (and assertions being tripped) with shapes that are too slender or too small, i'm guessing because you couldnt fit a core shape within them. I guess those issues are now gone?
Erin Catto
Posts: 2948
Joined: Thu Sep 06, 2007 12:34 am
### Re: Big Changes - Read Before Updating
dc443 wrote:I understand the concept of a "skin" of a certain thickness being wrapped around a polygon (with a thickness that i guess is what radius represents) but could you tell me a bit more about what things are now changed due to this? Before, if I create a 1x1 box it would create a core shape which is a (1-2*b2_toislop)x(1-2*b2_toislop) box. If I create a 1x1 box now, is it just basically equivalent to having a rounded out (1+2*b2_toislop)x(1+2*b2_toislop) box with the 1x1 unshrunken box geometry now treated as the core shape?
This is correct.
dc443 wrote:is the b2_toislop constant still there and being used in that way?
This is no longer used.
dc443 wrote:With the core shapes there were certain limitations (and assertions being tripped) with shapes that are too slender or too small, i'm guessing because you couldnt fit a core shape within them. I guess those issues are now gone?
This was one of the main reasons for using polygon skins. This also solves some problems with edge shapes. Going forward, edge shapes will just be polygons with two vertices.
dc443
Posts: 341
Joined: Fri May 16, 2008 10:09 am
### Re: Big Changes - Read Before Updating
Your responses are so concise it is mind-boggling. Thanks for everything, Erin!
mayobutter
Posts: 915
Joined: Fri Dec 14, 2007 8:07 pm
Contact:
### Re: Big Changes - Read Before Updating
Shapes can now be created without bodies and can be used in low level collision routines, indepedent of the physics system.
Can someone give an example of this? I'm working on an AS Alchemy port and am wondering whether it's worth it / makes sense to have b2Shape/b2Fixture distinction in AS.
skrekkur
Posts: 10
Joined: Thu Oct 29, 2009 7:58 am
### Re: Big Changes - Read Before Updating
Hi Im trying to branch my project to a newer version of box2d than I am currently using. Part of the changes are the fixtures.
Im starting to get the hang of them but I can't find anywhere a direct replacement of the b2ChainEdgeDef which I am using heavily for collision boundaries in my game, both as pure sensors and collidable terrain.
Is there some direct replacement I am missing, or do I have to manually create multiple 2 vertex edges?
Sincere regards,
Ævar Örn Kvaran
Edit: Found a way to to this, b2Loop2 is a b2Vec array filled in with vertices from my level editor.
Code: Select all
` b2BodyDef bd; b2Body* ground = world->CreateBody(&bd); ground->SetUserData(userData); b2PolygonShape shape; b2FixtureDef fd; fd.shape = &shape; fd.isSensor = isSensor; for (int32 i = 0; i < (vertexCount-1); i++) { shape.SetAsEdge(b2Loop2[i], b2Loop2[i+1]); ground->CreateFixture(&fd); }` | 1,110 | 4,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-09 | latest | en | 0.947322 |
http://www.jiskha.com/display.cgi?id=1363832550 | 1,498,407,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320539.37/warc/CC-MAIN-20170625152316-20170625172316-00144.warc.gz | 552,329,904 | 4,068 | # math
posted by .
Hello,
Is log_3 (5) equal to log_5 (3)? Explain your answer. Do not evaluate the logarithms.
is my question. What is a surefire way to show this is false.
• math -
let x = log (base3) 5
then 3^x = 5 , clearly x > 1 ,
because 3^1 = 3 and 3^2 = 25 ,so x must be between 1 and 2
let y = log(base5)^3
then 5^y = 3 , clearly y < 1,
because 5^0 = 1 and 5^1 = 5
so y has to be between 0 and 1
so x ≠ y
and the two log expressions cannot be equal
• math -
fantastic, makes so much sense now, thx
• math -
and btw 3^2 = 25 is typo | 204 | 551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-26 | latest | en | 0.902855 |
https://math.stackexchange.com/questions/736894/proof-that-lipschitz-condition-guarantees-well-posedness-of-initial-value-proble | 1,713,163,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00860.warc.gz | 352,098,066 | 36,441 | # Proof that Lipschitz condition guarantees well posedness of initial value problems
In the proof of the theorem which states that the Lipschitz condition guarantees well posed-ness of an initial value problem $y'=f(x,y)$, $y(x_0)=y_0$, I came across this
Let the perturbed problem be
$$z'=f(x,z) + \delta(t) \\ z(x_0)=y_0 + \epsilon_0$$
Let $\epsilon(x)=z(x)-y(x)$ and let $f$ satisfy the Lipschitz condition. Then we have
$$\epsilon '=z'-y'=f(x,z)-f(x,y)+\delta(x)\\ \implies |\epsilon'(x)| \leq |f(x,z)-f(x,y)| + |\delta(x)|\\$$
Let $\delta(x)$ have the maximum value $\epsilon$. Therefore, we get,
$$|\epsilon'(x)| \leq |f(x,z)-f(x,y)| + \epsilon\\ \implies |\epsilon'(x)| \leq L|\epsilon(x)| + \epsilon$$
I don't understand the next step i.e. how is the above inequality integrated to give
$$|\epsilon(x)| \leq \frac{\epsilon}{L} \left[ (L+1)e^{Lt}-1\right]$$
How is the inequality integrated?
• Is this in a homework? Or did you come up with it yourself? Apr 2, 2014 at 17:45
• @ellya this is not homework...came across this in a textbook... Gear(1971)...i'll update the exact title later Apr 2, 2014 at 18:43
It is more convenient to use the integral form: $$y(x)=y_0+\int_0^x f\big(s,y(s)\big)\,ds$$ and $$z(x)=y_0+\varepsilon_0+\int_0^x \big(\,f\big(s,z(s)\big)+\delta(s)\big)\,ds$$ and hence $$z(x)-y(x)=\varepsilon_0+\int_0^x\delta(s)\,ds+\int_0^x\big(f\big(s,z(s)\big)-f\big(s,y(s)\big)\big)\,ds,$$ which implies that (for $x>0$) $$\lvert z(x)-y(x)\rvert \le \lvert \varepsilon_0\rvert +\varepsilon x+ L\int_0^x\big|\,z(s)-y(s)\big|\,ds. \tag{1}$$ Now, let $w(x)=\int_0^x \big|\,z(s)-y(s)\big|\,ds$, so the above becomes $$w'(x) \le \lvert \varepsilon_0\rvert +\varepsilon x+ L w(x)$$ or $$\left(\mathrm{e}^{-Lx}w(x)\right)'=\mathrm{e}^{-Lx}\big(w'(x)-Lw(x)\big)\le \mathrm{e}^{-Lx}\big(\lvert\varepsilon_0\rvert +\varepsilon x\big).$$ Integrating the above in $[0,x]$ we obtain $$\mathrm{e}^{-Lx}w(x)-w(0)=\frac{\lvert\varepsilon_0\rvert}{L}(1-\mathrm{e}^{-Lx}) +\varepsilon\mathrm{e}^{-Lx}\left(-\frac{x}{L}+\frac{1}{L^2}\right)-\frac{\varepsilon}{L^2}$$ or $$\int_0^x\lvert z(s)-y(s)\rvert\,ds\le \frac{\lvert\varepsilon_0\rvert}{L}(\mathrm{e}^{Lx}-1) +\varepsilon\left(-\frac{x}{L}+\frac{1}{L^2}\right)-\frac{\varepsilon \mathrm{e}^{Lx}}{L^2}.$$ Using $(1)$ we obtain $$\lvert z(x)-y(x)\rvert \le \lvert\varepsilon_0\rvert(\mathrm{e}^{Lx}-1) +\varepsilon\left(-x+\frac{1}{L}\right)-\frac{\varepsilon \mathrm{e}^{Lx}}{L}+\lvert\varepsilon_0\rvert +\varepsilon x =\cdots.$$
Multiply through by $|\epsilon (x )|$ to get:
$|\epsilon'(x )\epsilon (x )|\leq L|\epsilon (x )|^2+\epsilon|\epsilon (x )|$
$\frac{1}{2}|(\epsilon^2(x))'|\leq L |\epsilon^2(x)|+\epsilon|\epsilon (x)|\leq L|\epsilon(x)^2|+\epsilon^2$
$\Rightarrow \frac {1}{2}\frac {d}{dx}(|\epsilon ^ 2( x ) | e ^ { -2Lx })\leq \epsilon^2e^{-2Lx}\Rightarrow\frac {1}{2}|\epsilon ^ 2( x ) | e ^ { -2Lx}\leq \frac{1}{2}|\epsilon ^ 2 ( 0) |+\frac{1}{2L}(\epsilon ^ 2)(1-e^{-2Lx})$
$\Rightarrow |\epsilon ^ 2( x ) |\leq|\epsilon ^ 2( 0) |e^{2Lx}+\frac{1}{L}(\epsilon^2)(e^{2Lx}-1)\leq\epsilon^2e^{2Lx}+\frac{1}{L}(\epsilon^2)(e^{2Lx}-1)=\frac{\epsilon^2}{L^2}(L^2+Le^{2Lx}-L)$
still working on this, but it's quite close to the desired estimate. | 1,316 | 3,230 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-18 | latest | en | 0.648671 |
http://ccl.northwestern.edu/cm/models/marketplace/info.html | 1,642,433,780,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00512.warc.gz | 11,464,019 | 4,938 | ```WHAT IS IT?
-----------
This model illustrates the Law of Supply and Demand which is the fundamental tool
of economic analysis. The law of supply and demand states that, in a free market,
the price of a good will move to a point where the quantity supplied and the
quantity demanded are equal. This point is known as 'equilibrium'. In theory, if
the functions determining the quantity supplied at a given price (the supply
function) and the quantity demanded at a given price (the demand function) are
well defined - as in this model - this law works very well. In practice, markets
tend to be much more complicated and application of the law is not as
straightforward. Real markets occasionally will disobey it, as there can be market
gluts and shortages, but for the most part the law of supply and demand is a fair
economic generalization.
This model demonstrates both the theory behind the law of supply and demand and
how the law is realized in a theoretical market of buyers and sellers. The supply
and demand functions can be computed and plotted ahead of time, thus yielding an
equilibrium price as predicted by theory. The model can then be run, and from the
interactions between the buyers and sellers we see that in trying to maximize
profit the sellers eventually narrow in on the equilibrium price.
As the model is run it will iteratively cycle through two phases: an interaction
phase (known as a 'round'), in which buyers visit sellers and purchase goods, and
a adjustment phase, in which prices and inventories are redetermined and buyer and
seller variables are reinitialized. It is important to note the activity of the
buyers and the sellers in one cycle in order to understand how an equilibrium
price is eventually reached.
The buyers buy one unit, if they are able to, each round. They will visit up to
ten different sellers searching for a price at which they can afford to buy the
good. If the seller has inventory on the item and the buyer has enough money, then
the purchase is made. The money in the buyers' wallets is equal to their income
per round, a random amount between five and fifty that is determined for each
buyer when the SETUP button in pressed. This money does not accumulate from round
to round, and so a buyers will always begin every round with the same amount of
money.
Sellers each have an individual price at which they will sell one unit of their
inventory. This price is initialized randomly to a value between 5 and 50 when the
SETUP button is pressed. New inventory is computed after each round by taking the
selling price and dividing it by the cost price (or wholesale price) of the good,
which is set as a constant at 5 dollars. This amount is then rounded down, and
added to any existing inventory (left over from the previous rounds). For example,
a seller with a price of 27 will restock 5 units (27/5 rounded down to 5) to the
existing inventory. When SETUP is pressed, a seller initialized with a price of 27
will begin the simulation with an inventory of 5 units.
A simple heuristic is used for the sellers to approximate the price at which their
profit will be maximized. If the sales from the previous round met or exceeded a
seller's inventory, then there is a ten percent chance that the seller will raise
the price. If the sales were less than the inventory, then the seller will lower
the price by one. In the real world such a strategy might not be very feasible,
but it is important to note that there is no magic formula for determining
pricing. As with real companies, sellers in this model have no access to an actual
demand function on which to base their decisions (sellers can NOT simply calculate
the equilibrium price!). This would require 'global' information about the market,
but sellers in this model have only the 'local' information about their individual
sales and inventories to work with. (Note that through surveys and polling of
potential customers, large companies today often do try to determine general
information about their market. However, as this model demonstrates, global
information about the market is not necessary for equilibrium to be reached.)
Before or during the running of the model, a plot of the supply and demand
functions can be made to predict the equilibrium price (using COMPUTE-CURVES). The
demand function is determined by calculating the total number of buyers with
incomes greater than or equal to each corresponding price on the y-axis. This is
the total quantity that will be sold at that price. The supply function is equal
to the total amount of inventory that will be available in the market if every
seller sets their price to the corresponding price on the y-axis. This is the
total quantity that will be produced at that price. The price at which these
curves intersect is the equilibrium price. Because of rounding down in
determination of supply at a given price, the supply function is actually a step
function - it has been smoothed to simplify interpretation.
There are many possibilities to extend this model beyond its basic framework to
provide an analysis of different economic factors. The law of supply and demand
can be used, among other options, to analyze unemployment, the value of the dollar
overseas, international trade, and environmental protection. See the 'EXTENDING
THE MODEL' section for a list of ideas.
HOW TO USE IT
-------------
The number of buyers and sellers can be adjusted on the interface panel. Choose a
ratio between the number of buyers and the number of sellers, or to start off try
a ratio of one hundred buyers to twenty sellers. Press the SETUP button, and then
press GO. Buyers are in green, Sellers are in blue. The average price after each
round is plotted over time on the second plot window. The first plot window is
used to predict what the equilibrium price will be based on the supply and demand
functions.
INTERFACE ITEMS:
SETUP: resets the simulation according to new settings
GO: runs the simulation.
COMPUTE-CURVES: computes and plots the supply and demand curves for the current
run (plot-window 2). the intersection of the these curves is the predicted
equilibrium price.
N-BUYERS: sets the number of buyers to create when RESET is pressed. Has no
effect during a simulation.
N-SELLERS: sets the number of sellers to create when RESET is pressed. Has no
effect during a simulation.
FAST-MODE: turns on/off the graphic display of the buyers entering the
marketplace. The simulation runs faster with the graphics turned off.
AVE-PRICE: shows the average selling price of all of the sellers.
AVE-SALES: shows the average number of sales for all of the sellers.
GROSS-PROFIT: shows the average gross-profit for all of the sellers.
PLOTS:
PREDICTED-EQUILIBRIUM-PRICE: To activate this plot, press the COMPUTE-CURVES
button. Shown in blue is the supply-function for the sellers in the
simulation. That is, corresponding to each quantity (on the x-axis)
is the price they are willing to sell one unit for. The supply function does
not change from run to run or as the settings are changed. Shown in green is
the demand function for the buyers in the simulation. That is, corresponding
to each price (on the y-axis) is the number of buyers who are willing to buy
at that price. As each buyer buys only one unit per round in this model, this
is equivalent to the quantity that will be bought at that price. The
demand function changes as the settings are changed and, because income is set
randomly for each buyer when the reset button is pressed, it will typically
shift slightly from between runs with the same settings. The price at which
the two curves intersect is known as the equilibrium price, which will be just
below the number indicated on the top of the y-axis. As the supply and demand
functions do not change during a simulation, sellers should - in theory -
maximize gross-profit by selling at the equilibrium price.
PRICE-OVER-TIME: tracks the average selling price over time.
SALES-OVER-TIME: tracks the average quantity sold (per seller) over time.
GROSS-PROFIT-OVER-TIME: tracks the average gross-profit (per-seller) over
time.
THINGS TO TRY
-------------
Try varying the ratio of buyers to sellers. What is the relation between this
ratio and the average price of the good?
What happens to the price when there is a scarcity or an abundance of the good?
After pressing SETUP, press COMPUTE-CURVES and look at the predicted equilibrium
price. Now run the model. Does the average price generated from the model fit the
prediction. Under what circumstances will the prediction fail? Why?
EXTENDING THE MODEL
-------------------
The demand and supply curves in this model are linear. This is because sellers
have no direct control over the quantity they stock - the amount of inventory is a
determinate, linear function of their selling price. Some things you can do to
change this: (1) Introduce a variable inventory so that sellers can control both
price AND quantity of inventory. (2) Make the mapping from price to amount of
inventory non-linear. Often supply functions (also demand functions) are modeled
as quadratic curves, as efficiency or interest in producing the good may rise or
fall with the selling price.
Introduce multiple goods into the system. Have buyers with variable demand for the
different goods.
