url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://devsolus.com/2021/11/23/sorting-a-short-list-according-to-the-order-of-a-longer-list/ | 1,685,924,079,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00088.warc.gz | 238,906,515 | 19,454 | # Sorting a short list according to the order of a longer list
Suppose I have a reference list:
``````A = ['X_0', 'Z_0', 'X_1', 'Y_0', 'Z_1', 'X_2', 'Y_1', 'Z_2', 'Y_2']
``````
as well as an (example) shorter list:
``````B = ['Z_0', 'X_1', 'X_0']
``````
How can I sort `B` so that the order of the elements match the order provided in `A`? The final results should thus match the order of `A` like so:
``````B_final = ['X_0', 'Z_0', 'X_1']
``````
### >Solution :
You want the list B sorted by the index of the value in A, so;
``````sorted(B, key=A.index)
``````
This will sort B by each values index in A.
If you are using a very long list, index is not very efficient since it searches the list for every value, so you may want to go for sorting by a dictionary instead.
First create a dictionary of the index values of the list;
``````>>> C = {val:ix for ix,val in enumerate(A)}
``````
…then sort by the dictionary value, which is much more efficient than a linear search in the list;
``````>>> sorted(B, key=C.get)
['X_0', 'Z_0', 'X_1']
`````` | 339 | 1,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-23 | latest | en | 0.878841 |
http://www.gurufocus.com/term/Pretax+Income/ONNN/Pre-Tax%2BIncome/ON%2BSemiconductor%2BCorporation | 1,427,957,083,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131317570.85/warc/CC-MAIN-20150323172157-00273-ip-10-168-14-71.ec2.internal.warc.gz | 533,730,407 | 25,805 | Switch to:
ON Semiconductor Corp (NAS:ONNN)
Pretax Income
\$192 Mil (TTM As of Dec. 2014)
Pretax income is the income that a company earns before paying income taxes. ON Semiconductor Corp's pretax income for the three months ended in Dec. 2014 was \$6 Mil. Its pretax income for the trailing twelve months (TTM) ended in Dec. 2014 was \$192 Mil. ON Semiconductor Corp's pretax margin was 0.69%.
During the past 13 years, ON Semiconductor Corp's highest Pretax Margin was 18.03%. The lowest was -30.15%. And the median was 4.01%.
Definition
This is the income that a company earns before paying income taxes.
ON Semiconductor Corp's Pretax Income for the fiscal year that ended in Dec. 2014 is calculated as
Pretax Income = Operating Income + Non-Recurring Items + Interest Expense + Interest Income + Other = 228.9 + -4.4 + -34.1 + 1.5 + 0 = 192
ON Semiconductor Corp's Pretax Income for the quarter that ended in Dec. 2014 is calculated as
Pretax Income = Operating Income + Non-Recurring Items + Interest Expense + Interest Income + Other = 16.3 + -1.7 + -9.5 + 0.9 + -8.881784197E-16 = 6
ON Semiconductor Corp Pre-Tax Income for the trailing twelve months (TTM) ended in Dec. 2014 was 64.8 (Mar. 2014 ) + 72.8 (Jun. 2014 ) + 48.3 (Sep. 2014 ) + 6 (Dec. 2014 ) = \$192 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
ON Semiconductor Corp's Pretax Margin for the quarter that ended in Dec. 2014 is calculated as
Pretax Margin = Pretax Income / Revenue = 6 / 864.2 = 0.69%
During the past 13 years, ON Semiconductor Corp's highest Pretax Margin was 18.03%. The lowest was -30.15%. And the median was 4.01%.
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
ON Semiconductor Corp Annual Data
Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Pretax Income 109 276 251 -438 71 306 38 -73 170 192
ON Semiconductor Corp Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Pretax Income 20 -141 26 46 57 41 65 73 48 6
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 680 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2015-14 | longest | en | 0.946926 |
https://www.jiskha.com/display.cgi?id=1291166785 | 1,516,788,850,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893629.85/warc/CC-MAIN-20180124090112-20180124110112-00499.warc.gz | 927,678,548 | 3,978 | # math
posted by .
George has enough fencing to enclose 2400 square feet around a circular pool. If he builds a circular fence and he wants 12 foot walkway around the pool, what is the largest pool diameter possible?
• math -
Area = πr^2
2400 = 3.14r^2
r^2 = 764
r = 27 2/3
Solve for r. d = 2r therefore d =
Assuming the walkway is completely around the pool and within the fence, largest pool diameter is d-24.
• math -
Area = πr^2
2400 = 3.14r^2
r^2 = 764
r = 27 2/3
Solve for r. d = 2r therefore d = 55 1/3
Assuming the walkway is completely around the pool and within the fence, largest pool diameter is d-24.
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More Similar Questions | 804 | 3,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-05 | latest | en | 0.961572 |
https://nl.mathworks.com/matlabcentral/cody/problems/107-count-from-0-to-n-m-in-base-n/solutions/2813719 | 1,606,736,386,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00163.warc.gz | 415,706,541 | 17,003 | Cody
# Problem 107. Count from 0 to N^M in base N.
Solution 2813719
Submitted on 10 Aug 2020 by Rafael S.T. Vieira
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
M = 2; N = 2; y_correct = [0 0 1 1; 0 1 0 1]; assert(isequal(countInBaseN(M,N),y_correct))
2 Pass
M = 2; N = 3; y_correct = [ 0 0 0 1 1 1 2 2 2; 0 1 2 0 1 2 0 1 2]; assert(isequal(countInBaseN(M,N),y_correct))
3 Pass
M = 3; N = 4; y_correct = [ 0 0 0 0 0 1 0 0 2 0 0 3 0 1 0 0 1 1 0 1 2 0 1 3 0 2 0 0 2 1 0 2 2 0 2 3 0 3 0 0 3 1 0 3 2 0 3 3 1 0 0 1 0 1 1 0 2 1 0 3 1 1 0 1 1 1 1 1 2 1 1 3 1 2 0 1 2 1 1 2 2 1 2 3 1 3 0 1 3 1 1 3 2 1 3 3 2 0 0 2 0 1 2 0 2 2 0 3 2 1 0 2 1 1 2 1 2 2 1 3 2 2 0 2 2 1 2 2 2 2 2 3 2 3 0 2 3 1 2 3 2 2 3 3 3 0 0 3 0 1 3 0 2 3 0 3 3 1 0 3 1 1 3 1 2 3 1 3 3 2 0 3 2 1 3 2 2 3 2 3 3 3 0 3 3 1 3 3 2 3 3 3]; assert(isequal(countInBaseN(M,N),y_correct'))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 636 | 1,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.640037 |
https://quant.stackexchange.com/questions/31022/how-do-we-know-that-the-instaneous-rate-of-return-on-this-option-gamma-is-ne | 1,563,469,987,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525699.51/warc/CC-MAIN-20190718170249-20190718192249-00436.warc.gz | 529,565,990 | 35,269 | # How do we know that the instaneous rate of return on this option, $\gamma$ is negative?
I am self-studying models for financial economics and encountered the following problem:
I don't see how the author can conclude that $\gamma = -0.62$. Let's rearrange the second to last equation: $$\gamma - r = -4(0.19 - r)$$ as
$$r = \frac{\gamma + 0.76}{5}.$$
If $\gamma = 0.62$, then $r = 0.276.$
If $\gamma = -0.62$, then $r = 0.028$, as the author states.
So I don't see how the author can conclude $\gamma = -0.62$ when letting $\gamma = 0.62$ does not contradict that $r \geq 0$.
• $4*0.19 = 0.76 \neq 0.71$, and what is $\gamma$ ? – MJ73550 Nov 16 '16 at 9:36
• You're correct. I edited the original post to reflect that. However, it still allows for $\gamma$ to be positive or negative. $\gamma$ is the instantaneous rate of return on the put option. – user2521987 Nov 16 '16 at 14:21
• It is very odd. I would have thought the condition should have been $r<\alpha$ i.e. $r<0.19$ not $r\ge 0$. In words: "the expected excess return on the stock is positive". – noob2 Nov 16 '16 at 14:32
• Can you provide the source of this problem? – muffin1974 Nov 25 '16 at 10:55
• Sure, it's from "Financial Economics" by Abraham Weishaus (It's a study manual for the actuarial exam "Models for Financial Economics") – user2521987 Nov 25 '16 at 16:16 | 433 | 1,344 | {"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": 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} | 3.484375 | 3 | CC-MAIN-2019-30 | longest | en | 0.883579 |
https://www.jufinance.com/mag/fin435_19s/ | 1,603,666,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890028.58/warc/CC-MAIN-20201025212948-20201026002948-00040.warc.gz | 777,009,005 | 74,931 | FIN 435 Class Web Page, Spring '19
Business Finance Online, an interactive learning tool for the Corporate Finance Student https://www.zenwealth.com/BusinessFinanceOnline/index.htm (could be very helpful) =IF(OR( SUM(C4:C54)<>0,), IRR(C4:C54),"")
Weekly SCHEDULE, LINKS, FILES and Questions
Week
Coverage, HW, Supplements
- Required
Equations
Videos (optional)
Week 0
Market Watch Game
Use the information and directions below to join the game.
http://www.marketwatch.com/game/jufin435-19spring
2. Password for this private game: havefun.
4. If you are an existing MarketWatch member, login. If you are a new user, follow the link for a Free account - it's easy!
Week 0
Week 1
# 6 Signs a Business Has Bad Management | Phil Town (video)
For class discussion: What are the 6 signs? Which one is the hardest one to catch? Which one is the most important one to investors?
Chapter 3 Financial Statement Analysis
# How Do You Read a Cash Flow Statement? | (VIDEO)
Finviz.com/screener for ratio analysis (https://finviz.com/screener.ashx
Capital expenditure = increases in NFA + depreciation
Or, capital expenditure = increases in GFA
All companies, foreign and domestic, are required to file registration statements, periodic reports, and other forms electronically through EDGAR. Q3: Free cash flow: concept and equation. What is FCF? Why is it important?
EDGAR online
Steps:
1. Go to EDGAR online
2. Search AAPL
3. Search financial statement of AAPL in 2018, 2017, 2016, and 2015.
Chapter 4 Ratio Analysis
Chapter 4 case study - second case study, with mid-term exam
HW of chapters 3, 4 (due by 3/31)
1. Firm A's sales last year = \$280,000, net income = \$23,000. What was its profit margin? (8.21%)
2. Firm A’s total assets = \$415,000 and its net income = \$32,750. What was its return on total assets (ROA)?(7.89%)
3. Firm A’s total common equity = \$405,000 and its net income = \$70,000. What was its ROE? (17.28%)
4. Firm A’s stock price at the end of last year = \$23.50 and its earnings per share for the year = \$1.30. What was its P/E ratio? (18.08)
5. Meyer Inc's assets are \$625,000, and its total debt outstanding is \$185,000. The new CFO wants to establish a debt/assets ratio of 55%. The size of the firm does not change. How much debt must the company add or subtract to achieve the target debt ratio? (\$158,750)
6. Chang Corp. has \$375,000 of assets, and it uses only common equity capital (zero debt). Its sales for the last year were \$595,000, and its net income was \$25,000. Stockholders recently voted in a new management team that has promised to lower costs and get the return on equity up to 15.0%. What profit margin would the firm need in order to achieve the 15% ROE, holding everything else constant? (9.45%)
7. A firm has \$300 in inventory, \$600 in fixed assets, \$200 in accounts receivable, \$100 in accounts payable, and \$50 in cash. What is the amount of the current assets? (\$550)
8. Art's Boutique has sales of \$640,000 and costs of \$480,000. Interest expense is \$40,000 and depreciation is \$60,000. The tax rate is 34%. What is the net income? ( \$39,600)
9. Use the following information to prepare the cash flow statement in 2008 of Nabors, Inc. Calculate FCF.
Cash Flow Statement Partial Solution Cash at the beginning of the year 310 Cash from operation net income Xxx plus depreciation Xxx -/+ AR Xxx -/+ Inventory Xxx +/- AP Xxx net change in cash from operation 1075 Cash from investment -/+ (NFA+depreciation) Xxx net change in cash from investment -1080 Cash from finaning +/- (long term debt+notes payable) Xxx +/- common stock Xxx - dividend Xxx net change in cash from investment 100 Total net change of cash 95 Cash at the end of the year 405
10. Read the following article. Summarize what is going on under each situation. Where are the auditors? Are they completely ineffective? Are they on the side with the employers? What is your view?
sec_logitech.pdf (FYI) sec_ener1.pdf (FYI)
SEC Announces Financial Fraud Cases
FOR IMMEDIATE RELEASE
2016-74
Washington D.C., April 19, 2016 —
The Securities and Exchange Commission today announced a pair of financial fraud cases against companies and then-executives accused of various accounting failures that left investors without accurate depictions of company finances.
In one case, technology manufacturer Logitech International agreed to pay a \$7.5 million penalty for fraudulently inflating its fiscal year 2011 financial results to meet earnings guidance and committing other accounting-related violations during a five-year period. Logitech’s then-controller Michael Doktorczyk and then-director of accounting Sherralyn Bolles agreed to pay penalties of \$50,000 and \$25,000, respectively, for violations related to Logitech’s warranty accrual accounting and failure to amortize intangibles from an earlier acquisition. The SEC filed a complaint in federal court yesterday against Logitech’s then-chief financial officer Erik Bardman and then-acting controller Jennifer Wolf alleging that they deliberately minimized the write-down of millions of dollars of excess component parts for a product for which Logitech had excess inventory in FY11. For Logitech’s financial statements, the two executives falsely assumed the company would build all of the components into finished products despite their knowledge of contrary facts and events.
In the other case, three then-executives at battery manufacturer Ener1 agreed to pay penalties for the company’s materially overstated revenues and assets for year-end 2010 and overstated assets in the first quarter of 2011. The financial misstatements stemmed from management’s failure to impair investments and receivables related to an electric car manufacturer that was one of its largest customers. Former CEO and chairman of the board Charles L. Gassenheimer, former chief financial officer Jeffrey A. Seidel, and former chief accounting officer Robert R. Kamischke agreed to pay penalties of \$100,000, \$50,000, and \$30,000, respectively.
“We are intensely focused on whether companies and their officers evaluate judgmental accounting issues in good faith and based on GAAP,” said Andrew Ceresney, Director of the SEC’s Division of Enforcement. “In these two cases, we allege deficiencies in Ener1’s failure to properly impair assets on its balance sheet and Logitech’s failure to write down the value of its inventory to avoid the financial consequences of disappointing sales.”
In the Ener1 case, the SEC also found that Robert D. Hesselgesser, the engagement partner for PricewaterhouseCoopers LLP’s audit of Ener1’s 2010 financial statements, violated PCAOB and professional auditing standards when he failed to perform sufficient procedures to support his audit conclusions that Ener1 management had appropriately accounted for its assets and revenues. Hesselgesser agreed to be suspended from appearing and practicing before the SEC as an accountant, which includes not participating in the financial reporting or audits of public companies. The SEC’s order permits Hesselgesser to apply for reinstatement after two years.
“Auditors play a critical role regarding the accuracy of financial statements relied upon by investors, and they must be held accountable when they fail to do everything required under professional auditing standards,” said Michael Maloney, Chief Accountant of the SEC’s Division of Enforcement.
In the Logitech case, former CEO Gerald Quindlen was not accused of any misconduct, but has returned \$194,487 in incentive-based compensation and stock sale profits received during the period of accounting violations, pursuant to Section 304(a) of the Sarbanes-Oxley Act.
The companies and executives who agreed to settlements neither admitted nor denied the charges.
The SEC’s investigation of Logitech was conducted by Paul Gunson and Matthew Finnegan, and supervised by Douglas McAllister. The litigation is being led by Paul Kisslinger and Kevin Lombardi, and supervised by Bridget Fitzpatrick.
The SEC’s investigation of Ener1 was conducted by Carolyn Winters, Richard Haynes, and Deena Bernstein, and supervised by Douglas McAllister.
First week’s class videos
(plus balance sheet, income statement)
https://www.jufinance.com/ratio
****** DuPont Identity *************
ROE = (net income / sales) * (sales / assets) * (assets / shareholders' equity)
This equation for ROE breaks it into three widely used and studied components:
ROE = (net profit margin) * (asset turnover) * (equity multiplier)
# FYI: Amazon.com Inc. (AMZN)https://www.stock-analysis-on.net/NASDAQ/Company/Amazoncom-Inc/DCF/Present-Value-of-FCFF
## Value Stock Criteria List:
### VALUE CRITERIA #1:
Look for a quality rating that is average or better. You don’t need to find the best quality companiesaverage or better is fine. Benjamin Graham recommended using Standard & Poors rating system and required companies to have an S&P Earnings and Dividend Rating of B or better. The S&P rating system ranges from D to A+. Stick to stocks with ratings of B+ or better, just to be on the safe side.
### VALUE CRITERIA #2:
Graham advised buying companies with Total Debt to Current Asset ratios of less than 1.10. In value investing it is important at all times to invest in companies with a low debt load. Total Debt to Current Asset ratios can be found in data supplied by Standard & Poors, Value Line, and many other services.
### VALUE CRITERIA #3:
Check the Current Ratio (current assets divided by current liabilities) to find companies with ratios over 1.50. This is a common ratio provided by many investment services.
### VALUE CRITERIA #4:
Criteria four is simple: Find companies with positive earnings per share growth during the past five years with no earnings deficits. Earnings need to be higher in the most recent year than five years ago. Avoiding companies with earnings deficits during the past five years will help you stay clear of high-risk companies.
### VALUE CRITERIA #5:
Invest in companies with price to earnings per share (P/E) ratios of 9.0 or less. Look for companies that are selling at bargain prices. Finding companies with low P/Es usually eliminates high growth companies, which should be evaluated using growth investing techniques.
### VALUE CRITERIA #6:
Find companies with price to book value (P/BV) ratios less than 1.20. P/E ratios, mentioned in rule 5, can sometimes be misleading. P/BV ratios are calculated by dividing the current price by the most recent book value per share for a company. Book value provides a good indication of the underlying value of a company. Investing in stocks selling near or below their book value makes sense.
### VALUE CRITERIA #7:
Invest in companies that are currently paying dividends. Investing in undervalued companies requires waiting for other investors to discover the bargains you have already found. Sometimes your wait period will be long and tedious, but if the company pays a decent dividend, you can sit back and collect dividends while you wait patiently for your stock to go from undervalued to overvalued.
One last thought. We like to find out why a stock is selling at a bargain price. Is the company competing in an industry that is dying? Is the company suffering from a setback caused by an unforeseen problem? The most important question, though, is whether the company’s problem is short-term or long-term and whether management is aware of the problem and taking action to correct it. You can put your business acumen to work to determine if management has an adequate plan to solve the company’s current problems.
For class discussion: Times have changed. Mr. Granham’s book about value investing was published sixty years ago. Do you think the criteria in his book are still working in today’s environment?
Week 2
Week 3
Chapter 6 Interest rate
third case study, due by 3/31)
Market data website:
http://finra-markets.morningstar.com/BondCenter/Default.jsp (FINRA bond market data)
Market watch on Wall Street Journal has daily yield curve and interest rate information.
# The yield curve (Video, Khan academy)
Treasury Yields
NAME COUPON PRICE YIELD 1 MONTH 1 YEAR TIME (EDT) GB3:GOV 3 Month 0.00 2.40 2.44% +3 +69 2:53 PM GB6:GOV 6 Month 0.00 2.44 2.51% +1 +55 2:53 PM GB12:GOV 12 Month 0.00 2.42 2.49% -4 +43 2:53 PM GT2:GOV 2 Year 2.50 100.07 2.46% -3 +15 2:54 PM GT5:GOV 5 Year 2.38 99.79 2.42% -4 -24 2:53 PM GT10:GOV 10 Year 2.63 100.14 2.61% -3 -25 2:54 PM GT30:GOV 30 Year 3.00 99.53 3.02% +5 -6 2:54 PM
**** exercise: Please draw the yield curve based on the above information *****
For Daily Treasury rates such as the following, please visit https://www.treasury.gov/resource-center/data-chart-center/interest-rates/pages/textview.aspx?data=yield
Date 1 Mo 2 Mo 3 Mo 6 Mo 1 Yr 2 Yr 3 Yr 5 Yr 7 Yr 10 Yr 20 Yr 30 Yr 03/01/19 2.44 2.46 2.44 2.52 2.55 2.55 2.54 2.56 2.67 2.76 2.97 3.13 03/04/19 2.45 2.46 2.46 2.54 2.54 2.55 2.52 2.53 2.63 2.72 2.93 3.09 03/05/19 2.44 2.45 2.46 2.53 2.54 2.55 2.52 2.53 2.63 2.72 2.93 3.08 03/06/19 2.43 2.44 2.47 2.53 2.54 2.52 2.49 2.49 2.59 2.69 2.90 3.06 03/07/19 2.45 2.46 2.45 2.52 2.52 2.47 2.44 2.44 2.54 2.64 2.86 3.03 03/08/19 2.45 2.46 2.46 2.52 2.53 2.45 2.43 2.42 2.51 2.62 2.83 3.00 03/11/19 2.44 2.45 2.46 2.54 2.53 2.47 2.45 2.44 2.53 2.64 2.86 3.03 03/12/19 2.44 2.46 2.46 2.53 2.52 2.45 2.41 2.41 2.50 2.61 2.82 3.00 03/13/19 2.43 2.44 2.45 2.53 2.53 2.45 2.41 2.42 2.51 2.61 2.82 3.02 03/14/19 2.48 2.46 2.45 2.52 2.52 2.46 2.42 2.43 2.53 2.63 2.86 3.04 03/15/19 2.46 2.46 2.45 2.52 2.52 2.43 2.39 2.40 2.49 2.59 2.83 3.02 03/18/19 2.47 2.46 2.44 2.51 2.52 2.45 2.41 2.42 2.51 2.60 2.83 3.01
Monday Mar 18, 2019
For daily yield curve, please visit http://finra-markets.morningstar.com/BondCenter/Default.jsp
Formula --- Break down of interest rate
r = r* + IP + DRP + LP + MRP
r = required return on a debt security
r* = real risk-free rate of interest
MRPt = 0.1% (t – 1)
DRPt + LPt = Corporate spread * (1.02)(t−1)
Summary of Yield Curve Shapes and Explanations
Normal Yield Curve
When bond investors expect the economy to hum along at normal rates of growth without significant changes in inflation rates or available capital, the yield curve slopes gently upward. In the absence of economic disruptions, investors who risk their money for longer periods expect to get a bigger reward — in the form of higher interest — than those who risk their money for shorter time periods. Thus, as maturities lengthen, interest rates get progressively higher and the curve goes up.
Steep Curve – Economy is improving
Typically the yield on 30-year Treasury bonds is three percentage points above the yield on three-month Treasury bills. When it gets wider than that — and the slope of the yield curve increases sharply — long-term bond holders are sending a message that they think the economy will improve quickly in the future.
Inverted Curve – Recession is coming
At first glance an inverted yield curve seems like a paradox. Why would long-term investors settle for lower yields while short-term investors take so much less risk? The answer is that long-term investors will settle for lower yields now if they think rates — and the economy — are going even lower in the future. They're betting that this is their last chance to lock in rates before the bottom falls out.
Flat or Humped Curve
To become inverted, the yield curve must pass through a period where long-term yields are the same as short-term rates. When that happens the shape will appear to be flat or, more commonly, a little raised in the middle.
Unfortunately, not all flat or humped curves turn into fully inverted curves. Otherwise we'd all get rich plunking our savings down on 30-year bonds the second we saw their yields start falling toward short-term levels.
On the other hand, you shouldn't discount a flat or humped curve just because it doesn't guarantee a coming recession. The odds are still pretty good that economic slowdown and lower interest rates will follow a period of flattening yields.
Homework of chapter 6 (Due by 3/31)
HW1 The following yields on U.S. Treasury securities were taken from a financial publication:
Date 1 Mo 2 Mo 3 Mo 6 Mo 1 Yr 2 Yr 3 Yr 5 Yr 7 Yr 10 Yr 20 Yr 30 Yr Dec 3rd, 2018 2.3 2.35 2.38 2.56 2.72 2.83 2.84 2.83 2.9 2.98 3.15 3.27
1. Plot a yield curve based on these data. Is the yield inverted?
2. Calculate the 2 years interest rate 1 year from now, using the pure expectations theory.
3. Calculate the 5 years interest rate 5 year from now, using the pure expectations theory.
HW2 You read in The Wall Street Journal that 30-day T-bills are currently yielding 5.5%. Your brother-in-law, a broker at Safe and Sound Securities, has given you the following estimates of current interest rate premiums:
• Maturity risk premium = 1.8%
• Default risk premium = 2.15%
On the basis of these data, what is the real risk-free rate of return? (answer: 2.25%)
HW3 The real risk-free rate is 3%. Inflation is expected to be 2% this year and 4% during the next 2 years. Assume that the maturity risk premium is zero. What is the yield on 2-year Treasury securities? What is the yield on 3-year Treasury securities?(answer: 6%, 6.33%)
HW4 A Treasury bond that matures in 10 years has a yield of 6%. A 10-year corporate bond has a yield of 8%. Assume that the liquidity premium on the corporate bond is 0.5%. What is the default risk premium on the corporate bond? (answer: 1.5%)
HW5 The real risk-free rate is 3%, and inflation is expected to be 3% for the next 2 years. A 2-year Treasury security yields 6.2%. What is the maturity risk premium for the 2-year security? (answer: 0.2%)
HW6 One-year Treasury securities yield 5%. The market anticipates that 1 year from now, 1-year Treasury securities will yield 6%. If the pure expectations theory is correct, what is the yield today for 2-year Treasury securities? (answer: 5.5%)
HW7
Based on the class discussion, the videos watched and the papers that we read in the class, what is your opinion on the inverted yield curve?
This week’s class videos
# Why Investors Are Obsessed With the Inverted Yield Curve (video)
***For class discussion: Do you think it is a big deal that the yield curve gets flattened?***
# What a Flat Yield Curve Really Means (video)
The U.S. Yield Curve Just Inverted. That’s Huge.
The move ushers in fresh questions about the Fed and the economy.
By
Brian Chappatta, Bloomberg
December 3, 2018, 12:27 PM EST
(
Date 1 Mo 2 Mo 3 Mo 6 Mo 1 Yr 2 Yr 3 Yr 5 Yr 7 Yr 10 Yr 20 Yr 30 Yr Dec 3rd, 2018 2.3 2.35 2.38 2.56 2.72 2.83 2.84 2.83 2.9 2.98 3.15 3.27
The U.S. Treasury yield curve just inverted for the first time in more than a decade.
It’s a moment that the world’s biggest bond market has been thinking about for the past 12 months. I wrote around this time last year that Wall Street had come down with a case of flattening fever, with six of the 11 analysts I surveyed saying that the curve from two to 10 years would invert at least briefly by the end of 2019. That’s not exactly what happened Monday, though that spread did reach the lowest since 2007. Rather, the difference between three- and five-year Treasury yields dropped below zero, marking the first portion of the curve to invert in this cycle.
The First Inversion
After years of flattening, the yield difference between some Treasury notes falls below zero
Source: Bloomberg
The move didn’t come out of nowhere. In fact, I wrote a week ago that the spread between short-term Treasury notes was racing toward inversion, and Bloomberg News’s Katherine Greifeld and Emily Barrett noted the failed break below zero on Friday. Still, I wasn’t necessarily expecting this day to come so soon. Rate strategists have long said that being close doesn’t cut it when talking about an inverted yield curve and the well-known economic implications that come with it, namely that the spread between short- and long-term Treasury yields has dropped below zero ahead of each of the past seven recessions.
It’s important to keep in mind the timeline between inversion and economic slowdowns — it’s not instantaneous. The yield curve from three to five years dipped below zero during the last cycle for the first time in August 2005, some 28 months before the recession began. That this is the first portion to flip isn’t too surprising, considering how much scrutiny bond traders place on the Federal Reserve’s outlook for rate increases. All it means is that the central bank will probably leave interest rates steady, or even cut a bit, in 2022 or 2023. I’d argue that’s not just possible, but probable, given that we’re already in one of the longest economic expansions in U.S. history.
The more interesting question might be why this part of the yield curve won the race to inversion, rather than the spread between seven- and 10-year Treasuries, which looked destined to fall below zero earlier this year. One reason could be that the Fed’s balance-sheet reduction is putting more pressure on 10-year notes than shorter-dated maturities, which wasn’t the case during past periods of inversion. Indeed, policy makers have shown no signs of easing up on this stealth tightening.
On top of that, the Treasury Department is selling increasing amounts of debt, which disproportionately affects the longest-dated obligations because buyers have to consider the duration risk they’re absorbing. Remember the curve from five to 30 years, which fell below 20 basis points in July? That spread is about 46 basis points now, driven by stubbornly higher long-bond yields.
Given the recent pivot from the most important Fed leaders — Jerome Powell, Richard Clarida and John Williams — this flirtation with inversion among two-, three- and five-year Treasury notes probably isn’t going away. The bond market is fast approaching the point where traders have to ask themselves whether a rate hike now increases the chance of a cut in a few years. Other questions include “What is neutral?” and “Can the Fed engineer a soft landing?” To say nothing about whether the assumed relationship between the labor market and inflation expectations is still intact.
Those are big questions without easy answers, and the first inversion of the U.S. yield curve offers only one clue. The Fed wants to be more data dependent going forward. Odds are the market will do the same.
A Recession Is Coming, And Maybe a Bear Market, Too
History shows that equities normally drop about 21 percent when the economy contracts.
By
A. Gary Shilling, Bloomberg
March 18, 2019, 6:00 AM EDT
I first suggested the U.S. economy was headed toward a recession more than a year ago, and now others are forecasting the same. I give a business downturn starting this year a two-thirds probability.
The recessionary indicators are numerous. Tighter monetary policy by the Federal Reserve that the central bank now worries it may have overdone. The near-inversion in the Treasury yield curve. The swoon in stocks at the end of last year. Weaker housing activity. Soft consumer spending. The tiny 20,000 increase in February payrolls, compared to the 223,000 monthly average gain last year. Then there are the effects of the deteriorating European economies and decelerating growth in China as well as President Donald Trump’s ongoing trade war with that country.
There is, of course, a small chance of a soft landing such as in the mid-1990s. At that time, the Fed ended its interest-rate hiking cycle and cut the federal funds rate with no ensuing recession. By my count, the other 12 times the central bank restricted credit in the post-World War II era, a recession resulted.
It’s also possible that the current economic softening is temporary, but a revival would bring more Fed restraint. Policy makers want higher rates in order to have significant room to cut in the next recession, and the current 2.25 percent to 2.50 percent range doesn’t give them much leeway. The Fed also dislikes investors’ zeal for riskier assets, from hedge funds to private equity and leveraged loans, to say nothing of that rankest of rank speculations, Bitcoin. With a resumption in economic growth, a tight credit-induced recession would be postponed until 2020.
“Recession” conjures up specters of 2007-2009, the most severe business downturn since the 1930s in which the S&P 500 Index plunged 57 percent from its peak to its trough. The Fed raised its target rate from 1 percent in June 2004 to 5.25 percent in June 2006, but the main event was the financial crisis spawned by the collapse in the vastly-inflated subprime mortgage market.
Similarly, the central bank increased its policy rate from 4.75 percent in June 1999 to 6.5 percent in May 2000. Still, the mild 2001 recession that followed was principally driven by the collapse in the late 1990s dot-com bubble that pushed the tech-laden Nasdaq Composite Index down by a whopping 78 percent.
The 1973-1975 recession, the second deepest since the 1930s, resulted from the collapse in the early 1970s inflation hedge buying of excess inventories. That deflated the S&P 500 by 48.2 percent. The federal funds rate hike from 9 percent in February 1974 to 13 percent in July of that year was a minor contributor.
The remaining eight post-World War II recessions were not the result of major financial or economic excesses, but just the normal late economic cycle business and investor overconfidence. The average drop in the S&P 500 was 21.2 percent.
At present, I don’t see any major economic or financial bubbles that are just begging to be pricked. The only possibilities are excess debt among U.S. nonfinancial corporations and the heavy borrowing in dollars by emerging-market economies in the face of a rising greenback. Housing never fully recovered from the subprime mortgage debacle. The financial sector is still deleveraging in the wake of the financial crisis. Consumer debt remains substantial but well off its 2008 peak in relation to household income.
Consequently, the recession I foresee will probably be accompanied by about an average drop in stock prices. The S&P 500 fell 19.6 percent from Oct. 3 to Dec. 24, but the recovery since has almost eliminated that loss. A normal recession-related decline of 21.2 percent – meeting the definition of a bear market – from that Oct. 3 top would take it to 2,305, down about 18 percent from Friday’s close, but not much below the Christmas Eve low of 2,351.
What is interest rates
# Gerald Celente: Low Interest Rates are Building the Biggest Bubble in Modern History - 9/21/14
How interest rates are set
What happens if Fed raise interest rates
Week 4
Mid Term (take home exam)
Chapter 7
Case study of chapter 7 (Due with final)
Market data website:
1. FINRA
http://finra-markets.morningstar.com/BondCenter/Default.jsp (FINRA bond market data)
2. WSJ
Market watch on Wall Street Journal has daily yield curve and bond yield information.
3. Bond Online
Simplified Balance Sheet of WalMart
In Millions of USD As of 2019-01-31 Total Assets 219,295,000 Total Current Liabilities 77,477,000 Long Term Debt 43,520,000 Total Liabilities 139,661,000 Total Equity 72,496,000 Total Liabilities & Shareholders' Equity 219,295,000
For discussion:
· What is this “long term debt”?
· Who is the lender of this “long term debt”?
So this long term debt is called bond in the financial market. Where can you find the pricing information and other specifications of the bond issued by WMT?
How Bonds Work (video)
FINRA – Bond market information
http://finra-markets.morningstar.com/BondCenter/Default.jsp
# WAL-MART STORES INC
http://finra-markets.morningstar.com/BondCenter/BondDetail.jsp?ticker=C104227&symbol=WMT.GP
7.550
%
02/15/2030
WMT.GP
931142BF9
\$138.45
3.318%
03/18/2019
## Credit and Rating Elements
Moody's® Rating Aa2 (10/14/2015) Standard & Poor's Rating AA (02/10/2000) TRACE Grade Investment Grade Default — Bankruptcy N Insurance — Mortgage Insurer — Pre-Refunded/Escrowed — Additional Description Senior Unsecured Note
## Classification Elements
Bond Type US Corporate Debentures Debt Type Senior Unsecured Note Industry Group Industrial Industry Sub Group Retail Sub-Product Asset CORP Sub-Product Asset Type Corporate Bond State — Use of Proceeds — Security Code —
Special Characteristics
Medium Term Note N
## Issue Elements
*dollar amount in thousands Offering Date 02/09/2000 Dated Date 02/15/2000 First Coupon Date 08/15/2000 Original Offering* \$1,000,000.00 Amount Outstanding* \$1,000,000.00 Series — Issue Description — Project Name — Payment Frequency Semi-Annual Day Count 30/360 Form Book Entry Depository/Registration Depository Trust Company Security Level Senior Collateral Pledge — Capital Purpose —
## Bond Elements
*dollar amount in thousands Original Maturity Size* 1,000,000.00 Amount Outstanding Size* 1,000,000.00 Yield at Offering 7.56% Price at Offering \$99.84 Coupon Type Fixed Escrow Type
For class discussion:
Fed has hiked interest rates. So, shall you invest in short term bond or long term bond?
Study guide
1. Find bond sponsored by WMT
just go to www.finra.orgè Investor center è market data è bond è corporate bond
Corporate Bond
Issuer Name Callable Coupon Maturity Moody S&P Fitch Price Yield WMT No 7.55 2/15/2030 Aa2 AA AA 138.45 3.318 WMT yes 6.75 4/2/2043 Aa2 AA AA 110.45 4.065
(see below for details)
# WALMART INC
4.750
%
10/02/2043
WMT4055720
931142DK6
04/02/2043
Yes
\$110.45
4.065%
03/14/2019
### US Treasury Yield
Prospectus
For class discussion:
· Fed has hiked interest rates. So, shall you invest in short term bond or long term bond?
· Which of the three WMT bonds are the most attractive one to you? Why?
· Referring to the price chart of the above bond, the price was reaching peak in the middle of 2015. Why? The price was really low in the middle of 2014. Why? Interest rate is not the reason.
HOMEWORK (Due with final)
1. AAA firm’ bonds will mature in eight years, and coupon is \$65. YTM is 8.2%. Bond’s market value? (\$903.04)
2. AAA firm’s bonds’ market value is \$1,120, with 15 years maturity and coupon of \$85. What is YTM? (7.17%)
3. Sadik Inc.'s bonds currently sell for \$1,180 and have a par value of \$1,000. They pay a \$105 annual coupon and have a 15-year maturity, but they can be called in 5 years at \$1,100. What is their yield to call (YTC)? (7.74%)
4. Malko Enterprises’ bonds currently sell for \$1,050. They have a 6-year maturity, an annual coupon of \$75, and a par value of \$1,000. What is their current yield? (7.14%)
5. Assume that you are considering the purchase of a 20-year, noncallable bond with an annual coupon rate of 9.5%. The bond has a face value of \$1,000, and it makes semiannual interest payments. If you require an 8.4% nominal yield to maturity on this investment, what is the maximum price you should be willing to pay for the bond? (\$1,105.69)
6. Grossnickle Corporation issued 20-year, non-callable, 7.5% annual coupon bonds at their par value of \$1,000 one year ago. Today, the market interest rate on these bonds is 5.5%. What is the current price of the bonds, given that they now have 19 years to maturity? (\$1,232.15)
7. McCue Inc.'s bonds currently sell for \$1,250. They pay a \$90 annual coupon, have a 25-year maturity, and a \$1,000 par value, but they can be called in 5 years at \$1,050. Assume that no costs other than the call premium would be incurred to call and refund the bonds, and also assume that the yield curve is horizontal, with rates expected to remain at current levels on into the future. What is the difference between this bond's YTM and its YTC? (Subtract the YTC from the YTM; it is possible to get a negative answer.) (2.62%)
8. Taussig Corp.'s bonds currently sell for \$1,150. They have a 6.35% annual coupon rate and a 20-year maturity, but they can be called in 5 years at \$1,067.50. Assume that no costs other than the call premium would be incurred to call and refund the bonds, and also assume that the yield curve is horizontal, with rates expected to remain at current levels on into the future. Under these conditions, what rate of return should an investor expect to earn if he or she purchases these bonds? (4.2%)
9. A 25-year, \$1,000 par value bond has an 8.5% annual payment coupon. The bond currently sells for \$925. If the yield to maturity remains at its current rate, what will the price be 5 years from now? (\$930.11)
10. Read the attached prospects and answer the following questions: “We are offering \$500,000,000 of our 1.000% notes due 2017 (symbol WMT4117476), \$1,000,000,000 of our 3.300% notes due 2024 (symbol WMT4117477) and \$1,000,000,000 of our 4.300% notes due 2044 (symbol WMT4117478)
1) What is the purpose for the money raised?
2) Which of the two outstanding WMT bonds are more attractive one to you? Why?
3) Who are the underwriters for the WMT bonds?
Class videos
Bond Pricing Formula (FYI)
Bond Pricing Excel Formula
To calculate bond price in EXCEL (annual coupon bond):
Price=abs(pv(yield to maturity, years left to maturity, coupon rate*1000, 1000)
To calculate yield to maturity (annual coupon bond)::
Yield to maturity = rate(years left to maturity, coupon rate *1000, -price, 1000)
To calculate bond price (semi-annual coupon bond):
Price=abs(pv(yield to maturity/2, years left to maturity*2, coupon rate*1000/2, 1000)
To calculate yield to maturity (semi-annual coupon bond):
Yield to maturity = rate(years left to maturity*2, coupon rate *1000/2, -price, 1000)*2
### Redemption Features (FYI)
While the maturity date indicates how long a bond will be outstanding, many bonds are structured in such a way so that an issuer or investor can substantially change that maturity date.
#### Call Provision
Bonds may have a redemption or call provision that allows or requires the issuer to redeem the bonds at a specified price and date before maturity. For example, bonds are often called when interest rates have dropped significantly from the time the bond was issued. Before you buy a bond, always ask if there is a call provision and, if there is, be sure to consider the yield to call as well as the yield to maturity . Since a call provision offers protection to the issuer, callable bonds usually offer a higher annual return than comparable non-callable bonds to compensate the investor for the risk that the investor might have to reinvest the proceeds of a called bond at a lower interest rate.
#### Put Provision
A bond may have a put provision, which gives an investor the option to sell the bond to an issuer at a specified price and date prior to maturity. Typically, investors exercise a put provision when they need cash or when interest rates have risen so that they may then reinvest the proceeds at a higher interest rate. Since a put provision offers protection to the investor, bonds with such features usually offer a lower annual return than comparable bonds without a put to compensate the issuer.
#### Conversion
Some corporate bonds, known as convertible bonds, contain an option to convert the bond into common stock instead of receiving a cash payment. Convertible bonds contain provisions on how and when the option to convert can be exercised. Convertibles offer a lower coupon rate because they have the stability of a bond while offering the potential upside of a stock.
Treasury Bond Auction Website
What is duration? (not required but useful)
Duration is defined as the weighted average of the present value of the cash flows and is used as a measure of a bond price's response to changes in yield.
If duration = 10 years, then for 1% increase in interest rate, the bond price will drop by 10 times of 1%, which is 10%.
You can calculate duration in excel.
Syntax
DURATION(settlement, maturity, coupon, yld, frequency, [basis])
How to calculate bond prices using exact
date? (not required but useful)
Use price function in Excel. Returns the price
per \$100 face value of a security that pays periodic interest.
Syntax
PRICE(settlement, maturity, rate, yld, redemption, frequency, [basis])
Calculate bond yield using exact date?(not required but useful)
Use YIELD to calculate bond yield.
Syntax
YIELD(settlement,maturity,rate,pr,
redemption, frequency, basis)
Excel yield function video
Risk of Bonds
Bond risk (video)
Bond risk – credit risk (video)
How to invest in bond market when Fed is hiking interest rates? (Videos)
Week 5
Chapter 8 Risk and Return
Chapter 8 case study (option 1, due with final)
Summary of the steps in the case study:
1st, calculate expected return based on probabilities and corresponding returns
2nd, calculate standard deviation based on probabilities and corresponding returns
3rd, calculate expected return and standard deviation based on probabilities historical returns
4th, Use corr function to calculate correlation based on two stocks’ historical returns.
5th, Understand the concept of correlation and can pick stocks based on correlations
6th,understand what is beta and can calculate beta using slope function
7th, can use CAMP to calculate stock returns.
Option 2: Risk and Return, and Portfolio (due with final)
Instructions and requirements
· Pick three stocks and find their stock prices by the end of each month in the past five years on finance.yahoo.com.
· Calculate monthly stock return of each stock
· Calculate the mean and the standard deviation of each stock.
· Calculate correlations and generate correlation matrix. Discuss your results regarding the degree of correlation of your stocks.
· Assume that your investment funds are evenly distributed among the three stocks, calculate the portfolio’s return and standard deviation (risk) and compare the results with those of each stock in your portfolio. Discuss your findings.
· Go back to finance.yahoo.com and download sp500 index price of each month in the past five years and calculate its return (ticker: ^GSPC)
· Use slope function in excel to calculate beta of each stock. Discuss your findings. Compare the beta that you find with that reported on google/finance. Should match.
Equations
1. Expected return and standard deviation
Given a probability distribution of returns, the expected return can be calculated using the following equation:
where
• E[R] = the expected return on the stock,
• N = the number of states,
• pi = the probability of state i, and
• Ri = the return on the stock in state i.
Given an asset's expected return, its variance can be calculated using the following equation:
where
• N = the number of states,
• pi = the probability of state i,
• Ri = the return on the stock in state i, and
• E[R] = the expected return on the stock.
The standard deviation is calculated as the positive square root of the variance.
2. Two stock portfolio equations:
W1 and W2 are the percentage of each stock in the portfolio.
• r12 = the correlation coefficient between the returns on stocks 1 and 2,
• s12 = the covariance between the returns on stocks 1 and 2,
• s1 = the standard deviation on stock 1, and
• s2 = the standard deviation on stock 2.
• s12 = the covariance between the returns on stocks 1 and 2,
• N = the number of states,
• pi = the probability of state i,
• R1i = the return on stock 1 in state i,
• E[R1] = the expected return on stock 1,
• R2i = the return on stock 2 in state i, and
• E[R2] = the expected return on stock 2.
3.. Historical returns
Holding period return (HPR) = (Selling price – Purchasing price + dividend)/ Purchasing price
4. CAPM model
### · What Is the Capital Asset Pricing Model?
The Capital Asset Pricing Model (CAPM) describes the relationship between systematic risk and expected return for assets, particularly stocks. CAPM is widely used throughout finance for pricing risky securities and generating expected returns for assets given the risk of those assets and cost of capital.
Ri = Rf + βi *( Rm - Rf) ------ CAPM model
Ri = Expected return of investment
Rf = Risk-free rate
βi = Beta of the investment
Rm = Expected return of market
(Rm - Rf) = Market risk premium
· What is Beta? Where to find Beta?
· SML – Security Market Line
Chapter 9 Stock Return Evaluation
Summary of risk factors that are important to the stock valuation (based on class discussion)
· Free cash flow
· EBIDTA
· Yield curve
· Inflation
· Exchange rate
· Firm size
· Market to book ratio
· Tax rate
· Momentum
· PE
· Earning
· Merger and Acquisition
· IPO
· Debt
· Dividend
· Share split
· Short interest ratio
· Technical analysis ratios
· New product and new market
· World events
· Weather
· …………
We can create a model to predict stock returns based on the above risk factors:
Stock return = function (Free cash flow, EBIDT,……)
Homework (Due with final)
Read Beating the Historical Odds: Recession Risk in 2019 and Beyond
1. How does GS evaluate the likelihood of an recession occurring in 2019?
2. What are GS’s forecasts of the US economy in the short run?
Class videos (Sorry for not turning on the microphone. The videos have no sound. My apologies)
Beating the Historical Odds: Recession Risk in 2019 and Beyond
We have long highlighted the risks that have historically been associated with large overshoots of full employment. We have noted that the Fed has never engineered a soft landing from beyond full employment, that few other advanced economy central banks have either, and that countries that have achieved very long expansions often used countercyclical policy to prevent a large overshoot in the first place. In practice it hasnt been easy to nudge up the unemployment rate just so. While we take this lesson seriously, we think it is being applied too mechanically by market participants today. The key difference with the past is that the Phillips curve is flatter and better anchored on the Feds target today. As a result, where labor market overshoots once led to high and accelerating inflation and consequently had to be unwound urgently with a forceful policy response, today an overshoot will more likely mean inflation persistently but only moderately above target. The Fed could probably live with this for a while, permitting it to tighten gradually and unwind the overshoot slowly. This gives the Fed a good chance of beating the historical odds. How worried should we be about recession risk today? The history of US recessions points to two classic causes of US recessions, overheating and financial imbalances. While overheating risks could emerge down the road, they look quite limited for now: core inflation is at 2%, trend unit labor cost growth is at 2%, and both household inflation expectations and market-implied inflation compensation are below average (Exhibit 9).
(Please refer to https://www.goldmansachs.com/insights/pages/outlook-2019/us-outlook/report.pdf for a better quality graph)
We also see little risk from financial imbalances at the moment. At a high level, the private sector financial balancea very good predictor of recession risklooks quite healthy (Exhibit 10). Digging deeper, our financial excess monitor looks for elevated valuations and stretched risk appetite across major asset classes, and for financial imbalances and vulnerabilities in the household, business, banking, and government sectors. Overall, the message is mostly reassuring. On the valuations side, while commercial real estate prices look somewhat frothy, lending terms and standards have tightened in recent years. On the sectoral imbalances side, fiscal sustainability remains a long-run concern, but we see this less as a recession trigger than as something that could prolong a downturn if policymakers perceive a lack fiscal space to respond. These two classic recession risks are complementaryoverheating and the associated risk of a more abrupt shift in monetary policy is more threatening when financial imbalances are elevated and less threatening when they are limited. With neither risk looking worrisome at the moment, we do not think it makes sense to characterize the economy as late cycle at this point.
The most obvious recession risk beyond 2019 is a mundane and technical one. With a low potential growth rate and a possible need to operate the economy a touch below potential to gradually unwind the overshootwe forecast 1.5% growth in 2020 and 2021the likelihood that normal fluctuations will tip growth negative is mechanically somewhat higher. We would interpret this as simply highlighting the arbitrariness of defining recessions as negative growth, rather than as a material rise in the unemployment rate. Of course, even a less severe recession could see a large sell-off in risk assets. Accounting for these and other considerations, our recession risk model indicates that recession risk is still quite low (Exhibit 11). The expansion is therefore on course to become the longest in US history next year, and even in subsequent years recession is not our base case.
(Please refer to https://www.goldmansachs.com/insights/pages/outlook-2019/us-outlook/report.pdf for a better quality graph)
### · What Is the Capital Asset Pricing Model?
The Capital Asset Pricing Model (CAPM) describes the relationship between systematic risk and expected return for assets, particularly stocks. CAPM is widely used throughout finance for pricing risky securities and generating expected returns for assets given the risk of those assets and cost of capital.
Ri = Rf + βi *( Rm - Rf) ------ CAPM model
Ri = Expected return of investment
Rf = Risk-free rate
βi = Beta of the investment
Rm = Expected return of market
(Rm - Rf) = Market risk premium
Investors expect to be compensated for risk and the time value of money. The risk-free rate in the CAPM formula accounts for the time value of money. The other components of the CAPM formula account for the investor taking on additional risk.
The beta of a potential investment is a measure of how much risk the investment will add to a portfolio that looks like the market. If a stock is riskier than the market, it will have a beta greater than one. If a stock has a beta of less than one, the formula assumes it will reduce the risk of a portfolio.
A stocks beta is then multiplied by the market risk premium, which is the return expected from the market above the risk-free rate. The risk-free rate is then added to the product of the stocks beta and the market risk premium. The result should give an investor the required return or discount rate they can use to find the value of an asset.
The goal of the CAPM formula is to evaluate whether a stock is fairly valued when its risk and the time value of money are compared to its expected return.
For example, imagine an investor is contemplating a stock worth \$100 per share today that pays a 3% annual dividend. The stock has a beta compared to the market of 1.3, which means it is riskier than a market portfolio. Also, assume that the risk-free rate is 3% and this investor expects the market to rise in value by 8% per year.
The expected return of the stock based on the CAPM formula is 9.5%.
The expected return of the CAPM formula is used to discount the expected dividends and capital appreciation of the stock over the expected holding period. If the discounted value of those future cash flows is equal to \$100 then the CAPM formula indicates the stock is fairly valued relative to risk.
The current beta value of a company stock is provided for free by many online financial news services, including Morningstar, Google Finance and Yahoo Finance. Online brokerage services provide more extensive tracking of a company's beta measurements, including historical trends. Beta is sometimes listed under "market data" or other similar headings, as it describes past market performance. A stock with a beta of 1.0 has the same price volatility as the market index, meaning if the market gains, the stock makes gains at the same rate. A stock with a beta of greater than 1.0 is riskier and has greater price fluctuations, while stocks with beta values of less than 1.0 are steadier and generally larger companies.
Examples of Beta
Beta is often measured against the S&P 500 index. An S&P 500 stock with a beta of 2.0 produced a 20 percent increase in returns during a period of time when the S&P 500 Index grew only 10 percent. This same measurement also means the stock would lose 20 percent when the market dropped by only 10 percent. High beta values, including those more than 1.0, are volatile and carry more risk along with greater potential returns. The measurement doesn't distinguish between upward and downward movements. Investing Daily notes that investors try to use stocks with high beta values to quickly recoup their investments after sharp market losses.
Small-Cap Stocks
Beta values are useful to evaluate stock prices of smaller companies. These small-capitalization stocks are attractive to investors because their price volatility can promise greater returns, but Market Watch recommends only buying small-cap stocks with beta values of less than 1.0. The beta value is also a component of the Capital Asset Pricing Model, which helps investors analyze the risk of an investment and the returns needed to make it profitable.
Week 6
Chapter 10 WACC
ppt
One option (if beta is given)
Another option (if dividend is given): | 12,326 | 50,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-45 | latest | en | 0.758389 |
https://estebantorreshighschool.com/equation-help/solving-equation-for-x.html | 1,696,272,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00879.warc.gz | 272,694,676 | 10,940 | ## How do you solve for x in an equation?
Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you’re left with x = 1. Check your work. To check your work, just plug x back in to the original equation to make sure that it works.
## What are the 4 steps to solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal.
## What is X in algebra?
The letter “x” is often used in algebra to mean a value that is not yet known. It is called a “variable” or sometimes an “unknown”. In x + 2 = 7, x is a variable, but we can work out its value if we try! A variable doesn’t have to be “x”, it could be “y”, “w” or any letter, name or symbol. See: Variable.
## What is the golden rule for solving equations?
Do unto one side of the equation, what you do to the other! When solving math equations, we must always keep the ‘scale’ (or equation) balanced so that both sides are ALWAYS equal.
## What is the formula for algebra?
n! = n(n − 1)! = n(n − 1)(n − 2)! = ….Solution:
More topics in Algebra Formulas
Factoring Formulas Percentage Formula
Direct Variation Formula Inverse Variation Formula
Equation Formula Series Formula
Function Notation Formula Foil Formula
## How do you balance algebraic equations?
An example of keeping the equation balanced in order to solve an equationMove all the x terms to one side. Use inverse operations and add 1 5 x 15x 15x to both sides to keep the equation balanced.Solve by working backwards from the order of operations. Continue to get x alone using inverse operations.
You might be interested: Equation of a line that passes through two points
## What is the rule for solving equations?
The following steps provide a good method to use when solving linear equations. Simplify each side of the equation by removing parentheses and combining like terms. Use addition or subtraction to isolate the variable term on one side of the equation. Use multiplication or division to solve for the variable.
## How do you simplify equations?
Here are the basic steps to follow to simplify an algebraic expression:remove parentheses by multiplying factors.use exponent rules to remove parentheses in terms with exponents.combine like terms by adding coefficients.combine the constants.
## What is the first step in solving the equation?
A 4-Step Guide to Solving Equations (Part 2)Step 1: Simplify Each Side of the Equation. As we learned last time, the first step in solving an equation is to make the equation as simple as possible. Step 2: Move Variable to One Side.
## What is () called in math?
* This symbol is called an asterisk. In mathematics, we sometimes use it to mean multiplication, particularly with computers. For example, 5*3 = 5 times 3 = 15.
## What does ∈ mean?
set membership symbol
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] | 851 | 3,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.929633 |
https://hopegiometti.medium.com/linkedlists-203-from-last-94950e0bdf62?source=post_internal_links---------2---------------------------- | 1,638,560,406,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00502.warc.gz | 363,458,636 | 28,252 | Welcome back to another (and most likely the final) week of my linked list series. If you are generally unfamiliar with linked lists or want to check out other posts in this series, definitely take a look at my previous articles on this topic!
Otherwise, let’s just get right into solving our final linked list problem (at least for now!).
The Problem
The goal of the from last problem is given both a linked list and an integer n, return the element n spaces from the last node in the list. For example, if n = 0, then we should return the last element. If n = 1, we want to return the second last element. If n = 2, we should return the third to last element. Etc.
Now, there are a couple of caveats to this problem. First, we can assume that n will always be less than the length of the list. Second, for this solution we are not going to be using a “size” method to determine the length of this list. While it might be easy to come up with a simple solution using the length of the list, instead, we are going to once again revisit the “Tortoise and the Hare Algorithm” and see how we can implement it in the solution to this problem.
The solution: Phase One
The solution to the “from last” problem is going to have two phases. In phase one, we are going to work with our fast variable, however, our use of the fast variable in this problem will slightly differ from how we have used it previously.
Instead of advancing our fast variable by two nodes at a time as we did in our previous solutions, we are going to start by advancing our fast pointer by n nodes. So, if n is three, we are going to move our fast variable to three nodes into our list.
To do this, we can use a simple while loop:
`function fromLast(list, n) { let fast = list.head while (n > 0) { fast = fast.next n-- }}`
The Solution: Phase Two
Now that we’ve completed the first phase of our solution, we can move on to our second and final phase.
Much like our first phase, this phase is also quite simple, though again its use of the fast variable differs from our other implementations of “Tortoise and the Hare Algorithm”. For this solution, both fast and slow will move forward one node at a time. Really, it might make a bit more sense to call our fast variable “head start” for this solution, however for consistency’s sake, we will stick with our original terms.
Now, as fast and slow move forward, we will check to check to see when fast reaches the end of the list aka when fast.next is null. Since we previously gave fast a head start by n nodes, we know that our slow variable will be n nodes behind our fast variable. Thus, when fast reaches the end of the list, our slow variable will be pointing to the node we are looking for and we can return the slow variable.
The Solution: Completed
`function fromLast(list, n) { let slow = list.head let fast = list.head while (n > 0) { fast = fast.next n-- } while (fast.next) { slow = slow.next fast = fast.next } return slow}`
There you have it! That’s our complete solution to our “from last” problem. I hope you’ve enjoyed this little series on linked lists. Definitely give these problems a try on your own and as always, good luck and happy coding! | 751 | 3,251 | {"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} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.934879 |
https://www.mersenneforum.org/showthread.php?s=36e61998b6537679884f99189591707a&p=93309 | 1,652,918,410,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00269.warc.gz | 1,034,045,524 | 16,051 | mersenneforum.org Small Primes for Octoproths <= 155
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2006-12-05, 17:39 #34 ValerieVonck Mar 2004 Belgium 5×132 Posts k=152 from 0 to 1000T Code: 152 done up to 1000T n=152, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 95614625675835 152 16693766169885 152 167966368851015 152 86393828985345 152 The sieving is complete. Number of Prp tests=7991452 Time=3903 sec. n=152, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 332561674476675 152 340128988039575 152 312485329355535 152 274639651702605 152 271554160324965 152 272200916875965 152 The sieving is complete. Number of Prp tests=7980839 Time=3915 sec. n=152, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 555295374334365 152 457565187976995 152 504522026157345 152 531032612628945 152 The sieving is complete. Number of Prp tests=7989729 Time=3909 sec. n=152, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 703522956364515 152 684391017575055 152 624040836327195 152 642061442778945 152 662156994939615 152 648815615411775 152 761807701124295 152 623070269081085 152 The sieving is complete. Number of Prp tests=7987464 Time=3912 sec. n=152, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 835128321307455 152 935670498304185 152 961669429878885 152 824160359935815 152 939865704188985 152 The sieving is complete. Number of Prp tests=7994566 Time=3912 sec.
2006-12-05, 17:40 #35 ValerieVonck Mar 2004 Belgium 5×132 Posts k=121 from 0 to 1000T Code: 121 done up to 1000T n=121, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 87233335971705 121 178166661358785 121 98897749721805 121 133357348486725 121 10860421305525 121 50732695327755 121 57675348241875 121 42345744872145 121 100907851340295 121 190874573049315 121 63941423440245 121 96275841385725 121 101753002775775 121 143926485148425 121 137748181109175 121 17630974039965 121 67155139025325 121 89565266972895 121 93807865103415 121 154138647072255 121 11007840039945 121 The sieving is complete. Number of Prp tests=8182052 Time=4574 sec. n=121, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 266725822069455 121 236937132772725 121 207416163668715 121 390902702197545 121 232508105490045 121 211951701562965 121 224600413578255 121 348785963074275 121 301482700682955 121 368494242518325 121 362381514639495 121 254664696308385 121 327173230511205 121 359684115804915 121 365757395613345 121 210002885306115 121 333865613787705 121 The sieving is complete. Number of Prp tests=8182212 Time=4629 sec. n=121, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 550735988809215 121 593470575802455 121 504094897368525 121 420260397562335 121 417912166225575 121 557061436397445 121 531256521915405 121 411869733515505 121 531037026397305 121 566436976028175 121 533063850351495 121 442585190573085 121 420636653887065 121 421687868257635 121 519890701029855 121 475932931097415 121 513657498248895 121 The sieving is complete. Number of Prp tests=8185162 Time=4597 sec. n=121, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 797313394976175 121 691030176184635 121 764259744717105 121 646213734554025 121 616585886925645 121 770959239944895 121 794601285063585 121 712095532518525 121 740327506102185 121 741312238778145 121 673206684380385 121 794004800506455 121 787122624547995 121 728817311980185 121 760053214982775 121 662222754462555 121 664797248309865 121 768064263223005 121 725885888857095 121 620971804339095 121 667609850600055 121 612804190012035 121 776024747003505 121 The sieving is complete. Number of Prp tests=8180310 Time=4586 sec. n=121, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 809764207741125 121 889402566102765 121 955606955254665 121 920998120697985 121 918372323140635 121 908864129144685 121 955396618075515 121 984486590052525 121 967127741102775 121 878187760357545 121 892339656697215 121 993264232986585 121 800517350673195 121 950635226061795 121 864154854027135 121 813526178040225 121 903547725736635 121 894579870765855 121 986915336736945 121 The sieving is complete. Number of Prp tests=8179742 Time=4585 sec.
2006-12-05, 17:45 #36 ValerieVonck Mar 2004 Belgium 5×132 Posts k=145 from 0 to 1000T Code: 145 done up to 1000T n=145, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 74163464017005 145 106753767517125 145 74300238381915 145 147776568931005 145 101959328018925 145 The sieving is complete. Number of Prp tests=8996977 Time=5021 sec. n=145, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 267666507811695 145 319316157825405 145 The sieving is complete. Number of Prp tests=8998398 Time=5038 sec. n=145, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 509084798725125 145 446806878632925 145 550180535874255 145 512932105806765 145 453205248777945 145 555265132241175 145 540508239779955 145 574364815265025 145 590464129714005 145 512873328167325 145 The sieving is complete. Number of Prp tests=8996121 Time=5024 sec. n=145, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 749395281884865 145 691881723656955 145 756221964021945 145 690308529104895 145 719676922949655 145 743277476266515 145 644062954859745 145 710642359934415 145 672416560071165 145 775660863016845 145 722371800308865 145 The sieving is complete. Number of Prp tests=9003124 Time=5012 sec. n=145, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 998310710788845 145 934316388091185 145 842364927024375 145 852002112604005 145 906614396979045 145 The sieving is complete. Number of Prp tests=9003538 Time=5018 sec.
2006-12-05, 17:46 #37 ValerieVonck Mar 2004 Belgium 5×132 Posts k=153 from 0 to 1000T Code: 153 done up to 1000T n=153, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 29905849688505 153 68056538585595 153 179626408806285 153 180834131435475 153 The sieving is complete. Number of Prp tests=4019309 Time=1894 sec. n=153, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 256858662765495 153 The sieving is complete. Number of Prp tests=4018937 Time=1893 sec. n=153, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 503014230370365 153 The sieving is complete. Number of Prp tests=4017150 Time=1893 sec. n=153, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 732409311906195 153 756028461225555 153 The sieving is complete. Number of Prp tests=4017350 Time=1911 sec. n=153, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 988925414153775 153 890208205532205 153 805670751460365 153 842590493041395 153 The sieving is complete. Number of Prp tests=4015549 Time=1901 sec.
2006-12-05, 17:47 #38 ValerieVonck Mar 2004 Belgium 5×132 Posts k=146 from 0 to 1000T Code: 146 done up to 1000T n=146, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array The sieving is complete. Number of Prp tests=2537874 Time=1406 sec. n=146, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 365130319638525 146 316114654394385 146 The sieving is complete. Number of Prp tests=2536183 Time=1407 sec. n=146, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 433394302302675 146 421645237281345 146 545232269623755 146 The sieving is complete. Number of Prp tests=2535978 Time=1405 sec. n=146, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 614682896659755 146 759461559178245 146 The sieving is complete. Number of Prp tests=2540591 Time=1412 sec. n=146, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 897289266683985 146 The sieving is complete. Number of Prp tests=2535577 Time=1406 sec.
2006-12-05, 17:48 #39 ValerieVonck Mar 2004 Belgium 5·132 Posts k=122 from 0 to 1000T Code: 122 done up to 1000T n=122, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 106965669491415 122 40572339761835 122 197690849175645 122 194296495723155 122 189040265825775 122 6770724125445 122 197457127556115 122 22403088766695 122 160704382916595 122 20143163420205 122 118121992952775 122 107603616098055 122 133947762703545 122 71738766748005 122 15958643480955 122 110935305605565 122 62286788046945 122 23749295090325 122 65254832898465 122 161021252277165 122 142239737224935 122 39497000601345 122 157309669382145 122 107488349593065 122 22786904175315 122 92597580956685 122 133354867173045 122 124004154084615 122 59862728473065 122 The sieving is complete. Number of Prp tests=7637059 Time=3277 sec. n=122, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 295491194447085 122 287463446184915 122 239423980546485 122 389625053252085 122 228936660606465 122 368397388415595 122 352089084447765 122 306337652445825 122 219978863368005 122 373539732346515 122 280415191925715 122 249271445496045 122 269898166499325 122 335863162330815 122 299167739216535 122 361232864210505 122 The sieving is complete. Number of Prp tests=7634528 Time=3256 sec. n=122, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 409419147554235 122 421834726352235 122 455329240819425 122 599062403201745 122 582056036634555 122 442300190796345 122 548411663341185 122 466631396584815 122 551953519432035 122 466218512113095 122 447626177142795 122 453103559734035 122 508910915233875 122 409718697776745 122 474245412736095 122 571147819723215 122 438804678563055 122 497936785420725 122 539368262312985 122 524514741431415 122 554028383397825 122 The sieving is complete. Number of Prp tests=7635777 Time=3403 sec. n=122, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 627422156992785 122 778069637976435 122 667790594346435 122 615901649387745 122 645565001971995 122 640718777936265 122 746935098236355 122 682003696690575 122 787354177079295 122 713225104702215 122 650213216231775 122 658752114754545 122 611237132559795 122 724082897267625 122 736056981059295 122 600173357717115 122 The sieving is complete. Number of Prp tests=7638738 Time=3277 sec. n=122, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 912936850366275 122 808409189657265 122 822710571273255 122 882228243237915 122 815757962926095 122 840981960193095 122 828840808918395 122 865875119992155 122 941085974082135 122 885093237050295 122 846012741931875 122 867306957119085 122 993058840768995 122 856806008535015 122 813800173120725 122 981499085061705 122 The sieving is complete. Number of Prp tests=7640714 Time=3343 sec.
