url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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https://www.oreilly.com/library/view/introduction-to-digital/9780470900550/chap3-sec004.html | 1,611,240,925,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524858.74/warc/CC-MAIN-20210121132407-20210121162407-00492.warc.gz | 914,905,922 | 9,924 | ## 3.4 BOOLEAN AXIOMS AND THEOREMS
The basic logic operations include logic sum, logic product, and logic complement. If a logic variable is true, its logic complement is false. The following set of logic expressions illustrates the axioms of Boolean algebra:
• 0 * 0 = 0
• 0 + 0 = 0
• 1 * 1 = 1
• 1 + 1 = 1
• 0 * 1 = 1 * 0 = 0
• 0 + 1 = 1 + 0 = 1
• if x = 0, then = 1
• if x = 1; then = 0
The character (*) represents the AND logic product, and the character (+) stands for the OR logic sum. A bar over a character represents the NOT logic. From these logic axioms, basic Boolean identities were formulated. The following expressions illustrate these identities.
• x + 0 = x
• x * 1 = x
• x + 1 = 1
• x * 0 = 0
• x + x = x
• x * x = x
• x + = 1
• x * = 0
### Commutative Property
• x + y = y + x
• x * y = y * x
### Associative Property
• x + (y + z) = (x + y) + z
• x * (y * z) = (x * y) * z
### Distributive Property
• x * (y + z) = (x * y) + (x * z)
• x + (y * ...
Get Introduction to Digital Systems: Modeling, Synthesis, and Simulation Using VHDL now with O’Reilly online learning.
O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers. | 410 | 1,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.80361 |
https://www.flashcardmachine.com/chapter-6and8forcesmotionandworkmachines.html | 1,685,243,457,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00780.warc.gz | 857,253,374 | 6,409 | # Shared Flashcard Set
## Details
Unit 8-Forces/Motion and Work/Machines
vocab cards
14
Science
6th Grade
02/26/2013
## Cards Return to Set Details
Term
1. air resistance
Definition
A force that opposes the motion of objects that are falling through the air. (air friction)
Term
vacuum
Definition
A space where there is no matter. Outer space is an example of this.
Term
inertia
Definition
the tendency of an object to stay at rest or in motion (objects just want to keep on doing whatever it is doing)
Term
Newton's 1st Law of Motion
Definition
an object at rest will remain at rest, an object in motion will remain in motion UNLESS acted on by an UNBALANCED force.
Term
Newton's 2nd Law of Motion
Definition
The acceleration of an object depends on the mass of the object and the amount of force applied. F = m x a
Term
formula for force
Definition
F= ma (Force= mass x acceleration)
Term
Newton's 3rd Law of Motion
Definition
For every action there is an equal and opposite reaction. Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. Forces always work in pairs. Example-a rocket shipping blasting off. The gases push down and the rocket pushes up.
Term
force pairs
Definition
All forces act in pairs. When a force is exerted, there is always a reaction force. Example-Hitting a baseball bat, the bat exerts a force on the ball and sends the ball flying.
Term
momentum
Definition
A property of all moving objects. Momentum is determined by two factors: mass and velocity Momentum = mass x velocity
Term
Law of Conservation of Momentum
Definition
Momentum can be transferred but cannot be lost within a system. When one moving object collides with another moving object, the motion of both objects changes.
Term
work
Definition
The transfer of energy to an object using a force that causes the object to move in the direction of the force.
Term
Formula for Work
Definition
W= F x d
Term
joule (J)
Definition
The unit/label for work
Term
machine
Definition
A device that helps do work by either overcoming a force or changing the direction of the applied force.
Supporting users have an ad free experience! | 517 | 2,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.870376 |
https://www.futurestarr.com/blog/mathematics/a-35-percent-of-1000 | 1,660,295,118,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571597.73/warc/CC-MAIN-20220812075544-20220812105544-00398.warc.gz | 661,547,019 | 47,222 | FutureStarr
A 35 Percent of 1000
35 Percent of 1000
via GIPHY
Our first stop for the day was another bird sanctuary at a hill-top nearby. And the hill! It sloped so steeply up and away from the water that for a moment the entry was obscured by a dense tangle of trees and vines. It was like the forest had gathered on top of the cliffs and tried to climb up to the sky.
Value
In calculating 35% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,35% off, 35% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.1000 dollar to pound = 660 pound
Here is a calculator to solve percentage calculations such as 35 is what percent of 1000. You can solve this type of calculation with your values by entering them into the calculator's fields, and click 'Calculate' to get the result and explanation. (Source: percentage-calculator.net)
Percentage
Percentage calculator tool can be used by first entering the fractional value you want to calculate. For example 5% of 20, which is the same thing as fraction x/100 * 20=5%. To find the value of x enter 5 in the first box and 20 in the second box and the answer 1 will be shown in the result box.
We think that illustrating multiple ways of calculating 35 percent of 1000 will give you a comprehensive understanding of what 35% of 1000 means, and provide you with percent knowledge that you can use to calculate any percentage in the future. (Source: percent.info)
Problem
50 percent of 500000 equals 250000. To get this answer, multiply 0.50 by 500000. You may need to know this answer when solving a math problem that multiplies both 50% and 500000. Perhaps a product worth 500000 dollars, euros, or pounds is advertised as 50% off. Knowing the exact amount discounted from the original price
You may need to know this answer when solving a math problem that multiplies both 35% and 1000. Perhaps a product worth 1000 dollars, euros, or pounds is advertised as 35% off. Knowing the exact amount discounted from the original price of 1000 can help you make a more informed decision on whether or not it is a good deal. (Source: thenextgenbusiness.com)
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August 12, 2022 | Bushra Tufail | 905 | 3,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-33 | latest | en | 0.933864 |
https://brainmass.com/math/graphs-and-functions/graph-proofs-digraphs-225073 | 1,638,838,560,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00552.warc.gz | 217,894,828 | 75,412 | Explore BrainMass
# Graph proofs
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
I attached few description pages of Chapter 1.
Could you please send solved solution from attached questions.
Exercise Problem See Page 11.
1.2 & 1.3
Len n be a given positive integer, and let r and s be nonnegative integers....
https://brainmass.com/math/graphs-and-functions/graph-proofs-digraphs-225073
#### Solution Preview
Hello
Please find the solution in the attached file.
1.3 In this problem given that
r + s = n
where r and s are non negative integer with s is even.
Since s is even ...
#### Solution Summary
This provides examples of working with proofs regarding graphs and nonisomorphic graphs.
\$2.49 | 195 | 857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-49 | latest | en | 0.890791 |
http://www.cram.com/flashcards/high-yield-biostatistics-445411 | 1,527,343,411,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00233.warc.gz | 355,192,791 | 21,544 | • Shuffle
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### 35 Cards in this Set
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What is a Case-control study? compares a group w/ a disease to a group w/o to yield an odds ratio. What is a co-hort study? compares a group w/ a risk factor to a group w/o to yield a relative risk. What is a cross-sectional study? data taken from a group of people to assess the frequency of a disease/related risk factors at a given time. measures prevalence. What is a meta-analysis? pools data from a lot of studies; limitations include quality of studies studied & publication bias. What is prevalence? total cases in pop at a given time/ toltal pop at risk What is incidence? new cases in pop in given time period/ total pop at risk at that time What is sensitivity? percent of people w/ a disease that test +; high value SNOUT= SeNsitivity Rules OuT. What is specificity? percent of people w/o a disease that test -; high value SPIN= Specificity rules IN. What is a positive predictive value? probability that the person testing + actually has the disease; PPV= a/a+b What is a negative predictive value? probability that the person testing - truely doesn't have the disease. NPV= d/d+c What is an odds ratio? calculated in case control studies; odds of having a disease in the exposed group vs the unexposed group; a/b/c/d What is relative risk? calculated for cohort studies; probability of getting a disease in the exposed group vs the unexposed group; a/a+b/c/c+d What is attributable risk? difference in risk b/w exposed & unexposed groups; a/a+b - c/c+d What is precision vs accuracy? precision is consistency & reproducibility ( reliability) of a test while accuracy is the trueness of the test measurments (validity. What is meant by a positively skewed curve? asymmetry w/tail on the right; mean> median> mode. What is meant by a negatively skewed curve? asymmetry w/tail on the left; mode> median> mean. What is a type I error? saying that there is a relationship when there's not; false rejection of Null hypothesis. alpha level < .05 for most studies must not exceed 5%. What is type II error? saying that there is not a relationship when there is one; false rejection of alternative hypothesis (H1). beta is the probability of this error occuring in a study. What is power? power is the probability that your hypothesis (Null or H1) is true. power= 1-beta (type II error) What is a t-test? difference in means of 2 groups. What is ANOVA? difference in means of 3 or more groups. What is a chi-square test? compares proportions (not means) of 2 or more groups. What is an APGAR score? score is indicative of level of "liveliness". 0-10 score based on appearance, pulse, grimace, activity, & respiration; taken at 1 min & 5 min. What is a normal BMI? 18.5- 24.9kg How often does REM sleep occur? q90min. What is positive reinforcement? when a desired reward increases beh; operant conditioning. What is negative reinforcement? when an undesired stimulus is removed in order to increase a beh; operant conditioning. When does Babinski reflex disappear? 12-14m. When do kids start walking? 15m. When do kids tiolet train? 30-36m. What are beta waves? eyes open, awake waves; highest freq/ lowest amplitude. alpha waves are awakes w/eyes shut. What are theta waves? light sleep; approx 5% of sleep time. Where is most time spent during sleep? 45% of sleep is in stage 2 sleep w/ spindles/ K complexes. Where does the deepest non-REM sleep occur? stage 3-4 (approx 25% of sleep time); delta waves; night terrors, bed-wetting, sleep walking. this is slow wave sleep (as opposed to wakefulness). How much sleep time is REM? approx 25%. | 1,010 | 4,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-22 | latest | en | 0.903863 |
https://pastecode.io/s/vzot6wgd | 1,721,611,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00890.warc.gz | 382,949,439 | 59,539 | Untitled
unknown
python
7 months ago
1.1 kB
5
Indexable
Never
```class Solution:
def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
n = len(gas)
if len(gas) == 1:
if gas[0] >= cost[0]:
return 0
else:
return -1
for i in range(n):
curr = i # index to start
tank = gas[curr] # current tank availability
transition = n # how many stations visited
while transition > 1:
# determine the next index
if curr == n - 1:
next = 0
else:
next = curr + 1
# check whether the cost is untenable
if tank < cost[curr]:
break
# update the tank availability
tank = tank + gas[next] - cost[curr]
transition -= 1
# update the current index
curr = next
if transition == 1 and tank >= cost[curr]:
return i
return -1``` | 216 | 722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.563121 |
https://iphone.apkpure.com/mathboard/com.palasoftware.mathboard | 1,569,314,431,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572896.15/warc/CC-MAIN-20190924083200-20190924105200-00332.warc.gz | 513,463,402 | 12,055 | MathBoard