This model doesn't take geography into account - set up conditions so that
different areas of the screen lead to buyers and sellers with different
characteristics. One way to this would be to have sellers 'forage' for goods
instead of purchasing them wholesale. Goods could sprout from patches, and various
areas could have different growth characteristics or grow different products. What
kinds of regional differences can you get to emerge (i.e. lower priced areas,
high-income areas)?
Test the pricing strategy of the sellers - are they maximizing their gross-profit?
Why do you think that the price setting heuristic works as well as it does?
Discover and implement a better price-setting strategy - perhaps one that isn't a
probabilistic heuristic, but that has explicit rules.
Where do the sellers' profits go? Have a model that conserves the total amount of
money in the system and/or resources in the system to demonstrate a true flow of
goods.
How would government policies affect the model? What happens when a sales tax is
imposed? What is the effect of a federally imposed price floor, such as the
minimum wage price on labor? Play the role of government and test your policies on
the model.
Can variables be introduced which take into account the externalities of
production, i.e. pollution?
The model can be used more generally to consider international trade between an
importer and an exporter. How do tariffs and quotas affect the model? What are the
benefits to the nation that imports?
``` | 2,314 | 10,847 | {"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-2022-05 | latest | en | 0.961154 |
https://www.coursehero.com/file/6214646/ch16-p086/ | 1,490,304,907,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187206.64/warc/CC-MAIN-20170322212947-00660-ip-10-233-31-227.ec2.internal.warc.gz | 879,954,433 | 114,375 | # ch16-p086 - 86. Repeating the steps of Eq. 16-47 Eq. 16-53,...
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86. Repeating the steps of Eq. 16-47 Eq. 16-53, but applying cos cos 2cos cos 22 α β αβ +− §· += ¨¸ ©¹ (see Appendix E) instead of Eq. 16-50, we obtain [0.10cos ]cos4 y xt π , with SI units understood. (a) For non-negative x , the smallest value to produce cos π x = 0 is x = 1/2, so the answer is x = 0.50 m. (b) Taking the derivative, [] () 0.10cos 4 sin4 dy ux t dt == π
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## This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
Ask a homework question - tutors are online | 250 | 779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-13 | longest | en | 0.867073 |
https://www.convertunits.com/from/siriometer/to/U | 1,638,815,613,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00413.warc.gz | 806,859,711 | 12,867 | ## ››Convert siriometre to U
siriometer U
How many siriometer in 1 U? The answer is 2.9712989897517E-19.
We assume you are converting between siriometre and U.
You can view more details on each measurement unit:
siriometer or U
The SI base unit for length is the metre.
1 metre is equal to 6.684587153547E-18 siriometer, or 22.497187851519 U.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between siriometers and U.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of siriometer to U
1 siriometer to U = 3.3655313835771E+18 U
2 siriometer to U = 6.7310627671541E+18 U
3 siriometer to U = 1.0096594150731E+19 U
4 siriometer to U = 1.3462125534308E+19 U
5 siriometer to U = 1.6827656917885E+19 U
6 siriometer to U = 2.0193188301462E+19 U
7 siriometer to U = 2.3558719685039E+19 U
8 siriometer to U = 2.6924251068616E+19 U
9 siriometer to U = 3.0289782452193E+19 U
10 siriometer to U = 3.3655313835771E+19 U
## ››Want other units?
You can do the reverse unit conversion from U to siriometer, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 518 | 1,658 | {"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} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.785708 |
https://howtodunia.com/mcq-questions-for-class-11-physics-chapter-10/ | 1,685,483,143,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00106.warc.gz | 357,564,333 | 32,057 | # MCQ Questions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids with Answers
Here you will find NCERT MCQ Questions for Class 11 Physics with Answers PDF Free Download based on the important concepts and topics given in the textbook as per CBSE new exam pattern. This may assist you to understand and check your knowledge about the chapters. Students also can take a free test of the Multiple Choice Questions of Class 11 Physics. Each question has four options followed by the right answer. These MCQ Questions are selected supported by the newest exam pattern as announced by CBSE.
## Q1. One end of a towel dips into a bucket full of water and other end hangs over the bucket. It is found that after some time the towel becomes fully wet. It happens
(a) Because viscosity of eater is high
(b) Because of the capillary action of cotton threads
(c) Because of gravitational force
(d) Because of evaporation of water.
(b) Because of the capillary action of cotton threads
(a) Acute angle
(b) Obtuse angle
(c) 90°
(d) 0°
(d) 0°
(a) M
(b) 2M
(c) M/2
(d) 4M
(b) 2M
(a) 1.5 g
(b) 10 g
(c) 5 g
(d) 15 g
(b) 10 g
## Q5. The height of a liquid in a fine capillary tube
(a) Increases with an increase in the density of a liquid
(b) Decreases with a decrease in the diameter of the tube
(c) Decreases with an increase in the surface tension
(d) Increases as the effective value of acceleration due to gravity is decreased
(d) Increases as the effective value of acceleration due to gravity is decreased
## Q6. At critical temperature, the surface tension of a liquid
(a) Is zero
(b) Is infinity
(c) Is the same as that at any other temperature
(d) Can not be determined
(a) Is zero
(a) 2 cm
(b) 3 cm
(c) 6 cm
(d) 9 cm
(c) 6 cm
## Q8. When the angle of contact between a solid and a liquid is 90°, then
(a) Cohesive force > Adhesive force
(b) Cohesive force < Adhesive force
(c) Cohesive force = Adhesive force
(d) Cohesive force >> Adhesive force
(c) Cohesive force = Adhesive force
(a) 4 mm
(b) 1 mm
(c) 3 mm
(d) 1 cm
(b) 1 mm
(a) Acute angle
(b) Obtuse angle
(c) 90°
(d) 0°
(d) 0°
## Q11. Liquid pressure depends upon
a) height of the liquid column
b) directions
c) shape of the liquid surface
d) area of the liquid surface
a) height of the liquid column
(a) 1 cm
(b) 0.1 cm
(c) 2 cm
(d) 4 cm
(a) 1 cm
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 2 cm
(c) 8 cm
(a) Plane
(b) concave
(c) convex
(d) Both b and c
(b) concave
## Q15. Pressure inside two soap bubbles is 1.01 and 1.02 atmospheres. ratio between their volume is
(a) 102 : 101
(b) (102)3 : (101)3
(c) 8 : 1
(d) 2 : 1
(c) 8 : 1
## Q16. The pressure at the bottom of a tank containing a liquid does not depend on
a) area of the bottom surface
b) nature of the liquid
c) height of the liquid column
d) acceleration due to gravity
a) area of the bottom surface
## Q17. Specific gravity of a body is numerically equal to
a) relative density of the body
b) density of body in water
c) weight of the body in water
d) weight of the body in air
a) relative density of the body
## Q18. Assertion: A small iron needle sinks in water while a large iron ship floats.Reason: The shape of iron needle is like a flat surface while the shape of a ship is that which makes it easier to float.
a) Assertion is correct, reason is incorrect
b) Assertion is incorrect, reason is correct
c) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
d) Assertion is correct, reason is correct; reason is a correct explanation for assertion
a) Assertion is correct, reason is incorrect
a) 11%
b) 8%
c) 3%
d) 1.8%
a) 11%
a) rises
b) remains same
c) lowers
d) None of these
a) rises
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 2 cm
(c) 8 cm
## Class 11 Physics MCQs Questions with Answers Chapter Wise
Practicing NCERT Physics MCQs With Answers Pdf Class 11 is one of the best ways to prepare for the CBSE Class 11 board exam. There is no substitute for consistent practice whether one wants to understand a concept thoroughly or one wants to score better. If you have any queries regarding The CBSE Class 11 Physics MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.
Scroll to Top | 1,220 | 4,236 | {"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} | 3.265625 | 3 | CC-MAIN-2023-23 | latest | en | 0.829727 |
https://www.jiskha.com/questions/1846644/to-describe-a-sequence-of-transformations-that-maps-triangle-abc-onto-triangle-abc | 1,632,387,825,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00562.warc.gz | 846,087,360 | 4,877 | # Math
To describe a sequence of transformations that maps triangle ABC onto triangle A''B''C'', a student starts
with a reflection over the x-axis. How should the student complete the sequence of transformations to
map triangle ABC onto triangle A''B''C''?
Answer: The set of modifications to map triangle ABC to triangle A'B'C' by a student beginning with a reflection over the x-axis can be completed by a student starting with a reflection over the x-axis. The student completes the series of transformations by converting the units of Figure 2 to the right to map triangle ABC to triangle A'B 'C
1. 👍
2. 👎
3. 👁
1. Is that the right answer?
1. 👍
2. 👎
2. IS IT?
1. 👍
2. 👎
3. To describe a sequence of transformations that maps triangle ABC onto triangle A"B"C", a student starts with a reflection over the x-axis. The student student complete the sequence of transformations to map triangle ABC to triangle A"B"C" is by translating the figure 2 units to the right. Translate the figure 6 units up.
1. 👍
2. 👎
## Similar Questions
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Triangle ABC is reflected over the y-axis what are the coordinates of the reflected triangle describe in words what happens to the x coordinates and the y coordinates of the original triangle's vertices as a result of this | 966 | 3,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-39 | latest | en | 0.840398 |
http://www.stata.com/statalist/archive/2013-06/msg01179.html | 1,501,203,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436316.91/warc/CC-MAIN-20170728002503-20170728022503-00388.warc.gz | 559,014,342 | 4,009 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: Data Management question
From Nick Cox To "statalist@hsphsun2.harvard.edu" Subject Re: st: Data Management question Date Tue, 25 Jun 2013 18:02:58 +0100
```You said
"I could do this the long way with repeated gen and replace commands".
Actually that sounds a nightmare without some small tricks. Here are mine:
egen ABCDE = concat(A B C D E)
For consecutive 1s, the trick is only that somewhere in the string
concatenation we will find "111" or "11".
gen X = strpos(ABCDE, "111") > 0
gen Y = strpos(ABCDE, "11") > 0
Your third definition splits into two
For one non-consecutive 1, the condition "non-consecutive" is superfluous.
For two non-consecutive 1s, there are in total two 1s and they must
not be consecutive. The variable -Y- already codes two consecutive
1s.
Either way, try
gen Z = inlist(A + B + C + D + E, 1, 2) & !Y
Nick
njcoxstata@gmail.com
On 25 June 2013 17:46, Hitesh Chandwani <hchandwani.stata@gmail.com> wrote:
> I have a data management issue. I need to generate a couple of
> indicator variables from variables that already exist in the dataset.
>
> Let's call the required indicator vars X, Y, and Z with values of 1 or
> 0. Whether the value of these vars is 1 or 0 depends on the value of 5
> other indicator variables (A, B, C, D, E). A-E are chronologically
> ordered (i.e., A indicates an event that happened before B, which
> happened before C, and so on).
>
> For each observation:
> X = 1 if at least 3 "consecutive" vars from A-E are equal to 1.
> Y = 1 if at least 2 "consecutive" vars from A-E are equal to 1.
> Z = 1 if only 1 or 2 "non-consecutive" vars from A-E are equal to 1.
>
> I could do this the long way with repeated gen and replace commands
> but I'm sure there's a better way to do it (only I don't know how).
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
``` | 613 | 2,063 | {"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} | 2.78125 | 3 | CC-MAIN-2017-30 | latest | en | 0.89916 |
http://www.dpi.inpe.br/terralib/html/v410/class_te_graph_network.html | 1,368,883,350,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382398/warc/CC-MAIN-20130516092622-00047-ip-10-60-113-184.ec2.internal.warc.gz | 442,426,581 | 5,291 | TerraLib 4.1
TeGraphNetwork Class Reference
class for handling networks More...
`#include <TeNetwork.h>`
List of all members.
## Public Member Functions
TeGraphNetwork ()
Empty constructor.
TeGraphNetwork (TeLineSet &ls)
Create a graph from TeLineSet; the line size is considered the cost. In this case, graph non-directed.
TeGraphNetwork (TeLineSet &ls, map< string, double > &line_costs)
Create a graph from TeLineSet; the cost is given by the map.
TeGraphNetwork (TeSTElementSet &stos, string &attrName)
Create a network from a set of line objects; the cost is given by the attrName sto property. The graph is non-directed.
TeGraphNetworkoperator= (TeGraphNetwork &other)
Assignment operator.
bool Add (TeLineSet &ls, map< string, double > &line_costs)
Add lineset to graph. Useful specially for directed graphs (lines must be entered in both directions. It has to be tested for existing graphs.
bool minimumPath (TeNode &n1, TeNodeSet &set, vector< double > &dist)
Calculate the minimun path.
bool getNode (int i, TeNode &node)
Get the i-th node.
bool nearestNodePoint (TeCoord2D &p1, int &lindex, TeCoord2D &pinter, double &distance, double tol=0.0)
Get the nearest network node of a specific point (p1)
bool nearestNetworkPoint (TeCoord2D &p1, int &lindex, TeCoord2D &pinter, double &distance, double tol=0.0)
Get the nearest network lines point from a specific point (p1)
TeLineSet getLineSet ()
Get line Set.
map< string, double > getLineCosts ()
Get line costs.
void insertLine (TeLine2D &line, const double &attr)
Insert a new line.
bool breakLineSet (TeNode &node, int i)
Insert a new node.
virtual ~TeGraphNetwork ()
Destructor.
## Protected Attributes
br_stl::Graph< TeNode, double > graph_
Set of nodes and the cost of each edge.
TeLineSet line_set_
Set of geometric representation of edges.
map< string, double > line_cost_
A map to associate each edge (line object_id) to its cost.
## Detailed Description
class for handling networks
## Constructor & Destructor Documentation
TeGraphNetwork::TeGraphNetwork ( ) ` [inline]`
Empty constructor.
TeGraphNetwork::TeGraphNetwork ( TeLineSet & ls )
Create a graph from TeLineSet; the line size is considered the cost. In this case, graph non-directed.
TeGraphNetwork::TeGraphNetwork ( TeLineSet & ls, map< string, double > & line_costs )
Create a graph from TeLineSet; the cost is given by the map.
TeGraphNetwork::TeGraphNetwork ( TeSTElementSet & stos, string & attrName )
Create a network from a set of line objects; the cost is given by the attrName sto property. The graph is non-directed.
virtual TeGraphNetwork::~TeGraphNetwork ( ) ` [inline, virtual]`
Destructor.
## Member Function Documentation
bool TeGraphNetwork::Add ( TeLineSet & ls, map< string, double > & line_costs )
Add lineset to graph. Useful specially for directed graphs (lines must be entered in both directions. It has to be tested for existing graphs.
bool TeGraphNetwork::breakLineSet ( TeNode & node, int i )
Insert a new node.
map TeGraphNetwork::getLineCosts ( ) ` [inline]`
Get line costs.
TeLineSet TeGraphNetwork::getLineSet ( ) ` [inline]`
Get line Set.
bool TeGraphNetwork::getNode ( int i, TeNode & node )
Get the i-th node.
void TeGraphNetwork::insertLine ( TeLine2D & line, const double & attr )
Insert a new line.
bool TeGraphNetwork::minimumPath ( TeNode & n1, TeNodeSet & set, vector< double > & dist )
Calculate the minimun path.
bool TeGraphNetwork::nearestNetworkPoint ( TeCoord2D & p1, int & lindex, TeCoord2D & pinter, double & distance, double tol = `0.0` )
Get the nearest network lines point from a specific point (p1)
bool TeGraphNetwork::nearestNodePoint ( TeCoord2D & p1, int & lindex, TeCoord2D & pinter, double & distance, double tol = `0.0` )
Get the nearest network node of a specific point (p1)
TeGraphNetwork& TeGraphNetwork::operator= ( TeGraphNetwork & other ) ` [inline]`
Assignment operator.
## Member Data Documentation
br_stl::Graph TeGraphNetwork::graph_` [protected]`
Set of nodes and the cost of each edge.
map TeGraphNetwork::line_cost_` [protected]`
A map to associate each edge (line object_id) to its cost.
TeLineSet TeGraphNetwork::line_set_` [protected]`
Set of geometric representation of edges.