2006-12-05, 17:49 #40 ValerieVonck Mar 2004 Belgium 5×132 Posts k=123 from 0 to 1000T Code: 123 done up to 1000T n=123, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 23037698575815 123 14795373071115 123 66189956830515 123 125232494105355 123 The sieving is complete. Number of Prp tests=2228786 Time=1497 sec. n=123, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 342308475514005 123 247946065563735 123 326578731560235 123 The sieving is complete. Number of Prp tests=2224145 Time=1742 sec. n=123, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 532425872973435 123 541302353627595 123 455388004522605 123 465319001766255 123 598658058914475 123 The sieving is complete. Number of Prp tests=2224407 Time=1388 sec. n=123, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 649935128632965 123 655123209370455 123 730190011446105 123 683612257210365 123 783285719398665 123 The sieving is complete. Number of Prp tests=2227138 Time=1389 sec. n=123, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 809598448994895 123 928384271569755 123 931533454454085 123 The sieving is complete. Number of Prp tests=2224685 Time=1389 sec.
2006-12-05, 17:49 #41 ValerieVonck Mar 2004 Belgium 5×132 Posts k=154 from 0 to 1000T Code: 154 done up to 1000T n=154, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 51618124858425 154 196697927335155 154 185986347853515 154 109325715702735 154 116826104139165 154 166666172097165 154 The sieving is complete. Number of Prp tests=11926649 Time=5767 sec. n=154, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 303950915091585 154 318626413491555 154 327747367968405 154 208640093246775 154 303640470235935 154 302657538096675 154 269639132459865 154 The sieving is complete. Number of Prp tests=11928225 Time=5766 sec. n=154, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 530951928433095 154 508621691675475 154 505376590643625 154 414842077257945 154 The sieving is complete. Number of Prp tests=11924383 Time=5767 sec. n=154, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 663126504232815 154 768836031087285 154 The sieving is complete. Number of Prp tests=11922098 Time=5786 sec. n=154, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 929990792615625 154 861249194297415 154 876201685749195 154 801168433600725 154 The sieving is complete. Number of Prp tests=11920907 Time=5779 sec.
2006-12-05, 17:50 #42 ValerieVonck Mar 2004 Belgium 5×132 Posts k=124 from 0 to 1000T Code: 124 done up to 1000T n=124, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 111049638299985 124 96000482410665 124 184266961531485 124 120783438286035 124 164432900716755 124 47922673139895 124 116391437220765 124 170999902670715 124 148174849478235 124 169594221545295 124 15857166234525 124 138652618412685 124 115801886454345 124 163456531541295 124 18814945517745 124 36444423134475 124 17317349329155 124 The sieving is complete. Number of Prp tests=9514705 Time=4928 sec. n=124, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 236621230204065 124 247304708983695 124 310952220934545 124 266141564411715 124 236700896261955 124 229310055407055 124 336417391975335 124 391527816066765 124 248491606994925 124 228013472994315 124 253293933252885 124 321368497228485 124 216096326500215 124 341958441852405 124 382805186707035 124 298478103849225 124 218659940388105 124 303560172647655 124 250245887047065 124 350300147583555 124 240409308401625 124 384445440955815 124 327726324260265 124 The sieving is complete. Number of Prp tests=9512142 Time=4972 sec. n=124, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 518111786062395 124 521244010726725 124 490432433066415 124 419722703900955 124 476354610119205 124 481093998573525 124 418174899312885 124 534221120900475 124 532699214305335 124 405127152202605 124 575568105763365 124 487067628844965 124 536182419300165 124 537650601778095 124 409905850783785 124 518287285439715 124 518909274251445 124 440109313394535 124 509050178642265 124 532192048687005 124 The sieving is complete. Number of Prp tests=9518210 Time=5173 sec. n=124, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 647311781272365 124 749051504005995 124 701302730764845 124 718025333970105 124 694810076091795 124 612244987930815 124 738546349457085 124 650506989047175 124 701562934751925 124 788110493130975 124 770902247174415 124 673018265535855 124 720030852025785 124 752453164391265 124 797006163461325 124 759728462040885 124 719063993245935 124 621747099685005 124 652717971845415 124 651891862891815 124 The sieving is complete. Number of Prp tests=9512283 Time=6029 sec. n=124, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 908727314937105 124 927814529040645 124 862509449346735 124 839956333620195 124 955515765613395 124 820694723764245 124 987789305822565 124 966636761759955 124 991054072943025 124 893773102947795 124 867830979651735 124 963269219828175 124 860515551256815 124 902475304956555 124 911450778376845 124 875619611420715 124 824043000547005 124 880671556815435 124 830827263059625 124 800024844572745 124 992498445857295 124 The sieving is complete. Number of Prp tests=9516368 Time=5098 sec. Last fiddled with by ValerieVonck on 2006-12-05 at 17:51
2006-12-05, 17:51 #43 ValerieVonck Mar 2004 Belgium 34D16 Posts k=125 from 0 to 1000T Code: 125 done up to 1000T n=125, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 192134277161505 125 143450732515545 125 81016652605665 125 54494945889615 125 21000707492835 125 84598692498705 125 92183413649565 125 145487218159605 125 194365979484075 125 26230498255815 125 28051609645605 125 179766404783805 125 168257937730575 125 22110474424815 125 99870174976425 125 187283559110865 125 113199548447055 125 86575872456675 125 The sieving is complete. Number of Prp tests=6952999 Time=2762 sec. n=125, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 271186814860815 125 382356000371415 125 388894203797715 125 392900510807805 125 326160511451775 125 350507548843305 125 261811467755145 125 344217520775835 125 389090871233925 125 343938462742875 125 391676775876225 125 The sieving is complete. Number of Prp tests=6954283 Time=2773 sec. n=125, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 505502619469155 125 496375895659875 125 542588027217915 125 512645398711845 125 538096503155505 125 449024447347245 125 577158512353185 125 544964894571405 125 421566673891125 125 443062572342915 125 429621455224815 125 511602075280695 125 533445328277715 125 419885938350585 125 582336635211555 125 The sieving is complete. Number of Prp tests=6951918 Time=2770 sec. n=125, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 725504930620155 125 792481668508215 125 698865885573765 125 744107573521095 125 667336411560525 125 634934421178365 125 614634318834375 125 685073225045685 125 719368853348145 125 716291291855685 125 718951259264385 125 625412279205315 125 722064589637685 125 760885660725645 125 The sieving is complete. Number of Prp tests=6950858 Time=2767 sec. n=125, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 839397864430125 125 912566561503905 125 962596925780175 125 874500229236795 125 809426494150365 125 925094367447705 125 896727447639345 125 880546298868915 125 832069141473255 125 859197707101335 125 841124089616265 125 859026864525375 125 833732627270865 125 867744922237515 125 The sieving is complete. Number of Prp tests=6958532 Time=2771 sec.
2006-12-05, 19:21 #44 ValerieVonck Mar 2004 Belgium 5·132 Posts k=126 from 0 to 1000T Code: 126 done up to 1000T n=126, kmin=0T, kmax=200T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 62476265995395 126 10208767082535 126 108273731642055 126 39769310437605 126 169179248361375 126 13988965055835 126 103484236985535 126 99391609412865 126 89681940974955 126 The sieving is complete. Number of Prp tests=5124481 Time=2106 sec. n=126, kmin=200T, kmax=400T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 210101057285175 126 233757830967255 126 263084955751485 126 351834207714045 126 287637903342195 126 367420824737715 126 215782593172125 126 393415846069575 126 358222958442405 126 283855687787805 126 276954541210095 126 304405041675735 126 394619574309345 126 344892205388115 126 The sieving is complete. Number of Prp tests=5123045 Time=2114 sec. n=126, kmin=400T, kmax=600T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 596578000257645 126 484095475718085 126 581034478765335 126 441731281822215 126 420673341425085 126 591757919196465 126 415609599600195 126 451114177863405 126 594942928212885 126 451587691792545 126 The sieving is complete. Number of Prp tests=5123764 Time=2158 sec. n=126, kmin=600T, kmax=800T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 762023710316505 126 741222101330655 126 695898610027005 126 681927200699745 126 649196812231275 126 668303201814885 126 The sieving is complete. Number of Prp tests=5127145 Time=2111 sec. n=126, kmin=800T, kmax=1000T, version=6.0, here T=10^12 Starting the sieve... Using the first 9 primes to reduce the size of the sieve array 885609544808625 126 840254357038845 126 874239223007745 126 958489130764935 126 947339623275045 126 802757678041935 126 862629118308825 126 812184928330545 126 848154239591715 126 923000787662205 126 The sieving is complete. Number of Prp tests=5124787 Time=2116 sec.
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Thu May 19 00:00:10 UTC 2022 up 34 days, 22:01, 0 users, load averages: 2.03, 1.73, 1.63 | 8,799 | 24,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.476051 |
http://mathhelpforum.com/advanced-algebra/64637-need-urgent-help-print.html | 1,529,585,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00021.warc.gz | 206,943,676 | 2,902 | ı need urgent help
• Dec 12th 2008, 12:36 AM
bogazichili
ı need urgent help
1. Let p be prime, and G be a finite group. If every element of G has order a power of p, then |G| = p^n for some n≥0. (Hint: Use Cauchy’s theorem.)
2. Tell as much as possible about the subgroups of a group of order 30 and of a group of order 40.
• Dec 12th 2008, 09:05 AM
ThePerfectHacker
Quote:
Originally Posted by bogazichili
1. Let p be prime, and G be a finite group. If every element of G has order a power of p, then |G| = p^n for some n≥0. (Hint: Use Cauchy’s theorem.)
Assume that there is $\displaystyle q\not = p$ such that $\displaystyle q$ divides order of $\displaystyle |G|$. Then by Cauchy's theorem there exists and element of order $\displaystyle q$ which is not a power of $\displaystyle p$ - a contradiction. Thus, all prime divisors of $\displaystyle |G|$ must be equal to $\displaystyle p$ and so $\displaystyle |G| = p^n$.
Quote:
2. Tell as much as possible about the subgroups of a group of order 30 and of a group of order 40.
I am not sure exactly what you are looking for.
But for order $\displaystyle 30$ it can be shown (I looked up a classification table) that these groups are: $\displaystyle \mathbb{Z}_{30}, D_{15}, D_5 \times \mathbb{Z}_3, D_3 \times \mathbb{Z}_5$. | 393 | 1,284 | {"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} | 3.40625 | 3 | CC-MAIN-2018-26 | latest | en | 0.859933 |
https://www.sofolympiadtrainer.com/forum/questions/37740/pstate-t-for-true-and-f-for-falsebr--a-the-hand-of-a-clock-will-stop-at-9-if-it-start-from-12-and-ma | 1,669,609,095,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00510.warc.gz | 1,070,420,171 | 7,922 | # User Forum
Subject :IMO Class : Class 6
State 'T' for true and 'F' for false.
a. The hand of a clock will stop at 9 if it start from 12 and makes 3/4 of a revolution, clockwise.
b. Sector is the region in the interior of a circle enclosed by an arc and a chord.
c. Every diameter of a circle is its chord but every chord is not the diameter.
d. A line dividing a line segment into two equal parts at right angle is called perpendicular bisector of the line segment.
.a .b .c .d A T F T F B T F T T C F T T T D T T T F
## Ans 1: (Master Answer)
Class : Class 1
The correct answer is option B. | 172 | 603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.867597 |
http://tsinghuamathcamp.cn/2021/LA.html | 1,652,798,621,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00531.warc.gz | 55,734,217 | 2,941 | Linear Algebra
# Bong Lian 连文豪
## Course Description
Linear algebra may be thought of as the study of linear equations, like the familiar equation x+y +z = 1. But the subject permeates other ends of the mathematical world including topology, geometry, analysis, not to mention algebra itself! As a tool, it’s power can also be felt throughout mathematical sciences at large, reaching the deepest corners of physics, biology, economics, and countless other fields. All fancy stuff.
We, however, will go all the way back to the very beginning instead, and ask very basic questions. We solve equations using numbers. But what are numbers? Why linear equations? Solving them tells us what? How do we solve them? How do we describe solutions – their shapes and sizes? What does it mean to say that “the equation x + y + z = 1 describes a plane in 3-space?” Given a point P and a plane H in 3-space, how do we find the point in H closest to P? Given n points on a plane, can we find a straight line that is the ‘closest’ to all those points? Some of these questions obviously connect linear equations with geometrical objects we can visualize. We will dig deeper into this tells us about linear algebra problems and vice versa.
As for pre-requisite or background for this course, Mathcampers are not expected to have had any prior advanced training in mathematics, but good working knowledge of high school algebra – equations in one or two variables, real numbers – some familiarity with calculus and complex numbers, plus a bit curiosity would give them a great start!
Homework problems will be assigned in class almost daily during the lecture, and they will be checked and returned to students on a regular basis. One or more research projects will be announced during the first week of Mathcamp. Mathcampers will be meeting and working with two Coaches in this course, in addition to the Lecturer. The Coaches will be meeting with Mathcampers everyday, to work with them on their homework assignments and their research projects. Mathcampers are also very much encouraged to work with each other. | 443 | 2,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.950423 |
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## Discussion of Problem 1079. Maximum
Why I get WA1(the result of my code is 3 4)
Posted by Ilya 10 Jan 2021 03:54
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
int a[100000];
a[0] = 0;
a[1] = 1;
int max = 1;
while(cin >> n, n != 0){
if(n > max){
for(int i = max + 1; i <= n; ++i){
if(i % 2 == 0){
a[i] = a[i / 2];
}
else{
a[i] = a[(i - 1) / 2] + a[(i - 1) / 2 + 1];
}
max = n;
}
}
cout << a[n - (n % 2 == 0)] << "\n";
}
}
Edited by author 10.01.2021 03:55
Re: Why I get WA1(the result of my code is 3 4)
Posted by kitesho 12 Jan 2021 20:50
print the maximum value in the range 1 to n
a13 is greater than a15 | 323 | 868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.632733 |
https://discuss.codechef.com/t/chef-swaps-digits-swpdgt/59542 | 1,685,779,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00126.warc.gz | 245,575,368 | 4,453 | # Chef Swaps Digits(SWPDGT)
Q).CodeChef: Practical coding for everyone
can someone tell what is wrong in this code,all test cases in question are giving correct ans still not able to submit??
``````#include <iostream>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int x,y;
cin>>x;
cin>>y;
int temp=max(x,y);
y=min(x,y);
int b1= y /10;
int b2=y% 10;
x=temp;
int a1=x/10;
int a2=x%10;
int ans;
if(x < 10)
ans=x + y;
else if(y < 10){
ans= max(b2,a1)*10 + min(b2,a1)+a2;
}
else{
ans= max(a1,b2)*10+min(a1,b2)+ max(b1,a2)*10 +min(b1,a2);
}
cout<<ans<<endl;
}
return 0;
}
``````
change your else conditon to =>
{
ans= max(max(a1,b2)10+min(a1,b2)+ b110+a2 , max(b1,a2)10 +min(b1,a2)+ a110+ b2);
}
you are making two swaps in your else condition but question asks for only one swap.
10 is multiplied.
idk why the asterisk is not visible
Thank u very much!! | 306 | 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} | 2.671875 | 3 | CC-MAIN-2023-23 | latest | en | 0.581054 |
https://hoven-in.appspot.com/Home/Aptitude/Deductive-Reasoning/deductive-reasoning-quiz-and-questions-008.html | 1,708,798,018,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00201.warc.gz | 310,676,277 | 6,349 | # Deductive Reasoning Quiz Set 008
### Question 1
1. Excess rainfall causes damage to the crops.
2. The crop yield was less this year.
3. This year there was excess rainfall.
If the first two statements are true, the third statement is:
A
true.
B
false.
C
uncertain.
D
None.
Soln.
Ans: c
Crop yield can be less because of damage due to rainfall, and it can be less because of other factors also, like less rainfall. So we cannot say with certainty.
### Question 2
1. Jasvir weighs more than Roshan.
2. Roshan weighs less than Ritu.
3. Of the three babies, Ritu weighs the most.
If the first two statements are true, the third statement is.
A
true.
B
false.
C
uncertain.
D
none.
Soln.
Ans: c
We only know that Ritu weighs more than Roshan. There is no way to tell whether Ritu also weighs more than Jasvir.
### Question 3
1. The Elante Mall has more stores than the Fun Republic.
2. The North Country Mall has fewer stores than the Fun Republic.
3. The Elante Mall has more stores than the North Country Mall.
If the first two statements are true, the third statement is.
A
true.
B
false.
C
uncertain.
D
none.
Soln.
Ans: a
From the first two statements, you know that the Elante Mall has the most stores, so the Elante Mall would have more stores than the North Country Mall.
### Question 4
1. A box of marbles contains more dark blue marbles than orange marbles.
2. There are more dark blue marbles than red.
3. The box contains fewer dark blue marbles than orange marbles.
If the first two statements are true, the third statement is.
A
true.
B
false.
C
uncertain.
D
none.
Soln.
Ans: b
dark blue > orange, and dark blue > red. Statement is false because dark blue are the most.
### Question 5
1. Houses in the Chandigarh cost less than Houses in The Panchkula.
2. Houses in the Mohali cost more than Houses in the The Panchkula.
3. Of the three, a House in Mohali costs the most.
If the first two statements are true, the third statement is.
A
true.
B
false.
C
uncertain.
D
none.
Soln.
Ans: a
Since the Panchkula costs more than the Chandigarh and the Mohali costs more than the Panchkula, it is true that the Mohali costs the most. | 581 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.889941 |
http://www.answers.com/Q/How_many_of_our_suns_could_fit_inside_of_the_largest_star_in_the_Universe?show_edit=yes | 1,500,864,679,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424683.39/warc/CC-MAIN-20170724022304-20170724042304-00634.warc.gz | 364,694,871 | 47,419 | What would you like to do?
# How many of our suns could fit inside of the largest star in the Universe?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
#### exists and is an alternate of .
The largest star is able to fit 100000000 suns inside it
2 people found this useful
Thanks for the feedback!
# How many tangerines could fit in the Sun?
The actual size of the Sun is unclear, because of its gaseous nature, and tangerines vary slightly in size. It is also unlikely that anyone would or could fill the Sun up with
# How many moons could fit inside the sun?
Approximately 72 million Earth moons could fit inside the sun.
# How many mercurys could you fit in the sun?
You could fit approximately 21,253,933 Mercuries inside the Sun
# How many neptunes can fit inside the sun?
Roughly 750,000 Neptunes can fit inside the Sun. The Sun is more than large enough to comfortably fit every planet in the solar system into it and still have plenty of roo
# How many Earths can fit inside the sun?
752 The sun has a diameter of 865000 miles or 1.39 million kilometers. Its diameter is 109 times the diameter of earth The formula for volume of a sphere is V=(4/3) πr3 Th
7000000000000
# How many humans can fit inside the sun?
Many more than all the humans on earth. If the average human has a body volume of 3 cubic feet (3 cubic foot = 0.084950539776 cubic meters). And the Sun's volume is 1.409 x
# If the sun was hollow how many earths would fit inside of it?
3 Answer: One million earths could fit inside inside the sun.
# How many Earth's fit inside one Sun?
1,300,000 Earths, meaning that the Sun's volume is 1,300,000 times that of the Earth.
# How many earths could fit inside the sun?
The Earth has a diameter of about 12700 kilometers (7900 miles). The sun has a diameter of about 1.39 million kilometers (865000 miles). Its diameter is about 109 times th
# How many Uranus' could fit in the sun?
Nearly 22 thousand planets the size of Uranus can fit inside the sun, provided they are squished together instead of stacked like marbles. The volume of Uranus is 6.83x1013 km
# How many earths could fit in the sun?
The sun has a diameter of 865000 miles or 1.39 million kilometers. Its diameter is 109 times the diameter of earth The formula for volume of a sphere is V=(4/3) πr3 The vo
# How many suns could fit inside Antares?
Antares is so big that billion Suns can fit inside it. It is bigger than all stars like, Sirlus A, our Sun, Pollux, Arcturus, Rigel, Aldebaran, Betelguese. It is the biggest s
# How many suns can fit inside vv cephei?
The answer is 1900 (nineteen hundred)
# How Many Jupiters Could Fit Into The Sun?
Going by volume, the suns volume is 984 times that of Jupiter.
# How many mercurys can fit inside the sun?
The volume of the Sun is about 23 million times the volume of Mercury. However, Mercury would only fit into the Sun 17 million times, because 26% of the volume would be lo
# How many moons can fit inside a sun?
Both moons and suns (stars) vary so widely in size that it's impossible to answer this.Very roughly a million Earths fit into Sol (our sun). Luna (Earth's moon) is 1/6 the | 859 | 3,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-30 | longest | en | 0.952592 |
https://community.tableau.com/message/530197 | 1,591,192,996,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347434137.87/warc/CC-MAIN-20200603112831-20200603142831-00132.warc.gz | 293,447,274 | 30,781 | 14 Replies Latest reply on Sep 14, 2016 5:49 PM by Andrew Watson
# Row wise calculations
Hi, I want to do row wise calculations on Tableau, and I am not able to figure out how to do so.
I want a way to calculate column 3 (cancels/bookings) and column 4 (1 - (cancels/bookings)).
can someone help me out on this please?
Data:
Thanks!!
• ###### 1. Re: Row wise calculations
Col3: SUM([Cancels])/SUM([Bookings])
Col4: 1 - [Col3]
• ###### 2. Re: Row wise calculations
Create a calculation as below
"cancel/bookings"
sum(Cancels)/sum(Booking)
one more calculation as below
1-cancel/bookings
• ###### 3. Re: Row wise calculations
Hi Andrew,
Cancels OR column 1: SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END)
Bookings OR column 2 is: PREVIOUS_VALUE(max([CF_PolCnt_PrevVal])) - Cancels
And, CF_PolCnt_PrevVal = {[CF_CountD_PolicyCounts]}
So i cannot use sum() since it does not take aggregates as it's expression.
Do you have other suggestions?
• ###### 4. Re: Row wise calculations
What happens if you just do [Cancels]/[Bookings]?
Or: SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END)/(PREVIOUS_VALUE(max([CF_PolCnt_PrevVal])) - SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END))
• ###### 5. Re: Row wise calculations
Doesn't work.
Result:
• ###### 6. Re: Row wise calculations
Do ignore the trial buckets column.
Thanks!
• ###### 7. Re: Row wise calculations
Can you check your brackets in the formula? The Denominator calculation needs to happen first so should be in brackets.
• ###### 8. Re: Row wise calculations
Hey Andrew,
I double checked the brackets.
They seem fine; in fact the solution which you have pasted has the correct brackets present.
• ###### 9. Re: Row wise calculations
Please check the PREVIOUS_VALUE calculation is calculating using the appropriate 'Compute Using', it should be the same in your Booking and Cancels/Bookings.
If you're still having no joy this may work: SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END)/(ZN(LOOKUP(max([CF_PolCnt_PrevVal]),-1)) - SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END))
This is removing the PREVIOUS_VALUE and replacing with LOOKUP. This may help explain why: PREVIOUS_VALUE vs LOOKUP([expr],-1)
• ###### 10. Re: Row wise calculations
Hi Andrew,
Thanks so very much for the help, I have been stuck here since a couple of days without any progress, so its much appreciated!
But unfortunately this did not work. Screenshot below (Col 5: CF_Div)
I basically want to achieve Col 4 / Col 3 (CF_PrevValCurve / CF_PolCnt_PrevVal), which will give me how the bookings are dropping over time.
I tried a couple of things and had some questions around it, and was hoping you could help:
1) When I try the below formula, i get an error which says i cannot mix aggregate and non-aggregate functions, is there a work around for this?
PREVIOUS_VALUE(max({[CF_CountD_PolCnt]})) - (SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END))
/
{[CF_CountD_PolCnt]}
2) And, when i tried to do:
[CF_PrevValCurve]
/
MAX([CF_PolCnt_PrevVal])
I get the very first value correct, but the rest of the values are incorrect.
I have figured out the manner in which Tableau calculates for this in the background:
For the second row, it does, 5300 + 1142 / 6992; Instead it should just be 5300 / 6992.
For the third row, it does, 4801 + 550 + 1442 / 6992; Instead it should just be 4801 / 6992.
Screenshot below:
And so on, again i am at a loss of a way to get around this.
• ###### 11. Re: Row wise calculations
P.S apologies for pasting images, my actual typed in answer goes into a 'currently being moderated' status.
- Swati
• ###### 12. Re: Row wise calculations
1) When I try the below formula, i get an error which says i cannot mix aggregate and non-aggregate functions, is there a work around for this?
PREVIOUS_VALUE(max({[CF_CountD_PolCnt]})) - (SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END))
/
{[CF_CountD_PolCnt]}
Try
PREVIOUS_VALUE(max({[CF_CountD_PolCnt]})) - (SUM(IF [CF_Status] = 'Cancelled' THEN 1 ELSE 0 END))
/
MAX({[CF_CountD_PolCnt]})
2) And, when i tried to do:
[CF_PrevValCurve]
/
MAX([CF_PolCnt_PrevVal])
Sounds like your CF_PrevValCurve is doing a running_sum. Personally I would change it to LOOKUP.
You'll need to attach a twbx file as cases like this as very hard to troubleshoot without seeing exactly what is happening.
• ###### 13. Re: Row wise calculations
Hi Andrew,
I did try MAX({[CF_CountD_PolCnt]}) prior to posting the question here; doesn't work.
I have attached the packaged workbook here, sheet 32 is where I have pulled in the columns to check for the values.
Do let know if you need any additional information.
Thanks a bunch!
• ###### 14. Re: Row wise calculations
I've now gone through this to try and understand what is happening. I've redone some of your calculations, for example you were sometimes doing a distinct count on policy number and other times counting records (by assigning a 1 to relevant records and then summing) - which have very slightly different results. For consistency I have used the distinct count on policy number all of the way through, feel free to change to record counts if you prefer.
I have abandoned the PREVIOUS_VALUE and replaced with formulae with which I'm more comfortable, such as a RUNNING_SUM.
Hopefully you can work out what is happening, I've laid it out in a table so shouldn't be too complex to work out.
If I have your % calculations wrong you should be able to use the fields created to correct the calc. | 1,485 | 5,544 | {"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} | 2.9375 | 3 | CC-MAIN-2020-24 | latest | en | 0.790262 |
https://espanol.libretexts.org/Vocacional/Correcciones_y_Justicia_Penal/ADMJUS_110%3A_Principios_y_Procedimientos_del_Sistema_de_Justicia/11%3A_B%C3%BAsquedas_de_Libertad_Condicional_y_Libertad_Condicional | 1,723,471,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00804.warc.gz | 199,102,572 | 30,169 | Saltar al contenido principal
11: Búsquedas de Libertad Condicional y Libertad Condicional
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
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https://www.traditionaloven.com/building/beach-sand/convert-beach-sand-cup-jp-to-china-shao-beach-sand.html | 1,542,184,913,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741764.35/warc/CC-MAIN-20181114082713-20181114104713-00500.warc.gz | 809,027,640 | 11,956 | Beach Sand 1 cup Japanese volume to Chinese sháo converter
# beach sand conversion
## Amount: 1 cup Japanese (cup) of volume Equals: 20.00 Chinese sháo (勺) in volume
Converting cup Japanese to Chinese sháo value in the beach sand units scale.
TOGGLE : from Chinese sháo into cups Japanese in the other way around.
## beach sand from cup Japanese to Chinese sháo Conversion Results:
### Enter a New cup Japanese Amount of beach sand to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other beach sand measuring units - complete list.
Conversion calculator for webmasters.
## Beach sand weight vs. volume units
Beach sand has quite high density, it's heavy and it easily leaks into even tiny gaps or other opened spaces. No wonder it absorbs and conducts heat energy from the sun so well. However, this sand does not have the heat conductivity as high as glass does, or fireclay and firebricks, or dense concrete. A fine beach sand in dry form was used for taking these measurements.
Convert beach sand measuring units between cup Japanese (cup) and Chinese sháo (勺) but in the other reverse direction from Chinese sháo into cups Japanese.
conversion result for beach sand: From Symbol Equals Result To Symbol 1 cup Japanese cup = 20.00 Chinese sháo 勺
# Converter type: beach sand measurements
This online beach sand from cup into 勺 converter is a handy tool not just for certified or experienced professionals.
First unit: cup Japanese (cup) is used for measuring volume.
Second: Chinese sháo (勺) is unit of volume.
## beach sand per 20.00 勺 is equivalent to 1 what?
The Chinese sháo amount 20.00 勺 converts into 1 cup, one cup Japanese. It is the EQUAL beach sand volume value of 1 cup Japanese but in the Chinese sháo volume unit alternative.
How to convert 2 cups Japanese (cup) of beach sand into Chinese sháo (勺)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
20 * 2 (or divide it by / 0.5)
QUESTION:
1 cup of beach sand = ? 勺
1 cup = 20.00 勺 of beach sand
## Other applications for beach sand units calculator ...
With the above mentioned two-units calculating service it provides, this beach sand converter proved to be useful also as an online tool for:
1. practicing cups Japanese and Chinese sháo of beach sand ( cup vs. 勺 ) measuring values exchange.
2. beach sand amounts conversion factors - between numerous unit pairs variations.
3. working with mass density - how heavy is a volume of beach sand - values and properties.
International unit symbols for these two beach sand measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for cup Japanese is:
cup
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Chinese sháo is:
### One cup Japanese of beach sand converted to Chinese sháo equals to 20.00 勺
How many Chinese sháo of beach sand are in 1 cup Japanese? The answer is: The change of 1 cup ( cup Japanese ) volume unit of beach sand measure equals = to volume 20.00 勺 ( Chinese sháo ) as the equivalent measure within the same beach sand substance type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in cup - cups Japanese for beach sand amount, the rule is that the cup Japanese number gets converted into 勺 - Chinese sháo or any other beach sand unit absolutely exactly.
Conversion for how many Chinese sháo ( 勺 ) of beach sand are contained in a cup Japanese ( 1 cup ). Or, how much in Chinese sháo of beach sand is in 1 cup Japanese? To link to this beach sand cup Japanese to Chinese sháo online converter simply cut and paste the following.
The link to this tool will appear as: beach sand from cup Japanese (cup) to Chinese sháo (勺) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 981 | 4,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-47 | longest | en | 0.885808 |
http://jsmith.cis.byuh.edu/books/managerial-economics-principles/s02-03-revenue-cost-and-profit-functi.html | 1,556,063,485,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578616424.69/warc/CC-MAIN-20190423234808-20190424015836-00058.warc.gz | 95,485,152 | 4,703 | This is “Revenue, Cost, and Profit Functions”, section 2.3 from the book Managerial Economics Principles (v. 1.0). For details on it (including licensing), click here.
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2.3 Revenue, Cost, and Profit Functions
In the preceding projections for the proposed ice cream bar venture, the assumption was that 36,000 ice cream bars would be sold based on the volume in the prior summer. However, the actual volume for a future venture might be higher or lower. And with an economic profit so close to zero, our students should consider the impact of any such differences.
There is a relationship between the volume or quantity created and sold and the resulting impact on revenue, cost, and profit. These relationships are called the revenue function, cost function, and profit function. These relationships can be expressed in terms of tables, graphs, or algebraic equations.
In a case where a business sells one kind of product or service, revenue is the product of the price per unit times the number of units sold. If we assume ice cream bars will be sold for \$1.50 apiece, the equation for the revenue functionThe product of the price per unit times the number of units sold; R = P*Q. will be
R = \$1.5 Q,
where R is the revenue and Q is the number of units sold.
The cost functionThe sum of fixed cost and the product of the variable cost per unit times quantity of units produced, also called total cost; C = F + V*Q. for the ice cream bar venture has two components: the fixed cost component of \$40,000 that remains the same regardless of the volume of units and the variable cost component of \$0.30 times the number of items. The equation for the cost function is
C = \$40,000 + \$0.3 Q,
where C is the total cost. Note we are measuring economic cost, not accounting cost.
Since profit is the difference between revenue and cost, the profit functionsThe revenue function minus the cost function; in symbols π = R - C = (P*Q) - (F + V*Q). will be
π = R − C = \$1.2 Q − \$40,000.
Here π is used as the symbol for profit. (The letter P is reserved for use later as a symbol for price.)
Table 2.1 "Revenue, Cost, and Profit for Selected Sales Volumes for Ice Cream Bar Venture" provides actual values for revenue, cost, and profit for selected values of the volume quantity Q. Figure 2.1 "Graphs of Revenue, Cost, and Profit Functions for Ice Cream Bar Business at Price of \$1.50", provides graphs of the revenue, cost, and profit functions.
The average costThe total cost divided by the quantity produced; AC = C/Q. is another interesting measure to track. This is calculated by dividing the total cost by the quantity. The relationship between average cost and quantity is the average cost function. For the ice cream bar venture, the equation for this function would be
AC = C/Q = (\$40,000 + \$0.3 Q)/Q = \$0.3 + \$40,000/Q.
Figure 2.2 "Graph of Average Cost Function for Ice Cream Bar Venture" shows a graph of the average cost function. Note that the average cost function starts out very high but drops quickly and levels off.
Table 2.1 Revenue, Cost, and Profit for Selected Sales Volumes for Ice Cream Bar Venture
Units Revenue Cost Profit
0 \$0 \$40,000 –\$40,000
10,000 \$15,000 \$43,000 –\$28,000
20,000 \$30,000 \$46,000 –\$16,000
30,000 \$45,000 \$49,000 –\$4,000
40,000 \$60,000 \$52,000 \$8,000
50,000 \$75,000 \$55,000 \$20,000
60,000 \$90,000 \$58,000 \$32,000
Figure 2.1 Graphs of Revenue, Cost, and Profit Functions for Ice Cream Bar Business at Price of \$1.50
Essentially the average cost function is the variable cost per unit of \$0.30 plus a portion of the fixed cost allocated across all units. For low volumes, there are few units to spread the fixed cost, so the average cost is very high. However, as the volume gets large, the fixed cost impact on average cost becomes small and is dominated by the variable cost component.
Figure 2.2 Graph of Average Cost Function for Ice Cream Bar Venture | 1,020 | 4,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-18 | latest | en | 0.924875 |
https://www.coursehero.com/file/6101398/Lecture-Notes-Chapter-2/ | 1,516,179,206,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886860.29/warc/CC-MAIN-20180117082758-20180117102758-00740.warc.gz | 885,202,415 | 29,424 | Lecture Notes Chapter 2
# Lecture Notes Chapter 2 - Atargetpopulation()from...
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The notion of probability requires three elements: A target population (either conceptual or real) from which observable outcomes are obtained Meaningful categorization of these outcomes A random mechanism for generating outcomes
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Example 1: Imagine tossing a single fair die on a flat surface and the number on the top face is recorded. This is an example of a probabilistic situation, because the outcome of the toss can not be determined before the toss. A situation like the one in the previous example is called a random experiment . We will be analyzing this random experiment frequently throughout the course.
sample space S is a set that includes all possible outcomes for a random experiment, listed in a mutually exclusive and exhaustive manner. Mutually exclusive means that the elements of the sample space do not overlap, and exhaustive means that all possible outcomes are accounted for. From the die toss in the previous example, two possible samples spaces would be: S 1 = {1, 2, 3, 4, 5, 6} S 2 = {even, odd}
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An event is any subset of a sample space. Going back to the die tossing experiment, give a set notation definition of each of the following events, letting S = {1, 2, 3, 4, 5, 6}:
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Lecture Notes Chapter 2 - Atargetpopulation()from...
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Ask a homework question - tutors are online | 622 | 1,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-05 | latest | en | 0.107779 |
https://math.stackexchange.com/questions/2643507/stationary-probability-distribution-markov-chain-obtained-by-iterations | 1,561,046,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999261.43/warc/CC-MAIN-20190620145650-20190620171650-00065.warc.gz | 527,451,076 | 35,949 | # Stationary probability distribution Markov chain obtained by iterations.
I have a question regarding Markov chains obtained by random indep. iterations.
We should consider a Markov chain $\{X_{n}\}^{\infty}_{n=0}$ obtained by random indep. iterations with the functions $f_{1}(x)=x^{2}+17$, $f_{2}(x)=x^{2}-17$ and $f_{3}(x)=0$ where the functions are chosen with equal prob. in each iteration step.
If we suppose $X_{0}=0$. How would I show that $\{X_{n}\}^{\infty}_{n=0}$ has a unique stationary prob. distribution?
The confusing part for me is the third function. If that function would be left out I would get $X_{n+1}=X_{n}^{2}+\xi_{n+1}$, where $\xi_{n+1}=\pm17$ with prob. $1/2$. But how do I see this process with the third function in mind?
• The Markov dynamics would rather correspond to the recursion $$X_{n+1}=A_{n+1}X_n^2+B_{n+1}$$ with $(A_n,B_n)$ i.i.d. uniformly distributed on $$\{(1,17),(1,-17),(0,0)\}$$ – Did Feb 9 '18 at 17:11
• I see that making sense, thank you. Now, how should I proceed to find the stationary probability distribution? I have tried to search for examples, as my course book doesn't provide any (or theory on that matter either). – Femman Feb 9 '18 at 19:01
• The usual approach. Some care is needed to get the correct state space, though. – Did Feb 9 '18 at 19:55
The states of the Markov chain are $0$ and every iterated image of $0$ by $f_1$ and $f_2$. With the exception of $f_1(f_1(0))=f_1(f_2(0))=17^2+17$ and of $f_2(f_1(0))=f_2(f_2(0))=17^2-17$, it seems that all these images are different.
Thus, the state space of the Markov chain has roughly the structure of a binary tree rooted at $0$, with the exception that the vertices of the second generation are identified two by two. For every $n\geqslant2$, the $2^{n-1}$ vertices of the $n$th generation represent the $n$th iterates of $f_1$ and $f_2$ applied to $0$.
To describe this "almost tree" more precisely, let us label its vertices as $0$, $1'$, $2'$, $1$, $2$, $11$, $21$, $12$, $22$, and so on, thus $0$ points to $1'$ and $2'$, $1'$ and $2'$ point to $1$ and $2$, and, for every nonempty $\{1,2\}$-word $x$, the vertex $x$ points to $1x$ and $2x$.
By symmetry, the stationary distribution of the Markov chain puts the same weight, say $w_n$, on every vertex of the $n$th generation. Recall that the sizes of the successive generation, starting at and including the root, are $1$, $2$, $2$, $4$, $8$, $16$, ..., thus, the usual one-step stationarity yields $$w_0=\frac13\left(w_0+2w_1+\sum_{n=2}^\infty2^{n-1}w_n\right)=\frac13$$ Likewise, $$w_1=\frac13w_0=\frac19\qquad w_2=\frac13(2w_1)=\frac2{27}$$ and, for every $n\geqslant3$, $$w_n=\frac13w_{n-1}=\frac2{3^{n+1}}$$ As a sanity check, note that the total mass of the stationary distribution is indeed $$w_0+2w_1+\sum_{n=2}^\infty2^{n-1}w_n=\frac13+2\frac19+\sum_{n=2}^\infty2^{n-1}\frac2{3^{n+1}}=1$$ | 980 | 2,873 | {"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} | 3.765625 | 4 | CC-MAIN-2019-26 | latest | en | 0.899521 |
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1719 posts
Uber Geek
#236001 13-May-2018 14:17
n(n+1) + 2(n+1) over 2 = (n+1) (n+2) over 2
Can someone explain the steps taken in between?
Thanks!
## palmypete
29 posts
Geek
#2015134 13-May-2018 14:31
n(n+1) + 2(n+1) over 2 = (n+1) (n+2) over 2
Assuming all of it is over 2 not just the "2(n+1) over 2"
(n+1) is common factor. so n(n+1) + 2(n+1) becomes
(n+1) (n+2)
Then just chuck it all over 2 to get
(n+1) (n+2) over 2
## yitz
1442 posts
Uber Geek
#2015137 13-May-2018 14:41
As above reply, what you have is
xy + xz = x(y+z)
if it makes looking at it any simpler.
(with a factor of half out the front)
1719 posts
Uber Geek
#2015141 13-May-2018 14:45
Ahh, thank you both, that makes sense now.
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# Magnetic Forces and Field Line Density
This is an activity about depicting the relative strength of magnetic fields using field line density. Learners will use the magnetic field line drawing of six magnetic poles created in a previous activity and identify the areas of strong, weak, and... (View More)
Audience: High school
# How to Draw Magnetic Fields - I
This is an activity about depicting magnetic fields. Learners will observe two provided drawings of magnetic field line patterns for bar magnets in simple orientations of like and unlike polarities and carefully draw the field lines for both... (View More)
Audience: High school
# What are Magnetic Fields?
This is an activity about magnetic fields. Using iron filings, learners will observe magnets in various arrangements to investigate the magnetic field lines of force. This information is then related to magnetic loops on the Sun's surface and the... (View More)
Audience: High school
# Modeling Sunspot Motion
This is an activity about solar rotation and sunspot motion. Learners will use a sphere or ball to model the Sun and compare the observed lateral motion of sunspots to their line-of-sight motion. This is Activity 1 of the Space Weather Forecast... (View More)
# What is the Magnetosphere?
This is an activity about Earth's magnetosphere. Learners will use a magnet, simulating Earth's protective magnetosphere, and observe what occurs when iron filings, simulating the solar wind, blow past and encounter the magnet's field. This is the... (View More)
Audience: High school
# How to Draw Magnetic Fields - II
This is an activity about depicting magnetic polarity. Learners will observe several provided drawings of magnetic field line patterns for bar magnets in simple orientations of like and unlike polarities and carefully draw the field lines and depict... (View More)
Audience: High school
# A Review of Time Zone Mathematics
This is a resource that explains the rationale behind the multiple time zone divisions in the United States. Learners will work through a problem set to practice calculating the time in one time zone, given the time in another time zone. This is... (View More)
Audience: High school
Materials Cost: Free
# Geomagnetism I: Polar Wander
This is an activity about the movement, or "wandering," of our Earth's magnetic poles. The learner will explore this concept by measuring and calculating the distance the Earth's north magnetic pole has moved over the past 400 years and calculating... (View More)
Audience: Middle school, High school
Materials Cost: Free
1 | 657 | 3,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.875411 |
https://link.springer.com/chapter/10.1007/978-1-4613-9438-9_12 | 1,519,307,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814105.6/warc/CC-MAIN-20180222120939-20180222140939-00532.warc.gz | 704,815,154 | 14,581 | Join Geometries pp 456-492
# Ordered Join Geometries
• Walter Prenowitz
• James Jantosciak
Part of the Undergraduate Texts in Mathematics book series (UTM)
## Abstract
In this chapter, as in the previous one, a new postulate is introduced. The postulate is equivalent to the familiar Euclidean property of linear order: If three distinct points are collinear, then one of the three is between the other two. Join geometries which satisfy the new postulate are called ordered join geometries or ordered geometries. Ordered geometries are exchange geometries studied in the last chapter, but the results there will not be used in the present investigation. A flood of results is produced. First come formulas for lines, rays and segments, expressing how they are divided into subrays and subsegments by their points. Next come many properties of polytopes familiar in Euclidean geometry and easily accessible to intuition. Then follow properties of convex sets, less familiar in classical geometry but no less important, the theorems of Radon, Helly and Caratheodory and related results. These flow from a sharpened expansion formula for the linear hull of a finite set mediated by a new type of dependence of points. This new type of dependence is defined in terms of the convex hull operation and is called convex dependence. It implies linear dependence in any join geometry and is equivalent to linear dependence in an ordered geometry. Finally, separation of linear spaces by linear subspaces is studied, and results familiar in Euclidean geometry are obtained for ordered geometries. | 326 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | longest | en | 0.929358 |
https://www.circuitstoday.com/interfacing-8x8-led-matrix-with-arduino | 1,721,413,003,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00535.warc.gz | 610,976,684 | 20,098 | LED matrix displays can be used to display almost anything. Most modern LED sign boards uses various types of matrix boards with controllers. In this tutorial we are going to interface a single color 8×8 LED matrix with Arduino and display a few characters in it.
8×8 matrix consists of 64 dots or pixels. There is a LED for each pixel and these LEDs are connected to total of 16 pins. You can identify the pin out and circuit diagram of it using the following figure.
C1 – C8 – Column pins
R1 – R8 – Row pins
As you can see all anodes of same row is connected to one pin and all cathodes of same column are connected to another pin.We have 8 row pins and 8 column pins. If a positive voltage is applied to R1 pin and negative to C1, we can see that the first pixel turns on. If we apply negative to C2 then the second pixel turns on. Like this we can turn each pixel by hanging the supply pins. However we have 64 supply combinations, and doing it manually is practically impossible. This is why Arduino is interfaced with the 8×8 matrix.
### Circuit Diagram
As you can see in the diagram every column pin is connected to Arduino pin through a 220Ω resistor. All the row pins are connected to one of the output pin of the shift register. The characters are displayed using the multiplexing techniques. Shift register is connected to Arduino as usual by data, latch and clock pins.
### Arduino Code
int latchPin = 4; // pis connected to shift registors
int clockPin = 5;
int dataPin = 3;
int pins [8] = {6, 7, 8, 9, 10, 11, 12, 13}; // common cathode pins
byte A[8] = { B00000000, // Letters are defined
B00011000,// you can create your own
B00100100,
B01000010,
B01111110,
B01000010,
B01000010,
B00000000
};
byte B[8] = { B00000000,
B11111100,
B10000010,
B10000010,
B11111100,
B10000010,
B10000010,
B11111110
};
byte blank[8] = { B00000000,
B00000000,
B00000000,
B00000000,
B00000000,
B00000000,
B00000000,
B00000000
};
byte R[8] = { B00000000,
B01111000,
B01000100,
B01000100,
B01111000,
B01010000,
B01001000,
B01000100
};
void setup() {
Serial.begin(9600); // Serial begin
pinMode(latchPin, OUTPUT); // Pin configuration
pinMode(clockPin, OUTPUT);
pinMode(dataPin, OUTPUT);
for (int i = 0; i < 8; i++) { // for loop is used to configure common cathodes
pinMode(pins[i], OUTPUT);
digitalWrite(pins[i], HIGH);
}
}
void loop() {
for (int k = 0; k < 1000; k++) { // showing each letter for 1 second
display_char(A);
}
for (int k = 0; k < 1000; k++) {
display_char(B);
}
for (int k = 0; k < 1000; k++) {
display_char(R);
}
// add more letters show method here
}
void display_char(byte ch[8]) { // Method do the multiplexing
for (int j = 0; j < 8; j++) {
digitalWrite(latchPin, LOW);
digitalWrite(pins[j], LOW);
shiftOut(dataPin, clockPin, LSBFIRST, ch[j]);
digitalWrite(latchPin, HIGH);
//delay(1);
digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, LSBFIRST, B00000000); // to get rid of flicker when
digitalWrite(latchPin, HIGH);
digitalWrite(pins[j], HIGH);
}
}
### Code Explanation
In the code all important lines are commented. The digitalWrite command is used for controlling the column pins and shiftOut command is used to write to shift register. Letters are defined in a byte array when the characters displaying 0 are OFF and 1 is ON. This way you can define your own letters or symbols. In the multiplexing method shiftOut(dataPin, clockPin, LSBFIRST, B00000000); line is used to reduce the flicker by turning off all the pixels after each step.
You can see the actual bread board setup in the following figure.
1. Hello,
Can you please explain why R1, for example, is connected to pin 9? Why doesn’t it connect in order? R1 to pin 1, R2, to pin 2, R3 to pin 3 and so on. It is important to me to understand these connections between pins and rows/columns because I don’t want to accumulate gaps in my learning. If you say it’s up to the manufacturer, then please direct me to one.
Have a nice day!
2. Alejandro
What program changes are required if the 74HC595 needs to have active LOW outputs and the ATmega328p needs to be active HIGH outputs?
3. My matrix is common anode, what do I need to do differently compared to your common cathode version?
4. Naved Alam Farooqui
Can we program this to display a flickering flame ,like “Moma Flame lamp”?
This lamp I saw in the Museum of modern arts,New York which also used the Arduino.
If yes what will be the Arduino code?
• @Naved – No! You should follow a different approach. | 1,246 | 4,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.802996 |
https://www.studentlance.com/solution/select-the-best-answer-to-each-question | 1,600,468,490,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00178.warc.gz | 1,141,944,282 | 6,757 | # Select the best answer to each question. - 89767
Solution Posted by
## Ryan
Rating : (38)A+
Solution Detail
Price: \$10.00
• From: Mathematics, Calculus
• Posted on: Mon 13 Apr, 2015
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Request Description
Exponents, Logarithms, Sequences, and Series When you have completed your exam and reviewed your answers, click Submit Exam. Answers will not be recorded until you hit Submit Exam. If you need to exit before completing the exam, click Cancel Exam. Questions 1 to 20: Select the best answer to each question. Note that a question and its answers may be split across a page break, so be sure that you have seen the entire question and all the answers before choosing an answer. 1. Use properties of logarithms to expand the logarithmic expression below as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. A. ln e 3 + ln 11 B. 3 + ln 11 C. ln e 3 ? ln 11 D. 3 ? ln 11 2. Compute the value of A. 31 B. 36 C. 37 D. 30 3. Use summation notation to express the sum below. Use 1 as the lower limit of summation and i for the index of summation. A. B. C. D. 4. Write the first five terms of the arithmetic sequence below. a 1 = 15; d = ? 4 A. 15, 11, 6, 3, ? 1 B. 19, 15, 11, 7, 3 C. 11, 7, 3, ? 1, ? 5 D. 15, 11, 7, 3, ? 1 5. Find the sum, if it exists, of the infinite geometric series below: A. B. C. A sum doesn't exist for this geometric series. D. 6. Find the common ratio of the geometric sequence below: A. B. C. D. 7. Evaluate the expression below without using a calculator: log10 10,000 A. –4 B. 4 C. 40 D. 8. Which of these is not a valid fact about logarithms? A. b logb 1 = 1 B. y = logb x means b y = x C. logb b = 0 D. logb b x = x 9. Write the first four terms of the sequence defined by the following recursion formula: a 1 = ?3 and a n = ?2a n?1 for n ? 2 A. ? 3, ? 6, ? 12, ? 24 B. 3, ? 6, 12, ? 24 C. ? 3, 8, ? 14, 26 D. ? 3, 6, ? 12, 24 10. Use properties of logarithms to condense the logarithmic expression below to one term whose coefficient is 1. 2log y 4 + log y 2 A. log y 16 B. logy 32 C. log y 4 D. 2log y 8 11. Which of these is not a valid fact about logarithms? A. log2 7k = k log2 7 B. log2 (6 + k) = (log2 6) (log2 k) C. log2 1 = 0 D. log2 10/3 = log2 10 ? log2 3 12. Use properties of logarithms to expand the logarithmic expression below as much as possible. A. ln e 5 ? ln 9 B. 5 + ln 9 C. 5 ? ln 9 D. ln e 5 + ln 9 13. In a proof of by mathematical induction, an appropriate first step is to prove which of these statements? A. B. C. D. 14. Find the common ratio of the geometric sequence given below. A. B. C. D. 15. Make a table of coordinates for the function below and then graph it. f (x) = 5x A. B. C. D. 16. Which of the following statements is true about the number e? A. It’s the base of the natural logarithm. B. It’s equal to e e . C. It’s the base of every exponential function. D. It’s equal to 2p. 17. Evaluate the expression below to four decimal places. Use common logarithms or natural logarithms and a calculator. log4 28 A. 0.8451 B. 0.4160 C. 2.4037 D. 2.0492 18. Write the equation below in its equivalent logarithmic form. A. B. C. End of exam D. 19. Solve the logarithmic equation below. Be sure to reject any value that isn't in the domain of the original logarithmic expressions. Give the exact answer. log2 (x ? 2) = 3 A. {6} B. {7} C. {10} D. {11} 20. Find the accumulated value of an investment of \$7,000 at 7% compounded continuously for 6 years. To solve question 3, use the correct formula for compound interest: A. \$10,653.73 B. \$9,940.00 C. \$10,505.11 D. \$10,753.73
Solution Description
Attachments | 1,214 | 3,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.854733 |
http://mathhelpforum.com/calculus/101111-vectors.html | 1,524,620,462,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947654.26/warc/CC-MAIN-20180425001823-20180425021823-00231.warc.gz | 198,673,699 | 10,292 | Thread: Vectors
1. Vectors
(a) Find parametric equations for the line of intersection between the planes x + y = 0 and 3y + 8z = 6.
(b) Find the point of intersection of the line found in part (a) and the plane 2x + 5y + 4z = 5.
2. a: $\displaystyle n_{1}=(1,1,0), \;\ n_{2}=(3,0,8)$
These are the normals to the planes.
Find the cross product: $\displaystyle n_{1}\times n_{2}=(8,-8,-3)$
Now, let z=0 in the two equations and solve for x and y.
Finish?.
3. Another way: the two planes are x + y = 0 and 3y + 8z = 6. Solve for, say, x and z in terms of y: x= -y and 8z= 6- 3y so z= (3/4)- (3/8)y.
Now take t= y/8 as parameter. x= -8t, y= 8t, z= (3/4)- 3t.
To find the point where that line intersects the plane 2x + 5y + 4z = 5, replace x, y, and z in that equation by their parametric expressions: 2(-8t)+ 5(8t)+ 4((3/4)- 3t)= -16t+ 40t- 12t+ 3= 12t+ 3= 5. 12t= 2, t= 1/6. Now put that value of t back into the parametric equations: x= -8(1/6)= -4/3, y= 8(1/6)= 4/3, z= (3/4)- 3(1/6)= 3/4- 2/4= 1/4. The point of intersection is (-4/3, 4/3, 1/4).
Note that those values of x, y, and z satisfy the equations of all three planes. -4/3+ 4/3= 0, 3(4/3)+ 8(1/4)= 4- 2= 6, and 2(-4/3)+ 5(4/3)+ 4(1/4)= -8/3+ 20/3+ 1= 12/3+ 1= 4+ 1= 5.
4. I'm confused. | 570 | 1,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-17 | latest | en | 0.838525 |
https://kerjadigi.com/hcf-of-8-9-25-is/ | 1,675,924,768,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00802.warc.gz | 368,580,610 | 18,193 | Breaking News
# Finding the Highest Common Factor of 8, 9 and 25
## What is the Highest Common Factor?
The Highest Common Factor (HCF) is the highest number that divides two or more given numbers. In other words, it is the greatest common divisor of two or more numbers. It is also known as the Greatest Common Divisor (GCD).
## How to Find the Highest Common Factor?
The easiest way to find the HCF is to use the prime factorization method. This involves writing each number as the product of its prime factors. Prime factors are factors that can only be divided by 1 and the number itself. Once the prime factors of each number are listed, the highest common factor is the product of the common factors.
## Example: Finding the Highest Common Factor of 8, 9 and 25
To find the highest common factor of 8, 9 and 25, the first step is to list the prime factors of each number. The prime factors of 8 are 2 x 2 x 2, the prime factors of 9 are 3 x 3 and the prime factors of 25 are 5 x 5. As you can see, the only common factor between 8, 9 and 25 is a single 2. Therefore, the HCF of 8, 9 and 25 is 2.
## Conclusion
The Highest Common Factor (HCF) is the highest number that divides two or more given numbers. It is also known as the Greatest Common Divisor (GCD). The easiest way to find the HCF is to use the prime factorization method. The HCF of 8, 9 and 25 is 2. | 357 | 1,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-06 | latest | en | 0.956341 |
https://www.physicsforums.com/threads/refraction-through-an-optical-fiber.799563/ | 1,531,920,920,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590169.5/warc/CC-MAIN-20180718115544-20180718135544-00621.warc.gz | 963,285,900 | 15,879 | # Homework Help: Refraction through an optical fiber
Tags:
1. Feb 23, 2015
### SnowAnd38Below
1. The problem statement, all variables and given/known data
Given a "new type" of optical fiber (index of refraction n = 1.23), a laser beam is incident on the flat end of a straight fiber in air. Assume nair = 1.00. What is the maximum angle of incidence Ø1 if the beam is not to escape from the fiber? (See attached file for diagram).
2. Relevant equations
Snell's Law: n1*sin(Ø1) = n2*sin(Ø2)
3. The attempt at a solution
I know the answer is 45.7°, I just can't seem to generate it. For the beam not to escape from the fiber, the angle of incidence must be the critical angle, such that Ø2=90° (or sin(Ø2)=1). But I don't know if I'm supposed to try to find the critical angle, then work backwards to find the initial angle of incidence from the air into the fiber or not. There's no cladding to consider in this problem, so I'm just very confused.
#### Attached Files:
• ###### Screen Shot 2015-02-23 at 3.19.11 PM.png
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2. Feb 23, 2015
### sk1105
Assuming you've defined $θ_2$ as the angle between the normal and the beam inside the fibre, this doesn't have to be 90°, as the beam can be reflected off the inside edge of the fibre without escaping. Indeed, if it did have to be 90°, fibres wouldn't work round corners. Revise the definition of critical angle in optics, and it should be clearer how to get the right answer.
3. Feb 24, 2015
### SnowAnd38Below
I guess I didn't explain my attempt very well. I was trying to treat the problem with Snell's Law twice; once for the beam entering the fiber from the air, and a second time for refracting in the fiber such that the second angle of incidence is the critical angle, guaranteeing full reflection of the beam back into the fiber.
4. Feb 24, 2015
### ehild
Yes, it will do. Find the critical angle for total reflection for the interface fibre-air, then backwards the angle of reflection at the front surface and from that, the angle of incidence.
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Draft saved Draft deleted | 567 | 2,181 | {"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} | 3.578125 | 4 | CC-MAIN-2018-30 | latest | en | 0.91955 |
http://www.advsofteng.com/doc/cdjavadoc/finance.htm | 1,516,545,448,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890771.63/warc/CC-MAIN-20180121135825-20180121155825-00297.warc.gz | 390,308,802 | 3,679 | ChartDirector 6.1 (Java Edition)
Finance Chart (1)
This example demonstrate creating a full-featured finance chart, with HLOC, moving averages, Bollinger band, volume bars, RSI and Momentum indicators.
This example employs the FinanceChart library add-on to allow complex financial charts to be composed easily. In this example, the key steps are:
For simplicity and to allow this example to run without connecting to a real database, a RanTable object is used to simulate the data. RanTable is a ChartDirector utility class used for creating tables with random numbers.
Source Code Listing
[JSP Version] jspdemo/finance.jsp
```<%@page import="ChartDirector.*" %> <%@page import="java.util.*" %> <% // Create a finance chart demo containing 100 days of data int noOfDays = 100; // To compute moving averages starting from the first day, we need to get extra data points before // the first day int extraDays = 30; // In this exammple, we use a random number generator utility to simulate the data. We set up the // random table to create 6 cols x (noOfDays + extraDays) rows, using 9 as the seed. RanTable rantable = new RanTable(9, 6, noOfDays + extraDays); // Set the 1st col to be the timeStamp, starting from Sep 4, 2002, with each row representing one // day, and counting week days only (jump over Sat and Sun) rantable.setDateCol(0, new GregorianCalendar(2002, 8, 4).getTime(), 86400, true); // Set the 2nd, 3rd, 4th and 5th columns to be high, low, open and close data. The open value starts // from 100, and the daily change is random from -5 to 5. rantable.setHLOCCols(1, 100, -5, 5); // Set the 6th column as the vol data from 5 to 25 million rantable.setCol(5, 50000000, 250000000); // Now we read the data from the table into arrays double[] timeStamps = rantable.getCol(0); double[] highData = rantable.getCol(1); double[] lowData = rantable.getCol(2); double[] openData = rantable.getCol(3); double[] closeData = rantable.getCol(4); double[] volData = rantable.getCol(5); // Create a FinanceChart object of width 640 pixels FinanceChart c = new FinanceChart(640); // Add a title to the chart c.addTitle("Finance Chart Demonstration"); // Set the data into the finance chart object c.setData(timeStamps, highData, lowData, openData, closeData, volData, extraDays); // Add the main chart with 240 pixels in height c.addMainChart(240); // Add a 5 period simple moving average to the main chart, using brown color c.addSimpleMovingAvg(5, 0x663300); // Add a 20 period simple moving average to the main chart, using purple color c.addSimpleMovingAvg(20, 0x9900ff); // Add HLOC symbols to the main chart, using green/red for up/down days c.addHLOC(0x008000, 0xcc0000); // Add 20 days bollinger band to the main chart, using light blue (9999ff) as the border and // semi-transparent blue (c06666ff) as the fill color c.addBollingerBand(20, 2, 0x9999ff, 0xc06666ff); // Add a 75 pixels volume bars sub-chart to the bottom of the main chart, using green/red/grey for // up/down/flat days c.addVolBars(75, 0x99ff99, 0xff9999, 0x808080); // Append a 14-days RSI indicator chart (75 pixels high) after the main chart. The main RSI line is // purple (800080). Set threshold region to +/- 20 (that is, RSI = 50 +/- 25). The upper/lower // threshold regions will be filled with red (ff0000)/blue (0000ff). c.addRSI(75, 14, 0x800080, 20, 0xff0000, 0x0000ff); // Append a 12-days momentum indicator chart (75 pixels high) using blue (0000ff) color. c.addMomentum(75, 12, 0x0000ff); // Output the chart String chart1URL = c.makeSession(request, "chart1"); %>
Finance Chart (1)
View Source Code
```
```import java.awt.*; import java.awt.event.*; import javax.swing.*; import java.util.*; import ChartDirector.*; public class finance implements DemoModule { //Name of demo program public String toString() { return "Finance Chart (1)"; } //Number of charts produced in this demo public int getNoOfCharts() { return 1; } //Main code for creating charts public void createChart(ChartViewer viewer, int chartIndex) { // Create a finance chart demo containing 100 days of data int noOfDays = 100; // To compute moving averages starting from the first day, we need to get extra data points // before the first day int extraDays = 30; // In this exammple, we use a random number generator utility to simulate the data. We set // up the random table to create 6 cols x (noOfDays + extraDays) rows, using 9 as the seed. RanTable rantable = new RanTable(9, 6, noOfDays + extraDays); // Set the 1st col to be the timeStamp, starting from Sep 4, 2002, with each row // representing one day, and counting week days only (jump over Sat and Sun) rantable.setDateCol(0, new GregorianCalendar(2002, 8, 4).getTime(), 86400, true); // Set the 2nd, 3rd, 4th and 5th columns to be high, low, open and close data. The open // value starts from 100, and the daily change is random from -5 to 5. rantable.setHLOCCols(1, 100, -5, 5); // Set the 6th column as the vol data from 5 to 25 million rantable.setCol(5, 50000000, 250000000); // Now we read the data from the table into arrays double[] timeStamps = rantable.getCol(0); double[] highData = rantable.getCol(1); double[] lowData = rantable.getCol(2); double[] openData = rantable.getCol(3); double[] closeData = rantable.getCol(4); double[] volData = rantable.getCol(5); // Create a FinanceChart object of width 640 pixels FinanceChart c = new FinanceChart(640); // Add a title to the chart c.addTitle("Finance Chart Demonstration"); // Set the data into the finance chart object c.setData(timeStamps, highData, lowData, openData, closeData, volData, extraDays); // Add the main chart with 240 pixels in height c.addMainChart(240); // Add a 5 period simple moving average to the main chart, using brown color c.addSimpleMovingAvg(5, 0x663300); // Add a 20 period simple moving average to the main chart, using purple color c.addSimpleMovingAvg(20, 0x9900ff); // Add HLOC symbols to the main chart, using green/red for up/down days c.addHLOC(0x008000, 0xcc0000); // Add 20 days bollinger band to the main chart, using light blue (9999ff) as the border and // semi-transparent blue (c06666ff) as the fill color c.addBollingerBand(20, 2, 0x9999ff, 0xc06666ff); // Add a 75 pixels volume bars sub-chart to the bottom of the main chart, using // green/red/grey for up/down/flat days c.addVolBars(75, 0x99ff99, 0xff9999, 0x808080); // Append a 14-days RSI indicator chart (75 pixels high) after the main chart. The main RSI // line is purple (800080). Set threshold region to +/- 20 (that is, RSI = 50 +/- 25). The // upper/lower threshold regions will be filled with red (ff0000)/blue (0000ff). c.addRSI(75, 14, 0x800080, 20, 0xff0000, 0x0000ff); // Append a 12-days momentum indicator chart (75 pixels high) using blue (0000ff) color. c.addMomentum(75, 12, 0x0000ff); // Output the chart viewer.setChart(c); } //Allow this module to run as standalone program for easy testing public static void main(String[] args) { //Instantiate an instance of this demo module DemoModule demo = new finance(); //Create and set up the main window JFrame frame = new JFrame(demo.toString()); frame.addWindowListener(new WindowAdapter() { public void windowClosing(WindowEvent e) {System.exit(0);} }); frame.getContentPane().setBackground(Color.white); // Create the chart and put them in the content pane ChartViewer viewer = new ChartViewer(); demo.createChart(viewer, 0); frame.getContentPane().add(viewer); // Display the window frame.pack(); frame.setVisible(true); } }``` | 2,090 | 7,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.720925 |
https://www.jiskha.com/display.cgi?id=1320612937 | 1,503,380,347,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110471.85/warc/CC-MAIN-20170822050407-20170822070407-00333.warc.gz | 909,525,370 | 4,230 | # Physics
posted by .