# MathBoard
## PalaSoftware Inc.
\$4.99
0
0 Ratings
Release Date
2018-11-07
Size
24.6 MB
### Screenshots for iPhone
MathBoard Description
MathBoard now supports iCloud Syncing, along with a host of other newer technologies, like Text to Speech as well as Speech Input, the Apple Pencil, Slide Over and Split Views and more.
MathBoard® is a highly configurable math app appropriate for all school aged children. Beginning in kindergarten, with simple addition and subtraction problems, through elementary school where learning multiplication and division can be a real challenge. MathBoard will allow you to configure the app to best match the abilities of your individual child/student.
More than just standard drills, MathBoard encourages students to actually solve problems, and not just guess at answers. This is done by providing multiple answer styles, as well as a scratchboard area where problems can be worked out by hand. Students can also turn to MathBoard's Problem Solver for further help. This powerful teaching feature walks students through the steps required to solve addition, subtraction, multiplication, and division equations. Additionally, the included quick reference tables serve as a valuable learning tool.
Math Activities. These activities include Find the Sign, Equality/Inequality, and Match Math (a memory game). Each activity is based upon the current settings level, so the difficulty will vary based on the student’s knowledge.
WHAT THE EXPERTS SAY:
“This is the best integer practice app or program I have found in 34 years of teaching math.”- Tim Seiber, Math Teacher
“MathBoard has charm without losing track of its fundamental purpose (Arithmetic drills).” - MacWorld (4 out of 5 mice)
“MathBoard is a stunning app for teaching addition, subtraction, multiplication and division problems. The number of customization options available makes this a brilliant app for making math fun for elementary school children.” - androapple
FEATURES:
- Random equation generation for Addition, Subtraction, Multiplication, Division, Squares, Cubes, and Square Root problems.
- Number ranges are configurable, including the ability to require certain numbers to be in each problem and the ability to omit negative answers.
- Number and order of displayed digits can be limited, allowing for equations to conform to certain learning levels (e.g. 2 digit numbers over 1 digit numbers).
- Generates simple equations, as well as single step algebraic equations. (e.g. 6+x=12; x-8=2; 5x=25).
- Intelligent problem and "wrong answer" generation makes guessing more difficult.
- Includes both multiple choice, as well as, fill in the blank style questions.
- Activities and quizzes can be timed, either as a countdown timer or elapsed time.
- Equation configuration settings can be saved for future use as well as shared with others.
- Multiple student profiles are supported. Includes the ability to save, review, and share the results for activities and quizzes with others.
- Problem Solver will outline the steps needed to solve addition, subtraction, multiplication, and division problems.
- In addition to standard drills, activities are included (Find the Sign, Equality/Inequality and Match Math).
- Quick reference math tables for counting, addition, subtraction, and multiplication.
- iCloud (syncing) support.
- Support for Slide Over and Split View.
- Support for the Apple Pencil.
- Keyboard commands.
Price:
\$4.99
Version:
3.4.2
Size:
24.6 MB
Genre:
Education
Release Date:
2018-11-07
Developer:
PalaSoftware Inc.
Language:
Chinese Dutch English French German Italian Japanese Korean Portuguese Spanish Thai
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Developer Apps | 770 | 3,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-39 | latest | en | 0.9276 |
https://fr.slideserve.com/robinsont/portions-what-s-fair-powerpoint-ppt-presentation | 1,653,363,722,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00701.warc.gz | 318,897,532 | 20,592 | Portions: What’s Fair?
# Portions: What’s Fair?
Télécharger la présentation
## Portions: What’s Fair?
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Portions: What’s Fair? MOM! He got more than me!! http://www.foodnetwork.com/healthy-eating/10-ways-to-measure-perfect-portion-sizes/pictures/index.html
2. What do we already know about dimensions, portions and ratios? Dimension - A measurement of length in one direction. Examples: width, depth and height are dimensions. Area – the space inside a shape. Portion/Yield – number of servings per dish
3. What affects pizza portion size? • Shape of slice • Size of pan/pizza • Number of slices per pizza • Number of people • Available ingredients • Crust thickness
4. Comparing Pan Dimensions Circular Square/Rectangular
5. Here’s the deal: • We are making pizza tomorrow. • How will you make sure that your group will have the same portion size as the rest of the groups, when you have different pan sizes, i.e. different dimensions of pans?
6. How will we use the dimensions to compare the size of the pizzas, i.e. how will we know who got more (or less, or the same)? • We need to use some math…sigh. • What math concepts/ideas should we use? • Area of circles and rectangles… • Circle formula Rectangle formula • A = A =
7. Circular Pizza Find the area: Area = π •r2 Example: Find the area of a 10” pizza. A = 3.14 x (5 x 5) A = 78.5 sq. inches Square/Rectangular Pizza Find the area: A = length x width A = l x w Example: Find the area of a 10” x 15” pan. A = 10 x 15 A = 150 sq. inches
8. Portion/Yield Portion/yield is: Area divided by number of slices, s. Area ÷ s = Portion size
9. Now complete the table on your worksheet to figure out how many slices are needed to get approximately 20 sq. inches per serving. Circular Area = 78.5 sq. inches A ÷ s = Portion/yield 78.5 ÷ 4 = 19.6 sq. in. Portion size by slice ≈ 20 sq. in. Square/Rectangular Area = 150 sq. in. A ÷ s = Portion/yield 150 ÷ __s_ = 150 ÷ __s_ = 150 ÷ __s_ = Portion size by slice ≈ 20 sq. in.
10. What if it wasn’t a pizza? • Examples: • Cutting fabric into multiple patterns/shapes, how many shapes can you fit in limited yardage? • Making a cake with a cake pan of different dimensions than are called for in the recipe. • Complete the second part of the worksheet, using area and “portion sizes” and conversion factors.
11. So, how do you respond to your sibling or friend who says, “You got a bigger piece of pizza than I did!”? You have to know the dimensions and determine the area, then divide by the number of slices in order to figure out if they are right or not. | 750 | 2,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-21 | latest | en | 0.824026 |
http://www.java2s.com/Code/VBA-Excel-Access-Word/Excel/R1C1Style.htm | 1,723,459,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00092.warc.gz | 43,649,176 | 2,934 | # R1C1 Style : Cell Reference « Excel « VBA / Excel / Access / Word
R1C1 Style
```
Sub R1C1Style()
FinalRow = cells(Rows.count, 2).End(xlUp).row
range("D4:D" & FinalRow).FormulaR1C1 = "=RC[-1]*RC[-2]"
range("F4:F" & FinalRow).FormulaR1C1 = "=IF(RC[-1],ROUND(RC[-2]*R1C2,2),0)"
range("G4:G" & FinalRow).FormulaR1C1 = "=+RC[-1]+RC[-3]"
cells(FinalRow + 1, 1).value = "Total"
cells(FinalRow + 1, 6).Formula = "=SUM(G4:G" & FinalRow & ")"
End Sub
```
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1 A1 Style References 2 Write a formula in the first 10 columns of row 21 in a worksheet that calculates the standard deviation of the values in rows 2 through 20 3 The Cells property is best suited for use with loops because it accepts numerical parameters representing the row and column indices in a worksheet 4 Entering A1 Versus R1C1 in VBA 5 R1C1 Style References 6 Use loop to select cells | 297 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.526603 |
https://www.esaral.com/q/which-of-the-following-statements-are-true-and-which-are-false-52598 | 1,721,266,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00272.warc.gz | 673,495,839 | 11,584 | # Which of the following statements are true and which are false?
Question:
Which of the following statements are true and which are false?
(i) $\frac{-3}{5}$ lies to the left of 0 on the number line.