The documentation for this class was generated from the following files:
• E:/Projetos_Primeiro_Semestre_2012/TerraView/terralib/src/terralib/kernel/TeNetwork.h
• E:/Projetos_Primeiro_Semestre_2012/TerraView/terralib/src/terralib/kernel/TeNetwork.cpp | 1,192 | 4,496 | {"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-2013-20 | latest | en | 0.658893 |
https://us.metamath.org/nfeuni/19.21tOLD.html | 1,716,747,113,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00545.warc.gz | 496,499,998 | 3,840 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > 19.21tOLD GIF version
Theorem 19.21tOLD 1863
Description: Obsolete proof of 19.21t 1795 as of 30-Dec-2017. (Contributed by NM, 27-May-1997.) (Revised by Mario Carneiro, 24-Sep-2016.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
19.21tOLD (Ⅎxφ → (x(φψ) ↔ (φxψ)))
Proof of Theorem 19.21tOLD
StepHypRef Expression
1 id 19 . . . 4 (Ⅎxφ → Ⅎxφ)
21nfrd 1763 . . 3 (Ⅎxφ → (φxφ))
3 alim 1558 . . 3 (x(φψ) → (xφxψ))
42, 3syl9 66 . 2 (Ⅎxφ → (x(φψ) → (φxψ)))
5 nfa1 1788 . . . . . 6 xxψ
65a1i 10 . . . . 5 (Ⅎxφ → Ⅎxxψ)
71, 6nfimd 1808 . . . 4 (Ⅎxφ → Ⅎx(φxψ))
87nfrd 1763 . . 3 (Ⅎxφ → ((φxψ) → x(φxψ)))
9 sp 1747 . . . . 5 (xψψ)
109imim2i 13 . . . 4 ((φxψ) → (φψ))
1110alimi 1559 . . 3 (x(φxψ) → x(φψ))
128, 11syl6 29 . 2 (Ⅎxφ → ((φxψ) → x(φψ)))
134, 12impbid 183 1 (Ⅎxφ → (x(φψ) ↔ (φxψ)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 176 ∀wal 1540 Ⅎwnf 1544 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-6 1729 ax-11 1746 This theorem depends on definitions: df-bi 177 df-ex 1542 df-nf 1545 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 655 | 1,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-22 | latest | en | 0.319229 |
https://mcdonaldsalesandmarketing.biz/category/the-problem-with-lecturing/ | 1,722,701,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00835.warc.gz | 311,265,807 | 83,399 | Learning: The Problem with Lecturing is that very Little Learning Occurs
Learning: “The introductory classes only seem to be really working for about 10 percent of the students. These are the 10 percent who would learn it even without the instructor.” -David Hestenes
Back in the late 1970s a colleague came to David Hestenes with a problem.
The two of them were physics professors at Arizona State University. Hestenes was teaching mostly graduate students, but his colleague was teaching introductory physics, and the students in his classes were not doing well. Semester after semester, the class average on his exams never got above about 40 percent.
“And I noted that the reason for that was that his examination questions were mostly qualitative, requiring understanding of the concepts,” says Hestenes.
Most professors didn’t test for this kind of understanding; students just had to solve problems to pass the exams.
This observation prompted a series of conversations between Hestenes and his colleague about the difference between being able to solve problems and really understanding the concepts behind those problems. They had a sneaking suspicion students were just learning the problem-solving part and never really getting the concepts.
Testing Understanding
Hestenes and one of his graduate students, Ibrahim Halloun, decided to test this theory by coming up with a way to probe students’ conceptual understanding of physics.
They developed a multiple-choice test, now known as the Force Concept Inventory, or FCI. Here is the first and probably easiest question on that test:
Two metal balls are the same size but one weighs twice as much as the other. The balls are dropped from the roof of a two-story building at the same instant. The time it takes the balls to reach the ground will be
1. about half as long for the heavier ball as for the lighter one.
2. about half as long for the lighter ball as for the heavier one.
3. about the same time for both balls.
4. considerably less for the heavier ball but not necessarily half as long.
5. considerably less for the lighter ball but not necessarily half as long.
This is a fundamental concept in physics but even some people who have taken physics courses get this question wrong. The correct answer is not what intuition suggests (we’ll get to the right answer in a minute).
Hestenes and Halloun gave their test to about 1000 students in introductory physics courses taught by seven different instructors at two different schools.
Each class was taught in a traditional lecture mode, though the instructors had different styles. One did a lot of elaborate demonstrations. One emphasized problem-solving. Another was a theoretical physicist who devoted a lot of time to talking about the conceptual structure of physics.
Students in these seven different classes took the test at the beginning of the semester. Perhaps not surprisingly, they didn’t do very well (though many of them had already taken high school physics; Hestenes and Halloun expected them to do better).
The students took the test again at the end of the semester. And they still didn’t do very well. Their scores went up by
David Hestenes
only about 14 percent, meaning that, after an entire semester, they understood only about 14 percent more about the fundamental concepts of physics than they had at the beginning.
The results were published in a series of articles in the American Journal of Physics in 1985.
Hestenes says the editor of the American Journal of Physics was fascinated by the findings, but there was not a lot of immediate interest from his fellow physics professors.
Even the ones who’d participated in the experiment “didn’t pay any attention” to the results, says Hestenes. “One reason may be that they really didn’t know what to do.”
Lecturing was the way just about everyone taught introductory physics. To think there was something wrong with the lecture meant physics instructors would “have to really change the way they do things,” says Hestenes.
A lot of them ignored his study and kept teaching the way they always had. They insisted their lectures were working just fine.
Taking It to Heart
But Eric Mazur was unusual, says Hestenes. “He was the first one who took it to heart.”
Mazur is a physics professor at Harvard University. He came across Hestenes’s articles in 1990, five years after they’d been published.
To understand why the articles had such a big impact on Mazur you have to know some things about his history.
Mazur grew up dreaming of becoming an astronomer.
“When I was five years old I fell in love with the universe,” he says. “I tried to get my hands on to every accessible book on astronomy. I was so excited by the world of science.”
But when Mazur got to university, he hated the astronomy classes.
“It was all sitting in the lecture, and then scribbling down notes and cramming those notes and parroting them back on
Eric Mazur
the exam,” he says. “Focusing on the details, focusing on memorizing and regurgitation, the whole beauty of astronomy was lost.”
So he switched to physics. It wasn’t as heartbreaking for him to sit in a physics lecture and memorize things.
Mazur eventually got a Ph.D. in physics and a job at Harvard University. Like most Ph.D.s, Mazur never got any training in how to teach.
“I just mimicked what my instructors had done to me. I think that’s what we all do. So, I lectured.”
Turns out he loved lecturing. It’s a lot more fun being on stage delivering a lecture than it is sitting in the audience watching. And that’s exactly what a lecture is, says Mazur: a performance. He decided to make it fun.
“Thanks to the setup we have here at Harvard, it was very flashy, like a Hollywood show,” he says. “Attention-grabbing demos, me shooting through the lecture hall in a rocket car.”
Mazur’s students apparently loved it. His classes were full and he got great evaluations from the students at the end of every semester.
“For a long while, I thought I was doing a really, really good job,” he says.
Not My Students
Then Mazur read the articles by Hestenes and Halloun. Mazur’s first instinct was to dismiss the results. The test covered such basic material; he was sure his students were learning this stuff.
But what if they weren’t? How boring it would be to learn physics and never really understand the fundamental concepts that make physics so fascinating. Mazur thought back to his own experience with astronomy; if his students were just memorizing information and solving problems, he had to know, and he had to do something about it.
So he gave them the FCI, and he was shocked.
“They didn’t do much better,” he says. “In fact, when they looked at the test that I gave to them some students asked me, ‘How should I answer these questions? According to what you taught me, or according to the way I usually think about these things?’ That’s when it started to dawn on me that something was really amiss.”
What Mazur and other physicists have come to understand is that one reason it’s hard for students to learn physics is that they come into class with a very strong set of intuitive beliefs about how the physical world works.
“We can function quite well using these intuitive beliefs,” says Mazur. “We can push a chair on the floor, we can throw a ball, even though we’ve never studied parabolic trajectories and even though we’ve never really understood forces and friction.”
But it turns out that many of these intuitive notions do not square with what physicists have discovered about how things actually work.
Most people’s intuition tells them if you drop two balls of different weights from the second story of a building, the heavier ball will reach the ground first.
But it doesn’t. The answer to the question above is #3: The balls reach the ground at the same time.
This is a very difficult concept for most students to understand because they already have a concept in their mind that’s in conflict with this new concept.
“Once you understand physics you can connect those two concepts and you can see everything as part of a coherent set of laws and framework of laws,” says Mazur.
But according to the results of the Force Concept Inventory test, which has now been given to tens of thousands of students around the world, most people who take conventional lecture-based courses don’t end up with a good understanding of the fundamental concepts of physics.
“I think what many students in their introductory physics courses do is they retain their intuitive notions,” says Mazur. They memorize what the professor tells them and “parrot it back” on the exam but they never really connect what they are learning to what they already think about how the physical world works.
The way he really learned physics was to teach himself.
“I am what I am not because of my education but probably in spite of it,” he says.
“They Learn It on Their Own.”
David Hestenes says this is true for a lot of students.
“If you look at what’s happening in the introductory classes, even at the best schools, the classes only seem to be really working for about 10 percent of the students,” he says. “And I think all the evidence indicates that these 10 percent are the 10 percent of students that would learn it even without the instructor. They essentially learn it on their own.”
A whole field of education research has emerged from what physicists have learned about the problems with the traditional lecture. There are now Physics Education Research groups at dozens of universities and a long list of peer-reviewed studies that confirm what they have found. It’s not just physics where lectures fail; the traditional lecture is not an effective way to teach any subject.
“Students have to be active in developing their knowledge,” says Hestenes. “They can’t passively assimilate it.”
The idea that people learn better when they’re actively engaged is one of the central findings from an explosion of cognitive research conducted over the last several decades.
“Students come to the classroom with preconceptions about how the world works,” according to a major report by the National Research Council. “If their initial understanding is not engaged, they may fail to grasp the new concepts and information that are taught, or they may learn for the purposes of a test but revert to their preconceptions outside the classroom.”
Changing the Way Professors Teach
Researchers and instructors have developed a number of “interactive-engagement” techniques in recent years that have proven more effective than lectures for teaching large classes. Mazur uses an approach that he calls “peer instruction.” It’s hard to know how many instructors are using these approaches, but experts say most large, introductory classes â especially in the sciences â are still taught using the conventional lecture method.
Mazur says even now, years after the first articles about the FCI and the establishment of dozens of Physics Education Research programs, a lot of professors still have a hard time getting their heads around the idea that there is anything wrong with lecturing.
“Most of the people who are teaching are products of this approach to teaching,” he says. “They were successful in this approach.”
He says they are the ones, like him, who taught themselves, because they were interested and motivated to learn. But Mazur says professors need to do better, because the economy is changing; more people need to succeed in college now.
Mazur believes the challenge for educators in the 21st century is to find new ways to reach more students, and to help them learn better than they ever have before. | 2,384 | 11,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.984632 |
http://mathhelpforum.com/calculus/109377-cauchy-riemman-equations-print.html | 1,516,440,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00592.warc.gz | 235,964,475 | 3,326 | # Cauchy-Riemman equations
• Oct 20th 2009, 09:10 PM
devouredelysium
Cauchy-Riemman equations
Ok. I am used to see written everywhere the following Cauchy-Rieamman equations for complex functions:
and
Now, today I saw this other notation
http://upload.wikimedia.org/math/0/9...98334feee9.png and I'd like to understand how it maps to the first. It must be pretty simple, I guess?
Thanks
• Oct 20th 2009, 09:30 PM
mr fantastic
Quote:
Originally Posted by devouredelysium
Ok. I am used to see written everywhere the following Cauchy-Rieamman equations for complex functions:
and
Now, today I saw this other notation
http://upload.wikimedia.org/math/0/9...98334feee9.png and I'd like to understand how it maps to the first. It must be pretty simple, I guess?
Thanks
f = u + iv.
• Oct 20th 2009, 09:34 PM
tonio
Quote:
Originally Posted by devouredelysium
Ok. I am used to see written everywhere the following Cauchy-Rieamman equations for complex functions:
and
Now, today I saw this other notation
http://upload.wikimedia.org/math/0/9...98334feee9.png and I'd like to understand how it maps to the first. It must be pretty simple, I guess?
Thanks
Yes it is, but only after one understands what's going on here: you have a complex variable function $f(z)$, but we can put $z=x+iy=(x,y)$, where $i=\sqrt{-1}$ and $(x,y)$ is the representation of a complex number in the complex plane.
Thus we can write $f(z)=f(x,y)=u(x,y)+iv(x,y)$ , with the usual division of the function in its real and imaginary functions $u , v$
Now, finally, applying the rule for partial derivatives of multivariable functions we get:
$i\frac{\partial f}{\partial x}=i\left(\frac{\partial u}{\partial x}+i\,\frac{\partial v}{\partial x}\right)=$ $\left(\frac{\partial u}{\partial y}+i\,\frac{\partial v}{\partial y}\right)=\frac{\partial f}{\partial y}$
Now just compare real and imaginary parts in both sides.
Tonio
• Oct 20th 2009, 09:38 PM
devouredelysium
Hmm was almoooost there. Now I get it! Thanks! | 580 | 1,989 | {"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": 8, "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-2018-05 | longest | en | 0.847249 |
algotron.medium.com | 1,721,062,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00477.warc.gz | 75,090,785 | 27,441 | --
# How to Find Highly Correlated Assets and Trade Them Using CCXT Library
In the world of cryptocurrency trading, finding highly correlated assets and trading them against each other can be a profitable strategy. In this blog post, we will discuss how to find highly correlated assets, how to calculate trading fees, and how to trade them using the CCXT library.
# Finding Highly Correlated Assets
The first step in implementing a statistical arbitrage trading strategy is to find highly correlated assets. To do this, you can use historical price data to calculate the correlation coefficient between two or more crypto assets. The correlation coefficient is a statistical measure that indicates the strength of the relationship between two variables. In this case, we are interested in finding crypto assets that move in the same direction most of the time.
To calculate the correlation coefficient between two crypto assets, you can use a spreadsheet program like Microsoft Excel or Google Sheets. Simply input the price data for each asset into separate columns, and then use the CORREL function to calculate the correlation coefficient.
Alternatively, you can use Python and the Pandas library to calculate the correlation coefficient. Here’s an example code snippet:
`import ccxtimport pandas as pdexchange = ccxt.binance()symbol_1 = 'BTC/USDT'symbol_2 = 'ETH/USDT'ohlcv_1 = exchange.fetch_ohlcv(symbol_1, '1d')ohlcv_2 = exchange.fetch_ohlcv(symbol_2, '1d')df_1 = pd.DataFrame(ohlcv_1, columns=['timestamp', 'open', 'high', 'low', 'close', 'volume'])df_2 = pd.DataFrame(ohlcv_2, columns=['timestamp', 'open', 'high', 'low', 'close', 'volume'])df_1.set_index('timestamp', inplace=True)df_2.set_index('timestamp', inplace=True)df = pd.concat([df_1['close'], df_2['close']], axis=1)df.columns = ['BTC', 'ETH']correlation_coefficient = df.corr().iloc[0][1]`
In this code snippet, we are using the CCXT library to fetch historical price data for two crypto assets (BTC and ETH) from the Binance exchange. We then use the Pandas library to calculate the correlation coefficient between the two assets.
When implementing a trading strategy, it is important to factor in the trading fees into your calculations. Most cryptocurrency exchanges charge fees for each trade that you make. These fees can vary depending on the exchange and the trading pair that you are using. Some exchanges offer reduced fees for high volume traders or for users who hold a certain amount of the exchange’s native token.
To calculate the fees associated with a trade on the Binance exchange using the CCXT library, you can use the following code snippet:
`import ccxtexchange = ccxt.binance()symbol = 'BTC/USDT'amount = 0.1buy_price = exchange.fetch_ticker(symbol)['ask']sell_price = exchange.fetch_ticker(symbol)['bid']buy_fee = exchange.calculate_fee(symbol, 'limit', 'buy', amount, buy_price, {})sell_fee = exchange.calculate_fee(symbol, 'limit', 'sell', amount, sell_price, {})# Calculate net profitprofit = (sell_price - buy_price) * amount - buy_fee['cost'] - sell_fee['cost']pyt`
In this code snippet, we are using the `calculate_fee` method provided by the CCXT library to calculate the fees associated with a buy and sell order for a given trading pair. We then subtract the fees from the expected profit to calculate the net profit.
Once you have identified a pair of highly correlated crypto assets, you need to come up with a profitable trading strategy to trade them against each other. Here are some common strategies:
# Mean Reversion
In a mean reversion strategy, you would buy the asset that is trading at a lower price than its historical average and sell the asset that is trading at a higher price than its historical average. This strategy assumes that the prices of the two assets will eventually converge back to their historical averages.