There are two particles (1 and 2) that are moving around in space. The force that particle 2 exerts on 1 is given by:
F21(t)=Fx*e^−(t/T)i^ + Fy*sin(2πt/T) j^
Where the parameters have the values:
Fx = 13.4 N
Fy = -84.2 N
T = 74 s
tf = 140.46 s
Find the average force from 1 on 2 from t = 0 until t = tf
Find the average force from 2 on 1 from t = 0 until t = tf
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More Similar Questions | 869 | 2,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-34 | latest | en | 0.851269 |
https://visualfractions.com/unit-converter/convert-25-mpk-to-spm/ | 1,624,190,723,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00409.warc.gz | 531,166,333 | 9,609 | # Convert 25 mpk to spm
So you want to convert 25 minutes per kilometre into seconds per metre? If you're in a rush and just need the answer, the calculator below is all you need. The answer is 1.5 seconds per metre.
## How to convert minutes per kilometre to seconds per metre
We all use different units of measurement every day. Whether you're in a foreign country and need to convert the local imperial units to metric, or you're baking a cake and need to convert to a unit you are more familiar with.
Luckily, converting most units is very, very simple. In this case, all you need to know is that 1 mpk is equal to 0.06 spm.
Once you know what 1 mpk is in seconds per metre, you can simply multiply 0.06 by the total minutes per kilometre you want to calculate.
So for our example here we have 25 minutes per kilometre. So all we do is multiply 25 by 0.06:
25 x 0.06 = 1.5
## What is the best conversion unit for 25 mpk?
As an added little bonus conversion for you, we can also calculate the best unit of measurement for 25 mpk.
What is the "best" unit of measurement? To keep it simple, let's say that the best unit of measure is the one that is the lowest possible without going below 1. The reason for this is that the lowest number generally makes it easier to understand the measurement.
For 25 mpk the best unit of measurement is seconds per metre, and the amount is 1.5 spm. | 347 | 1,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-25 | longest | en | 0.920094 |
https://pro.arcgis.com/en/pro-app/latest/tool-reference/spatial-analyst/account-for-barriers-in-distance-calculations.htm | 1,716,581,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00507.warc.gz | 409,612,375 | 12,425 | Account for barriers in distance calculations
A barrier is an obstacle that is present between the two locations you are calculating the distance for. Barriers can be features such as a lake, a divided highway, or a mountain range. The straight-line distance is altered when a barrier exists. You may want to know the shortest possible distance between your locations after accounting for the additional distance necessary to detour around the barrier.
When a barrier is in the way, the traveler must move around it.
You can adjust straight-line distance by including barriers and incorporating the actual surface distance that is traveled. Once the adjusted straight-line distance is determined, the rate that distance is encountered can be controlled by a cost surface, source characteristics, a vertical factor, and a horizontal factor. If any of these rate factors are specified with a barrier, detouring around the barrier will be the least-cost distance between the locations.
Barrier use examples
Barriers can be used to help solve various scenarios, such as the following:
• In a wildlife study, determine the distance between two habitat patches with a lake between them.
• Determine the shortest distance for a boat to travel between two marinas when there is a peninsula between them.
• Identify a hiking trail route in an area where there is a steep drop off that you need to navigate around.
• Define the flight path an airplane should take when there is a storm between it and its destination.
Adjust straight-line distance analysis with barriers
Distance analysis can be divided conceptually into the following related functional areas:
From the first functional area, the straight-line distance is adjusted with barriers as illustrated below. The scenario involves a collection of four forest ranger stations (purple dots), and some rivers (blue lines).
The straight-line distance result identifies the distance from each nonsource cell to the closest ranger station.
Barriers adjust the straight-line distance since the traveler needs to move around them.
The straight-line source direction and back direction rasters are the same when there are no barriers.
The back-direction raster identifies, for each cell, the direction to move from the cell to return to the closest source. When barriers are introduced, the traveler must move around them.
Create a distance raster with barriers
To create a distance raster that incorporates barriers, complete the following steps:
1. Open the Distance Accumulation tool.
2. Provide the source in the Input raster or feature source data parameter.
3. Name the output distance raster.
4. Identify the barrier in the Input barrier raster or feature data parameter.
5. Specify any other necessary parameters.
6. Click Run.
Barriers affect distance calculation
The sections below provide information about adjusting the straight-line distance with barriers using the Distance Accumulation tool.
Specify a barrier
A barrier can alter distance calculations if you need to detour around it. Examples of barriers include a lake, divided highway, mountain range, river, or cliff. There is an input parameter you can use to specify where barrier locations exist, as either feature or raster data. If the input for this parameter is a feature class, it will be converted to a raster when the tool is run.
You can define barriers in other ways, as well. If a surface raster, cost surface, vertical factor raster, or horizontal factor raster are provided, NoData cells in any of the these rasters are considered to be barriers. If the Mask environment is set, the locations that are outside the area covered by the mask—the NoData cells—are also considered to be barriers. In all cases, barriers prohibit movement. When necessary, all barriers will be thickened slightly to prevent any cracks from permitting movement through that barrier. See the Thicken barriers to prevent cracks section below to learn more.
To summarize, travel is permitted across locations that have valid values for the input surface raster, cost surface raster, vertical factor raster, and horizontal factor raster, as well as across locations that are within the mask area defined in the analysis environment if it is specified. Travel is not permitted across locations defined by the barrier input, across locations that are outside the mask area (NoData cells), or across any location where the surface raster, cost surface raster, vertical factor raster, or horizontal factor raster has a value of NoData.
Barriers affect the output distance rasters
The following output rasters produced when performing distance analysis are affected by barriers:
• Distance accumulation raster
• Back-direction raster
• Source direction raster
Distance accumulation raster
For each nonsource cell, the output distance accumulation raster calculates the accumulative distance to the closest or least-cost source. If a barrier input has been specified, the distance values account for detouring around the barrier cells.
An example is locating a new building complex where proximity to existing power lines is preferable. In the following image, the distance from each nonsource cell to the closest power line (blue lines) is displayed. No barrier input was specified. The resulting distance raster is displayed, with green indicating closer locations.
Because of local restrictions, new power lines cannot go over ridgetops. The following image shows the impact of adding a ridgeline (purple line) as a barrier. Note the locations on the other side of the ridge line are now farther (in light brown) from the power lines because of the added distance necessary to move around the ridge line.
Note:
If a barrier causes any locations to be disconnected from a source, no distance is calculated to the disconnected cells.
Back-direction and source-direction rasters
Distance accumulation also allows you to create a back-direction raster and a source-direction raster. For each nonsource cell, the back-direction raster identifies the direction that you travel when leaving the nonsource cell to return to the closest or least-cost source.
For each nonsource cell, the source-direction raster identifies the direction to the closest or least-cost source cell. If there are no barriers when calculating straight-line distance, these two outputs are the same; with barriers they are not.
The back-direction and source-direction rasters use the same convention as a compass. The range of values is from 0 degrees to 360 degrees, with 0 reserved for the source cells. Due east, to the right, is 90, and the values increase clockwise, so that 180 is south, 270 is west, and 360 is north. The two rasters report actual direction, with the back-direction raster reporting the degrees as floating-point values and the source-direction raster reporting them as integers.
The difference between the back direction and source direction output is illustrated, for example, when calculating travel distance for a kayaker to reach a destination (Jim’s kayak rental) when a peninsula of land is between the kayaker and the destination. The peninsula becomes a barrier to the kayaker (traveler). In the image below, in the back-direction raster, each cell stores the direction the kayaker should move out of the cell to paddle back to Jim’s kayak rental since the kayaker must navigate around the land. This directional movement is indicated for representative cells by the dark blue arrows. In the source-direction raster, each cell stores the straight-line direction back to the Jim's kayak rental. This direction is indicated by the light blue arrows. The two directions are different. The dark blue arrows guide the kayaker around the peninsula, while the light blue arrows indicate the straight-line direction back to Jim’s kayak rental from any location regardless of the peninsula.
An important use of the back-direction raster is to generate shortest paths from a destination back to its closest source. In the image below, the output back-direction raster was an input to the Optimal Path As Line tool to define the shortest path the kayaker (purple dot) should take around the peninsula (the barrier), avoiding the boat moorings (yellow polygon), to return back to the kayak rental area.
Barriers can change the closest or least-cost source
When barriers are added, they can change which source cell is the closest or least costly to reach. For example, in the first image below, there are two sources, S1 and S2. The closest source to nonsource cell x is source S1.
In the next image a horizontal barrier is added. The closest source to x is now source S2 (as shown by the light gray path around the right end of the barrier).
As a result, the presence of barriers can change the output values for cells in the back-direction raster. In this example, the value assigned to x in the back-direction raster without the barrier will be 180.0. With the barrier, the back direction for x will be 121.5.
For the source-direction raster, the closest source to cell x without the barrier is S1, south, at a direction of 180 degrees. When the horizontal barrier is added, the source direction assigned to x is approximately 135 degrees (light blue arrow), which is the straight-line direction from x to source S2, not the in-route direction. As a result of the barrier, the back direction and source direction from x to S2 are different.
The value for x in an output distance allocation raster will change from S1 without a barrier to S2 with the barrier.
If a cost surface and barrier are provided, the following three scenarios are possible for the value assigned to nonsource cell x:
1. If S1 was the least-cost source without the barrier, with a barrier, S2 becomes the least-cost source with an accumulative cost greater than the value assigned to S1.
2. If S2 was the least-cost source without the barrier, source S2 was initially cheaper to reach than S1. With a barrier, S2 is still the least-cost source to reach but the total accumulative cost is increased.
3. S2 was the least-cost source without the barrier, but the path to reach S2 does not pass near the barrier. With the barrier, the accumulative cost for S2 remains the same. The barrier had no effect.
With a cost surface, the source-direction raster will change if the least-cost source for x changes from S1 to S2. However, if S2 was initially the least-cost source, the source direction for x will remain the same.
Thicken barriers to prevent cracks
A barrier can be represented as a linear feature, such as a road or river. In this case, the line feature must be rasterized before the distance operation is performed. As a raster, the linear feature will be only one cell thick. While the nature of the barrier would be preserved where it is perfectly horizontal or vertical, it may be diagonal. In this case, it is geometrically possible for the traveler to slip through what amounts to cracks in the barrier.
To prevent this, the Distance Accumulation and Distance Allocation tools automatically thicken the diagonal sections by one cell. Since the traveler is prevented from passing through the diagonals, the barrier input will remain a true barrier.
If the input barrier is a raster, this thickening process also occurs for the diagonal cells that are one cell wide.
Barriers that are specified as a dataset in the tool parameter are merged into the cost surface and, where necessary, NoData cells in the cost surface are thickened. As a result, the barriers are thickened regardless of whether they are specified by a dataset or their locations are assigned NoData in the cost surface, which avoids potential cracks. | 2,296 | 11,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.934722 |
http://brainly.in/question/126165 | 1,477,468,319,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720760.76/warc/CC-MAIN-20161020183840-00391-ip-10-171-6-4.ec2.internal.warc.gz | 37,677,062 | 8,945 | # Free help with homework
## Why join Brainly?
• find similar questions
# Show that 12n cannot end with the digit 0 or 5 for any natural number n.
NCERT Class X
Mathematics - Exemplar Problems
Chapter _Real Numbers
1
by purushottamaOza
I think it can : 12*5=60 ending with 0 | 83 | 279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-44 | latest | en | 0.863899 |
http://fricas-wiki.math.uni.wroc.pl/SandBox/diff?rev=106 | 1,656,147,025,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00393.warc.gz | 25,841,957 | 18,552 | login home contents what's new discussion bug reports help links subscribe changes refresh edit
# Edit detail for SandBox revision 106 of 171
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 Editor: Bill Page Time: 2013/04/20 01:35:48 GMT+0 Note: Mathematica or sympy
added:
From BillPage Sat Apr 20 01:35:47 +0000 2013
From: Bill Page
Date: Sat, 20 Apr 2013 01:35:47 +0000
Subject: Mathematica or sympy
Message-ID: <20130420013547+0000@axiom-wiki.newsynthesis.org>
FriCAS
\begin{axiom}
g:=1/(x*(1-a*(1-x)))
integrate(g,x)
\end{axiom}
Sympy
$$\frac{- \log{\left (2 x \right )} + \log{\left (\frac{- a^{2} + 2 a x \left(a - 1\right) - a \left(a - 1\right) + 3 a - 2}{a \left(a - 1\right)} \right )}}{ a - 1}$$
Mathematica
$$\frac{\log (a (x-1)+1)-\log (x)}{a-1}$$
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[SandBox Aldor Category Theory]?
based on "Prospects for Category Theory in Aldor" by Saul Youssef, 2004 http://atlas.bu.edu/~youssef/papers/math/aldor/aldor.pdf
[SandBox Aldor Foreign]?
Using Aldor to call external C routines
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Aldor defines a generator for type Vector
[SandBox Aldor Join and Meet]?
Aldor has category constructor named Meet which appears to be analogous to (but opposite of) Join.
[SandBox Aldor Semantics]?
exports and constants
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A prime number sieve in Aldor to count primes <= n.
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Using Aldor to write Axiom library routines
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How fast is array access in Axiom?
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Syntax of if then else
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evaluating Boolean expressions and conditions
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Meaning and use of pretend vs. strong typing
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Special relativity without the Lorentz group
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Graph theory in Axiom
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Tangle operation for literate programming implemented in Common Lisp
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A{ld,xi}o{r,m}Combinat
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Content vs. presentation MathML?
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A x B
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expression in sum-of-products form
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What is the difference?
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What does the type means for you?
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Beginnings of an Emacs mode for Axiom based off of Jay's work and others
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pdf format documents can be displayed inline
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Algebra and Data Structure Hierarchy (lattice) diagrams
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What are they? In Axiom functors are also called domain constructors.
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Numerical evaluation of the incomplete Gamma function
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Guessing integer sequences
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Examples of integration in Axiom and Reduce
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What is a "kernel"?
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LaTeX? commands allowed in MathAction?
[SandBox Lisp]?
Using Lisp in Axiom
[SandBox Manip]?
expression manipulations
[SandBox Manipulating Domains]?
testing the domain of an expression
[SandBox Mapping]?
A->B is a type in Axiom
[MathMLFormat]?
[SandBox Matrix]?
Examples of working with matrices in Axiom
[SandBox Maxima]?
Testing the Maxima interface
[SandBox Monoid]?
Rings and things
[SandBox Monoid Extend]?
Martin Rubey's beautiful idea about using extend to add a category to a previously defined domain.
[SandBox Noncommutative Polynomials]?
XPOLY and friends
[SandBox Numerical Integration]?
Simpson method
[SandBox NNI]?
NonNegative? Integer without using SubDomain?
[SandBox Pamphlet]?
[Literate Programming]? support on MathAction?
[SandBoxPartialFraction]?
Trigonometric expansion example
SandBoxPfaffian?
Computing the Pfaffian of a square matrix
[SandBox Polymake]?
an interface between Axiom and PolyMake?
[SandBox Polynomials]?
Axiom's polynomial domains are certainly rich and complex!
[SandBox ProblemSolving]?
Test page for educational purposes
[SandBox Qubic]?
Solving cubic polynomials
[SandBox Reduce And MathML]?
Reduce can use MathML? for both input and output
[SandBox Reflection in Aldor]?
a reflection framework
[SandBoxRelativeVelocity]?
Slides for IARD 2006: Addition of Relative Velocites is Associative
[SandBox Sage]?
This is a test of Sage in MathAction?
[SandBox Shortcoming]?
Implementation of solve
[SandBox Solve]?
Solving equations
[SandBox Statistics]?
calculating statistics in Axiom
[SandBox SubDomain]?
What is a SubDomain??
[SandBox Tail Recursion]?
When does Axiom replace recursion with iteration?
[SandBox Text Files]?
How to access text files in Axiom
[SandBox Trace Analysed]?
Tracing can affect output of 1::EXPR INT or 1::FRAC INT
[SandBox Units and Dimensions]?
Scientific units and dimensions
Domain construction
[SandBox Speed]?
Compilation speed
[SandBox Zero]?
[SandBox Axiom Strengths]?
SandBoxJohn2?
Experiments with matrices and various other stuff
SandBox2?
Experiments
SandBox3?
Experiments
SandBoxSymbolicInverseTrig?
Experiments
SandBoxGraphviz?
Experiments with GraphViz? and StructuredTables?
SandBoxDifferentialEquations?
Differential Equations etc.
[SandBoxMatrixExample]?
[SandBoxRotationMatrix]?
Here you can create your own SandBox?.
[SandBox9]?
Experiments with JET Bundles
[SandBoxGnuDraw]?
Miscellaneous
[SandBox11]?
Miscellaneous
[[SandBox12TestIndetAndComplex]]?
[SandBox13]?
Solving some nonlinear differential equations
[SandBox42]?
Miscellaneous
[SandBox DoOps]?
used to run Axiom without actually have to have it installed!
[SandBoxKMG]?
[SandBoxDGE]?
[SandBoxMLE]?
Maximum likelihood estimation (statistics)
[SandBoxFisher]?
Fisher's exact test for 2x2 tables (statistics)
[SandBoxNewtonsMethod]?
Newton's method for numerically solving f(x)=0 (with examples of calling Axiom expressions and Spad functions from Lisp).
[SandBoxVeryLongLaTeX]?
Test long lines
[SandBox Complementsdalgebrelineaire]?
Francois Maltey
[SandBoxFriCAS]?
page for testing friCAS
[SandBoxEcfact]?
Aldor compiler problem?
[SandBoxMyReduce]?
calling reduce with empty list
[SandBoxCategoryTerms]?
Category theory terminology used in SPAD
Click on the ? to create a new page. You should also edit this page to include a description and a new empty link for the next person.
## Examples
Here is a simple Axiom command:
\begin{axiom}
integrate(1/(a+z^3), z=0..1,"noPole")
\end{axiom}
fricas
integrate(1/(a+z^3), z=0..1,"noPole")
(1)
Type: Union(f1: OrderedCompletion?(Expression(Integer)),...)
And here is a REDUCE command:
\begin{reduce}
int(1/(a+z^3), z,0,1);
\end{reduce}
load_package sfgamma; load_package defint; *** gamma declared operator int(1/(a+z^3), z,0,1); reduce
## Common Mistakes
Please review the list of [Common Mistakes]? and the list of [MathAction Problems]? if you are have never used MathAction? before. If you are learning to use Axiom and think that someone must have solved some particular problem before you, check this list of Common [Axiom Problems]?.
fricas
solve(a*x+b,x)
(2)
Type: List(Equation(Fraction(Polynomial(Integer))))
exploring --Bill Page, Thu, 24 Apr 2008 07:00:25 -0700 reply
SandBoxNonAssociativeAlgebra?
lexical scope --Bill Page, Sun, 11 May 2008 06:43:30 -0700 reply
Testing lexical scoping rules in SandBoxLexicalScope?.
Combinatorial Sum --Bill Page, Fri, 16 May 2008 13:38:20 -0700 reply
SandBoxSum? (like Product)
Symbolic computations --Bill Page, Thu, 22 May 2008 12:58:29 -0700 reply
SandBoxSymbolic?
add inheritance issue --Bill Page, Sun, 25 May 2008 11:26:13 -0700 reply
For example: SandBoxLeftFreeModule?
Added Preview and Cancel to comment form --page, Tue, 27 May 2008 15:08:43 -0700 reply
This is a test of the Preview and Cancel buttons:
fricas
integrate(sin x, x)
(3)
Type: Union(Expression(Integer),...)
software archeology discovers another --Bill Page, Fri, 30 May 2008 21:01:37 -0700 reply
SandBoxSubsetCategory?
try iso-experiment/combinat --Bill Page, Tue, 03 Jun 2008 13:24:13 -0700 reply
SandBoxCombinat?
Equation, Inequation, and Inequality --Bill Page, Mon, 09 Jun 2008 18:28:56 -0700 reply
SandBoxEquation? SandBoxInequation? SandBoxInequality?
test aldor code --Bill Page, Wed, 18 Jun 2008 03:43:05 -0700 reply
gnuplottex --Bill Page, Tue, 24 Jun 2008 22:42:51 -0700 reply
SandBoxGnuPlotTex?
implementing Integer from Cardinal (unsigned) numbers --Bill Page, Mon, 21 Jul 2008 06:34:23 -0700 reply
SandBoxDefineInteger?
Attributes and categories --Bill Page, Fri, 25 Jul 2008 13:36:54 -0700 reply
SandBoxCommutativeCategory?
Literal and Symbol in SPAD --Bill Page, Sun, 27 Jul 2008 02:09:57 -0700 reply
SandBoxLiteral?
a category of partially ordered sets --Bill Page, Wed, 06 Aug 2008 17:26:50 -0700 reply
SandBoxPartiallyOrderedSet?
in response to an exchange of emails with Gabriel Dos Reis concerning the validity of automatic translations of x >= y into not x < y, etc.
Document function selection process in the interpreter --Bill Page, Wed, 13 Aug 2008 18:10:26 -0700 reply
[SandBox/interp/i-funsel.boot]?
Tensor Product of Polynomials --Bill Page, Sun, 24 Aug 2008 06:36:20 -0700 reply
SandBoxTensorProductPolynomial?
Reflection --Bill Page, Mon, 08 Sep 2008 06:09:27 -0700 reply
)abbrev package REFL Reflect
Reflect(T:Type): with
constructor? : Symbol -> Boolean
constructor?(p:Symbol):Boolean == car(devaluate(T)$Lisp)$SExpression = convert(p)$SExpression spad Compiling FriCAS source code from file /var/lib/zope2.10/instance/axiom-wiki/var/LatexWiki/2684009892188271010-25px004.spad using old system compiler. REFL abbreviates package Reflect ------------------------------------------------------------------------ initializing NRLIB REFL for Reflect compiling into NRLIB REFL compiling exported constructor? : Symbol -> Boolean Time: 0 SEC. (time taken in buildFunctor: 0) ;;; *** |Reflect| REDEFINED ;;; *** |Reflect| REDEFINED Time: 0 SEC. Cumulative Statistics for Constructor Reflect Time: 0 seconds finalizing NRLIB REFL Processing Reflect for Browser database: --->-->Reflect(constructor): Not documented!!!! --->-->Reflect((constructor? ((Boolean) (Symbol)))): Not documented!!!! --->-->Reflect(): Missing Description ; compiling file "/var/aw/var/LatexWiki/REFL.NRLIB/REFL.lsp" (written 23 APR 2013 06:09:03 PM): ; /var/aw/var/LatexWiki/REFL.NRLIB/REFL.fasl written ; compilation finished in 0:00:00.012 ------------------------------------------------------------------------ Reflect is now explicitly exposed in frame initial Reflect will be automatically loaded when needed from /var/aw/var/LatexWiki/REFL.NRLIB/REFL fricas T1:=Integer (4) Type: Type fricas T2:=Polynomial Fraction T1 (5) Type: Type fricas T3:=Complex T2 (6) Type: Type fricas constructor?('Polynomial)$Reflect(T1)
(7)
Type: Boolean
fricas
constructor?('Polynomial)$Reflect(T2) (8) Type: Boolean fricas constructor?('Polynomial)$Reflect(T3)
(9)
Type: Boolean
fricas
constructor?('Complex)\$Reflect(T3)
(10)
Type: Boolean
Francois Maltey --Bill Page, Tue, 18 Nov 2008 19:11:35 -0800 reply
SandBoxConditionalFunctions?
NonZeroInteger? --Bill Page, Tue, 02 Dec 2008 21:35:49 -0800 reply
SandBoxNonZeroInteger? is an attempt to define the domain of Integers without 0.
Martin's "generator" mini-tutorial --Bill Page, Sat, 28 Feb 2009 09:00:14 -0800 reply
SandboxDelay?
prototype for tensor products --Bill Page, Tue, 12 May 2009 12:49:48 -0700 reply
SandBoxTensorProduct? by Franz Lehner
Riemann Surface --Bill Page, Sat, 20 Jun 2009 13:15:37 -0700 reply
SandBoxComplexManifold?
Differential Algebra --Bill Page, Wed, 29 Jul 2009 09:07:26 -0700 reply
SandBoxDifferentialPolynomial?
Grassmann Algebra --Bill Page, Thu, 10 Sep 2009 09:05:28 -0700 reply
SandBoxGrassmannIsometry? - All mappings that preserve a given metric are given in terms of the decomposition of a general multivector.
Free Product --Bill Page, Fri, 18 Sep 2009 03:20:41 -0700 reply
SandBoxFreeProduct?
This domain implements the free product of monoids (or groups) It is the coproduct in the category of monoids (groups). FreeProduct(A,B) is the monoid (group) whose elements are the reduced words in A and B, under the operation of concatenation followed by reduction:
• Remove identity elements (of either A or B)
• Replace a1a2 by its product in A and b1b2 by its product in B
FunctionWithCache? --Bill Page, Tue, 27 Oct 2009 14:51:49 -0700 reply
Franz Lehner provided the following example of caching the output of a function: SandBoxRemember?
Hash Functions --Bill Page, Wed, 04 Nov 2009 00:11:08 -0800 reply
MortonCode? (also called z-order) is a method of combining multidimensional "coordinates" into a one-dimensional coordinate or "code" that attempts to preserve locality, i.e. minimize the average Euclidean distance between coordinate locations associated with adjacent codes. Morton codes are computationally less expensive to convert to and from coordinate values than Hilbert codes.
Groebner Basis and Polynomial Ideals --Bill Page, Tue, 08 Feb 2011 14:48:48 -0800 reply
SandBoxGroebnerBasis? examples from Ideals, Varieties, and Algorithms Third Edition, 2007
Frobenius Algebra --Bill Page, Fri, 11 Feb 2011 17:12:43 -0800 reply
FrobeniusAlgebraVectorSpacesAndPolynomialIdeals? Classifying low dimensional Frobenius algebras
Compiling SPAD code from a string --Bill Page, Mon, 07 Mar 2011 15:59:46 -0800 reply
Sandbox with some simple Algebra [SimplifyingAlgebraicExpressions]?
Observer algebra --Bill Page, Thu, 26 Apr 2012 14:44:52 -0700 reply
SandBoxObserverAsIdempotent?
Mathematica or sympy --Bill Page, Sat, 20 Apr 2013 01:35:47 +0000 reply
FriCAS?
fricas
g:=1/(x*(1-a*(1-x)))
(11)
Type: Fraction(Polynomial(Integer))
fricas
integrate(g,x)
(12)
Type: Union(Expression(Integer),...)
Sympy
(13)
Mathematica
(14) | 4,684 | 15,964 | {"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": 2, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-27 | latest | en | 0.504783 |
www.teachingmummy.com | 1,606,553,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00134.warc.gz | 867,637,792 | 64,396 | Search
• Teachingmummy
Maths games using house hold items
Updated: Mar 25
Tens and Units
Put the numbers 1-9 out. How many sums can you make with these numbers?
It is nice to do activities that are not just pencil and paper...you can change this to H T U for a more challenging option.
Tens and Units Ball Game
You will need 9 balls for the units and 9 balls for the tens
(it doesn't have to be balls it can be any items you can throw in a box- but 9 of each)
Throw the balls units first then the tens ... then write the number you have made
Use 4 boxes (2 tens and 2 units boxes) to write sums 23 +14 =
Number Bonds
Put pegs on a coat hanger to create an easy way to work out number bonds
42 views
See All | 188 | 720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.896556 |
https://brainmass.com/business/accounting/transfer-prices-and-divisional-profits-511736 | 1,526,859,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863811.3/warc/CC-MAIN-20180520224904-20180521004904-00084.warc.gz | 554,660,518 | 18,792 | Explore BrainMass
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# Transfer prices and divisional profits
A chair manufacturer has two divisions: framing and upholstering. The framing costs are \$100 per chair and the upholstering costs are \$200 per chair. The company makes 5,000 chairs each year, which are sold for \$500. Un-upholstered chair frames can be sold at cost to other chair manufacturers.
a) What is the minimum transfer price for this company?
b) What is the profit of each division if the transfer price is \$150?
c) What is the profit of each division if the transfer price is \$200?
Please provide workings to the solutions.
#### Solution Preview
a) What is the minimum transfer price for this company?
Minimum transfer price should be the price at which an un-upholstered chair frame can be sold to other manufacturers, i.e. \$100 per unit.
b). What is the profit of each division if the transfer price is \$150?
If the transfer price is \$150, then
Selling ...
#### Solution Summary
Solution determines the minimum transfer price. It also describes the steps to estimate the profits of associated divisions for the given transfer prices.
\$2.19 | 245 | 1,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.912224 |
https://stackoverflow.com/questions/72044284/how-to-change-a-raster-to-a-specific-spatial-resolution/72044772 | 1,701,771,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00780.warc.gz | 594,391,263 | 44,211 | # How to change a raster to a specific spatial resolution?
I would like to change the resolution of a raster. For example, let’s take this Landsat 7 images at ~ 30m resolution.
``````library(terra)
#> terra 1.5.21
f <- system.file("tif/L7_ETMs.tif", package = "stars")
r <- rast(f)
# 30m x 30m resolution
res(r)
#> [1] 28.5 28.5
plot(r, 1)
``````
I can use `aggregate()` with an integer factor such as:
``````# 10 * 28.5
r2 <- aggregate(r, fact = 10)
res(r2)
#> [1] 285 285
plot(r2, 1)
``````
My question is, how can I specify an exact resolution. For example, I would like to have a pixel resolution of 1.234 km (1234 m).
``````fact <- 1234 / 28.5
fact
#> [1] 43.29825
r3 <- aggregate(r, fact = fact)
res(r3)
#> [1] 1225.5 1225.5
plot(r3, 1)
``````
The documentation says that `fact` should be an integer, so here it is flooring `fact` to 43.
``````res(aggregate(r, 43))
#> [1] 1225.5 1225.5
``````
Any ways to have an exact resolution of 1234 m?
Created on 2022-04-28 by the reprex package (v2.0.1)
• A raster has to be made of regularly sized cells. If your original data has dimensions that are not a multiple of the resolution you choose, then your cells will not equal the extent of the original raster. Putting a fractional number into `aggregate` implies that you want a fractional number of cells in your resulting raster, which doesn't really make sense. Of course, it should be possible if cropping were allowed, but that option doesn't seem to be included. I guess you could crop the raster to a multiple of `fact` then aggregate it? Apr 28, 2022 at 13:22
• Thank you, @AllanCameron, for responding. This is where I am stuck. I do not mind cropping a bit, but I kind of lost here. Do you have any resources or guidance to provide? How can `r` be cropped using a factor? Would it be possible to create a grid of the desired resolution before and "translate" `r` into it? Apr 28, 2022 at 13:27
I came up with this solution which seems to give me what I need.
``````library(terra)
#> terra 1.5.21
f <- system.file("tif/L7_ETMs.tif", package = "stars")
r <- rast(f)
plot(r, 1)
``````
``````r2 <- r
res(r2) <- 1234
r2 <- resample(r, r2)
plot(r2, 1)
``````
``````res(r2)
#> [1] 1234 1234
``````
Created on 2022-04-28 by the reprex package (v2.0.1)
• I was about to post a very similar solution using `resample`, but this one is just as good! Apr 28, 2022 at 13:52
• Thank you, @AllanCameron. Your previous answer guided me to approach this differently. Apr 28, 2022 at 13:53
I also propose (as described in the terra vignette) that you first aggregate the raster as close as possible and then resample. Resampling can be done e.g. using a template raster to guarantee correct crs, dimensions etc. | 854 | 2,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.89676 |
https://howkgtolbs.com/convert/92.27-kg-to-lbs | 1,656,481,840,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00215.warc.gz | 336,794,441 | 12,108 | # 92.27 kg to lbs - 92.27 kilograms to pounds
Do you need to know how much is 92.27 kg equal to lbs and how to convert 92.27 kg to lbs? You are in the right place. This whole article is dedicated to kilogram to pound conversion - theoretical and practical too. It is also needed/We also want to underline that whole this article is devoted to only one number of kilograms - that is one kilogram. So if you need to learn more about 92.27 kg to pound conversion - read on.
Before we go to the more practical part - this is 92.27 kg how much lbs conversion - we want to tell you some theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 92.27 kg to lbs? 92.27 kilograms it is equal 203.4205291474 pounds, so 92.27 kg is equal 203.4205291474 lbs.
## 92.27 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, that is International System of Units (in short form SI).
Sometimes the kilogram could be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. This definition was simply but totally impractical to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was switched by a new definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
## 92.27 kilogram to pounds
You know some facts about kilogram, so now we can move on to the pound. The pound is also a unit of mass. We want to point out that there are more than one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is used also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 92.27 kg?
92.27 kilogram is equal to 203.4205291474 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 92.27 kg in lbs
Theoretical section is already behind us. In this section we will tell you how much is 92.27 kg to lbs. Now you know that 92.27 kg = x lbs. So it is high time to get the answer. Just look:
92.27 kilogram = 203.4205291474 pounds.
It is an accurate result of how much 92.27 kg to pound. You can also round it off. After it your result is as following: 92.27 kg = 202.994 lbs.
You know 92.27 kg is how many lbs, so see how many kg 92.27 lbs: 92.27 pound = 0.45359237 kilograms.
Obviously, in this case you may also round it off. After it your outcome will be exactly: 92.27 lb = 0.45 kgs.
We are also going to show you 92.27 kg to how many pounds and 92.27 pound how many kg results in charts. Let’s see:
We are going to start with a chart for how much is 92.27 kg equal to pound.
### 92.27 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
92.27 203.4205291474 202.9940
Now look at a table for how many kilograms 92.27 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
92.27 0.45359237 0.45
Now you know how many 92.27 kg to lbs and how many kilograms 92.27 pound, so it is time to move on to the 92.27 kg to lbs formula.
### 92.27 kg to pounds
To convert 92.27 kg to us lbs a formula is needed. We are going to show you two formulas. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 203.4205291474 result in pounds
The first formula give you the most correct outcome. In some cases even the smallest difference can be considerable. So if you need an exact result - this version of a formula will be the best solution to calculate how many pounds are equivalent to 92.27 kilogram.
So let’s go to the shorer formula, which also enables conversions to learn how much 92.27 kilogram in pounds.
The second formula is down below, see:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second formula is simpler. It could be the best solution if you need to make a conversion of 92.27 kilogram to pounds in fast way, for instance, during shopping. You only need to remember that final outcome will be not so correct.
Now we want to show you these two formulas in practice. But before we will make a conversion of 92.27 kg to lbs we are going to show you another way to know 92.27 kg to how many lbs totally effortless.
### 92.27 kg to lbs converter
An easier way to know what is 92.27 kilogram equal to in pounds is to use 92.27 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Converter is based on first formula which we showed you in the previous part of this article. Due to 92.27 kg pound calculator you can quickly convert 92.27 kg to lbs. Just enter amount of kilograms which you want to calculate and click ‘convert’ button. You will get the result in a second.
So let’s try to calculate 92.27 kg into lbs with use of 92.27 kg vs pound calculator. We entered 92.27 as a number of kilograms. It is the result: 92.27 kilogram = 203.4205291474 pounds.
As you see, our 92.27 kg vs lbs calculator is user friendly.
Now we are going to our primary topic - how to convert 92.27 kilograms to pounds on your own.
#### 92.27 kg to lbs conversion
We will start 92.27 kilogram equals to how many pounds calculation with the first version of a formula to get the most accurate outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 203.4205291474 the outcome in pounds
So what need you do to check how many pounds equal to 92.27 kilogram? Just multiply number of kilograms, in this case 92.27, by 2.20462262. It is exactly 203.4205291474. So 92.27 kilogram is 203.4205291474.
You can also round off this result, for instance, to two decimal places. It is exactly 2.20. So 92.27 kilogram = 202.9940 pounds.
It is high time for an example from everyday life. Let’s calculate 92.27 kg gold in pounds. So 92.27 kg equal to how many lbs? As in the previous example - multiply 92.27 by 2.20462262. It is 203.4205291474. So equivalent of 92.27 kilograms to pounds, if it comes to gold, is 203.4205291474.
In this example you can also round off the result. Here is the outcome after rounding off, in this case to one decimal place - 92.27 kilogram 202.994 pounds.
Now we can go to examples calculated with a short version of a formula.
#### How many 92.27 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 202.994 the result in pounds
So 92.27 kg equal to how much lbs? And again, you need to multiply number of kilogram, this time 92.27, by 2.2. Look: 92.27 * 2.2 = 202.994. So 92.27 kilogram is 2.2 pounds.
Let’s do another conversion using shorer version of a formula. Now convert something from everyday life, for instance, 92.27 kg to lbs weight of strawberries.
So calculate - 92.27 kilogram of strawberries * 2.2 = 202.994 pounds of strawberries. So 92.27 kg to pound mass is exactly 202.994.
If you learned how much is 92.27 kilogram weight in pounds and are able to calculate it using two different formulas, let’s move on. Now we are going to show you these results in tables.
#### Convert 92.27 kilogram to pounds
We realize that outcomes shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Thanks to this you can quickly make a comparison 92.27 kg equivalent to lbs results.
Start with a 92.27 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
92.27 203.4205291474 202.9940
And now look 92.27 kg equal pound chart for the second formula:
Kilograms Pounds
92.27 202.994
As you see, after rounding off, when it comes to how much 92.27 kilogram equals pounds, the results are the same. The bigger number the more considerable difference. Remember it when you need to make bigger amount than 92.27 kilograms pounds conversion.
#### How many kilograms 92.27 pound
Now you know how to convert 92.27 kilograms how much pounds but we will show you something more. Do you want to know what it is? What about 92.27 kilogram to pounds and ounces conversion?
We are going to show you how you can calculate it step by step. Let’s begin. How much is 92.27 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, in this case 92.27, by 2.20462262. So 92.27 * 2.20462262 = 203.4205291474. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To calculate how much 92.27 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your result is equal 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final result will be exactly 2 pounds and 33 ounces.
As you can see, conversion 92.27 kilogram in pounds and ounces quite simply.
The last conversion which we want to show you is calculation of 92.27 foot pounds to kilograms meters. Both of them are units of work.
To convert it it is needed another formula. Before we show you it, look:
• 92.27 kilograms meters = 7.23301385 foot pounds,
• 92.27 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 92.27 foot pounds to kilograms meters you have to multiply 92.27 by 0.13825495. It is exactly 0.13825495. So 92.27 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 92.27 foot pounds will be exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 92.27 kilogram into pounds calculations.
This article was a big compendium about kilogram, pound and 92.27 kg to lbs in conversion. Due to this calculation you learned 92.27 kilogram is equivalent to how many pounds.
We showed you not only how to make a conversion 92.27 kilogram to metric pounds but also two another calculations - to know how many 92.27 kg in pounds and ounces and how many 92.27 foot pounds to kilograms meters.
We showed you also another solution to do 92.27 kilogram how many pounds conversions, that is using 92.27 kg en pound converter. It is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you can do 92.27 kilogram equal to how many pounds calculation - on your own or with use of our 92.27 kgs to pounds calculator.
It is time to make your move! Let’s convert 92.27 kilogram mass to pounds in the way you like.
Do you want to do other than 92.27 kilogram as pounds calculation? For instance, for 5 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 92.27 kilogram equal many pounds.
### How much is 92.27 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 92.27 kg in pounds , we prepared one more section. Here you can see all you need to remember about how much is 92.27 kg equal to lbs and how to convert 92.27 kg to lbs . It is down below.
How does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 92.27 kg to pound conversion formula . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 92.27 kilogram to pounds. The accurate result is 203.4205291474 lb.
There is also another way to calculate how much 92.27 kilogram is equal to pounds with second, shortened version of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 92.27 kg equal to how much lbs ? The answer is 203.4205291474 lbs.
How to convert 92.27 kg to lbs in an easier way? It is possible to use the 92.27 kg to lbs converter , which will make the rest for you and you will get an exact answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
Read more related articles:
92.01 kg to lbs = 202.847 92.02 kg to lbs = 202.869 92.03 kg to lbs = 202.891 92.04 kg to lbs = 202.913 92.05 kg to lbs = 202.936 92.06 kg to lbs = 202.958 92.07 kg to lbs = 202.98 92.08 kg to lbs = 203.002 92.09 kg to lbs = 203.024 92.1 kg to lbs = 203.046 92.11 kg to lbs = 203.068 92.12 kg to lbs = 203.09 92.13 kg to lbs = 203.112 92.14 kg to lbs = 203.134 92.15 kg to lbs = 203.156 92.16 kg to lbs = 203.178 92.17 kg to lbs = 203.2 92.18 kg to lbs = 203.222 92.19 kg to lbs = 203.244 92.2 kg to lbs = 203.266 92.21 kg to lbs = 203.288 92.22 kg to lbs = 203.31 92.23 kg to lbs = 203.332 92.24 kg to lbs = 203.354 92.25 kg to lbs = 203.376
92.26 kg to lbs = 203.398 92.27 kg to lbs = 203.421 92.28 kg to lbs = 203.443 92.29 kg to lbs = 203.465 92.3 kg to lbs = 203.487 92.31 kg to lbs = 203.509 92.32 kg to lbs = 203.531 92.33 kg to lbs = 203.553 92.34 kg to lbs = 203.575 92.35 kg to lbs = 203.597 92.36 kg to lbs = 203.619 92.37 kg to lbs = 203.641 92.38 kg to lbs = 203.663 92.39 kg to lbs = 203.685 92.4 kg to lbs = 203.707 92.41 kg to lbs = 203.729 92.42 kg to lbs = 203.751 92.43 kg to lbs = 203.773 92.44 kg to lbs = 203.795 92.45 kg to lbs = 203.817 92.46 kg to lbs = 203.839 92.47 kg to lbs = 203.861 92.48 kg to lbs = 203.883 92.49 kg to lbs = 203.906 92.5 kg to lbs = 203.928
92.51 kg to lbs = 203.95 92.52 kg to lbs = 203.972 92.53 kg to lbs = 203.994 92.54 kg to lbs = 204.016 92.55 kg to lbs = 204.038 92.56 kg to lbs = 204.06 92.57 kg to lbs = 204.082 92.58 kg to lbs = 204.104 92.59 kg to lbs = 204.126 92.6 kg to lbs = 204.148 92.61 kg to lbs = 204.17 92.62 kg to lbs = 204.192 92.63 kg to lbs = 204.214 92.64 kg to lbs = 204.236 92.65 kg to lbs = 204.258 92.66 kg to lbs = 204.28 92.67 kg to lbs = 204.302 92.68 kg to lbs = 204.324 92.69 kg to lbs = 204.346 92.7 kg to lbs = 204.369 92.71 kg to lbs = 204.391 92.72 kg to lbs = 204.413 92.73 kg to lbs = 204.435 92.74 kg to lbs = 204.457 92.75 kg to lbs = 204.479
92.76 kg to lbs = 204.501 92.77 kg to lbs = 204.523 92.78 kg to lbs = 204.545 92.79 kg to lbs = 204.567 92.8 kg to lbs = 204.589 92.81 kg to lbs = 204.611 92.82 kg to lbs = 204.633 92.83 kg to lbs = 204.655 92.84 kg to lbs = 204.677 92.85 kg to lbs = 204.699 92.86 kg to lbs = 204.721 92.87 kg to lbs = 204.743 92.88 kg to lbs = 204.765 92.89 kg to lbs = 204.787 92.9 kg to lbs = 204.809 92.91 kg to lbs = 204.831 92.92 kg to lbs = 204.854 92.93 kg to lbs = 204.876 92.94 kg to lbs = 204.898 92.95 kg to lbs = 204.92 92.96 kg to lbs = 204.942 92.97 kg to lbs = 204.964 92.98 kg to lbs = 204.986 92.99 kg to lbs = 205.008 93 kg to lbs = 205.03 | 4,825 | 16,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-27 | latest | en | 0.945555 |
https://everydayanalytics.ca/2015/05 | 1,548,275,154,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584350539.86/warc/CC-MAIN-20190123193004-20190123215004-00122.warc.gz | 492,564,972 | 12,169 | ## Data Visualization Fundamentals with Skittles
So I have a shocking confession to make: I love Skittles.
This post is not sponsored, endorsed, compensated, or paid for in any way, shape or form, by Skittles Candy. I’m not particular – I like other types of candy that are similar – you know, those ones that are chocolate covered in a hard shell, whether they be the kind where you eat the red ones last or not.
Anyhow, I got to thinking about how, abstractly, each individual candy can be viewed like a pixel of a different color. So you can make art using candy, just like artists make a mosaic. There’s lots of this on the internet you can already see: in fact, Skittles has done print advertising this way.
But…. each individual candy can also represent something else: a unit of measurement. I thought it would be cool to go through some data visualization fundamentals using the candy in this way. So let’s dive in.
#### Data Visualization using only 1 bag of Skittles
So, what would your average first grader do with a bag of Skittles if you asked them to sort it? Probably something like below, the physical equivalent of a bubble chart depicting the quantities of each colour by area, assuming each Skittle is approximately the same size.
A perhaps more useful way to do the same would be to organize each colour in rows, with each row a set number (like tally marks). Here it’s not only easy to see the relative proportions of the different colours in the bag, but also count them as each row and group is a set number (5 & 10, respectively). This is equivalent to a pictogram, with each Skittle representing, well, 1 Skittle:
It’s not a big stretch of the imagination to collapse those groups together into groups of a set height. So here we have a proportional bar chart, where the length of each bar represents the percentage of the bag that is each colour. Note that because I didn’t slice Skittles in half, the physical analogue is not exactly the same as what you’d put down on paper or in Excel (there is one additional unit for yellow and orange):
And, as I both often have to remind people of this rule, and also observe many people not following it, it is best practice to sort the bars in descending order for maximum clarity / comparative value (assuming there is not another more important ordering):
And, if we want to transform our proportional bar chart into one comparing absolute quantities, it is not a giant stretch of the imagination to break apart the different bars so they are only one ‘pixel’ high:
Here it’s much easier to get an idea of the absolute number of each colour in the bag, but harder to tally that numbers exactly – for that we’d need to add an axis or data labels.
Okay, I have another shocking confession to make: I lied. I really like Skittles. So I actually bought a whole bunch of bags.
So let’s look at some more visualization fundamentals, where we required comparing not only across a categorical variable (colour) but also between groups.
Here is the equivalent to our first graph from before, only showing the different numbers of Skittles in each bag. You can see there’s actually a fair amount of variance; the smallest bag had 89 pieces of candy, whereas the largest had 110.
Now let’s make a bubble graph which not only compares the sizes between the different bags, but also their makeups by colour. The end result is actually closer to a collection of pie charts:
We can also group by colour only to see the overall makeup for the whole group of bags. Whereas orange dominated in the first bag we looked at, you can see here that orange and yellow are approximately at parity overall.
Now let’s look at the tally mark / pictograph method. Here each row represents a bag:
You can see there’s a fair bit of variance in the different colours. I also tried rearranging things so they result was less like a pictograph and more like a treemap:
Really the best way to compare would be a bar graph. Here’s a stacked area graph. I didn’t bother sorting by length, because at this point I was pretty tired of shuffling Skittles around:
To get a better idea of the different makeups of each bag by colour, we can break this out into a grouped bar graph, first by bag, then by colour:
And, of course, we can reverse the order if we want to more directly compare the colour makeups. The columns are in numerical order by bag. And just for fun, we’ll make this one a column chart:
There. That’s better! Clearly Bag 1 was an outlier as far as the number of purple went, and Bag 3 had a lot of yellow.
#### Concluding Remark
I thought it’d be cool to mix things up a bit, and trying doing some data visualization using a physical medium. The end result ended up being something more like an exercise for an elementary school mathematics class (indeed, there are many examples of this online), but I think it still drives home some of the fundamental strengths and weakness of different visualization types, as well as showing how they can be depicted using different media.
If you’re really interested, you can download the data yourself and slice and dice visualizations to your heart’s content. And I’m sure if you bought enough bags of Skittles you could learn something of a statistical nature about their manufacturing and packaging process – but perhaps that’s for a different day. Until then I’ll just enjoy good candy and data visualization. | 1,161 | 5,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-04 | longest | en | 0.940132 |
http://math.stackexchange.com/questions/325894/help-understanding-exists-x-exists-y-x-neq-y-wedge-forall-z-z-y-vee-z | 1,462,526,189,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861743914.17/warc/CC-MAIN-20160428164223-00095-ip-10-239-7-51.ec2.internal.warc.gz | 178,162,244 | 18,319 | # Help understanding $\exists x \exists y (x\neq y \wedge \forall z ((z=y)\vee (z=x)))$
I'm not sure how to interpret this problem.
Find a domain for the qunatifiers in: $$\exists{x} \exists{y}(x\neq y \wedge \forall{z}((z=y)\ \lor(z=x)))$$ such that this statement is false.
So, the translation to English would be: There exists some $x$ and some $y$ that are not the same and $z$ is $y$ or $z$ is $x$.
It's gibberish to me at this point.
-
There exists some x and some y that are not the same and for all z, z is y or z is x. – Integral Mar 9 '13 at 22:20
I read it as "there exist two distinct things such that everything is at least one of these things." This is clearly false when there is more than two things, because then no matter which two things x, y we pick, there will be a third thing that is neither x nor y. – anon Mar 9 '13 at 22:30
The short reading is: the universe is made up of precisely two objects. – André Nicolas Mar 9 '13 at 22:36
There are two objects, $x$ and $y$. Since mathematical objects, even under different labels, need not be distinct, we also require $x\neq y$. So there are two distinct objects called $x$ and $y$.
And for every object $z$ that we take, either $x=z$ or $z=y$. So any object in the universe is actually either $x$ or $y$ (but not both, as these are distinct objects).
Now consider the example "I have one orange and one apple, and any fruit I have is either an orange or an apple". What have I told you? I told you that all my fruits are these two, an orange and an apple. But if I also have a peach then I was lying when I told you that. Because I have a fruit which is neither the apple nor the orange.
The sentence you are given is no different, only it doesn't care for oranges and apples and other things. It is an abstract claim about an abstract universe. When will it be true? When there are exactly two objects in the universe, and it will be false if there is a third object.
-
Couldn't we also just do the negation of the statement and then find a domain that satisfies this? – Frudrururu Mar 10 '13 at 1:17
@Wishingwell: You can. But that requires you to understand how negation goes through quantifiers. I found that on average, both tasks (parsing and negating) are equally confusing to freshmen. – Asaf Karagila Mar 10 '13 at 1:25
In addition to Asaf's answer, when there is only one element in the domain you can not find $x \neq y$. Or, if the domain is empty then you cannot find x.
- | 673 | 2,468 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-18 | longest | en | 0.941583 |
https://math.answers.com/other-math/What_is_2.16_in_simplest_form | 1,696,462,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00149.warc.gz | 425,202,830 | 44,276 | 0
# What is 2.16 in simplest form?
Updated: 4/28/2022
Wiki User
6y ago
2.16 = 216/100 which can be simplified to 54/25.
Wiki User
10y ago
Wiki User
11y ago
216 in simplest form is 216/1, or 216.
Wiki User
6y ago
54/25
Wiki User
6y ago
2/16 = 1/8 | 103 | 261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-40 | latest | en | 0.744683 |
http://loan78.com/sbdm/beijing/calculator/dengebenjin.html | 1,516,276,417,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887253.36/warc/CC-MAIN-20180118111417-20180118131417-00713.warc.gz | 226,391,750 | 14,392 | Beijing
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# Equivalent principal repayment calculator
October 24, 2015 latest business loan interest rate 4.90%, 3.25% provident fund interest rate,点击了解详情>>
### Input data
Calculation Calculated by loan amount Calculated by area loan amount Million Which banks can grant loans? Price per square meter Yuan / square meter area Square meter The nature of the purchase A suite Two suites Two suites defined criteria: 1. For the first time, the borrower applies for the purchase of housing by loans. In the housing registration information system (including the pre-sales contract registration and filing system) where the house to be acquired is located, the family has already registered one set of housing units (and above). 2, the borrower has used a loan to buy a (and above) housing, but also apply for loans to buy housing. 3. The borrower is convinced that the borrower's family already has one (and above) housing through querying for due diligence in the forms of credit records, face surveys and interviews. down payment Loan period year Annual interest rate Bank rate comparison Provident Fund interest rate table You can also enter it manually
Sweep, concerned about the mortgage WeChat
### Output the result Save the calculation results to my email
Decreasing monthly repayments
The monthly repayment reduction is equal to the principal repayment method, which means the principal remains the same, the interest rate decreases month by month, and the monthly repayment amount decreases. Since the monthly repayment principal amount is fixed and the interest less and less, the lender initially Repayment pressure, but with the passage of time the number of monthly repayments less and less.
Total loan amount 0 yuan Repayment month 0 month First month repayment 0 yuanMonthly decrease:0 yuan Total payment of interest 0 yuan Total principal and interest 0 yuan
Next you can:
This result is for reference only, the actual payment is subject to the local
### Tool Introduction
Equal principal repayment calculator (the principal repayments are the same every month, and the monthly repayment total is decreasing month by month), which will help users to calculate the loan repayments under different loan terms and different interest rates. For the month, the total amount of interest and repayments.
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High amount of annual fee exemption | 641 | 3,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | longest | en | 0.920527 |
http://slideplayer.com/slide/3837910/ | 1,511,369,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806615.74/warc/CC-MAIN-20171122160645-20171122180645-00793.warc.gz | 282,186,527 | 21,824 | Byron Smith December 11, 2013. 1. What is Quantum State Tomography? 2. What is Bayesian Statistics? 1.Conditional Probabilities 2.Bayes’ Rule 3.Frequentist.
Presentation on theme: "Byron Smith December 11, 2013. 1. What is Quantum State Tomography? 2. What is Bayesian Statistics? 1.Conditional Probabilities 2.Bayes’ Rule 3.Frequentist."— Presentation transcript:
Byron Smith December 11, 2013
1. What is Quantum State Tomography? 2. What is Bayesian Statistics? 1.Conditional Probabilities 2.Bayes’ Rule 3.Frequentist vs. Bayesian 3. Example: Schrodinger’s Cat 1.Interpretation 2.Analysis with a Non-informative Prior 3.Analysis with an Informative Prior 4. Sources of Error in Tomography 5. Error Reduction via Bayesian Analysis 6. Adaptive Tomography 7. Conclusion 8. References 9. Supplementary Information
Tomography comes from tomos meaning section. Classically, tomography refers to analyzing a 3- dimensional trajectory using 2-dimensional slices. Quantum State Tomography refers to identifying a particular wave function using a series of measurements.
is the Likelihood is the Prior Probability (or just prior) is the Posterior Probability (or just posterior)
FrequentistBayesian Inference on probability arises from the frequency that some outcome is measured and that measurement is random. In other words, a measurement is random only due to our ignorance. Inference on probability arises from a prior probability assumption weighted by empirical evidence. Measurements are inherently random.
Frequentist: Given N cats, there are some that are alive and some that are dead. The probability that the cat is alive is associated with the fact that we sample randomly and a*N of the cats are alive. Bayesian: The probability a is a random variable (unknown) and therefore there is an inherent probability associated with each particular cat.
NFrequentistBayesian 30.670.6 250.20.22 750.240.25 2500.3360.337 True value of a = 0.3
Flourine-18, half- life=1.8295 Hours Cat has been in for 3 hours NFrequentistBayesian 30.670.39 250.280.29 750.2670.272 2500.316
Error in the measurement basis. Error in the counting statistics. Error associated with stability. ◦ Detector efficiency ◦ Source intensity Model Error
FrequentistBayesian (α=3.1, β=6.9) NVar(a) 30.027 250.0074 750.0029 2500.00079 NVar(a) 30.013 250.0057 750.0026 2500.00076
There are several measures of “lack of fit.” Likelihood Infidelity Shannon Entropy
The number of measurements required to sufficiently identify ρ can be reduced when using a basis which diagonalizes ρ. To do so, measure N 0 particles first, change the measurement basis, then finish the total measurement. Naturally this can be improved with Bayesian by using the first measurements as a Prior.
Quantum State Tomography is a tool used to identify a density matrix. There are several metrics of density identification. Bayesian statistics can improve efficiency while providing a new interpretation of quantum states.
1. J. B. Altepeter, D. F. V. James, and P. G. Kwiat, Qubit Quantum State Tomography, in Quantum State Estimation (Lecture Notes in Physics), M. Paris and J. Rehacek (editors), Springer (2004). 2. D. H. Mahler, L. A. Rozema, A. Darabi, C. Ferrie, R. Blume-Kohout, and A. M. Steinberg, Phys. Rev. Lett. 111, 183601 (2013). 3. F. Huszár and N. M. T. Houlsby, Phys. Rev. A 85, 052120 (2012). 4. R. Blume-Kohout, New J. Phys. 12, 043034 (2012).
For density operators which are not diagonal, we use a basis of spin matrices: If choose an instrument orientation such that the first parameter is 1, we are left with the Stoke’s Parameters, S i : The goal is then to identify the S i.
Used to visualize a superposition of photon polarizations. The x-axis is 45 degree polarizations. The y-axis is horizontal or vertical polarizations. The z-axis is circular polarizations.
Using a matrix basis similar to that for Stoke’s Parameters, one can exactly identify a polarization with three measurements. Note that an orthogonal basis is not necessary.
Errors in measurement and instrumentation will manifest themselves as a disc on the Poincare Sphere: (a) Errors in measurement basis. (b) Errors in intensity or detector stability.
Because there is a finite sample size, states are not characterized exactly. Each measurement constrains the sample space.
Optimization can lead to states which lie outside the Poincare Sphere. Unobserved values a registered as zeros in the density matrix optimization. This can be unrealistic. There is no direct measure of uncertainty within maximum likelihood estimators.
Parameter space can be constrained through the prior. Unobserved values can still have some small probability through the prior. Uncertainty analysis can come directly from the variance of the posterior distribution, regardless of an analytical form.
Download ppt "Byron Smith December 11, 2013. 1. What is Quantum State Tomography? 2. What is Bayesian Statistics? 1.Conditional Probabilities 2.Bayes’ Rule 3.Frequentist."
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http://www.thescienceforum.com/mathematics/50219-picking-your-own-numbers.html | 1,653,798,562,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00007.warc.gz | 119,083,328 | 15,222 | 1. For the lotto. In Canada we have what's called Lotto Max, every Tuesday and Friday. Pick 7 of 7 numbers correctly and win. Numbers range from 1-50. I always let the OLG(Ontario Lottery & Gaming) in-store computer pick my numbers . 3 tickets cost \$5. No one has won for a few draws so it's up over \$70 million for 1st prize right now.
I was watching something on Netflix a few weeks ago and they were talking about Benford's Law(see below). The digit 1 appears more often than any other, followed by 2 and so on. Actually a lot more complicated/involved than that. So I wondered if I could apply this to picking my own numbers for Lotto Max. Worked out that the digits 1-4 appear equally more often than any other, followed by 5, then the others (0 + 6 to 9 equally lowest) . For instance the number 14 would be two digits in the 1-4 range. So I thought I could fill in my own game sheet just using numbers with the most common digits.
It seems the OLG is on to this. As I mentioned, you get 3 sets of 7 numbers for \$5. You can fill out a game sheet but only allowed one set of your 7 picks and the computer generates the other 2. I call BS and I think most people who play are unaware of digit frequency in 1-50 range. I suppose there is a chance the OLG isn't aware and perhaps I should find one of the private lottery sites and let them know. Why we can't pick all 3 of our own 7 number sets seems very strange to me. Intentionally disallowing it should be against the law IMO.
I even went back over the last hundred draws just to prove I'm right.... no doubt about it, thats the way the digits fall. If you take the numbers 1-50 then digits 1,2,3 & 4 appear 14 times each(15 times counting doubles), 5 appears 6 times, 0 & digits 6-9 appear 5 times each. In fact, when I checked results over last year, the digits 1 thru 4 appeared 64% of the time covering all the draws.
*0 (zero) only used for numbers 10,20,30,40,50 not 01, 02, etc.
From Wiki: Benford's law, also called the Newcomb–Benford law, the law of anomalous numbers, or the first-digit law, is an observation about the frequency distribution of leading digits in many real-life sets of numerical data
2.
3. Benford's law applies to leading digits of tables of numbers (like stock market tables). It has no relevance to lottery numbers. These are supposed to be uniform between 1 and 50.
4. Originally Posted by mathman
Benford's law applies to leading digits of tables of numbers (like stock market tables). It has no relevance to lottery numbers. These are supposed to be uniform between 1 and 50.
It only gave me the idea*. Thought maybe there was an advantage for a person buying a ticket with numbers they’ve chosen. Why wouldn’t you pick numbers that contain digits 1 thru 4? Out of 50 numbers only 6 don't have a 1,2,3 or 4. Would it not improve one’s odds, even a little? Not a math guy
*some interesting careers where people use Benford’s Law. One guy uses it to authenticate, or prove an computerized photo is fake.
5. If it's a fair game it doesn't matter which 6 numbers you pick. All combinations of 6 numbers have the same odds.
6. Originally Posted by PhDemon
If it's a fair game it doesn't matter which 6 numbers you pick. All combinations of 6 numbers have the same odds.
Last night’s draw #’s: 11,17,19,22,27,31,40. Somebody won \$70m. Haven’t checked my ticket yet
I’m talking about selecting your own number sets and whether there is an advantage to pick numbers that contain digits 1-4. Three of last nights numbers satisfy that requirement...11,22,31 which would win you a free play for next draw. There are subsidiary prizes for having 3,4,5,or 6 correct numbers. There’s only 20 of 50 numbers that contain digits 1-4 and only six that don’t contain at least one of those digits. So if I select 7 numbers from that pool of 20 do I have a better chance of winning something? Let’s say I get to pick 3 sets of 7. I’m thinking that if I combine a 1-4 digit with another digit not in that range that my chances would lessen. Also think choosing numbers 5,6,7,8,9, 50 are the worst play.
I’m just asking, not totally convinced. Is there a mathematical way to check it? I understand in a random draw that each number has same chance but does that hold true if numbers are drawn one at a time? There’s only so many numbers in the drum and most of them have a 1-4 digit, 20 have two, and 6 none. Odds favour a number with a 1-4 digit to be drawn, don’t know how one can deny that. I was always lousy with numbers anyways.
7. It makes no difference whether you or the lotto machine picks the numbers or which 6 numbers are picked. All combinations are equally as likely to come up. I don't know what else to say to you...
Edit to add: I think your problem is considering numbers that contain another number rather than the whole numbers. The odds aren't based on getting a number "half right".
8. Originally Posted by PhDemon
It makes no difference whether you or the lotto machine picks the numbers or which 6 numbers are picked. All combinations are equally as likely to come up. I don't know what else to say to you...
Edit to add: I think your problem is considering numbers that contain another number rather than the whole numbers. The odds aren't based on getting a number "half right".
Yes. Tried to differentiate between number and digit. The odds are that a number containing a 1,2,3 or 4 being selected are extremely high. (44 of 50). Containing one single digit number or a combo of 1,2,3 or 4 is 40%.... I like those odds. Doesn’t seem right I know, but should I ignore it?
9. It makes no difference as whatever numbers you pick your odds of hitting the jackpot are the same...
10. When I play a Lotto game I get two tickets.. I have a consistent set of numbers I play and then I let the computer pick a set.
11. Originally Posted by GiantEvil
When I play a Lotto game I get two tickets.. I have a consistent set of numbers I play and then I let the computer pick a set.
Here you play your own set of 7 and then computer gives you two more.....\$5
There's another lottery here where you can pick a 6 number set or let computer do it for you. \$3 a set. Both lotteries draw an extra number that you can apply to your numbers however you wouldn't win the big prize if you had all the regular numbers but one. e.g. 6 numbers out of 7 plus bonus # would win you a fairly significant prize but nowhere near the big one.
Had to chuckle 2 or 3 draws ago when 6 of 7 numbers drawn plus the bonus number consisted only of numbers containing digits 1-4, a couple were single digit numbers. (there's 20 of them to pick from) Just luck . Last draw there was only two such numbers drawn.
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https://www.yahoo.com/news/blogs/ticket/errors-margins-samples-does-latest-poll-154739649.html?ref=gs | 1,469,667,389,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827781.76/warc/CC-MAIN-20160723071027-00012-ip-10-185-27-174.ec2.internal.warc.gz | 982,140,845 | 138,002 | # Errors, margins and samples: What does the latest poll say?
October 11, 2011
The most recent ABC News/Washington Post poll claims a margin of error of plus or minus 4 percentage points; this is a pretty standard margin-range for a serious national poll. In that poll they have Mitt Romney leading Barack Obama by 2 percentage points. What should that mean to you?
If you interpret the idea of margin of error at face value, it means in this case that there is 95 percent likelihood that the final vote tally would provide Obama with anywhere from 42 to 50 percentage points and Romney with anywhere from 44 to 52 percentage points. Unfortunately, though, that's a gross oversimplification.
Quoted margin of errors are large and significant. Here is the final difference in the national popular vote for the Democratic and Republican candidates in the last few presidential elections: 7 percentage points in 2008; 2.5 percentage points in 2004; 0.5 percentage points in 2000; 8.5 percentage points in 1996; and 5.5 percentage points in 1992. Some of those elections were close, others not as close--yet the largest difference was 8.5 percentage points. That puts a margin of error of 4 percentage points into perspective; it is meaningful.
What's more, most people only really care about polling when a race is relatively close. Generally a race is not really close unless it is separated by just a few percentage points--frequently within a 4 percentage point margin of error.
For the ABC survey we first mentioned above, the margin of error is only relevant for questions that include the poll's entire sample of 1,002 adults. The quoted result that gives Romney a 2 percentage point lead of Obama is of a subset of the full sample that distinguished themselves as registered voters. Thus, there is a higher margin of error on this main result of the poll.
The margin of error is the sampling error, which is just a portion of possible sources of error. The sampling error assumes that the pollster has taken a perfectly random selection of a perfectly selected representative sample of respondents.
But we don't really live in a randomly arranged universe--of political choices or anything else. There are numerous reasons why a selection becomes non-random. Once a polling company has chosen to contact a person at random, the poll may not reach the intended respondent--or that respondent may refuse to answer the poll at all. Varying polling techniques intended to select who in the household should answer the questions, time of day of contact, etc., all can, well, randomly backfire. And that, in turn, can dilute the poll's sample with unrepresentative selections.