(ii) $\frac{-12}{7}$ lies to the right of 0 on the number line.
(iii) The rational numbers $\frac{1}{3}$ and $\frac{-5}{2}$ are on opposite sides of 0 on the number line.
(iv) The rational number $\frac{-18}{-13}$ lies to the left of 0 on the number line.
Solution:
(i) True
A negative number always lies to the left of 0 on the number line.
(ii) False
A negative number always lies to the left of 0 on the number line.
(iii) True
Negative and positive numbers always lie on the opposite sides of 0 on the number line.
(iv) False
The negative sign cancels off and the number becomes $\frac{18}{13}$; it lies to the right of 0 on the number line. | 228 | 857 | {"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} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.856908 |
https://www.spartanpoker.com/spin-wheel | 1,643,269,071,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00367.warc.gz | 1,002,889,909 | 234,339 | # Spin Wheel
## What is Spin Wheel Game?
To play the spin the wheel game, the dealer asks the players to choose a symbol to bet on. Once the players have decided on the symbol and placed their bet, the dealer spins the wheel. The players expect the wheel to stop on the symbol on which the bet was placed.
• If the money wheel stops on the symbol on which the player placed their bet, then the player would win the money.
• If the money wheel does not stop on the symbol on which the player placed their bet, no one would win the spin wheel game.
Here are some of the most often used terms when playing the spin the wheel game.
• The Wheel — Without the wheel, you can't play the spin the wheel game. It's the center of the wheel game, and it's how you'll figure out your payouts, wins, and chances.
• Wedges - These are the sections of the wheel that you will see. The amount and total multiplier for your wager are displayed on the wedges.
• The wheel's action is referred to as "spin." One round is represented by each spin. You place your wager before the wheel spins and then wait for the wheel to stop spinning to see how much money you've won or lost.
• Table: This is where you'll place your wagers. The available guesses are displayed on the table for you to choose from. You can't take your bet off the table after the spin wheel has started spinning.
• Bet: This is the amount you wager on a guess. Depending on your luck, this will determine whether you win or lose.
• Play High or Play Low: The amount you are willing to pay each round is determined by this option. If you're feeling confident, you can stake the maximum amount on each round.
• Clear Bet: If you want to clear the table of your prior bet, this is an option. You can then re-set your bets before spinning the wheel.
These are just a few of the terms you'll come across while playing the spin the wheel game
### Spin Wheel Online
Spin Wheel is a simple game to learn, and most variations are very similar. The spin wheel online game is divided into several pieces, each of which is separated by spokes or pins. There are several possible winning amounts, which may be in Rupees but vary by online casino and locale, or may include unique symbols, such as a joker, that payout even better odds.
• Place a wager on which section the spin game online wheel pointer will land in.
• The wheel is spun by the dealer in some online casinos, bets are allowed even while the spin the wheel game online wheel is spinning.
• The wheel will come to a halt, and the pointer will indicate the winning segment.
• You win if you bet on that part!
• The spin the wheel game online is quite simple, which is why players adore it!
### Money Wheel Game
You'll never grow tired of the gameplay since it combines amazing odd and probability principles with a colorful design and intriguing and interesting music. Most versions are determined by the number of sections on the wheel. However, some online variations differ slightly from the typical money wheel:
1. The Big Six
Although Big Six appears to be a traditional money wheel, it is distinguished by its sections being divided into six payout levels. They may differ from one online casino to the next, but they are typical as follows: even, 2-1, 5-1, 10-1, 20-1, and 40-1. There's a lot of money to be won in Big Six!
2. Dice Wheel
Dice Wheel is a money wheel that incorporates features of Sic Bo. The portions of the wheel depict numerous three-dice combinations. You make your wagers according to the numbers that appear on the dice, that is, 1-6.
You will be paid 1-1 if the wheel pointer falls on a portion with a die showing the number you bet on. Let's say you bet on 3. If your number was shown more than once in the section, your reward increases to 2-1 or even 3-1!
### Options for Betting
Now that you've mastered the fundamentals, it's time to decide how much and how often you'd like to win. Let's have a look at the various betting possibilities accessible when playing the money wheel game.
Each spin or round of the money wheel allows you to place different wagers. For each round, you can put a minimum or maximum amount on all of them. In other words, if you place a wager on each choice on the table, you can increase your chances of covering the correct number or symbol on every spin. You can practically see every spin or round if you have total coverage. You can, on the other hand, concentrate on a single number and wager on it every round. The quantity of victories and payouts is entirely dependent on how much you wager.
### What is The Best Way to Play Money Wheel?
Money Wheel game is a simple game to play. It is centered on your betting and playing experience. All you have to do now is play the game and concentrate on winning large on each spin. The spinning wheel is at the center of everything.
As previously stated, this is a simple game. All you have to do now is put your wager. To begin, identify the symbols that appear on the wheel. Feel what you believe will happen and place your wager. It's all about predicting where the arrow will land. You win if the arrow lands on the color and symbol that you bet on; otherwise, you lose. The wheel will then reset for another round or spin, and you can place another bet. That's all there is to it.
You can choose from seven distinct bets, each with its own set of colors, letters, and numbers. On each round or spin, you can bet on all of them or any of them. You can also use a different color chip for each of the options on the table. You have six chip selections to pick from, ranging in size from 0.5 to 100. It is all up to you. You can only bet a certain amount on each of them, and there is a limit to how much you may bet on each of them.
After you've put your wager, you'll be able to begin spinning the wheel. If you are dissatisfied with the betting table, you can reset it and clear off the bets you have put. You have plenty of time to figure out how much you want to bet and how much you want to win on the spin-to-win wheel.
## Money Wheel Strategy
1. Money Wheel Is a Chance Game
Money Wheel or spin wheel game online is a game of chance, and you have no control over the outcome. Keep that in mind and focus on the thrill of the ride.
2. Organize Your Funds
When playing Money Wheel, don't go overboard with your spending. We recommend not betting more than 5% of your total bankroll on a single bet. You may play for longer and hopefully win more if you manage your cash.
3. Seek Out the Best Odds.
You can't change the spin wheel, but you can choose a wheel with higher odds. Do your homework before playing to identify Money Wheel games with a lower house edge, which provide players a better chance of winning.
• The money wheel is a basic game that takes no expertise.
• Simply place your wager before the spinning of the big six money wheel.
• After all, bets have been put, the dealer grabs one side of the wheel and spins it down with a hard pull and push.
• The wheel then performs a few more spins before grinding to a halt.
• When the wheel stops, the marked number the indicator points to is the winning number and reward amount.
• You can place a wager on one or more numbers at once on the spin to win wheel.
### Play Poker Online and Win Real Money
Online poker is a game of skills. You can play online poker games with your friends and family and win real money. The various types of poker games you can play are Texas Hold’em, Omaha Poker, Omaha Hi-Lo, Chinese Poker, and others. Register on the Spartan Poker platform to play online poker games and win up to ? 30,000 daily. So, what are you waiting for? Go ahead and play poker game of your choice now!
Download the Spartan Poker app on your mobile phone and play real money games now! You can also play poker on your desktop or use the flash version to play these cash games. Make sure you have fun when you play poker games!
### Spin Wheel Game FAQs
Q.1) How do you play the Spin the wheel game?
The players choose an option on the spin wheel to place a bet. Once the bet is placed, the dealer spins the win. If the win stops on the option on which the bet is placed, then the player wins the spin the wheel game.
Q.2) How does the money wheel work?
The money wheel is divided into pieces, each of which has its symbol. On the wheel and the table, the odds for each sector are indicated, ranging from even money to 47 to 1. The goal is to bet on the symbol you believe will be lucky on the next spin.
Q.3) How do you play the Money Wheel game?
The dealer asks the players to choose a symbol to bet. Once the players have decided on the symbol and placed their bet, the dealer spins the wheel. The players expect the wheel to stop on the symbol on which the bet was placed. If the money wheel stops on the symbol on which the player placed their bet, then the player wins the money. | 2,004 | 8,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.963506 |
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The tag has no usage guidance.
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https://www.tutorhunt.com/questions/view/345/ | 1,591,506,618,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00273.warc.gz | 912,945,407 | 13,879 | # Tutor Hunt Questions
The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =t2+ 100t, for 0 ≤ t ≤ 20.
When does the rocket reach its maximum height above the ground? What is its maximum height?
2 years ago
Maths Question asked by Mrs
The rocket keeps going up for 20 secs so put t = 20
01/03/2018 06:48:04 | comment by Martin
- 1
I am not sure if you have written the equation correctly.
07/03/2018 13:54:37 | comment by Yusuf
- 1
Differentiate the equation and put it equal to zero to find the time t for the maximum height. To find the maximum height, put that value of t into the original equation.
22/03/2018 21:35:41 | comment by Binny
+ 6
Hello, I can answer this question but would need to clarify something. There appears to be a formatting problem in the question. There is a symbol before the "t2" which appears an an empty square in my browser. Is it supposed to read "h(t) = -t^2 + 100t"? (t^2 means t squared - the ^ is used to indicate a superscri pt in typed mathematics).
27/03/2018 12:00:35 | comment by Chris
- 2
the symbol before the `t` appears as a box, what is it meant to be showing? and does `t2` mean t squared? I think the formatting on my Chrome is incorrect
14/10/2018 19:15:38 | comment by Peter
- 2
Please enter your response to the question below. The student will get a notification as soon your response has been approved by our moderation team.