# Momentum
In a momentum strategy, you would buy the asset that is showing strong positive momentum and sell the asset that is showing strong negative momentum. This strategy assumes that the prices of the two assets will continue to move in the same direction in the short term.
In a pairs trading strategy, you would buy the underperforming asset and sell the outperforming asset. This strategy assumes that the prices of the two assets will eventually converge back to their historical relationship.
To execute these strategies using CCXT library, you can use the following code snippet:
`import ccxtexchange = ccxt.binance()symbol_1 = 'BTC/USDT'symbol_2 = 'ETH/USDT'amount = 0.1buy_price = exchange.fetch_ticker(symbol_1)['ask']sell_price = exchange.fetch_ticker(symbol_2)['bid']buy_fee = exchange.calculate_fee(symbol_1, 'limit', 'buy', amount, buy_price, {})sell_fee = exchange.calculate_fee(symbol_2, 'limit', 'sell', amount, sell_price, {})spread = (sell_price - buy_price) * amountnet_profit = spread - buy_fee['cost'] - sell_fee['cost']# Mean Reversionif spread > mean_spread: order_1 = exchange.create_limit_buy_order(symbol_2, amount, sell_price) order_2 = exchange.create_limit_sell_order(symbol_1, amount, buy_price)elif spread < -mean_spread: order_1 = exchange.create_limit_buy_order(symbol_1, amount, buy_price) order_2 = exchange.create_limit_sell_order(symbol_2, amount, sell_price)# Momentumif momentum_1 > momentum_2: order_1 = exchange.create_limit_buy_order(symbol_1, amount, buy_price) order_2 = exchange.create_limit_sell_order(symbol_2, amount, sell_price)else: order_1 = exchange.create_limit_buy_order(symbol_2, amount, sell_price) order_2 = exchange.create_limit_sell_order(symbol_1, amount, buy_price)# Pairs Tradingif spread > std_spread: order_1 = exchange.create_limit_buy_order(symbol_2, amount, sell_price) order_2 = exchange.create_limit_sell_order(symbol_1, amount, buy_price)elif spread < -std_spread: order_1 = exchange.create_limit_buy_order(symbol_1, amount, buy_price) order_2 = exchange.create_limit_sell_order(symbol_2, amount, sell_price)`
In this code snippet, we are using the CCXT library to execute trades on the Binance exchange. We first fetch the current ask price for asset 1 (BTC/USDT) and the current bid price for asset 2 (ETH/USDT). We then use the spread between the two assets to determine which trading strategy to use. The `mean_spread` and `std_spread` variables are calculated based on historical data. The trading fees are calculated using the `calculate_fee` method and are subtracted from the spread to calculate the net profit.
# Conclusion
In this blog post, we have discussed how to find highly correlated crypto assets, how to calculate trading fees, and how to trade them against each other using the CCXT library. By following these steps and implementing a profitable trading strategy that takes trading fees into account, you can take advantage of statistical arbitrage opportunities in the world of cryptocurrency trading.
--
-- | 1,560 | 6,906 | {"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-2024-30 | latest | en | 0.800558 |
https://carmencincotti.com/2022-07-04/cloth-representation-for-simulations/ | 1,718,862,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00153.warc.gz | 137,585,012 | 124,703 | # Cloth: Representation for Simulations
There are many ways to model cloth! We'll explore continuous and discrete options to do so. We'll also dive deep into a discrete representation by reviewing how we can simulate cloth with particles and forces.
Keywords: cloth simulation, fabrics, numerical methods
Let’s continue our conversation about cloth after having already discussed the materials used to construct it. Although each piece of fabric is different, we can model the internal structures the same way.
According to scientific papers, cloth can be modeled either continuously or discretely. In this article, we focus on a discrete representation since we already have learned what mass-spring systems are ⚖️.
With this choice, we will dive into how to model fabric in a 3D environment - including the usage of triangles, particles and forces.
⚠️ My goal is to create a visually pleasing simulation. Exact physical accuracy to predict real-world scenarios is not included in this objective.
## Modeling - continuous vs discrete
We need to first choose how to model our fabric, as this will influence our entire simulation! Most research models fabric as either a continuous or discrete material.
### Continuous
A continuous representation ignores the fact that there are atoms and holes between the threads (between warps and wefts). In other words, the fabric is like a single body where a force or deformation acts entirely on this body. Consequently, if I choose a point anywhere on the fabric, it could exist because there are no discontinuities.
This approach opens the door to using continuum mechanics to simulate our cloth, which is described by Wikipedia as the following:
Continuum mechanics is a branch of mechanics that deals with the mechanical behavior of materials modeled as a continuous mass rather than as discrete particles.
This branch of mechanics allows for a more authentic and realistic simulation. However, the calculations are expensive and complex because of the systems of partial differential equations! 🧮
Additionally, during the simulation it would be necessary to discretize our model in time and space. To do this, one generally uses a finite element discretization using the finite element method (FEM).
The idea of FEM is to discretize, or to mesh, a continuous surface into simple elements, or meshes.
⚠️ The discussion on this subject is too vast for this blog. Here are some resources that can help you.
### Discrete
A discrete representation is based on the fact that a piece of cloth is not continuous at all.
From our conversation from last week, we know that cloth is made up of intersecting threads (warps and wefts). Thus, if we zoom in on a piece of fabric, we will see its true structure 🧵 :
🤔 Can you recognize the warp and weft weaves? If not, take a look at the article from last week to review!
As the fabric moves, threads collide, bend, etc due to the intertwining of the warps and wefts. It is therefore possible to model the threads as individual discrete elements. However, modeling all of these discrete elements is complex and impractical.
In this case, scientific researchers recommend a particle representation:
This representation zooms in on the discrete element - the particle - which represents a crossing point between a weft and a warp. The mechanical interactions (the forces of stretching, compression, bending, and twisting) are calculated on the basis of the relations between neighboring local particles.
## The discrete representation of fabric
I am choosing to move forward with a discrete fabric representation since we’ve already implemented a particle system in the past. So, let’s dive a little deeper into this type of representation.
### Triangles
Our model will be a triangular mesh, where each vertex will be our representation of a particle. Here is an example:
by Obi
The blue points that are visible on the fabric are particles in this mesh! The fabric I just created in Blender is close to this idea:
When I export it, I’ll choose to triangulate the mesh to ensure I’m working with triangles.
### Particles
The behavior of particles is governed by the laws of physics which describe material properties. In other words, the particles must react to the forces acting on them.
To do this, the position and velocity of the particles must change over time. The combination of position and velocity in a data structure is called phase space. We have already seen this representation of particles a few months ago during the article on the creation of a system of particles.
### The mass
In the next part, we will see that particles are part of a mass-spring system. That said, how is the mass of a given particle determined? 🤔
Since the fabric is formed by a mesh of triangles, one option for calculating the mass is to first define a density for the fabric. Recall that the unit of density is kg/m². Therefore, we need to find the area of a triangle to find its mass. Finally, we divide this total mass by three to find the mass of a vertex.
Let’s see a fun trick to find the area of a triangle using the cross product. The magnitude of the product u × v is by definition the area of the parallelogram. Therefore, we can use the cross product to calculate the area of a triangle formed by three points A, B, and C in space.
Suppose we know the position of each vertex. It is thus possible to solve the area as follows:
$A = \frac{1}{2}|\vec{AB} \times \vec{AC}|$
### The forces
Although there are external forces that affect fabric such as gravity, wind and collisions with other bodies, internal forces are responsible for the behavior of cloth as we know it.
The internal forces of interest that will act on the particles in our model are stretch, bend and shear. The idea is that these forces will cause the particles to move in a manner similar to how a fabric deforms and drapes.
The process of simulating such forces is the same as the 2D simulation we did together a few months ago.
We can model the interactions between particles as a mass-spring system. Here is an example of such a system:
We just need to calculate the position and the velocity of each particle after taking into account all the forces. We must use a numerical method to solve this system as we have already seen in a previous article - Implicit Euler method.
In any case, you must choose the numerical method that meets your needs.
## Next time
We’re going to build off this knowledge and take a deeper dive into a potential way to solve this cloth simulation, with position based dynamics (PBD).
## Comments for Cloth: Representation for Simulations
Written by Carmen Cincotti, computer graphics enthusiast, language learner, and improv actor currently living in San Francisco, CA. Follow @CarmenCincotti
## Contribute
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# The Patterns of the Pandemic
In calculus, continuity and discontinuity play significant roles in analyzing functions and their behaviour. These concepts are not confined to mathematical equations alone, they are relevant to real world phenomena. The covid pandemic, with its dynamic and evolving nature, provides an intriguing example that demonstrates the concepts of continuity and discontinuity in action. Continuity refers to the smooth, uninterrupted nature of a function or a process. In calculus, it signifies that a function can be drawn without lifting the pen. In the context of covid, continuity can be observed in regions with consistent infection rates, where the number of cases increases steadily over time, without sudden spikes or drops. This continuity indicates a predictable and manageable progression of the disease. However, the pandemic has also revealed instances of discontinuity. Discontinuities occur when there are sudden and significant disruptions in the pattern of a function. In the case of covid, discontinuities arise when unexpected events, such as superspreader events or the emergence of new variants, lead to sudden surges or declines in infection rates. These disruptions challenge the smoothness and predictability of the pandemic's progression, making it challenging to model accurately. Mathematically, continuity and discontinuity are vital for understanding the behaviour of functions. Similarly, in the context of the pandemic, recognizing the presence of these concepts helps us comprehend the unpredictable nature of infectious diseases. By studying these patterns, epidemiologists can identify critical points of discontinuity, enabling them to adjust public health measures accordingly and minimize the impact of sudden spikes in infection rates. The covid pandemic serves as a real-life demonstration of calculus concepts, with continuity and discontinuity providing insights into the progression of the disease. Continuity represents a predictable pattern of infection rates, while discontinuity arises from unexpected events that disrupt the smoothness of the pandemic's trajectory. By applying mathematical principles to understand these patterns, we can improve our preparedness, response, and mitigation strategies for future health crises. Just as calculus allows us to navigate complex mathematical functions, it also empowers us to navigate the intricate dynamics of the world around us, including the ongoing battle against covid. | 436 | 2,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-40 | longest | en | 0.910529 |
ankitnitjsr13.medium.com | 1,701,700,817,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00378.warc.gz | 119,630,723 | 49,829 | # Math behind SVM (Support Vector Machine)
--
SVM is one of the most popular, versatile supervised machine learning algorithm. It is used for both classification and regression task.But in this thread we will talk about classification task. It is usually preferred for medium and small sized data-set.
The main objective of SVM is to find the optimal hyperplane which linearly separates the data points in two component by maximizing the margin .
Don’t panic with the word ‘hyperplane’ and ‘margin’ that you have seen above. It is explained in details below.
## Points to be covered:
1. Basic linear algebra
2. Hyper-plane
3. What if data points is not linearly separable ?
4. Optimal Hyperplane
5. How to choose optimal Hyperplane ?
6. Mathematical Interpretation of Optimal Hyperplane
7. Basic optimization term
8.SVM Optimization
## Vectors
Vectors are mathematical quantity which has both magnitude and direction. A point in the 2D plane can be represented as a vector between origin and the point.
## Length of Vectors
Length of vectors are also called as norms. It tells how far vectors are from the origin.
## Direction of vectors
Direction of vector .
## Dot Product
Dot product between two vectors is a scalar quantity . It tells how to vectors are related.
## Hyper-plane
It is plane that linearly divide the n-dimensional data points in two component. In case of 2D, hyperplane is line, in case of 3D it is plane.It is also called as n-dimensional line. Fig.3 shows, a blue line(hyperplane) linearly separates the data point in two components.
In the Fig.3, hyperplane is line divides data point into two classes(red & green), written as
## What if data points is not linearly separable ?
Look Fig.4 how can we separate the data-points linearly? This type of situation comes very often in machine learning world as raw data are always non-linear here.So, Is it do-able? yes!!. we will add one extra dimension to the data points to make it separable.
so , the above process of making non-linearly separable data point to linearly separable data point is also known as Kernel Trick, which we will cover later in details.
## Optimal Hyperplane
If you look above Fig there are numbers of hyperplane that can separate the data points in two components. So optimal hyperplane is one which divides the data points very well. So question is why it is needed to choose optimal hyper plane?
So if you choose sub-optimal hyperplane, no doubt after number of training iteration , training error will decrease but during testing when an unseen instance will come, it will result in high test error. In that case it is must to choose an optimal plane to get good accuracy.
## Margin and Support Vectors
Let’s assume that solid black line in above Fig.7 is optimal hyperplane and two dotted line is some hyperplane, which is passing through nearest data points to the optimal hyperplane. Then distance between hyperplane and optimal hyperplane is know as margin, and the closest data-points are known as support vectors. Margin is an area which do not contains any data points. There will be some cases when we have data points in margin area but right now we stick to margin as no data points lands.
So, when we are choosing optimal hyperplane we will choose one among set of hyperplane which is highest distance from the closest data points. If optimal hyperplane is very close to data points then margin will be very small and it will generalize well for training data but when an unseen data will come it will fail to generalize well as explained above. So our goal is to maximize the margin so that our classifier is able to generalize well for unseen instances.
So, In SVM our goal is to choose an optimal hyperplane which maximizes the margin.
— — — — — — —
Since covering entire concept about SVM in one story will be very confusing. So i will be dividing the tutorial into three parts.
1. Linear separable data points.
2. Linear separable data points II.
3. Non-linear separable data-points.
So, In this story we will assume that data points(training data) are linearly separable.Lets start,
## Mathematical Interpretation of Optimal Hyperplane
we have l training examples where each example x are of D dimension and each have labels of either y=+1 or y= -1 class, and our examples are linearly separable. Then, our training data is form ,
We consider D=2 to keep explanation simple and data points are linearly separable, The hyperplane w.x+b=0 can be described as :
Support vectors examples are closest to optimal hyperplane and the aim of the SVM is to orientate this hyperplane as far as possible from the closest member of the both classes.
From the above Fig , SVM problem can be formulated as,
From the Fig.8 we have two hyperplane H1 and H2 passing through the support vectors of +1 and -1 class respectively. so
w.x+b=-1 :H1
w.x+b=1 :H2
And distance between H1 hyperplane and origin is (-1-b)/|w| and distance between H2 hyperplane and origin is (1–b)/|w|. So, margin can be given as
M=(1-b)/|w|-(-1-b)/|w|
M=2/|w|
Where M is nothing but twice of the margin. So margin can be written as 1/|w|. As, optimal hyperplane maximize the margin, then the SVM objective is boiled down to fact of maximizing the term 1/|w|,
## Unconstrained optimization
Example will be more intuitive in explaining the concept,
So, it is same that we used to do in higher school in calculus for finding maxima and minima of a function. Only difference is at that time we were calculating for univariate variable, but now we are calculating for multivariate variables.
## Constrained optimization
Again it will become clear with an example,
So basically we calculate the maxima and minima of a function by taking into the consideration of given constraints on the variables.
## Primal and Dual Concept
Shocked!! what the heck is this now???
Don’t worry, it ‘s theory only !!
Lets not get deep into these concept. Any optimization problem can be formulated into two way, primal and dual problem. First we use primal formulation for optimization algorithm, but if it does not yield any solution, we go for dual optimization formulation, which is guaranteed to yield solution.
## Optimization
This part will be more mathematical, some terms are very high level concept of mathematics, but don’t worry i will try to explain each one by one in layman term.
To make you comfortable, Learning algorithms of SVM are explained with pseudo code explain below. This is very abstract concept in SVM optimization. For below code assume x is data point and y is its corresponding labels.
Just above you see learning process of SVM. But here is a catch, how do we update w and b ??
Gradient descent of course!!! — -BIG ‘NO’.
SVM optimization problem is a case of constrained optimization problem, and it is always preferred to use dual optimization algorithm to solve such constrained optimization problem. That’s why we don’t use gradient descent.
Since it is constrained optimization problem Lagrange multipliers are used to solve it, which is described below, It looks like , will be more mathematical but it is not, its just few steps of finding gradient. We will divide the complete formulation into three parts.
1. In first we will formulate SVM optimization problem Mathematically
2. we will find gradient with respect to learning parameters.
3. we will find the value of parameters which minimizes ||w||
The above equation is Primal optimization problem.Lagrange method is required to convert constrained optimization problem into unconstrained optimization problem. The goal of above equation to get the optimal value for w and b.
May be above equation is looking tricky? but it is not, its just high school math of finding minima with respect to variable.