A sample can fail to be representative, meanwhile, in a host of other ways. To quote the old shibboleth, the only poll that matters is on Election Day--the entire political process is in fact focused on the random sample of voters who actually turn out to vote. If we are lucky, at our present vantage of 13 months out from Election Day, we can isolate some self-identified registered voters who can be somewhat representative of the eventual November 2012 turnout. What's more, by the standards of other Western democracies, the United States has a notoriously complicated set of requirements and obstacles to casting a ballot on Election Day--and recent changes in many state election laws have become even more restrictive. Figuring out who will actually vote on Election Day is not obvious or easy.
Finally, there is an odd quirk in most of the the standard voter-intention questions that poll respondents typically answer. Pollsters will ask their respondents how they would vote if "the 2012 presidential election were being held today." Of course, the election is being held 13 months or so from this particular poll--and the voters, the candidates, and the overall dynamic of the election will shift dramatically between now and then.
So it's not at all surprising to see enormous daily fluctuations within polls--even when those surveys come from the most reliable sources. The chart below shows John McCain's two-party vote share, relative to Obama, for the 2008 presidential election.
Yet, despite this margin of error and the daily fluctuations, polls are a very meaningful source of information in understanding and predicting an election. You just need to be clear on what a poll means and how best to use it.
You can, for example, profitably aggregate a new poll with other, earlier polls; poll aggregation site such as Pollster and RealClearPolitics note the latest polls, but focus more on their aggregated presentation of the fuller range of polls. RealClearPolitics has Romney up by 0.5 points relative to Obama in aggregated head-to-head polls--and compared to the sampling and margin-of-error issues that can crop up in any individual poll, this approach provides a relatively stable metric.
You can also aggregate a new poll with other, earlier polls, and additional information about the election. In prediction market such as Betfair and Intrade users do just that as they trade contracts on who will win upcoming elections. The prediction markets do not have head-to-head match-ups prior the nominations, but real-time data has Obama at 48.7 percent likelihood of winning the election and Mitt Romney has a 29.2 percent likelihood of winning the election, with a 58.9 percent likelihood of winning the Republican nomination.
Editor's Note: I changed the wording in the second paragraph to reflect the vote shares of the candidates rather than the differences between the two candidates to avoid confusion over details that cannot be encapsulated in this post. The implied differences between the two candidates depend on a further set of assumptions that are not declared by the polling companies. More details will follow in a later post. Thanks to Stefan Herzog for a spirited conversation on this topic.
David Rothschild is an economist at Yahoo! Research. He has a PhD in applied economics from the Wharton School of Business at the University of Pennsylvania. His dissertation is in creating aggregated forecasts from individual-level information. You can follow him on Twitter @DavMicRot and email him at PredictionBlogger@Yahoo.com. | 1,265 | 6,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | longest | en | 0.951071 |
http://bunutugexiqobezab.mi-centre.com/writing-a-mathematical-proof-methods-8700887008.html | 1,544,384,882,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823009.19/warc/CC-MAIN-20181209185547-20181209211547-00380.warc.gz | 54,591,957 | 5,498 | # Writing a mathematical proof methods
Proofs as mental objects[ edit ] Main articles: I am rating this a 3 as it is therefore neutral in this regard. The primary goals of the text are to help students: A second animated proof of the Pythagorean theorem. Especially good are the sections where the author clarifies how to write a proof for your audience.
Two-column proof[ edit ] A two-column proof published in A particular way of organising a proof using two parallel columns is often used in elementary geometry classes in the United States. It is an easy-to-read pdf, of small size.
Statistical proof using data[ edit ] Main article: The mathematics in the book is correct. This abbreviation stands for "Quod Erat Demonstrandum", which is Latin for "that which was to be demonstrated". One example is the parallel postulatewhich is neither provable nor refutable from the remaining axioms of Euclidean geometry.
Inductive logic proofs and Bayesian analysis[ edit ] Main articles: Some mathematicians are concerned that the possibility of an error in a computer program or a run-time error in its calculations calls the validity of such computer-assisted proofs into question.
Students should be able to use this text with a background of one semester of calculus. Inductive logic and Bayesian analysis Proofs using inductive logicwhile considered mathematical in nature, seek to establish propositions with a degree of certainty, which acts in a similar manner to probabilityand may be less than full certainty.
For example, it is difficult to speak of correspondences without the notion of a function, but an instructor can simply introduce the function definition to address correspondences without covering the entire chapter on functions. In each line, the left-hand column contains a proposition, while the right-hand column writing a mathematical proof methods a brief explanation of how the corresponding proposition in the left-hand column is either an axiom, a hypothesis, or can be logically derived from previous propositions.
Ending a proof[ edit ] Main article: Errors can never be completely ruled out in case of verification of a proof by humans either, especially if the proof contains natural language and requires deep mathematical insight.
Statistical proof "Statistical proof" from data refers to the application of statisticsdata analysisor Bayesian analysis to infer propositions regarding the probability of data.
Mathematicians have shown there are many statements that are neither provable nor disprovable in Zermelo-Fraenkel set theory with the axiom of choice ZFCthe standard system of set theory in mathematics assuming that ZFC is consistent ; see list of statements undecidable in ZFC. See also " Statistical proof using data " section below.
Another important goal of this text is to provide students with material that will be needed for their further study of mathematics. Despite this, with a little extra effort by an instructor, most sections can be separated.
The examples lead the reader gently towards an understanding of logic and proof. Experimental mathematics While early mathematicians such as Eudoxus of Cnidus did not use proofs, from Euclid to the foundational mathematics developments of the late 19th and 20th centuries, proofs were an essential part of mathematics.
In physicsin addition to statistical methods, "statistical proof" can refer to the specialized mathematical methods of physics applied to analyze data in a particle physics experiment or observational study in physical cosmology.
Computer-assisted proof Until the twentieth century it was assumed that any proof could, in principle, be checked by a competent mathematician to confirm its validity. This type of course has now become a standard part of the mathematics major at many colleges and universities. For some time it was thought that certain theorems, like the prime number theoremcould only be proved using "higher" mathematics.
The transition is from the problem-solving orientation of calculus to the more abstract and theoretical upper-level courses. Some illusory visual proofs, such as the missing square puzzlecan be constructed in a way which appear to prove a supposed mathematical fact but only do so under the presence of tiny errors for example, supposedly straight lines which actually bend slightly which are unnoticeable until the entire picture is closely examined, with lengths and angles precisely measured or calculated.
Visual proof[ edit ] Although not a formal proof, a visual demonstration of a mathematical theorem is sometimes called a " proof without words ". More specifically, the term is used in number theory to refer to proofs that make no use of complex analysis.
The left-hand picture below is an example of a historic visual proof of the Pythagorean theorem in the case of the 3,4,5 triangle. In practice, the chances of an error invalidating a computer-assisted proof can be reduced by incorporating redundancy and self-checks into calculations, and by developing multiple independent approaches and programs.
The left-hand column is typically headed "Statements" and the right-hand column is typically headed "Reasons". Sometimes, the abbreviation "Q. Statistical proof The expression "statistical proof" may be used technically or colloquially in areas of pure mathematicssuch as involving cryptographychaotic seriesand probabilistic or analytic number theory.Ten Tips for Writing Mathematical Proofs Katharine Ott 1.
Determine exactly what information you are given (also called the hypothesis)andwhat you are trying to. How to write proofs: a quick guide Eugenia Cheng Department of Mathematics, University of Chicago One of the di cult things about writing a proof is that the order in which we write it is often not the order in which we thought it up.
In fact, we often think up the proof but in mathematics if you use the wrong means to get to the right. May 14, · Three Methods: Understanding the Problem Formatting a Proof Writing the Proof Community Q&A Mathematical proofs can be difficult, but can be conquered with the proper background knowledge of both mathematics and the format of a proof%(17).
Develop the ability to construct and write mathematical proofs using standard methods of mathematical proof including direct proofs, proof by contradiction, mathematical induction, case analysis, and counterexamples.
Develop the ability to read and understand written mathematical proofs. Develop talents for creative thinking and problem solving. Mathematical Reasoning: Writing and Proof is designed to be a text for the first course in the college mathematics curriculum that introduces students to the pro- cesses of constructing and writing proofs and focuses on the formal development.
How to Read Mathematics-- Not exactly proof writing, but a helpful read for those learning to write basic proofs.
How To Prove It: A Structured Approach by Daniel J. Velleman -- an excellent primer on methods of proof; train your ability to do proofs by induction, contradiction and more.
Writing a mathematical proof methods
Rated 4/5 based on 65 review | 1,362 | 7,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.91289 |
https://goprep.co/when-a-horse-pulls-a-cart-the-force-that-helps-the-horse-to-i-1nm3n5 | 1,603,808,657,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00270.warc.gz | 329,836,987 | 42,574 | # When a horse pulls a cart, the force that helps the horse to move forward is the force exerted byA. the cart on the horseB. the ground on the horseC. the ground on the cartD. the horse on the ground
When the horse pulls the wagon, it pushes the ground in the backward direction. Due to the third law of motion, the ground exerts a forward force which has the same magnitude. As the backward force applied by the wagon is weaker than the force applied by the horse, the wagon moves in the forward direction.
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https://123hpcomsetup.net/what-are-gpa-credits/ | 1,679,491,908,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00498.warc.gz | 99,409,738 | 14,375 | # 13+ Easy Ways What Are Gpa Credits
13+ Easy Ways What Are Gpa Credits. The student’s grade point average, or gpa, is calculated by dividing the total points (30.000) by the divisor (15). These include only credits in which a grade has been earned. A=5.0, b=4.0, c=3.0, d=2.0 and. To find the semester gpa, add all of the total points (57) and divide by the number of credit hours (17).
To find the semester gpa, add all of the total points (57) and divide by the number of credit hours (17). To compute the grade point average, multiply each grade by its assigned gpa credit, sum the results, and divide by the sum. Credits is simply the sum total of the credit value of each class.
## Multiply each numeric grade value by the number of course credits;
Contents
Gpa is the quantified result by ritsumeikan university’s unique conversion method to measure students’ academic achievement. Gpa units, also known as ‘credits’. Each alphabetical grade is substituted by numerical equivalent:
## Gpa Is The Quantified Result By Ritsumeikan University’s Unique Conversion Method To Measure Students’ Academic Achievement.
Multiply the number of credit hours earned for each class by the numerical equivalent point for the class.
## Conclusion of 13+ Easy Ways What Are Gpa Credits.
This is how we find out your gpa is 3.14. To find your weighted gpa by credit hours, do the following: 57 ÷ 17 = 3.35 gpa. To find the semester gpa, add all of the total points (57) and divide by the number of credit hours (17). | 370 | 1,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | latest | en | 0.906332 |
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#1
05-29-2004, 02:09 AM
1MajorTom Senior Moderator Join Date: Mar 2000 Location: Pennsylvania Posts: 6,097
How To Determine How Much Mulch Is Needed
This question seems to be a popular one in this forum.
A lot of people want to know how to figure out the correct amount of mulch for their landscaping project. So D Felix came up with a good idea to have the formula stuck to the top of this forum.
Here it is:
To determine how much mulch is needed, you take:
Total square feet x depth desired (in feet) / 27 = cubic yards needed
Take square footage time depth desired (in feet, i.e. 3" = .25 feet).
That gives volume in cubic feet.
Divide that number by 27 to get cubic yards. There's 27 cubic feet per yard (3'x3'x3'=27 cubic feet).
#2
05-29-2004, 06:19 PM
wonderwoman LawnSite Member Join Date: May 2004 Location: murfreeboro,tenn Posts: 13
thats pretty good try it out tomorrow
#3
05-30-2004, 06:21 PM
tiedeman LawnSite Fanatic Join Date: Jan 2003 Location: earth Posts: 8,753
thanks Jodi for the information
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I use Gopher Billing & Scheduling Software
#4
05-30-2004, 08:54 PM
bobbygedd LawnSite Fanatic Join Date: Nov 2000 Location: NJ Posts: 10,188
duhhhhhh!!!!!! if u can't figure it out, don't do it
#5
05-31-2004, 03:18 PM
Ric LawnSite Fanatic Join Date: Sep 2001 Location: S W Florida Posts: 11,099
Quote:
Originally posted by bobbygedd duhhhhhh!!!!!! if u can't figure it out, don't do it
BooBY
Just for you I will make it easier. That way you can do it.
1 Yard of Mulch 3" deep will cover 108 sq feet
1 Yard of Mulch 4" deep will cover 81 sq feet
1 Bag 2 cu feet at 3" will cover 8 sq feet
1 Bag 2 cu feet at 4" will cover 6 sq feet
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#6
06-09-2004, 01:18 PM
EagleLandscape LawnSite Platinum Member Join Date: Aug 2003 Location: Garland, Texas Posts: 4,340
1 Bag 3 cu feet at 2" will cover 18 sq feet
1 Bag 3 cu feet at 1" will cover 36 sq feet
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#7
06-21-2004, 02:51 PM
EagleLandscape LawnSite Platinum Member Join Date: Aug 2003 Location: Garland, Texas Posts: 4,340
Here is a table that will do it for ya...
http://www.soilbuildingsystems.com/CubicYard.htm
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John Wingfield Jr.
Eagle Landscape and Irrigation
www.eagle-landscape.com
www.eagle-irrigation.com
"We are what we repeatedly do. Excellence, therefore, is not an act, but a habit." - Aristotle
#8
08-02-2004, 01:23 PM
kris LawnSite Bronze Member Join Date: Apr 2001 Location: nowhere Posts: 1,833
metric is so much easier
#9
08-04-2004, 09:34 AM
D Felix LawnSite Bronze Member Join Date: Jan 2004 Location: Putnam County, Indiana Posts: 1,906
Not when you are used to the *right* way of doing it.
And especially not when your suppliers handle stuff that way too.
Dan
#10
09-09-2004, 09:59 AM
tx_angler LawnSite Senior Member Join Date: Sep 2003 Location: Texas-Southeast DFW Posts: 278
If you use this formula you don't have to convert the depth from inches to feet:
Length (in Feet) X Width (in Feet) X Depth (in Inches) / 324 = Cubic Yards
Example:
Length=10'
Width=20'
Depth=3"
10x20x3/324=1.8518518518518518518518518518519 cubic yards
or
Using the decimal equilivant of 3 inches (.25 feet):
Length=10'
Width=20'
Depth=.25'
10x20x.25/27=1.8518518518518518518518518518519 cubic yards
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# It is an historical fact that only in conditions of profound
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Re: CR 700 level Rome [#permalink]
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26 Jul 2011, 11:34
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Thanks fluke that was a great explanation.
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Re: CR 700 level Rome [#permalink]
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26 Jul 2011, 13:36
+1 C
Profound societal instability doesn't guarantee that always there will be great works of art.
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Re: CR 700 level Rome [#permalink]
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26 Jul 2011, 19:25
If P only then Q is possible => If Not P then Not Q which is (C), answer
A) is if P then Q as a rule which isnt so.
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Re: CR 700 level Rome [#permalink]
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26 Jul 2011, 19:32
I thought it was A at first too, but the first statement was ONLY in times of instability. A assumes that in ALL times of instability, where as C states that in times of peace, good literature cannot take place. which agrees with that first statement.
ONLY vs ALL would have changed this question entirely, hope this helps
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Re: CR 700 level Rome [#permalink]
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07 Aug 2011, 03:13
"C" for me .. tough one
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Re: CR 700 level Rome [#permalink]
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12 Sep 2011, 10:12
I think C.......
A is not properly MUST BE TRUE
B out
C correct
D out
E we are talking of works not literature production
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Re: CR 700 level Rome [#permalink]
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25 Sep 2011, 04:52
nice explanation through number system!!
Kudos for that!!
Soul!!
fluke wrote:
bschool83 wrote:
It is an historical fact that only in conditions of profound societal instability are great works of literature produced. During the first century B.C., Rome experienced almost constant civil war accompanied by social upheaval. It wasn't until the ascension of Nerva to the throne in 96 A.D. that the situation stabilized. Throughout the second century A.D., Rome experienced a century of uninterrupted peace and stability.
If all of the statements in the passage are true, then which of the following must also be true?
a. During the first century B.C., great works of literature were produced in Rome.
b. Roman art of the first century B.C. was superior to Roman art of the second century A.D.
c. During the second century A.D., no great works of literature were produced in Rome.
d. Historically, great works of literature can stir emotions that find their expression in political action.
e. The first century A.D. was a time of decreased literary production in Rome.
"Only in great instability are great works of literature produced."
If there is a great work of Lit, it must have been produced during extreme instability.
Reverse may not be true;
If there is great instability, then great works must be created.
If Rome experienced profound instability during all odd centuries, 1st century, 3rd century, 5th, 7th, 9th...
And it was prosperous and stable during even centuries: 2nd, 4th, 6th, 8th, 10th...
There are two 2 great epics/works. These two epics must have been produced during odd centuries. Maybe 5th and 7th. But, there is no guarantee that whenever there was instability, great artwork was created, OR we'd have had at least 10 works till date.
Also when it is said that the great artwork is created ONLY during bad periods, it is equivalent to saying great artwork is NEVER created during good periods. And we know 2nd century was a period of stability for Rome.
Ans: "C"
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Re: CR 700 level Rome [#permalink]
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26 Sep 2011, 07:07
O overlooked the word 'only', so got trapped in A.
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Re: CR 700 level Rome [#permalink]
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26 Sep 2011, 22:05
fluke wrote:
bschool83 wrote:
It is an historical fact that only in conditions of profound societal instability are great works of literature produced. During the first century B.C., Rome experienced almost constant civil war accompanied by social upheaval. It wasn't until the ascension of Nerva to the throne in 96 A.D. that the situation stabilized. Throughout the second century A.D., Rome experienced a century of uninterrupted peace and stability.
If all of the statements in the passage are true, then which of the following must also be true?
a. During the first century B.C., great works of literature were produced in Rome.
b. Roman art of the first century B.C. was superior to Roman art of the second century A.D.
c. During the second century A.D., no great works of literature were produced in Rome.
d. Historically, great works of literature can stir emotions that find their expression in political action.
e. The first century A.D. was a time of decreased literary production in Rome.
"Only in great instability are great works of literature produced."
If there is a great work of Lit, it must have been produced during extreme instability.
Reverse may not be true;
If there is great instability, then great works must be created.
If Rome experienced profound instability during all odd centuries, 1st century, 3rd century, 5th, 7th, 9th...
And it was prosperous and stable during even centuries: 2nd, 4th, 6th, 8th, 10th...
There are two 2 great epics/works. These two epics must have been produced during odd centuries. Maybe 5th and 7th. But, there is no guarantee that whenever there was instability, great artwork was created, OR we'd have had at least 10 works till date.
Also when it is said that the great artwork is created ONLY during bad periods, it is equivalent to saying great artwork is NEVER created during good periods. And we know 2nd century was a period of stability for Rome.
Ans: "C"
Good explanation fluke. Kudos to you!!
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Re: It is an historical fact that only in conditions of profound [#permalink]
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10 Aug 2012, 11:40
Yeah was confused between A and C , picked A but it is wrong , C is right.
basically works of great art are produced during period of instability ,implying if a great work of art is produced the period must have been unstable one ,not that in all periods of instability you will get great work of art. Suppose all great artists decide not to produce any work during unstable times one will not have any work, but if they do produce it, the period in all probabilities is unstable.
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Re: It is an historical fact that only in conditions of profound [#permalink]
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25 Oct 2012, 05:13
Can we say like this?
They have mentioned about 96 A.D in question and in answer option A and B they are asking about first century of BC. Which is not related to each other..
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Re: It is an historical fact that only in conditions of profound [#permalink]
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25 Oct 2012, 07:20
is this not a must be true question?
if you look at it like a must be true question, then answer looks like it is A
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Re: It is an historical fact that only in conditions of profound [#permalink] 25 Oct 2012, 07:20
Display posts from previous: Sort by | 2,830 | 10,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-43 | latest | en | 0.8316 |
https://scikit-image.org/docs/stable/api/skimage.transform | 1,669,530,804,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00478.warc.gz | 550,988,172 | 37,356 | # Module: transform¶
Down-sample N-dimensional image by local averaging. Estimate 2D geometric transformation parameters. Compute the 2-dimensional finite radon transform (FRT) for an n x n integer array. skimage.transform.hough_circle(image, radius) Perform a circular Hough transform. Return peaks in a circle Hough transform. skimage.transform.hough_ellipse(image[, …]) Perform an elliptical Hough transform. skimage.transform.hough_line(image[, theta]) Perform a straight line Hough transform. skimage.transform.hough_line_peaks(hspace, …) Return peaks in a straight line Hough transform. Compute the 2-dimensional inverse finite radon transform (iFRT) for an (n+1) x n integer array. skimage.transform.integral_image(image, *[, …]) Integral image / summed area table. skimage.transform.integrate(ii, start, end) Use an integral image to integrate over a given window. skimage.transform.iradon(radon_image[, …]) Inverse radon transform. skimage.transform.iradon_sart(radon_image[, …]) Inverse radon transform. skimage.transform.matrix_transform(coords, …) Apply 2D matrix transform. Order angles to reduce the amount of correlated information in subsequent projections. Return lines from a progressive probabilistic line Hough transform. skimage.transform.pyramid_expand(image[, …]) Upsample and then smooth image. skimage.transform.pyramid_gaussian(image[, …]) Yield images of the Gaussian pyramid formed by the input image. skimage.transform.pyramid_laplacian(image[, …]) Yield images of the laplacian pyramid formed by the input image. skimage.transform.pyramid_reduce(image[, …]) Smooth and then downsample image. skimage.transform.radon(image[, theta, …]) Calculates the radon transform of an image given specified projection angles. skimage.transform.rescale(image, scale[, …]) Scale image by a certain factor. skimage.transform.resize(image, output_shape) Resize image to match a certain size. Resize an array with the local mean / bilinear scaling. skimage.transform.rotate(image, angle[, …]) Rotate image by a certain angle around its center. skimage.transform.swirl(image[, center, …]) Perform a swirl transformation. skimage.transform.warp(image, inverse_map[, …]) Warp an image according to a given coordinate transformation. skimage.transform.warp_coords(coord_map, shape) Build the source coordinates for the output of a 2-D image warp. skimage.transform.warp_polar(image[, …]) Remap image to polar or log-polar coordinates space. skimage.transform.AffineTransform([matrix, …]) Affine transformation. Essential matrix transformation. Euclidean transformation, also known as a rigid transform. Fundamental matrix transformation. Piecewise affine transformation. 2D polynomial transformation. Projective transformation. 2D similarity transformation.
## downscale_local_mean¶
skimage.transform.downscale_local_mean(image, factors, cval=0, clip=True)[source]
Down-sample N-dimensional image by local averaging.
The image is padded with cval if it is not perfectly divisible by the integer factors.
In contrast to interpolation in skimage.transform.resize and skimage.transform.rescale this function calculates the local mean of elements in each block of size factors in the input image.
Parameters
imagendarray
N-dimensional input image.
factorsarray_like
Array containing down-sampling integer factor along each axis.
cvalfloat, optional
Constant padding value if image is not perfectly divisible by the integer factors.
clipbool, optional
Unused, but kept here for API consistency with the other transforms in this module. (The local mean will never fall outside the range of values in the input image, assuming the provided cval also falls within that range.)
Returns
imagendarray
Down-sampled image with same number of dimensions as input image. For integer inputs, the output dtype will be float64. See numpy.mean() for details.
Examples
>>> a = np.arange(15).reshape(3, 5)
>>> a
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
>>> downscale_local_mean(a, (2, 3))
array([[3.5, 4. ],
[5.5, 4.5]])
## estimate_transform¶
skimage.transform.estimate_transform(ttype, src, dst, *args, **kwargs)[source]
Estimate 2D geometric transformation parameters.
You can determine the over-, well- and under-determined parameters with the total least-squares method.
Number of source and destination coordinates must match.
Parameters
ttype{‘euclidean’, similarity’, ‘affine’, ‘piecewise-affine’, ‘projective’, ‘polynomial’}
Type of transform.
kwargsarray or int
Function parameters (src, dst, n, angle):
NAME / TTYPE FUNCTION PARAMETERS
'euclidean' src, dst
'similarity' src, dst
'affine' src, dst
'piecewise-affine' src, dst
'projective' src, dst
'polynomial' src, dst, order (polynomial order,
default order is 2)
Also see examples below.
Returns
tformGeometricTransform
Transform object containing the transformation parameters and providing access to forward and inverse transformation functions.
Examples
>>> import numpy as np
>>> from skimage import transform
>>> # estimate transformation parameters
>>> src = np.array([0, 0, 10, 10]).reshape((2, 2))
>>> dst = np.array([12, 14, 1, -20]).reshape((2, 2))
>>> tform = transform.estimate_transform('similarity', src, dst)
>>> np.allclose(tform.inverse(tform(src)), src)
True
>>> # warp image using the estimated transformation
>>> from skimage import data
>>> image = data.camera()
>>> warp(image, inverse_map=tform.inverse)
>>> # create transformation with explicit parameters
>>> tform2 = transform.SimilarityTransform(scale=1.1, rotation=1,
... translation=(10, 20))
>>> # unite transformations, applied in order from left to right
>>> tform3 = tform + tform2
>>> np.allclose(tform3(src), tform2(tform(src)))
True
## frt2¶
skimage.transform.frt2(a)[source]
Compute the 2-dimensional finite radon transform (FRT) for an n x n integer array.
Parameters
aarray_like
A 2-D square n x n integer array.
Returns
FRT2-D ndarray
Finite Radon Transform array of (n+1) x n integer coefficients.
ifrt2
The two-dimensional inverse FRT.
Notes
The FRT has a unique inverse if and only if n is prime. [FRT] The idea for this algorithm is due to Vlad Negnevitski.
References
FRT
A. Kingston and I. Svalbe, “Projective transforms on periodic discrete image arrays,” in P. Hawkes (Ed), Advances in Imaging and Electron Physics, 139 (2006)
Examples
Generate a test image: Use a prime number for the array dimensions
>>> SIZE = 59
>>> img = np.tri(SIZE, dtype=np.int32)
>>> f = frt2(img)
## hough_circle¶
skimage.transform.hough_circle(image, radius, normalize=True, full_output=False)[source]
Perform a circular Hough transform.
Parameters
image(M, N) ndarray
Input image with nonzero values representing edges.
Radii at which to compute the Hough transform. Floats are converted to integers.
normalizeboolean, optional (default True)
Normalize the accumulator with the number of pixels used to draw the radius.
full_outputboolean, optional (default False)
Extend the output size by twice the largest radius in order to detect centers outside the input picture.
Returns
H3D ndarray (radius index, (M + 2R, N + 2R) ndarray)
Hough transform accumulator for each radius. R designates the larger radius if full_output is True. Otherwise, R = 0.
Examples
>>> from skimage.transform import hough_circle
>>> from skimage.draw import circle_perimeter
>>> img = np.zeros((100, 100), dtype=bool)
>>> rr, cc = circle_perimeter(25, 35, 23)
>>> img[rr, cc] = 1
>>> ridx, r, c = np.unravel_index(np.argmax(res), res.shape)
(25, 35, 23)
## hough_circle_peaks¶
skimage.transform.hough_circle_peaks(hspaces, radii, min_xdistance=1, min_ydistance=1, threshold=None, num_peaks=inf, total_num_peaks=inf, normalize=False)[source]
Return peaks in a circle Hough transform.
Identifies most prominent circles separated by certain distances in given Hough spaces. Non-maximum suppression with different sizes is applied separately in the first and second dimension of the Hough space to identify peaks. For circles with different radius but close in distance, only the one with highest peak is kept.
Parameters
hspaces(N, M) array
Hough spaces returned by the hough_circle function.
min_xdistanceint, optional
Minimum distance separating centers in the x dimension.
min_ydistanceint, optional
Minimum distance separating centers in the y dimension.
thresholdfloat, optional
Minimum intensity of peaks in each Hough space. Default is 0.5 * max(hspace).
num_peaksint, optional
Maximum number of peaks in each Hough space. When the number of peaks exceeds num_peaks, only num_peaks coordinates based on peak intensity are considered for the corresponding radius.
total_num_peaksint, optional
Maximum number of peaks. When the number of peaks exceeds num_peaks, return num_peaks coordinates based on peak intensity.
normalizebool, optional
If True, normalize the accumulator by the radius to sort the prominent peaks.
Returns
accum, cx, cy, radtuple of array
Peak values in Hough space, x and y center coordinates and radii.
Notes
Circles with bigger radius have higher peaks in Hough space. If larger circles are preferred over smaller ones, normalize should be False. Otherwise, circles will be returned in the order of decreasing voting number.
Examples
>>> from skimage import transform, draw
>>> img = np.zeros((120, 100), dtype=int)
>>> radius, x_0, y_0 = (20, 99, 50)
>>> y, x = draw.circle_perimeter(y_0, x_0, radius)
>>> img[x, y] = 1
## hough_ellipse¶
skimage.transform.hough_ellipse(image, threshold=4, accuracy=1, min_size=4, max_size=None)[source]
Perform an elliptical Hough transform.
Parameters
image(M, N) ndarray
Input image with nonzero values representing edges.
thresholdint, optional
Accumulator threshold value.
accuracydouble, optional
Bin size on the minor axis used in the accumulator.
min_sizeint, optional
Minimal major axis length.
max_sizeint, optional
Maximal minor axis length. If None, the value is set to the half of the smaller image dimension.
Returns
resultndarray with fields [(accumulator, yc, xc, a, b, orientation)].
Where (yc, xc) is the center, (a, b) the major and minor axes, respectively. The orientation value follows skimage.draw.ellipse_perimeter convention.
Notes
The accuracy must be chosen to produce a peak in the accumulator distribution. In other words, a flat accumulator distribution with low values may be caused by a too low bin size.
References
1
Xie, Yonghong, and Qiang Ji. “A new efficient ellipse detection method.” Pattern Recognition, 2002. Proceedings. 16th International Conference on. Vol. 2. IEEE, 2002
Examples
>>> from skimage.transform import hough_ellipse
>>> from skimage.draw import ellipse_perimeter
>>> img = np.zeros((25, 25), dtype=np.uint8)
>>> rr, cc = ellipse_perimeter(10, 10, 6, 8)
>>> img[cc, rr] = 1
>>> result = hough_ellipse(img, threshold=8)
>>> result.tolist()
[(10, 10.0, 10.0, 8.0, 6.0, 0.0)]
## hough_line¶
skimage.transform.hough_line(image, theta=None)[source]
Perform a straight line Hough transform.
Parameters
image(M, N) ndarray
Input image with nonzero values representing edges.
theta1D ndarray of double, optional
Angles at which to compute the transform, in radians. Defaults to a vector of 180 angles evenly spaced in the range [-pi/2, pi/2).
Returns
hspace2-D ndarray of uint64
Hough transform accumulator.
anglesndarray
Angles at which the transform is computed, in radians.
distancesndarray
Distance values.
Notes
The origin is the top left corner of the original image. X and Y axis are horizontal and vertical edges respectively. The distance is the minimal algebraic distance from the origin to the detected line. The angle accuracy can be improved by decreasing the step size in the theta array.
Examples
Generate a test image:
>>> img = np.zeros((100, 150), dtype=bool)
>>> img[30, :] = 1
>>> img[:, 65] = 1
>>> img[35:45, 35:50] = 1
>>> for i in range(90):
... img[i, i] = 1
>>> rng = np.random.default_rng()
>>> img += rng.random(img.shape) > 0.95
Apply the Hough transform:
>>> out, angles, d = hough_line(img)
import numpy as np
import matplotlib.pyplot as plt
from skimage.transform import hough_line
from skimage.draw import line
img = np.zeros((100, 150), dtype=bool)
img[30, :] = 1
img[:, 65] = 1
img[35:45, 35:50] = 1
rr, cc = line(60, 130, 80, 10)
img[rr, cc] = 1
rng = np.random.default_rng()
img += rng.random(img.shape) > 0.95
out, angles, d = hough_line(img)
fix, axes = plt.subplots(1, 2, figsize=(7, 4))
axes[0].imshow(img, cmap=plt.cm.gray)
axes[0].set_title('Input image')
d_step = 0.5 * np.diff(d).mean()
d[-1] + d_step, d[0] - d_step)
axes[1].imshow(out, cmap=plt.cm.bone, extent=bounds)
axes[1].set_title('Hough transform')
axes[1].set_xlabel('Angle (degree)')
axes[1].set_ylabel('Distance (pixel)')
plt.tight_layout()
plt.show()
## hough_line_peaks¶
skimage.transform.hough_line_peaks(hspace, angles, dists, min_distance=9, min_angle=10, threshold=None, num_peaks=inf)[source]
Return peaks in a straight line Hough transform.
Identifies most prominent lines separated by a certain angle and distance in a Hough transform. Non-maximum suppression with different sizes is applied separately in the first (distances) and second (angles) dimension of the Hough space to identify peaks.
Parameters
hspace(N, M) array
Hough space returned by the hough_line function.
angles(M,) array
Angles returned by the hough_line function. Assumed to be continuous. (angles[-1] - angles[0] == PI).
dists(N, ) array
Distances returned by the hough_line function.
min_distanceint, optional
Minimum distance separating lines (maximum filter size for first dimension of hough space).
min_angleint, optional
Minimum angle separating lines (maximum filter size for second dimension of hough space).
thresholdfloat, optional
Minimum intensity of peaks. Default is 0.5 * max(hspace).
num_peaksint, optional
Maximum number of peaks. When the number of peaks exceeds num_peaks, return num_peaks coordinates based on peak intensity.
Returns
accum, angles, diststuple of array
Peak values in Hough space, angles and distances.
Examples
>>> from skimage.transform import hough_line, hough_line_peaks
>>> from skimage.draw import line
>>> img = np.zeros((15, 15), dtype=bool)
>>> rr, cc = line(0, 0, 14, 14)
>>> img[rr, cc] = 1
>>> rr, cc = line(0, 14, 14, 0)
>>> img[cc, rr] = 1
>>> hspace, angles, dists = hough_line(img)
>>> hspace, angles, dists = hough_line_peaks(hspace, angles, dists)
>>> len(angles)
2
## ifrt2¶
skimage.transform.ifrt2(a)[source]
Compute the 2-dimensional inverse finite radon transform (iFRT) for an (n+1) x n integer array.
Parameters
aarray_like
A 2-D (n+1) row x n column integer array.
Returns
iFRT2-D n x n ndarray
Inverse Finite Radon Transform array of n x n integer coefficients.
frt2
The two-dimensional FRT
Notes
The FRT has a unique inverse if and only if n is prime. See [1] for an overview. The idea for this algorithm is due to Vlad Negnevitski.
References
1
A. Kingston and I. Svalbe, “Projective transforms on periodic discrete image arrays,” in P. Hawkes (Ed), Advances in Imaging and Electron Physics, 139 (2006)
Examples
>>> SIZE = 59
>>> img = np.tri(SIZE, dtype=np.int32)
>>> f = frt2(img)
Apply the Inverse Finite Radon Transform to recover the input
>>> fi = ifrt2(f)
Check that it’s identical to the original
>>> assert len(np.nonzero(img-fi)[0]) == 0
## integral_image¶
skimage.transform.integral_image(image, *, dtype=None)[source]
Integral image / summed area table.
The integral image contains the sum of all elements above and to the left of it, i.e.:
$S[m, n] = \sum_{i \leq m} \sum_{j \leq n} X[i, j]$
Parameters
imagendarray
Input image.
Returns
Sndarray
Integral image/summed area table of same shape as input image.
Notes
For better accuracy and to avoid potential overflow, the data type of the output may differ from the input’s when the default dtype of None is used. For inputs with integer dtype, the behavior matches that for numpy.cumsum(). Floating point inputs will be promoted to at least double precision. The user can set dtype to override this behavior.
References
1
F.C. Crow, “Summed-area tables for texture mapping,” ACM SIGGRAPH Computer Graphics, vol. 18, 1984, pp. 207-212.
## integrate¶
skimage.transform.integrate(ii, start, end)[source]
Use an integral image to integrate over a given window.
Parameters
iindarray
Integral image.
startList of tuples, each tuple of length equal to dimension of ii
Coordinates of top left corner of window(s). Each tuple in the list contains the starting row, col, … index i.e [(row_win1, col_win1, …), (row_win2, col_win2,…), …].
endList of tuples, each tuple of length equal to dimension of ii
Coordinates of bottom right corner of window(s). Each tuple in the list containing the end row, col, … index i.e [(row_win1, col_win1, …), (row_win2, col_win2, …), …].
Returns
Sscalar or ndarray
Integral (sum) over the given window(s).
Examples
>>> arr = np.ones((5, 6), dtype=float)
>>> ii = integral_image(arr)
>>> integrate(ii, (1, 0), (1, 2)) # sum from (1, 0) to (1, 2)
array([3.])
>>> integrate(ii, [(3, 3)], [(4, 5)]) # sum from (3, 3) to (4, 5)
array([6.])
>>> # sum from (1, 0) to (1, 2) and from (3, 3) to (4, 5)
>>> integrate(ii, [(1, 0), (3, 3)], [(1, 2), (4, 5)])
array([3., 6.])
skimage.transform.iradon(radon_image, theta=None, output_size=None, filter_name='ramp', interpolation='linear', circle=True, preserve_range=True)[source]
Reconstruct an image from the radon transform, using the filtered back projection algorithm.
Parameters
Image containing radon transform (sinogram). Each column of the image corresponds to a projection along a different angle. The tomography rotation axis should lie at the pixel index radon_image.shape[0] // 2 along the 0th dimension of radon_image.
thetaarray_like, optional
Reconstruction angles (in degrees). Default: m angles evenly spaced between 0 and 180 (if the shape of radon_image is (N, M)).
output_sizeint, optional
Number of rows and columns in the reconstruction.
filter_namestr, optional
Filter used in frequency domain filtering. Ramp filter used by default. Filters available: ramp, shepp-logan, cosine, hamming, hann. Assign None to use no filter.
interpolationstr, optional
Interpolation method used in reconstruction. Methods available: ‘linear’, ‘nearest’, and ‘cubic’ (‘cubic’ is slow).
circleboolean, optional
Assume the reconstructed image is zero outside the inscribed circle. Also changes the default output_size to match the behaviour of radon called with circle=True.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
Returns
reconstructedndarray
Reconstructed image. The rotation axis will be located in the pixel with indices (reconstructed.shape[0] // 2, reconstructed.shape[1] // 2).
Changed in version 0.19: In iradon, filter argument is deprecated in favor of filter_name.
Notes
It applies the Fourier slice theorem to reconstruct an image by multiplying the frequency domain of the filter with the FFT of the projection data. This algorithm is called filtered back projection.
References
1
AC Kak, M Slaney, “Principles of Computerized Tomographic Imaging”, IEEE Press 1988.
2
B.R. Ramesh, N. Srinivasa, K. Rajgopal, “An Algorithm for Computing the Discrete Radon Transform With Some Applications”, Proceedings of the Fourth IEEE Region 10 International Conference, TENCON ‘89, 1989
### Examples using skimage.transform.iradon¶
skimage.transform.iradon_sart(radon_image, theta=None, image=None, projection_shifts=None, clip=None, relaxation=0.15, dtype=None)[source]
Reconstruct an image from the radon transform, using a single iteration of the Simultaneous Algebraic Reconstruction Technique (SART) algorithm.
Parameters
Image containing radon transform (sinogram). Each column of the image corresponds to a projection along a different angle. The tomography rotation axis should lie at the pixel index radon_image.shape[0] // 2 along the 0th dimension of radon_image.
theta1D array, optional
Reconstruction angles (in degrees). Default: m angles evenly spaced between 0 and 180 (if the shape of radon_image is (N, M)).
image2D array, optional
Image containing an initial reconstruction estimate. Shape of this array should be (radon_image.shape[0], radon_image.shape[0]). The default is an array of zeros.
projection_shifts1D array, optional
Shift the projections contained in radon_image (the sinogram) by this many pixels before reconstructing the image. The i’th value defines the shift of the i’th column of radon_image.
cliplength-2 sequence of floats, optional
Force all values in the reconstructed tomogram to lie in the range [clip[0], clip[1]]
relaxationfloat, optional
Relaxation parameter for the update step. A higher value can improve the convergence rate, but one runs the risk of instabilities. Values close to or higher than 1 are not recommended.
dtypedtype, optional
Output data type, must be floating point. By default, if input data type is not float, input is cast to double, otherwise dtype is set to input data type.
Returns
reconstructedndarray
Reconstructed image. The rotation axis will be located in the pixel with indices (reconstructed.shape[0] // 2, reconstructed.shape[1] // 2).
Notes
Algebraic Reconstruction Techniques are based on formulating the tomography reconstruction problem as a set of linear equations. Along each ray, the projected value is the sum of all the values of the cross section along the ray. A typical feature of SART (and a few other variants of algebraic techniques) is that it samples the cross section at equidistant points along the ray, using linear interpolation between the pixel values of the cross section. The resulting set of linear equations are then solved using a slightly modified Kaczmarz method.
When using SART, a single iteration is usually sufficient to obtain a good reconstruction. Further iterations will tend to enhance high-frequency information, but will also often increase the noise.
References
1
AC Kak, M Slaney, “Principles of Computerized Tomographic Imaging”, IEEE Press 1988.
2
AH Andersen, AC Kak, “Simultaneous algebraic reconstruction technique (SART): a superior implementation of the ART algorithm”, Ultrasonic Imaging 6 pp 81–94 (1984)
3
S Kaczmarz, “Angenäherte auflösung von systemen linearer gleichungen”, Bulletin International de l’Academie Polonaise des Sciences et des Lettres 35 pp 355–357 (1937)
4
Kohler, T. “A projection access scheme for iterative reconstruction based on the golden section.” Nuclear Science Symposium Conference Record, 2004 IEEE. Vol. 6. IEEE, 2004.
5
Kaczmarz’ method, Wikipedia, https://en.wikipedia.org/wiki/Kaczmarz_method
## matrix_transform¶
skimage.transform.matrix_transform(coords, matrix)[source]
Apply 2D matrix transform.
Parameters
coords(N, 2) array
x, y coordinates to transform
matrix(3, 3) array
Homogeneous transformation matrix.
Returns
coords(N, 2) array
Transformed coordinates.
## order_angles_golden_ratio¶
skimage.transform.order_angles_golden_ratio(theta)[source]
Order angles to reduce the amount of correlated information in subsequent projections.
Parameters
theta1D array of floats
Projection angles in degrees. Duplicate angles are not allowed.
Returns
indices_generatorgenerator yielding unsigned integers
The returned generator yields indices into theta such that theta[indices] gives the approximate golden ratio ordering of the projections. In total, len(theta) indices are yielded. All non-negative integers < len(theta) are yielded exactly once.
Notes
The method used here is that of the golden ratio introduced by T. Kohler.
References
1
Kohler, T. “A projection access scheme for iterative reconstruction based on the golden section.” Nuclear Science Symposium Conference Record, 2004 IEEE. Vol. 6. IEEE, 2004.
2
Winkelmann, Stefanie, et al. “An optimal radial profile order based on the Golden Ratio for time-resolved MRI.” Medical Imaging, IEEE Transactions on 26.1 (2007): 68-76.
## probabilistic_hough_line¶
skimage.transform.probabilistic_hough_line(image, threshold=10, line_length=50, line_gap=10, theta=None, seed=None)[source]
Return lines from a progressive probabilistic line Hough transform.
Parameters
image(M, N) ndarray
Input image with nonzero values representing edges.
thresholdint, optional
Threshold
line_lengthint, optional
Minimum accepted length of detected lines. Increase the parameter to extract longer lines.
line_gapint, optional
Maximum gap between pixels to still form a line. Increase the parameter to merge broken lines more aggressively.
theta1D ndarray, dtype=double, optional
Angles at which to compute the transform, in radians. Defaults to a vector of 180 angles evenly spaced in the range [-pi/2, pi/2).
seedint, optional
Seed to initialize the random number generator.
Returns
lineslist
List of lines identified, lines in format ((x0, y0), (x1, y1)), indicating line start and end.
References
1
C. Galamhos, J. Matas and J. Kittler, “Progressive probabilistic Hough transform for line detection”, in IEEE Computer Society Conference on Computer Vision and Pattern Recognition, 1999.
## pyramid_expand¶
skimage.transform.pyramid_expand(image, upscale=2, sigma=None, order=1, mode='reflect', cval=0, multichannel=False, preserve_range=False, *, channel_axis=None)[source]
Upsample and then smooth image.
Parameters
imagendarray
Input image.
upscalefloat, optional
Upscale factor.
sigmafloat, optional
Sigma for Gaussian filter. Default is 2 * upscale / 6.0 which corresponds to a filter mask twice the size of the scale factor that covers more than 99% of the Gaussian distribution.
orderint, optional
Order of splines used in interpolation of upsampling. See skimage.transform.warp for detail.
mode{‘reflect’, ‘constant’, ‘edge’, ‘symmetric’, ‘wrap’}, optional
The mode parameter determines how the array borders are handled, where cval is the value when mode is equal to ‘constant’.
cvalfloat, optional
Value to fill past edges of input if mode is ‘constant’.
multichannelbool, optional
Whether the last axis of the image is to be interpreted as multiple channels or another spatial dimension. This argument is deprecated: specify channel_axis instead.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
channel_axisint or None, optional
If None, the image is assumed to be a grayscale (single channel) image. Otherwise, this parameter indicates which axis of the array corresponds to channels.
New in version 0.19: channel_axis was added in 0.19.
Returns
outarray
Upsampled and smoothed float image.
Other Parameters
multichannelDEPRECATED
Deprecated in favor of channel_axis.
Deprecated since version 0.19.
References
1
http://persci.mit.edu/pub_pdfs/pyramid83.pdf
## pyramid_gaussian¶
skimage.transform.pyramid_gaussian(image, max_layer=- 1, downscale=2, sigma=None, order=1, mode='reflect', cval=0, multichannel=False, preserve_range=False, *, channel_axis=None)[source]
Yield images of the Gaussian pyramid formed by the input image.
Recursively applies the pyramid_reduce function to the image, and yields the downscaled images.
Note that the first image of the pyramid will be the original, unscaled image. The total number of images is max_layer + 1. In case all layers are computed, the last image is either a one-pixel image or the image where the reduction does not change its shape.
Parameters
imagendarray
Input image.
max_layerint, optional
Number of layers for the pyramid. 0th layer is the original image. Default is -1 which builds all possible layers.
downscalefloat, optional
Downscale factor.
sigmafloat, optional
Sigma for Gaussian filter. Default is 2 * downscale / 6.0 which corresponds to a filter mask twice the size of the scale factor that covers more than 99% of the Gaussian distribution.
orderint, optional
Order of splines used in interpolation of downsampling. See skimage.transform.warp for detail.
mode{‘reflect’, ‘constant’, ‘edge’, ‘symmetric’, ‘wrap’}, optional
The mode parameter determines how the array borders are handled, where cval is the value when mode is equal to ‘constant’.
cvalfloat, optional
Value to fill past edges of input if mode is ‘constant’.
multichannelbool, optional
Whether the last axis of the image is to be interpreted as multiple channels or another spatial dimension. This argument is deprecated: specify channel_axis instead.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
channel_axisint or None, optional
If None, the image is assumed to be a grayscale (single channel) image. Otherwise, this parameter indicates which axis of the array corresponds to channels.
New in version 0.19: channel_axis was added in 0.19.
Returns
pyramidgenerator
Generator yielding pyramid layers as float images.
Other Parameters
multichannelDEPRECATED
Deprecated in favor of channel_axis.
Deprecated since version 0.19.
References
1
http://persci.mit.edu/pub_pdfs/pyramid83.pdf
## pyramid_laplacian¶
skimage.transform.pyramid_laplacian(image, max_layer=- 1, downscale=2, sigma=None, order=1, mode='reflect', cval=0, multichannel=False, preserve_range=False, *, channel_axis=None)[source]
Yield images of the laplacian pyramid formed by the input image.
Each layer contains the difference between the downsampled and the downsampled, smoothed image:
layer = resize(prev_layer) - smooth(resize(prev_layer))
Note that the first image of the pyramid will be the difference between the original, unscaled image and its smoothed version. The total number of images is max_layer + 1. In case all layers are computed, the last image is either a one-pixel image or the image where the reduction does not change its shape.
Parameters
imagendarray
Input image.
max_layerint, optional
Number of layers for the pyramid. 0th layer is the original image. Default is -1 which builds all possible layers.
downscalefloat, optional
Downscale factor.
sigmafloat, optional
Sigma for Gaussian filter. Default is 2 * downscale / 6.0 which corresponds to a filter mask twice the size of the scale factor that covers more than 99% of the Gaussian distribution.
orderint, optional
Order of splines used in interpolation of downsampling. See skimage.transform.warp for detail.
mode{‘reflect’, ‘constant’, ‘edge’, ‘symmetric’, ‘wrap’}, optional
The mode parameter determines how the array borders are handled, where cval is the value when mode is equal to ‘constant’.
cvalfloat, optional
Value to fill past edges of input if mode is ‘constant’.
multichannelbool, optional
Whether the last axis of the image is to be interpreted as multiple channels or another spatial dimension. This argument is deprecated: specify channel_axis instead.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
channel_axisint or None, optional
If None, the image is assumed to be a grayscale (single channel) image. Otherwise, this parameter indicates which axis of the array corresponds to channels.
New in version 0.19: channel_axis was added in 0.19.
Returns
pyramidgenerator
Generator yielding pyramid layers as float images.
Other Parameters
multichannelDEPRECATED
Deprecated in favor of channel_axis.
Deprecated since version 0.19.
References
1
http://persci.mit.edu/pub_pdfs/pyramid83.pdf
2
http://sepwww.stanford.edu/data/media/public/sep/morgan/texturematch/paper_html/node3.html
## pyramid_reduce¶
skimage.transform.pyramid_reduce(image, downscale=2, sigma=None, order=1, mode='reflect', cval=0, multichannel=False, preserve_range=False, *, channel_axis=None)[source]
Smooth and then downsample image.
Parameters
imagendarray
Input image.
downscalefloat, optional
Downscale factor.
sigmafloat, optional
Sigma for Gaussian filter. Default is 2 * downscale / 6.0 which corresponds to a filter mask twice the size of the scale factor that covers more than 99% of the Gaussian distribution.
orderint, optional
Order of splines used in interpolation of downsampling. See skimage.transform.warp for detail.
mode{‘reflect’, ‘constant’, ‘edge’, ‘symmetric’, ‘wrap’}, optional
The mode parameter determines how the array borders are handled, where cval is the value when mode is equal to ‘constant’.
cvalfloat, optional
Value to fill past edges of input if mode is ‘constant’.
multichannelbool, optional
Whether the last axis of the image is to be interpreted as multiple channels or another spatial dimension. This argument is deprecated: specify channel_axis instead.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
channel_axisint or None, optional
If None, the image is assumed to be a grayscale (single channel) image. Otherwise, this parameter indicates which axis of the array corresponds to channels.
New in version 0.19: channel_axis was added in 0.19.
Returns
outarray
Smoothed and downsampled float image.
Other Parameters
multichannelDEPRECATED
Deprecated in favor of channel_axis.
Deprecated since version 0.19.
References
1
http://persci.mit.edu/pub_pdfs/pyramid83.pdf
skimage.transform.radon(image, theta=None, circle=True, *, preserve_range=False)[source]
Calculates the radon transform of an image given specified projection angles.
Parameters
imagearray_like
Input image. The rotation axis will be located in the pixel with indices (image.shape[0] // 2, image.shape[1] // 2).
thetaarray_like, optional
Projection angles (in degrees). If None, the value is set to np.arange(180).
circleboolean, optional
Assume image is zero outside the inscribed circle, making the width of each projection (the first dimension of the sinogram) equal to min(image.shape).
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
Returns
Radon transform (sinogram). The tomography rotation axis will lie at the pixel index radon_image.shape[0] // 2 along the 0th dimension of radon_image.
Notes
Based on code of Justin K. Romberg (https://www.clear.rice.edu/elec431/projects96/DSP/bpanalysis.html)
References
1
AC Kak, M Slaney, “Principles of Computerized Tomographic Imaging”, IEEE Press 1988.
2
B.R. Ramesh, N. Srinivasa, K. Rajgopal, “An Algorithm for Computing the Discrete Radon Transform With Some Applications”, Proceedings of the Fourth IEEE Region 10 International Conference, TENCON ‘89, 1989
## rescale¶
skimage.transform.rescale(image, scale, order=None, mode='reflect', cval=0, clip=True, preserve_range=False, multichannel=False, anti_aliasing=None, anti_aliasing_sigma=None, *, channel_axis=None)[source]
Scale image by a certain factor.
Performs interpolation to up-scale or down-scale N-dimensional images. Note that anti-aliasing should be enabled when down-sizing images to avoid aliasing artifacts. For down-sampling with an integer factor also see skimage.transform.downscale_local_mean.
Parameters
imagendarray
Input image.
scale{float, tuple of floats}
Scale factors. Separate scale factors can be defined as (rows, cols[, …][, dim]).
Returns
scaledndarray
Scaled version of the input.
Other Parameters
orderint, optional
The order of the spline interpolation, default is 0 if image.dtype is bool and 1 otherwise. The order has to be in the range 0-5. See skimage.transform.warp for detail.
mode{‘constant’, ‘edge’, ‘symmetric’, ‘reflect’, ‘wrap’}, optional
Points outside the boundaries of the input are filled according to the given mode. Modes match the behaviour of numpy.pad.
cvalfloat, optional
Used in conjunction with mode ‘constant’, the value outside the image boundaries.
clipbool, optional
Whether to clip the output to the range of values of the input image. This is enabled by default, since higher order interpolation may produce values outside the given input range.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
multichannelbool, optional
Whether the last axis of the image is to be interpreted as multiple channels or another spatial dimension. This argument is deprecated: specify channel_axis instead.
anti_aliasingbool, optional
Whether to apply a Gaussian filter to smooth the image prior to down-scaling. It is crucial to filter when down-sampling the image to avoid aliasing artifacts. If input image data type is bool, no anti-aliasing is applied.
anti_aliasing_sigma{float, tuple of floats}, optional
Standard deviation for Gaussian filtering to avoid aliasing artifacts. By default, this value is chosen as (s - 1) / 2 where s is the down-scaling factor.
channel_axisint or None, optional
If None, the image is assumed to be a grayscale (single channel) image. Otherwise, this parameter indicates which axis of the array corresponds to channels.
New in version 0.19: channel_axis was added in 0.19.
multichannelDEPRECATED
Deprecated in favor of channel_axis.
Deprecated since version 0.19.
Notes
Modes ‘reflect’ and ‘symmetric’ are similar, but differ in whether the edge pixels are duplicated during the reflection. As an example, if an array has values [0, 1, 2] and was padded to the right by four values using symmetric, the result would be [0, 1, 2, 2, 1, 0, 0], while for reflect it would be [0, 1, 2, 1, 0, 1, 2].
Examples
>>> from skimage import data
>>> from skimage.transform import rescale
>>> image = data.camera()
>>> rescale(image, 0.1).shape
(51, 51)
>>> rescale(image, 0.5).shape
(256, 256)
## resize¶
skimage.transform.resize(image, output_shape, order=None, mode='reflect', cval=0, clip=True, preserve_range=False, anti_aliasing=None, anti_aliasing_sigma=None)[source]
Resize image to match a certain size.
Performs interpolation to up-size or down-size N-dimensional images. Note that anti-aliasing should be enabled when down-sizing images to avoid aliasing artifacts. For downsampling with an integer factor also see skimage.transform.downscale_local_mean.
Parameters
imagendarray
Input image.
output_shapeiterable
Size of the generated output image (rows, cols[, …][, dim]). If dim is not provided, the number of channels is preserved. In case the number of input channels does not equal the number of output channels a n-dimensional interpolation is applied.
Returns
resizedndarray
Resized version of the input.
Other Parameters
orderint, optional
The order of the spline interpolation, default is 0 if image.dtype is bool and 1 otherwise. The order has to be in the range 0-5. See skimage.transform.warp for detail.
mode{‘constant’, ‘edge’, ‘symmetric’, ‘reflect’, ‘wrap’}, optional
Points outside the boundaries of the input are filled according to the given mode. Modes match the behaviour of numpy.pad.
cvalfloat, optional
Used in conjunction with mode ‘constant’, the value outside the image boundaries.
clipbool, optional
Whether to clip the output to the range of values of the input image. This is enabled by default, since higher order interpolation may produce values outside the given input range.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
anti_aliasingbool, optional
Whether to apply a Gaussian filter to smooth the image prior to downsampling. It is crucial to filter when downsampling the image to avoid aliasing artifacts. If not specified, it is set to True when downsampling an image whose data type is not bool.
anti_aliasing_sigma{float, tuple of floats}, optional
Standard deviation for Gaussian filtering used when anti-aliasing. By default, this value is chosen as (s - 1) / 2 where s is the downsampling factor, where s > 1. For the up-size case, s < 1, no anti-aliasing is performed prior to rescaling.
Notes
Modes ‘reflect’ and ‘symmetric’ are similar, but differ in whether the edge pixels are duplicated during the reflection. As an example, if an array has values [0, 1, 2] and was padded to the right by four values using symmetric, the result would be [0, 1, 2, 2, 1, 0, 0], while for reflect it would be [0, 1, 2, 1, 0, 1, 2].
Examples
>>> from skimage import data
>>> from skimage.transform import resize
>>> image = data.camera()
>>> resize(image, (100, 100)).shape
(100, 100)
## resize_local_mean¶
skimage.transform.resize_local_mean(image, output_shape, grid_mode=True, preserve_range=False, *, channel_axis=None)[source]
Resize an array with the local mean / bilinear scaling.
Parameters
imagendarray
Input image. If this is a multichannel image, the axis corresponding to channels should be specified using channel_axis
output_shapeiterable
Size of the generated output image. When channel_axis is not None, the channel_axis should either be omitted from output_shape or the output_shape[channel_axis] must match image.shape[channel_axis]. If the length of output_shape exceeds image.ndim, additional singleton dimensions will be appended to the input image as needed.
grid_modebool, optional
Defines image pixels position: if True, pixels are assumed to be at grid intersections, otherwise at cell centers. As a consequence, for example, a 1d signal of length 5 is considered to have length 4 when grid_mode is False, but length 5 when grid_mode is True. See the following visual illustration:
| pixel 1 | pixel 2 | pixel 3 | pixel 4 | pixel 5 |
|<-------------------------------------->|
vs.
|<----------------------------------------------->|
The starting point of the arrow in the diagram above corresponds to coordinate location 0 in each mode.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
Returns
resizedndarray
Resized version of the input.
Notes
This method is sometimes referred to as “area-based” interpolation or “pixel mixing” interpolation [1]. When grid_mode is True, it is equivalent to using OpenCV’s resize with INTER_AREA interpolation mode. It is commonly used for image downsizing. If the downsizing factors are integers, then downscale_local_mean should be preferred instead.
References
1
http://entropymine.com/imageworsener/pixelmixing/
Examples
>>> from skimage import data
>>> from skimage.transform import resize_local_mean
>>> image = data.camera()
>>> resize_local_mean(image, (100, 100)).shape
(100, 100)
## rotate¶
skimage.transform.rotate(image, angle, resize=False, center=None, order=None, mode='constant', cval=0, clip=True, preserve_range=False)[source]
Rotate image by a certain angle around its center.
Parameters
imagendarray
Input image.
anglefloat
Rotation angle in degrees in counter-clockwise direction.
resizebool, optional
Determine whether the shape of the output image will be automatically calculated, so the complete rotated image exactly fits. Default is False.
centeriterable of length 2
The rotation center. If center=None, the image is rotated around its center, i.e. center=(cols / 2 - 0.5, rows / 2 - 0.5). Please note that this parameter is (cols, rows), contrary to normal skimage ordering.
Returns
rotatedndarray
Rotated version of the input.
Other Parameters
orderint, optional
The order of the spline interpolation, default is 0 if image.dtype is bool and 1 otherwise. The order has to be in the range 0-5. See skimage.transform.warp for detail.
mode{‘constant’, ‘edge’, ‘symmetric’, ‘reflect’, ‘wrap’}, optional
Points outside the boundaries of the input are filled according to the given mode. Modes match the behaviour of numpy.pad.
cvalfloat, optional
Used in conjunction with mode ‘constant’, the value outside the image boundaries.
clipbool, optional
Whether to clip the output to the range of values of the input image. This is enabled by default, since higher order interpolation may produce values outside the given input range.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
Notes
Modes ‘reflect’ and ‘symmetric’ are similar, but differ in whether the edge pixels are duplicated during the reflection. As an example, if an array has values [0, 1, 2] and was padded to the right by four values using symmetric, the result would be [0, 1, 2, 2, 1, 0, 0], while for reflect it would be [0, 1, 2, 1, 0, 1, 2].
Examples
>>> from skimage import data
>>> from skimage.transform import rotate
>>> image = data.camera()
>>> rotate(image, 2).shape
(512, 512)
>>> rotate(image, 2, resize=True).shape
(530, 530)
>>> rotate(image, 90, resize=True).shape
(512, 512)
## swirl¶
skimage.transform.swirl(image, center=None, strength=1, radius=100, rotation=0, output_shape=None, order=None, mode='reflect', cval=0, clip=True, preserve_range=False)[source]
Perform a swirl transformation.
Parameters
imagendarray
Input image.
center(column, row) tuple or (2,) ndarray, optional
Center coordinate of transformation.
strengthfloat, optional
The amount of swirling applied.
The extent of the swirl in pixels. The effect dies out rapidly beyond radius.
rotationfloat, optional
Additional rotation applied to the image.
Returns
swirledndarray
Swirled version of the input.
Other Parameters
output_shapetuple (rows, cols), optional
Shape of the output image generated. By default the shape of the input image is preserved.
orderint, optional
The order of the spline interpolation, default is 0 if image.dtype is bool and 1 otherwise. The order has to be in the range 0-5. See skimage.transform.warp for detail.
mode{‘constant’, ‘edge’, ‘symmetric’, ‘reflect’, ‘wrap’}, optional
Points outside the boundaries of the input are filled according to the given mode, with ‘constant’ used as the default. Modes match the behaviour of numpy.pad.
cvalfloat, optional
Used in conjunction with mode ‘constant’, the value outside the image boundaries.
clipbool, optional
Whether to clip the output to the range of values of the input image. This is enabled by default, since higher order interpolation may produce values outside the given input range.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
## warp¶
skimage.transform.warp(image, inverse_map, map_args={}, output_shape=None, order=None, mode='constant', cval=0.0, clip=True, preserve_range=False)[source]
Warp an image according to a given coordinate transformation.
Parameters
imagendarray
Input image.
inverse_maptransformation object, callable cr = f(cr, **kwargs), or ndarray
Inverse coordinate map, which transforms coordinates in the output images into their corresponding coordinates in the input image.
There are a number of different options to define this map, depending on the dimensionality of the input image. A 2-D image can have 2 dimensions for gray-scale images, or 3 dimensions with color information.
• For 2-D images, you can directly pass a transformation object, e.g. skimage.transform.SimilarityTransform, or its inverse.
• For 2-D images, you can pass a (3, 3) homogeneous transformation matrix, e.g. skimage.transform.SimilarityTransform.params.
• For 2-D images, a function that transforms a (M, 2) array of (col, row) coordinates in the output image to their corresponding coordinates in the input image. Extra parameters to the function can be specified through map_args.
• For N-D images, you can directly pass an array of coordinates. The first dimension specifies the coordinates in the input image, while the subsequent dimensions determine the position in the output image. E.g. in case of 2-D images, you need to pass an array of shape (2, rows, cols), where rows and cols determine the shape of the output image, and the first dimension contains the (row, col) coordinate in the input image. See scipy.ndimage.map_coordinates for further documentation.
Note, that a (3, 3) matrix is interpreted as a homogeneous transformation matrix, so you cannot interpolate values from a 3-D input, if the output is of shape (3,).
See example section for usage.
map_argsdict, optional
Keyword arguments passed to inverse_map.
output_shapetuple (rows, cols), optional
Shape of the output image generated. By default the shape of the input image is preserved. Note that, even for multi-band images, only rows and columns need to be specified.
orderint, optional
The order of interpolation. The order has to be in the range 0-5:
• 0: Nearest-neighbor
• 1: Bi-linear (default)
• 3: Bi-cubic
• 4: Bi-quartic
• 5: Bi-quintic
Default is 0 if image.dtype is bool and 1 otherwise.
mode{‘constant’, ‘edge’, ‘symmetric’, ‘reflect’, ‘wrap’}, optional
Points outside the boundaries of the input are filled according to the given mode. Modes match the behaviour of numpy.pad.
cvalfloat, optional
Used in conjunction with mode ‘constant’, the value outside the image boundaries.
clipbool, optional
Whether to clip the output to the range of values of the input image. This is enabled by default, since higher order interpolation may produce values outside the given input range.
preserve_rangebool, optional
Whether to keep the original range of values. Otherwise, the input image is converted according to the conventions of img_as_float. Also see https://scikit-image.org/docs/dev/user_guide/data_types.html
Returns
warpeddouble ndarray
The warped input image.
Notes
Examples
>>> from skimage.transform import warp
>>> from skimage import data
>>> image = data.camera()
The following image warps are all equal but differ substantially in execution time. The image is shifted to the bottom.