Cannot answer as this is what appears on my screen:
“The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =;;t2+ 100t, for 0 ≤;; t ≤;; 20. When does the rocket reach its maximum height above the ground? What is its maximum height?
Can you please make sure I can see the whole equation without any unreadable symbol? I suppose it is t squared, please confirm too.
Many thanks
Answered by Romain | 2 years ago
(continued) So, we start by finding the value of t at the turning point. To do this, we take the derivative of h with respect to t i.e.
t = -t^2 + 100t
dh/dt= -2t + 100
By setting the derivative to 0, we find the value of t at the turning point. So, if
dh/dt = 0 = -2t + 100, then 2t = 100, so t = 50.
We note that 50 > 20 and thus out of our range. We also note that because of our deduction earlier as our extreme value of t=20 is closest to the value of t at the turning point, this must be where our maximum occurs.
Substitute t=20 into our original equation for h:
h= -20^2 + 100*20 = 1600
As a final check, use the other extreme value in our range (t=0) to see if we made a mistake.
h=0 when t=0.
Thus our rocket reaches its max. height when t=20 and h=1600.
Suppose instead that h = -5^2 + 100t:
-similar graph
-dh/dt= -10t + 100 implies t=10 which is in range of values
-sub. t=10 into h to find our max. height.
If h(t) = 5t^2 - 100t why would the max value of h not occur at t=10?
Answered by Jack | 2 years ago
20 sec later.
2400ft.
Answered by Foday | 20 months ago
This equation is not clearly written out; it looks like it is trying to be a simple differential equation, however h(t) doesn’t really make sense. Are we multiplying the height by the time? In which case the equation will need to be rearranged to get height in terms of t. Moreover is t being squared? Again the question isn’t clear. And what is the coefficient in front of t? So at present question is impossible to answer
Answered by Caroline | 2 years ago
maximum height will occur when t=20, therefore plug in 20 instead of t to achieve the max height, answer should be 2400 feet
Answered by Akbar | 2 years ago
you want to find the maximum height, that is when h(t) is the biggest value. the value of h increases as t increases. as t can only be a maximum of 20 seconds, to find the max height sub t=20 into the equation to get the value for h.
Answered by Nikhil | 20 months ago
Q: h(t) = 5t^2 + 100t, 0 <= t <= 20,
All parts involving t are positive. The range of t is in the positive realm i.e. from 0 to 20. Therefore the rocket reaches maximum height when t is largest i.e. t=20,
Answer done in stages of working, substituting t for 20:
h(t=20) = 5(20)^2 + 100(20),
h(t=20) = 5(400) + 100(20),
h(t=20) = 2000 + 2000,
h(t=20) = 4000.
Answered by [Deleted Member]
Simply differentiate the equation and let it equal to 0.
From here you can find the time at max height (when the gradient of the curve = 0).
Input this identified t into the original equation to find the max height.
I can`t answer the equation given to me as I believe the coefficient before the t^2 is missing
Answered by David | 2 years ago
I think this questions seems hard because of the whole rocket thing and t instead of x. Let`s rephrase it:
a) Solve the equation x^2 + 100x = 0 to find where the graph of y = x^2 + 100x crosses the x axis
b) Sketch the graph of y= x^2 + 100x using the fact that a quadratic is U shaped and goes through the numbers you found in part (a) (ie, 0 and -100)
c) on your sketch, get rid of (scribble out) the bits where x is negative or bigger than 20. We only want to look at the graph from x=0 to x=20
d) look at the sketch - can you see that it is highest when x = 20? Find out how high by subbing x=20 into y = x^2 + 100x to get the y coordinate
They made this look a lot harder by using t instead of x and also talking about a rocket instead of saying whats the highest point on the graph
Answered by Ronnie | 2 years ago
The number in front of the t^2 is missing, therefore I am unable to give you a numerical answer. However, the method that you should use is to devide the range of t by 2 and it will give you 10 since (20-0)/2=10. Now you just need to insert 10 instead of t in your equation which will give you the maximum height in feet (if the question is given in feet). I hope that answered your question.
Answered by Ahmed | 2 years ago
Summary, when finding the MAX/MIN of the dependent variable (h) in a QUADRATIC EQUATION within a RANGE on the independent variable (t), we know it occurs at either an extreme value of the independent variable (t) or at a turning point, which we find by taking the derivative and setting this equal to 0. This is possible to do without sketching a graph. However, it is a good habit to get into and will be invaluable when confronted with cubic equations.
Answered by Jack | 2 years ago
Ignoring wind resistance and assuming the rocket returns to the ground at t=20 it will follow the path of a parabola which is symmetrical. The maximum height will therefore be when t=10. Substitute t=10 into the function to get the height. However, the function in the question cannot be correct as the number before the t^2 must be negative.
Answered by Ian | 20 months ago
You probably mean h(t) = -5(t^2) + 100t
To find the maxima (or minima) you need o find where the gradient of h(t) is zero. To do this, you need to differentiate h(t), see below:
dh / dt = -10t + 100 = 100 - 10t = 10(10 - t)
There is a turning point (a maxima or minima) when the differential is zero (i.e. when the gradient is zero). i.e. when dh/dt = 0
t = 10 seconds
I`m going to assume this is a maxima (if we needed to prove it was a maxima rather than a minima, we`d need to differentiate again, to see if it gave us a positive or negative result, but that is not part of the question)
Answered by Mike | 2 years ago
The maximum height is reached when the speed of the rocket is 0.
so differentiate h(t) and set this equal to 0. Then solve for t, and finally sub it back into
h(t) =t^(2)+100t for the maximum height.
I can show you the answer if you tell me what h(t) is properly.
Answered by Santhosh | 2 years ago
its a parabola opening downwards.
The maximum point we need to find is the vertex,where x coordinate represents the time it reaches the maximum height and y coordinate represents the the maximum height it reaches.
x = -b/2a (axis symmetry formula)
x = -100/-2
= 50 seconds
y = -(50)^2 + 100(50)
= -2500 + 5000
= 2500 feet
Answered by Syed Saad | 2 years ago
I cannot see all of the equation, so, for now, I will assume that the equation given is:
h(t)= -t^2 + 100t, 0 \$leq\$ t \$leq\$ 20
If the coefficients are different the principles are still the same, so don`t sweat.
A good place to start when presented with a quadratic equation is to sketch its corresponding graph. To sketch the graph:
-solve the equation for h=0 to see where the graph crosses the x-axis
-set t=0 to see where the graph crosses the y-axis
-check: is the coefficient of the t^2 term positive or negative?
-if positive our graph shape is a smiley face
-if negative - sad face (which is what we are working with here)
We observe our graph. We notice that the highest point of the graph is at the turning point and that the further we move from this point along the x-axis the smaller the value our graph has against the y-axis. Hence, either the value of t we are looking for is at the turning point or it is the closest value within our range to the turning point. (You may wish to re-read the last two sentences and look at your graph to convince yourself that this is true). (continued)
Answered by Jack | 2 years ago
Differentiate and equate to zero, rearrange for t to find t which gives max height. Plug same value of t back into original equation to find greatest height.
Answered by Jaafar | 20 months ago
(Sorry for lack of formatting there appears to be no way to enter a return character, put a return at each "CR")CRCR
The question hasn`t shown all characters so I guess the full equation was;CRCR
h(t) = -0.5at^2+ 100t, for 0 < t < 20. CRCR
That is a standard equation from physics. a is acceleration from gravity. The minus sign recognises that the initial velocity is upwards and gravity pulls it down.CRCR
Early in the flight (low values of t) 100t dominates and the rocket ascends. Late in the flight (high values of t) -0.5at^2 dominates and the rocket descends.CRCR
The "correct" solution is to differentiate height to get velocity;CRCR
V(t) = d h(t)/dt = -at + 100CRCR
At the highest point the rocket changes direction so v(t)=0. Assuming that a=10 m/s^2 (gravity on Earth);CRCR
0 = 100 - at CR
t = 100 / 10 = 10 s CRCR
maximum height is; CRCR
h(t) = 100t - 0.5at^2 CR
= 1000 - 0.5 x 10 x 100 CR
= 500m CRCR
The "cheat" way to do it is to notice that the equation is specified for the time 0 < t < 20. If the rocket flight is symmetrical (true for unpowered flight in a vacuum) it will ascend for the same time that it descends and the maximum height is at the mid point t=10s. The latter is not the textbook solution but allows a quick check on your answer.
Answered by Jarvis | 2 years ago
Think about what the gradient of the curve will be at the maximum height. Use differentiation and solve for t. This will give you the time when it happens. Use this to find the height.
Answered by Miguel | 19 months ago
The maximum height of the rocket will occur when it`s vertical velocity component is zero (at the apex of it`s flight path).
Find the rate of change of height h`(t) by differentiation:
h`(t) = -2t + 100 ( the formatting is not clear but I assume the first term if h(t) is -t^2 )
Find the time, t, when h`(t) = 0
-2t + 100 = 0
t = 50 seconds.
Now use the h(t) formula to find the maximum height:
h(50) = -(50)^2 + 100(50)
= 2500 m
Answered by Thomas | 2 years ago
Hi, I can`t see what character you`ve put before the t2 (t-squared). Let`s call it k.