As,from the above formulation we only able to find the optimal value of w and that is to dependent on λ, so we need to find the optimal value of λ also. And finding optimal value of b needs both w and λ. So finding the value of λ will be the important for us.
so how do we find the value of λ???
Above formulation is itself a optimization algorithm, but it not helpful to find the optimal value. It is primal optimization problem. As we read above that if Primal optimization doesn’t result in solution, we should use dual optimization formulation, which has guaranteed solution. Also when we move from primal to dual formulation we switch minimizing to maximizing the loss function. Again we will divide the complete formulation into three parts to easier to understand.
1. Problem formulation and substitution value from primal
2. Simplify the loss function equation after substitution
3. final optimization formulation to get the value of λ
Above equation is a dual optimization problem. The equation are looking scarier because of substitution of value of w.
It is simplified equation of above dual optimization problem.
So this is the final optimization problem, to find the maximum value of λ. Here one more term K is there, which is nothing but dot product of input variable x. (This K will be very important in future when we will learn about kernel trick and non-linear data points).
Now How do we Solve the above problem??
Above maximization operation can be solved with the SMO ( sequential minimization optimization) algorithms . There are also the various library support online for this optimization. Once we get the value of λ we can get w from below equation
and using value of w , λ we will calculate b as following,
As we now have the value for both w and b , then optimal hyperplane that can separates the data points can be written as,
w.x + b = 0
And a new example x_ can be classified as sign(w.x_ +b)
Thanks for reading, this is the PART1 of SVM, in PART 2 we will discuss about how to deal with a case when data is not fully linearly separable. | 2,114 | 9,890 | {"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} | 3.765625 | 4 | CC-MAIN-2023-50 | latest | en | 0.912788 |
https://calculat.io/en/weight/bmi-metric/159-cm--61-kg | 1,721,510,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00203.warc.gz | 134,311,403 | 26,499 | # Body Mass Index (BMI) for 159 cm and 61 kg
## What is BMI for male and female with height 159cm and weight 61kg?
Answer: BMI for Height 159 cm and Weight 61 kg is
24.13
(Normal/Healthy Weight)
Suggested* Healthy Weight Range for Height 159 cm is
Weight 61 is in Healthy Zone. You may lose -14.23 kg or gain 2.2 kg.
## How to Calculate BMI
BMI Formula: BMI = weightkg ÷ (heightm)2
According to Body Mass Index (BMI) Formula if you want to calculate BMI for Weight 61 kg and Height 159 cm you have to divide Weight by squared Height in meters (159 cm = 1.59 m)
Here is the complete solution:
61kg ÷ (1.59m)2
=
61 ÷ 2.5281
=
24.13
## BMI Table
For adults over the age of 20 years, BMI values are grouped* into the following weight status categories:
BMIWeight Category
< 16Underweight/Severe thinness
16 - 16.9Underweight/Moderate thinness
17 - 18.4Underweight/Mild thinness
18.5 - 24.9Normal/Healthy Weight
25 - 29.9Overweight/Pre-obese
30 - 34.9Obese/Class I
35 - 39.9Obese/Class II
≥ 40Morbid Obesity/Class III
## BMI Table for 159 cm Height for Man and Woman
Height and WeightBMICategory
Underweight/Mild thinness
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Normal/Healthy Weight
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
Overweight/Pre-obese
## BMI Calculator (Metric)
This calculator will help you to calculate Body Mass Index (BMI) for a given weight (kg) and height (cm). For example, it can help you find out what is BMI for male and female with height 159cm and weight 61kg? (The answer is: 24.13). Enter height (e.g. 159 cm) and weight (e.g. 61 kg) and hit the 'Calculate' button.
## FAQ
### What is BMI for male and female with height 159cm and weight 61kg?
BMI for Height 159 cm and Weight 61 kg is 24.13 (Normal/Healthy Weight)
### Is 61 kg underweight for 159 cm?
No, 61 kg 159 cm is considered Normal/Healthy Weight
### Is 61 kg overweight for 159 cm?
No, 61 kg 159 cm is considered Normal/Healthy Weight | 713 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.812494 |
https://nuprl.org/wip/Standard/arithmetic/rem_2_to_1.html | 1,653,330,367,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00432.warc.gz | 490,738,476 | 2,477 | ### Nuprl Lemma : rem_2_to_1
`∀[a:{...0}]. ∀[n:ℕ+]. ((a rem n) = (-(-a rem n)) ∈ ℤ)`
Proof
Definitions occuring in Statement : int_lower: `{...i}` nat_plus: `ℕ+` uall: `∀[x:A]. B[x]` remainder: `n rem m` minus: `-n` natural_number: `\$n` int: `ℤ` equal: `s = t ∈ T`
Definitions unfolded in proof : uall: `∀[x:A]. B[x]` member: `t ∈ T` int_lower: `{...i}` subtype_rel: `A ⊆r B` nat_plus: `ℕ+` int_nzero: `ℤ-o` so_lambda: `λ2x.t[x]` so_apply: `x[s]` uimplies: `b supposing a` prop: `ℙ` all: `∀x:A. B[x]` implies: `P `` Q` nequal: `a ≠ b ∈ T ` not: `¬A` false: `False` guard: `{T}` true: `True` squash: `↓T` iff: `P `⇐⇒` Q` and: `P ∧ Q` rev_implies: `P `` Q` subtract: `n - m` top: `Top`
Lemmas referenced : nat_plus_wf int_lower_wf subtype_rel_sets less_than_wf nequal_wf less_than_transitivity1 le_weakening less_than_irreflexivity equal_wf equal-wf-base int_subtype_base squash_wf true_wf rem_to_div iff_weakening_equal subtract_wf minus-one-mul mul-associates minus-one-mul-top mul-commutes one-mul minus-add minus-minus div_2_to_1
Rules used in proof : sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut hypothesis extract_by_obid sqequalRule sqequalHypSubstitution isect_memberEquality isectElimination thin hypothesisEquality axiomEquality because_Cache natural_numberEquality intEquality setElimination rename applyEquality lambdaEquality independent_isectElimination setEquality lambdaFormation dependent_functionElimination independent_functionElimination voidElimination baseClosed minusEquality imageElimination equalityTransitivity equalitySymmetry universeEquality imageMemberEquality productElimination multiplyEquality divideEquality voidEquality addEquality
Latex:
\mforall{}[a:\{...0\}]. \mforall{}[n:\mBbbN{}\msupplus{}]. ((a rem n) = (-(-a rem n)))
Date html generated: 2017_04_14-AM-07_18_22
Last ObjectModification: 2017_02_27-PM-02_52_32
Theory : arithmetic
Home Index | 656 | 1,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.305884 |
https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=120228 | 1,513,429,899,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588072.75/warc/CC-MAIN-20171216123525-20171216145525-00096.warc.gz | 757,122,286 | 4,054 | # Hydraulics Facts and Figures
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The flashcards below were created by user Jonnick112 on FreezingBlue Flashcards.
1. Weight of one cubic ft of water
62.4
2. One cubic ft = ___ cubic inches
1728
3. One gallon of water = ___ cubic inches
231
4. One gallon of water weighs ___lbs
8.34
5. One cubic foot of water = ___ gallons
7.48
6. The area of a circle =
.7854 D2
7. Atmoshperic pressure at sea level
14.7
8. One pound of pressure will raise water ___ft
2.304
9. 14.7 pounds of pressure will raise water ___
14.7 x 2.304 = 33.9ft
10. A perfect vacuum would lift water ___ ft
33.9ft
11. A column of water 1" square and 1" high exerts ___ psi at its base
.434
12. Friction Loss ______ as the volume of water ______
Increases ..... Increases
## Card Set Information
Author: Jonnick112 ID: 120228 Filename: Hydraulics Facts and Figures Updated: 2011-12-02 06:41:00 Tags: Hydraulics Facts Figures SKFR FIRE Folders: Description: Hydraulics Facts and Figures Show Answers:
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A276992 First 2-digit number to appear n times in the decimal expansion of Pi. 9
31, 26, 93, 62, 82, 28, 28, 28, 48, 48, 48, 48, 48, 9, 9, 81, 17, 17, 95, 95, 95, 95, 95, 95, 95, 19, 21, 21, 21, 19, 95, 9, 9, 9, 95, 46, 95, 59, 9, 9, 9, 95, 95, 95, 95, 59, 59, 59, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 14, 14, 14, 9, 9, 9, 9, 14, 9, 9 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n) is the 2-digit number that appears in Pi n times before any other 2-digit number appears in Pi n times. Note that the sequence contains elements whose number of digits is 2 or 1, see examples. - Omar E. Pol, Oct 05 2016 Comment from N. J. A. Sloane, Mar 08 2023 (Start) Make a table T[0,0], T[0,1], ...,T[9,9], with 100 columns, labeled 0,0 to 9,9. Scan the digits of pi = 3.14159.... First you see 3, 1 so increment the count for 3,1; next you see 1,4, so increment the count for 1,4. Then you see 4,1 so increment the count for 4,1. Do this for ever. The first time any count hits 6, say T[3,8] = 6, then a(6) = 38. If it happens that T[0,9] hits 6 first, then a(6) would be 09, but we would drop the 0, and write a(6) = 9. (End) Comment from Alois P. Heinz, Mar 08 2023 (Start) Initially, "09" is very often the first to occur n times, while other 2-digit substrings fall behind. They can show up later. This is not strange, this is Pi. In the first 10000 terms we see "09" 40 times, "14" 33 times, and so on. Here is the complete list: [40, 9], [33, 14], [2, 17], [13, 19], [3, 21], [1, 26], [892, 27], [3, 28], [1, 31], [144, 34], [107, 35], [179, 39], [2594, 46], [5, 48], [127, 54], [1387, 55], [4, 59], [6, 62], [41, 65], [671, 71], [19, 74], [3406, 76], [1, 81], [1, 82], [94, 85], [1, 93], [211, 94], [14, 95]. 67 of the two-digit strings never show up in the first 10000 terms. It does not mean that they do not appear in Pi. Indeed they do. It only means that they are never the first to reach some count. They may be behind by only a small amount. (End) The fact that 09 is ahead so often is an example of the Arcsine Law Paradox at work. See for example Feller, Volume I, Chapter III. As Feller says, "[the conclusions] are not only unexpected but actually come as a shock to intuition and common sense." Of course the same phenomenon occurs with single digits of Pi, see A096567, where 5 seems to be ahead most of the time. - N. J. A. Sloane, Mar 09 2023 REFERENCES William Feller, An Introduction to Probability Theory and Its Applications, Vol. I, Chapter III, Wiley, 3rd Ed., Corrected printing 1970. LINKS Alois P. Heinz, Table of n, a(n) for n = 1..10000 EXAMPLE a(2) = 26 because 26 is the first 2-digit number to appear 2 times in the decimal expansion of Pi = 3.14159(26)5358979323846(26)... a(14) = 9 because "09" is the first 2-digit number to appear 14 times in the decimal expansion of Pi. MATHEMATICA spi = ToString[Floor[10^100000 Pi]]; f[n_] := Block[{k = 2}, While[Length@ StringPosition[ StringTake[spi, k], StringTake[spi, {k - 1, k}]] != n, k++]; ToExpression@ StringTake[spi, {k - 1, k}]]; Apply[f, 72] (* Robert G. Wilson v, Oct 05 2016 *) CROSSREFS Cf. A000796, A096567, A276686, A276993, A277171, A277270, A291599, A291600. Sequence in context: A340744 A361315 A291471 * A008685 A162156 A141529 Adjacent sequences: A276989 A276990 A276991 * A276993 A276994 A276995 KEYWORD nonn,base AUTHOR Bobby Jacobs, Sep 24 2016 EXTENSIONS a(21)-a(40) from Bobby Jacobs, Oct 01 2016 More terms from Alois P. Heinz, Oct 02 2016 STATUS approved
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Last modified June 13 22:21 EDT 2024. Contains 373391 sequences. (Running on oeis4.) | 1,477 | 4,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.824616 |
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# simultaneous equations
0
49
1
3x plus 2y equals 8 2x plus 5y equals -2
Guest Mar 21, 2017
Sort:
#1
+75344
+1
3x + 2y = 8 multiply through by 2 → 6x + 4y = 16 (1)
2x + 5y = -2 multiply through by -3 → -6x - 15y = 6 (2)
Add (1) and (2) and we get
-11y = 22 divide both sides by (-11) and y = - 22/11 = -2
Sub this into 3x + 2y = 8 to find x
3x + 2(-2) = 8
3x - 4 = 8 add 4 to both sides
3x = 12 divide both sides by 3
x = 4
CPhill Mar 21, 2017
### 29 Online Users
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http://www.notesmela.com/addition-of-vectors-by-rectangular-components/ | 1,513,189,881,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948530668.28/warc/CC-MAIN-20171213182224-20171213202224-00012.warc.gz | 419,229,546 | 15,794 | # Addition of Vectors by Rectangular Components
by • 04/07/2012 • GeneralComments (0)746
Addition of Vectors by Rectangular Components
Consider two vectors A1 and A2 making angles θ1 and θ2 with x-axis respectively as shown in figure. A1 and A2 are added by using head to tail rule to give the resultant vector A.
The addition of two vectors A1 and A2 mentioned in the above figure consists of following four steps.
Step 1
For the x-components of A, we add the x-components of A1 and A2 which are A1x and A2x. If the x-components of A is denoted by Ax then
Ax = A1x + A2x
Taking magnitudes only
Ax = A1x + A2x
Or
Ax = A1 cos θ1 + A2 cos θ2 …………….. (1)
Step 2
For the y-components of A, we add the y-components of A1 and A2 which are A1y and A2y. If the y-components of A is denoted by Ay then
Ay = A1y + A2y
Taking magnitudes only
Ay = A1y + A2y
Or
Ay = A1 sin θ1 + A2 sin θ2 …………….. (2)
Step 3
Substituting the value of Ax and Ay from equations (1) and (2) respectively in equation (3) below, we get the magnitude of the resultant A
A = |A| = √ (Ax)2 + (Ay)2 ……………… (3)
Step 4
By applying the trigonometric ratio of tangent θ on triangle OAB, we can find the direction of the resultant vector A i.e. angle θ which A makes with the positive x-axis.
tan θ = Ay / Ax
θ = tan-1 [Ay / Ax]
Here four cases arise
(a) If Ax and Ay are both positive, then
θ = tan-1 |Ay / Ax|
(b) If Ax is negative and Ay is positive, then
θ = 180º – tan-1 |Ay / Ax|
(c) If Ax is positive and Ay is negative, then
θ = 360º – tan-1 |Ay / Ax|
(d) If Ax and Ay are both negative, then
θ = 180º + tan-1 |Ay / Ax|
Contributions to periodical literature, find out this here and a leading newspaper? | 529 | 1,699 | {"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} | 4.5625 | 5 | CC-MAIN-2017-51 | latest | en | 0.88121 |
https://www.kodytools.com/units/massflow/from/tnmetricps/to/dgps | 1,716,039,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00473.warc.gz | 746,352,628 | 16,199 | # Ton/Second [Metric] to Decigram/Second Converter
1 Ton/Second [Metric] = 10000000 Decigram/Second
## One Ton/Second [Metric] is Equal to How Many Decigram/Second?
The answer is one Ton/Second [Metric] is equal to 10000000 Decigram/Second and that means we can also write it as 1 Ton/Second [Metric] = 10000000 Decigram/Second. Feel free to use our online unit conversion calculator to convert the unit from Ton/Second [Metric] to Decigram/Second. Just simply enter value 1 in Ton/Second [Metric] and see the result in Decigram/Second.
Manually converting Ton/Second [Metric] to Decigram/Second can be time-consuming,especially when you don’t have enough knowledge about Mass Flow units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Ton/Second [Metric] to Decigram/Second converter tool to get the job done as soon as possible.
We have so many online tools available to convert Ton/Second [Metric] to Decigram/Second, but not every online tool gives an accurate result and that is why we have created this online Ton/Second [Metric] to Decigram/Second converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Ton/Second [Metric] to Decigram/Second (tn/s to dg/s)
By using our Ton/Second [Metric] to Decigram/Second conversion tool, you know that one Ton/Second [Metric] is equivalent to 10000000 Decigram/Second. Hence, to convert Ton/Second [Metric] to Decigram/Second, we just need to multiply the number by 10000000. We are going to use very simple Ton/Second [Metric] to Decigram/Second conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Ton/Second [Metric]} = 1 \times 10000000 = \text{10000000 Decigram/Second}$$
## What Unit of Measure is Ton/Second [Metric]?
Ton/Second or Ton per Second [Metric] is a unit of measurement for mass flow rate. It is defined as flow or movement of one ton [metric] of mass in one second.