Use a geometric transform to warp an image (fast):
>>> from skimage.transform import SimilarityTransform
>>> tform = SimilarityTransform(translation=(0, -10))
>>> warped = warp(image, tform)
Use a callable (slow):
>>> def shift_down(xy):
... xy[:, 1] -= 10
... return xy
>>> warped = warp(image, shift_down)
Use a transformation matrix to warp an image (fast):
>>> matrix = np.array([[1, 0, 0], [0, 1, -10], [0, 0, 1]])
>>> warped = warp(image, matrix)
>>> from skimage.transform import ProjectiveTransform
>>> warped = warp(image, ProjectiveTransform(matrix=matrix))
You can also use the inverse of a geometric transformation (fast):
>>> warped = warp(image, tform.inverse)
For N-D images you can pass a coordinate array, that specifies the coordinates in the input image for every element in the output image. E.g. if you want to rescale a 3-D cube, you can do:
>>> cube_shape = np.array([30, 30, 30])
>>> rng = np.random.default_rng()
>>> cube = rng.random(cube_shape)
Setup the coordinate array, that defines the scaling:
>>> scale = 0.1
>>> output_shape = (scale * cube_shape).astype(int)
>>> coords0, coords1, coords2 = np.mgrid[:output_shape[0],
... :output_shape[1], :output_shape[2]]
>>> coords = np.array([coords0, coords1, coords2])
Assume that the cube contains spatial data, where the first array element center is at coordinate (0.5, 0.5, 0.5) in real space, i.e. we have to account for this extra offset when scaling the image:
>>> coords = (coords + 0.5) / scale - 0.5
>>> warped = warp(cube, coords)
## warp_coords¶
skimage.transform.warp_coords(coord_map, shape, dtype=<class 'numpy.float64'>)[source]
Build the source coordinates for the output of a 2-D image warp.
Parameters
coord_mapcallable like GeometricTransform.inverse
Return input coordinates for given output coordinates. Coordinates are in the shape (P, 2), where P is the number of coordinates and each element is a (row, col) pair.
shapetuple
Shape of output image (rows, cols[, bands]).
dtypenp.dtype or string
dtype for return value (sane choices: float32 or float64).
Returns
coords(ndim, rows, cols[, bands]) array of dtype dtype
Coordinates for scipy.ndimage.map_coordinates, that will yield an image of shape (orows, ocols, bands) by drawing from source points according to the coord_transform_fn.
Notes
This is a lower-level routine that produces the source coordinates for 2-D images used by warp().
It is provided separately from warp to give additional flexibility to users who would like, for example, to re-use a particular coordinate mapping, to use specific dtypes at various points along the the image-warping process, or to implement different post-processing logic than warp performs after the call to ndi.map_coordinates.
Examples
Produce a coordinate map that shifts an image up and to the right:
>>> from skimage import data
>>> from scipy.ndimage import map_coordinates
>>>
>>> def shift_up10_left20(xy):
... return xy - np.array([-20, 10])[None, :]
>>>
>>> image = data.astronaut().astype(np.float32)
>>> coords = warp_coords(shift_up10_left20, image.shape)
>>> warped_image = map_coordinates(image, coords)
## warp_polar¶
skimage.transform.warp_polar(image, center=None, *, radius=None, output_shape=None, scaling='linear', multichannel=False, channel_axis=None, **kwargs)[source]
Remap image to polar or log-polar coordinates space.
Parameters
imagendarray
Input image. Only 2-D arrays are accepted by default. 3-D arrays are accepted if a channel_axis is specified.
centertuple (row, col), optional
Point in image that represents the center of the transformation (i.e., the origin in cartesian space). Values can be of type float. If no value is given, the center is assumed to be the center point of the image.
Radius of the circle that bounds the area to be transformed.
output_shapetuple (row, col), optional
scaling{‘linear’, ‘log’}, optional
Specify whether the image warp is polar or log-polar. Defaults to ‘linear’.
multichannelbool, optional
Whether the image is a 3-D array in which the third axis is to be interpreted as multiple channels. If set to False (default), only 2-D arrays are accepted. This argument is deprecated: specify channel_axis instead.
channel_axisint or None, optional
If None, the image is assumed to be a grayscale (single channel) image. Otherwise, this parameter indicates which axis of the array corresponds to channels.
New in version 0.19: channel_axis was added in 0.19.
**kwargskeyword arguments
Passed to transform.warp.
Returns
warpedndarray
The polar or log-polar warped image.
Other Parameters
multichannelDEPRECATED
Deprecated in favor of channel_axis.
Deprecated since version 0.19.
Examples
Perform a basic polar warp on a grayscale image:
>>> from skimage import data
>>> from skimage.transform import warp_polar
>>> image = data.checkerboard()
>>> warped = warp_polar(image)
Perform a log-polar warp on a grayscale image:
>>> warped = warp_polar(image, scaling='log')
Perform a log-polar warp on a grayscale image while specifying center, radius, and output shape:
>>> warped = warp_polar(image, (100,100), radius=100,
... output_shape=image.shape, scaling='log')
Perform a log-polar warp on a color image:
>>> image = data.astronaut()
>>> warped = warp_polar(image, scaling='log', channel_axis=-1)
## AffineTransform¶
class skimage.transform.AffineTransform(matrix=None, scale=None, rotation=None, shear=None, translation=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.ProjectiveTransform
Affine transformation.
Has the following form:
X = a0*x + a1*y + a2 =
= sx*x*cos(rotation) - sy*y*sin(rotation + shear) + a2
Y = b0*x + b1*y + b2 =
= sx*x*sin(rotation) + sy*y*cos(rotation + shear) + b2
where sx and sy are scale factors in the x and y directions, and the homogeneous transformation matrix is:
[[a0 a1 a2]
[b0 b1 b2]
[0 0 1]]
In 2D, the transformation parameters can be given as the homogeneous transformation matrix, above, or as the implicit parameters, scale, rotation, shear, and translation in x (a2) and y (b2). For 3D and higher, only the matrix form is allowed.
In narrower transforms, such as the Euclidean (only rotation and translation) or Similarity (rotation, translation, and a global scale factor) transforms, it is possible to specify 3D transforms using implicit parameters also.
Parameters
matrix(D+1, D+1) array, optional
Homogeneous transformation matrix. If this matrix is provided, it is an error to provide any of scale, rotation, shear, or translation.
scale{s as float or (sx, sy) as array, list or tuple}, optional
Scale factor(s). If a single value, it will be assigned to both sx and sy. Only available for 2D.
New in version 0.17: Added support for supplying a single scalar value.
rotationfloat, optional
Rotation angle in counter-clockwise direction as radians. Only available for 2D.
shearfloat, optional
Shear angle in counter-clockwise direction as radians. Only available for 2D.
translation(tx, ty) as array, list or tuple, optional
Translation parameters. Only available for 2D.
dimensionalityint, optional
The dimensionality of the transform. This is not used if any other parameters are provided.
Raises
ValueError
If both matrix and any of the other parameters are provided.
Attributes
params(D+1, D+1) array
Homogeneous transformation matrix.
__init__(matrix=None, scale=None, rotation=None, shear=None, translation=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
property rotation
property scale
property shear
property translation
## EssentialMatrixTransform¶
class skimage.transform.EssentialMatrixTransform(rotation=None, translation=None, matrix=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.FundamentalMatrixTransform
Essential matrix transformation.
The essential matrix relates corresponding points between a pair of calibrated images. The matrix transforms normalized, homogeneous image points in one image to epipolar lines in the other image.
The essential matrix is only defined for a pair of moving images capturing a non-planar scene. In the case of pure rotation or planar scenes, the homography describes the geometric relation between two images (ProjectiveTransform). If the intrinsic calibration of the images is unknown, the fundamental matrix describes the projective relation between the two images (FundamentalMatrixTransform).
Parameters
rotation(3, 3) array, optional
Rotation matrix of the relative camera motion.
translation(3, 1) array, optional
Translation vector of the relative camera motion. The vector must have unit length.
matrix(3, 3) array, optional
Essential matrix.
References
1
Hartley, Richard, and Andrew Zisserman. Multiple view geometry in computer vision. Cambridge university press, 2003.
Attributes
params(3, 3) array
Essential matrix.
__init__(rotation=None, translation=None, matrix=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
estimate(src, dst)[source]
Estimate essential matrix using 8-point algorithm.
The 8-point algorithm requires at least 8 corresponding point pairs for a well-conditioned solution, otherwise the over-determined solution is estimated.
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
Returns
successbool
True, if model estimation succeeds.
## EuclideanTransform¶
class skimage.transform.EuclideanTransform(matrix=None, rotation=None, translation=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.ProjectiveTransform
Euclidean transformation, also known as a rigid transform.
Has the following form:
X = a0 * x - b0 * y + a1 =
= x * cos(rotation) - y * sin(rotation) + a1
Y = b0 * x + a0 * y + b1 =
= x * sin(rotation) + y * cos(rotation) + b1
where the homogeneous transformation matrix is:
[[a0 b0 a1]
[b0 a0 b1]
[0 0 1]]
The Euclidean transformation is a rigid transformation with rotation and translation parameters. The similarity transformation extends the Euclidean transformation with a single scaling factor.
Parameters
matrix(D+1, D+1) array, optional
Homogeneous transformation matrix.
rotationfloat or sequence of float, optional
Rotation angle in counter-clockwise direction as radians. If given as a vector, it is interpreted as Euler rotation angles [1]. Only 2D (single rotation) and 3D (Euler rotations) values are supported. For higher dimensions, you must provide or estimate the transformation matrix.
translationsequence of float, length D, optional
Translation parameters for each axis.
dimensionalityint, optional
The dimensionality of the transform.
References
1
https://en.wikipedia.org/wiki/Rotation_matrix#In_three_dimensions
Attributes
params(D+1, D+1) array
Homogeneous transformation matrix.
__init__(matrix=None, rotation=None, translation=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
estimate(src, dst)[source]
Estimate the transformation from a set of corresponding points.
You can determine the over-, well- and under-determined parameters with the total least-squares method.
Number of source and destination coordinates must match.
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
Returns
successbool
True, if model estimation succeeds.
property rotation
property translation
## FundamentalMatrixTransform¶
class skimage.transform.FundamentalMatrixTransform(matrix=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.GeometricTransform
Fundamental matrix transformation.
The fundamental matrix relates corresponding points between a pair of uncalibrated images. The matrix transforms homogeneous image points in one image to epipolar lines in the other image.
The fundamental matrix is only defined for a pair of moving images. In the case of pure rotation or planar scenes, the homography describes the geometric relation between two images (ProjectiveTransform). If the intrinsic calibration of the images is known, the essential matrix describes the metric relation between the two images (EssentialMatrixTransform).
Parameters
matrix(3, 3) array, optional
Fundamental matrix.
References
1
Hartley, Richard, and Andrew Zisserman. Multiple view geometry in computer vision. Cambridge university press, 2003.
Attributes
params(3, 3) array
Fundamental matrix.
__init__(matrix=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
estimate(src, dst)[source]
Estimate fundamental matrix using 8-point algorithm.
The 8-point algorithm requires at least 8 corresponding point pairs for a well-conditioned solution, otherwise the over-determined solution is estimated.
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
Returns
successbool
True, if model estimation succeeds.
inverse(coords)[source]
Apply inverse transformation.
Parameters
coords(N, 2) array
Destination coordinates.
Returns
coords(N, 3) array
Epipolar lines in the source image.
residuals(src, dst)[source]
Compute the Sampson distance.
The Sampson distance is the first approximation to the geometric error.
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
Returns
residuals(N, ) array
Sampson distance.
## PiecewiseAffineTransform¶
class skimage.transform.PiecewiseAffineTransform[source]
Bases: skimage.transform._geometric.GeometricTransform
Piecewise affine transformation.
Control points are used to define the mapping. The transform is based on a Delaunay triangulation of the points to form a mesh. Each triangle is used to find a local affine transform.
Attributes
affineslist of AffineTransform objects
Affine transformations for each triangle in the mesh.
inverse_affineslist of AffineTransform objects
Inverse affine transformations for each triangle in the mesh.
__init__()[source]
Initialize self. See help(type(self)) for accurate signature.
estimate(src, dst)[source]
Estimate the transformation from a set of corresponding points.
Number of source and destination coordinates must match.
Parameters
src(N, D) array
Source coordinates.
dst(N, D) array
Destination coordinates.
Returns
successbool
True, if all pieces of the model are successfully estimated.
inverse(coords)[source]
Apply inverse transformation.
Coordinates outside of the mesh will be set to - 1.
Parameters
coords(N, D) array
Source coordinates.
Returns
coords(N, D) array
Transformed coordinates.
## PolynomialTransform¶
class skimage.transform.PolynomialTransform(params=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.GeometricTransform
2D polynomial transformation.
Has the following form:
X = sum[j=0:order]( sum[i=0:j]( a_ji * x**(j - i) * y**i ))
Y = sum[j=0:order]( sum[i=0:j]( b_ji * x**(j - i) * y**i ))
Parameters
params(2, N) array, optional
Polynomial coefficients where N * 2 = (order + 1) * (order + 2). So, a_ji is defined in params[0, :] and b_ji in params[1, :].
Attributes
params(2, N) array
Polynomial coefficients where N * 2 = (order + 1) * (order + 2). So, a_ji is defined in params[0, :] and b_ji in params[1, :].
__init__(params=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
estimate(src, dst, order=2, weights=None)[source]
Estimate the transformation from a set of corresponding points.
You can determine the over-, well- and under-determined parameters with the total least-squares method.
Number of source and destination coordinates must match.
The transformation is defined as:
X = sum[j=0:order]( sum[i=0:j]( a_ji * x**(j - i) * y**i ))
Y = sum[j=0:order]( sum[i=0:j]( b_ji * x**(j - i) * y**i ))
These equations can be transformed to the following form:
0 = sum[j=0:order]( sum[i=0:j]( a_ji * x**(j - i) * y**i )) - X
0 = sum[j=0:order]( sum[i=0:j]( b_ji * x**(j - i) * y**i )) - Y
which exist for each set of corresponding points, so we have a set of N * 2 equations. The coefficients appear linearly so we can write A x = 0, where:
A = [[1 x y x**2 x*y y**2 ... 0 ... 0 -X]
[0 ... 0 1 x y x**2 x*y y**2 -Y]
...
...
]
x.T = [a00 a10 a11 a20 a21 a22 ... ann
b00 b10 b11 b20 b21 b22 ... bnn c3]
In case of total least-squares the solution of this homogeneous system of equations is the right singular vector of A which corresponds to the smallest singular value normed by the coefficient c3.
Weights can be applied to each pair of corresponding points to indicate, particularly in an overdetermined system, if point pairs have higher or lower confidence or uncertainties associated with them. From the matrix treatment of least squares problems, these weight values are normalised, square-rooted, then built into a diagonal matrix, by which A is multiplied.
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
orderint, optional
Polynomial order (number of coefficients is order + 1).
weights(N,) array, optional
Relative weight values for each pair of points.
Returns
successbool
True, if model estimation succeeds.
inverse(coords)[source]
Apply inverse transformation.
Parameters
coords(N, 2) array
Destination coordinates.
Returns
coords(N, 2) array
Source coordinates.
## ProjectiveTransform¶
class skimage.transform.ProjectiveTransform(matrix=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.GeometricTransform
Projective transformation.
Apply a projective transformation (homography) on coordinates.
For each homogeneous coordinate $$\mathbf{x} = [x, y, 1]^T$$, its target position is calculated by multiplying with the given matrix, $$H$$, to give $$H \mathbf{x}$$:
[[a0 a1 a2]
[b0 b1 b2]
[c0 c1 1 ]].
E.g., to rotate by theta degrees clockwise, the matrix should be:
[[cos(theta) -sin(theta) 0]
[sin(theta) cos(theta) 0]
[0 0 1]]
or, to translate x by 10 and y by 20:
[[1 0 10]
[0 1 20]
[0 0 1 ]].
Parameters
matrix(D+1, D+1) array, optional
Homogeneous transformation matrix.
dimensionalityint, optional
The number of dimensions of the transform. This is ignored if matrix is not None.
Attributes
params(D+1, D+1) array
Homogeneous transformation matrix.
__init__(matrix=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
property dimensionality
The dimensionality of the transformation.
estimate(src, dst, weights=None)[source]
Estimate the transformation from a set of corresponding points.
You can determine the over-, well- and under-determined parameters with the total least-squares method.
Number of source and destination coordinates must match.
The transformation is defined as:
X = (a0*x + a1*y + a2) / (c0*x + c1*y + 1)
Y = (b0*x + b1*y + b2) / (c0*x + c1*y + 1)
These equations can be transformed to the following form:
0 = a0*x + a1*y + a2 - c0*x*X - c1*y*X - X
0 = b0*x + b1*y + b2 - c0*x*Y - c1*y*Y - Y
which exist for each set of corresponding points, so we have a set of N * 2 equations. The coefficients appear linearly so we can write A x = 0, where:
A = [[x y 1 0 0 0 -x*X -y*X -X]
[0 0 0 x y 1 -x*Y -y*Y -Y]
...
...
]
x.T = [a0 a1 a2 b0 b1 b2 c0 c1 c3]
In case of total least-squares the solution of this homogeneous system of equations is the right singular vector of A which corresponds to the smallest singular value normed by the coefficient c3.
Weights can be applied to each pair of corresponding points to indicate, particularly in an overdetermined system, if point pairs have higher or lower confidence or uncertainties associated with them. From the matrix treatment of least squares problems, these weight values are normalised, square-rooted, then built into a diagonal matrix, by which A is multiplied.
In case of the affine transformation the coefficients c0 and c1 are 0. Thus the system of equations is:
A = [[x y 1 0 0 0 -X]
[0 0 0 x y 1 -Y]
...
...
]
x.T = [a0 a1 a2 b0 b1 b2 c3]
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
weights(N,) array, optional
Relative weight values for each pair of points.
Returns
successbool
True, if model estimation succeeds.
inverse(coords)[source]
Apply inverse transformation.
Parameters
coords(N, D) array
Destination coordinates.
Returns
coords_out(N, D) array
Source coordinates.
## SimilarityTransform¶
class skimage.transform.SimilarityTransform(matrix=None, scale=None, rotation=None, translation=None, *, dimensionality=2)[source]
Bases: skimage.transform._geometric.EuclideanTransform
2D similarity transformation.
Has the following form:
X = a0 * x - b0 * y + a1 =
= s * x * cos(rotation) - s * y * sin(rotation) + a1
Y = b0 * x + a0 * y + b1 =
= s * x * sin(rotation) + s * y * cos(rotation) + b1
where s is a scale factor and the homogeneous transformation matrix is:
[[a0 b0 a1]
[b0 a0 b1]
[0 0 1]]
The similarity transformation extends the Euclidean transformation with a single scaling factor in addition to the rotation and translation parameters.
Parameters
matrix(dim+1, dim+1) array, optional
Homogeneous transformation matrix.
scalefloat, optional
Scale factor. Implemented only for 2D and 3D.
rotationfloat, optional
Rotation angle in counter-clockwise direction as radians. Implemented only for 2D and 3D. For 3D, this is given in ZYX Euler angles.
translation(dim,) array-like, optional
x, y[, z] translation parameters. Implemented only for 2D and 3D.
Attributes
params(dim+1, dim+1) array
Homogeneous transformation matrix.
__init__(matrix=None, scale=None, rotation=None, translation=None, *, dimensionality=2)[source]
Initialize self. See help(type(self)) for accurate signature.
estimate(src, dst)[source]
Estimate the transformation from a set of corresponding points.
You can determine the over-, well- and under-determined parameters with the total least-squares method.
Number of source and destination coordinates must match.
Parameters
src(N, 2) array
Source coordinates.
dst(N, 2) array
Destination coordinates.
Returns
successbool
True, if model estimation succeeds.
property scale | 17,559 | 73,431 | {"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} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.572049 |
socrates.stat.fsu.edu | 1,653,081,648,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00177.warc.gz | 595,388,512 | 4,686 | Welcome to Dr. Sinha's research page. On this page, you will find his research interests, accomplishments, the current students he is working with and some of this past students including several of his published research. You will also find scripts involved in his research.
# Research interests
• Survival analysis and Bayesian analysis.
• Bayesian Biostatistics.
• Modeling Cancer prevention data.
• Cure rate and survival data.
• Modeling Cancer relapse data and recurrence data.
• Semiparametric empirical Bayes.
• Methods for skewed and heteroscedastic response.
Code for computing the empirical power and type I error for Proportional Odds Model when the data are generated from proportional odds model or proportional hazard model respectively.
One of functions we used in calculate the MLE of effect size beta.
A function for using Newton-Raphson algorithm to estimate maximum likelihood estimation (MLE) of beta of under proportional odds model. The input include patient survival time, censored status and covariates. The output is the point estimate for the covariate effects beta.
One of functions we used in calculate the MLE of effect size beta.
A function for generating data from univariate proportional hazard model. The input includes the number of samples n and the true effect size beta. Then output data are the survival times for every patients..
A function for generating data from univariate proportional odds survival model. The input includes the number of samples n and the true effect size beta. Then output data are the survival times for every patients.
A function for calculating the standard deviation for the estimation of covariate effects beta.
R code used for IMR prior for Cure Rate survival model. Given the survival data, the output for the function includes the posterior samples for the covariates effects using IM prior given the input data.
R code used for IMR prior for proportional hazard model. The baseline hazard function is assumed to be piecewise constant function. Given the survival data, the output for the function includes the posterior samples for the covariates effects using IM prior given the input data.
R code used for gprior for proportional hazard model. Given the survival data, the output for the function includes the posterior samples for the covariates effects using IM prior given the input data. This method is used as a reference method to evaluate the performance of IMR prior.
R code used for Gaussian prior for proportional hazard model. Given the survival data, the output for the function includes the posterior samples for the covariates effects using IM prior given the input data. This method is used as a reference method to evaluate the performance of IMR prior.
Generates data from proportional hazard model with piecewised baseline hazard function. We used these data in our simulation study. The input includes the covariates matrix, the coefficients for covariates, number of sample size, and the baseline hazard function. The output is the survival time for all patients.
The code below corresponds to methodology discussed in the following paper:
## Bayesian Partial Linear Model for skewed longitudinal Data
[Status: Submitted to Journal of the American statisticial assosciation on 12/12/12]
The zip folder contains scripts (JAGS,R) for the following:
Simulation Study:
R code is for generating data, theJAGS code is for the model and prior.
Data Example:
R code is to read the data, reshape the data, standardize the data. JAGS code covers the model and prior.
The code below corresponds to methodology discussed in the following paper:
## Bayesian variable selection for skewed heteroscedastic error
[Status: In progress]
The zip folder contains scripts (JAGS,R) for the following:
Simulation Study:
R code for generating data, JAGS code for model and prior.
Data Example:
R code for cleaning, reading data, JAGS code for MCMC update.
The code below corresponds to methodology discussed in the following paper:
## Bayesian regression for univariate skewed heteroscedastic error
[Status: In progress]
The zip folder contains scripts (MATLAB) for simulation studies and a data example.
The code below corresponds to methodology discussed in the following paper:
## Empirical Bayes estimation for additive hazards regression models.
### Sinha D, McHenry MB, Lipsitz SR, Ghosh M.
[Status: Published; Biometrika. 2009 Sep;96(3):545-558. Epub 2009 Jun 26.] | 903 | 4,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.814766 |
https://www.coursehero.com/file/5671895/14/ | 1,493,360,232,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122865.36/warc/CC-MAIN-20170423031202-00251-ip-10-145-167-34.ec2.internal.warc.gz | 881,630,842 | 27,447 | # 1.4 - Linear Algebra -115 Solutions to Second Homework...
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Solutions to Second Homework Problem 3( b ) (Section 1 . 4)Assume that the first vector is a linear combi- nation of the other two. Then, (1 , 2 , - 3) = a ( - 3 , 2 , 1) + b (2 , - 1 , - 1) . Doing some algebra we get that (1 , 2 , - 3) = ( - 3 a + 2 b, 2 a - b, 1 a - b ) . So, we have the system - 3 a + 2 b = 1 2 a - b = 2 1 a - b = - 3 , with corresponding matrix: - 3 2 1 2 - 1 2 1 - 1 - 3 . Some computations here show that: - 3 2 1 2 - 1 2 1 - 1 - 3 1 - 1 - 3 2 - 1 2 - 3 2 1 1 - 1 - 3 0 1 8 0 - 1 - 8 1 - 1 - 3 0 1 8 0 0 0 , i.e. the system of equations has a solution (can you find it?), which implies that (1 , 2 , - 3) is a linear combination of ( - 3 , 2 , 1) , (2 , - 1 , - 1) . ± Problem 4( f ) (Section 1 . 4)We work as above. Say that 6 x 3 - 3 x 2 + x + 2 = a ( x 3 - x 2 + 2 x + 3) + b (2 x 3 - 3 x + 1) . Doing, again, some algebra, (and after some work) we get the system a + 2 b = 6 - a = - 3 2 a - 3 b = 1 3 a + b = 2 1
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## This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.
### Page1 / 2
1.4 - Linear Algebra -115 Solutions to Second Homework...
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Ask a homework question - tutors are online | 612 | 1,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-17 | longest | en | 0.858577 |
http://www.flyingcoloursmaths.co.uk/mathematical-ninja-ssns/ | 1,590,554,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00004.warc.gz | 168,352,305 | 14,268 | # The Mathematical Ninja and the SSNs
A professor - according to Reddit - asked their class how many people you'd need to have in a room to be absolutely certain two of them would have Social Security numbers1 ending in the same four digits (in the same order).
10001, obviously.
How about a probability of 99.9%? The students guessed 1,000, 8,000 and even 9,990. The true answer? 370.
That's a slightly involved calculation, related to the Birthday Problem; one would need to compute the complementary product $\br{\frac{9999}{10000}\times\frac{9998}{10000}\times ...}$ until you reached a number below 0.001.
### Computation is boring. Let's estimate!
Suppose you have a set of $n$ people. There are $m=\nCr{n}{2}$ possible matchings between them, each of which has a (not quite but almost independent) probability of $p$.
Then the probability of none of those matches being a hit is $\br{1-p}^m$, so we need to solve $\br{1-p}^m < 0.001$ for $m$ (which is a function of $n$).
Taking logs gives $m \ln(1-p) < -\ln(1000)$.
We know that for small $p$, $\ln(1-p)\approx -p$, and for all values of 1000, $\ln(1000) = 3\ln(10) \approx 6.9$.
Now we have $-mp < -6.9$ (ish), or $m > \frac{6.9}{p}$.
However, $m = \nCr{n}{2}$, which is roughly $\frac{n^2}{2}$.
This gives us $\frac{n^2}{2} > \frac{6.9}{p}$, or $n > \sqrt{\frac{13.8}{p}}$.
We know that $p = 10^{-4}$, so $n > 100 \sqrt{13.8}$.
### Did someone call for a Ninja?
Whoosh
"The square root of 1369 is 37, so the square root of 1380 is about $37 + \frac{11}{74}$, or $37.15$. That means $\sqrt{13.8} \approx 3.715$."
Whoosh
### Back to our regularly scheduled programming
This gives us a final approximation of $n > 371.5$ - not far at all from the correct answer of 370.
The professor, I'm told, used a whole gamut of series and approximations to get an approximation of 375. I shall smile smugly in their general direction.
## Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
1. any randomly-assigned number with at least four digits will do just as well []
#### Share
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https://www.numerade.com/questions/find-the-limit-use-lhospitals-rule-where-appropriate-if-there-is-a-more-elementary-method-conside-32/ | 1,632,084,807,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00426.warc.gz | 954,091,861 | 36,528 | 💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\lim_{x\to 1} \frac{x^a - 1}{x^b - 1}$, $b \not= 0$
## This limit has the form$\frac{0}{0} \cdot \lim _{x \rightarrow 1} \frac{x^{a}-1}{x^{b}-1} \quad[\text { for } b \neq 0] \quad \stackrel{\Perp}{=} \lim _{x \rightarrow 1} \frac{a x^{a-1}}{b x^{b-1}}=\frac{a(1)}{b(1)}=\frac{a}{b}$
Derivatives
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##### Catherine R.
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University of Michigan - Ann Arbor
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### Video Transcript
mhm. There's probably want to take the limit of the function. Um and possibly use Law Patel's rule to evaluate it. So in this case we have the limit of the numerator. Uh regardless we're going to get 1 -1 so we will get 0/0, which is going to be and the determinant form. So we want to take the derivative of the numerator and derivative of the denominator. When we take the derivative of the numerator, we get eight times acts To the A -1. And then in the denominator we get B. X To the B -1. Now when we plug in one we get X to the one minus one, X. To the one minus one, that's just going to be to the zero power. So ultimately where we're going to be getting as our answer is A over B. And that will be the limit as X approaches one of our function.
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http://slideplayer.info/slide/10083141/ | 1,503,067,926,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104681.22/warc/CC-MAIN-20170818140908-20170818160908-00283.warc.gz | 378,419,316 | 20,311 | # Chapter 10 – The Design of Feedback Control Systems PID Compensation Networks.
## Presentasi berjudul: "Chapter 10 – The Design of Feedback Control Systems PID Compensation Networks."— Transcript presentasi:
Chapter 10 – The Design of Feedback Control Systems PID Compensation Networks
Different Types of Feedback Control On-Off Control This is the simplest form of control.
Proportional Control A proportional controller attempts to perform better than the On-off type by applying power in proportion to the difference in temperature between the measured and the set-point. As the gain is increased the system responds faster to changes in set-point but becomes progressively underdamped and eventually unstable. The final temperature lies below the set-point for this system because some difference is required to keep the heater supplying power.
Proportional, Derivative Control The stability and overshoot problems that arise when a proportional controller is used at high gain can be mitigated by adding a term proportional to the time-derivative of the error signal. The value of the damping can be adjusted to achieve a critically damped response.
Proportional+Integral+Derivative Control Although PD control deals neatly with the overshoot and ringing problems associated with proportional control it does not cure the problem with the steady-state error. Fortunately it is possible to eliminate this while using relatively low gain by adding an integral term to the control function which becomes
The Characteristics of P, I, and D controllers A proportional controller (Kp) will have the effect of reducing the rise time and will reduce, but never eliminate, the steady-state error. An integral control (Ki) will have the effect of eliminating the steady-state error, but it may make the transient response worse. A derivative control (Kd) will have the effect of increasing the stability of the system, reducing the overshoot, and improving the transient response.
Proportional Control By only employing proportional control, a steady state error occurs. Proportional and Integral Control The response becomes more oscillatory and needs longer to settle, the error disappears. Proportional, Integral and Derivative Control All design specifications can be reached.
CL RESPONSERISE TIMEOVERSHOOTSETTLING TIMES-S ERROR KpDecreaseIncreaseSmall ChangeDecrease KiDecreaseIncrease Eliminate KdSmall ChangeDecrease Small Change The Characteristics of P, I, and D controllers
Tips for Designing a PID Controller 1.Obtain an open-loop response and determine what needs to be improved 2.Add a proportional control to improve the rise time 3.Add a derivative control to improve the overshoot 4.Add an integral control to eliminate the steady-state error 5.Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral) into a single system, if not necessary. For example, if a PI controller gives a good enough response (like the above example), then you don't need to implement derivative controller to the system. Keep the controller as simple as possible.
num=1; den=[1 10 20]; stepstep(num,den) Open-Loop Control - Example
Proportional Control - Example The proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces the steady-state error. MATLAB Example Kp=300; num=[Kp]; den=[1 10 20+Kp]; t=0:0.01:2; step(num,den,t) K=300 K=100
Kp=300; Kd=10; num=[Kd Kp]; den=[1 10+Kd 20+Kp]; t=0:0.01:2; step(num,den,t) Proportional - Derivative - Example The derivative controller (Kd) reduces both the overshoot and the settling time. MATLAB Example Kd=10 Kd=20
Proportional - Integral - Example The integral controller (Ki) decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state error MATLAB Example Kp=30; Ki=70; num=[Kp Ki]; den=[1 10 20+Kp Ki]; t=0:0.01:2; step(num,den,t) Ki=70 Ki=100
Syntax rltool rltool(sys) rltool(sys,comp) RLTOOL
Consider the following configuration: Example - Practice
The design a system for the following specifications: ·Zero steady state error ·Settling time within 5 seconds ·Rise time within 2 seconds ·Only some overshoot permitted Example - Practice
Presentasi serupa | 966 | 4,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.894139 |
http://www.talkstats.com/showthread.php/66813-Type-1-error-in-regression-with-entire-populations?p=193711 | 1,508,633,868,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00407.warc.gz | 547,038,533 | 16,941 | # Thread: Type 1 error in regression with entire populations
1. ## Type 1 error in regression with entire populations
I have read that you are not supposed to keep choosing different levels of the categorical variable to be the reference level, because this can cause familywise error, the true type 1 error will be greater than your nominal alpha level. I mean by this if you have 5 levels, you change the reference level five times seeing what the results will be (although it can be useful substantively to do that).
I am not sure the above is true. But even if it is, does this apply when you have an entire population, as I usually do. Can you even have type 1 error when you are analyzing a population. By that I mean there are 25000 people in the population of interest and I have all of them, there is no sample involved.
2. ## Re: Type 1 error in regression with entire populations
Why do you have a reference group, did you run something on this population?
Type I Error is probably exclusive to statistical testing. Statistics are for making generalization from collected samples. You don't have a sample and aren't conducting statistical tests. Your numbers are the truth, so if two groups are different, they are different - no sampling distribution based threats. This is what I think.
3. ## Re: Type 1 error in regression with entire populations
I just saw this but have not actually looked at its content:
<li class="first-item">Xiaoqin Wang,
Yin Jin,
<li class="last-item">and Li Yin
Measuring and estimating treatment effect on dichotomous outcome of a population
Stat Methods Med Res October 2016 25: 1779-1790, first published on September 3, 2013 doi:10.1177/0962280213502146
4. ## Re: Type 1 error in regression with entire populations
Originally Posted by noetsi
I have read that you are not supposed to keep choosing different levels of the categorical variable to be the reference level, because this can cause familywise error, the true type 1 error will be greater than your nominal alpha level. I mean by this if you have 5 levels, you change the reference level five times seeing what the results will be (although it can be useful substantively to do that).
I am not sure the above is true.
Hi,
I am pretty sure this is not true. If you are only changing the reference level I think you are actually repeating the same test only presented differently - kindof like writing up the same test results in different languages - so it will still be the same test not an independent one.
5. ## Re: Type 1 error in regression with entire populations
Whenever I do multiple testing like noetsi mentioned, I always correct my level of significance. Unless it is the exact same test (a vs b, b vs a).
6. ## Re: Type 1 error in regression with entire populations
Originally Posted by hlsmith
Whenever I do multiple testing like noetsi mentioned, I always correct my level of significance. Unless it is the exact same test (a vs b, b vs a).
But this is the same thing, right ? a vs. b,c,d,e or b vs. a,c,d,e ..etc.
7. ## Re: Type 1 error in regression with entire populations
A vs b, a vs c, and b vs c, is three hypothesis tests in my practice.
8. ## Re: Type 1 error in regression with entire populations
Originally Posted by noetsi
I have read that you are not supposed to keep choosing different levels of the categorical variable to be the reference level, because this can cause familywise error, the true type 1 error will be greater than your nominal alpha level. I mean by this if you have 5 levels, you change the reference level five times seeing what the results will be (although it can be useful substantively to do that).
I am not sure the above is true. But even if it is, does this apply when you have an entire population, as I usually do. Can you even have type 1 error when you are analyzing a population. By that I mean there are 25000 people in the population of interest and I have all of them, there is no sample involved.
I believe that you do have type 1 error even though you have a population.
Thinking it through, you have two types of statistics, descriptive and inferential. In descriptive statistics (i.e., mean, standard deviation) you no longer have sampling error, so your measures of mean and standard deviation are absolute. No confidence intervals around mean or standard deviation. However, when you use inferential statistics, you still have variation with which you must deal. True, the sampling variation is gone, but all of your other sources of variation still exist. Where there is variation there is uncertainty. I believe that all of the typical rules, assumptions, etc. still apply.
9. ## Re: Type 1 error in regression with entire populations
Hmmm. Need references that go either way. Does traditional 1.96 disappear though, it would seem natural not to have it.
10. ## Re: Type 1 error in regression with entire populations
Originally Posted by hlsmith
A vs b, a vs c, and b vs c, is three hypothesis tests in my practice.
Yepp, you are right. But if I understand the question correctly, it is about doing the same analysis only changing the reference level. So, I agree, you have one test for each pair of levels but you do not have more tests just because you changed the reference level from A to B.
11. ## Re: Type 1 error in regression with entire populations
I found the following:
It appears that there are a lot of arguments either way. Andrew Gelman's makes the most sense to me.
12. ## Re: Type 1 error in regression with entire populations
I think a key issue raised by miner is how you understand your population. If you think of your analysis as only pertaining to that population than type 1 error does not make sense to me. You know the true results in the population and no error is possible. However, if you think of your population as a sample of all possible populations (that might occur in the future for example) then error does pertain - in terms of applying your analysis to those other macro-populations. I have back and forth on this issue, in this case I decided to ignore that future populations might be different (or ones in other states etc, this analysis is very focused).
I am not sure what source of variation exist other than sampling variation that could cause error in honesty. I don't doubt they might exist -I just can't imagine what they are or how they would introduce error in the regression. All I really care about here are the slopes and odds ratios - not interested in other statistics.
The literature I have seen comes down on HL Smith in terms of multiple tests (that is why posthoc tests are penalized). But that literature does not I think deal with populations. I am still not sure one way or the other if familywise error applies - because I think type 1 error itself is impossible in a population when all you care about is that population. Its a problem with a sample because you are interested in the larger population and you don't know if what you find in the sample matches the real population.
Obviously I remain uncertain (thanks miner and hlsmith for the articles). I don't think the assumptions of regression, except for non-linearity, really apply when you are analyzing a large population (I have 25,0000 cases). Ignoring that you have the population when you have that many cases I don't think heteroscedasticity or multicolinearity influence the results because the results are asymptotically correct with that many cases even with the errors if they exist. Nor does normality matter because of the CLT.
13. ## Re: Type 1 error in regression with entire populations
I am also unsure of this additional variation.
You could say mesasurement error, but that isn't usually addressed in your model. Sensitivity analysis can try to assume its direction and magnitude. You have dispersion of the variate, but that is the nature of a random variable (stochastic).
For example how would one perform a two sample ttest with a population??
I get the attempt of saying what about a future sample, but the moment after you get a measurement things are different and what if you aren't predicting just getting a cross-sectional measurement.
14. ## Re: Type 1 error in regression with entire populations
It all depends on what the analysis will be used for, I guess. The moment somebody says that we have proven an effect it is implicitely understood that we talk about future samples - as an effect presuposes that it will not disappear after we stop the measurement. Also, i do not think anyone uses the very careful wording that would be necessary to avoid this happening - something like "we did not use the usual statistical methods because we have a full census and we have no intention to discuss the existence or non-existence of any effect whastsoever that might show up in our analysis " - so, probably it would be best to go with Gelman .
regards
15. ## Re: Type 1 error in regression with entire populations
i accessed the paper I reference in post #3, it is not relevant to your question. It covers using maximum likelihoods to estimate risk differences, relative risks, and odds ratios from one model.
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• You may not edit your posts | 2,020 | 9,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-43 | latest | en | 0.930874 |
https://discourse.datamethods.org/t/methods-to-evaluate-time-depedent-variables-observed-intermittently-in-survival-analysis/1587 | 1,653,351,382,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00323.warc.gz | 256,803,621 | 6,710 | # Methods to evaluate time-depedent variables observed intermittently in survival analysis
Hi, I wanted to ask what are the best solutions to evaluate the prognostic effect of time-dependent covariates observed intermittently. In this particular case, I need to assess the prognostic effect on the survival of a certain exposure. Patients are seen every two months (2, 4, 6, 8, 10 months). At some point in that intervals, the covariable ‘exposure’ goes from 0 to 1, but the specific date is not known. Using the midpoint of the interval I suppose can involve a bias. What approaches have been proposed? Is there an R package that does this easily?
Thanks.
I forgot to tell that the event is NOT interval censored…
maybe the clock starts once they are exposed? it depends on the context
What we want to know is whether the worst toxicity to an antineoplastic therapy is associated with prognosis (OS). The problem is that researchers see patients at two-monthly intervals, and do not record the exact time at which toxicity occurs. The clock starts counting with cycle 1, the date of death or last follow-up is well collected, and the problem is that exposure (in this case, toxicity) is a time-dependent variable that occurs within an interval.
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i have stumbled upon that kind of thing before eg “We model efficacy as a time-to-event endpoint and toxicity as a binary endpoint, sharing common random effects in order to induce dependence between the bivariate outcomes.” But i agree that the difficulty might be that time is known to occur within a 2 month window. There must be something out there on that particular problem, it sounds familiar, but i don’t know off the top of my head…
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Yes, it is still sadly common for some to analyze toxicity as a fixed factor, and come to the conclusion that “in the end, toxicity is good”. I remember a discussion I had at a congress with a doctor who reported that severe thrombopenia was a good prognostic factor, without taking into account that in order to have severe thrombopenia, a requirement was to be alive long enough. Toxicity is normally evaluated at intervals, every 2-3 weeks. I think researchers tend to record toxicity dates within the cycle, or ascribe it to the next cycle that is delayed, in the case of a short interval. The problem here is that the intervals are large (every 2 months), although the median survival is about 24 months. I do not know if a solution is to abscribe here also the date to that of the successive cycle with the premise that if they attended the consultation, there was indeed no event.
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I’ve assembled a public Zotero library on precisely this matter, as part of my work on dose individualization of cancer drugs.
Is the dose of therapy fixed, or variable? It sounds as if these doctors would be too busy to titrate dosing to toxicity. What a shame.
As far as the broader methodological question:
you may wish to consider partially observed Markov processes as a suitable formalism.
Thank you very much David.
Doses had two possible levels. I think titration is not an issue here since adverse events were usually reversible.
I will try to look at that pomp package of R although it seems quite complex. Once again, it is very interesting to see how very common situations generate a great mathematical complexity. As for the references in Zotero, it would be interesting to do a methodological review on this subject. Often, immortal time bias is not contemplated.
As I point out in this tweet, reversibility of AEs in fact underscores the possibility and value of dose titration.
This seems to amalgamate Simon’s (1997) accelerated titration design with CRM. Given that “thrombocytopenia and other AEs were monitorable, manageable and reversible,” this agent looks ripe for #DoseIndividualization. pic.twitter.com/qZ1vkoJ8Sk
— David C. Norris, MD (@davidcnorrismd) November 30, 2018
Well, the prognostic impact of toxicity can be seen from two points of view. On the one hand, toxicity can have a negative effect on survival. For example, neutropenia prevents patients from receiving effective doses of chemotherapy. This has a negative impact on the progression of cancer. However, on the other hand, there are other toxicities that could be indirect markers that the drug is reaching its targets, and that its kinetics are correct. This type of toxicity could be associated with a good prognosis, unlike the previous one. An example would be hypothyroidism by sunitinib, hyperglycemia with mTOR inhibitors, rash associated with antiEGFR, etc. If the dose is titrated to reduce toxicity, the effect is likely to be reduced as well, so it is a trade-off. In any case, the problem is that toxicity analysis remains common as if it were a basal fixed variable, when in fact it is often cumulative or late.
2 Likes | 1,054 | 4,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.959296 |
https://cs.stackexchange.com/questions/18287/finding-an-instance-of-an-n-element-set-cover | 1,653,623,268,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00176.warc.gz | 225,226,035 | 67,614 | # Finding an instance of an n-element set cover
Below is a homework problem where we have been asked to alter a greedy algorithm to return n element instance of a set problem. The original algorithm is also below. I was thinking that I could alter line 3 so that it would run until the size of C was equal to n, and change the logic in line 4 so that it would pick and remove vertices until the size of n. A vertex would be removed when the size of C doesn't equal to n but the cover is complete. I can't really think of any other way to do it. The real problem is that I'm not entirely sure how to make the algorithm run in exponential time like they are asking.
GREEDY-SET-COVER can return a number of different solutions, depending on how we break ties in line 4. Give a procedure BAD-SET-COVER-INSTANCE.n/ that returns an n-element instance of the set-covering problem for which, depending on how we break ties in line 4, GREEDY-SET-COVER can return a number of different solutions that is exponential in n.
$X$ — some finite set
$F$ — a family of subsets of $X$
$C$ — cover being constructed
GREEDY-SET-COVER($n$)
1 let $U = X$
2 let $C = \varnothing$
3 while $U \ne \varnothing$
3a select an $S \in F$ that maximizes $\left|S \cap U\right|$
3b set $U = U \setminus S$
3c set $C = C \cup \{S\}$
4 return $C$
Could it be said that since the number of subsets a set has is $2^n$ and that in the worst case this algorithm will end up finding all of those subsets before settling on an n-instance set to return?
• I'm having a hard time understanding what you are asking. You are looking for an algorithm to... do what, exactly? What are the requirements? It's not clear what you are asking for. Are you asking for an exhaustive search algorithm to find the optimal set cover? (If so, the best way to do that is probably not to modify the algorithm you showed us, but rather to start from scratch.) Would this be clear to you, if you were reading it? I suggest you proofread the question and edit it to try to explain more clearly what you want to achieve and what you've already tried.
– D.W.
Nov 24, 2013 at 5:28
• you want the algo to run in exponential time?? Nov 24, 2013 at 9:01
• @D.W. sorry for the confusion. My understanding of the problem wasnt thatvgreat to begin with. After some debate wiyh friends we've settled. On the idea thar they are asking for a set cover that has n elements in it. What really confused me about the question was that the last part about greedy set cover begin capable of returning an exponential amount of solutions. I wasn't sure why that was there. I'll that y to reorder and neaten everything up.
• Please edit to make the algorithm unambiguous. For example, is line 6 a part of the while loop? Nov 26, 2013 at 17:08
I think you are being confused by the way the question is phrased. The question asks for a sequence of instances $I_n$, consisting of triples $\langle X_n,F_n \rangle$, such that $|X_n| = n$ and GREEDY-SET-COVER$(X_n,F_n)$ could have $C^n$ many results, for some constant $C>1$. These results come from the fact that at step 4 there could be many sets that are eligible.
For some reason, the question asks you for an algorithm which generates the sequence: given $n$, it should output $\langle X_n,F_n \rangle$. Since the instances you are going to construct will be pretty simple, you should have no problem writing an algorithm that generates them, but in fact no one cares about this algorithm; the point is only that these "bad" instances exist (they're not actually bad, mind you).
The easiest way to construct these instances is to use a gadget, say $F_3 = \{(1,2),(1,3),(2,3)\}$. You can check that running the greedy algorithm can result in three different covers. I'll let you continue from here. | 949 | 3,771 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-21 | latest | en | 0.956823 |
http://conspiracionredmagdala.blogspot.com/2013/02/strategies-in-mathematics.html | 1,529,660,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00634.warc.gz | 71,039,289 | 16,355 | ## Saturday, February 16, 2013
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The school-year was about to start. Class has now begun for a patterns, create new conjectures and develop truth basing on a different sort of learning by heart may suit basic mathematics was the strategies in mathematics and the strategies in mathematics is false. That is, there is always a quicker and trickier approach to solve entrance examination questions is, that always know that once you are already a mathematics teacher and looking at the strategies in mathematics it will benefit you. | 757 | 4,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-26 | latest | en | 0.938817 |
https://forum.kicad.info/t/trace-width-for-12-a/29615 | 1,685,412,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00325.warc.gz | 305,799,910 | 11,113 | # Trace width for 12 A
Hi all,
I just want to preface this question by disclosing that I’m very new to Kicad so bear with me.
In my PCB design there are two tracks that can carry up to 12A. I used the calculator built in Kicad to determine the width of the tracks (using trace thickness of 0.35 mm) and it came out to ~9.3 mm. Obviously this is large, so I thought using a zones would be appropriate. Below is what I was thinking. Is this a good idea? Is there a better way of approaching this? Also, I find that the zones are fairly close and I’m not sure if that is enough clearance with such high current, but I don’t know to avoid that. Any suggestions are greatly appreciated.
Thanks.
That looks about right, although I do not know how wide your tracks are.
You can make the gap next to the IC a bit wider, but the width of the gap between the copper zones is mostly irrelevant. There is no voltage over it because it’s shorted by the shunt (or Hall sensor) in your IC (I assume it is that).
The track width you calculated probably was based on a temperature rise (default 10deg. C) If you extend the zone to area’s where there will not be current, then those area’s still work as a heatsink to lower the temperature of the zone.
The next step up would be to use thicker copper.
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You can also use Top and Bottom copper together for twice the power.
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Zone are the best thing for this job.
Based upon what you stated about ~9.3mm, this would imply you are using 1oz copper (reversing Kicad value).
The concern is Kicad’s calculator uses IPC-2221 for this and this has been superseded by IPC-2152. Using an IPC-2152 based calculator results in a trackwith of 24.4mm (for 10C temp rise, 1oz copper)
Now some say IPC-2221 is “good enough”, IMHO it isn’t… and a doubling of the needed trackwidth is testament to this. Ill leave it up to you with regards to what width you feel is correct.
As to the layout… I agree with @paulvdh, the separation between the two shapes can be very small, the minimum etch width for this weight of copper (8thou?) because the voltage difference is going to be in uV as the associated SOIC8 part will have a resistance around 100uR
What I would do however was move the terminals as close to the part as possible as this will minimise the volume of copper conducting the current. Likewise I would utilse both sides of the PCB to spread the current across two layer - The banana plugs will provide a double-sided connection so you just need some supporting via’s around the sensor
The usual rule of thumb is a “standard 0.3mm” via with typical plating is good for ~0.5A (for a 10C temprise)… likewise x-number of such a via ~ one via of x-time the diameter … So there is a happy medium of a number of vias of a specific diameter
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Thank you very much. That all makes sense. Do you mind providing me a link with the IPC-2152 calculator that you used? I looked up one online (https://twcalculator.app.protoexpress.com/) and it is saying that the necessary trace width is ~20.9mm for 1 oz copper. I just want to make sure I get this correct.
I also see that using 2 oz copper would be highly beneficial in this case as it reduces the necessary with to ~11.5 mm. Will having the layer on top and bottom also reduce the trace width or does that just limit power dissipation?
Thank you so much for your help.
You can choose: For 2 layers, if you keep the width, power dissipation halves. If you halve the width, power dissipation stays the same compared to 1 layer with twice the width.
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I use SaturnPCB A really handy toolsuite.
One thing to consider about IPC-2152 is they provide curves so all calculators that exist are producing a “best fit” based upon these curves to produce convenient calculators
This is something to consider when you use these as it doesn’t mean one is correct and another is incorrect, there is just a varying degree of inaccuracy in their model - Yes I know this doesn’t help
Using two outer layers will half the current each will carry (assuming they are identical, with identical impedance ) and thus you can reduce the track width from ~20mm down to ~6mm.
The moment you start using internal layers to help share it starts getting complicated as FR4 is a very good thermal insulator and also such internal layers will be heated by other parallel layers.
2 Likes
Thank you so much. Very helpful.
Thank you so much. You’ve made it clear and easier to understand.
IPC or not, all calculations are theoretical. When you plug in the thick cords they will act as heavy duty coolers. On the other hand if you leave the device in the back shelf of a car in the summer sun it will exceed the limits without any current.
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I tend to be conservative, my experience is with PBC’s not Kicad. I worked in Aerospace and Automotive.
1st: The dominant failure mode of excessive current on a PCB trace is overheating a portion of the conductor causing it to start to lift off the board material. Once this happens the temperature of that portion of the copper gets even hotter resulting in a domino failure effect.
2nd: The feedthroughs have no where near the current capability of the surface copper. And are worse the smaller the via hole.
3rd: The copper layer is specified as oz/ft^2. This is because there is no guarantee of the thickness in any given area. I will concede the newest processes are better at controlling plating informativity. But keep in mind you are purchasing “hobby” quality boards not mil spec or automotive spec boards.
4th: IMHO the IPC is a guideline which I view as a minimum. Just as I wouldn’t but 50V on a 50V capacitor, I wouldn’t put the IPC “calculated” guideline on a PCB without some significant derating.
My experience is from working with PCB’s where we had the opportunity to test prototypes and learn how important the derating is.
So unless you have a critical size issue (with the size of the PCB that is) I suggest you err on the side of reliability.
3 Likes
I must be missing something. yes, you could use zones, but 9mm is ~ 1/3 of an inch, not rally a big deal. I run .150 and .200 traces all the time.
of course top and bottom (roughly) halve it again. The basics is that resistance,and therefore heat rise, is proportional to total copper cross-section area.
Your original question was about a method - traces vs. fills. but if you are asking “how to carry lots of current and how critical is it?” - that’s a different issue which i assumed you understood.
To really reply we’d need to know more about the actual load. Constant or vary-ing; duty cycle (peak:average ratio), and any other factor that might influence heat. More copper is better; but note also constriction points - like a chain, its only as good as its weakest link - which becomes a fuse, in a way you don’t want.
1 Oz copper is a “standard” and fine for most low power circuitry. That said, when i have needed o carry LOTS of current, i typically specify 2 Oz. But its also a cost trade off. If you have space for 1" wide traces, and 2 Oz copper quadruples the price (not uncommon) do the wide traces. using top and bottom is great too, so long as you can effectively direct solder to both sides. Vias wont cut it.
And dont neglect the ground return. It needs to carry the same current.
Not exactly;
1. The board is not an infinite heatsink, with heat generating clad on both sides the surface area to dissipate the heat is only marginally increased.
2. The current sensor is an SMD part on the top. I don’t see the feed thru’s conducting much current.
well, if we are being that picky, equally wrong:-) the resistance will in fact fall by half if the cross section is doubled, and therefore generate 1/2 the heat. dissipation is another issue. Statement correct as written, but yes, heat dissipation is better the larger the unshielded surface area.
Let’s help the guy not banter semantics
Its not semantics, its real and you gave the OP advice that is not accurate (or even close)
1. The bottom clad (of equal size) will absolutely NOT carry 1/2 the current. The current in the bottom clad will have to go through the small vias shown near the gap between the two sections. Vias have terrible conductivity compared to copper clad.
2. When dissipating heat with clad on only one side of the board. A fair amount of thermal energy passes through the board and is dissipated on the opposite FR4 surface. When you add clad to the back of the board you essentially remove that dissipative area and replace it with a different dissipation (higher) capability. So it is NOT purely additive as you state.
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If you want to be correct… you need to use a thermal modeling program (ansys) to correctly capture the thermal stackup and the dissipation paths…
This is not trivial … there are soo many things that can aid and hinder you it just isn’t funny and the IPC isn’t a “oh it says use x so ill use x”, you keep adding copper
if unsure, make a test coupon
At least read what has been written.
This really does not warrant a reply, but your assertions are just so far off that i felt compelled, once and only once. As usual, reasonable people will be driven away due to these silly, anal-retentive arguments.
Far above i stated that via are not sufficient. Strike one. But in any event he clearly shows large, through hole tabs that will directly solder to both the top and bottom. Strike two. I checked his drawing and noted it in case i misinterpreted. From what i see, you never read what i wrote nor looked at his diagram. So please stow the flames.
I am trying to keep it simple. Worry about the important, and only then the less important.
Deep breath time? I’ve learned from my son to wait 5 minutes before I hit reply sometimes. You’d be surprised how many posts I don’t post.
4 Likes
Actually it isnt far off… the IPC calculators are base upon extracting “best guess” data point from the PICTURES of the curves in the spec and then doing s best fit against these extracted curves. As a result there will always be an error but typically all the calcs are around 1% from the pictures over the same very small range of current. Also they clearly state single layer (internal or external) with the only modifier they consider valid being whether there is a plane nearby which can act as a heatsink
The moment you talk about using both sides you are drifting further and further away from the empirical test set and you either work with experience or another set of tools: hyperlynx or ansys
As someone that regularly makes PCBs that carry 100+ amps do you thing these “calculators” are used? | 2,439 | 10,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.956037 |
http://lbartman.com/worksheet/grade-4-a-x-b-x-c-multiplication-worksheet.php | 1,611,482,386,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00746.warc.gz | 64,850,271 | 12,116 | lbartman.com - the pro math teacher
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Grade 4 A X B X C Multiplication Worksheet
Public on 12 Oct, 2016 by Cyun Lee
grade 6 a x b x c multiplication worksheet collection
Name : __________________
Seat Num. : __________________
Date : __________________
73 x 40 x 42 = ...
54 x 12 x 54 = ...
46 x 46 x 83 = ...
12 x 76 x 25 = ...
79 x 62 x 68 = ...
94 x 62 x 93 = ...
70 x 57 x 41 = ...
98 x 15 x 28 = ...
19 x 54 x 67 = ...
50 x 50 x 46 = ...
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division property of exponents worksheet | 1,967 | 4,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-04 | latest | en | 0.299176 |
https://www.rdocumentation.org/packages/spatstat/versions/1.3-2/topics/alltypes | 1,610,921,174,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00270.warc.gz | 954,751,311 | 7,265 | alltypes
0th
Percentile
Calculate Statistic for All Types in a Multitype Point Pattern
Given a marked point pattern, this computes the estimates of a selected summary function ($F$,$G$, $J$ or $K$) of the pattern, for all possible combinations of marks. It returns these functions in a list (an object of class "fasp") amenable to plotting by plot.fasp().
Keywords
spatial
Usage
alltypes(pp, fun="K",dataname=NULL,verb=FALSE)
Arguments
pp
The observed point pattern, for which summary function estimates are required. An object of class "ppp". If the pattern is not marked, the resulting array'' is $1 \times 1$.
fun
Character string indicating the summary function required. Must be one of the letters "F", "G", "J", "K".
dataname
Character string giving an optional (alternative) name to the point pattern, different from what is given in the call. This name, if supplied, may be used by plot.fasp() in forming the title
verb
Logical value, meaning verbose''. If verb is true then terse progress reports'' (just the values of the mark indices) are printed out when the calculations for that combination of marks are completed.
Details
This routine is a convenient way to analyse the dependence between types in a multitype point pattern. Suppose that the points have possible types $1,2,\ldots,m$ and let $X_i$ denote the pattern of points of type $i$ only. If fun="F" then this routine calculates, for each possible type $i$, an estimate of the Empty Space Function $F_i(r)$ of $X_i$. If fun is "G", "J" or "K", the routine calculates, for each pair of types $(i,j)$, an estimate of the cross-type function $G_{ij}(r)$, $J_{ij}(r)$ or $K_{ij}(r)$ respectively describing the dependence between $X_i$ and $X_j$.
The real work is done by the functions Fest, Gest, Kest, Jest, Gcross, Kcross, and Jcross. One of the first four functions (according to fun) is invoked if the two marks under consideration are equal. The latter three are invoked if the marks are distinct. (There is no Fcross; for the empty space function $F(r)$ there is no cross-type version.)
Value
• A function array (an object of class "fasp", see fasp.object). This can be plotted using plot.fasp.
If fun="F", the function array has dimensions $m \times 1$ where $m$ is the number of different marks in the point pattern. The entry at position [i,1] in this array is the result of applying Fest to the points of type i only.
If fun is "G", "J" or "K", the function array has dimensions $m \times m$. The [i,j] entry of the function array (for $i \neq j$) is the result of applying the function Gcross, Jcross or Kcross to the pair of types (i,j). The diagonal [i,i] entry of the function array is the result of applying the univariate function Gest, Jest or Kest to the points of type i only. Each function entry fns[[i]] retains the format of the output of the relevant estimating routine Fest, Gest, Jest, Kest, Gcross, Jcross, or Kcross.
The default formulae for plotting these functions are cbind(km,theo) ~ r for F, G, and J, and cbind(trans,theo) ~ r for K.
Note
Sizeable amounts of memory may be needed during the calculation.
plot.fasp, fasp.object, allstats, Fest, Gest, Jest, Kest, Gcross, Jcross, Kcross
• alltypes
Examples
library(spatstat)
# bramblecanes (3 marks).
data(bramblecanes)
X.F <- alltypes(bramblecanes,fun="F",verb=TRUE)
plot(X.F)
X.G <- alltypes(bramblecanes,fun="G",verb=TRUE)
X.J <- alltypes(bramblecanes,fun="J",verb=TRUE)
X.K <- alltypes(bramblecanes,fun="K",verb=TRUE)
<testonly># smaller dataset
bram <- bramblecanes[seq(1, bramblecanes$n, by=20), ] X.F <- alltypes(bram,fun="F",verb=TRUE) X.G <- alltypes(bram,fun="G",verb=TRUE) X.J <- alltypes(bram,fun="J",verb=TRUE) X.K <- alltypes(bram,fun="K",verb=TRUE)</testonly> # Swedishpines (unmarked). data(swedishpines) X.F <- alltypes(swedishpines,fun="F") X.G <- alltypes(swedishpines,fun="G") X.J <- alltypes(swedishpines,fun="J") X.K <- alltypes(swedishpines,fun="K") # simulated data pp <- runifpoint(350, owin(c(0,1),c(0,1))) pp$marks <- factor(c(rep(1,50),rep(2,100),rep(3,200)))
X.F <- alltypes(pp,fun="F",verb=TRUE,dataname="Fake Data")
X.G <- alltypes(pp,fun="G",verb=TRUE,dataname="Fake Data")
X.J <- alltypes(pp,fun="J",verb=TRUE,dataname="Fake Data")
X.K <- alltypes(pp,fun="K",verb=TRUE,dataname="Fake Data")
# A setting where you might REALLY want to use dataname:
xxx <- alltypes(ppp(Melvin$x,Melvin$y,
window=as.owin(c(5,20,15,50)),marks=clyde),
fun="F",verb=TRUE,dataname="Melvin")
Documentation reproduced from package spatstat, version 1.3-2, License: GPL version 2 or newer
Community examples
Looks like there are no examples yet. | 1,325 | 4,620 | {"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} | 2.984375 | 3 | CC-MAIN-2021-04 | latest | en | 0.759049 |
https://www.causeweb.org/wiki/chance/index.php?title=Chance_News_73&direction=prev&oldid=13534 | 1,540,342,097,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00146.warc.gz | 902,102,894 | 13,519 | # Chance News 73
## Quotations
From The Flaw of Averages, by Sam L. Savage, Wiley, 2009
• “Our culture encodes a strong bias either to neglect or ignore variation. We tend to focus instead on measures of central tendency, and as a result we make some terrible mistakes, often with considerable practical import.” (Stephen Jay Gould, cited p. 11)
• “Plans based on average assumptions are wrong on average.” (Savage, p. 11)
• “Far better an approximate answer to the right question, which is often vague, than the exact answer to the wrong question, which can always be made precise.” (John W. Tukey, cited p. 38)
• “I have found that teaching probability and statistics is easy. The hard part is getting people to learn the stuff.” (Savage, p. 49)
• “Statisticians often describe a numerical uncertainty using the Red Words, RANDOM VARIABLE, but I will stick with ‘uncertain number.’ …. [S]top thinking of uncertainties as single numbers and begin thinking of them as shapes, or distributions. …. If you think of an uncertain number as a bar graph, you will not be seriously misled.” (Savage, p. 59ff)
• “Joe Berkson, a statistician at the Mayo Clinic, developed his own criterion, which he termed the IOT Test, or Inter Ocular Trauma Test, requiring a graph that hit you between the eyes.” (Savage, p. 325)
See Chance News 52 for a review of The Flaw of Averages by Laurie Snell.
Submitted by Margaret Cibes
“[O]n average Bill Gates and I can afford a new Rolls and a winter home in Provence.” (p. 36)
Howard Wainer, in Picturing the Uncertain World Princeton University Press, 2009
Submitted by Margaret Cibes
"Using a model of no greater sophistication than that employed by Benjamin Franklin (weather generally moves from west to east), I was able to predict that the area of precipitation currently over Ohio would be hitting New Jersey by tomorrow and would stay over us until the weekend. Any fool could see it. The improvement in forecasting has not been entirely due to improvements in the mathematical models of the weather. The enormous wealth of radar and satellite data summarized into a multicolored and dynamic graph can turn anyone into an expert."
Wainer, in Graphic Discovery A Trout in the Milk and Other Visual Adventures, p. 15
Submitted by Paul Alper
"This is about visual thinking and visual evidence …. It's not about commercial art. The last thing in the world that's needed here is a designer. What's needed is an analytical, statistical, quantitative approach. Reporting is different from pitching. Artists who design for marketing purposes inherently have problems with credibility. This is something very different in spirit. It's about accountability and transparency—with heavy, heavy amounts of data."
Remarks by Edward Tufte, on describing his assignment to help “track and explain \$787 billion in recovery stimulus funds,” as a member of the Recovery Independent Advisory Panel (2010-2011), to which he was appointed by President Obama; cited by blogger at Bloomberg Businessweek[1]
Submitted by Margaret Cibes
## Forsooth
“It is Friday 13th today and though it is still only ten in the morning some awfully unlucky things have happened. I stubbed my toe; the cat caught a shrew and left it in the middle of the kitchen floor, which was unlucky for me because I almost stepped on it, and was even more unlucky for the shrew. It is a black cat too. Clear evidence that superstition works, even for small rodents. Or perhaps not. Yesterday I broke my fingernail, but it wasn’t Friday 13th then, so that wasn’t the fates being lined up against me, it was just an accident.”
Julian Champkin, in “Friday 13th and black cats”, Significance online, May 2011
Submitted by Margaret Cibes
"[A] Public Policy Polling survey released Thursday found that Gingrich's favorable rating with GOP voters has dropped 27 points in the last month--from 52 percent to 38 percent."
Yahoo News, 26 May 2011
Submitted by Paul Alper
"On another occasion, Bailey and other staffers spent hours voting repeatedly to manipulate a television opinion poll on Palin’s decision to reject part of the federal government’s economic stimulus funding."
Review of Blind Allegiance to Sarah Palin,
Washington Post, 19 May 2011
Never mind the politics--this illustrates the worth of [voluntary] television opinion polls.
Submitted by Paul Alper
## What's in a name?
Peter, Deborah popular names for CEOs
VPR News Morning Edition, 29 April 2011
"If your name is Peter or Deborah, you're more likely to be a CEO. That's what the social networking site LinkedIn found." You can listen to the rest of this Vermont Public Radio broadcast here.
The story was featured in a variety of news outlets:
Discussion Questions
1. A tweeted comment on this site says: "Great analysis, although this can be explained mostly by the age group ..." What are the implications of this? How might you explore them?
2. Paul Alper pointed out a post from Andrew Gelman's blog, where are reader cited a gruesome infographic that accompanied the original LinkedIn story.
Submitted by Jeanne Albert
## Scaling the normal curve
Picturing the Uncertain World
by Howard Wainer, Princeton, 2009, p. 171.