Now, h = kt^2 + 100t => dh/dt = 2kt + 100, and when the rocket reaches its maximum height, the rate of change of height with time is zero.
Therefore the relevant time t = -50/k
And now use this value of t in the original equation to work out that its height will then be
k (-50/k)^2 + 100 (-50/k) = 2500/k^2 - 5000/k
Answered by Dan | 2 years ago
The rocket reaches its maximum height when it`s at the top of its curve, when it is momentarily still before it comes back down. when it is momentarily still, the gradient of its curve is 0, i.e. when d(h)/d(t)= 0
d(h)/d(t) = -10t + 100 = 0
10t = 100
t= 10 so it reaches its maximum height at 10 seconds after lift-off.
Put this back into the original eq to get the height:
h= -5(10)^2 + 100(10) = -500 + 1000 = 500m.
Answered by Madeleine | 2 years ago
The formatting of the equation in you question appears corrupted, but is an equation of motion of the form d = u.t + 1/2.a.t^2 (where d is the distance, and in this case height, u is the initial speed, t is the time elapsed, and a is the acceleration which in this case is due to gravity, i.e. 9.8m/s/s). I rather think the corrected form would be d = 100 x t + 1/2 x 9.8 x t^2.
Although the equation comes from physics, the real point of the question is about turning points for a curve, which is mathematics.
The simplest, but more time consuming solution, is to tabulate values and draw a graph. However, if you have done some calculus you should be able to differentiate the equation to determine the gradient of the curve at any time t. As the maximum height will be where the differential of height with respect to time is zero you can use this to solve for when this occurs. (Note: You will need to be careful of the sign convention you use, i.e. consider which direct the distance and acceleration increases in a apply a sign accordingly.)
Answered by David | 20 months ago
This is a poorly framed question. It gives no information about what happens after 20 seconds. It is curious that the unit of distance in the question is feet: in UK Maths, for many decades, distance has been measured in metres.
The narrow answer is: within the 20-second window, the rocket reaches maximum height when t = 20 and the height at that time is 2400 feet (20^2 + 100 x 20).
However, a broader answer must consider what happens beyond the 20-second window. At t = 20, the rocket will be rising at 140 feet per second. Assuming that there is no upward acceleration after that time, and assuming deceleration due to gravity of 32 feet per second per second (and assuming that air resistance is negligible), the velocity will become zero at t = 24.375; at that time, the maximum height of the rocket`s trajectory will be reached, which is 3626.25 feet.
Answered by Mark | 2 years ago
h(t) = t2 + 100t
so dh/dt= 2t + 100
h is max when dh/dt = 0 that is, 2t+100= 0
so, t=50 but 0 ≤;; t ≤;; 20
hence t=20s when h is max
so max h = 20^2 + 100x20= 400 + 2000 = 2400ft
Answered by Suman | 2 years ago
If you consider the graph of the function which defines the height of the rocket, it had roots at both -100 and 0. We can however eliminate any aspects of the graph which are outside the range of x is less than or equal to 0 and x is more than or equal to 20. this means that the the maximum point on the graph (of which represents the maximum height of the rocket) can be clearly seen on the graph at t = 20. the maximum height of this point is h(20) = 2400.
Answered by [Deleted Member]
- 1
Considering the parabolic motion of the rocket, with the maximum height being half of the maximum time, 20 seconds, we can conclude that the rocket must reach its maximum height at 10 seconds.
Using the given function, h(10)=(10)^2+100(10), we can calculate the maximum height:
It gives us 100ft+1000ft, which equals 1,100ft.
Hence, Max height = 1,100ft at time 10 seconds.
Answered by Philip | 2 years ago
- 1
if (h*t)=((t*2)+(100*t))
t max=20 seconds
(20*2) + (100*20)=(h*t)
((20*2)+(100*20))/20=h
h=102 feet, at either 20 seconds, if this is the point at which the rocket reaches its highest point. If however, 20 seconds is the time at which the rocket has returned back to the ground, we can presume that the rocket reaches its peak at 20/2=10 seconds.
Answered by Claudia | 20 months ago
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# following gives you table of number of days it takes to
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following gives you table of number of days it takes to [#permalink] 12 May 2008, 06:01
1
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BOOKMARKED
following gives you table of number of days it takes to deliver the mail.
Days - number of mails
1 - 10
2 - 15
3 - 20
4 - x
5 - 30
whats the value of x so that average number of days it takes to deliver the mails is 3.
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Re: weighted average question [#permalink] 12 May 2008, 07:24
shmistry wrote:
following gives you table of number of days it takes to deliver the mail.
Days - number of mails
1 - 10
2 - 15
3 - 20
4 - x
5 - 30
whats the value of x so that average number of days it takes to deliver the mails is 3.
Wow, who can understand and solve this?
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Re: weighted average question [#permalink] 12 May 2008, 09:20
something wrong with the question?
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Re: weighted average question [#permalink] 12 May 2008, 09:31
I tried to look at it a couple different ways but the either way the missing number would have to be negative for the average to be 3. So I'm not sure what it's asking.
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Re: weighted average question [#permalink] 12 May 2008, 09:40
I am not sure about the numbers here. You can use bigger numbers and tell me the approach.
Thanks
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Re: weighted average question [#permalink] 12 May 2008, 16:17
shmistry wrote:
I am not sure about the numbers here. You can use bigger numbers and tell me the approach.
Thanks
Are you missing some probability weighting?
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Re: weighted average question [#permalink] 12 May 2008, 18:49
no..it wasnt percentage.. i am 99% sure this is what i saw .
Last edited by shmistry on 12 May 2008, 19:38, edited 1 time in total.
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Re: weighted average question [#permalink] 12 May 2008, 19:03
Re: weighted average question [#permalink] 12 May 2008, 19:03
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https://www.daniweb.com/programming/software-development/threads/423657/class-polygon-c | 1,656,859,512,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00530.warc.gz | 776,771,434 | 16,477 | Hi again and thanks in advance ! Last question !
I have to create a class called polygon (not normal) that has in private the number of sides and the length of each side and in public
that has 2 constructors (the one has as variables the number and each length (how is that possible), getlength (perimeter) , setpolygon , getedge (of a k side) .
Also then i have to make a subclass called rectangle that has constructor with length,height
getlength again ,getarea ...
What does rectangle inherits from polygon ?
Here's the code of the polygon class
but i have no idea what i'm supposed to create in that way dynamically
here's the code !
``````#ifndef POLYGON_H
#define POLYGON_H
class Polygon
{
private :
int size;
int *len;
public:
Polygon();
Polygon(int k,int *l);
~Polygon();
int getLength();
void setPolygon(int k,int *l);
int getEdge(int k);
};
#endif
``````
polygon.cpp
``````#include "polygon.h"
Polygon::Polygon()
{
size=0;
}
Polygon::~Polygon()
{
delete [] len;
}
Polygon::Polygon(int k,int *l)
{
size=k;
len=new int[size];
for(int i=0;i<size;i++)len[i]=l[i];
}
int Polygon::getLength()
{
int sum=0;
for(int i=0;i<size;i++) sum+=len[i];
}
void Polygon::setPolygon(int k,int *l)
{
size=k;
len=new int[size];
for(int i=0;i<size;i++) len[i]=l[i];
}
int Polygon::getEdge(int k)
{
return len[k];
}
``````
main.cpp
``````#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
system("PAUSE");
return EXIT_SUCCESS;
}
``````
## All 2 Replies
A rectangle is a polygon with four sides where the opposite sides are equal. I imagine it would inherit everything that polygon has, plus add a constructor taking a length and a width and some getters and setters. Who wrote this code?
but i have no idea what i'm supposed to create in that way dynamically
I have no idea what you mean by this.
The task you have seems quite unusual. with regards to
2 constructors (the one has as variables the number and each length (how is that possible),
There a few options for this as the constructor will need a different number of arguments for each polygon type.
One option is to take from printf or iostream and use either `va_arg`s or `istream` as your argument. this allows you the felxibility to have any length of arguments. or you can use an object like `vector<T>` or a special object with a similar feature set say something like
`````` typedef SideType_t float;
typedef std::vector<SideType_t> SideData_t;
class ClPolygonSideDescriptor
{
public:
size_t GetNumSides( void );
SideType_t GetSide( size_t _side );
private:
SideData_t m_sideData;
}
``````
I think any of these approaches would be ok. The method keeps the interface class generic and expandable (im assuming thats what polygon will be )
Maybe its a push to far but i dont like seeing variable arguments passed with `int size, int* data` in C++ as its not really OOP design and its not exactly the safest code and makes it easy to make mistakes. using something like STL if it is available would be better in my opinion but thats just my take on it.
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sevekpetrol.com | 1,656,836,832,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00646.warc.gz | 553,283,272 | 13,441 | # Why are gasoline and oil boiling point measurements so unreliable?
Gasoline and other petroleum products can burn hotter than normal, but the pressure on the gasoline, oil, and water will always be too high for that to occur.
This is called a boiling point.
Gasoline and water at different boiling points will react in different ways.
For example, at a gas station, it’s the boiling point that makes the difference between being able to buy a gallon of gasoline and being able only to buy 1 gallon of water.
At home, it will depend on how much water you have and how often you use the pump.