## What is the Symbol of Ton/Second [Metric]?
The symbol of Ton/Second [Metric] is tn/s. This means you can also write one Ton/Second [Metric] as 1 tn/s.
## What Unit of Measure is Decigram/Second?
Decigram/Second or Decigram per Second is a unit of measurement for mass flow rate. It is defined as flow or movement of one decigram of mass in one second.
## What is the Symbol of Decigram/Second?
The symbol of Decigram/Second is dg/s. This means you can also write one Decigram/Second as 1 dg/s.
## How to Use Ton/Second [Metric] to Decigram/Second Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Ton/Second [Metric] and in the first input field, enter a value.
• From the second dropdown, select Decigram/Second.
• Instantly, the tool will convert the value from Ton/Second [Metric] to Decigram/Second and display the result in the second input field.
## Example of Ton/Second [Metric] to Decigram/Second Converter Tool
Ton/Second [Metric]
1
Decigram/Second
10000000
# Ton/Second [Metric] to Decigram/Second Conversion Table
Ton/Second [Metric] [tn/s]Decigram/Second [dg/s]Description
1 Ton/Second [Metric]10000000 Decigram/Second1 Ton/Second [Metric] = 10000000 Decigram/Second
2 Ton/Second [Metric]20000000 Decigram/Second2 Ton/Second [Metric] = 20000000 Decigram/Second
3 Ton/Second [Metric]30000000 Decigram/Second3 Ton/Second [Metric] = 30000000 Decigram/Second
4 Ton/Second [Metric]40000000 Decigram/Second4 Ton/Second [Metric] = 40000000 Decigram/Second
5 Ton/Second [Metric]50000000 Decigram/Second5 Ton/Second [Metric] = 50000000 Decigram/Second
6 Ton/Second [Metric]60000000 Decigram/Second6 Ton/Second [Metric] = 60000000 Decigram/Second
7 Ton/Second [Metric]70000000 Decigram/Second7 Ton/Second [Metric] = 70000000 Decigram/Second
8 Ton/Second [Metric]80000000 Decigram/Second8 Ton/Second [Metric] = 80000000 Decigram/Second
9 Ton/Second [Metric]90000000 Decigram/Second9 Ton/Second [Metric] = 90000000 Decigram/Second
10 Ton/Second [Metric]100000000 Decigram/Second10 Ton/Second [Metric] = 100000000 Decigram/Second
100 Ton/Second [Metric]1000000000 Decigram/Second100 Ton/Second [Metric] = 1000000000 Decigram/Second
1000 Ton/Second [Metric]10000000000 Decigram/Second1000 Ton/Second [Metric] = 10000000000 Decigram/Second
# Ton/Second [Metric] to Other Units Conversion Table
ConversionDescription
1 Ton/Second [Metric] = 86400000 Kilogram/Day1 Ton/Second [Metric] in Kilogram/Day is equal to 86400000
1 Ton/Second [Metric] = 3600000 Kilogram/Hour1 Ton/Second [Metric] in Kilogram/Hour is equal to 3600000
1 Ton/Second [Metric] = 60000 Kilogram/Minute1 Ton/Second [Metric] in Kilogram/Minute is equal to 60000
1 Ton/Second [Metric] = 1000 Kilogram/Second1 Ton/Second [Metric] in Kilogram/Second is equal to 1000
1 Ton/Second [Metric] = 86400000000000 Milligram/Day1 Ton/Second [Metric] in Milligram/Day is equal to 86400000000000
1 Ton/Second [Metric] = 3600000000000 Milligram/Hour1 Ton/Second [Metric] in Milligram/Hour is equal to 3600000000000
1 Ton/Second [Metric] = 60000000000 Milligram/Minute1 Ton/Second [Metric] in Milligram/Minute is equal to 60000000000
1 Ton/Second [Metric] = 1000000000 Milligram/Second1 Ton/Second [Metric] in Milligram/Second is equal to 1000000000
1 Ton/Second [Metric] = 86400000000 Gram/Day1 Ton/Second [Metric] in Gram/Day is equal to 86400000000
1 Ton/Second [Metric] = 3600000000 Gram/Hour1 Ton/Second [Metric] in Gram/Hour is equal to 3600000000
1 Ton/Second [Metric] = 60000000 Gram/Minute1 Ton/Second [Metric] in Gram/Minute is equal to 60000000
1 Ton/Second [Metric] = 1000000 Gram/Second1 Ton/Second [Metric] in Gram/Second is equal to 1000000
1 Ton/Second [Metric] = 8.64e-8 Exagram/Day1 Ton/Second [Metric] in Exagram/Day is equal to 8.64e-8
1 Ton/Second [Metric] = 3.6e-9 Exagram/Hour1 Ton/Second [Metric] in Exagram/Hour is equal to 3.6e-9
1 Ton/Second [Metric] = 6e-11 Exagram/Minute1 Ton/Second [Metric] in Exagram/Minute is equal to 6e-11
1 Ton/Second [Metric] = 1e-12 Exagram/Second1 Ton/Second [Metric] in Exagram/Second is equal to 1e-12
1 Ton/Second [Metric] = 0.0000864 Petagram/Day1 Ton/Second [Metric] in Petagram/Day is equal to 0.0000864
1 Ton/Second [Metric] = 0.0000036 Petagram/Hour1 Ton/Second [Metric] in Petagram/Hour is equal to 0.0000036
1 Ton/Second [Metric] = 6e-8 Petagram/Minute1 Ton/Second [Metric] in Petagram/Minute is equal to 6e-8
1 Ton/Second [Metric] = 1e-9 Petagram/Second1 Ton/Second [Metric] in Petagram/Second is equal to 1e-9
1 Ton/Second [Metric] = 0.0864 Teragram/Day1 Ton/Second [Metric] in Teragram/Day is equal to 0.0864
1 Ton/Second [Metric] = 0.0036 Teragram/Hour1 Ton/Second [Metric] in Teragram/Hour is equal to 0.0036
1 Ton/Second [Metric] = 0.00006 Teragram/Minute1 Ton/Second [Metric] in Teragram/Minute is equal to 0.00006
1 Ton/Second [Metric] = 0.000001 Teragram/Second1 Ton/Second [Metric] in Teragram/Second is equal to 0.000001
1 Ton/Second [Metric] = 86.4 Gigagram/Day1 Ton/Second [Metric] in Gigagram/Day is equal to 86.4
1 Ton/Second [Metric] = 3.6 Gigagram/Hour1 Ton/Second [Metric] in Gigagram/Hour is equal to 3.6
1 Ton/Second [Metric] = 0.06 Gigagram/Minute1 Ton/Second [Metric] in Gigagram/Minute is equal to 0.06
1 Ton/Second [Metric] = 0.001 Gigagram/Second1 Ton/Second [Metric] in Gigagram/Second is equal to 0.001
1 Ton/Second [Metric] = 86400 Megagram/Day1 Ton/Second [Metric] in Megagram/Day is equal to 86400
1 Ton/Second [Metric] = 3600 Megagram/Hour1 Ton/Second [Metric] in Megagram/Hour is equal to 3600
1 Ton/Second [Metric] = 60 Megagram/Minute1 Ton/Second [Metric] in Megagram/Minute is equal to 60
1 Ton/Second [Metric] = 1 Megagram/Second1 Ton/Second [Metric] in Megagram/Second is equal to 1
1 Ton/Second [Metric] = 864000000 Hectogram/Day1 Ton/Second [Metric] in Hectogram/Day is equal to 864000000
1 Ton/Second [Metric] = 36000000 Hectogram/Hour1 Ton/Second [Metric] in Hectogram/Hour is equal to 36000000
1 Ton/Second [Metric] = 600000 Hectogram/Minute1 Ton/Second [Metric] in Hectogram/Minute is equal to 600000
1 Ton/Second [Metric] = 10000 Hectogram/Second1 Ton/Second [Metric] in Hectogram/Second is equal to 10000
1 Ton/Second [Metric] = 8640000000 Dekagram/Day1 Ton/Second [Metric] in Dekagram/Day is equal to 8640000000
1 Ton/Second [Metric] = 360000000 Dekagram/Hour1 Ton/Second [Metric] in Dekagram/Hour is equal to 360000000
1 Ton/Second [Metric] = 6000000 Dekagram/Minute1 Ton/Second [Metric] in Dekagram/Minute is equal to 6000000
1 Ton/Second [Metric] = 100000 Dekagram/Second1 Ton/Second [Metric] in Dekagram/Second is equal to 100000
1 Ton/Second [Metric] = 864000000000 Decigram/Day1 Ton/Second [Metric] in Decigram/Day is equal to 864000000000
1 Ton/Second [Metric] = 36000000000 Decigram/Hour1 Ton/Second [Metric] in Decigram/Hour is equal to 36000000000
1 Ton/Second [Metric] = 600000000 Decigram/Minute1 Ton/Second [Metric] in Decigram/Minute is equal to 600000000
1 Ton/Second [Metric] = 10000000 Decigram/Second1 Ton/Second [Metric] in Decigram/Second is equal to 10000000
1 Ton/Second [Metric] = 8640000000000 Centigram/Day1 Ton/Second [Metric] in Centigram/Day is equal to 8640000000000
1 Ton/Second [Metric] = 360000000000 Centigram/Hour1 Ton/Second [Metric] in Centigram/Hour is equal to 360000000000
1 Ton/Second [Metric] = 6000000000 Centigram/Minute1 Ton/Second [Metric] in Centigram/Minute is equal to 6000000000
1 Ton/Second [Metric] = 100000000 Centigram/Second1 Ton/Second [Metric] in Centigram/Second is equal to 100000000
1 Ton/Second [Metric] = 86400000000000000 Microgram/Day1 Ton/Second [Metric] in Microgram/Day is equal to 86400000000000000
1 Ton/Second [Metric] = 3600000000000000 Microgram/Hour1 Ton/Second [Metric] in Microgram/Hour is equal to 3600000000000000
1 Ton/Second [Metric] = 60000000000000 Microgram/Minute1 Ton/Second [Metric] in Microgram/Minute is equal to 60000000000000
1 Ton/Second [Metric] = 1000000000000 Microgram/Second1 Ton/Second [Metric] in Microgram/Second is equal to 1000000000000
1 Ton/Second [Metric] = 190479394.53 Pound/Day1 Ton/Second [Metric] in Pound/Day is equal to 190479394.53
1 Ton/Second [Metric] = 7936641.44 Pound/Hour1 Ton/Second [Metric] in Pound/Hour is equal to 7936641.44
1 Ton/Second [Metric] = 132277.36 Pound/Minute1 Ton/Second [Metric] in Pound/Minute is equal to 132277.36
1 Ton/Second [Metric] = 2204.62 Pound/Second1 Ton/Second [Metric] in Pound/Second is equal to 2204.62
1 Ton/Second [Metric] = 3047670312.44 Ounce/Day1 Ton/Second [Metric] in Ounce/Day is equal to 3047670312.44
1 Ton/Second [Metric] = 126986263.02 Ounce/Hour1 Ton/Second [Metric] in Ounce/Hour is equal to 126986263.02
1 Ton/Second [Metric] = 2116437.72 Ounce/Minute1 Ton/Second [Metric] in Ounce/Minute is equal to 2116437.72
1 Ton/Second [Metric] = 35273.96 Ounce/Second1 Ton/Second [Metric] in Ounce/Second is equal to 35273.96
1 Ton/Second [Metric] = 95239.7 Ton/Day [Short]1 Ton/Second [Metric] in Ton/Day [Short] is equal to 95239.7
1 Ton/Second [Metric] = 3968.32 Ton/Hour [Short]1 Ton/Second [Metric] in Ton/Hour [Short] is equal to 3968.32
1 Ton/Second [Metric] = 66.14 Ton/Minute [Short]1 Ton/Second [Metric] in Ton/Minute [Short] is equal to 66.14
1 Ton/Second [Metric] = 1.1 Ton/Second [Short]1 Ton/Second [Metric] in Ton/Second [Short] is equal to 1.1
1 Ton/Second [Metric] = 86400 Ton/Day [Metric]1 Ton/Second [Metric] in Ton/Day [Metric] is equal to 86400
1 Ton/Second [Metric] = 3600 Ton/Hour [Metric]1 Ton/Second [Metric] in Ton/Hour [Metric] is equal to 3600
1 Ton/Second [Metric] = 60 Ton/Minute [Metric]1 Ton/Second [Metric] in Ton/Minute [Metric] is equal to 60 | 3,727 | 11,603 | {"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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.933363 |
https://www.theverge.com/2016/12/25/14071214/santa-claus-christmas-physics-science-space-time-continuum-toys | 1,702,221,932,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00080.warc.gz | 1,122,842,325 | 31,866 | # Can Santa Claus bend the space-time continuum?
/
## Santa employs some fancy physics in his annual chimney-hopping marathon.
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Okay, so here’s the thing about Santa Claus. Every year, this elderly, overweight man somehow manages to lug giant bags of toys down millions of chimneys. He does this in the span of a couple hours, without being spotted, and without toppling off icy roofs.
It’s an impressive — if suspicious — series of feats. And Claus does, after all, have a history of suspicious behavior, including a series of alleged hit-and-runs involving grandmas.
So, how does he accomplish this annual chimney-hopping marathon? Claus could not be reached for comment. So instead, we quizzed two physics professors about Claus’s possible methods.
“I know magic is magic and everything, but last we checked, energy needs to be conserved.”
The main thing Claus needs is a serious source of energy, says Dave Custer, a physics and writing lecturer at MIT. From Custer’s back-of-the-envelope calculations, he estimates that Claus visits maybe a million households in the US in the span of approximately six hours. That means that each visit — including getting down the chimney, arranging presents, and getting back up again — can only take 0.0002 seconds.
If each of those households has a chimney that averages 16 feet in height, that means on Christmas Eve, Santa Claus climbs the equivalent of Mount Everest over 50,000 times. And if he weighs a healthy 175 pounds these days, then it would take the equivalent of a 24,000-horsepower engine, or the engines of more than 160 Ford F150 pickup trucks, to propel him that distance. That requires as much energy as MIT’s sustainable power plant produces, he says.
“We’re getting closer to being able to explain how Santa Claus actually does it.”
“That’s a pretty beefy power expenditure. I know magic is magic and everything, but last we checked energy needs to be conserved,” Custer says. At least, if Claus is abiding by the law of conservation of energy, which says energy is neither created nor destroyed — it just changes from one form to another.
So if Claus is powering his efforts with cookies, he’d need to consume a steady stream of them at each house he visits. Even then, he’s probably be in a caloric deficit. “How on earth does Santa manage to keep that paunch?” he wrote in an email. So cookies are probably insufficient fuel, but a power plant is too big to lug around and to fit through chimneys.
Custer’s not certain how Claus fits, or how he gets back up the chimney. “I can’t claim to have been in a lot of chimneys, but I’ve looked up a few, and they’re not handhold friendly,” Custer says. He thinks that the inside of a chimney is probably too smooth to scale, but if Claus were ascending back up his rope as fast as he needs to, he’d probably burn up the rope.
Photograph from Pixabay (CC0)
Larry Silverberg, a physics professor at North Carolina State University, suspects Claus manages this by altering the space-time continuum. “Basically what we’ve learned in relativity is two things: one is that that time can be stretched — that’s called time dilation. And space can be contracted,” Silverberg explains. Claus could be creating little pockets where he can control time and space, called relativity clouds. Inside the cloud, Claus can shrink himself to the size of a Christmas ornament, so he could easily fit down the chimney.
That’s probably how he manages to propel himself back up the chimney, too. When he shrinks himself in the relativity cloud, he decreases his total mass. Shrink an object or bearded humanoid’s mass, and it can accelerate more rapidly, according to Newton’s second law of motion.
“He can basically bounce his way back up the chimney.”
“With the same force the resulting acceleration of the body increases dramatically when the mass decreases, making him, essentially very bouncy,” Silverberg told The Verge in an email. “He can basically bounce his way back up the chimney.”
Inside these relativity clouds, Claus can also stretch and distort time so that six months elapse inside the cloud, but only seconds pass outside of it. Six months would be plenty of time for Claus to complete the deliveries, assuming that he has an army of 500 or so elves assisting him, Silverberg says.