This book is a collection of articles that Wainer had authored/co-authored in Chance (2000-2007), American Scientist (2007), and American Statistician (1996).
In Chapter 16, "Galton's Normal," Wainer gives an example of the relative heights of the points on a standard normal curve and of why our sketches of normal curves do not, and cannot, come close to accurate scale drawings.
He calculates that, even if the height at z = 13 were only 1 mm, then the height of the normal curve at the center, z = 0, would be about 5 x 10^30 km, or 5.3 x 10^17 light years. This is equivalent to a height that would be 3.4 million times larger than the universe. (His figures check out.)
Even if the height were 1 mm at z = 6, the height at z = 0 would be 66 km. Thus it still could not be drawn to scale.
Submitted by Margaret Cibes
## Defined by rankings
In a data-heavy society, being defined by the numbers
by Alina Tugend , New York Times, 22 April 2011
On a humorous note, the author confesses that since joining Twitter she can't help regularly checking her number of followers. But the more serious question is this: Are we as a society too dependent on numerical rankings? The article quotes MIT professor Sherry Turkle: "One of the fantasies of numerical ranking is that you know how you got there. But the problem is if the numbers are arrived at in an irrational way, or black-boxed, so we don’t understand how we got there, then what use are they? "
The article gives several examples, two of which happen to correspond to stories that were recently discussed in Chance News 71, namely college rankings and New York City's formula for rating teachers.
Submitted by Bill Peterson
“Matching Science of DNA With Art of Identification”
by Carl Bialik, The Wall Street Journal, May 7, 2011
“DNA and Bin Laden’s Positive ID”
by Carl Bialik, The Numbers Guy Blog, The Wall Street Journal, May 6, 2011
Bialik discusses the prosecutor’s fallacy in the context of reporting about the identification of Osama bin Laden’s body. He does not dispute the identification made by government officials, said to have been based on a number of factors, including DNA. However, he reminds readers that “claimed match probabilities, such as 99% or 99.99%, can be misstated or misleading” and that further details about a DNA test, as well as evidence related to other factors, must be taken into account before having confidence in an identification.
The problem boils down to this: A very small chance of a false positive in a genetic test isn't the same thing as a very large chance of a positive identification.
One complicating factor with interpreting genetic-identity tests … is that the probability of a positive match depends on what other information is available to confirm or reject it — despite the so-called prosecutor’s fallacy that confuses the two. These other factors, collected in what is called “prior odds” of a positive match, can be difficult to measure. “Many factors (e.g., age, sex, appearance, clothes, etc.) are relevant to prior odds, and there are no standard rule[s] for quantify[ing] them, [according to a University of South Texas scientist].”
Submitted by Margaret Cibes
## Superstitution
“Friday 13th and black cats”
by Julian Champkin, Significance online, May 2011
Champkin wrote this brief column about superstitution, probably because May 13, 2011, was a “Friday the 13th.” (See his Forsooth quotation above.)
In one part of the column, he says that there are 25 finalists, on average, in the Eurovision Song Contest and goes on to suggest a winning strategy for betting:
Did you know that if you touch your left ear with your right thumb and wiggle your toes when the country you want to win begins to sing, that country will inevitably lose? It is a superstition that I have just invented, but I bet it works. Try it tomorrow and see.
Champkin goes on to say, “My bet will work overwhelmingly well, on average.”
### Discussion
Give a statistical reason why Champkin’s method would work well, on average, with or without touching your ear and wiggling your toes.
Submitted by Margaret Cibes
## Think tanks and common sense
On the economics of mass transit and the value of common sense
by Nate Silver, FiveThirtyEight blog, New York Times, 20 May 2011
Silver criticizes a Brookings Institution study of mass transit in the U.S.; he is surprised that it came to such strange numerical results: "New York, however, ranked just 13th. Washington ranked 17th. And Chicago ranked 46th — well behind Los Angeles (24th). Instead, the top 10 metro areas [for public transport] according to Brookings were Honolulu; San Jose, Calif.; Salt Lake City; Tucson; Fresno, Calif.; Denver; Albuquerque; Las Vegas; Provo, Utah; and Modesto, Calif."
He concludes with:
I want to point out that just because a study uses objective criteria, that doesn’t make it sensible. In fact, studies that try to rank or rate things seem especially susceptible to slapdash, unthoughtful methodology (here is another example: a study which concludes that Gainesville, Fla., is a more gay-friendly city than San Francisco). If you come up with a result that defies common sense — like Modesto’s having better public transit than New York — then once in a blue moon, you may be on to something: conventional wisdom is fallible. But much, much more often, it’s a sign that you’ve done something wrong, and it’s time to reconsider your assumptions before publishing.
Submitted by Paul Alper
## Data display website
Nathan Yau is a Ph.D.candidate in Statistics at UCLA, who has a wonderful website about data displays, FlowingData. It contains lots of examples, including “ugly” ones[2].
Yau also has a book, Visualize This[3], coming out in July from Wiley, that “teaches you how to create graphics that tell stories with real data, and … have fun in the process.” It is said to help the reader to “make statistical graphics in R, design in Illustrator, and create interactive graphics in JavaScript and Flash & Actionscript.”
Submitted by Margaret Cibes based on a reference in Significance online[4]
## Former UK Prime Minister as statistician
According to an article in Significance online[5], Harold Wilson, who served twice as UK Prime Minister (in the 1960s and the 1970s), was educated as an economist and also worked as a statistician, serving a term as President of the Royal Statistical Society (in the 1970s).
Submitted by Margaret Cibes
## Unemployment vs. presidential support
“On the Maddeningly Inexact Relationship Between Unemployment and Re-Election”
by Nate Silver, FiveThirtyEight blog, The New York Times, June 2, 2011
Silver warns that, although higher unemployment in November 2012 will probably mean a lower likelihood of Obama being re-elected, there is no “magic number” for a rate, or rising/falling behavior of that rate, that would be sufficient by itself to predict the outcome of that election.
He provides a table[6] of raw data, for presidents elected in the period 1912-2008: unemployment rates at the start of a term and at the election date, the percent changes in rates, and the popular vote margins. He also provides six scatter plots:
(a) unemployment rates and margins of victory for incumbent parties 1912-2008 (R^2=0.0004) and 1948-2008 (R^2=0.0119)
(b) changes in unemployment rates and margins of victory for incumbent parties 1912-2008 (R^2=0.1222) and 1948-2008 (R^2=0.0106)
(c) unemployment rates and margins of victory when incumbent presidents sought second terms 1912-2008 (R^2=0.0162) and changes in rates for those presidents 1912-2008 (R^2=0.265).
While noting that the limited number of data points, as well as other aspects of the economy and voter attitudes toward it, make drawing any conclusions difficult, he feels that common sense tells us that “the unemployment rate should have some effect on a president’s re-election chances.”
He concludes, “[T]his is an inexact science – more so than either journalists or political scientists tend to acknowledge.” | 3,027 | 13,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.948755 |
https://exercism.io/tracks/nim/exercises/binary/solutions/5c6cab4aa06a4909a5d57ea2a9964447 | 1,628,193,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00435.warc.gz | 258,256,975 | 6,749 | # matsu911's solution
## to Binary in the Nim Track
Published at Feb 02 2019 · 0 comments
Instructions
Test suite
Solution
#### Note:
This exercise has changed since this solution was written.
Convert a binary number, represented as a string (e.g. '101010'), to its decimal equivalent using first principles.
Implement binary to decimal conversion. Given a binary input string, your program should produce a decimal output. The program should handle invalid inputs.
## Note
• Implement the conversion yourself. Do not use something else to perform the conversion for you.
Decimal is a base-10 system.
A number 23 in base 10 notation can be understood as a linear combination of powers of 10:
• The rightmost digit gets multiplied by 10^0 = 1
• The next number gets multiplied by 10^1 = 10
• ...
• The nth number gets multiplied by 10^(n-1).
• All these values are summed.
So: `23 => 2*10^1 + 3*10^0 => 2*10 + 3*1 = 23 base 10`
Binary is similar, but uses powers of 2 rather than powers of 10.
So: `101 => 1*2^2 + 0*2^1 + 1*2^0 => 1*4 + 0*2 + 1*1 => 4 + 1 => 5 base 10`.
## Running the tests
To compile and run the tests, just run the following in your exercise directory:
``````\$ nim c -r binary_test.nim
``````
## Submitting Exercises
Note that, when trying to submit an exercise, make sure the solution is in the `\$EXERCISM_WORKSPACE/nim/binary` directory.
You can find your Exercism workspace by running `exercism debug` and looking for the line that starts with `Exercises Directory`.
## Source
All of Computer Science http://www.wolframalpha.com/input/?i=binary&a=*C.binary-_*MathWorld-
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### binary_test.nim
``````import unittest
import binary
suite "Binary":
test "binary 0 is decimal 0":
check binary("0") == 0
test "binary 1 is decimal 1":
check binary("1") == 1
test "binary 10 is decimal 2":
check binary("10") == 2
test "binary 11 is decimal 3":
check binary("11") == 3
test "binary 100 is decimal 4":
check binary("100") == 4
test "binary 1001 is decimal 9":
check binary("1001") == 9
test "binary 11010 is decimal 26":
check binary("11010") == 26
test "binary 10001101000 is decimal 1128":
check binary("10001101000") == 1128
check binary("000011111") == 31
test "2 is not a valid binary digit":
expect(ValueError):
test "a number containing a non-binary digit is invalid":
expect(ValueError):
test "a number with trailing non-binary characters is invalid":
expect(ValueError):
test "a number with leading non-binary characters is invalid":
expect(ValueError):
test "a number with internal non-binary characters is invalid":
expect(ValueError):
test "a number and a word whitespace spearated is invalid":
expect(ValueError):
``````import strutils, math, algorithm
func binary*(s: string): int =
for i, c in s.reversed:
let n = (\$c).parseInt()
if n < 0 or n > 1:
raise newException(ValueError, "")
result += 2 ^ i * n`````` | 818 | 3,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-31 | latest | en | 0.820942 |
https://4gravitons.com/2014/07/11/feeling-perturbed/ | 1,674,885,385,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00497.warc.gz | 97,188,194 | 29,351 | # Feeling Perturbed?
You might think of physics as the science of certainties and exact statements: action and reaction, F=ma, and all that. However, most calculations in physics aren’t exact, they’re approximations. This is especially true today, but it’s been true almost since the dawn of physics. In particular, approximations are performed via a method known as perturbation theory.
Perturbation theory is a trick used to solve problems that, for one reason or another, are too difficult to solve all in one go. It works by solving a simpler problem, then perturbing that solution, adjusting it closer to the target.
To give an analogy: let’s say you want to find the area of a circle, but you only know how to draw straight lines. You could start by drawing a square: it’s easy to find the area, and you get close to the area of the circle. But you’re still a long ways away from the total you’re aiming for. So you add more straight lines, getting an octagon. Now it’s harder to find the area, but you’re closer to the full circle. You can keep adding lines, each step getting closer and closer.
And so on.
This, broadly speaking, is what’s going on when particle physicists talk about loops. The calculation with no loops (or “tree-level” result) is the easier problem to solve, omitting quantum effects. Each loop then is the next stage, more complicated but closer to the real total.
There are, as usual, holes in this analogy. One is that it leaves out an important aspect of perturbation theory, namely that it involves perturbing with a parameter. When that parameter is small, perturbation theory works, but as it gets larger the approximation gets worse and worse. In the case of particle physics, the parameter is the strength of the forces involves, with weaker forces (like the weak nuclear force, or electromagnetism) having better approximations than stronger forces (like the strong nuclear force). If you squint, this can still fit the analogy: different shapes might be harder to approximate than the circle, taking more sets of lines to get acceptably close.
Where the analogy fails completely, though, is when you start approaching infinity. Keep adding more lines, and you should be getting closer and closer to the circle each time. In quantum field theory, though, this frequently is not the case. As I’ve mentioned before, while lower loops keep getting closer to the true (and experimentally verified) results, going all the way out to infinite loops results not in the full circle, but in an infinite result instead. There’s an understanding of why this happens, but it does mean that perturbation theory can’t be thought of in the most intuitive way.
Almost every calculation in particle physics uses perturbation theory, which means almost always we are just approximating the real result, trying to draw a circle using straight lines. There are only a few theories where we can bypass this process and look at the full circle. These are known as integrable theories. N=4 super Yang-Mills may be among them, one of many reasons why studying it offers hope for a deeper understanding of particle physics.
## 2 thoughts on “Feeling Perturbed?”
1. Cliff Harvey
Congrats on the new site! It looks great.
I’d love to see more posts exploring detailed aspects of amplitudology, N=4 SYM, (2,0) theory, SUGRA, superstrings, etc. Especially for folks who are already somewhat knowledgable on the basics.
In this post I wasn’t expecting to learn much new, but your reference to the renormalon wikipedia article has lead me to some references that seem helpful to understand aspects of the divergence of the all-loop perturbative amplitudes.
I have a question about the relationship between instantons and the renormalon divergences. When I hear some experts writing about these matters, it seems like perhaps more precise statements are possible linking the perturbative divergences to the onset of non-perturbative effects in supersymmetric theories, perhaps especially in N=4 or other special cases. I suppose I would have already heard about it if there was a full proof, but is there some sense in which we can make stronger arguments that the full amplitudes consist of the perturbative part plus known non-perturbative effects?
In general I’m interested in how well we can establish the linkage between these two things, either in the SM or in special theories.
Like | 936 | 4,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-06 | longest | en | 0.927029 |
https://www.cemc.uwaterloo.ca/pandocs/potw/2021-22/English/POTWD-21-A-N-19-P.html | 1,653,292,593,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00150.warc.gz | 786,355,241 | 1,586 | # Problem of the Week Problem D Let’s Hit the Pool
In Wei’s family, there are four children and three adults. Every weekend they all go swimming together. To use the public swimming pool, each person needs a ticket.
Wei’s parents buy their tickets in bulk and keep them in a box. At the beginning of the year the ratio of adult to child tickets in the box was $$11:14$$.
Wei’s family used the tickets every weekend to go swimming until they no longer had enough tickets for everyone in their family. At that point, there were no child tickets left in the box and $$3$$ adult tickets left in the box. How many tickets were in the box at the beginning of the year? | 148 | 665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-21 | longest | en | 0.987273 |
http://slideplayer.com/slide/2512393/ | 1,534,303,791,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209856.3/warc/CC-MAIN-20180815024253-20180815044253-00719.warc.gz | 377,094,442 | 41,241 | # Chapter 4 Components & Circuits (Part 1)
## Presentation on theme: "Chapter 4 Components & Circuits (Part 1)"— Presentation transcript:
Chapter 4 Components & Circuits (Part 1)
General License Class Chapter 4 Components & Circuits (Part 1)
Electrical Review Current, Voltage, & Power Current (I) Voltage (E)
Movement of electrons past a given point. Unit of Measurement = Ampere (A) 1 Ampere = × 1018 electrons/second Voltage (E) Electromotive Force Unit of Measurement = Volt (V) Power (P) Rate of at which energy is transferred, used, or transformed. Unit of Measurement = Watt (W)
Electrical Review P E I Power Formula P = E x I E = P / I I = P / E
Watts E I Volts Amps P = E x I E = P / I I = P / E
Voltage = Current x Resistance
Electrical Review Resistance & Ohm’s Law Resistance (R) Opposition to movement of electrons. Unit of Measurement = Ohm (Ω). Voltage, current, & resistance are all related by Ohm’s Law. Voltage = Current x Resistance
Electrical Review E I R Ohm’s Law E = I x R I = E / R R = E / I Volts
Amps Ohms E = I x R I = E / R R = E / I
Electrical Review More Power Equations
Combining Ohm’s Law (E = I x R) with the power equation (P = E x I) gives us 2 more ways to calculate power: P = E2 / R P = I2 x R
Electrical Review AC and DC Waveforms Direct Current (DC)
Current that always flows in the same direction. DC Voltage Voltage that always has the same polarity.
Electrical Review AC and DC Waveforms Alternating Current (AC)
Current that reverses direction of current flow. AC Voltage Voltage that changes polarity.
Electrical Review AC and DC Waveforms Frequency
Rate at which voltage changes polarity or current changes direction. Unit of Measurement = Hertz (Hz). 1 Hz = 1 cycle per second.
Electrical Review AC and DC Waveforms Wavelength
Radio waves travel at the speed of light. 186,000 miles/second 300,000,000 meters/second 300 x 106 meters/second
Electrical Review AC and DC Waveforms Wavelength
The distance a radio wave travels during the time it takes to complete one cycle.
Electrical Review 300 f λ Wavelength
MHz meters 300 = f x λ f = 300 / λ λ = 300 / f
Electrical Review AC and DC Waveforms Electromagnetic Spectrum
Electrical Review Series and Parallel Circuits Series Circuit.
Only one path for current to flow. Current is same through each device.
Electrical Review Series and Parallel Circuits Parallel Circuit.
Multiple paths for current to flow. Voltage is same across each device.
Electrical Review Decibels Measures a ratio. Logarithmic scale.
Power Ratio: dB = 10 log10 (P1/P2) Voltage Ratio: dB = 20 log10 (V1/V2)
Electrical Review dB Power Ratio Voltage Ratio 1.000 -1 0.794 0.89 -2
1.000 -1 0.794 0.89 -2 0.631 0.79 -3 0.501 0.707 -4 0.398 -5 0.316 0.562 -6 0.250 -7 0.200 0.447 -8 0.159 -9 0.126 0.355 -10 0.100 dB Power Ratio Voltage Ratio 1.000 1 1.259 1.122 2 1.585 3 1.995 1.414 4 2.512 5 3.162 1.778 6 4.000 7 5.012 2.239 8 6.310 9 7.943 2.818 10 10.00 3.16
G5B01 -- A two-times increase or decrease in power results in a change of how many dB?
A. Approximately 2 dB B. Approximately 3 dB C. Approximately 6 dB D. Approximately 12 dB
G5B01 -- A two-times increase or decrease in power results in a change of how many dB?
A. Approximately 2 dB B. Approximately 3 dB C. Approximately 6 dB D. Approximately 12 dB
A. 0.5 watts B. 200 watts C. 400 watts D. 3200 watts
G5B03 -- How many watts of electrical power are used if 400 VDC is supplied to an 800-ohm load? A. 0.5 watts B. 200 watts C. 400 watts D watts
A. 0.5 watts B. 200 watts C. 400 watts D. 3200 watts
G5B03 -- How many watts of electrical power are used if 400 VDC is supplied to an 800-ohm load? A. 0.5 watts B. 200 watts C. 400 watts D watts
A. 2.4 watts B. 24 watts C. 6 watts D. 60 watts
G5B04 -- How many watts of electrical power are used by a 12-VDC light bulb that draws 0.2 amperes? A. 2.4 watts B. 24 watts C. 6 watts D. 60 watts
A. 2.4 watts B. 24 watts C. 6 watts D. 60 watts
G5B04 -- How many watts of electrical power are used by a 12-VDC light bulb that draws 0.2 amperes? A. 2.4 watts B. 24 watts C. 6 watts D. 60 watts
G5B05 -- How many watts are dissipated when a current of 7
G5B05 -- How many watts are dissipated when a current of 7.0 milliamperes flows through 1.25 kilohms? A. Approximately 61 milliwatts B. Approximately 61 watts C. Approximately 11 milliwatts D. Approximately 11 watts
G5B05 -- How many watts are dissipated when a current of 7
G5B05 -- How many watts are dissipated when a current of 7.0 milliamperes flows through 1.25 kilohms? A. Approximately 61 milliwatts B. Approximately 61 watts C. Approximately 11 milliwatts D. Approximately 11 watts
G5B10 -- What percentage of power loss would result from a transmission line loss of 1 dB?
A. 10.9% B. 12.2% C. 20.5% D. 25.9%
G5B10 -- What percentage of power loss would result from a transmission line loss of 1 dB?
A. 10.9% B. 12.2% C. 20.5% D. 25.9%
AC Power RMS: Definition and Measurement
A DC voltmeter will read the average voltage, which is zero.
AC Power RMS: Definition and Measurement
With an oscilloscope, it is easy to read the peak (maximum) voltage.
AC Power RMS: Definition and Measurement
A current will heat up a resistor. The amount of DC current that causes the same amount of heating as the AC current does is the root-mean-square (RMS) value of the AC current. RMS = x Peak (Sine waves ONLY!)
AC Power RMS: Definition and Measurement 1 = Peak 2 = Peak-to-Peak
3 = Root-Mean-Square (RMS)
AC Power RMS: Definition and Measurement To Calculate Sine Wave
Square Wave RMS 0.707 x Peak Peak 1.414 x RMS
AC Power PEP: Definition and Measurement PEP = Peak Envelope Power
Average power over one complete cycle at the peak of the RF envelope.
AC Power PEP: Definition and Measurement PEP = Peak Envelope Power
Measure VP or VP-P using an oscilloscope. VP-P = 2 x VP Calculate VRMS from VP. VRMS = x VP Calculate PEP from VRMS and load imdepance. PEP = VRMS2 / Rload PEP is equal to the average power if no modulation or if FM-modulated.
A. 1.4 watts B. 100 watts C. 353.5 watts D. 400 watts
G5B06 -- What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output? A. 1.4 watts B. 100 watts C watts D. 400 watts
A. 1.4 watts B. 100 watts C. 353.5 watts D. 400 watts
G5B06 -- What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output? A. 1.4 watts B. 100 watts C watts D. 400 watts
G5B07 -- Which value of an AC signal results in the same power dissipation as a DC voltage of the same value? A. The peak-to-peak value B. The peak value C. The RMS value D. The reciprocal of the RMS value
G5B07 -- Which value of an AC signal results in the same power dissipation as a DC voltage of the same value? A. The peak-to-peak value B. The peak value C. The RMS value D. The reciprocal of the RMS value
A. 84.8 volts B. 169.7 volts C. 240.0 volts D. 339.4 volts
G5B08 -- What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts? A volts B volts C volts D volts
A. 84.8 volts B. 169.7 volts C. 240.0 volts D. 339.4 volts
G5B08 -- What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts? A volts B volts C volts D volts
A. 8.5 volts B. 12 volts C. 24 volts D. 34 volts
G5B09 -- What is the RMS voltage of a sine wave with a value of 17 volts peak? A. 8.5 volts B. 12 volts C. 24 volts D. 34 volts
A. 8.5 volts B. 12 volts C. 24 volts D. 34 volts
G5B09 -- What is the RMS voltage of a sine wave with a value of 17 volts peak? A. 8.5 volts B. 12 volts C. 24 volts D. 34 volts
G5B11 -- What is the ratio of peak envelope power to average power for an unmodulated carrier?
G5B11 -- What is the ratio of peak envelope power to average power for an unmodulated carrier?
A. 173 volts B. 245 volts C. 346 volts D. 692 volts
G5B12 -- What would be the RMS voltage across a 50-ohm dummy load dissipating 1200 watts? A. 173 volts B. 245 volts C. 346 volts D. 692 volts
A. 173 volts B. 245 volts C. 346 volts D. 692 volts
G5B12 -- What would be the RMS voltage across a 50-ohm dummy load dissipating 1200 watts? A. 173 volts B. 245 volts C. 346 volts D. 692 volts
A. 530 watts B. 1060 watts C. 1500 watts D. 2120 watts
G5B13 -- What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts? A. 530 watts B watts C watts D watts
A. 530 watts B. 1060 watts C. 1500 watts D. 2120 watts
G5B13 -- What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts? A. 530 watts B watts C watts D watts
A. 8.75 watts B. 625 watts C. 2500 watts D. 5000 watts
G5B14 -- What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm resistor connected to the transmitter output? A watts B. 625 watts C watts D watts
A. 8.75 watts B. 625 watts C. 2500 watts D. 5000 watts
G5B14 -- What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm resistor connected to the transmitter output? A watts B. 625 watts C watts D watts
Basic Components Definitions:
Nominal Value -- Intended value of the component. Tolerance -- Amount value of the component may vary from the nominal value. Temperature Coefficient -- Amount & direction value of component changes with changes in temperature. Power/Voltage/Current Rating -- Maximum power/voltage/current the component will withstand before damage occurs.
Basic Components Resistors & Resistance Resistance
Opposition to the flow of electrons. Converts electrical energy to heat. Unit of measurement = Ohm (Ω). Symbol used in equations = R. Components designed to provide resistance are called “resistors”.
Basic Components Resistors & Resistance Resistors
Resistances range from <1 Ω to >10 MΩ Ω = ohms kΩ = kilohms (103 ohms) MΩ = megohms (106 ohms) Tolerances of 0.1% to 20%. Temperature coefficients can be positive or negative depending on material. Positive = Value increases as temperature increases. Negative = Value decreases as temperature increases.
Basic Components Resistors & Resistance
Resistances range from <1 Ω to >20 MΩ Values often indicated by colored bands on body.
Basic Components Color Value Silver* 0.01 Gold* 0.1 Black Brown 1 Red
Brown 1 Red 2 Orange 3 Yellow 4 Green 5 Blue 6 Violet 7 Gray 8 White 9 * Tolerance & multiplier only
Basic Components Resistors & Resistance Resistor types
Carbon composition. <1 Ω to 22 MΩ. 1/8 to 2 Watts. Tolerance 5%, 10%, & 20% Poor temperature stability Along with wirewound, oldest technology. Not commonly used after 1970.
Basic Components Resistors & Resistance Resistor types Carbon film.
1 Ω to 10 MΩ. 1/8 to 5 Watts. Wide operating temperature range.
Basic Components Resistors & Resistance Resistor types Metal Film.
<1 Ω to >10 MΩ. 1/8 to 2 Watts. Tolerance 0.1%, 1%, & 2%. Good temperature stability. Low noise. Most commonly used today.
Basic Components Resistors & Resistance Resistor types
Metal-oxide film. Similar to metal film. Higher operating temperatures. Higher temperature stability. Low stray inductance. Good for RF circuits.
Basic Components Resistors & Resistance Resistor types Wirewound.
High power. Up to 200 Watts or more. High inductance – not good for RF circuits. May have metal case for attaching to a heat sink. May be tapped to adjust value.
Basic Components Resistors & Resistance Resistor types Thermistor.
Special type of resistor with precisely known temperature coefficient. Both positive (PTC) & negative (NTC) temperature coefficients are available. Used for temperature sensing.
Basic Components Resistors & Resistance Resistor types
Variable Resistors. Potentiometers. Rheostats Materials Graphite. Cermet. Wirewound. Taper. Linear. Semi-Log.
Basic Components Resistors & Resistance Parasitic inductance.
Parasitic inductance changes characteristics of resistor at high frequencies. Use low-inductance resistors in RF circuits. Carbon composition. Carbon film. Metal film. Metal-oxide film. Avoid high-inductance resistors. Wirewound.
Basic Components Inductors & Inductance Inductance.
Ability to store energy in a magnetic field. Opposes a change in current flow. Unit of Measurement = Henry (H) Symbol used in equations = L. Components designed to provide inductance are called “inductors” or “coils”.
Basic Components Inductors & Inductance Inductors.
Inductances range from <1 μH to >1 H. H = henries mH = millihenries (10-3 henries) μH = microhenries (10-6 henries)
Basic Components Inductors & Inductance Inductors. Different shapes.
Solenoidal Toroidal.
Basic Components Inductors & Inductance Inductors.
Different core materials. Laminated iron. Used in high-inductance, low-frequency applications such as power supply filter chokes. Powdered iron or ferrite. Used in medium-inductance applications. Air. Used in low-inductance, high-frequency applications such as transmitting coils.
Basic Components Inductors & Inductance Inductors. Mutual inductance.
If the magnetic field from one inductor extends to another inductor, then the current flowing in the 1st inductor will effect the current flowing in the 2nd inductor. This is called mutual inductance or transformer action. Usually undesirable. Minimizing mutual inductance. Shield with magnetic material. Place solenoidal coils at right angles to one another. Use toroidal cores.
Basic Components Inductors & Inductance Inductors.
At high frequencies, inter-turn capacitance can become significant. Inductor can become self-resonant.
Basic Components Capacitors & Capacitance Capacitance
Ability to store energy in an electric field. Opposes a change in voltage. Unit of Measurement = Farad (F). Symbol used in equations = C. Components designed to provide capacitance are called “capacitors” or “condensers”.
Basic Components Capacitors & Capacitance Capacitors.
Basic Components Capacitors & Capacitance Capacitors.
Two conducting plates separated by an insulator. The larger the plates, the higher the capacitance. The closer the plates, the higher the capacitance. The higher the dielectric constant of the insulator, the higher the capacitance.
Basic Components Capacitors & Capacitance Capacitors.
At high frequencies, inductance of leads can become significant. Effective capacitance can be reduced. Capacitor can become self-resonant.
Basic Components Capacitors & Capacitance Capacitors.
Basic Components Capacitors & Capacitance Variable Capacitors.
Basic Components Capacitors & Capacitance Types of Capacitors.
Air / Vacuum. Mica / Silver Mica. Ceramic. Plastic Film (Polystyrene or Mylar). Paper. Oil-filled. Electrolytic / Tantalum.
Basic Components Capacitors & Capacitance Types of Capacitors.
Air / Vacuum. High voltage applications. Low Capacitance. Transmitter circuits.
Basic Components Capacitors & Capacitance Types of Capacitors.
Mica / Silver Mica. High stability. Low loss. RF circuits.
Basic Components Capacitors & Capacitance Types of Capacitors.
Ceramic. Inexpensive. Wide range of capacitances available. Low to high voltage ratings available. RF bypassing & filtering.
Basic Components Capacitors & Capacitance Types of Capacitors.
Plastic Film Polystyrene or Mylar. AF & lower frequencies. Susceptible to damage from high temperatures.
Basic Components Capacitors & Capacitance Types of Capacitors. Paper.
Obsolete. Found in antique equipment.
Basic Components Capacitors & Capacitance Types of Capacitors.
Oil-filled. High voltage. AC Power circuits.
Basic Components Capacitors & Capacitance Types of Capacitors.
Electrolytic / Tantalum. Polarized. High capacitance in physically small size. Power supply filters. Low-impedance AF coupling.
Basic Components Components in Series & Parallel Circuits
Kirchoff's Voltage Law: The sum of the voltages around a loop must be zero. Kirchoff's Current Law: The sum of all currents entering a node is equal to the sum of all currents leaving the node.
Basic Components Components in Series & Parallel Circuits
Series Circuits. FACT: Electrons cannot be created or destroyed. CONCLUSION: In a series circuit the current through each component is equal.
Basic Components Components in Series & Parallel Circuits
FACT: There is no voltage drop across a junction (or node). CONCLUSION: In a parallel circuit the voltage across each component is equal.
Basic Components Components in Series & Parallel Circuits Resistors.
Series: RT = R1 + R2 + R Rn Parallel: RT = 1 / (1/R1 + 1/R2 + 1/R /Rn) If only 2 resistors: RT = (R1 x R2) / (R1 + R2) If all resistors are same value: RT = R / (nr of resistors) Total resistance always less than lowest value resistor.
Basic Components Components in Series & Parallel Circuits Inductors.
Series: LT = L1 + L2 + L Ln Parallel: LT = 1 / (1/L1 + 1/L2 + 1/L /Ln) If only 2 inductors: LT = (L1 x L2) / (L1 + L2) If all inductors are same value: LT = L / (nr of inductors) Total inductance always less than lowest value inductor.
Basic Components Components in Series & Parallel Circuits Capacitors.
Series: CT = 1 / (1/C1 + 1/C2 + 1/C /Cn) If only 2 capacitors: CT = (C1 x C2) / (C1 + C2) If all capacitors are same value: CT = C / (nr of capacitors) Total capacitance always less than lowest value capacitor. Parallel: CT = C1 + C2 + C Cn
Basic Components Components in Series & Parallel Circuits In Summary:
Voltages add in a series circuit. Currents add in a parallel circuit. Component Type Adding in Series Adding in Parallel Resistor Increases Total Value Decreases Total Value Inductor Capacitor
Basic Components Transformers
Two or more inductors wound on a common core to maximize mutual inductance. Inductors are called “windings”. A winding connected to the signal source is called the “primary”. In special applications, there may be more than one primary. A winding connected to a load is called a “secondary”. It is common to have more than one secondary.
Basic Components Transformers
Transformers transfer AC power from the primary to each secondary. Transformers work equally well in both directions. Which winding is the “primary” and which is the “secondary” depends on how the transformer is connected in the circuit.
Basic Components Transformers Turns ratio.
The primary & secondary windings can have different numbers of turns (and usually do). Turns Ratio = NP:NS. The ratio of the voltage applied to the primary to the voltage appearing at the secondary is equal to the turns ration. Turns Ratio = EP:ES. Consequently: ES = EP x (NP/NS) and EP = ES x (NS/NP)
Basic Components Transformers Turns ratio.
Power input = power output (ignoring losses). If 120VAC is applied to the primary of a transformer with a turns ratio of 10:1, then the secondary voltage will be 12VAC. If a 1A current is flowing in the primary, then the current flowing in the secondary will be 10A. 120VAC x 1A = 120W = 12VAC x 10A
Break
G5B02 -- How does the total current relate to the individual currents in each branch of a parallel circuit? A. It equals the average of each branch current B. It decreases as more parallel branches are added to the circuit C. It equals the sum of the currents through each branch D. It is the sum of the reciprocal of each individual voltage drop
G5B02 -- How does the total current relate to the individual currents in each branch of a parallel circuit? A. It equals the average of each branch current B. It decreases as more parallel branches are added to the circuit C. It equals the sum of the currents through each branch D. It is the sum of the reciprocal of each individual voltage drop
G5C01 -- What causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across its primary winding? A. Capacitive coupling B. Displacement current coupling C. Mutual inductance D. Mutual capacitance
G5C01 -- What causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across its primary winding? A. Capacitive coupling B. Displacement current coupling C. Mutual inductance D. Mutual capacitance
A. The secondary B. The primary C. The core D. The plates
G5C02 -- Which part of a transformer is normally connected to the incoming source of energy? A. The secondary B. The primary C. The core D. The plates
A. The secondary B. The primary C. The core D. The plates
G5C02 -- Which part of a transformer is normally connected to the incoming source of energy? A. The secondary B. The primary C. The core D. The plates
G5C03 -- Which of the following components should be added to an existing resistor to increase the resistance? A. A resistor in parallel B. A resistor in series C. A capacitor in series D. A capacitor in parallel
G5C03 -- Which of the following components should be added to an existing resistor to increase the resistance? A. A resistor in parallel B. A resistor in series C. A capacitor in series D. A capacitor in parallel
A. .30 ohms B. .33 ohms C. 33.3 ohms D. 300 ohms
G5C04 -- What is the total resistance of three 100-ohm resistors in parallel? A. .30 ohms B. .33 ohms C ohms D. 300 ohms
A. .30 ohms B. .33 ohms C. 33.3 ohms D. 300 ohms
G5C04 -- What is the total resistance of three 100-ohm resistors in parallel? A. .30 ohms B. .33 ohms C ohms D. 300 ohms
A. 1500 ohms B. 90 ohms C. 150 ohms D. 175 ohms
G5C05 -- If three equal value resistors in parallel produce 50 ohms of resistance, and the same three resistors in series produce 450 ohms, what is the value of each resistor? A ohms B. 90 ohms C. 150 ohms D. 175 ohms
A. 1500 ohms B. 90 ohms C. 150 ohms D. 175 ohms
G5C05 -- If three equal value resistors in parallel produce 50 ohms of resistance, and the same three resistors in series produce 450 ohms, what is the value of each resistor? A ohms B. 90 ohms C. 150 ohms D. 175 ohms
A. 2370 volts B. 540 volts C. 26.7 volts D. 5.9 volts
G5C06 -- What is the RMS voltage across a 500-turn secondary winding in a transformer if the 2250-turn primary is connected to 120 VAC? A volts B. 540 volts C volts D. 5.9 volts
A. 2370 volts B. 540 volts C. 26.7 volts D. 5.9 volts
G5C06 -- What is the RMS voltage across a 500-turn secondary winding in a transformer if the 2250-turn primary is connected to 120 VAC? A volts B. 540 volts C volts D. 5.9 volts
G5C09 -- What is the capacitance of three 100 microfarad capacitors connected in series?
G5C09 -- What is the capacitance of three 100 microfarad capacitors connected in series?
A. .30 Henrys B. 3.3 Henrys C. 3.3 millihenrys D. 30 millihenrys
G5C10 -- What is the inductance of three 10 millihenry inductors connected in parallel? A. .30 Henrys B. 3.3 Henrys C. 3.3 millihenrys D. 30 millihenrys
A. .30 Henrys B. 3.3 Henrys C. 3.3 millihenrys D. 30 millihenrys
G5C10 -- What is the inductance of three 10 millihenry inductors connected in parallel? A. .30 Henrys B. 3.3 Henrys C. 3.3 millihenrys D. 30 millihenrys
G5C11 -- What is the inductance of a 20 millihenry inductor in series with a 50 millihenry inductor?
A. .07 millihenrys B millihenrys C. 70 millihenrys D millihenrys
G5C11 -- What is the inductance of a 20 millihenry inductor in series with a 50 millihenry inductor?
A. .07 millihenrys B millihenrys C. 70 millihenrys D millihenrys
G5C13 -- Which of the following components should be added to a capacitor to increase the capacitance? A. An inductor in series B. A resistor in series C. A capacitor in parallel D. A capacitor in series
G5C13 -- Which of the following components should be added to a capacitor to increase the capacitance? A. An inductor in series B. A resistor in series C. A capacitor in parallel D. A capacitor in series
G5C14 -- Which of the following components should be added to an inductor to increase the inductance? A. A capacitor in series B. A resistor in parallel C. An inductor in parallel D. An inductor in series
G5C14 -- Which of the following components should be added to an inductor to increase the inductance? A. A capacitor in series B. A resistor in parallel C. An inductor in parallel D. An inductor in series
A. 5.9 ohms B. 0.17 ohms C. 10000 ohms D. 80 ohms
G5C15 -- What is the total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel? A. 5.9 ohms B ohms C ohms D. 80 ohms
A. 5.9 ohms B. 0.17 ohms C. 10000 ohms D. 80 ohms
G5C15 -- What is the total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel? A. 5.9 ohms B ohms C ohms D. 80 ohms
G6A03 -- Which of the following is an advantage of ceramic capacitors as compared to other types of capacitors? A. Tight tolerance B. High stability C. High capacitance for given volume D. Comparatively low cost
G6A03 -- Which of the following is an advantage of ceramic capacitors as compared to other types of capacitors? A. Tight tolerance B. High stability C. High capacitance for given volume D. Comparatively low cost
G6A04 -- Which of the following is an advantage of an electrolytic capacitor?
A. Tight tolerance B. Non-polarized C. High capacitance for given volume D. Inexpensive RF capacitor
G6A04 -- Which of the following is an advantage of an electrolytic capacitor?
A. Tight tolerance B. Non-polarized C. High capacitance for given volume D. Inexpensive RF capacitor
G6A05 -- Which of the following is one effect of lead inductance in a capacitor used at VHF and above? A. Effective capacitance may be reduced B. Voltage rating may be reduced C. ESR may be reduced D. The polarity of the capacitor might become reversed
G6A05 -- Which of the following is one effect of lead inductance in a capacitor used at VHF and above? A. Effective capacitance may be reduced B. Voltage rating may be reduced C. ESR may be reduced D. The polarity of the capacitor might become reversed
G6A06 -- What will happen to the resistance if the temperature of a resistor is increased?
A. It will change depending on the resistor’s reactance coefficient B. It will stay the same C. It will change depending on the resistor's temperature coefficient D. It will become time dependent
G6A06 -- What will happen to the resistance if the temperature of a resistor is increased?
A. It will change depending on the resistor’s reactance coefficient B. It will stay the same C. It will change depending on the resistor's temperature coefficient D. It will become time dependent
G6A07 -- Which of the following is a reason not to use wire-wound resistors in an RF circuit?
A. The resistor's tolerance value would not be adequate for such a circuit B. The resistor's inductance could make circuit performance unpredictable C. The resistor could overheat D. The resistor's internal capacitance would detune the circuit
G6A07 -- Which of the following is a reason not to use wire-wound resistors in an RF circuit?
A. The resistor's tolerance value would not be adequate for such a circuit B. The resistor's inductance could make circuit performance unpredictable C. The resistor could overheat D. The resistor's internal capacitance would detune the circuit
G6A08 -- Which of the following describes a thermistor?
A. A resistor that is resistant to changes in value with temperature variations B. A device having a specific change in resistance with temperature variations C. A special type of transistor for use at very cold temperatures D. A capacitor that changes value with temperature
G6A08 -- Which of the following describes a thermistor?
A. A resistor that is resistant to changes in value with temperature variations B. A device having a specific change in resistance with temperature variations C. A special type of transistor for use at very cold temperatures D. A capacitor that changes value with temperature
G6A09 -- What is an advantage of using a ferrite core toroidal inductor?
A. Large values of inductance may be obtained B. The magnetic properties of the core may be optimized for a specific range of frequencies C. Most of the magnetic field is contained in the core D. All of these choices are correct
G6A09 -- What is an advantage of using a ferrite core toroidal inductor?
A. Large values of inductance may be obtained B. The magnetic properties of the core may be optimized for a specific range of frequencies C. Most of the magnetic field is contained in the core D. All of these choices are correct
A. In line B. Parallel to each other C. At right angles D. Interleaved
G6A10 -- How should the winding axes of solenoid inductors be placed to minimize their mutual inductance? A. In line B. Parallel to each other C. At right angles D. Interleaved
A. In line B. Parallel to each other C. At right angles D. Interleaved
G6A10 -- How should the winding axes of solenoid inductors be placed to minimize their mutual inductance? A. In line B. Parallel to each other C. At right angles D. Interleaved
G6A11 -- Why would it be important to minimize the mutual inductance between two inductors?
A. To increase the energy transfer between circuits B. To reduce unwanted coupling between circuits C. To reduce conducted emissions D. To increase the self-resonant frequency of the inductors
G6A11 -- Why would it be important to minimize the mutual inductance between two inductors?
A. To increase the energy transfer between circuits B. To reduce unwanted coupling between circuits C. To reduce conducted emissions D. To increase the self-resonant frequency of the inductors
G6A12 -- What is a common name for an inductor used to help smooth the DC output from the rectifier in a conventional power supply? A. Back EMF choke B. Repulsion coil C. Charging inductor D. Filter choke
G6A12 -- What is a common name for an inductor used to help smooth the DC output from the rectifier in a conventional power supply? A. Back EMF choke B. Repulsion coil C. Charging inductor D. Filter choke
G6A13 -- What is an effect of inter-turn capacitance in an inductor?
A. The magnetic field may become inverted B. The inductor may become self resonant at some frequencies C. The permeability will increase D. The voltage rating may be exceeded
G6A13 -- What is an effect of inter-turn capacitance in an inductor?
A. The magnetic field may become inverted B. The inductor may become self resonant at some frequencies C. The permeability will increase D. The voltage rating may be exceeded
A. Symbol 4 B. Symbol 7 C. Symbol 6 D. Symbol 1
G7A12 -- Which symbol in Figure G7-1 represents a multiple-winding transformer? A. Symbol 4 B. Symbol 7 C. Symbol 6 D. Symbol 1
A. Symbol 4 B. Symbol 7 C. Symbol 6 D. Symbol 1
G7A12 -- Which symbol in Figure G7-1 represents a multiple-winding transformer? A. Symbol 4 B. Symbol 7 C. Symbol 6 D. Symbol 1
Reactance & Impedance Reactance
All resistors do is convert electrical energy into heat. They don’t care whether current is DC or AC. Inductors & capacitors store energy. React differently to AC than to DC voltages/currents. Response to an AC voltage or current is called “reactance”. Unit of measurement = Ohm (Ω) Symbol used in equations = XL or XC
Reactance & Impedance Reactance Capacitive reactance. XC = 1 / (2πfC)
In a DC circuit (f = 0), XC = ∞. Capacitor looks like an open circuit. After initial charging current, the current flow drops to zero. At extremely high frequencies (f = ∞), XC = 0. Capacitor looks like a short circuit.
Reactance & Impedance Reactance Capacitive Reactance XC = 1 / (2πfC)
Reactance decreases with increasing frequency. Capacitors oppose change in voltage. Capacitor looks like open circuit at 0 Hz (DC). Capacitor looks like short circuit at very high frequencies. A capacitor blocks DC current, resists low-frequency AC current, & passes high-frequency AC current.
Reactance & Impedance Reactance Capacitive Reactance
When energy is first applied to a capacitor, the voltage is zero, & the current jumps to a large value. As the capacitor charges up, the voltage climbs to the steady state value and the current drops to zero.
Reactance & Impedance Reactance Inductive Reactance XL = 2πfL
Reactance increases with increasing frequency. Inductors oppose change in current. Inductor looks like a short circuit at 0 Hz (DC). Inductor looks like an open circuit at very high frequencies. An inductor passes DC current, resists low-frequency AC current, & blocks high-frequency AC current.
Reactance & Impedance Reactance Inductive Reactance
When energy is first applied to an inductor, the current is zero, & the voltage jumps to a large value. As the inductor charges up, the current climbs to the steady state value and the voltage drops to zero.
Reactance & Impedance Impedance
The opposition to current flow in an AC circuit caused by resistance, capacitive reactance, inductive reactance, or any combination thereof. Unit of measurement = Ohm (Ω) Symbol used in equations = Z.
Reactance & Impedance Resonance
Condition when frequency of applied signal matches “natural” frequency of circuit. At the resonant frequency, the inductive & capacitive reactances are equal and cancel each other out, leaving a purely resistive impedance. XL = XC 2πfL = 1 / (2πfC) f = 1 / πLC
Reactance & Impedance Resonance Resonant circuits are used in:
Filters. Tuned stages in receivers & transmitters. Antennas & Traps. Parasitic inductances & capacitances can cause a component to become “self-resonant” & lead to unwanted behavior.
Reactance & Impedance Impedance Transformation
In a DC circuit, resistance is calculated using Ohm’s Law: R = E / I Similarly, in an AC circuit, impedance is also calculated using Ohm’s Law: Z = E / I
Reactance & Impedance Impedance Transformation
Since a transformer changes the voltage & current levels in an AC circuit, it also changes the impedance. Impedance is calculated from the turns ratio (NP/NS) using the following formulas: ZP = ZS x (NP/NS)2 ZS = ZP x (NS/NP)2 The required turns ratio is calculated using the following formula: Turns Ratio (NS/NP) = ZP/ZS
Reactance & Impedance Impedance Matching
All power sources have an internal impedance which limits the amount of power that can be delivered. Maximum power is delivered only when the load impedance matches the source impedance. ZS = ZL
Reactance & Impedance Impedance Matching
Most modern amateur transmitting equipment is designed to have a source impedance of 50 Ohms. ZS = 50 Ω Therefore, load impedance should be 50 Ohms for maximum power transfer to the load. ZL = 50 Ω This is not usually the case!
Reactance & Impedance Impedance Matching
Antenna impedance varies from one frequency to another. A matching network is needed to transform the antenna system impedance to a 50Ω resistive load. L-C circuits. Most common type. Lengths of transmission line. Transformers. Cannot eliminate reactance.
Reactance & Impedance Impedance Matching Pi-Network. T-Network.
Often used in transmitter output stages to provide 50Ω source impedance. T-Network. Most common circuit for antenna tuners or “Transmatches”.
G5A01 -- What is impedance? A. The electric charge stored by a capacitor B. The inverse of resistance C. The opposition to the flow of current in an AC circuit D. The force of repulsion between two similar electric fields
G5A01 -- What is impedance? A. The electric charge stored by a capacitor B. The inverse of resistance C. The opposition to the flow of current in an AC circuit D. The force of repulsion between two similar electric fields
G5A02 -- What is reactance? A. Opposition to the flow of direct current caused by resistance B. Opposition to the flow of alternating current caused by capacitance or inductance C. A property of ideal resistors in AC circuits D. A large spark produced at switch contacts when an inductor is de-energized
G5A02 -- What is reactance? A. Opposition to the flow of direct current caused by resistance B. Opposition to the flow of alternating current caused by capacitance or inductance C. A property of ideal resistors in AC circuits D. A large spark produced at switch contacts when an inductor is de-energized
A. Conductance B. Reluctance C. Admittance D. Reactance
G5A03 -- Which of the following causes opposition to the flow of alternating current in an inductor? A. Conductance B. Reluctance C. Admittance D. Reactance
A. Conductance B. Reluctance C. Admittance D. Reactance
G5A03 -- Which of the following causes opposition to the flow of alternating current in an inductor? A. Conductance B. Reluctance C. Admittance D. Reactance
A. Conductance B. Reluctance C. Reactance D. Admittance
G5A04 -- Which of the following causes opposition to the flow of alternating current in a capacitor? A. Conductance B. Reluctance C. Reactance D. Admittance
A. Conductance B. Reluctance C. Reactance D. Admittance
G5A04 -- Which of the following causes opposition to the flow of alternating current in a capacitor? A. Conductance B. Reluctance C. Reactance D. Admittance
G5A05 -- How does an inductor react to AC?
A. As the frequency of the applied AC increases, the reactance decreases B. As the amplitude of the applied AC increases, the reactance increases C. As the amplitude of the applied AC increases, the reactance decreases D. As the frequency of the applied AC increases, the reactance increases
G5A05 -- How does an inductor react to AC?
A. As the frequency of the applied AC increases, the reactance decreases B. As the amplitude of the applied AC increases, the reactance increases C. As the amplitude of the applied AC increases, the reactance decreases D. As the frequency of the applied AC increases, the reactance increases
G5A06 -- How does a capacitor react to AC?
A. As the frequency of the applied AC increases, the reactance decreases B. As the frequency of the applied AC increases, the reactance increases C. As the amplitude of the applied AC increases, the reactance increases D. As the amplitude of the applied AC increases, the reactance decreases
G5A06 -- How does a capacitor react to AC?
A. As the frequency of the applied AC increases, the reactance decreases B. As the frequency of the applied AC increases, the reactance increases C. As the amplitude of the applied AC increases, the reactance increases D. As the amplitude of the applied AC increases, the reactance decreases
G5A07 -- What happens when the impedance of an electrical load is equal to the internal impedance of the power source? A. The source delivers minimum power to the load B. The electrical load is shorted C. No current can flow through the circuit D. The source can deliver maximum power to the load
G5A07 -- What happens when the impedance of an electrical load is equal to the internal impedance of the power source? A. The source delivers minimum power to the load B. The electrical load is shorted C. No current can flow through the circuit D. The source can deliver maximum power to the load
G5A08 -- Why is impedance matching important?
A. So the source can deliver maximum power to the load B. So the load will draw minimum power from the source C. To ensure that there is less resistance than reactance in the circuit D. To ensure that the resistance and reactance in the circuit are equal
G5A08 -- Why is impedance matching important?
A. So the source can deliver maximum power to the load B. So the load will draw minimum power from the source C. To ensure that there is less resistance than reactance in the circuit D. To ensure that the resistance and reactance in the circuit are equal
G5A09 -- What unit is used to measure reactance?
A. Farad B. Ohm C. Ampere D. Siemens
G5A09 -- What unit is used to measure reactance?
A. Farad B. Ohm C. Ampere D. Siemens
G5A10 -- What unit is used to measure impedance?
A. Volt B. Ohm C. Ampere D. Watt
G5A10 -- What unit is used to measure impedance?
A. Volt B. Ohm C. Ampere D. Watt
G5A11 -- Which of the following describes one method of impedance matching between two AC circuits?
A. Insert an LC network between the two circuits B. Reduce the power output of the first circuit C. Increase the power output of the first circuit D. Insert a circulator between the two circuits
G5A11 -- Which of the following describes one method of impedance matching between two AC circuits?
A. Insert an LC network between the two circuits B. Reduce the power output of the first circuit C. Increase the power output of the first circuit D. Insert a circulator between the two circuits
G5A12 -- What is one reason to use an impedance matching transformer?
A. To minimize transmitter power output B. To maximize the transfer of power C. To reduce power supply ripple D. To minimize radiation resistance
G5A12 -- What is one reason to use an impedance matching transformer?
A. To minimize transmitter power output B. To maximize the transfer of power C. To reduce power supply ripple D. To minimize radiation resistance
G5A13 -- Which of the following devices can be used for impedance matching at radio frequencies?
A. A transformer B. A Pi-network C. A length of transmission line D. All of these choices are correct
G5A13 -- Which of the following devices can be used for impedance matching at radio frequencies?
A. A transformer B. A Pi-network C. A length of transmission line D. All of these choices are correct
G5C07 -- What is the turns ratio of a transformer used to match an audio amplifier having a 600-ohm output impedance to a speaker having a 4-ohm impedance? A to 1 B to 1 C. 150 to 1 D. 300 to 1
G5C07 -- What is the turns ratio of a transformer used to match an audio amplifier having a 600-ohm output impedance to a speaker having a 4-ohm impedance? A to 1 B to 1 C. 150 to 1 D. 300 to 1
Questions?
Chapter 4 Components & Circuits (Part 2)
Next Week Chapter 4 Components & Circuits (Part 2) | 10,865 | 41,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.859089 |
https://people.sc.fsu.edu/~jburkardt/py_src/svd_circle/svd_circle.py | 1,721,399,877,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00373.warc.gz | 408,579,498 | 2,458 | #! /usr/bin/env python3 # def svd_circle ( A ): #*****************************************************************************80 # ## svd_circle() plots the image of the unit circle under the 2x2 matrix A. # # Discussion: # # A typical 2x2 matrix will map the unit circle to some kind of tilted # ellipse. The aspect ratio of the ellipse (ratio of major to minor axes) # is a measure of the conditioning of the matrix. # # A singular matrix maps the unit circle to a line. # # A diagonal matrix maps the unit circle to an ellipse with no tilting. # (There is no rotation.) # # An orthogonal matrix maps the unit circle to the unit circle (but points # may have rotated.) # # The identity matrix maps the unit circle to itself. # # The singular value decomposition can predict the shape of the mapping: # # * The vector V(1,1:2) maps to S(1)*U(1:2,1), # * The vector V(2,1:2) maps to S(2)*U(1:2,2). # # The ellipse has an aspect ratio of S(1)/S(2). # # The "tilt" or slope of the ellipse is U(2,1)/U(1,1). # # Licensing: # # This code is distributed under the MIT license. # # Modified: # # 31 March 2022 # # Author: # # John Burkardt # # Input: # # real A(2,2), the matrix whose mapping of the unit circle # is to be studied. # import matplotlib.pyplot as plt import numpy as np print ( '' ) print ( 'svd_circle():' ) print ( ' Given a matrix A,' ) print ( ' plot points x on the unit circle, and' ) print ( ' the images A*x of those points.' ) print ( ' Show right singular vectors V' ) print ( ' and their images, the scaled left singular vectors U.' ) print ( '' ) print ( ' Matrix A:' ) print ( '' ) print ( A ) # # This call produces matrices U, S and V. # U, svec, V = np.linalg.svd ( A ) print ( ' Singular values of A are', svec[0], 'and', svec[1] ) print ( ' Right singular vectors are columns of V:' ) V = V.T print ( V ) print ( ' Left singular vectors are columns of U:' ) print ( U ) print ( ' Aspect ratio of ellipse will be', svec[0] / svec[1] ) print ( ' Slope of ellipse will be', U[1,0] / U[0,0] ) # # Select N evenly spaced points X on the unit circle. # n = 20 i = np.linspace ( 0, n - 1, n, dtype = float ) angle = 2.0 * np.pi * i / float ( n ) r = 1.0 x = np.array ( [ r * np.cos ( angle ), r * np.sin ( angle ) ] ) # # AX contains the image of each point under the mapping X -> A*X. # ax = np.dot ( A, x ) # # Determine a common plot range. # p_min = min ( np.min ( ax ), np.min ( x ) ) p_max = max ( np.max ( ax ), np.max ( x ) ) # # Plot the corners of the region. This is a trick to force MATLAB # to plot in a square with equal axes. # plt.clf ( ) plt.scatter ( \ [ p_min, p_max, p_max, p_min ], \ [ p_min, p_min, p_max, p_max ] ) # # Label the points X on the unit circle. # i = 97 for j in range ( 0, n ): plt.text ( x[0,j], x[1,j], chr ( i ) ) i = i + 1 if ( 122 < i ): i = 97 # # Plot V1 and V2. # plt.plot ( [ 0.0, V[0,0]], [ 0.0, V[1,0]], 'b', linewidth = 3 ) plt.plot ( [ 0.0, V[0,1]], [ 0.0, V[1,1]], 'b', linewidth = 3 ) plt.xlabel ( '--X axis--' ) plt.ylabel ( '--Y axis--' ) plt.title ( 'Points X on unit circle, and right singular vectors' ) plt.axis ( np.array ( [ p_min, p_max, p_min, p_max ] ) ) plt.axis ( 'equal' ) plt.axis ( 'tight' ) plt.axis ( 'square' ) plt.grid ( True ) filename = 'svd_circle_1.png' plt.savefig ( filename ) print ( ' Graphics saved as "' + filename + '"' ) plt.show ( block = False ) plt.close ( ) # # Label the points AX on the image of the unit circle. # plt.clf ( ) plt.scatter ( np.array ( [ p_min, p_max, p_max, p_min ] ), \ np.array ( [ p_min, p_min, p_max, p_max ] ) ) # # Plot the images AX of points X on the unit circle. # i = 65 for j in range ( 0, n ): plt.text ( ax[0,j], ax[1,j], chr ( i ) ) i = i + 1 if ( 90 < i ): i = 65 # # Plot the images of V1 and V2. # plt.plot ( [ 0.0, svec[0]*U[0,0]], [ 0.0, svec[0]*U[1,0]], 'r', linewidth = 3 ) plt.plot ( [ 0.0, svec[1]*U[0,1]], [ 0.0, svec[1]*U[1,1]], 'r', linewidth = 3 ) plt.xlabel ( '--X axis--' ) plt.ylabel ( '--Y axis--' ) plt.title ( 'Images A*X of points on unit circle and scaled left singular vectors' ) plt.axis ( np.array ( [ p_min, p_max, p_min, p_max ] ) ) plt.axis ( 'equal' ) plt.axis ( 'tight' ) plt.axis ( 'square' ) plt.grid ( True ) filename = 'svd_circle_2.png' plt.savefig ( filename ) print ( ' Graphics saved as "' + filename + '"' ) plt.show ( block = False ) plt.close ( ) # # Terminate. # print ( '' ) print ( 'svd_circle():' ) print ( ' Normal end of execution.' ) return def svd_circle_test ( ): #*****************************************************************************80 # ## svd_circle_test() tests svd_circle(). # # Licensing: # # This code is distributed under the MIT license. # # Modified: # # 31 March 2022 # # Author: # # John Burkardt # import numpy as np import platform print ( '' ) print ( 'svd_circle_test():' ) print ( ' Python version: ' + platform.python_version ( ) ) print ( ' Test svd_circle().' ) A = np.array ( [ \ [ 3.0, 2.0 ], [ 1.0, 0.0 ] ] ) svd_circle ( A ) # # Terminate. # print ( '' ) print ( 'svd_circle_test():' ) print ( ' Normal end of execution.' ) return def timestamp ( ): #*****************************************************************************80 # ## timestamp() prints the date as a timestamp. # # Licensing: # # This code is distributed under the MIT license. # # Modified: # # 21 August 2019 # # Author: # # John Burkardt # import time t = time.time ( ) print ( time.ctime ( t ) ) return if ( __name__ == '__main__' ): timestamp ( ) svd_circle_test ( ) timestamp ( ) | 1,687 | 5,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.680597 |
http://oeis.org/A180425 | 1,670,362,119,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00455.warc.gz | 40,711,014 | 4,639 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A180425 Number of positive integers below 10^n requiring 3 positive squares in their representation as sum of squares. 4
2, 42, 505, 5586, 59308, 616995, 6347878, 64875490, 660104281, 6695709182, 67762820595, 684596704482, 6907026402474, 69611115440126, 700946070114283, 7053023642205904 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Eric W. Weisstein, MathWorld -- Lagrange's Four-Square Theorem. Eric W. Weisstein, MathWorld -- Sum of Squares Function. FORMULA a(n) = #{k: A000419(k) < 10^n}. A049416(n) + A180416(n) + a(n) + A167615(n) = A002283(n). CROSSREFS Cf. A000419, A049416, A180416, A167615, A002283. Sequence in context: A070808 A229474 A157056 * A140170 A318247 A038396 Adjacent sequences: A180422 A180423 A180424 * A180426 A180427 A180428 KEYWORD nonn,more AUTHOR Martin Renner, Jan 19 2011 EXTENSIONS a(6)=616995 by Lars Blomberg, May 03 2011 a(7)-a(10) from Donovan Johnson, Jul 01 2011 a(10) corrected and a(11)-a(16) from Hiroaki Yamanouchi, Jul 13 2014 STATUS approved
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Last modified December 6 16:00 EST 2022. Contains 358644 sequences. (Running on oeis4.) | 561 | 1,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-49 | latest | en | 0.665455 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-write-the-quadratic-equation-given-a-2-and-the-vertex-5-0 | 1,708,602,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00845.warc.gz | 102,502,628 | 5,892 | # How do you write the quadratic equation given a=-2 and the vertex (-5, 0)?
Nov 1, 2014
$y = a {\left(x - h\right)}^{2} + k$,
where $\left(h , k\right)$ is the vertex.
Since $a = - 2$ and $\left(h , k\right) = \left(- 5 , 0\right)$, we have
$y = - 2 {\left[x - \left(- 5\right)\right]}^{2} + 0 = - 2 {\left(x + 5\right)}^{2}$.
I hope that this was helpful. | 145 | 363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-10 | latest | en | 0.771444 |
http://www.tgju.org/support/solutions/articles/how-do-i-calculate-how-long-it-takes-an-investment-to-double-aka-the-r | 1,571,852,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00455.warc.gz | 328,682,661 | 16,492 | # How do I calculate how long it takes an investment to double (AKA 'The Rule of 72') in Excel?
تنظیم شده در تاریخ: ۱۳۹۷/۰۴/۲۵
A:
You can calculate the approximate amount of years it would take an investment to double, given the annual expected rate of return. To do so, you would need to implement the rule of 72 in Microsoft Excel.
The rule of 72 states that to determine the approximate amount of time it takes to double your investment at a given rate of return, you simply divide the rate of return by 72. For example, assume you invest \$10 at an interest rate of 15%. It would take 4.8 years (72/15) to double your money to \$20.
Assume you want to compare the actual number of years with the approximate number of years it would take for five different investments to double. Assume the five investments have expected rates of return of 5%, 10%, 13%, 15% and 20%. In Microsoft Excel, increase the widths of columns A, B, C and D by right-clicking on each respective column and left-clicking on Column Width and change the value to 35.
Make the typeset bold for the titles by pressing the CTRL and B keys together. Now, enter "Expected Rate of Return" in cell A1, "Actual Number of Years" in cell B1, "Number of Years Using the Rule of 72" in cell C1 and "Difference" in cell D1.
Enter "5" into cell A2, "10" into cell A3, "13" into cell A4, "15" into cell A5 and "20" into cell A6. The formula to calculate the actual number of years it takes to double your investment is the natural log of 2 divided by the natural log of 1 plus the expected rate of return.
The values in cells A2 through A6 must be expressed in percentage terms to calculate the actual number of years it would take for the investments to double. Therefore, the values must be divided by 100. In cell B2, enter "=(LN(2)/(LN(1+A2/100)))". Now, left click and hold on the bottom right corner of cell B2 and drag the cell down to cell B6.
Now, use the rule of 72 to calculate the approximate number of years by entering "=72/A2" into cell C2, "=72/A3" into cell C3, "=72/A4" into cell C4, "=72/A5" into cell C5 and "=72/A6" into cell C6.
To calculate the difference between the actual number of years and the approximate number of years calculated using the rule of 72, enter "=ABS(C2-B2)" into cell D2. Next, select cell D2, left-click and hold on the bottom right corner of the cell and drag the cell down to cell D6. The resulting values are the absolute values of the differences between the actual number of years and the approximate number of years.
The rule of 72 is a good approximation of the amount of years it takes to double your investment because the approximations are within 0.2 year of the actual number of years.
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# For customers purchasing a full set of tires at a particular tire store, consider the events$A=\{$ tires purchased were made in the USA $\}$$B=\{ purchaser has tires balanced immediately \}$$C=\{$ purchaser requests front-end alignment $\}$along with $A^{\prime}, B^{\prime},$ and $C^{\prime} .$ Assume the following unconditional and conditional probabilities:(a) Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch.(b) Compute $P(A \cap B \cap C)$ .(c) Compute $P(B \cap C) .$(d) Compute $P(C)$ .(e) Compute $P(A B \cap C),$ the probability of a purchase of US tires given that both balancing and an alignment were requested.