The boiling point of a gas or oil is determined by a chemical called its free energy.
As you can see, the free energy of a compound changes as the temperature rises.
So, if you have a chemical compound that has a boiling temperature of 200 degrees Fahrenheit, the molecule is stable in its stable form at 200 degrees F. However, if the compound is in a stable form, it won’t react with its free energies to create energy.
The free energy in a molecule is usually expressed as a number.
It’s called a chemical formula.
You can think of this as the formula for the chemical formula of water, which has a specific boiling point at 212 degrees F: 212 – 3.18 x 10-6 = 7.6.
When you have these numbers in mind, you can calculate the boiling points of many products: oil, gas, and gasoline.
And, for some products, it is also possible to calculate the free energies of these products.
In other words, you’re not just trying to get the boiling temperature for one particular product, you are trying to find the free electrons of a molecule that have been converted into free energy for a specific product.
Here’s a picture of a chemical reaction: The diagram shows how these free electrons are converted into energy for the product in question.
On the left is a molecule of petroleum, and on the right is a molecular structure of an enzyme.
(click to enlarge) For this example, we’ll assume you know how to calculate free energies.
But there are many other ways to find out the boiling temperatures of various products.
For example: How can you tell which fuel is being used?
Some chemicals are unstable at high temperatures, and the boiling level will rise with the temperature of the fuel.
Similarly, the boiling rate of some solvents is dependent on the concentration of the solvent.
An example of this is in gasoline, where the temperature changes in the mixture of fuel and water that you burn.
How long will it take to get to the boiling boil?
It depends on how hot the fuel is and how many solvants are present.
That’s because there are two different types of solvators: water and oil.
One type is called the water solvator, and it has a surface area of one atom and a specific energy of 0.
There are two types of oil solvulators: gasoline and diesel.
Both types of the same type of solver have different surface areas and different specific energies.
This is because gasoline can be heated at high heat, and diesel at low heat.
What is the boiling range for gasoline?
To find out how hot a gasoline will be when it’s being heated, you take the specific energy at which the product is in the boiling mode.
From that energy, you get the temperature.
To calculate this, you divide the specific temperature by the temperature in degrees Fahrenheit.
If you’re using a thermometer, divide by 60.
Finally, you add the specific heat. | 732 | 3,453 | {"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} | 3.828125 | 4 | CC-MAIN-2022-27 | longest | en | 0.951972 |
https://physics.stackexchange.com/questions/339737/is-general-relativity-only-applicable-where-the-curvature-is-small | 1,618,781,517,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038860318.63/warc/CC-MAIN-20210418194009-20210418224009-00276.warc.gz | 554,280,444 | 43,572 | # Is general relativity only applicable where the curvature is small?
In Eddington Finkelstein Coordinates: Lecture By Emil Akhmedov at 7:10 (transcript below the video) the instructor mentions in passing that on general grounds, Einstein's theory of general relativity can only be expected to apply where the curvature is small, but he doesn't go into further detail.
Why is that? Is there a physical reason why we should expect that?
EDIT What he says is this (slightly adapted from the transcript):
... this tensor measures the strength of tidal forces, and they become enormous as $r$ goes to zero. That's the reason that this metric is applicable only beyond this point. Moreover, one can say that Einstein's general theory of relativity is not applicable as $r$ goes to zero anymore, because higher terms and powers of curvature are becoming relevant. So the curvature becomes strong. And Einstein's theory on general grounds can be expected to be applicable only if curvature is small, sufficiently small.
He seems to be saying that the Schwarzschild geometry ceases to be applicable for $r\to 0$ not only because $R_{\mu\nu\alpha\beta}R^{\mu\nu\alpha\beta} \to \infty$, but also because more generally the theory is only applicable when the curvature is small. The fact that he speaks about higher powers of the curvature seems to indicate that general relativity is just a first order approximation of a more general theory.
EDIT 2
I asked on the course forum, where I only got two views but one of them was the lecturer's. Essentially what he was alluding to was what @gj255 already said in the comments:
It is an assumption, if you think about it more carefully. We have assumed the simplest invariant - $R$ - as the Lagrangian density, but why not $R^2$ e.g.? If you look at this, you will realize that $R^2$, or even higher powers of Ricci scalar or tensor or Riemann tensor, become more relevant, as $r\to 0$, because then curvature tends to infinity...
To give some context: in the derivation of the Einstein equations in this course, an invariant Lagrangian density depending on the metric was constructed. The simplest possible term, $\sqrt{|g|}$ cannot account for any dynamics. Adding the term $\sqrt{|g|}R$ we get dynamical equations of motion and the relative coefficient between these terms is the cosmological constant. This is where he stopped, but apparently there is no a reason why we should, and apparently we should expect non-zero higher order contributions.
• In the context given, the comment does not make too much sense. GR is only an approximation once we include quantum effects, but he doesn't seem to be talking about that. Once we include "quantum effects" (whatever that means), then we have $$\text{large curvature}\leftrightarrow\text{large energies}\leftrightarrow\text{quantum effects}.$$ GR cannot describe the singularity of the black hole because that's not even a part of the spacetime manifold. (There are ways of including it, but strictly speaking that's not GR.) – Ryan Unger Jun 16 '17 at 20:04
• Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Also Emil deserves the credit. – Qmechanic Jun 16 '17 at 20:20
• I don't agree with the comment in the video that G.R. is only applicable when the curvature is small. In very rough terms, if "curvature" is small, I..e, negligible, then, you don't have G.R. at all, your spacetime is Minkowski, and you are in the realm of special relativity. What I think he meant, and it is sort of clear, is that he is computing the Kretschmann scalar, $K = R^{abcd}R_{abcd}$, which for Schwarzschild is $~1/r^6$, so, G.R. breaks down when $r \to 0$. This was my reading of it. – Dr. Ikjyot Singh Kohli Jun 16 '17 at 20:42
• @IkjyotSinghKohli I don't think he means to say "negligible", but maybe that its square is negligible. I edited to include his text, in which he seems to say that there is a reason in addition to the Kretschmann scalar diverging. – doetoe Jun 16 '17 at 21:35
• I think the point the lecturer was trying to get across is that the Einstein-Hilbert action, $\mathcal{L} = R$, is in some sense the lowest-order gravitational action you can write down. Just as Hooke's Law, $F = k x$, is only the first term in a Taylor expansion of the force-extension relation of an elastic medium, and breaks down for large forces, so we might expect that the Einstein-Hilbert action is corrected by other terms, such as $R_{\mu \nu \rho \sigma}R^{\mu \nu \rho \sigma}$, at large curvatures. This has nothing to do with quantum mechanics. – gj255 Jun 16 '17 at 23:52
Maybe you should for a moment forget the narrative of the lecture and consider the following:
Einstein constructed relativity as a more or less "unique" theory of gravity. However, this theory has been robustly proven to exhibit singularities such as the one you see at $r=0$ in the Schwarzschild black hole. Many people don't take singularities seriously enough; when a singularity occurs, the theory has no prediction whatsoever for the behaviour at that point. In other words, the "one and only" Einstein's theory of gravity tells you that it cannot, even in principle, apply in the curvature singularity in the center of the black hole and is thus necessarily an incomplete theory.
This is an extremely clear statement on which the vast majority of theorists agree. What they don't agree on, though, is any kind of resolution or even the bare framing in which an alternate theory should be developed.
Let me sketch what kind of problems we have even with framing how Einstein's gravity should be changed. The curvature tensor $R^{\mu}_{\;\nu\kappa\lambda}$ can be understood as a $4 \times 4\times 4\times 4$ table of numbers (of dimension length$^{-2}$) and some of them blow up near the singularity in a black hole. In principle you could try to define a transition to some modification of relativity when any of the components of $R^{\mu}_{\;\nu\kappa\lambda}$ grow beyond a certain number, lets say one over Planck length squared $1/l_p^2$.
But this is not a coordinate-covariant statement. You could make a funny coordinate transformation even in a weak gravitational field and the components of $R^{\mu}_{\;\nu\kappa\lambda}$ may look large. Or you can make coordinate transform near the black hole singularity and make $R^{\mu}_{\;\nu\kappa\lambda}$ components arbitrarily small when arbitrarily close to the singularity.
What relativists usually do is that they show that coordinate invariants such as the Kretschmann scalar $K=R^{\mu\nu\kappa \lambda} R_{\mu \nu \kappa \lambda}$ blow up. And that truly is a valid, coordinate-independent criterion. But when you go into more complicated cases such as the spinning black hole, you will find e.g. the Kretschmann invariant to be zero arbitrarily close to the central singularity (see e.g. Cherubini et al. 2003). This is because $R^{\mu\nu\kappa\lambda}R_{\mu\nu\kappa\lambda}$ has the character of $F^{\mu\nu}F_{\mu\nu} = E^2 - B^2$ in electromagnetism and becomes zero at points where the gravitomagnetic effects balance the gravitoelectric ones. When you dig around some more you find out that there is really no true "scalar strength" of Einstein's gravity, very much in analogy to the fact that there is no "scalar strength" of the electromagnetic field.
So, there is no easy way to classify what exactly do we expect from the corrections to Einstein's gravity, and this is also the part of the conundrum of contemporary theoretical physics. In other words, the phrase about "small curvature" or "corrections at higher curvature" will be mentioned in many relativity lectures but it will never be given any concrete development because it does not have one, at least not one supported by a wide consensus.