The kind of technology necessary to alter the space-time continuum like this is still 200 years away for us, Silverberg estimates. But in the North Pole’s frigid equivalent of Silicon Valley, harsh conditions accelerated technological development.
“People have made that claim that it’s impossible,” Silverberg says. “But with technology over the last couple of years getting stronger and stronger, we’re getting closer to being able to explain how Santa Claus actually does it.”
We will update this story if Claus replies to requests for comment. | 1,046 | 4,954 | {"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 | 0.940491 |
https://brainmass.com/engineering/electrical-engineering/bjt-voltage-noise-347066 | 1,550,852,495,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247518497.90/warc/CC-MAIN-20190222155556-20190222181556-00288.warc.gz | 501,474,955 | 19,338 | Explore BrainMass
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# BJT Circuit
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Please see the attachment for circuit diagram, where
R1 = 91k ohms
R2 = 20k ohms
Re = 1k ohms
Rc = 5k ohms.
Connect two identical amplifier stages like the one above in cascade (one after the another). What are the voltage gain and input and output impedances of the combination? Assume that the circuit is driven from a 50 Ohm source resistance generator. What is the (rms) voltage noise (referred to the input of the first stage) for a bandwidth from 1 kHz to 11 kHz frequency? What is the total noise referred to the output (of the second stage)? Your calculation of noise should be accurate to about 10%.
Use spice if necessary. | 193 | 785 | {"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-2019-09 | longest | en | 0.902745 |
https://the-biggest.net/people/what-is-the-largest-unit-of-weight.html | 1,624,390,759,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00122.warc.gz | 480,073,444 | 20,834 | # What Is The Largest Unit Of Weight?
Contents
Mass (weight) Units
1 gigatonne (Gt) =1 000 000 000 000 000 g
1 megatonne (Mt) =1 000 000 000 000 g
1 tonne (t) =1 000 000 g
1 kilogram (kg) =1 000 g
1 gram (g) =1 g
4 more rows
## What is the largest unit?
The millimeter (mm) is the smallest metric measure of length and equals 1/1000 of a meter. The centimeter (cm) is the next largest unit of length and equals 1/100 of a meter. The decimeter (dm) is the next largest unit of length and equals 1/10 of a meter.
## What is the biggest weight?
List of heaviest people
Name Country Peak weight
Jon Brower Minnoch United States 635 kg (1,400 lb; 100 st 0 lb)
Khalid bin Mohsen Shaari Saudi Arabia 610 kg (1,345 lb; 96 st 1 lb)
Manuel Uribe Mexico 597 kg (1,316 lb; 94 st 0 lb)
Juan Pedro Franco Mexico 595 kg (1,312 lb; 93 st 10 lb)
24 more rows
## What is the largest unit of mass?
Some units such as:1 metric ton = 1000 kg, Quintal = 100 kg, Slug = 14.57 kg are commonly used. The largest practical unit of mass is Chandrasekhar limit (CSL). 1 CSL = 1.4 times the mass of sun. Atomic or nuclear masses are made in terms of a unit called amu (atomic mass unit).
READ Which state has largest district in India?
## What unit is bigger than a ton?
An ounce is the smallest unit for measuring weight, a pound is a larger unit, and a ton is the largest unit. Whales are some of the largest animals in the world. Some species can reach weights of up to 200 tons–that’s equal to 400,000 pounds.
## What is the largest unit of time?
The largest unit is the supereon, composed of eons. Eons are divided into eras, which are in turn divided into periods, epochs and ages.
## What is 10 cm called?
The metric system is based on 10s. For example, 10 decimeters make a meter (39.37 inches). Deci- means 10; 10 decimeters make a meter. Centi- means 100; 100 centimeters make a meter.
## How heavy is the heaviest thing in the world?
The heaviest object ever directly weighed was the Revolving Service Structure (RSS) of launch pad 39B at NASA’s Kennedy Space Center, Florida, USA. The structure was lifted up on 21 jacking points which, between them, measured the mass of the RSS as 2,423 tonnes (5,342,000 lbs).
## Who is the fattest president of the United States?
William Howard Taft
## Is milligrams bigger than grams?
Answers: There are 1,000 milligrams (mg) in 1 gram (g). To get your answer, simply multiply x by 1000.
## Is meter bigger than a yard?
It’s because a yard is shorter than a meter. A yard is 36 inches. A meter is 39.4 inches. If your question is “”Is an Imperial Yard bigger than a meter?”…
## What is the smallest unit of mass?
What is the smallest unit of measurement of mass. The metric measure units of weight are based on the gram. The milligram (mg) is the smallest metric measure of weight and equals 1/1000 of a gram.
## Does 2000 pounds equal 1 ton?
Ton, unit of weight in the avoirdupois system equal to 2,000 pounds (907.18 kg) in the United States (the short ton) and 2,240 pounds (1,016.05 kg) in Britain (the long ton). The metric ton used in most other countries is 1,000 kg, equivalent to 2,204.6 pounds avoirdupois.
READ Question: What College Has The Biggest Library?
## What’s the difference between a short ton and a long ton?
The British ton is the long ton, which is 2240 pounds, and the U.S. ton is the short ton which is 2000 pounds. Both tons are actually defined in the same way. There is also an third type of ton called the metric ton, equal to 1000 kilograms, or approximately 2204 pounds. The metric ton is officially called tonne.
## Is a ton or a kiloton bigger?
1 Kilotonne or metric kiloton (unit of mass) is equal to 1000 metric tons. A metric ton is exactly 1000 kilograms (SI base unit) making a kilotonne equal to 1000000 kilograms. 1 kt = 1000000 kg.
## How long is a jiffy?
In astrophysics and quantum physics a jiffy is, as defined by Edward R. Harrison, the time it takes for light to travel one fermi, which is approximately the size of a nucleon. One fermi is 10−15 m, so a jiffy is about 3 × 10−24 seconds.
## What is smaller than a Yoctosecond?
Yocto comes from the Latin/Greek word octo/οκτώ, meaning “eight”, because it is equal to 1000−8. Yocto is the smallest official SI prefix. A yoctosecond is the shortest lifetime measured, so far. The elementary particle Z-boson has a mean lifetime of less than a yoctosecond, 0.26 ys.
## What is a Zeptosecond?
Noun. zeptosecond (plural zeptoseconds) A unit of time equal to 0.000 000 000 000 000 000 001 seconds, that is, 10−21 second, and with symbol zs.
## What is the smallest length?
The smallest possible size for anything in the universe is the Planck Length, which is 1.6 x10-35 m across.
## What is one meter long?
Meters. 1 m is equivalent to 1.0936 yards, or 39.370 inches. Since 1983, the metre has been officially defined as the length of the path travelled by light in a vacuum during a time interval of 1/299,792,458 of a second.
## What do cm mean?
Acronym Definition
cm Configuration Management
cm Chief Minister
cm Chemical
cm Corporate (Data) Model
235 more rows
## Who is the tallest person alive in 2018?
Share. The tallest man living is Sultan Kösen (Turkey, b.10 December 1982) who measured 251 cm (8 ft 2.8 in) in Ankara, Turkey, on 08 February 2011.
## What are the last words in the Bible?
In the end, each one chooses whether to reject the grace of Lord Jesus Christ or not. This is because we have freewill. Amen! The fact that the word “amen” is the last word in the Bible is significant.
## What height was the tallest human ever recorded?
The tallest man in medical history for whom there is irrefutable evidence is Robert Pershing Wadlow (USA) (born 6:30 a.m. at Alton, Illinois, USA on 22 February 1918), who when last measured on 27 June 1940, was found to be 2.72 m (8 ft 11.1 in) tall.
## What is the smallest unit of capacity?
Standard Unit of Capacity
• The capacity of a vessel is also called it’s the inner volume.
• Liter is the standard unit of capacity and the smaller unit of capacity is milliliter.
• Liter (l) and milliliter (ml) are related to each other as:
• 1 liter = 1000 milliliter (1 l = 1000 ml).
## What are common weight units?
Customary units of weight. Weight is a quantity of heaviness. For example, you are measuring the weight of your body when you step on to a scale. In the customary system of measurement, the most common units of weight are the ounce (oz) and pound (lb).
## What is a GT in weight?
Gross tonnage (GT) is a function of the volume of all ship’s enclosed spaces (from keel to funnel) measured to the outside of the hull framing. Lightship or Lightweight measures the actual weight of the ship with no fuel, passengers, cargo, water, etc. on board.
Photo in the article by “Wikipedia” `https://en.wikipedia.org/wiki/Body_mass_index` | 1,893 | 6,871 | {"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-2021-25 | latest | en | 0.875089 |
https://discourse.mc-stan.org/t/calculating-abundance-from-marginalized-n-mixture-model/12751 | 1,656,223,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00173.warc.gz | 276,258,791 | 12,120 | # Calculating abundance from marginalized N-mixture model
I am running an N-mixture model (binomial mixture model) to calculate the abundance of animals from repeated counts at many sites. I have posted about this analysis before (https://discourse.mc-stan.org/t/calculating-the-likelihood-for-loo-in-a-binomial-mixture-model/4787/4) relating to loo, but I have a new question regarding the best approach to calculating abundance (N) from the marginalized form (latent N not sampled as done with BUGS/JAGS/Nimble formulation).
The full code is below but the relevant parts are the likelihood:
// Likelihood
for (i in 1:R) {
vector[K[i] - max_y[i] + 1] lp;
for (j in 1:(K[i] - max_y[i] + 1))
lp[j] = poisson_log_lpmf(max_y[i] + j - 1 | log_lambda[i])
+ binomial_logit_lpmf(y[i] | max_y[i] + j - 1, logit_p[i]); // vectorized over T
target += log_sum_exp(lp);
}
}
where R is the number of sites each visited T times. K is the maximum abundance to iterate through to marginalize N out of the likelihood. max_y is the maximum number of individuals observed at a particular site on any of the T visits. This can also be considered the minimum population size, since we know there are at least that many individuals but likely many more that we missed since they are hard to observe.
This all works great. We are in part interested in the true abundance N (also about the coefficients on the variables affecting log_lambda and logit_p). To calculate N, I was using:
N[i] = poisson_log_rng(log_lambda[i]);
But because it’s a random Poisson draw from lambda it can be a large range above and below that for large lambda. In some cases this means that N[i] is less than max_y[i], the number we observed on a single visit (no chance of double counting with our methods). This is fine to get the expected abundance for any plot at that site and most of the posterior predictive checks look good. There is slight overestimate at low counts near zero and slight underestimate at very high counts when predicting new observations, but generally things look good. However, we know that at that plot, N should be at least the number observed so I changed the code to:
for (i in 1:R) {
N_rng[i] = poisson_log_rng(log_lambda[i]);
// Restrict N to be at least as big as the number of animals observed on a single night
if(N_rng[i] < max_y[i])
N[i] = max_y[i];
else
N[i] = N_rng[i];
p[i, 1:T] = inv_logit(logit_p[i, 1:T]);
}
At some sites this pushes all the lower estimates of N across iterations up to max_y but obviously doesn’t change the rest of the distribution. Maybe this isn’t an issue or maybe it depend about the inference I want to make. My question is if anyone thinks this is a bad idea or has a better approach to estimating N in this model?
Thanks for any thoughts or ideas! The data and everything is at https://github.com/djhocking/GSMNP-Elevation, sorry I don’t have a simple reproducible example at the moment. I am planning to turn this into a tutorial for ecologists eventually with simulated data.
Full Stan model:
// Binomial mixture model with covariates
data {
int<lower=0> R; // Number of transects
int<lower=0> T; // Number of temporal replications
int<lower=0> nsites; // Number of sites
int<lower=1> sites[R]; // vector of sites
int<lower=0> y[R, T]; // Counts
vector[R] elev; // Covariate
vector[R] elev2; // Covariate
vector[R] litter; // Covariate
vector[R] twi; // Covariate
vector[R] stream; // Covariate
matrix[R, T] RH; // Covariate
matrix[R, T] precip; // Covariate
matrix[R, T] temp; // Covariate
matrix[R, T] temp2; // Covariate
vector[R] gcover; // Covariate
vector[R] gcover2; // Covariate
int<lower=0> K[R]; // Upper bound of population size
}
transformed data {
int<lower=0> max_y[R];
int<lower=0> N_ll;
int tmp[R];
for (i in 1:R) {
max_y[i] = max(y[i]);
tmp[i] = K[i] - max_y[i] + 1;
}
N_ll = sum(tmp);
}
parameters {
real alpha0;
real alpha1;
real alpha2;
real alpha3;
real alpha4;
real alpha5;
real alpha6;
real beta0;
real beta1;
real beta2;
real beta3;
real beta4;
real beta5;
real beta6;
vector[nsites] eps; // Random site effects
real<lower=0> sd_eps;
matrix[R, T] delta; // Random transect-visit effects
real<lower=0> sd_p;
}
transformed parameters {
vector[R] log_lambda; // Log population size
matrix[R, T] logit_p; // Logit detection probability
for (i in 1:R) {
log_lambda[i] = alpha0 + alpha1 * elev[i] + alpha2 * elev2[i] + alpha3 * twi[i] + alpha4 * litter[i] + alpha5 * gcover[i] + alpha6 * stream[i] + eps[sites[i]] * sd_eps; // non-centered formulation of random effect (see Monnahan et al. 2017)
for (t in 1:T) {
logit_p[i,t] = beta0 + beta1 * temp[i,t] + beta2 * temp2[i,t] + beta3 * precip[i,t] + beta4 * gcover[i] + beta5 * gcover2[i] + beta6 * RH[i,t] + delta[i, t] * sd_p; // non-centered formulation
}
}
}
model {
// Priors
// Improper flat priors are implicitly used on alpha0, alpha1, beta0 etc. if not specified
alpha0 ~ normal(0, 10);
alpha1 ~ normal(0, 10);
alpha2 ~ normal(0, 10);
alpha3 ~ normal(0, 10);
alpha4 ~ normal(0, 10);
alpha5 ~ normal(0, 10);
alpha6 ~ normal(0, 10);
beta0 ~ normal(0, 2);
beta1 ~ normal(0, 2);
beta2 ~ normal(0, 2);
beta3 ~ normal(0, 2);
beta4 ~ normal(0, 2);
beta5 ~ normal(0, 2);
beta6 ~ normal(0, 2);
eps ~ normal(0, 1);
sd_eps ~ cauchy(0, 2.5);
for (i in 1:R) {
for (t in 1:T) {
delta[i,t] ~ normal(0, 1);
}
}
sd_p ~ cauchy(0, 2.5); // even this might be more heavily tailed than we want on p. maybe half normal with sd = 3-5?
// Likelihood
for (i in 1:R) {
vector[K[i] - max_y[i] + 1] lp;
for (j in 1:(K[i] - max_y[i] + 1))
lp[j] = poisson_log_lpmf(max_y[i] + j - 1 | log_lambda[i])
+ binomial_logit_lpmf(y[i] | max_y[i] + j - 1, logit_p[i]); // vectorized over T
target += log_sum_exp(lp);
}
}
generated quantities {
int<lower=0> N[R]; // Abundance (must be at least max_y)
int<lower=0> N_rng[R]; // Abundance proposed by random draw from lambda
int totalN;
vector[R] log_lik;
real mean_abundance;
real mean_detection;
vector[R] mean_p;
real fit = 0;
real fit_new = 0;
matrix[R, T] p;
matrix[R, T] eval; // Expected values
int y_new[R, T];
int y_new_sum[R];
matrix[R, T] E;
matrix[R, T] E_new;
for (i in 1:R) {
vector[K[i] - max_y[i] + 1] ll;
for (j in 1:(K[i] - max_y[i] + 1)) {
ll[j] = poisson_log_lpmf(max_y[i] + j - 1 | log_lambda[i]) // remake lp because can't use from model statement in generated quantities
+ binomial_logit_lpmf(y[i] | max_y[i] + j - 1, logit_p[i]);
}
log_lik[i] = log_sum_exp(ll); // for use in loo and multimodel comparison
}
for (i in 1:R) {
N_rng[i] = poisson_log_rng(log_lambda[i]);
// Restrict N to be at least as big as the number of animals observed on a single night
if(N_rng[i] < max_y[i])
N[i] = max_y[i];
else
N[i] = N_rng[i];
p[i, 1:T] = inv_logit(logit_p[i, 1:T]);
}
// Bayesian p-value fit - Diff than JAGS but all parameters about the same. Not a great test anyway. Problem with RNG for poisson rather than sampling N directly. Makes a big diff with the chi-sq and other Bayesian P-value metrics. But other posterior predictive checks look great.