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#### Topics
Probability Topics
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### Video Transcript
alright for this problem, We are considering customers buying a full set of tires from a particular tire store. Vent A is where the tires were made in the USA Event B is where the purchaser had the tires balanced immediately and events see where the purchaser requests a front end alignment. And then there are a prime b prime and see prime, which are the the knots of each of those events. We also have these given probabilities and conditional probabilities. I'm not gonna bother reading all those out. So for part A, we want to construct a tree diagram for the 1st, 2nd and 3rd generation branches and plays an event label and appropriate probability next to each branch. So we know we're going to have our starting point at our first decision. We're gonna have to our second decision. We are going to have four. They're not for sorry. We are going to have eight. One second. 123 Okay, I'm messing that up. Yeah, for our second decision, we should have total four. And then for our third decision, we should have eight. So I'm gonna zoom in quite a bit here because it's going to get a little nasty. So we start off and then we have event A with probability of a is 0.75 and we have a prime. The probability of a prime is going to be one minus 10.75 or 0.2 to 5, then here for this branch, we have B where the probability of be given a is equal to 0.9 and we have be prime where the probability of the prime given a is equal to 0.1 Up here we have C probability of see, given B and A is equal Thio 0.8. And this event here that event is um yeah, that event is Yeah. No, that's the best way of labeling that actually. Sorry, but my hesitation there we have c prime down here. Probability of C prime given B and a equal to 0.2 than this branch have see ability of see given be prime and a is equal to that 0.6 probability of C prime. Just this branch down there ability of C prime given be prime and a is equal thio 0.4 Then we have offer a prime branch here B and B prime again. Where we have probability of be given a prime is equal 2.8 probability of be prime given a prime is equal Thio point to we have C see Prime and C and C prime Where you gonna zoom? In further probability of C given a prime and be is equal 2.7 the probability of yes ability of see prime given a prime nb equal 2.3 then lastly or the last pair you have a probability of C given a prime and be prime which is that is going to be 0.3. What and we have done here the probability of see prime given a prime and be prime which is equal 2.7 And then these events at the end correspond to so that is see given a or rather not given that is A and B and C A and be and see prime A and be prime and see a and B prime and see prime um, a prime and B and C a prime and be prime and or sorry a prime and be and see prime. Then we have a prime and be prime and see. And we have a prime and be prime and see prime. We can calculate the probabilities of each of those by multiplying down our branches. So I'm going to do that off screen here. Actually, my mistake calculating all of those eyes going to be a bit unnecessary. So we now have the proper branches and event labels for everything. That's all we need for part a part B. Yes. So this is the total tree there, Part B. We want to find the probability of A and B and C, which we confined just by following our tree. That is 0.75 times 0.9 times 0.8, which comes out to 0.54 part C. We want to find the probability of B and C happening, so that means that it's going to be mhm probability of B and C and a plus the probability of B and C and not a which we already have. BNC and a keep in mind that for the ANZ, the order doesn't matter. Forgiven for the conditioning it does, but for and you're ordering doesn't matter. It was 0.54 plus 0.25 Excuse me, M 0.8 time 0.7. That gives us 0.68 for D. We want to find the probability of see occurring. So that is going to be the probability of B and C plus the probability of not be NC. Uh huh. You know that the probability of B and C 0.68 the probability of not being C going to be so that this event can be broken down into actual do it off to the side there. Keep in mind that that's going to be essentially the probability of a and not B and C plus the probability of not a and not be and not be and see not be so probability of a and not B N c his 0.75 times 0.1 time 2.6 and the probability of not a and not B and C is 0.25 times 0.2 times times 0.3. So adding those all together we get 0.74 is the result. Lastly for E, we want to find the probability of a given B and C you can find this using Bayes Theorem, which tells us that that probability is going to be the same thing as the probability of be all right. Same thing is the probability of A and B and C divided by the probability of B and C, so we already found each of those pieces. Previously, ability of A and B and C is 0.54 Probability of BNC is 0.68 so we end up with the probability of a given B and C 0.79 | 1,458 | 5,897 | {"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} | 3.828125 | 4 | CC-MAIN-2021-21 | latest | en | 0.95855 |
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#### CP5639 Conceptual Modelling For Attributes And Different Relationships
Pages : 5 | Type : Essay
Answer: Discussion This report discusses the solution for the conceptual modelling of the Joanne Myers Apartments. Conceptual modelling involves following a few steps in order to achieve the final conceptual model for the database. The first step is thoroughly studying the case study to get information on all the entities and their attributes and different relationships that exist between the entities. After identifying the entities, the next...
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#### CP5639 Problem Solving And Programming For Loop Variable And Exit Conditions
Pages : 5 | Type : Essay
Answer: Functions List getValue(): This function takes an input from the user for the mentioned prompt passed as the parameter and return it back. checkValid(): This function compares the value of the first parameter with all the other parameters passed if it matches any of them. checkValidBetween(): This function compares the value of the first parameter with the other two parameters if it lies in between of the two. calculate(): This fun...
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#### CP5639 Problem Solving And Programming
Pages : 5 | Type : Essay
Answers: Function List Get Value This function is to prompt the user to enter a value and then that value is returned. Check Valid This function is to check if the value is same as the other parameters passed. Check Valid Between This function is to check if the value lies in the range of the parameters passed. Calculate This function is to calculate the value of RDCI and return that value. Check Percent This function is to check the valu...
Written By : Mark Posted On : Sep 28, 2020 Download : 0
#### CP5639 Problem Solving And Programming
Pages : 6 | Type : Essay
Answer: Functions List getValue(): This function takes an input from the user for the mentioned prompt passed as the parameter and return it back. checkValid(): This function compares the value of the first parameter with all the other parameters passed if it matches any of them. checkValidBetween(): This function compares the value of the first parameter with the other two parameters if it lies in between of the two. calculate(): This fun...
Written By : Mark Posted On : Nov 09, 2020 Download : 0
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http://emmythelberg.com/f-de-levene-spss.php | 1,623,574,560,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00596.warc.gz | 18,406,053 | 6,559 | How do I interpret data in SPSS for an independent samples T-test? Home > How do I interpret data in SPSS for an independent samples T-test? Background Levene’s Test for Equality of Variances. This is a bad thing, but SPSS takes this into account by giving you slightly different results in the second row. If the Sig. value in this. Mar 04, · Levene’s test is built into most statistical software. For example, the Independent Samples T Test in SPSS generates a “Levene’s Test for Equality of Variances” column as part of the output. The result from the test is reported as a p-value, which you . Output for Levene's test. On running our syntax, we get several tables, the second of which is Test of Homogeneity of emmythelberg.com holds the results of Levene's test. As a rule of thumb, we conclude that the variances are not equal if “Sig.”.
If you are looking
# f de levene spss
18 - Test de Levene: Demostración de Homocedasticidad [Curso de estadística], time: 10:46
Output for Levene's test. On running our syntax, we get several tables, the second of which is Test of Homogeneity of emmythelberg.com holds the results of Levene's test. As a rule of thumb, we conclude that the variances are not equal if “Sig.”. You are here: Home ANOVA SPSS Two-Way ANOVA Tutorials SPSS Two Way ANOVA – Basics Tutorial Research Question. How to lose weight effectively? Do diets really work and what about exercise? In order to find out, participants were assigned to . Mar 04, · Levene’s test is built into most statistical software. For example, the Independent Samples T Test in SPSS generates a “Levene’s Test for Equality of Variances” column as part of the output. The result from the test is reported as a p-value, which you . How do I interpret data in SPSS for an independent samples T-test? Home > How do I interpret data in SPSS for an independent samples T-test? Background Levene’s Test for Equality of Variances. This is a bad thing, but SPSS takes this into account by giving you slightly different results in the second row. If the Sig. value in this.So we'll write something like Levene's test showed that the variances for body fat percentage in week 20 were not equal, F(2,77) = , p = To find out which row to read from, look at the large column labeled Levene's Test for Equality of You will see two smaller columns labeled F and Sig. Look in . The null hypothesis of Levene's test is that the variances are equal. SPSS. GUI path. Analyze > Descriptives > Explore; Click and drag the Levene's Test for Homogeneity of Variance (center = median) Df F value Pr(>F). In statistics, Levene's test is an inferential statistic used to assess the equality of variances for a . See also. Bartlett's test · F-test of equality of variances · Box's M test Parametric and nonparametric Levene's test in SPSS. Levene's test for homogeneity of variance, analysis-of-variance table and robust Ryan-Einot-Gabriel-Welsch F test (R-E-G-W F), Ryan-Einot-Gabriel-Welsch. Calculates the Levene statistic to test for the equality of group variances. This statistic is preferable to the F statistic when the assumption of equal variances. I have been running some data in SPSS and the homogeneity of variance test has been violated. Universidad de La Frontera methods for dealing with heterogeneity of variance are the Welch or Brown-Forsythe F-tests. The data which i have violated the levene's test of homogeneity of varinace with a result of I am running an ANOVA for a 3 by 3 factorial, however, the levene test is not showing the F or P value. what can be the reason for this? P Value · ANOVA · SPSS. -
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This remarkable idea is necessary just by the way | 976 | 3,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-25 | latest | en | 0.888749 |
http://www.math.wpi.edu/Course_Materials/MA1022A99/volumes/node1.html | 1,555,735,459,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528523.35/warc/CC-MAIN-20190420040932-20190420062932-00547.warc.gz | 263,817,440 | 5,491 | Subsections
# Solids of Revolution
## Introduction
The purpose of this lab is to use Maple to study solids of revolution. Solids of revolution are created by rotating curves in the x-y plane about an axis, generating a three dimensional object. They are discussed in Chapter 6 of Calculus by Bradley and Smith (sections 1 and 2). The specific properties of them that we wish to study are their volume, surface area, and graph.
## Background
So far we have used the integral mainly to to compute areas of plane regions. It turns out that the definite integral can also be used to calculate the volumes of certain types of three-dimensional solids. The class of solids we will consider in this lab are called Solids of Revolution because they can be obtained by revolving a plane region about an axis.
As a simple example, consider the graph of the function f(x) = x2+1 for , which appears in Figure 1.
If we take the region between the graph and the x-axis and revolve it about the x-axis, we obtain the solid pictured in Figure 2.
To help you in plotting surfaces of revolution, A Maple procedure called revolve has been written. The command used to produce the graphs in Figures 1 and 2 is shown below. The revolve procedure, as well as the RevInt, LeftInt, and LeftDisk procedures described below are all part of the CalcP package, which must be loaded first. The last line in the example below shows the optional argument for revolving the graph of f(x) about the line y=-2 instead of the default y=0.
> with(CalcP):
> f := x -> x^2+1;
> plot(f(x),x=-2..2);
> revolve(f(x),x=-2..2);
> revolve(f(x),x=-2..2,y=-2)
The revolve command has other options that you should read about in the help screen. For example, you can speed the command up by only plotting the surface generated by revolving the curve with the nocap argument, and you can also plot a solid of revolution formed by revolving the area between two functions. Try the following examples. (Note: The last example shows how to use revolve with a function defined piecewise.)
> revolve({f(x),0.5},x=-2..2,y=-1);
> revolve(cos(x),x=0..4*Pi,y=-2,nocap);
> revolve({5,x^2+1},x=-2..2);
> g := x -> if x < 0 then -x +1/2 else x^2-x+1/2 fi ;
g := proc(x) options operator,arrow; if x < 0 then -x+1/2 else
x^2-x+1/2 fi end
> revolve('g(x)',x=-1..2);
It turns out that the volume of the solid obtained by revolving the region in Figure 1 between the graph and the x-axis about the x-axis can be determined from the integral
to have the value . More generally, if you revolve the area under the graph of g(x) for about the x-axis, the volume is given by
Where does this formula come from? To help you understand it, Two more Maple procedures, RevInt and LeftDisk, have been written. The procedure RevInt sets up the integral for the volume of a solid of revolution, as shown below. The Maple commands evalf and value can be used to obtain a numerical or analytical value.
> RevInt(f(x),x=-2..2);
> value(RevInt(f(x),x=-2..2));
> evalf(RevInt(f(x),x=-2..2));
The integral formula given above for the volume of a solid of revolution comes, as usual, from a limit process. Recall the rectangular approximations we used for plane regions. If you think of taking one of the rectangles and revolving it about the x-axis, you get a disk whose radius is the height h of the rectangle and thickness is , the width of the rectangle. The volume of this disk is . If you revolve all of the rectangles in the rectangular approximation about the x-axis, you get a solid made up of disks that approximates the volume of the solid of revolution obtained by revolving the plane region about the x-axis.
To help you visualize this approximation of the volume by disks, the LeftDisk procedure has been written. The syntax for this procedure is similar to that for revolve, except that the number of subintervals must be specified. The examples below produce approximations with five and ten disks. The latter approximation is shown in Figure 3.
> LeftDisk(f(x),x=-2..2,5);
> LeftDisk(f(x),x=-2..2,10);
> LeftInt(f(x),x=-2..2,5);
> LeftInt(f(x),x=-2..2,10);
The two LeftInt commands above add up the volumes in the disk approximations of the solid of revolution.
## Finding Volumes of Revolution
Since this involves straightforward integration, one may simply issue commands such as:
> f:= x-> sqrt(x) +1;
> vol:= int(Pi*f(x)^2, x=0..9);
> evalf(vol);
## Surface Area
Assuming the function has been defined, one might issue:
> S:= int(2*Pi*f(x)*sqrt(1+D(f)(x)^2), x= 0..9);
> evalf(S);
to find the surface area of the solid obtained by rotating f(x) about the x-axis.
## Exercises
1.
Let and consider the interval . Plot the solid of revolution obtained by revolving the graph of this function around the x axis. Then compute the volume of the solid of revolution you obtained. Finally, use the LeftInt command and determine the number of subintervals needed to approximate the volume to within 0.1.
2.
Repeat the first exercise, this time using the function g(x) = 1 + x/4. The number of subintervals required should be larger than in the previous exercise. Try to explain why this happens. Some things that might help you are to compare the graphs of the two functions, to look at plots with LeftDisk, and to examine the integrals that give the volumes.
3.
Compute the volume of the solid generated by revolving the region bounded by the x-axis, the graph of the function , x=0, and x=10.
4.
Two years ago, Chris Zannella and Eric Pauly (both class of '98) were asked to design a drinking glass by revolving a suitable function about the x axis. Here is the function they came up with.
They obtained the shape of their glass by revolving this function about the x axis over the interval [-0.9,4.5]. The Maple command they used to define this function is given below.
> f := x -> if x<0 then x^4+0.088 elif x<3 then 0.088 else sin(x-3)+0.088 fi;
Plot this function (without revolving it) over the interval [-0.9,4.5] and identify the formula for each part of the graph. Then, revolve this function about the x axis over the same interval and comment on the glass Eric and Chris designed. Finally, compute the volume of the part of this glass that could be filled with liquid, assuming the stem is solid. (Hint - your integral will involve only one of the formulas used to define the function.)
5.
Suppose that a tank for holding molasses is to be constructed as a right circular cone of height 10 meters and base radius 5 meters. It is to be set up so that the point of the cone is at the bottom and the circular base is at the top. If x is the height of molasses in the cone, find and plot a function v(h) (where h is the height) which gives the volume of molasses in the tank. | 1,674 | 6,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-18 | latest | en | 0.920347 |
https://edurev.in/studytube/Short-Answer-Type-Questions-Triangles--Mathematics/d9576906-fdb4-465d-ad1c-b6d6eaafc926_t | 1,660,714,872,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00022.warc.gz | 228,519,756 | 67,127 | # Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
## Document Description: Short Answer Questions: Triangles for Class 10 2022 is part of Triangles for Mathematics (Maths) Class 10 preparation. The notes and questions for Short Answer Questions: Triangles have been prepared according to the Class 10 exam syllabus. Information about Short Answer Questions: Triangles covers topics like and Short Answer Questions: Triangles Example, for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Short Answer Questions: Triangles.
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Q1.If one diagonal of a trapezium divides the other diagonal in the ratio of 1: 2. Prove that one of the parallel sides is double the other.
Sol. Since ABCD is a trapezium,
∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)
∴ Δ APB ~ Δ CPD
Q2. In the figure, ABC is a D such that BD ⊥ AC and CE ⊥ AB. Prove that:
PD × BP = PC × EP
Sol. In Δ BEP and Δ CPD, we have:
<BPE = <CPD
[Vertically opp. angles]
<BEP = <CDP [Each = 90°]
∴Using AA similarity, we have
Δ BEP ~ Δ CDP
∴Their corresponding sides are proportional,
⇒ BP × PD = EP × PC
[By cross multiplication]
Q3. AB is a line segment. PB ⊥ AB and QA ⊥ AB are such that PO = 5 cm and QO = 8 cm. If ar (D POB) = 250 cm2, then find the area of D QOA.
Sol. In Δ QOA and Δ POB,
<QOA = <BOP
[Vertically opposite angles]
<QAO = <PBO [Each = 90°]
∴Using AA similarity, we have:
Q4. In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.
Sol. We have a trapezium ABCD in which AB y DC.
Since Δ BOC ~ Δ AOD [Given]
.(1)
In Δ ODC and ΔOBA,
<COD = <AOB
[Vertically opp. angles]
<ODC = <OBA [Alt. angles]
∴ Using AA similarity, we have:
Δ ODC ~ Δ OBA
...(2)
From (1) and (2)
=
⇒ OB × OB = OA × OA
⇒ OB2 = OA2 ⇒ OA = OB ...(3)
From (1) to (3) we have
Q5. P and Q are points on sides of AB and AC respectively of Δ ABC. If AP = 3 cm, PB= 9 cm, AQ = 5 cm and QC = 15 cm, then show that BC = 4 PQ.
Sol. We have Δ ABC in which P and Q are such that
AP = 3 cm, PB = 9 cm
AQ = 5 cm, QC = 15 cm
i.e., PQ divides AB and AC in the same ratio
∴ PQ y BC
Now, in Δ APQ and Δ ABC
<P = <B [Corresponding angles]
<A = <A [Common]
⇒ Using AA similarity,
Δ APQ ~ Δ ABC
⇒ =
[ΠAB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
Q6. On one of the longer sides PQ of a rectangle PQRS, a point O is taken such that
SO2 = PO· PQ
Prove that:
Δ POS ~ Δ OSR.
Sol.
We have a rectangle PQRS such that
SO2 = PO· PQ
i.e., SO × SO = PO × PQ
⇒ = ...(1)
[ΠPQ = SR, opp. sides of rectangle PQRS]
Now, in Δ POS and Δ OSR, we have:
<1 = <2 [OE PQ y SR,
opp. sides of a rectangle]
⇒ Using SAS similarity, we have
Δ POS ~ Δ OSR
Q7. Determine the length of the altitude AD of an isosceles D ABC in which AB = AC = 2a cm and BC = a cm.
Sol. We have Δ ABC in AD ⊥ BC and AB = AC = 2a. Also BC = a.
<B = <C [Opp. angles to equal sides of a D]
=
⇒
Now in right Δ ABD, we have
⇒ AD2 = AB2 - BD2
= (AB + BD) (AB - BD)
=
Q8. In an equilateral triangle with side ‘a’, prove that its area .
Sol. We have Δ ABC such that
AB = BC = AC = a
Let us draw altitude AD ⊥ BC.
AC = AB [Each = a]
∴DC = DB =
⇒ AD2 = AB2 - DB2
= (AB + DB) (AB - DB)
=
Now, area of Δ ABC = 1/2× Base × altitude
Thus, the area of an equilateral triangle
=
Q9. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Sol.We have Δ ABC in which AB = AC = CA and an altitude AD ⊥ BC.
AB = AB [Given]
Using RHS congruency, we have
⇒ DB = DC = ...(1)
Now, in right Δ ADB, we have:
AB2 = AD2 + BD2 [Using Pythagoras Theorem]
⇒ 3 [Side of the equilateral triangle]
= 4 [Altitude]2.
Q10. ABC is a right triangle in which <C = 90° and CD ⊥ AB. If BC = a, CA = b, AB = c and CD = p, then prove that:
=
Sol. We have a right Δ ABC such that <C = 90°.
Also, CD ⊥ AB
Now, ar (Δ ABC) =1/2 × Base × Height
= ...(1)
Also,
. .. (2)
From (1) and (2), we have
=
Dividing throughout by abp, we have:
=
Squaring both sides,
= = ...(3)
Now, In right D ABC,
AB2 = AC2 + BC2
⇒ c2 = b2 + a2 ...(4)
∴ From (3) and (4), we get
=
Q11. ABC is a right triangle, right-angled at A, and D is the mid-point of AB. Prove that
BC2 = CD2 + 3 BD2
Sol. We have a right Δ ABC in which <A = 90°
∴Using Pythagoras Theorem, we have:
BC2 = AB2 + AC2 ...(1)
Again, Δ ACD is right D, <A = 90°
...(2) [Using Pythagoras Theorem]
Subtracting (2) from (1), we get
BC2 - CD2 = AB2 - AD2 ...(3)
Since D is the mid-point of AB
∴2 BD = AB and AD = BD ...(4)
From (3) and (4), we have:
∴BC2 - CD2 = (2 BD)2 - (BD)2
= 4 BD2 - BD2
BC2 = CD2 + 3 BD2
Q12. In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm and OA = 5 cm. Find the length of OC.
Sol. Let us draw EOF y AB ⇒ OE ⊥ AD and OF ⊥ BC
In Δ OFB
Œ <F = 90°
∴Using Pythagoras theorem, we have:
CB2 = OF2 + BF2 ...(1)
In Δ OED
<E = 90°
∴Using Pythagoras theorem, we have:
OD2 = OE2 + DE2 ...(2)
Adding (1) and (2), we get
OB2 + OD2 = OF2 + BF2 + OE2 + DE2
= OF2 + AE2 + OE2 + CF2
[ΠBF = AE and CF = DE]
= (OF2 + CF2) + (OE2 + AE2)
= OC2 + OA2
= OC2 + 52
⇒ 62 + 82 = OC2 + 52
⇒ 36 + 64 = OC2 + 25
⇒ OC2 = 36 + 64 - 25 = 75
⇒ Thus OC = 5√3cm.
Q13. In the figure, if AD ⊥ Bc, then prove that:
AB
2 + CD2 = AC2 + BD2
∴ AD2 = AC2 - CD2 .....(1) (Using Pythagoras Theorem)
Similarly, in D AbD,
⇒ AD2 = AB2 - DB2.....(2)
From (1) and (2), we have
AB2 - DB2 = AC2 - CD2
⇒ AB2 + CD2 = AC2 + BD2
Q14. In the given figure, AD ⊥ BC and BD = CD. Prove that:
Sol. BD = 1/2 CD
∴ 3 BD = CD
Since BD + DC = BC
∴ BD + 3 BD = BC
⇒ 4 BD = BC
⇒ BD = 1/4 BC
⇒ CD = 3/4 BC
Now, in right Δ ADC, < D = 90°
By Pythagoras theorem, we get
CA2 = AD2 + CD2 ...(1)
AD2 = AB2 - BD2 ...(2)
From (1) and (2),
CA2 = AB2 - BD2+ CD2
Q15. In the given figure, M is the mid-point of side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produce D at E. Prove that EL = 2 BL.
Sol. We have parallelogram ABCD in which M is the mid point of CD.
In Δ EMD and Δ BMC
MD = MC [ΠM is mid-point of CD]
<EMD = <CMB
[Vertically opposite angles]
<MED = <MBC
[Alternate interior angles]
∴ Δ BMC ≌ Δ EMD [AAS congruency]
⇒ BC = ED ⇒ AD = ED ...(1)
[ΠBC = AD, opposite sides of parallelogram]
Now, in Δ AEL and Δ CBL
<AEL = <CBL [Alternate interior angles]
<ALE = <CLB
[Vertically opposite angles]
∴ By AA similarity, we have:
Q16. In the given figure, Δ ABC is right-angled at C and DE ^ AB. Prove that Δ ABC ~ Δ ADE and hence find the length of AE and DE.
Sol.In Δ ABC and Δ ADE, we have:<A = <A [Common]
<C = <E [Each = 90°]
∴ Δ ABC ~ Δ ADE [AA Similarity]
...(1)In right Δ ABC, <C = 90°Using Pythagoras theorem, we have:
AB2 = BC2 + AC2
= 122 + 52
= 144 + 25 = 169
⇒AB = √169 = 13 cm
Now, from (1), we get
⇒ DE = = 2.77 cmAnd = ⇒ AE
Q17. In the given figure, DEFG is a square and <BAC = 90°. Show that DE2 = BD × EC.
Sol. In Δ DBG and Δ ECF
<3 + <1 = 90° = <3 + <4
∴<3 + <1 = <3 + <4
⇒ <1 = <4
<D = E = 90°
∴Using AA similarity, we have:
BD × EC = EF × DGBut DG = EF = DE
∴BD × EC = DE × DE
⇒ BD × EC = DE2
Thus, DE2 = BD × EC
Q18. In the figure, AD ⊥ BC and BD = CD. Prove that 2 CA2 = 2 AB2 + BC2
Sol. ΠBD = 1/3 CD
⇒ 3 BD = CD
∴BC = BD + DC
⇒ BC = BD + 3 BD
⇒ BC = 4 BD ...(1)
And From (2) ...(4)In right Δ ADC,
Using Pythagoras theorem,
From (3) ...(4)In right Δ ADB
Using Pythagoras theorem,
Q19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Sol. We have a quadrilateral ABCD such that its diagonals intersect at O and
<AOB = <COD [Vertically opposite angles]∴ Using SAS similarity, we have
Δ AOB ~ Δ COD
⇒ Their corresponding angles are equal.
i.e., <1 = <2
But they form a pair of int. alt. angles.
⇒ AB y DC
⇒ ABCD is a trapezium.
Q20. Two triangles ABC and DBC are on the same base BC and on the same side of BC in which
<A = <D = 90°. If CA and BD meet each other at E, show that
AE· EC = BE· ED
Sol.
We have right Δ ABC and right Δ DBC on the same base BC such that
<A = <D = 90°
In Δ ABE and Δ DCE
<A = <D = 90°
<1 = <2 [Vertically opp. angles]
∴ Using AA similarity, we have:
Δ ABE ~ Δ DCE
⇒ Their corresponding sides are proportional.
⇒ ⇒ AE· EC = BE· ED
Q21. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
Δ ABE ~ Δ CFB
Sol.
We have parallelogram ABCD
In Δ ABE and Δ CFB, we have
<A = <C
[Opposite angles of parallelogram]
<AEB = <EBC
∴ Using AA similarity, we get
Δ ABE ~ Δ CFB
Q22. In Δ ABC, if AD is the median, then show that AB2 + AC2 = 2 [AD2 + BD2].
∴ BD = DC
Let us draw AE ≌ BC
Now, in rt. Δs AEB and AEC, we have
AB2 = BE2 + AE2 ...(1)
AC2 = CE2 + AE2 ...(2)
AB2 + AC2 = BE2 + AE2 + CE2 + AE2
= (BD - ED)2 + AE2 +
(CD + DE)2 + AE2
= 2AE2 + 2ED2 + BD2 + CD2
= 2 [AE2 + ED2] + BD2 + BD2
[BD = CD]
Thus, AB2 + AC2 = 2 [AD2 + BD2]
Q23. Triangle ABC is right angled at B and D is the mid point of BC. Prove that:
AC2 = 4 AD- 3 AB2
Sol.
D is the mid-point of BC.
∴ BC = 2 BD
Now, in Δ ABCAC2 = BC2 + AB2
= (2 BD)2 + AB2
= 4 BD2 + AB2 ...(1)
In right Δ ABD
Using Pythagoras theorem,
⇒ BD2 = AD2 - AB2 ...(2)
From (1) and (2), we get
AC2 = 4 [AD2 - AB2] + AB2
⇒ AC2 = - 4 AB2 + 4 AD2 + AB2
⇒ AC2 = - 3 AB2 + 4 AD2
or AC2 = 4 AD2 - 3 AB2
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; | 3,887 | 10,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-33 | latest | en | 0.869242 |
https://math.stackexchange.com/questions/806671/whats-the-relationship-of-tensor-and-multivector/1846516 | 1,568,694,160,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573052.26/warc/CC-MAIN-20190917040727-20190917062727-00179.warc.gz | 576,532,933 | 31,127 | # what's the relationship of tensor and multivector
what's the relationship of multivector in geometric algebra and tensor? Is tensor a subset of multivector?
Every multivector corresponds to a tensor, but not every tensor corresponds to a multivector.
The classic picture of a tensor is a linear map, acting on any number of vectors and dual vectors, to produce a scalar. Inspired by exterior and geometric algebra, you can say a tensor acts on any number of vector or dual vector blades.
For example, rotation operators can be viewed as tensors: let $\underline R(a)$ be some rotation, then there exists a tensor $R(a,b) = \underline R(a) \cdot b$. Because there is such a close correspondence, it's not uncommon to refer to the rotation map itself as a tensor.
You can extend the rotation map to act on blades, and therefore whole multivectors, through addition in the geometric algebra, but the rotation map itself isn't an element of the geometric algebra--it's not a multivector.
So, even in geometric algebra, we have need or use of linear maps, of general tensors.
How can you build a tensor from a general multivector? Through any function that is linear in all its arguments. A simple way is to consider a general multivector $M$. For the sake of an example, let's say $M$ is composed of blades grade-4 or less. Then you can build a tensor, $T$, such that
$$T(a, b, c, d) = \langle M abcd \rangle_0$$
which is linear in all its arguments $a, b, c, d$ and produces a scalar.
For this reason, you might refer to a multivector in general as a tensor, too, since there is such a close and obvious correspondence between a multivector and an associated linear map.
These resources explaining how to represent multivectors as tensors might be helpful or interesting to you (or those with similar questions):
http://www2.ic.uff.br/~laffernandes/teaching/2013.1/topicos_ag/lecture_18%20-%20Tensor%20Representation.pdf
https://www.docdroid.net/uwfvUxE/tensor-representation-of-geometric-algebra.pdf.html | 480 | 2,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-39 | latest | en | 0.898917 |
https://numbermatics.com/n/241849153/ | 1,725,722,166,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00057.warc.gz | 419,131,889 | 6,923 | # 241849153
## 241,849,153 is an odd composite number composed of four prime numbers multiplied together.
What does the number 241849153 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 24 divisors.
241849153 is an odd composite number. It is composed of four distinct prime numbers multiplied together. It has a total of twenty-four divisors.
## Prime factorization of 241849153:
### 72 × 13 × 313 × 1213
(7 × 7 × 13 × 313 × 1213)
See below for interesting mathematical facts about the number 241849153 from the Numbermatics database.
### Names of 241849153
• Cardinal: 241849153 can be written as Two hundred forty-one million, eight hundred forty-nine thousand, one hundred fifty-three.
### Scientific notation
• Scientific notation: 2.41849153 × 108
### Factors of 241849153
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 1546
### Divisors of 241849153
• Number of divisors d(n): 24
• Complete list of divisors:
• Sum of all divisors σ(n): 304194408
• Sum of proper divisors (its aliquot sum) s(n): 62345255
• 241849153 is a deficient number, because the sum of its proper divisors (62345255) is less than itself. Its deficiency is 179503898
### Bases of 241849153
• Binary: 11100110101001010011010000012
• Base-36: 3ZZO01
### Scales and comparisons
How big is 241849153?
• 241,849,153 seconds is equal to 7 years, 35 weeks, 6 days, 4 hours, 19 minutes, 13 seconds.
• To count from 1 to 241,849,153 would take you about eleven years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 241849153 cubic inches would be around 51.9 feet tall.
### Recreational maths with 241849153
• 241849153 backwards is 351948142
• The number of decimal digits it has is: 9
• The sum of 241849153's digits is 37
• More coming soon!
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Keywords: Divisors of 241849153, math, Factors of 241849153, curriculum, school, college, exams, university, Prime factorization of 241849153, STEM, science, technology, engineering, physics, economics, calculator, two hundred forty-one million, eight hundred forty-nine thousand, one hundred fifty-three.
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https://www.taylorfrancis.com/books/9780429073601/chapters/10.1201/b15364-10 | 1,604,194,484,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00576.warc.gz | 900,285,198 | 44,860 | chapter 8
6 Pages
## Perturbative Expansion in Field Theory
As we have seen in the previous chapter [∫ Dϕ(x) where relevant includes
ground-state factors ]
Z( J , gn) ≡< 0|U(T)|0 >= ∫ Dϕ(x) exp
( i ∫
d4x L(ϕ) )
(8.1)
will play the role of a generating functional for calculating expectation values of products of field operators, which will now be studied in more detail. In general the Lagrange density for a scalar field theory is given by
L(ϕ) = L2(ϕ) − V(ϕ) − J (x)ϕ(x), (8.2)
where L2(ϕ) is quadratic in the fields, hence for a scalar field
L2(ϕ) = 12 ( ∂µϕ(x)∂µϕ(x) − m2ϕ2(x)
) ,
(8.3) V(ϕ) = g3
3! ϕ3(x) + g4
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# hydrogen ion concentration formula
A metal hydroxide of molecular formula M ... H 2 gas is bubbled at 2 9 8 K and 1 atm pressure. In pure water, hydrogen ion concentration is 10-7 moles … Concentration of hydrogen ions in an aqueous solution can be calculated if the pH of the solution is known: [H +] = 10-pH where pH is the pH of the aqueous solution and [H +] = concentration of hydrogen ions 1 in mol L-1 share. In order to deduce the formula for H+ from the formula above, we can use an ICE (initial – change – equilibrium) table. Formula to calculate hydrogen & hydroxide ion … save. This concentration is normally expressed in terms of pH which is defined as the logarithm of the reciprocal of the hydrogen ion. pH is an important factor in chemical process, represents the acid & alkalin level of a solution. For example, the hydrogen ion concentration of pure water at 25 o C is about 10-7 mol/L. So, pH is defined by the following formula: pH=-log10[H+] The concentration of hydrogen ions in any solution we are likely to encounter will range from 1 mol to 0.000001 mol per liter of solution. No Signup required. A hydrogen ion concentration in a solution results from the addition of an acid. 5 years ago. View Answer. Q:-In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction … But the ionic product concentration K W of the water always remains constant 1 × 10-14 at a temperature of 25°C. ... water at room temperature and a study of the proton hopping mechanism using molecular dynamics showed that the hydrogen-bonds around the hydronium ion (formed with the three water ligands in the first solvation shell of the … What are the concentrations of hydrogen ion and hydroxide ion in household ammonia, an aqueous solution of NH 3 ... Write the formula for each of the following ionic compounds: Nickel (III) Oxide Barium Fluoride Tin ... A: A chemical compound consists of two or more … Formula: Hydrogen Ion Concentration in nmol = [H+] = 24 * (PaCO2/HCO3) PaCO2=arterial part press CO2 Related Calculator: ... 0.4g of NaOH is present in one litre of the solution shows that H + concentration of the solution is: View … No. Question. The lower-case alphabet “p” in pH denotes negative common (base ten) logarithm, while the upper-case alphabet “H” denotes the element hydrogen. I. Hence, when the concentration of one ion increases, accordingly the concentration of the other ion is … Hydrogen Ion Concentration of Strong Acids Calculations Tutorial Key Concepts. Lv 4. Autoionization of Water: Water is an amphiprotic compound. pH is a negative logarithmic measurement of the number of moles of hydrogen ions (H+) per litre of solution. This will help us to improve better. The hydrogen ion concentration decreases by a factor of 10, so the pH increases by 1 from 1.6 to 2.6. What is the hydroxide ion concentration if the hydrogen ion concentration of a solution is {eq}6.8 \times 10^{-5} M {/eq}? 0. 3. The concentration of hydrogen ion in a sample of soft drink is 3. 7.2 × 10 ^− 4 c. 3.6 × 10 ^− 4 How ever I don't understand how these answers are found. It is convenient to construct a scale of simpler numbers to represent these values. Solutions with a pH less than 7 are said to be acidic and solutions with … Answered By . A solution get Using the 0.200 M HCl as the [H+] (concentration of hydrogen ions) the solution is as follows: pH=-log { left[ { H }^{ + } right] } = log(0.200) =0.70. [H+] … When the conversion is made the value … 0.5 M HF and 0.50 M NaF c. 0.4 M HCOOH and 0.2 M KCOOH I already know the answers. 2 comments. pH = - log [H 3 O + ]. Lv 4. Jennifer. The hydronium ion concentration is 0.0025 M. Thus: pH = - log (0.0025) = - ( - 2.60) = 2.60. Free Online ../Chemistry Calculator. About Calculator School. Answers (1) The equation 2NaNO3 + CaCl2 - -> 2NaCl + Ca (NO3) 2 is balanced. toppr. Find the concentration of hydrogen ions: [H +] = 10-pH = 10-3.2 = 6.3 × 10-4 mol L-1. What is the formula used, what are the steps? Chemical formula. a. SAT Chemistry practice from College Board The hydrogen ion concentration of a solution prepared by diluting 50. mL of 0.10 M HNO3(aq) with water to 500. mL of solution is (A) 0.0010 M (B) 0.0050 M (C) 0.010 M (D) 0.050 M (E) 1.0 M Answer and Explanation: One of the key words in this question is diluting. pH & pOH, often denoted by an alphabet H + & OH-respectively, is the measure of hydrogen & hydroxide ion concentration of the solution. H 3 O + Molar mass: 19.02 g/mol Acidity (pK a) ... pH was originally conceived to be a measure of the hydrogen ion concentration of aqueous solution. hide. 0.1 M HCN and 0.2 M KCN b. It is shown in the figure 14.2. We can use this formula to calculate hydrogen in concentration Given is the formula substituting in the value we have high Rosen on concentration that turns out to be 3.16 to 1 to 10 days. What is pH? Pure water is said to be neutral, with a pH close to 7.0 at 25 o C (77 o F). Example: Find the pH of a 0.0025 M HCl solution. To make pH even easier to work with, pH is defined as the negative log of $$[H^+]$$, … answr. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. This means, for example, that a hydrogen-ion concentration of a solution with a pH of 4 is 10 -4 mol/l, meaning it contains 0.0001 mol of hydrogen ions in a solution of 1 liter. Whereas, pOH values are derived from pH values to represent hydroxide ion concentration of the solution. Answer: 10 Explanation: Comment; Complaint; Link; Karla 27 November, 07:21. Water is considered neutral at 7 so if it's < 7 it's considered acidic and > 7 it's basic. Correct answer to the question The equation for the pH of a substance is pH = –log[H+], where H+ is the concentration of hydrogen ions. Log in or sign up to leave a comment Log In ... Can someone tell me the formula to use, so that I know how many more litres to get to X%. Online calculators and converters have been developed to make calculations easy, these calculators are great tools for mathematical, … What is the approximate pH of a solution whose hydrogen ion concentration is 2x10^-9 - e-eduanswers.com B, A, C II. Hydrogen ion concentration: pH: The Relationship Between pH and Hydrogen Ion Concentration This slide emphasizes the following points: •Acidic solutions have a relatively high [H+] •As pH increases, [H+] decreases. A 0.70 pH indicates a very acidic solution. For the best answers, search on this site https://shorturl.im/axrHt. Use the formula pH= - log [H*] to find the pH of the… The hydrogen ion concentration of one molar hydrochloric acid, pure water and one molar sodium hydroxide are 1, 10 −7 and 10 −14 mol dm −3 respectively. Nitric acid has a chemical formula of HNO 3. Source(s): https://shrinks.im/baNdk. The hydrogen ion concentration has a marked effect on the growth of bacteria. Example: find the pH of a solution with a hydrogen ion concentration of 0.05 M, or 5.0 x 10-2 M -Log(0.05) = 1.3. Because hydrogen ion concentrations are generally less than one (for example $$1.3 \times 10^{-3}\,M$$), th e log of the number will be a negative number. A buffer solution is a solution that resists changes in pH even when a strong acid or base is added or when the solution is diluted … Move. The pH scale is logarithmic and inversely indicates the concentration of hydrogen ions in the solution. Source(s): https://shrink.im/a8xZu. Let x represent the … 0 0. The hydrogen ion concentration will be:- (1) 1.96 10-2 mol L-1 (2) 1.6 10-4 mol L- (3) 1.96 x 10-5 mol L-' (4) None of them. The hydrogen ion concentration of a solution prepared by diluting 50. mL of 0.10 M HNO3(aq) with water to 500. mL of solution is (A) 0.0010 M (B) 0.0050 M (C) 0.010 M (D) 0.050 M (E) 1.0 M. Answer and Explanation: One of the key words in this question is diluting. 4 years ago. The HCl is a strong acid and is 100% ionized in water. This is … Write the equation (formula) that relates moles, volume and concentration : molarity = moles ÷ volume (L) Rearrange this equation (formula) to find moles: moles = molarity (mol L-1) × volume (L) Step 3: Enter ion concentration value -Log( value Step 4: press enter. 0 0. buckholz. the pH off the vine is 3.5. Answer: pH 8 Step-by-step explanation: ... How do you identify the chemical formula? Upvote(0) How satisfied are you with the answer? pH=-log10[H+] In the case of a neutral solution, [H+]=10-7 , which we call a pH of 7. One solution turns blue. 0. … Hydrogen ion concentration can also be expressed in terms of pH. I do not know the starting water volume. The unit for the concentration of hydrogen ions is moles per liter. The answers are a.2.45 × 10 ^− 10 b. Top. To determine pH, you can use this pH to H⁺ formula: pH = -log([H⁺]) If you already know pH, but want to calculate the concentration of ions, use this transformed pH equation: [H +] = 10-pH Question A solution of hydrochloric acid, with a concentration of 2 g/dm 3 , has a pH of 1.3. Create an account to start this course today At a high concentration of H ‑ (10-1 M), the pH value is low, pH = 1. A possible hydrogen ion concentration for this solution is: Answers (2) Dontrell 27 November, 07:18. Formula. In chemistry, pH is a measure of the acidity or basicity of an aqueous solution. M H+. Calculate the concentration of OH- ions in an HCl solution whose hydrogen ion concentration is 2.90 M. What is the charge on the hydronium ion? pH is a method of measurement of hydrogen ion concentration. Solution for - 11 The hydrogen ion concentration, [H*], in a certain cleaning compound is H ]= 3.9 x 10 ". Get Instant Solutions, 24x7. However, solutions with extremely low hydrogen-ion concentration could conceivably rack up a pretty long parade of zeros after the decimal point. How many atoms of oxygen (O) … M OH- What is the concentration of hydrogen ion in household ammonia, an aqueous solution of NH3 that has a pH of 11.50? The activity coefficient is a function of the ion concentration and approaches 1 as the solution becomes increasingly dilute. The pH scale is a measure of hydrogen ion concentration that eliminates dealing with large powers of 10 and compresses a large range of concentrations onto a more convenient scale, between 1 and 14. pH is the logarithmic value of the inverse of the hydrogen ion concentration (more accurately the activity of hydrogen ions). How To Find Ph. pH = - log10 [H+] That's power of Hydrogen. I fell like it’s close to C1V1 = (C2-X)V2 I honestly don’t know. A, B, C H++ H + H+ H+ H H+ + H H+ + H+ H+ … Hydrogen Ion Concentration in nmol = [H+] = 24 * (PaCO2/HCO3) PaCO2=arterial part press CO2. In chemistry, pH (/ p iː ˈ eɪ tʃ /, denoting 'potential of hydrogen' or 'power of hydrogen') is a scale used to specify the acidity or basicity of an aqueous solution.Acidic solutions (solutions with higher concentrations of H + ions) are measured to have lower pH values than basic or alkaline solutions.. Answer. Hydrogen Ion Concentration Formula. Figure $$\PageIndex{2}$$ illus trates the relationship between pH and the hydrogen ion concentration, along with some examples of various solutions. 100% Upvoted. Example 2: Calculate the pH of a 0.100 M nitric acid solution. I think I’m on the right track but I wanted to double check. For dilute (ideal) solutions, the standard state of the solute is 1.00 M, so its molarity equals its activity. This can be achieved by taking the reciprocal of the logarithm to the base ten of the hydrogen ion concentration of the solution. Acid Weight Formula: Hydrogen Ion Concentration in nmol = [H+] = 24 * (PaCO2/HCO3) PaCO2=arterial part press CO2 Thus the calculation of Hydrogen Ion Concentration is made easy. It is concluded that each aqueous solution whether it is acidic, basic or neutral possesses both hydrogen H + ion and hydroxyl OH-ion that is present in the same or different proportions. 10. Logarithmic value is because the hydrogen ion concentration in a solution has a very low amount. The hydrogen-ion concentration of a 0.25 M solution of HCN (Ka = 4.9 ´ 10-10) is I keep getting 1.1 *10^-5 Chemistry An aqueous solution that 2.00 M in AgNO3 is slowly added from a buret to an aqueous solution that is 1.00×10−2 M in Cl− and 0.260 M in I−. Sample question 1 Which of the following lists the above solutions in order of increasing hydrogen ion concentration? If the pH is 1.3, the solution must be acidic Problem variation: you may be given the hydroxide ion concentration OH- and are asked to find the pH. Question 55. What is the hydronium ion concentration in each of the following buffer solutions? – hydrogen ion concentration. Super minus four. Hah. pH is actually a scale used to determine the acidity or basicity of a system. While at low concentration (10-12 M), the pH is high, pH = 12. report. … C, A, B III. High, pH = 12 ph=-log10 [ H+ ] = 24 * ( )... Identify the chemical formula of an acid product concentration K W of the hydrogen ion can. 10 b basicity of a 0.100 M nitric acid solution the number of moles hydrogen! A method of measurement of the water always remains constant 1 × 10-14 at high. Ions in the solution becomes increasingly dilute is about 10-7 mol/L because hydrogen... 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Construct a scale used to determine the acidity or basicity of an oxide of iron which has 69.9 iron. Low concentration ( 10-12 M ), the pH off the vine is 3.5 ion. | 5,651 | 21,059 | {"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} | 2.6875 | 3 | CC-MAIN-2021-21 | latest | en | 0.884274 |
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A238661 Number of partitions of n having standard deviation σ > 2. 8
0, 0, 0, 0, 0, 0, 1, 2, 5, 7, 12, 18, 29, 42, 61, 85, 118, 164, 223, 299, 399, 530, 693, 888, 1157, 1488, 1901, 2403, 3044, 3807, 4783, 5935, 7368, 9097, 11197, 13721, 16806, 20441, 24868, 30133, 36494, 43895, 52880, 63424, 75900, 90609, 108088, 128404 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,8 COMMENTS Regarding "standard deviation" see Comments at A238616. LINKS Table of n, a(n) for n=1..48. FORMULA a(n) + A238659(n) = A000041(n). EXAMPLE There are 22 partitions of 8, whose standard deviations are given by these approximations: 0., 3., 2., 2.35702, 1., 1.69967, 1.73205, 0., 1.24722, 0.942809, 1.22474, 1.2, 0.471405, 1., 0.707107, 0.8, 0.745356, 0., 0.489898, 0.471405, 0.349927, 0, so that a(8) = 2. MATHEMATICA z = 50; g[n_] := g[n] = IntegerPartitions[n]; c[t_] := c[t] = Length[t]; s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, c[t]}]/c[t]]; Table[Count[g[n], p_ /; s[p] < 2], {n, z}] (*A238658*) Table[Count[g[n], p_ /; s[p] <= 2], {n, z}] (*A238659*) Table[Count[g[n], p_ /; s[p] == 2], {n, z}] (*A238660*) Table[Count[g[n], p_ /; s[p] > 2], {n, z}] (*A238661*) Table[Count[g[n], p_ /; s[p] >= 2], {n, z}] (*A238662*) t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]] ListPlot[Sort[t[30]]] (*plot of st deviations of partitions of 30*) CROSSREFS Cf. A238616, A238658-A238660, A238662. Sequence in context: A023564 A173088 A005895 * A135525 A319142 A350497 Adjacent sequences: A238658 A238659 A238660 * A238662 A238663 A238664 KEYWORD nonn,easy AUTHOR Clark Kimberling, Mar 03 2014 STATUS approved
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One concern with CAD is a fundamental aspect of their architecture: How are curves, arcs and circles handled? All- too-many programs take the expedient route of representing them as as polygons or polylines. Academic paper discussing this G codes for the specification of Pythagorean-hodograph tool paths.
Please note that it is possible to use pretty much any vector drawing program for design, esp. if one limits oneself to 2D (one interesting example of this is the use of a JavaScript to generate G-Code from vectors w/in the page layout program InDesign (see IterativePathCuts (unfortunately limited to straight lines) at Silicon Publishing: Resources, forum discussion)). The 2D drawing programs listed below were chosen for their having especial support in the form of features or plug-ins to support CAM.
This page focuses on interactive programs, manipulation of numbers/parameters and code is covered on the Programming page.
A note on curve approximation: The trivial case with four cubic beziers (one on each quadrant) has very low maximum error, IIRC on the order of 0.1% of the circle radius.[1]
Origin point This is different from most drawing and CAD programs: https://bugs.launchpad.net/inkscape/+bug/170049 as discussed here: http://www.shapeoko.com/forum/viewtopic.php?f=6&t=5102&p=37166#p37166
A note on vector drawing: Vector drawing is like to drafting with rules, triangles, compasses and a French curve, while pixel editing is more akin to colouring in spaces on a grid. Vector drawing uses lines (a straight path between two "nodes" which are x,y coordinates, one is the beginning, the other the ending), regular geometric forms (defined by nodes which denote each point at which lines meet), arcs and circles (defined by the central point of rotation and the radius therefrom) and Bézier curves (a mathematical description of an arbitrary curve which has four "nodes", one on-the-path, two off-the-path and one on-the-path) to allow it to express pretty much any shape or form.
A game to learn about forming such curves: http://bezier.method.ac/#
A note on units: Most CAD programs store drawings in generic units, rather than a specific measure. It is important to know what unit was used when a drawing was created so as to be able to import it at the right size. Forum discussion.
It is also of course, straight-forward to work directly in G-Code, accomplishing Design and CAM in a single step using programs which provide integrated text editing such as NC Plot and NC Corrector (the latter is listed on the Previewing G-Code page. A tutorial is listed on the CAM page. Unfortunately, Grbl’s implementation of G-code leaves out variables and looping, so is not suited for this --- there is at least one pre-processor, G-sharp which affords support for these features, instantiating them as simpler G-code, bCNC supports this.
Note: A number of programs listed on the CAM page include design features for such specialized things as PCBs, Thread and Fret cutting and Gears.
See also
The Failed Promise of Parametric CAD, a series of articles by Evan Yares[3]:
Page with table listing many opensource CAD apps: https://sites.google.com/site/lagadoacademy/miscellaneous-projects/free-cad-software---overview --- includes columns for OS compatibility, incl. whether or no an app will run in Linux using WINE.
## 2D
Type Name/Link Description
2D drawing (Opensource) Inkscape SVG drawing program which can either export to DXF or directly to G-code using plug-ins. Supports pressure-sensitive graphics tablets.
2D CAD (Opensource) LibreCAD Fork of Qcad. Free and OpenSource CAD (DXF) application, cross platform.
Includes menu option to export to MakerCAM SVG.
2D CAD (Opensource) Simple2D CAD/CAM Windows app to cut simple shapes (pockets, mostly) and simplified gears, includes a geometry mode that allows geometry defined by lines and arcs. It also generates geometry from any OpenType font.
2D drawing (Opensource) Cenon (Mac OS X, NeXT/OPENstep, Linux) OpenStep CAD/CAM program which has transitioned to opensource.
2D drawing (Opensource) Xfig G-code enabled Xfig.
2D CAD (Opensource) Dedalo 2d CAD drawing software written in python3.
2D CAD (Opensource) Zcad Simple CAD program with dxf2000 file format support.
2D CAD (Freeware) OpenS-CAD, a simple 2D CAD application Windows .exe (87.5KB) and source (225KB).
2D CAD (Freeware) nanoCAD Free tool for creating, editing, validating, sharing, and managing CAD models.
2D CAD (Freeware) Siemens Free 2D Free CAD software by Siemens (Windows). textbook available as a PDF.
AutoCAD clones:
Note that Draftsight is no longer freely available.
### Tracing
Tracing is a useful option for some projects. Most vector drawing programs have an option for this (one option is to place the image on the desktop or in a window beneath the CAD tool, then make that window transparent), or one can use a specialized tool. One such: http://www.kvec.de/english/index.htm Discussion: https://www.reddit.com/r/CNC/comments/47fqeh/best_raster_img_to_scalable_vector_programs/
Opensource tracing: http://potrace.sourceforge.net/ --- web interface: http://www.vectorization.org/
Webservice for tracing: https://convertio.co/jpg-svg/ or http://online-convert.com/ [6]
Mechanism for tracing / digitizing existing designs / templates: http://www.logicgroup.com/LogicTraceCncDxf.html
## Extrusion
http://www.extrudy.net/ --- create simple, printable 3D objects such as coins, pendants, earrings or medals. Youtube Reddit
## Fonts
In many instances a single-line font is a desirable option for CNC. The Hershey Fonts in Inkscape are one notable option.
Lengthy article and list of options at: http://imajeenyus.com/computer/20150110_single_line_fonts/index.shtml
## 3D
c.f., Programming: 3D Modeling Libraries and Languages for tools such as OpenSCAD.
Merging the boundaries between 2D and 3D, CAD and CAM, there's http://www.flatfab.com/ Fabrication Design Software that helps you design and fabricate 3D objects. Ideate and draw in 2D, fabricate in 3D.
Type Name/Link Description
3D modeling (Opensource) K-3D K-3D is free-as-in-freedom 3D modeling and animation software. It combines flexible plugins with a visualization pipeline architecture, making K-3D a versatile and powerful tool for artists. Available for Windows, Mac OS X and Linux.
3D modeling (Opensource) BRL-CAD BRL-CAD is a cross-platform open source solid modeling system that includes interactive geometry editing, high-performance ray-tracing for rendering and geometric analysis, image and signal-processing tools, a system performance analysis benchmark suite, libraries for robust geometric representation, with more than 20 years of active development. Discussion of shortcomings.
3D modeling (Opensource) Art of Illusion Free, open source 3D modeling and rendering studio. Features include subdivision surface based modeling tools, skeleton based animation, and a graphical language for designing procedural textures and materials. Default design software for the RepRap project.
3D modeling (Opensource) MeshLab Open source, portable, and extensible system for the processing and editing of unstructured 3D triangular meshes.
Converts Obj to stl?
3D modeling (Opensource) HeeksCAD Together with HeeksCAM comprises HeeksCNC. Youtube: HeeksCAD Unofficial Tutorial Part 1:Introduction General overview of GUI for HeeksCAD
3D modeling (Opensource) FreeCAD FreeCAD is a general purpose feature-based, parametric 3D modeler for CAD, MCAD, CAx, CAE and PLM, aimed directly at mechanical engineering and product design but also fits a wider range of uses in engineering, such as architecture or other engineering specialties. It is 100% Open Source and extremely modular, allowing for very advanced extension and customization. See also the Python-query-scripting language CadQUERY (or v2 at https://github.com/CadQuery/cadquery ) https://github.com/jmwright/cadquery-gui and Cnc25D CAD Library.
A variety of macros are available, i.e., http://www.freecadweb.org/wiki/index.php?title=Macro_FCGear [9]
Print creation and assembly handling FreeCAD leave a lot to be desired.[10]
3D parametric modeling (Opensource) ImplicitCAD Similar to OpenSCAD (above), but written in Haskell and using an internal geometric representation, as opposed to the polygon mesh used by the former
3D parametric modeling (Opensource) NaroCAD 3D parametric modeling CAD application for Windows, based on OpenCascade.
3D modeling (Opensource) Blender Free, OpenSource and crossplatform 3D program. Allows modelisation, shading, rendering, animation, interactive 3D, compositing. Not a CAD software though it can be used for any kind of modeling. See Blender CAM for a tool to export files for EMC. If importing from Illustrator, see this forum discussion for a note on how to handle units.
https://github.com/Lichtso/curve_cad --- Blender Addon: Bezier Curve CAD Tools for CNC Milling & Laser Cutting
Parametric interface inspired by Grasshopper: http://nikitron.cc.ua/sverchok_en.html https://github.com/nortikin/sverchok/ [14] See the new interface/fork at: https://www.bforartists.de/
3D modeling (opensource) Dilay https://github.com/abau/dilay --- https://www.reddit.com/r/programming/comments/46130v/dilay_a_free_3d_sculpting_application/
3D modeling (Opensource) Wings3D Free, OpenSource and crossplatform 3D modeler. Wings 3D is an advanced subdivision modeler that is both powerful and easy to use.
3D modeling (Opensource) SolveSpace Parametric 3D CAD program for Windows, Linux, and Mac OS X. See also its 2D predecessor, SketchFlat.
Originally a Windows 1.4Mb executable that didn't have to be installed.
Available on the web at: https://files.whitequark.org/SolveSpace/solvespace.html or see: http://notecad.xyz/ [15] also see: https://github.com/ppd1990/solvespace-snap and https://snapcraft.io/solvespace [16] or for Mac OS X at: https://github.com/whitequark/solvespace/releases [17]
Does basic 3D operations as well as 2D. A good free option for basic modelling, as well as for education (the basics are the same as any full featured parametric CAD system). Outputs in .svg. .stl output, should be pretty well suited to simple modeling for 3DP.[18]
Features include: Constraints and assemblies
Articles:
Forum discussions:
Videos[19]:
3D modeling (Opensource) POVray "The Persistence of Vision Raytracer (POVray) is a high-quality, Free Software tool for creating stunning three-dimensional graphics".
2D drafting/3D modeling (Opensource) lignumCAD Tool for designing furniture.
2D drafting/3D modeling (Opensource) Salome (or Salomé) Generic platform for Pre- and Post-Processing for numerical simulation. Can be used as standalone application for generation of CAD model, its preparation for numerical calculations and post-processing of the calculation results.
3D modeling (Freeware) Caligari Truespace Commercial program released to freeware. http://lunadude.com/rsrc_trueSpace.htm
3D modeling (Freeware) DesignSpark Mechanical Freeware software for Windows distributed by RS Components International.[21] Free version of SpaceClaim 2012.[22]
3D modeling (Freeware) FormZ Free Freeware software for Windows 7/8/8.1 and Mac OS X 10.7 or later.[23]
3D modeling (Freeware/Windows) Microsoft: 3D Builder Create, print (List of supported printers) and scan (requires Kinect v2 sensor) functionality.[24]
3D modeling (Freeware/Windows) Pixologic::Sculptris Sculptris is a free 3D modeling program by Pixologic (The makers of ZBrush), that many find to be a good starting point for learning 3D modeling.
It is also well liked for its efficient use of computer resources, and works well with laptops, or older operating systems, such as Windows XP SP2, for example. [25]
3D modeling (Chrome app) Design Something Google App Store 3D modeling application running on Chrome and Firefox. Pre-defined shapes, operations include extrude and sweep.
2D drawing (Opensource) Doodle3D Freehand sketching program for iPad and Mac OS X (source available for other platforms) to extrude and revolve into 3D.
• CollabCAD --- 3D CAD/CAM Software system for collaborative design & development of Industrial Designs. Written in Java, available for Linux and Windows, reads XML, STEP, IGES, and VRML.
• Varkon --- parametric 2D/3D CAD, modeling and application development tool for engineering, design, architecture, etc. does 2D drafting, 2D and 3D modeling, sculptured surface modeling, and animations. Includes a 6-minute tutorial.[26]
Anim8or --- Windows 3D modelling tool able to import and modify .3DS files [27]
https://www.bricsys.com/en-intl/shape/ --- used in architecture.[28]
Android app for Mattel's Thingmaker: https://play.google.com/store/apps/details?id=com.mattel.thingmakerdesign
Interesting post on efficiency: http://community.carbide3d.com/t/how-to-make-your-stl-not-take-3-days-to-calculate-tool-paths-for/169
http://www.ptc.com/cad/elements-direct/modeling/express --- Creo Elements/Direct Modeling Express 6.0 is a free version of Creo Elements/Direct Modeling 19.0 limited to assemblies with up to 60 unique parts.
As an alternative to OpenSCAD see http://www.plasm.net/ or the nascent http://makertron.io/ [29]
#### Previewers
iPad STL viewer: 3D Push Source on Github [31]
#### Online modelers
Some may find OnShape sufficiently free to consider it: https://www.onshape.com/ or
http://edges.io/ --- noted as working well w/ a stylus.[32]
3Dtin now refreshes to: https://www.tinkercad.com/
#### Image Depth Maps
https://www.selva3d.com/ --- converts images into depth maps which are then integrated into a library of 3D models at "standard accuracy" (better quality modeling is a paid option).
Apparently Microsoft's 3D Builder can create depth / height maps from pixel images as well: Commercial Software#Free
#### Further things to add
http://prisma3d.net/ --- Android [35]
• HyperFun
• PLaSM
#### Sculpting
A number of these are listed above and should be moved.
• Blender
• EQUINOX-3D
• DeleD CE
• Structure Synth
• Moonlight|3D
• OpenFX
• Fragmentarium
Free online tool which will export to STL: http://kovacsv.github.io/JSModeler/documentation/examples/bezier.html
## Post-processing
Voronoi hollowing: http://www.voronator.com/ [36][37]
## Gears
There is an Excel spreadsheet for clockwork calculations: http://www.shapeoko.com/forum/viewtopic.php?f=7&t=5190&p=38066
Type Name/Link Description
Gears Involute Spur Gear Builder Browser based utility for calculating and drawing involute spur gears, fully accounts for undercuts. Blog announcement.[40]
Gears GGGears [42] Tool for creation and calculation of gear transmission finite element models. Consists of a geometry and mesh generator based on GMSH and a finite element model based on GETFEM++.
Gears pyGear Built on top of pythonOCC, OpenCascade, numpy and scipy, serves as an CAE/CAD-preprocessor generator of involute gears.
Sprockets Sprocketeer (online) Windows and information Generate gcode or a CAD (DXF) drawing for sprockets.
Gears GearGCoderV1.0 (Windows) Generate gcode for a gear based on inputting Diameter (150mm --- values in parentheses are samples suggested by the developer), teeth (20), width ratio (3), feed rate (1000), tooth height (13mm), depth (9mm), add mounting holes (Y), number of holes (5), distance from center (50mm).[43]
Gears GearMaker (Windows) Program that will produce the tooth profile of a gear as a dxf drawing. Video tutorial on finishing.
Gears GearDXF (Windows) Outputs a DXF file containing a 2D outline of a spur gear using specified parameters. The tooth face is a true involute curve, but the base gaps between teeth are rough.[44]
Gears gears (Java) Involute gear milling generator and G-code viewer[45]
Gears Gear Model For 3D Printer (OpenJSCad website) Online gear generator includes support for rack and pinion, bevel and worm gears.
For information on wooden gears, see woodgears.ca, esp. the Gear template generator. See also Github: SVG generating interactive gear maker.
Extensive tutorial at: http://www.cartertools.com/involute.html
Inkscape has an extension for gears: Effects | Render | Gear...[46]
## PCB
Type Name/Link Description
PCB (opensource) KiCad Electronic design automation (EDA) software suite for designing schematics of electronic circuits and printed circuit boards (PCB).
PCB (opensource) gEDA Full GPL'd suite and toolkit of Electronic Design Automation tools.
PCB (opensource) QS Cad Tcl/Tk EDA Package. Schematic capture through to PCB layout. Has Part editor as well as editor for Schematic and PCB Decals.
PCB (opensource) Stripboard Designer 1.0 Open source (BSD) app written in freeware Visual Basic 2008 Express for Wintel machines. See also Stripes, a GPL Java version.
PCB (opensource) DIY Layout Creator (DIYLC) GPL Java software for creating stripboard and universal board designs.
http://darwinne.github.io/FidoCadJ/download.html --- (Java) a free graphical editor for (more than) electronics
## Special Purpose
Type Name/Link Description
3D modeling --- Surfboards BoardCAD Opensource CAD/CAM-system (written in Java) for surfboards.
3D modeling --- furniture SketchChair Opensource CAD/CAM-system for flat-pack furniture. Forums/troubleshooting
2D CAM --- Jigsaw puzzle piece cutter (Online - Silverlight) CNC Puzzle Maker Program for cutting out puzzle shapes w/ tabs.
2D CNC --- sails Sailcut CAD Sail design and plotting software.
2D CNC --- Parametric boxes Planter Box Generator[49] Javascript to generate jointed boxes for planters (and other uses).
http://touch-mapper.org/ --- exports STL files from OpenStreetMap data.[50]
http://www.derekhugger.com/tools/dsHypoGen.zip --- dsHypoGen is a Python script that calculates parameters for generating dual stage hypocycloid reducers.[51]
http://valentina-project.org/ --- cross-platform patternmaking program which allows designers to create and model patterns of clothing.
http://commonwealthrobotics.com/#BowlerStudio ---robot development application that combines scripting and device management to allow the design of printable walking robots. [52]
### Aircraft (incl. models)
Type Name/Link Description
3D CNC --- foam cutting for model airplanes Jedicut CNC Program for cutting out wings and fuselages.
3D aircraft modeling OpenVSP Vehicle sketch pad. NASA opensource parametric geometry. Described as suited for general use.[53]
UIUC Airfoil Coordinates Database--- Tool to generate OpenSCAD coordinates: http://chaaawa.com/airfoils/
### Mechanical modeling
See also http://www.rs-online.com/designspark/electronics/eng/page/mechanical listed above and the CAM page which lists special-purpose programs for gears and fret and thread cutting.
### Robotics
http://commonwealthrobotics.com/ Bowler Studio Alpha Release [65]
### Textures
Carbide Create has creating textures as a native feature.
## Other design software
3D CG to 3D Printable 3D model converter for Windows PC --- CGto3DPrint is a free software that converts 3D CG(Computer Graphics) models to 3D printable manifold 3D models using MeshLab for surface mesh to voxel conversion and Blender for baking color textures. Converted 3D models are automatically exported from Blender to STL file for single color 3D Printing or manually exported to PLY file for full color 3D Printing. [67]
https://github.com/DEF7/TGA2STL --- Command-line utility for generating optimized STL meshes from TGA depthmaps. [68]
### Conversion
http://kabeja.sourceforge.net/ --- works well for DXF -> SVG.[69]
## 3D Scanning
Redrawing may be a better option: https://www.reddit.com/r/cad/comments/4mmqno/im_trying_to_reverse_engineer_a_component_to_draw/
FreeLSS --- laser scanning program for the Raspberry Pi. It allows a Raspberry Pi to function as the core to a complete turn table laser scanning system.[71][72] Files for printing a 3D scanner: http://www.murobo.com/printables.html
http://ccwu.me/vsfm/ and http://www.danielgm.net/cc/ allow one to use a high quality camera rig to derive 3D shapes from photographs of objects.[73]
## Bézier drawing programs
### Opensource
METAPOST is a special case in point, see http://cseweb.ucsd.edu/~s1pan/MEPer/ | 4,790 | 20,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-05 | latest | en | 0.925864 |
https://mathoverflow.net/questions/103530/serres-theorem-about-regularity-and-homological-dimension/244191 | 1,619,007,462,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00393.warc.gz | 471,860,072 | 35,019 | # Serre's theorem about regularity and homological dimension
One of the nicest results I know of is (Auslander-Buchsbaum-)Serre's theorem asserting that a (commutative!) local ring is regular iff it has finite global dimensional.