• Good point, smallness only is a meaningful concept when we are dealing with coordinate independent quantities, like $R$ or $K$. – doetoe Jun 17 '17 at 10:15
The quotation is misleading, because "small" is meaningless unless you specify "small relative to what." He really means that general relativity can only be a good approximate description when the curvature is finite, i.e. bounded above. The curvature becomes arbitrarily large near a singularity, so we expect any quantum effects and/or higher-derivative corrections to the Einstein-Hilbert action to kick in near there, regardless of how small the length threshold is where they become significant.
• Strictly speaking you are right, but as I am sure you are aware, this is a very common manner of speech, even in mathematics. Usually it means either "sufficiently small", or "equal to first order". In neither case there is any absolute meaning attached to it. It could be that indeed he does mean that it only works when the curvature is finite, but from the way he says it seems that he says that there is an additional reason. Maybe the second part of your answer, that there must exist a quantum theory of gravity of which general relativity can only be a good approximation when R is small. – doetoe Jun 16 '17 at 21:59
• @doetoe Yes, I suspect he's thinking of quantum corrections that presumably kick in when the curvature scale approaches the Planck scale. – tparker Jun 16 '17 at 22:08
• @doetoe I'm sure he doesn't just mean "accurate to first order" though, because that corresponds to linearized GR, and we have good (though circumstantial) observational evidence that GR holds well beyond the linearized regime - for example, we have very strong observation evidence for the existence of black holes. – tparker Jun 16 '17 at 22:10
• Even if we didn't know about quantum mechanics, we would know that GR had problems at singularities. The problems become especially acute at a naked singularity, because you can't predict what will come out of it. – user4552 Jun 16 '17 at 22:11
But, I still don't understand what "small" curvature means. For example, take the FLRW metric for any $k = -1,0,1$, the Ricci scalar is:
$R = -6\left[\frac{\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 + \frac{k}{a^2}\right]$
The curvature is $< \infty$ as long as practically $a(t)$ does not go to zero, but this is a cosmological singularity anyways, where quantum gravity is expected to become relevant (unless you like Conformal Cyclic Cosmology a la Penrose, but that is a different story :) ), so $R < \infty$ in this case for all practical purposes, even if it is large, which means the scale factor can be small. G.R. is certainly valid in the region where $R < \infty$, i.e., as long as $a(t)$ is bounded. So, I don't know what exactly the lecturer is referring to.
• Yes, I guess you are right that general relativity itself doesn't break down at large R, only that for reasons beyond GR it is assumed/known that GR doesn't correctly describe nature anymore. – doetoe Jun 16 '17 at 22:03
• Hi. could you please explain this comment a bit further? – Dr. Ikjyot Singh Kohli Jun 17 '17 at 17:20
• I just meant to say that it is not general relativity as devised by Einstein that breaks down for large $R$, but that it may be that it ceases to provide a correct description of nature at large $R$. – doetoe Jun 18 '17 at 10:21 | 2,739 | 11,250 | {"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} | 2.671875 | 3 | CC-MAIN-2021-17 | latest | en | 0.959397 |
https://www.enotes.com/homework-help/how-prove-that-sin-2xcos-3x-sin-2x-sin-4-cosx-431738 | 1,485,043,036,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00441-ip-10-171-10-70.ec2.internal.warc.gz | 895,479,052 | 12,621 | # How to prove that: `sin^2xcos^3x=(sin^2x-sin^4x)cosx`
Asked on by farajallah
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The identity `sin^2xcos^3x=(sin^2x-sin^4x)cosx` has to be proved. | 87 | 222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-04 | longest | en | 0.856012 |
https://apboardsolutions.guru/ap-ssc-10th-class-maths-solutions-chapter-3-ex-3-3/ | 1,723,743,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00527.warc.gz | 79,614,802 | 11,626 | AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.3 Textbook Questions and Answers.
## AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.3
### 10th Class Maths 3rd Lesson Polynomials Ex 3.3 Textbook Questions and Answers
Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) x2 – 2x – 8
ii) 4s2 – 4s + 1
iii) 6x2 – 3 – 7x
iv) 4u2 + 8u
v) t2 – 15
vi) 3x2 – x – 4
i) Given polynomial is x2 – 2x – 8
We have x2 – 2x – 8 = x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
So, the value of x2 – 2x – 8 is zero
when x – 4 = 0 or x + 2 = 0 i.e.,
when x = 4 or x = -2
So, the zeroes of x2 – 2x – 8 are 4 and -2.
Sum of the zeroes = 4 – 2 = 2 Coefficient of ,x -(-2)
= – $$\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$$ = $$\frac{-(-2)}{1}$$ = 2
And product of the zeroes = 4 × (-2) = -8
= $$\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$$ = $$\frac{-8}{1}$$ = -8
ii) Given polynomial is 4s2 – 4s + 1
We have, 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
= 2s (2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
= (2s – 1)2
So, the value of 4s2 – 4s + 1 is zero
when 2s-1 = 0 or s = $$\frac{1}{2}$$
∴ Zeroes of the polynomial are $$\frac{1}{2}$$ and $$\frac{1}{2}$$
∴ Sum of the zeroes = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 1.
= – $$\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}$$ = –$$\frac{-4}{4}$$ = 1
And product of the zeroes = $$\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)$$ = $$\frac{1}{4}$$
= $$\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$$ = $$\frac{1}{4}$$
iii) Given polynomial is 6x2 – 3 – 7x
We have, 6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
The value of 6x2 – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0
i.e., when 3x + 1 = 0 and 2x – 3 = 0
3x = -1 and 2x = 3
x = $$\frac{-1}{3}$$ and x = $$\frac{3}{2}$$
∴ The zeroes of 6x2 – 3 – 7x = $$\frac{-1}{3}$$ and $$\frac{3}{2}$$
∴ Sum of the zeroes = $$\frac{1}{3}$$ + $$\frac{3}{2}$$ = $$\frac{7}{6}$$.
= – $$\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$$ = $$\frac{-(-7)}{6}$$ = $$\frac{7}{6}$$
And product of the zeroes = $$\left(\frac{-1}{3}\right) \times\left(\frac{3}{2}\right)$$ = $$\frac{-1}{2}$$
= $$\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$$ = $$\frac{-3}{6}$$ = $$\frac{-1}{2}$$
iv) Given polynomial is 4u2 + 8u
We have, 4u2 + 8u = 4u (u + 2)
The value of 4u2 + 8u is 0,
when the value of 4u(u + 2) = 0, i.e.,
when u = 0 or u + 2 = 0, i.e.,
when u = 0 (or) u = – 2
∴ The zeroes of 4u2 + 8u are 0 and – 2.
Therefore, sum of the zeroes = 0 + (-2) = -2
= – $$\frac{\text { Coefficient of } u}{\text { Coefficient of } u^{2}}$$ = $$\frac{-8}{4}$$ = -2
And product of the zeroes 0 . (-2) = 0
= $$\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}$$ = $$\frac{0}{4}$$ = 0
v) Given polynomial is t2 – 15.
We have, t2 – 15 = (t – √15 ) (t + √l5)
The value of t2 – 15 is 0,
when the value of (t – √15 ) (t + √l5) = 0, i.e.,
when t – √15 = 0 or t + √15 = 0, i.e.,
when t = √15 (or) t = -√15
∴ The zeroes of t2 – 15 are √15 and -√15.
Therefore, sum of the zeroes = √15 + (-√15) = 0
= – $$\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}$$ = –$$\frac{0}{1}$$ = 0
And product of the zeroes √15 × (-√15) = -15
= $$\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}$$ = $$\frac{-15}{1}$$ = -15
vi) Given polynomial is 3x2 – x – 4
we have, 3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
The value of 3x2 – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0.
i.e., when x + 1 = 0 or 3x – 4 = 0
i.e., when x = -1 or x = $$\frac{4}{3}$$
∴ The zeroes of 3x2 – x – 4 are -1 and $$\frac{4}{3}$$
Therefore, sum of the zeroes = -1 + $$\frac{4}{3}$$ = $$\frac{-3+4}{3}$$ = $$\frac{1}{3}$$
= – $$\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$$ = $$\frac{-(-1)}{3}$$ = $$\frac{1}{3}$$
And product of the zeroes -1 × $$\frac{4}{3}$$ = $$\frac{-4}{3}$$
= $$\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$$ = $$\frac{-4}{3}$$
Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.
i) $$\frac{1}{4}$$, -1
ii) √2, $$\frac{1}{3}$$
iii) 0, √5
iv) 1, 1
v) –$$\frac{1}{4}$$, $$\frac{1}{4}$$
vi) 4, 1
Let the polynomial be ax2 + bx + c
and its zeroes be α and β.
i) Here, α + β = $$\frac{1}{4}$$ and αβ = -1
Thus, the polynomial formed = x2 – (sum of the zeroes)x + product of the zeroes
= x2 – ($$\frac{1}{4}$$)x – 1
= x2 – $$\frac{x}{4}$$ – 1
The other polynomials are (x2 – $$\frac{x}{4}$$ – 1)
then the polynomial is 4x2 – x – 4.
ii) Here, α + β = √2 and αβ = $$\frac{1}{3}$$
Thus, the polynomial formed = x2 – (sum of the zeroes)x + product of the zeroes
= x2 – (√2)x + $$\frac{1}{3}$$
= x2 – √2x + $$\frac{1}{3}$$
The other polynomials are (x2 – √2x + $$\frac{1}{3}$$)
then the polynomial is 3x2 – 3√2x + 1.
iii) Here, α + β = 0 and αβ = √5
Thus, the polynomial formed = x2 – (sum of the zeroes)x + product of the zeroes
= x2 – (0)x + √5
= x2 + √5
iv) Let the polynomial be ax2 + bx + c and its zeroes be α and β.