// Initialize E and E_new
for (i in 1:1) {
for(j in 1:T) {
E[i, j] = 0;
E_new[i, j] = 0;
}
}
for (i in 2:R) {
E[i] = E[i - 1];
E_new[i] = E_new[i - 1];
}
for (i in 1:R) {
mean_p[i] = mean(p[i]);
for (j in 1:T) {
// Assess model fit using Chi-squared discrepancy
// Compute fit statistic E for observed data
eval[i, j] = p[i, j] * N[i];
E[i, j] = square(y[i, j] - eval[i, j]) / (eval[i, j] + 0.5);
// Generate replicate data and
// Compute fit statistic E_new for replicate data
y_new[i, j] = binomial_rng(N[i], p[i, j]);
E_new[i, j] = square(y_new[i, j] - eval[i, j]) / (eval[i, j] + 0.5);
}
y_new_sum[i] = sum(y_new[i]);
}
totalN = sum(N); // Total pop. size across all sites
for (i in 1:R) {
fit = fit + sum(E[i]);
fit_new = fit_new + sum(E_new[i]);
}
mean_abundance = exp(alpha0);
mean_detection = 1 / (1 + exp(-1 * beta0));
}
2 Likes
I am not a specialist. Take my points as some starting of points what you could do.
• I have used such an approach before for an industry/applied audience and it made so much sense to them. Statisticians were less enthusiastic.
• An alternative approach would be to use a rejection sampling approach to generate N_rng. Something like code below. This approach is more akin to using a truncated distribution instead of the censoring in your approach. My approach will affect the rest of the distribution. I don’t know whether that makes sense for your question though. Sidenote: the stan code can be improved to avoid getting into an infinite while loop.
for (i in 1:R) {
N_rng[i] = poisson_log_rng(log_lambda[i]);
while (N_rng[i] < max_y[i])
N_rng[i] = poisson_log_rng(log_lambda[i]);
}
• Another option would be more complicated where you use the full structure of the model. You would have to account for the mixture part of the likelihood. I would have been staring at this for a while and I can’t exactly work it out. I think the rough version is that you generate p first, than account for the observed y and max_y to reweight the poisson simulation. Maybe someone with a better understanding of these models can help you with this.
ll[j] = poisson_log_lpmf(max_y[i] + j - 1 | log_lambda[i]) // remake lp because can't use from model statement in generated quantities
+ binomial_logit_lpmf(y[i] | max_y[i] + j - 1, logit_p[i]);
2 Likes
Thanks so much! I always neglect the use of while control statements. I must have had a bad experience with infinite loops when I first started programming. This makes a lot of sense and makes for nice truncated distributions as you note. I’ll have to think about your third suggestion some more. It’ll be a good exercise when I have some time.
I’m going to leave this open a bit longer before accepting just to see if anyone else has any thoughts on it.
My new code is below. I only set the counter to 100 because R is large and the model is slow already. It seems unlikely to require more than 100 draws from a Poisson to get a value larger than max_y since in most cases max_y is much smaller than \lambda.
for (i in 1:R) {
N[i] = poisson_log_rng(log_lambda[i]);
counter[i] = 0;
while (N[i] < max_y[i]) {
N[i] = poisson_log_rng(log_lambda[i]);
counter[i] += 1;
if (counter[i] > 100) break;
}
}
Are you looking to recover some discrete parameters that you’ve previously marginalised out? If so, have you seen this article and does it help in your case?
“Recovering posterior mixture proportions”
1 Like
See https://github.com/betanalpha/knitr_case_studies/blob/master/principled_bayesian_workflow/sample_joint_ensemble4.stan for an example of how to efficiently sample from a discrete truncated probability mass function without infinite loops.
2 Likes
Thanks, I’m trying to wrap my head around this in how to apply it to my problem. This is what I’ve got so far:
// int<lower=0> N[R];
int N[R] = rep_array(0, R); // It didn't like it if I didn't initiate it.
for (i in 1:R) {
real sum_p = 0;
real u = uniform_rng(0, 1); //
lambda[i] = exp(log_lambda[i]); // because no poisson_log_lcdf
for (k in max_y[i]:K[i]) { //
sum_p = sum_p + exp(poisson_lpmf(k | lambda[i]) - poisson_lcdf(K[i] | lambda[i]));
if (sum_p >= u) {
N[i] = k;
break;
}
}
}
It seems to push the weight higher but doesn’t keep it all above max_y. I think there’s still something I’m missing in your approach.
Warning: Please double check everything I say because on the one hand I feel like I am on to something. On the other hand, I have no special skills in any of this type of modeling and I am pretty sure I messed up somewhere.
I am not really sure how this code can generate N < max_y if N = k and k is between max_y and K. Are you sure that you get N < max_y.
I think the code that you have is accounting for the fact that the poisson is trunctated at K and you cannot have any probability above K (- poisson_lcdf(K[i])). You have to make the same adjustment for the fact that the poisson is also truncated below at max_iter.
However - and this is the most speculative part -, the loglikelihood that is in the model is not only truncated. It’s also perturbed by the binomial. So to take that into account I think this is how you generate N equivalent to your model. The code is in R because it’s easier for me to proto type in R.
For each observation and mcmc draw, start with generating p according to the model. max_y, y, and K are data. lambda is an estimated parameter. So you have a number for each. The first likelihood is the binomial, the second is the poisson in your model (I hope I got the relation between y and max_y right here). Multiplying the likelihoods (or summing the log_likelihoods as in the stan model) gives you the perturbed poisson. Than standardize the likelihood to get the probabilities for each k (This is the step I am most unsure about). Sample N from the probabilities of all ks. If you don’t really need exact simulated N but you are more interested in the distribution of the N for each observation I think you could also calculate the expected value for N (meanN). Remember that is the expected value for each observation and each mcmc draw.
p <- .75
max_y <- 50
lambda <- 40
y <- 40
K <- 10
ks <- max_y + (0:K)
lik_binom <- dbinom(x = y, size = ks, p = p)
lik_poisson <- dpois(x = ks, lambda = lambda)
lik = lik_binom * lik_poisson
probs <- lik/sum(lik)
N <- sample(x = ks, size = 1, prob = probs)
meanN = sum(ks * probs)
@Daniel_Hocking does eq. 4 of Royle’s original Biometrics paper on N-mixture models provide a solution? Looks like he outlines how to compute a posterior for abundance, [N_i = k \mid y, \theta]:
[N = k \mid n_1, ..., n_T, \theta, p] = \dfrac{\text{Pr}(n_1, ..., n_T \mid N=k, p) \times \text{Pr}(N=k \mid \theta) }{\sum_{k=0}^\infty \text{Pr}(n_1, ..., n_T \mid N=k, p) \times \text{Pr}(N=k \mid \theta) }.
You could implement this in the generated quantities block (generating an R by K matrix, where the entry is the probability that N_r = k), and sample from the posterior with an inverse probability integral transform. This will respect the constraint that N must be greater than the maximum count, because there will be zero probability mass for values where this condition is not satisfied (i.e., in the paper’s notation \text{Pr}(n_1, ..., n_T \mid N=k, p) = 0 if \text{max}(n_1, ..., n_T) > k).
Edit: This is essentially the same idea that @stijn ust outlined above as well :)
2 Likes
Not going to lie, I am pretty happy that I was on the right track. Also hooray for the Stan Forums, we found the theory and some code!
2 Likes
Given the other responses it sounds like sampling from a truncated Poisson might be insufficient, but I’ll explain a bit more for the truncated Poisson case just in case.
Given truncations you have to correct for the missing probability outside of the truncations. If you have both lower and upper truncations then you can just compute the probability mass at each point within the truncations and sample from a categorical_rng. This also works if you have an upper bound that’s not too big because the discrete states are always positive so 0 acts as an implicit lower bound.
If you have only a lower bound, however, then this won’t work because you have an infinite number of states left. Even when you have a finite number of states you might have too many for the categorical approach to be practical.
The code I shared tries to sample the states sequentially. You compute the probability of the first state and the probability of all other states and simulate a probability. On success you take the first state and stop, and on failure you reject the first state and move onto the next state to see if the combined probabilities are greater than the sampled probability. You then repeat this process recursively until the summed probabilities are greater than the sampled probability.
For example let’s say that we have a lower bound, K. The first viable state is k = K, with probability mass p1 = exp(poisson_lpdf(K | lambda)) / poisson_ccdf(K | lambda), where the complementary cumulative distribution function accounts for the truncated normalization. If we sample u then we checked whether u <= p1; if it is then we continue and if it’s not then we stop taking k = K as our sampled state. If we continue then we compute p_sum = p1 + p2 and check u <= p_sum. At the final state p_sum has to be equal to one by construction, so the final state be taken if reached.
If you have both a lower bound and an upper bound then the truncated normalization has to account for both, p1 = exp(poisson_lpdf(K_lower | lambda)) / (poisson_ccdf(K_lower | lambda) - poisson_cdf(K_upper | lambda)) .
I’m guessing that you were missing the proper normalization so the sequential sampling was going past the highest value and then terminating the loop without changing the default value.
3 Likes | 4,724 | 17,032 | {"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} | 3.109375 | 3 | CC-MAIN-2022-27 | latest | en | 0.913765 |
http://book.caltech.edu/bookforum/showpost.php?s=078536ab7359764005c8e862c425c3e0&p=11723&postcount=3 | 1,582,111,106,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144111.17/warc/CC-MAIN-20200219092153-20200219122153-00355.warc.gz | 25,053,585 | 5,347 | View Single Post
#3
09-11-2014, 05:00 AM
magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 595
Re: Exercise 1.12 - Failing to make Ein(g) small enough
Just to elaborate a little on the last point in Yaser's answer.
Suppose your strategy is to use only if fails.
If fails and you use it is natural that you should pay the price for the VC bound implied by .
The interesting case is if succeeds and you get low . You cannot use the VC bound that applies for .
It is the option to use in the event that fails that complicates the matter.
This is why to get a correct theoretical bound, you must always specify your entire strategy first. The simplest strategy is to fix a hypothesis set. If it fails, it fails and you are done. If in the back of your mind you are thinking about the possibility of changing hypothesis sets if it fails, then this has to be taken into account in the theoretical analysis from the very begining, in particular, even if the first hypothesis set succeeds.
As mentioned by Yaser, one framework that is useful in analyzing such adaptive strategies is structural risk minimization.
Quote:
Originally Posted by yaser Let us say that is the hypothesis set that didn't work, and you want now to try another hypothesis set . The theoretical guarantees would still hold, but for the equivalent hypothesis set . Just because this uses a "hierarchy" of hypothesis sets (in this case the hierarchy being followed by upon failure of , folllowed by possibly other expansions if failed), there is in general an additional theoretical price to pay, but it is low. Look at structural risk minimization if you are further interested.
__________________
Have faith in probability | 373 | 1,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-10 | latest | en | 0.945942 |
http://teacherweb.com/OH/SylvaniaCitySchools/nneal/apt5.aspx | 1,368,978,494,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697745221/warc/CC-MAIN-20130516094905-00012-ip-10-60-113-184.ec2.internal.warc.gz | 248,929,113 | 4,537 | ## Math Facts
Strategies! Strategies! Strategies!
Doubles Clues
1+1 (eye) 2+2 (dice) 3+3 (ladybug) 4+4 (spider
5+5 (hand) 6+6 (egg) 7+7 (calendar) 8+8 (catepillar)
9+9 (truck)
Doubles +1
If I know my double facts I just add one more.
Ex: 7+6=?
Think: If I know 6+6=12, then I know 7+6 is one more at 13. | 133 | 308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2013-20 | longest | en | 0.502697 |
https://www.physicsforums.com/threads/help-with-part-c.296673/ | 1,519,442,137,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00786.warc.gz | 939,045,978 | 15,608 | # Help with part c?
1. Mar 2, 2009
### phy_
help with part c??
A worker of a moving company places a 252 kg trunk on a piece of carpeting and slides it across the floor at constant velocity by exerting a horizontal force of 425 N on the trunk.
a) what is the coeffient of kinetic friction
Fn = mg
= 252 kg x 9.8 N/kg
= 2469.6 N
kinetic coefficient equals 425 N/2469.6 N = 0.17
b) what happens to the coefficient of kinetic friction if another 56 kg trunk is splaced on top of the 252 kg trunk?
Fn = mg
=(308 kg) (9.8 N/kg)
=3018.4
kinetic coefficient equals 425 N/3018.4 N = 0.14
c) what horizontal force must the mover apply to move the combination of the two trunks at constant velocity
the answer is 5.2 x 10^2 N any help would be appreciated as to how to solve c
thank-you
Last edited: Mar 2, 2009
2. Mar 2, 2009
### Sumanyu
Re: help with part c??
what horizontal force must the mover apply to move the combination of the two trunks at constant velocity
force is not being applied to an object moving an constant velocity or sitting "still"....
force = mass x acceleration...
I don't see how you'd need a force to move an object at a constant velocity
3. Mar 2, 2009
### The Bob
Re: help with part c??
You need a force because, although moving at constant velocity, there is a coefficent of friction that has to be overcome.
Also, please don't post a question twice.
The Bob
4. Mar 2, 2009
### LowlyPion
Re: help with part c??
In b) can you explain how the coefficient of friction would change with more weight?
5. Mar 2, 2009
### phy_
Re: help with part c??
i am not sure other than i added 56 kg (i had made an error and written 5 earlier)
6. Mar 2, 2009
### phy_
Re: help with part c??
is b correct?
7. Mar 2, 2009
### LowlyPion
Re: help with part c??
If it has a coefficient of friction from a), what is the coefficient of friction?
Moving force/Normal force ?
Isn't it constant? If you change the normal force doesn't the moving force need to be bigger?
Isn't that what c) is about? | 577 | 2,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-09 | longest | en | 0.919917 |
https://builtin.com/articles/is-0-even | 1,723,745,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00341.warc.gz | 113,947,624 | 11,049 | Is 0 an Even Number?
Zero is an even number because it can be divided by two without producing a remainder. Learn how to prove 0 is even and why it isn’t also an odd number.
Image: Shutterstock / Built In
UPDATED BY
Matthew Urwin | Jun 13, 2024
Is zero even, or odd? Or is it both? Believe it or not, there is a right answer to this question: Zero is even — and only even.
Is 0 Even?
An even number in mathematics is a number that can be written as `n=2⋅k` for a whole number `k`. If `k=0`, the equation `0=2⋅k` remains accurate. This makes zero an even number.
To understand how we come to this conclusion, we first need to define what “even” means.
Why Is Zero Even?
An even number is a number `n` which can be written as `n=2⋅k` for a whole number `k`. That means if we divide an even number by two, there is no remainder left.
Can we write `0=2⋅k` for some `k`? Yep, we can if we choose `k=0`.
That means that zero is even.
More on StatisticsThe Pigeonhole Principle Explained
Is Zero Also Odd?
Again, we first need to define what “odd” means. The common definition is very similar to the definition of even.
An odd number is a number n which can be written as `n=2⋅k+1` for a whole number `k`.
Can we write `0=2⋅k+1` for some whole number k? Nope. We can’t because solving this equation for `k` yields `k=-1/2`, which is not a whole number. | 393 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-33 | latest | en | 0.9311 |
https://www.diagramelectric.co/2-1-mux-circuit-diagram/ | 1,686,080,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00242.warc.gz | 808,043,619 | 16,475 | # 2 1 Mux Circuit Diagram
Do you need to know the ins and outs of a 2-1 Mux circuit diagram? Whether you are in college studying electronics or looking to understand the basics of digital logic as an enthusiast, understanding a 2-1 Mux circuit can help you better understand how circuits work.
A 2-1 Multiplexer is a logic circuit that has two select inputs and one data input, producing one output. The output depends on the values of the two select inputs. This circuit is used when you want to control which of two signals is sent to a single output. For example, it could be used in a computer system to send data from two different sources to one destination.
To better understand how a 2-1 Mux works, we need to look at its circuit diagram. The diagram will provide all the information needed to interpret the circuit's function. We can see that the two select inputs are labeled as S0 and S1 and are connected to the output via a control logic gate. The data input is labeled as D and is connected to the output through an AND gate. When both select inputs are low, the AND gate is enabled, which causes the data input to be sent to the output. When either one of the select inputs is high, the output is disabled and the data input is not sent to the output.
Understanding the workings of a 2-1 Mux circuit can help you better understand digital logic and open up new possibilities in your projects. It may seem intimidating at first, but once you understand the basic principles, you can unlock the potential in your circuits. Knowing how the circuit diagram works can help you make the most of your design.
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