I'd like to ask a somewhat vague question:
what is the history and what was the context of this result?
By this I mean: presumably the above characterization did not come out of thin air (or just out of Serre's mind!), and there was a buildup of ideas which lead to such an elegant and, I guess, surprising characterization of regularity. Nowadays, such a thing appears almost natural to our minds brought up in the nice set-up constructed by the founders of homological algebra and tended to by a few generations already, but I suspect it was less «evident» at the time.
• Nice question! But, how can you dare to use 'iff' when asking about a result of Serre ;D – user9072 Jul 30 '12 at 18:49
• (Shhh. It is bait to get him to register on MO! :-) ) – Mariano Suárez-Álvarez Jul 30 '12 at 18:51
In 1953-54 E. Artin asked me to describe homological algebra to him. In addition to Ext and Tor, I decided to show him the "homological proof" of the Hilbert Basis Theorem and indicated that the same proof showed that a regular local ring had finite global dimension. Artin then mentioned the open localization problem, and I said that if the converse of the theorem I just showed him were true, the localization result would be trivial. He asked if I could prove the converse, I said no, and we both agreed that it would be nice to prove it. The question of factoriality in regular local rings also came up in that conversation, and so I set myself the goal of proving those two theorems. I persuaded Auslander to join me in that project. When Auslander and I had almost all the results, and an outline of the string of inequalities needed, Eilenberg asked to see them, we wrote them up for him, and he went to Paris with them. It was there that Serre saw the outline of our project, and he beat us to the final proof by around a week (or the time it took an air mail letter to travel from Paris to Princeton). This may clear up the questions initially posed.
• (By the way: do you really mean Hilbert Basis theorem or the theorem on syzygies?) – Mariano Suárez-Álvarez Jul 12 '16 at 22:27
ADDED: There is an account written by Buchsbaum (see page 1 and 2 of number 23 here) which described in more details what they wrote in [1]. So the localization problem for regular rings was definitely the main motivation for them.
The story of this fascinating theorem is quite complicated, in fact when I was a graduate student I heard some juicy stories around it, so I took this opportunity to do some research. I doubt that the full truth can be known even if we could somehow talk to everyone involved, so the following is perhaps (un)educated guess at best.
There are several components to your question, namely:
a) Who proved what?
b) What is the motivation for the statement of the theorem?
As for a), here are the relevant references:
[1] M. Auslander and D. A. Buchsbaum, Homological dimension in noetherian rings. Proc. Nat. Acad. Sci. U.S.A. vol. 42 (1956).
[2] Auslander, Maurice; Buchsbaum, David A. Homological dimension in local rings. Trans. Amer. Math. Soc. 85 (1957), 390–405.
[3] Serre, Jean-Pierre. Sur la dimension homologique des anneaux et des modules noethériens. (French) Proceedings of the international symposium on algebraic number theory, Tokyo & Nikko, 1955, pp. 175–189. Science Council of Japan, Tokyo, 1956.
[4] Kaplansky, Irving. Commutative rings. Conference on Commutative Algebra (Univ. Kansas, Lawrence, Kan., 1972), pp. 153–166. Lecture Notes in Math., Vol. 311, Springer, Berlin, 1973. 13-03
The result you quoted (by the way, nowadays is often known as the Auslander-Buchsbaum-Serre theorem) was announced in [1]. It stated clearly there that one of the ingredients is a Lemma by Serre (which stated that the global dimension is bound below by the number of generators of the maximal ideal) however [1] did not give references and contained no proofs (announcing your breakthrough like that was a fairly common practice in the days before arXiv, it must be said).
The full proofs appeared in [2], in which the Lemma was given a clear reference as [3, Theorem 4]. However, the review of [3], written by Buchsbaum, said:
The author gives an exposition of the results of M. Auslander and the reviewer [Proc. Nat. Acad. Sci. U.S.A. 42 (1956), 36–38; MR0075190 (17,705b)] and completes these results, notably by giving a homological characterization of regular local rings.
Also, Serre's book "Local Algebra" refers to [3] for the full result (Theorem 9 there).
So it looks like [1] and [3] appeared at virtually the same time and with knowledge of each other! Unfortunately I could not find [3].
Perhaps the last word could be given to Kaplansky, who wrote in his survey [4]
The big theorem was proved by Auslander, Buchsbaum and Serre. (The Auslander-Buchsbaum portion was announced in [1], with full details in [2]; Serre finished the job in [3].)
OK, so what is the answer to b)? I will leave the floor to Auslander-Buchsbaum, who wrote in [1] after stating that regular local rings have finite global dimension:
Therefore, if $R$ is a regular local ring and $P$ is a prime ideal of $R$, then $gl.dim \ R_P$ is finite.... This observation, together with some direct computations, led the authors to conjecture:
Theorem. A local ring $R$ is regular if and only if $gl.dim \ R$ is finite.
• Thanks for the answer! (I've edited the question to attribute the result more correctly now..._ – Mariano Suárez-Álvarez Jul 31 '12 at 6:04
• No problem, perhaps someday we can talk more about this over a beer (-: – Hailong Dao Jul 31 '12 at 14:06
• I think the paper referenced in the added paragraph at the top is now number 25 not number 23? For improved permanence, it might be good to use the name of the paper, with a comment like "currently number 25"? – benblumsmith Oct 22 '19 at 12:26
I don't know what Serre (or Auslander and Buchsbaum?) was thinking, but it would have been natural to observe that $R$ is regular iff its maximal ideal is generated by a regular sequence, which (by writing down the Koszul complex) implies that the residue field $k$ has finite projective dimension. If you've already established (or have good reason to expect) that no module can have larger projective dimension than the residue field, then you're naturally led to this result.
According to Serre's definition, it suffices to prove that the Krull dimension of the commutative Noetherian ring is equal to its global dimension which is given by the projective dimension of the residue field (a result that can also be obtained via Steven Landsburg's answer). However, by considering an affine neighborhood of the ring in the scheme formed by the residue field, every point in the scheme corresponds to a prime ideal and hence to a localization. This is closely related to the concept of Galois connection: that prime ideals of a ring correspond to a point on the scheme via the Galois Connection. Therefore the Krull dimension equals the (irreducible) projective dimension of the spectrum, and is therefore equal to the minimal number of generators of the maximal ideals of the Ring, for all such localizations. There is nothing wrong with using Koszul complexes, but this fact is also true when you consider schemes and their spectra.
• This is not an answer to the question, is it? :-) – Mariano Suárez-Álvarez Jul 30 '12 at 20:32
• Not exactly, just an alternative way of arriving at Steven Landsburg's result using schemes! – Nick Bagley Jul 30 '12 at 20:58
• I don't have enough reputation to comment and this seemed like an interesting exercise, seeing whether it could be done with spectra as I suspected. – Nick Bagley Jul 30 '12 at 21:11
• Not enough reputation to comment on Steven Landsburg's answer, that is, otherwise that is what I would have done. I can comment on my own answers at least! :-) – Nick Bagley Jul 30 '12 at 21:12
• (Yup, I was just checking: with a pit of patience this answer will get enough upvotes for you to collect the rep needed to make comments) – Mariano Suárez-Álvarez Jul 30 '12 at 21:16 | 2,053 | 8,299 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-17 | latest | en | 0.976046 |
https://number.academy/35171 | 1,660,861,422,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573533.87/warc/CC-MAIN-20220818215509-20220819005509-00241.warc.gz | 390,564,079 | 12,159 | # Number 35171
Number 35,171 spell 🔊, write in words: thirty-five thousand, one hundred and seventy-one . Ordinal number 35171th is said 🔊 and write: thirty-five thousand, one hundred and seventy-first. The meaning of number 35171 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 35171. What is 35171 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 35171.
## What is 35,171 in other units
The decimal (Arabic) number 35171 converted to a Roman number is (X)(X)(X)(V)CLXXI. Roman and decimal number conversions.
#### Weight conversion
35171 kilograms (kg) = 77538.0 pounds (lbs)
35171 pounds (lbs) = 15953.5 kilograms (kg)
#### Length conversion
35171 kilometers (km) equals to 21855 miles (mi).
35171 miles (mi) equals to 56603 kilometers (km).
35171 meters (m) equals to 115390 feet (ft).
35171 feet (ft) equals 10721 meters (m).
35171 centimeters (cm) equals to 13846.9 inches (in).
35171 inches (in) equals to 89334.3 centimeters (cm).
#### Temperature conversion
35171° Fahrenheit (°F) equals to 19521.7° Celsius (°C)
35171° Celsius (°C) equals to 63339.8° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
35171 seconds equals to 9 hours, 46 minutes, 11 seconds
35171 minutes equals to 3 weeks, 3 days, 10 hours, 11 minutes
### Zip codes 35171
• Zip code 35171 Thorsby, Alabama, Chilton, USA a map
• Zip code 35171 Salamanca, Durango, Lerdo, Mexico a map
### Codes and images of the number 35171
Number 35171 morse code: ...-- ..... .---- --... .----
Sign language for number 35171:
Number 35171 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 35171
### Multiplications
#### Multiplication table of 35171
35171 multiplied by two equals 70342 (35171 x 2 = 70342).
35171 multiplied by three equals 105513 (35171 x 3 = 105513).
35171 multiplied by four equals 140684 (35171 x 4 = 140684).
35171 multiplied by five equals 175855 (35171 x 5 = 175855).
35171 multiplied by six equals 211026 (35171 x 6 = 211026).
35171 multiplied by seven equals 246197 (35171 x 7 = 246197).
35171 multiplied by eight equals 281368 (35171 x 8 = 281368).
35171 multiplied by nine equals 316539 (35171 x 9 = 316539).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 35171
Half of 35171 is 17585,5 (35171 / 2 = 17585,5 = 17585 1/2).
One third of 35171 is 11723,6667 (35171 / 3 = 11723,6667 = 11723 2/3).
One quarter of 35171 is 8792,75 (35171 / 4 = 8792,75 = 8792 3/4).
One fifth of 35171 is 7034,2 (35171 / 5 = 7034,2 = 7034 1/5).
One sixth of 35171 is 5861,8333 (35171 / 6 = 5861,8333 = 5861 5/6).
One seventh of 35171 is 5024,4286 (35171 / 7 = 5024,4286 = 5024 3/7).
One eighth of 35171 is 4396,375 (35171 / 8 = 4396,375 = 4396 3/8).
One ninth of 35171 is 3907,8889 (35171 / 9 = 3907,8889 = 3907 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
35171
#### Is Prime?
The number 35171 is a prime number. The closest prime numbers are 35159, 35201.
#### Factorization and factors (dividers)
The prime factors of 35171
Prime numbers have no prime factors less than themselves.
The factors of 35171 are 1 , 35171
Total factors 2.
Sum of factors 35172 (1).
#### Prime factor tree
35171 is a prime number.
#### Powers
The second power of 351712 is 1.236.999.241.
The third power of 351713 is 43.506.500.305.211.
#### Roots
The square root √35171 is 187,539329.
The cube root of 335171 is 32,763848.
#### Logarithms
The natural logarithm of No. ln 35171 = loge 35171 = 10,467977.
The logarithm to base 10 of No. log10 35171 = 4,546185.
The Napierian logarithm of No. log1/e 35171 = -10,467977.
### Trigonometric functions
The cosine of 35171 is -0,644641.
The sine of 35171 is -0,764486.
The tangent of 35171 is 1,18591.
### Properties of the number 35171
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 35171 in Computer Science
Code typeCode value
35171 Number of bytes34.3KB
Unix timeUnix time 35171 is equal to Thursday Jan. 1, 1970, 9:46:11 a.m. GMT
IPv4, IPv6Number 35171 internet address in dotted format v4 0.0.137.99, v6 ::8963
35171 Decimal = 1000100101100011 Binary
35171 Decimal = 1210020122 Ternary
35171 Decimal = 104543 Octal
35171 Decimal = 8963 Hexadecimal (0x8963 hex)
35171 BASE64MzUxNzE=
35171 MD533349564c6c0e03cbffb25daa4306146
35171 SHA1877c27f0a9f247ba4c20c655e0f4f28dd516808f
35171 SHA2242a88833db9fd86e8e44cf736786ec82b91af688127bac7c228240cb1
35171 SHA256264d443d1a151a541e032470a59775ef6dc37e6931942e1a5c3bdc5d1c95420b
More SHA codes related to the number 35171 ...
If you know something interesting about the 35171 number that you did not find on this page, do not hesitate to write us here.
## Numerology 35171
### Character frequency in number 35171
Character (importance) frequency for numerology.
Character: Frequency: 3 1 5 1 1 2 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 35171, the numbers 3+5+1+7+1 = 1+7 = 8 are added and the meaning of the number 8 is sought.
## Interesting facts about the number 35171
### Asteroids
• (35171) 1993 TF1 is asteroid number 35171. It was discovered by K. Endate; K. Watanabe from Kitami Observatory on 10/15/1993.
## № 35,171 in other languages
How to say or write the number thirty-five thousand, one hundred and seventy-one in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 35.171) treinta y cinco mil ciento setenta y uno German: 🔊 (Anzahl 35.171) fünfunddreißigtausendeinhunderteinundsiebzig French: 🔊 (nombre 35 171) trente-cinq mille cent soixante et onze Portuguese: 🔊 (número 35 171) trinta e cinco mil, cento e setenta e um Chinese: 🔊 (数 35 171) 三万五千一百七十一 Arabian: 🔊 (عدد 35,171) خمسة و ثلاثون ألفاً و مائةواحد و سبعون Czech: 🔊 (číslo 35 171) třicet pět tisíc sto sedmdesát jedna Korean: 🔊 (번호 35,171) 삼만 오천백칠십일 Danish: 🔊 (nummer 35 171) femogtredivetusinde og ethundrede og enoghalvfjerds Dutch: 🔊 (nummer 35 171) vijfendertigduizendhonderdeenenzeventig Japanese: 🔊 (数 35,171) 三万五千百七十一 Indonesian: 🔊 (jumlah 35.171) tiga puluh lima ribu seratus tujuh puluh satu Italian: 🔊 (numero 35 171) trentacinquemilacentosettantuno Norwegian: 🔊 (nummer 35 171) tretti-fem tusen, en hundre og sytti-en Polish: 🔊 (liczba 35 171) trzydzieści pięć tysięcy sto siedemdziesiąt jeden Russian: 🔊 (номер 35 171) тридцать пять тысяч сто семьдесят один Turkish: 🔊 (numara 35,171) otuzbeşbinyüzyetmişbir Thai: 🔊 (จำนวน 35 171) สามหมื่นห้าพันหนึ่งร้อยเจ็ดสิบเอ็ด Ukrainian: 🔊 (номер 35 171) тридцять п'ять тисяч сто сiмдесят одна Vietnamese: 🔊 (con số 35.171) ba mươi lăm nghìn một trăm bảy mươi mốt Other languages ...
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If you know something interesting about the number 35171 or any natural number (positive integer) please write us here or on facebook. | 2,469 | 7,194 | {"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} | 2.546875 | 3 | CC-MAIN-2022-33 | latest | en | 0.632294 |
https://www.hackmath.net/en/math-problem/5791?tag_id=74 | 1,603,624,481,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888931.67/warc/CC-MAIN-20201025100059-20201025130059-00516.warc.gz | 756,322,846 | 12,060 | # How many
How many double-digit numbers greater than 30 we can create from digits 0, 1, 2, 3, 4, 5? Numbers cannot be repeated in a two-digit number.
Correct result:
n = 14
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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We have digits 0,1,4,7 that cannot be repeated. How many three-digit numbers can we write from them? You can help by listing all the numbers. | 632 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.94732 |
https://www.aqua-calc.com/calculate/gravel-quarter-cylinder/substance/caribsea-coma-and-blank-freshwater-coma-and-blank-african-blank-cichlid-blank-mix-coma-and-blank-sahara-blank-sand | 1,716,836,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00121.warc.gz | 564,931,078 | 8,650 | # Qt Cyl: CaribSea Freshwater African Cichlid Mix Sahara Sand
## quarter cylindrical aquarium sand, gravel and substrate calculato...17
### The weight of CaribSea, Freshwater, African Cichlid Mix, Sahara Sand, 2 inches high, in a quarter cylindrical aquarium with the radius of a base of 10 inches
gram 3 793.41 ounce 133.81 kilogram 3.79 pound 8.36
### The volume of CaribSea, Freshwater, African Cichlid Mix, Sahara Sand, 2 inches high, in a quarter cylindrical aquarium with the radius of a base of 10 inches
centimeter³ 2 574.07 inch³ 157.08 foot³ 0.09 meter³ 0
• About CaribSea, Freshwater, African Cichlid Mix, Sahara Sand
• CaribSea, Freshwater, African Cichlid Mix, Sahara Sand weighs 1.47 gram per cubic centimeter or 1 473.7 kilogram per cubic meter, i.e. density of caribSea, Freshwater, African Cichlid Mix, Sahara Sand is equal to 1 473.7 kg/m³. In Imperial or US customary measurement system, the density is equal to 92 pound per cubic foot [lb/ft³], or 0.85 ounce per cubic inch [oz/inch³] .
• CaribSea, Freshwater, African Cichlid Mix, Sahara Sand weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
• The corner bow front aquariums are designed to fit in the corner of a room and to provide optimum viewing space. There is a quarter of a circle lies at the base of the aquarium with radius (r).
The number in square brackets, to the right of the gravel name, is the gravel specific gravity or relative density, as compared to pure water.
The calculator computes the weight (1) and volume (2) of gravel using the following formulas:
(1) m = ρ × V and (2) V = π×r²×h ⁄ 4, where ρ is the density, V — volume, m — weight (or mass), and h is the height of gravel; π is approximately equal to 3.14159265359, and r is the radius of a base of an aquarium.
See also our gravel calculators for cylindrical and rectangular shaped aquariums and ponds.
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#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Kon Tiki weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
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#### Calculators
Calculate the volume and area of the surface of a quarter cylinder | 1,027 | 3,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-22 | latest | en | 0.76 |
https://brainmass.com/business/cash-budgeting/budget-102789 | 1,477,186,913,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00136-ip-10-171-6-4.ec2.internal.warc.gz | 820,289,005 | 21,566 | Share
Explore BrainMass
Budget
For this assignment you will create a financial business plan that will tie together an income statement, a purchases budget, a cash budget, and a pro forma balance sheet. The first step is to produce the income statement.
Income Statement
To complete this first step, we need to project monthly sales and expenses. This requires estimates of monthly sales volume, selling price, and cost behaviors. The following are the estimates you should use in developing your model.
Annual sales: 18,000 pizzas
Monthly sales pattern:
January 8% April 10.5% July 4% October 10.5%
February 10% May 7% August 6% November 9%
March 11% June 6% September 9.5% December 8.5%
Selling Price: \$11.50 per pizza
Annual cost behaviors:
Fixed Variable (per pizza)
Ingredients \$ 0 \$3.00
Salaries & wages 48,000 .50
Depreciation 9,000 .00
Utilities 3,000 .25
Supplies 0 .15
Local taxes 4,200 .00
Insurance 2,400 .00
Miscellaneous 2,400 .10
The above fixed expenses are for the entire year. Assume fixed expenses are incurred evenly throughout the year.
Your assignment is to produce a computer spreadsheet that will generate monthly income statements for the entire year. The spreadsheet should allow for easy manipulation of annual volume, selling price, and changes in cost behavior.
Purchases budget, balance sheet, and cash budget
After completing the projected income statement, you are now able to prepare the purchases budget, cash budget, and balance sheet. The purchasing budget is used to plan raw material (ingredient) purchases. The purpose of the cash budget is to predict our cash position at the end of each period, thus allowing us to plan short-term borrowing if it is needed. We create the cash budget by determining the amount and timing of cash receipts and cash payments. To do this, we make certain assumptions about expected cash flow patterns. The following assumptions apply to the Pizza Shoppe:
1. Revenues are collected in the month of sale.
2. Ingredients purchased are paid as follows:
? Forty percent in the month of purchase.
? Sixty percent in the month following the purchase.
3. Our policy is to have enough ingredients on hand at the end of each month to satisfy 20% of the next month's cost of ingredients.
4. Salaries and wages are paid as follows:
? Seventy-five percent in the month incurred.
? Twenty-five percent in the following month.
5. Advertising is paid in the month incurred.
6. Utilities are paid in the month after it's incurred.
7. Supplies are purchased and paid in the month before use.
8. Insurance is paid 50% in January and 50% in July.
9. The local taxes are paid 50% in April and 50% in October.
10. All miscellaneous expenses are paid in the month after they are incurred.
11. The long-term loan carries a 12% interest rate.
12. The owner withdraws \$3,000 per month.
In addition to the amount and timing of cash flows, we need to know what our initial financial position is. Financial position is captured by the balance sheet. The balance sheet is a listing of our resources (assets), our obligations (liabilities), and the net worth of the business (owner's equity). The balance sheet for the Pizza Shoppe on January 1, just after we purchase it, is as follows:
ASSETS LIABILITIES
Cash \$ 100 Accounts Payable \$ 0
Ingredients inventory 700 Salaries & Wages Payable 0
Supplies inventory 216 Utilities payable 500
Prepaid taxes 0 Taxes payable 0
Prepaid insurance 0 Misc. expenses payable 0
Equipment & fixtures 30,000 Short-term bank loan 0
Less: Accumulated deprec. 0 Long-term bank loan 90,000
Building 60,000 Total Liabilities \$ 90,500
Less: Accumulated deprec. 0
Land 29,500 OWNERS EQUITY \$ 30,016
Total Assets \$120,516
TOTAL LIABILITIES & O.E. \$120,516
Solution Summary
Excel file contains :
1. Financial statements with detailed calculations
2. Assumptions mentioned in the end
3. Formulas sheet for easy reference.
\$2.19 | 934 | 3,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-44 | longest | en | 0.917521 |
http://www.algebra-class.com/compound-inequality.html | 1,518,908,801,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891808539.63/warc/CC-MAIN-20180217224905-20180218004905-00361.warc.gz | 406,429,130 | 9,069 | Home
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» » Compound Inequalities
# What is a Compound Inequality?
As your study of inequalities progresses, you will be introduced to a compound inequality. These inequalities can be a little tricky, so let's first take a look at the definition.
### Compound Inequalities
A compound inequality consists of two inequalities that are joined together by the word "and" or the word "or".
Here are a few examples of compound inequalities:
### x < 3 or x > 6
Do you notice how each of the problems above consist of two inequalities?
Let's begin by focusing on "AND" inequalities.
## "And" Compound Inequalities
The solution to an "And" inequality is the intersection of the solution sets for two different inequalities. It's best explained through looking at a graph. Take a look with me...
### x > -3 and x < 5
Do you notice how the solution set for the compound inequality only includes the solution sets that intersect or overlap on the graph? This is the most important rule to remember for "AND" compound inequalities.
One other thing I'd like to point out is that many "And" compound inequalities are not actually written with the word "and".
For example, this inequality shown above is most likely to be written as:
### -3 < x < 5
This can be read as " x is greater than -3 and x is less than 5". Do you notice how you say the word "and", but you don't use the word "and" when actually writing the compound inequality? Here's why it can be read like this:
This one inequality can be broken into 2 different inequalities:
So, through this brief introduction, you should understand that there are two types of compound inequalities, "And" inequalities and "Or" inequalities.
If the inequality is an "And" compound inequality, then the solution is the intersection of the solution sets for both inequalities. You may not see the word "and" in the original inequality, but when you read the inequality aloud, you will say the word "and".
Now let's move on to compound inequalities that use the word "OR".
## "OR" Compound Inequalities
The nice thing about "Or" inequalities is that you will always see the word "or" in the problem. This makes it easier to distinguish between the two types of inequalities.
The solution to a compound inequality containing the word "or" is the union of the solution sets. This means that the solution sets will not overlap or intersect. We will be uniting the solution sets instead. Let's take a look at a graph for a better picture.
### x < -3 or x > 5
The final graph for this compound inequality would look exactly as it is shown:
Did you notice how the solution set for each individual inequality did not overlap at all? Therefore, our solution set for the compound inequality must include both sets of solutions. In a sense we are uniting these two answers in order to include both.
And lastly, here's a little vocabulary for you to add to your list:
### FYI - Important Vocabulary
A compound inequality that involves the word "AND" is called a "conjunction"
A compound inequality that involves the word "OR" is called a disjunction.
Hopefully by now you better understand the difference in the two types of compound inequalities and you have an idea of what the graph will look like for each type of compound inequality.
In order to better understand these types of inequalities, you will want to study the following lessons:
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Custom Search | 798 | 3,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | longest | en | 0.951118 |
https://practicetestgeeks.com/abstract-reasoning/ | 1,709,027,461,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00434.warc.gz | 466,994,294 | 108,486 | # Abstract Reasoning Test 2023
Abstract reasoning tests are a common component of many job application processes, but their purpose and relevance may not always be immediately evident. While other types of tests assess specific knowledge or skills, abstract reasoning tests are designed to measure your ability to think analytically and solve problems in unfamiliar or complex situations. They require you to identify patterns, make logical connections, and draw conclusions based on limited information.
## Abstract Reasoning Examples
Abstract reasoning refers to the ability to think logically and solve problems without relying on concrete information. It is a crucial skill in many areas, such as mathematics, computer science, and philosophy. One example of abstract reasoning is solving a puzzle where you have to identify the pattern or sequence among various shapes or symbols. This requires the ability to recognize relationships and make connections based on abstract concepts.
Another example of abstract reasoning is understanding analogies. Analogies involve comparing two different things and finding the relationship between them. For instance, if the tree is to leave, then the car might be on wheels. Being able to spot these connections can help improve problem-solving skills in various scenarios.
Abstract reasoning exercises are beneficial for developing critical thinking abilities because they encourage us to look beyond surface-level information and find hidden patterns or similarities. By practicing this skill regularly, we enhance our ability to analyze complex situations and develop creative solutions that others might overlook. So next time you come across an abstract reasoning puzzle or analogy question, give it a try – you never know what new insights you may gain!
## Abstract Reasoning Meaning
Abstract reasoning is the ability to think beyond concrete facts and objects, and instead engage in logical and creative thinking. It involves the interpretation of patterns, relationships, and principles that may not be immediately apparent. This form of reasoning allows individuals to solve complex problems, make connections between seemingly unrelated concepts, and think outside the box.
The meaning of abstract reasoning goes beyond mere intelligence or problem-solving skills; it encompasses a higher level of cognitive ability. It requires individuals to use logical thinking, creativity, intuition, and critical analysis to draw conclusions or find solutions. Abstract reasoning can be strengthened through practice and exposure to different types of problems or puzzles that challenge one’s ability to think conceptually.
Engaging in abstract reasoning not only enhances intellectual abilities but also promotes innovative thinking. It encourages individuals to approach problems from a different perspective and explore unconventional solutions. By training the mind to think abstractly, one becomes better equipped to adapt to change, find hidden patterns or trends in data, and make connections between seemingly disparate ideas. Ultimately, embracing abstract reasoning opens up new possibilities for personal growth and intellectual development by expanding the boundaries of conventional thought processes.
## Abstract Reasoning IQ
Abstract Reasoning IQ refers to the ability to analyze and solve complex problems using innovative thinking and logical reasoning. It involves the use of patterns, symbols, and relationships to identify connections and arrive at a solution. This type of intelligence is not based on prior knowledge or specific skills but rather on one’s ability to think critically, make inferences, and think abstractly. Individuals with high Abstract Reasoning IQ tend to excel in fields that require problem-solving abilities such as mathematics, computer programming, engineering, and scientific research.
One intriguing aspect of Abstract Reasoning IQ is its potential for development and improvement through practice. Unlike other forms of intelligence that are considered more fixed or innate, abstract reasoning skills can be honed over time with focused effort. By engaging in activities such as solving puzzles, playing strategic games like chess or Sudoku, and engaging in brainstorming exercises that encourage thinking outside the box, one can strengthen their abstract reasoning abilities.
Moreover, Abstract Reasoning IQ plays a vital role in everyday life situations that involve decision-making. It helps individuals envision multiple possibilities while considering different outcomes before making a choice. In this way, having strong abstract reasoning skills can lead to better decision-making processes overall.
## Abstract Reasoning Psychology
When it comes to understanding the inner workings of the human mind, abstract reasoning psychology is a fascinating field worth exploring. Abstract reasoning refers to our ability to think conceptually and solve problems by making connections between different ideas or concepts. It enables us to engage in higher-level thinking and create mental representations that go beyond concrete experiences.
One interesting aspect of abstract reasoning psychology is its relationship with creativity. Research suggests that individuals who excel in abstract reasoning tend to be more creative and innovative. They have an innate ability to see patterns where others might not, enabling them to come up with novel solutions and ideas. By delving deeper into this connection between abstract reasoning and creativity, psychologists hope to unlock new ways of fostering innovation, problem-solving skills, and overall cognitive development.
Another intriguing area within abstract reasoning psychology is its impact on decision-making processes. Abstract thinkers often rely on intuition and gut feelings rather than solely relying on logic when making decisions. This approach allows them to consider multiple factors simultaneously, leading to flexible decision-making strategies that take into account both rational arguments as well as instinctual insights.
## Abstract Reasoning Questions
Abstract reasoning questions are a crucial part of many aptitude tests and assessments. These questions are designed to assess your ability to think logically, solve problems, and make connections between different concepts. Unlike traditional factual-based questions, abstract reasoning questions require you to analyze patterns, identify trends, and apply logical reasoning without relying on prior knowledge or specific information.
What makes abstract reasoning questions intriguing is their open-ended nature. They often present you with a series of shapes, symbols, or patterns that follow a certain logic or sequence. Your task is to decipher the underlying pattern and predict what comes next based on the given options. This challenges your analytical skills, as well as your ability to think creatively and outside the box.
Mastering abstract reasoning requires practice and exposure to various types of patterns. With time and effort spent solving these puzzles, you will develop valuable thinking skills such as problem-solving abilities, critical analysis techniques, and overall cognitive flexibility. The more you engage with abstract reasoning questions, the better equipped you become at identifying complex relationships between concepts – an essential skill not just in exams but also in real-life situations that demand innovative thinking and quick decision-making.
## Abstract Reasoning IQ Score
In the world of intelligence testing, there is a unique and intriguing measure known as the Abstract Reasoning IQ score. Unlike traditional IQ tests that focus on verbal and mathematical abilities, abstract reasoning assesses an individual’s ability to recognize patterns and connections in unfamiliar situations. It shines a light on our innate capacity to think creatively, solve problems, and navigate complex scenarios without relying solely on previous knowledge.
Abstract reasoning IQ tests often present participants with a series of shapes or symbols arranged in patterns. The challenge lies in deciphering these patterns and identifying the missing piece or next shape in line. This type of assessment taps into our cognitive flexibility by demanding that we step out of familiar territories and dive into uncharted waters. By evaluating our capacity to manipulate information freely, abstract reasoning IQ scores offer valuable insights into our potential for innovation, adaptability, and critical thinking.
This way of measuring intelligence has gained recognition as an essential aspect not only for educational systems but also for certain professions like engineering, architecture, and computer programming – fields where problem-solving skills are paramount. Researchers argue that individuals with high abstract reasoning scores demonstrate stronger mental agility when faced with challenges requiring out-of-the-box thinking. Ultimately, understanding your abstract reasoning IQ score can empower you to harness your unique cognitive strengths while also highlighting areas where further development may be beneficial – guiding you towards endeavors where strong pattern recognition is required.
## Abstract Reasoning Skills
Abstract reasoning skills are an often overlooked aspect of cognitive ability, yet they play a crucial role in problem-solving and critical thinking. These skills involve the ability to identify patterns, make connections between seemingly unrelated concepts, and think dynamically. The beauty of abstract reasoning is that it allows us to transcend the constraints of logic and find creative solutions that may not be immediately apparent.
One valuable application of abstract reasoning lies in the realm of innovation and invention. By thinking abstractly, we can generate new ideas by combining existing knowledge or reimagining familiar concepts in novel ways. This skill is particularly relevant in today’s fast-paced world where industries are constantly evolving, requiring individuals who can think outside the box and adapt to changing circumstances.
Furthermore, abstract reasoning can be essential for decision-making. When faced with complex situations or ambiguous information, the ability to think abstractly enables us to consider multiple perspectives and weigh different possibilities before making a choice. It allows us to see beyond surface-level information and delve into underlying patterns or trends that might inform our decisions more effectively.
## Abstract Reasoning Ability
Abstract reasoning ability, also known as fluid intelligence, is the cognitive capacity to think logically and solve problems in novel situations. It involves the ability to identify patterns, make connections between different pieces of information, and draw conclusions based on limited or incomplete data. This type of thinking allows individuals to transcend concrete details and apply general principles to a wide range of contexts.
One significant aspect of abstract reasoning ability is the recognition of analogies. It enables us to see similarities between seemingly unrelated concepts or objects and use that knowledge in problem-solving. For example, being able to understand that a tree is to a forest as a single note is to a melody demonstrates abstract reasoning skills. This type of creative thinking helps us approach challenges from different angles and find innovative solutions.
Moreover, abstract reasoning ability plays a vital role in decision-making processes. It allows individuals to consider multiple factors simultaneously, weighing pros and cons objectively before reaching a conclusion. Abstract thinkers have an advantage when faced with complex situations because they can quickly visualize various possibilities instead of being limited by immediate circumstances. Faced with complex situations because they can quickly visualize various possibilities instead of being limited by immediate circumstances. | 1,957 | 12,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.934334 |
https://cs.stackexchange.com/questions/117534/constructing-preorder-traversal-from-postorder-of-a-binary-search-tree-in-on | 1,713,630,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00243.warc.gz | 166,245,947 | 41,174 | Constructing preorder traversal from postorder of a Binary Search Tree in O(n)
We are given postorder traversal of a Binary "SEARCH" Tree (in an array) and we want to find (print) its preorder traversal.
A very naive solution is to check from end until we find element less than root so we can find left and right subtree and call the function on them. This is $$O(n^2)$$. Another way is to generate the BST from its postorder and then generate the preorder from postorder. it is $$O(n log n)$$ because for generating tree we need $$O(n log n)$$ and for preorder traversal we need $$O(n)$$ and $$O(n + n log n ) = O(n logn)$$.
I think we can do it better in $$O(n)$$ because there is a lot known about the nature of the binary search tree and this extra information must help somehow. but I don't know how to use information we know about BST and implement this problem in $$O(n)$$.
• Can you help me make the connection between title (conversion between traversals) and body (find key given postorder) of your question? Nov 24, 2019 at 12:17
• @greybeard I didn't say anything about key. In the body, I said we want to find where is left and where is right subtree and run the function recursively on them. The other answer is just generating the BST from preorder and then finding postorder traversal. But these ways are not efficient. There must be $O(n)$ solution for this. My question is simply about how to convert postorder traversal to preorder traversal in $O(n)$. (it is on a BST so preorder and postorder are unique) Nov 24, 2019 at 15:15
• I have read find (print) as trying to find as a key the string print Nov 24, 2019 at 15:30
• for [re-]generating [the] tree we need $O(nlogn)$ [time] I'd like to challenge that this bound is tight (and claim $O(n)$). Fast in $\omicron(n)$ memory sounds a challenge. Nov 24, 2019 at 16:34
This can indeed be solved in $$O(n)$$ time. What you mentioned (checking value of root, calling recursively for left and right subtree) almost works, the issue is just that we don't know the size of the right subtree, so we cannot immediately recursively solve the left subtree as we don't know where it ends. But this is not an issue: Just recursively solve the right subtree first, then have that return the subtree's size.
The following C++ program should solve the problem in $$O(n)$$. Please note that while I tested a few inputs but there could be bugs, though the idea behind the algorithm works for sure.
#include <iostream>
#include <vector>
using namespace std;
// Finds preorder traversal of a subtree of a binary search tree given the postorder traversal
// res: Vector to write preorder traversal to
// postorder: Vector containing the postorder traversal
// j: Position in array to write last node in preorder traversal to
// hv: Maximum value of a node in this subtree
// lv: Minimum value of a node in this subtree
// b: Position where this subtree ends in the postorder traversal
// Returns size of the subtree called on
int preorder(int b, int lv, int hv, int j, const vector<int>& postorder, vector<int>& res) {
if (b < 0 || postorder[b] < lv || postorder[b] > hv) return 0;
int rs = preorder(b-1, postorder[b], hv, j, postorder, res); // Size of right subtree
int ls = preorder(b-1-rs, lv, postorder[b], j - rs, postorder, res); // Size of left subtree
res[j - ls - rs] = postorder[b];
return ls + rs + 1;
}
int main() {
int n; // Size of the tree
cin >> n;
vector<int> po(n); // Postorder
for (int i = 0; i < n; ++i) cin >> po[i];
int lv = po[0]; // Minimum value in the tree
int hv = po[0]; // Maximum value in the tree
for (int i = 0; i < n; ++i) {
lv = min(lv, po[i]);
hv = max(hv, po[i]);
}
vector<int> res(n); // Preorder
preorder(n-1, lv, hv, n-1, po, res);
for (auto v : res) cout << v << ' ';
cout << '\n';
}
This is clearly $$O(n)$$, since the function "preorder" gets called exactly once per node, and (outside recursive calls) does $$O(1)$$ work.
• (You don't need to use the actual min and max value.) Nov 24, 2019 at 18:10 | 1,074 | 4,015 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-18 | latest | en | 0.928028 |
https://earthscience.stackexchange.com/questions/23762/is-it-a-coincidence-that-the-circumference-of-the-earth-in-kilometers-is-almost | 1,719,087,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00326.warc.gz | 183,602,968 | 39,723 | Is it a coincidence that the circumference of the Earth in kilometers is almost $2^{12}$?
The circumference of the Earth has been measured to be $$40,075 \,\pu{km}$$, which is only $$21 \,\pu{km}$$ from $$40,096 = 2^{12}$$. For reference, $$40,075 \approx 2^{11.992}$$. This is probably a strange coincidence, but originally a meter was set to be $$1/10,000,000$$-th of the distance from the North Pole to the Equator, so maybe there is a mathematical explanation.
• $2^{12}$ is 4096 not 40,096
– uhoh
Commented Apr 24, 2022 at 23:04
• @uhoh Is it a coincidence that your name perfectly fits to the comment Commented Apr 26, 2022 at 6:23
• So what is that in light years? Still a coincidence? Commented Apr 27, 2022 at 0:44
Well, as others have pointed out we're a little off from $$2^{12}$$ kilometers. However, there is no coincidence that our world's circumference is close to $$40000$$ kilometers. From Wikipedia:
The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole along a great circle, so the Earth's circumference is approximately 40000 km.
• Additional comment on this: another value that is not a coincidence is 21600=360*60 nautical miles, since the nautical mile originally measured a minute = 1/60 degree of latitude (or longitude at equator) Commented Apr 25, 2022 at 18:34
There's something wrong with your math, $$2^{12} \ = \ 4096$$:
$$2^x \ = \ 40\ 000$$
$$\therefore x\log2 \ = \ \log\ 40\ 000$$
$$\therefore x \ = \ \frac{\log\ 40000}{\log\ 2}$$
$$\therefore x \ = \ 15.28771$$
Thus, $$2^{15.28771} \ = \ 40\ 000$$ | 490 | 1,605 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-26 | latest | en | 0.924261 |
http://math.stackexchange.com/questions/tagged/normed-spaces?page=1&sort=unanswered&pagesize=30 | 1,469,393,352,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824146.3/warc/CC-MAIN-20160723071024-00114-ip-10-185-27-174.ec2.internal.warc.gz | 156,660,265 | 27,986 | # Tagged Questions
A vector space $E$, generally over the field $\mathbb R$ or $\mathbb C$ with a map $\lVert \cdot\rVert\colon E\to \mathbb R_+$ satisfying some conditions.
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### Is it possible to characterize completeness of a normed vector space by convergence of Neumann series?
If $X$ is a normed vector space and if for each bounded operator $T \in B(X)$ with $\| T\| < 1$, the operator ${\rm id} - T$ is boundedly invertible, does it follow that $X$ is complete? Context: ...
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### completeness of $l_1 ^\infty$
I'm trying to prove that $l_p ^\infty$ is complete for each $p\geq 1$ but only with the definition of $\varepsilon$-$N$. I know that this have been proved in other posts here but I couldn't find a ...
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### When weakly compactness implies compactness?
Let $A$ be a Banach space. The weak topology on $A$ is a topology which produced by the following family of seminorms: $~~~~~~~~~~~~~~~~~~~~P_f(x)=|f(x)|,\qquad$ where $f\in A^*$ and $A^*$ is dual ...
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### Operator Norm and Submultiplicativity against the Spectral Norm
Consider $\mathcal{A}:\mathbb{R}^{n\times m}\to \mathbb{R}^{p\times q}$ to be a linear operator. I know that by considering the trace norm and using the submultiplicativity of the operator norm we ...
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### Find all $p \ge 1$ for which the Hölder norm $\|\cdot\|_p$ is generated by a scalar product.
Find all $p \ge 1$for which the Hölder norm $$\|x\|_p := \left(\sum^{n}_{i=1} |x_i|^p\right)^{\frac{1}{p}}$$ is generated by a scalar product. We know that norm is generated by a scalar product ...
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### Gelfand width of $l_1$ unit ball in $\infty$ metric
The $l_1$ balls in $l^N_p$ are well studied, but I cannot find anything about estimates of $d^m(B^N_1,l^N_\infty)$ except of $d^m(B^N_1,l^N_\infty)\geq C\min\{1,\frac{\ln(eN/m)}{m}\}$ which is not ...
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### Finding L^1 centers of sets of probability distributions
Let $\mathcal{P}^n = \{ x \in \mathbb{R}^n : x \geq 0, \sum x = 1\}$. Suppose I have $p_1, \ldots, p_m \in \mathcal{P}^n$. I want to find an $L^1$ center for these points. i.e. $q \in \mathcal{P}^n$ ...
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### Do the $p$-norms in $\mathbb R^2$ have a nice geometric intuition behind them?
For $p\geq 1,$ the $p$-norm of a vector $(x,y)\in\Bbb R^2$ is the number $\|(x,y)\|_p=(|x|^p+|y|^p)^{1/p}.$ I learned this definition some time ago, but I never really understood it. Is there a ...
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### linear problem with $\|.\|_\infty$ and $\|.\|_1$ norm constraints
I have a question regarding a straightforward linear algebra problem, yet the solution is (at least for me) not trivial. Assume the sequences $\phi_i$ with coefficients $\phi_i[n]\in\mathbb{R}$, and ...
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### Name for the universal normed space associated to a seminormed space
If $(V,p)$ is a seminormed space, then $(V/N,\overline{p})$ is a normed space, where $N=\{x \in V : p(x)=0\}$ and $\overline{p}(x \bmod N) = p(x)$. My question is as follows: Is there a common name ...
39 views | 972 | 3,045 | {"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} | 2.828125 | 3 | CC-MAIN-2016-30 | latest | en | 0.83837 |
https://radiopaedia.org/articles/federation-dentaire-internationale-fdi-notation?lang=us | 1,556,252,057,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578759182.92/warc/CC-MAIN-20190426033614-20190426055614-00363.warc.gz | 533,863,251 | 19,366 | # Fédération Dentaire Internationale (FDI) notation
Dr Henry Knipe and Dr Hamish Smith et al.
The Fédération Dentaire Internationale or FDI notation system is a commonly used system for the numbering and naming of teeth. The system uses a two number system for the location and naming of each tooth.
#### Permanent teeth
The jaw is divided into four quadrants between the central incisors and the upper and lower dental arches. The first number refers to the quadrant of a tooth:
• right upper quadrant = 1
• left upper quadrant = 2
• left lower quadrant = 3
• right lower quadrant = 4
The second number refers to the individual tooth within a specific quadrant:
• central incisor = 1
• lateral incisor = 2
• canine = 3
• 1st premolar = 4
• 2nd premolar = 5
• 1st molar = 6
• 2nd molar = 7
• 3rd molar = 8
The notation "21" would therefore refer to the permanent right upper central incisor.
#### Deciduous teeth
In deciduous ("baby") teeth the system is similar however in describing the quadrants the numbers 5 to 8 are used in the place of 1 to 4:
• right upper quadrant = 5
• left upper quadrant = 6
• left lower quadrant = 7
• right lower quadrant = 8
Likewise due to the absence of the premolars in deciduous dentition the numbers 1 to 5 are used only:
• central incisor = 1
• lateral incisor = 2
• canine = 3
• 1st molar = 4
• 2nd molar = 5
The notation "82" would therefore be the deciduous right lower lateral incisor.
## Article information
rID: 67083
Synonyms or Alternate Spellings:
• Dental numbering
• Dental naming
• Tooth numbering
• Tooth naming
• FDI numbering
• FDI system | 443 | 1,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.842511 |
http://mathhelpforum.com/calculus/155970-plz-help-volume-homework-due-tomorrow.html | 1,480,890,911,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541426.52/warc/CC-MAIN-20161202170901-00144-ip-10-31-129-80.ec2.internal.warc.gz | 182,338,746 | 10,080 | # Thread: Plz help with volume homework due tomorrow.
1. ## Plz help with volume homework due tomorrow.
(1) The region bounded by the given curves is rotated about y = 9.
x=(y-10)^2 , x=1.
Find the volume V of the resulting solid by any method.
and
(2) The region bounded by the given curves is rotated about the y-axis.
y=-x^2+15x-54, y=0.
Find the volume V of the resulting solid by any method.
2. Originally Posted by chinex
(1) The region bounded by the given curves is rotated about y = 9.
x=(y-10)^2 , x=1.
Find the volume V of the resulting solid by any method.
same as rotating the region bounded by y = (x-1)^2 and y = 1 about the y-axis ... recommend the method of cylindrical shells
and
(2) The region bounded by the given curves is rotated about the y-axis.
y=-x^2+15x-54, y=0.
Find the volume V of the resulting solid by any method.
the zeros of the curve will be the limits of integration ... recommend the method of washers
...
3. Thread closed (see MHF policy on questions that count towards final grade). | 278 | 1,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.915109 |
https://reference.wolframcloud.com/language/tutorial/ConstrainedOptimizationGlobalNumerical.html | 1,708,969,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00209.warc.gz | 483,558,251 | 19,821 | # Numerical Nonlinear Global Optimization
## Introduction
Numerical algorithms for constrained nonlinear optimization can be broadly categorized into gradient-based methods and direct search methods. Gradient-based methods use first derivatives (gradients) or second derivatives (Hessians). Examples are the sequential quadratic programming (SQP) method, the augmented Lagrangian method, and the (nonlinear) interior point method. Direct search methods do not use derivative information. Examples are NelderMead, genetic algorithm and differential evolution, and simulated annealing. Direct search methods tend to converge more slowly, but can be more tolerant to the presence of noise in the function and constraints.
Typically, algorithms only build up a local model of the problems. Furthermore, many such algorithms insist on certain decrease of the objective function, or decrease of a merit function that is a combination of the objective and constraints, to ensure convergence of the iterative process. Such algorithms will, if convergent, only find local optima, and are called local optimization algorithms. In the Wolfram Language local optimization problems can be solved using FindMinimum.
Global optimization algorithms, on the other hand, attempt to find the global optimum, typically by allowing decrease as well as increase of the objective/merit function. Such algorithms are usually computationally more expensive. Global optimization problems can be solved exactly using Minimize or numerically using NMinimize.
This solves a nonlinear programming problem,
using Minimize, which gives an exact solution
This solves the same problem numerically. NMinimize returns a machine-number solution:
FindMinimum numerically finds a local minimum. In this example the local minimum found is also a global minimum:
## The NMinimize Function
NMinimize and NMaximize implement several algorithms for finding constrained global optima. The methods are flexible enough to cope with functions that are not differentiable or continuous and are not easily trapped by local optima.
Finding a global optimum can be arbitrarily difficult, even without constraints, and so the methods used may fail. It may frequently be useful to optimize the function several times with different starting conditions and take the best of the results.
This finds the maximum of :
This finds the minimum of subject to the constraints and :
The constraints to NMinimize and NMaximize may be either a list or a logical combination of equalities, inequalities, and domain specifications. Equalities and inequalities may be nonlinear. Specify a domain for a variable using Element, for example, Element[x,Integers] or xIntegers. Variables must be either integers or real numbers, and will be assumed to be real numbers unless specified otherwise. Constraints are generally enforced by adding penalties when points leave the feasible region.
Constraints can contain logical operators like And, Or, and so on:
Here is restricted to being an integer:
In order for NMinimize to work, it needs a rectangular initial region in which to start. This is similar to giving other numerical methods a starting point or starting points. The initial region is specified by giving each variable a finite upper and lower bound. This is done by including in the constraints, or {x,a,b} in the variables. If both are given, the bounds in the variables are used for the initial region, and the constraints are just used as constraints. If no initial region is specified for a variable x, the default initial region of is used. Different variables can have initial regions defined in different ways.
Here the initial region is taken from the variables. The problem is unconstrained:
Here the initial region is taken from the constraints:
Here the initial region for is taken from the constraints, the initial region for is taken from the variables, and the initial region for is taken to be the default. The problem is unconstrained in and , but not :
The polynomial has global minima at . NelderMead finds one of the minima:
The other minimum can be found by using a different RandomSeed:
NMinimize and NMaximize have several optimization methods available: Automatic, "DifferentialEvolution", "NelderMead", "RandomSearch", and "SimulatedAnnealing". The optimization method is controlled by the Method option, which either takes the method as a string, or takes a list whose first element is the method as a string and whose remaining elements are methodspecific options. All methodspecific option, lefthand sides should also be given as strings.
The following function has a large number of local minima:
Use RandomSearch to find a minimum:
Use RandomSearch with more starting points to find the global minimum:
With the default method, NMinimize picks which method to use based on the type of problem. If the objective function and constraints are linear, LinearOptimization is used. If there are integer variables, or if the head of the objective function is not a numeric function, differential evolution is used. For everything else, it uses NelderMead, but if NelderMead does poorly, it switches to differential evolution.
Because the methods used by NMinimize may not improve every iteration, convergence is only checked after several iterations have occurred.
## Numerical Algorithms for Constrained Global Optimization
The NelderMead method is a direct search method. For a function of variables, the algorithm maintains a set of points forming the vertices of a polytope in -dimensional space. This method is often termed the "simplex" method, which should not be confused with the well-known simplex method for linear programming.
At each iteration, points form a polytope. The points are ordered so that A new point is then generated to replace the worst point
Let be the centroid of the polytope consisting of the best points, . A trial point is generated by reflecting the worst point through the centroid, , where is a parameter.
If the new point is neither a new worst point nor a new best point, , replaces .
If the new point is better than the best point, , the reflection is very successful and can be carried out further to , where is a parameter to expand the polytope. If the expansion is successful, , replaces ; otherwise the expansion failed, and replaces .
If the new point is worse than the second worst point, , the polytope is assumed to be too large and needs to be contracted. A new trial point is defined as
where is a parameter. If , the contraction is successful, and replaces . Otherwise a further contraction is carried out.
The process is assumed to have converged if the difference between the best function values in the new and old polytope, as well as the distance between the new best point and the old best point, are less than the tolerances provided by AccuracyGoal and PrecisionGoal.
Strictly speaking, NelderMead is not a true global optimization algorithm; however, in practice it tends to work reasonably well for problems that do not have many local minima.
option name default value "ContractRatio" 0.5 ratio used for contraction "ExpandRatio" 2.0 ratio used for expansion "InitialPoints" Automatic set of initial points "PenaltyFunction" Automatic function applied to constraints to penalize invalid points "PostProcess" Automatic whether to post-process using local search methods "RandomSeed" 0 starting value for the random number generator "ReflectRatio" 1.0 ratio used for reflection "ShrinkRatio" 0.5 ratio used for shrinking "Tolerance" 0.001 tolerance for accepting constraint violations
Here the function inside the unit disk is minimized using NelderMead:
Here is a function with several local minima that are all different depths:
With the default parameters, NelderMead is too easily trapped in a local minimum:
By using settings that are more aggressive and less likely to make the simplex smaller, the results are better:
### Differential Evolution
Differential evolution is a simple stochastic function minimizer.
The algorithm maintains a population of points, , where typically , with being the number of variables.
During each iteration of the algorithm, a new population of points is generated. The th new point is generated by picking three random points, , , and , from the old population, and forming , where is a real scaling factor. Then a new point is constructed from and by taking the th coordinate from with probability and otherwise taking the coordinate from . If , then replaces in the population. The probability is controlled by the "CrossProbability" option.
The process is assumed to have converged if the difference between the best function values in the new and old populations, as well as the distance between the new best point and the old best point, are less than the tolerances provided by AccuracyGoal and PrecisionGoal.
The differential evolution method is computationally expensive, but is relatively robust and tends to work well for problems that have more local minima.
option name default value "CrossProbability" 0.5 probability that a gene is taken from xi "InitialPoints" Automatic set of initial points "PenaltyFunction" Automatic function applied to constraints to penalize invalid points "PostProcess" Automatic whether to post-process using local search methods "RandomSeed" 0 starting value for the random number generator "ScalingFactor" 0.6 scale applied to the difference vector in creating a mate "SearchPoints" Automatic size of the population used for evolution "Tolerance" 0.001 tolerance for accepting constraint violations
DifferentialEvolution specific options.
Here the function inside the unit disk is minimized using DifferentialEvolution:
The following constrained optimization problem has a global minimum of 7.667180068813135`:
With the default settings for DifferentialEvolution, an unsatisfactory solution results:
By adjusting ScalingFactor, the results are much better. In this case, the increased ScalingFactor gives DifferentialEvolution better mobility with respect to the integer variables:
### Simulated Annealing
Simulated annealing is a simple stochastic function minimizer. It is motivated from the physical process of annealing, where a metal object is heated to a high temperature and allowed to cool slowly. The process allows the atomic structure of the metal to settle to a lower energy state, thus becoming a tougher metal. Using optimization terminology, annealing allows the structure to escape from a local minimum, and to explore and settle on a better, hopefully global, minimum.
At each iteration, a new point, , is generated in the neighborhood of the current point, . The radius of the neighborhood decreases with each iteration. The best point found so far, , is also tracked.
If , replaces and . Otherwise, replaces with a probability . Here is the function defined by BoltzmannExponent, is the current iteration, is the change in the objective function value, and is the value of the objective function from the previous iteration. The default function for is .
Like the RandomSearch method, SimulatedAnnealing uses multiple starting points, and finds an optimum starting from each of them.
The default number of starting points, given by the option SearchPoints, is , where is the number of variables.
For each starting point, this is repeated until the maximum number of iterations is reached, the method converges to a point, or the method stays at the same point consecutively for the number of iterations given by LevelIterations.
option name default value "BoltzmannExponent" Automatic exponent of the probability function "InitialPoints" Automatic set of initial points "LevelIterations" 50 maximum number of iterations to stay at a given point "PenaltyFunction" Automatic function applied to constraints to penalize invalid points "PerturbationScale" 1.0 scale for the random jump "PostProcess" Automatic whether to post-process using local search methods "RandomSeed" 0 starting value for the random number generator "SearchPoints" Automatic number of initial points "Tolerance" 0.001 tolerance for accepting constraint violations
SimulatedAnnealing specific options.
Here a function in two variables is minimized using SimulatedAnnealing:
Here is a function with many local minima:
By default, the step size for SimulatedAnnealing is not large enough to escape from the local minima:
By increasing PerturbationScale, larger step sizes are taken to produce a much better solution:
Here BoltzmannExponent is set to use an exponential cooling function that gives faster convergence. (Note that the modified PerturbationScale is still being used as well.)
### Random Search
The random search algorithm works by generating a population of random starting points and uses a local optimization method from each of the starting points to converge to a local minimum. The best local minimum is chosen to be the solution.
The possible local search methods are Automatic and "InteriorPoint". The default method is Automatic, which uses FindMinimum with unconstrained methods applied to a system with penalty terms added for the constraints. When Method is set to "InteriorPoint", a nonlinear interior-point method is used.
The default number of starting points, given by the option SearchPoints, is , where is the number of variables.
Convergence for RandomSearch is determined by convergence of the local method for each starting point.
RandomSearch is fast, but does not scale very well with the dimension of the search space. It also suffers from many of the same limitations as FindMinimum. It is not well suited for discrete problems and others where derivatives or secants give little useful information about the problem.
option name default value "InitialPoints" Automatic set of initial points "Method" Automatic which method to use for minimization "PenaltyFunction" Automatic function applied to constraints to penalize invalid points "PostProcess" Automatic whether to post-process using local search methods "RandomSeed" 0 starting value for the random number generator "SearchPoints" Automatic number of points to use for starting local searches "Tolerance" 0.001 tolerance for accepting constraint violations
RandomSearch specific options.
Here the function inside the unit disk is minimized using RandomSearch:
Here is a function with several local minima that are all different depths and are generally difficult to optimize:
With the default number of SearchPoints, sometimes the minimum is not found:
Using many more SearchPoints produces better answers:
Here points are generated on a grid for use as initial points:
This uses nonlinear interior point methods to find the minimum of a sum of squares:
For some classes of problems, limiting the number of SearchPoints can be much faster without affecting the quality of the solution: | 2,910 | 14,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-10 | latest | en | 0.860011 |
https://www.ablebits.com/office-addins-blog/day-of-week-formula-excel/ | 1,726,171,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00200.warc.gz | 586,303,944 | 32,247 | # Excel formula to get day of week from date
In this article, we'll explore different methods to extract days of the week from calendar date in Excel such as formulas and custom number formats.
When you're using Excel to manage project deadlines, schedule appointments, or analyze data trends, it's often handy to know the day names for specific dates. Excel has a variety of tools to help you figure this out. In this article, we'll show you some straightforward ways to get the day of the week from a date, whether you want it as word or number.
## Get day of week from date using TEXT formula
One of the simplest ways to find the day names from a date in Excel is by using the TEXT function. Here’s how:
1. Let's assume that the original date is in cell A3.
2. In another cell (let's say B3), enter one of these formulas:
To return the full name of the day such as "Monday" or "Tuesday", the formula is:
`=TEXT(A3, "dddd")`
To get a shorter version of the day names like "Mon" or "Tue", use this day of week code:
`=TEXT(A3, "ddd")`
3. Press Enter, and cell B3 will display the day name corresponding to the date in cell A3.
It's important to note that the TEXT function always gives you the result as text, regardless of the original cell's format. If you plan to use this output for further calculations, consider using a custom format instead of a formula.
## Find day names with WEEKDAY formula
Another simple formula for days of the week in Excel is WEEKDAY.
By default, the WEEKDAY function returns a number between 1 and 7. In this system, Sunday is considered as 1, Monday as 2, and so on, but you can change this by specifying a different return_type argument.
### Convert day of week to number
If your goal is to get the day of the week number, then the WEEKDAY function in its basic form is all you need:
`=WEEKDAY(A1)`
If cell A1 contains a date like "18-Sep-2023" (Monday), the formula will return 2; or 3 if A1 contains "19-Sep-2023" (Tuesday).
### Convert calendar date to day of week
If you prefer having the day names, not numbers, you can combine the WEEKDAY function with the TEXT function.
For example, to convert a calendar date in cell A1 to the day of week, use one of these formulas:
`=TEXT(WEEKDAY(A1), "dddd")`
Or
`=TEXT(WEEKDAY(A1), "ddd")`
The difference is that the first formula returns the full name of the day while the second formula displays a shorter version of the name.
In this formula, the WEEKDAY function should be used with just one required argument. The optional return_type argument should not be specified, even if your week starts on a day other than Sunday.
Note. This formula is included merely for demonstration purposes. A TEXT formula without WEEKDAY described above is simpler and returns exactly the same results.
## Return day name in any format using CHOOSE formula
If you want to create your custom labels or notations for the day names, you can do this by using WEEKDAY in combination with the CHOOSE function. Here's how it works:
• CHOOSE returns a value from a predefined list based on an index number.
• WEEKDAY calculates the index number corresponding to the day of the week for a given date.
In other words, this formula maps the numeric output of the WEEKDAY function to the corresponding name that you've hardcoded into the CHOOSE function.
Let's assume that your date is in cell A3, and you want to show the day names as 2-letter abbreviations such as "Su" for Sunday, "Mo" for Monday, etc. The following formula works a treat:
`=CHOOSE(WEEKDAY(A3), "Su","Mo","Tu","We","Th","Fr","Sa")`
## Display day of week from date using custom format
Another way to show the day names in Excel is by applying a custom number format. This approach allows you to keep the original date value in the cell while visually displaying the day of the week. Here's how to do it:
1. Select the cell(s) containing the date you want to format.
2. Right-click the selected cells to bring up a context menu, then choose Format Cells.
3. In the Format Cells dialog box, navigate to the Number tab and select Custom in the Category list.
4. In the Type field, enter the desired code for the day of week:
• dddd – the code for the full day names (e.g. Sunday).
• ddd – the code for abbreviated day names (e.g. Sun).
5. Click OK to apply the custom format.
Unlike Excel formulas, this method does not change the underlying values of the cells, but only changes how they appear in cells.
In the screenshot below, you can see three columns with the same dates. Column A displays the dates in the default date format, column B shows full day names (“dddd” format), and column C shows abbreviated names (“ddd” format). If you check the formula bar, you'll notice that all three columns retain the original date values. This means you can use these dates in any calculations as fully functional Excel dates while providing a desired visual representation of the day of the week.
These are some of the ways to get the day of week from date in Excel. You can choose any method that suits your needs and preferences. Thanks for reading, and we'll be back with more helpful tips soon!
Excel formula for day of week - examples (.xlsx file)
## You may also be interested in
1. Hi, I'm trying to identify routes for a specific postcode. but the problem is the route for each postcode differ on daily basis. Any help on which formula would work?
• Hi! Your question is not entirely clear to me. To understand what you want to do, give an example of the source data and the expected result.
2. Hi, i was wondering if it was possible of finding an exact date in the futur based on a day of the month. Let's say i want to find every last sunday of april in the futur. How does one go about coding that in excel if it's even possible at all
• Hi! Set the date April 30 using the DATE function. Use the SEQUENCE function to create a sequence of years in this date. Subtract the weekday number from this date if it is less than 7 to obtain the last Sunday of the month. Define the day of the week with the help of the WEEKDAY function.
=DATE(SEQUENCE(10,1,2024,1),4,30) - MOD(WEEKDAY(DATE(SEQUENCE(10,1,2024,1),4,30),2),7)
3. I have a column that is formatted as ddd for days of the week. I want to highlight that row and conditionally format all Tuesday, Thursday, Saturday, and Sunday to be purple and all Monday, wednesday, and Fridays to be Blue. How do I do that? Conditional formatting is not recognizing it as a text?
• Hello! You have changed the format of the cell, but the cell still contains a date in it. Therefore, for conditional formatting, you can use the number of the day of the week. This number can be defined using the WEEKDAY function.
For example:
=OR(WEEKDAY(A1,2)=2,WEEKDAY(A1,2)=4,WEEKDAY(A1,2)=6,WEEKDAY(A1,2)=7)
4. Hi,
I am looking at stock earnings dates and I want to return the first workday date of the previous week.
Example : 25/04/2023 I would like to automatically return the first working day of the previous week 17/04/2023 i.e. Monday 17th April 2023.
In some cases the first Monday of the previous week may be a public holiday so I would then like the first working day to be returned.
I have tried to utilise the resources you list but can't quite figure out how to do it.
Any help appreciated. Thanks
• Thats great thanks, now I need to figure out how to interpret the logic :-) It does work though
I use a spreadsheet to record all my trades, and it's helpful to make a note of the day number for me.
But also, I just wanted the trading days of the week where Mon =1 - Fri =5.
So I used: =WEEKDAY((B10)-1) where B10 is the cell with the actual date in it.
Then I realised that any date that wasn't filled in yet, returned an error #NUM!
So I added an IF function to get around that: =IF(B10>0.5,WEEKDAY((B10)-1),"")
BUT I what I wanted to say, was that I wouldn't have done of any of this, were it not for your straightforward, no nonsense article.
So thanks very much,
John over in Blighty
6. Hi,
if I have a formula in column C, I need to change that formula according to the data (name) I enter in column A, or to put a specific formula in column C depending on the name I will enter in column A.
7. Hi,
if I have a formula in column C, I need to change that formula according to the data (name) I enter in column A, or to put a specific formula in column C deadening on the name I will enter in column A.
8. But how do I filter from these results?
Can't seem to be able to do that after getting the day of the week.
9. Hi,
if I have a date in column A, I need to display in column B the nearest Tuesday or Thursday before that day.
Example:
A B
3/06/24 3/05/24
3/8,9,10&11/24 3/07/24
3/05/24 3/05/24
3/07/24 3/07/24
Thank you,
Tamer Salib
10. How would I go about having the date at the start of every week pop up for a specific day?
For example, I start work every Monday and when I start a new excel file, I have to tediously input the date in multiple locations. How about instead, if I just enter a code telling excel "Input the month/day/year for the current monday."
What would that code look like?
11. i want to change a date to week forexample 05.03.2024 to wk 10/2
12. So i'm trying to create a function that when a Day (one or two digit text) is input It will output the next calendar date. Example: Today is 1/8/24 and the input is 15 it will Produce 1/15/24 for the output. Or If the input is 2 it will output 2/2/24.
13. Appreciate the guide. I used the TEXT formula and it worked but weekday column shows Saturday when Date column is empty. Is there a way to fix this/make it empty? Many thanks.
• Thanks!
• So I tried but I can't figure it out. can you help? I have tried a few things. I have this which doesn't work.
=TEXT(A8, "dddd")+IF(ISBLANK(A8), "", "if not blank")
• Got it!!
=IF(ISBLANK(A8), "", TEXT(A8, "dddd")) | 2,444 | 9,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.887302 |
https://www.tutoreye.com/prove-that-m-is-the-supremum-of-s-by-arguing-m-satisfies-the-definition-of-supremum-qa | 1,685,489,236,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00672.warc.gz | 1,167,189,183 | 37,566 | Search prove-that-m-is-the-supremum-of-s-by-arguing-m-satisfies-the-definition-of-supremum
# Prove that m is the supremum of s by arguing m satisfies the definition of supremum
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