Then α + β = 1 = $$\frac{-(-1)}{1}$$ = $$\frac{-b}{a}$$ and
αβ = 1 = latex]\frac{1}{1}[/latex] = $$\frac{c}{a}$$
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x2 + x + 1.
v) Let the polynomial be ax2 + bx + c and its zeroes be α and β.
Then α + β = $$\frac{-1}{4}$$ = $$\frac{-b}{a}$$ and
αβ = $$\frac{1}{4}$$ = $$\frac{c}{a}$$
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x2 + x + 1.
vi) Let the polynomial be ax2 + bx + c and its zeroes be α and β.
Then α + β = 4 = $$\frac{-(-4)}{1}$$ = $$\frac{-b}{a}$$ and
αβ = 1 = $$\frac{1}{1}$$ = $$\frac{c}{a}$$
If a = 1, then b = -4 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is x2 – 4x + 1.
Question 3.
Find the quadratic polynomial, for the zeroes α, β given in each case.
i) 2, -1
ii) √3, -√3
iii) $$\frac{1}{4}$$, -1
iv) $$\frac{1}{2}$$, $$\frac{3}{2}$$
i) Let the polynomial be ax2 + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = 2 and β = – 1
Sum of the zeroes = α + β = 2 + (-l) = 1
Product of the zeroes = αβ = 2 × (-1) = -2
Therefore the quadratic polynomial ax2 + bx + c is x2 – (α + β)x + αβ = [x2 – x – 2]
the quadratic polynomial will be x2 – x – 2.
ii) Let the zeroes be α = √3 and β = -√3
Sum of the zeroes = α + β
= √3 + (-√3) = 0
Product of the zeroes = αβ
= √3 × (-√3) = -3
ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – 0.x + (-3)] = [x2 – 3]
the quadratic polynomial will be x2 – 3.
iii) Let the zeroes be α = $$\frac{1}{4}$$ and β = -1
Sum of the zeroes = α + β
= $$\frac{1}{4}$$ + (-1) = $$\frac{1+(-4)}{4}$$ = $$\frac{-3}{4}$$
Product of the zeroes = αβ
= $$\frac{1}{4}$$ × (-1) = $$\frac{-1}{4}$$
ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – $$\left(\frac{-3}{4}\right)$$.x + ($$\frac{-1}{4}$$)]
the quadratic polynomial will be 4x2 + 3x – 1.
iv) Let the zeroes be α = $$\frac{1}{2}$$ and β = $$\frac{3}{2}$$
Sum of the zeroes = α + β
= $$\frac{1}{2}$$ + $$\frac{3}{2}$$ = $$\frac{1+3}{2}$$ = $$\frac{4}{2}$$ = 2
Product of the zeroes = αβ
= $$\frac{1}{2}$$ × $$\frac{3}{2}$$ = $$\frac{3}{4}$$
ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – 2x + ($$\frac{3}{4}$$)]
the quadratic polynomial will be 4x2 – 8x + 3.
Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x3 + 3x2 – x – 3 and check the relationship between zeroes and the coefficients.
Given cubic polynomial
p(x) = x3 + 3x2 – x – 3
Comparing the given polynomial with ax3 + bx2 + cx + d, we get a = 1, b = 3, c = -1, d = -3
Futher given zeroes are 1,-1 and – 3
p(1) = (1)3 + 3(1)2 – 1 – 3
= 1 + 3 – 1 – 3 = 0
p(-1) = (-1)3 + 3(-1)2 – 1 – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)3 + 3(-3)2 – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore, 1, -1 and -3 are the zeroes of x3 + 3x2 – x – 3.
So, we take α = 1, β = -1 and γ = -3 Now,
α + β + γ = 1 + (-1) + (-3) = -3
αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1
= -1 + 3 – 3 = -1
= $$\frac{c}{a}$$ = $$\frac{-1}{1}$$ = -1
αβγ = 1 (-1) (-3) = 3 = $$\frac{-d}{a}$$ = $$\frac{-(-3)}{1}$$ = 3 | 3,754 | 8,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-33 | latest | en | 0.743474 |
https://convertilo.com/long-tons-to-milligrams | 1,716,915,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00458.warc.gz | 147,502,953 | 5,379 | # Long Tons to Milligrams - uk tn to mg
## How to convert from Long Tons to Milligrams
The conversion factor between Long Tons and Milligrams is 1016046908.8. To convert an amount in Long Tons in Milligrams, multiply the the desired amount in Long Tons by 1016046908.8:
Amount(Long Tons) × 1016046908.8(Conversion Factor) = Result(Milligrams)
## Definition of units
Let's see how both units in this conversion are defined, in this case Long Tons and Milligrams:
### Long Ton (uk tn)
A long ton is defined as exactly 2,240 pounds. The long ton arises from the traditional British measurement system: A long ton is 20 cwt, each of which is 8 stone (1 stone = 14 pounds). Thus a long ton is 20 × 8 × 14 lb = 2,240 lb. Long ton, also known as the imperial ton or displacement ton is the name for the unit called the "ton" in the avoirdupois or Imperial system of measurements standardised in the thirteenth century that is used in the United Kingdom
### Milligram (mg)
The milligram (abbreviation: mg) is a unit of mass, equal to 1/000 of a gram, and 1/10000000 of a kilogram (also written 1E-6 kg).
## Long Tons to Milligrams conversion table
Below is the conversion table you can use to convert from Long Tons to Milligrams
Long Tons (uk tn) Milligrams (mg)
1 Long Tons 1016046908.8 Milligrams
2 Long Tons 2032093817.6 Milligrams
3 Long Tons 3048140726.4 Milligrams
4 Long Tons 4064187635.2 Milligrams
5 Long Tons 5080234544 Milligrams
6 Long Tons 6096281452.8 Milligrams
7 Long Tons 7112328361.6 Milligrams
8 Long Tons 8128375270.4 Milligrams
9 Long Tons 9144422179.2 Milligrams
10 Long Tons 10160469088 Milligrams | 478 | 1,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.762651 |
https://gmatclub.com/forum/a-student-s-grade-in-a-course-is-determined-by-4-quizzes-and-1-exam-285027.html | 1,560,913,319,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998882.88/warc/CC-MAIN-20190619023613-20190619045613-00416.warc.gz | 458,990,492 | 144,249 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# A student’s grade in a course is determined by 4 quizzes and 1 exam.
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A student’s grade in a course is determined by 4 quizzes and 1 exam. [#permalink]
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24 Dec 2018, 21:24
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A student’s grade in a course is determined by 4 quizzes and 1 exam. If the exam counts twice as much as each of thequizzes, what fraction of the final grade is determined by the exam?
A. 1/6
B. 1/5
C. 1/3
D. 1/4
E. 1/2
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Re: A student’s grade in a course is determined by 4 quizzes and 1 exam. [#permalink]
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24 Dec 2018, 21:37
1
Given :
A student’s grade in a course is determined by 4 quizzes and 1 exam. (Final Grade = 4 Quizzes + 1 Exam)
If the exam counts twice as much as each of the quizzes (1 Exam grade unit = 2 Quiz grade units)
Question: what fraction of the final grade is determined by the exam? (Exam Grades / Final Grades =?)
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Re: A student’s grade in a course is determined by 4 quizzes and 1 exam. [#permalink]
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25 Dec 2018, 03:03
Bunuel wrote:
A student’s grade in a course is determined by 4 quizzes and 1 exam. If the exam counts twice as much as each of thequizzes, what fraction of the final grade is determined by the exam?
A. 1/6
B. 1/5
C. 1/3
D. 1/4
E. 1/2
score would be 1 points of each quiz and 2 points for 1 exam = 4*1+2*1 = 6
so
2/6= 1/3rd of the total score.
IMO C
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Re: A student’s grade in a course is determined by 4 quizzes and 1 exam. [#permalink] 25 Dec 2018, 03:03
Display posts from previous: Sort by | 768 | 2,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-26 | latest | en | 0.89525 |
http://www.jiskha.com/display.cgi?id=1328849487 | 1,493,508,440,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123632.58/warc/CC-MAIN-20170423031203-00576-ip-10-145-167-34.ec2.internal.warc.gz | 580,542,498 | 3,680 | # physics
posted by on .
A wheel of 40cm radius rotates on a stationary axle. It isuniformly speeded up from rest to a speed of 900 rpm in a time of20s. Find (a) the constant angular acceleration of the wheel and(b) the tangential acceleration of a point of a point on itsrim.
• physics - ,
(a) convert 900 rpm to radians per second and divide by 20s. That is usually called alpha, in radians/s^2
(b) tangential acceleration = R*(alpha)
This is basic stuff. Show your work if additional assistance is needed | 128 | 513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-17 | latest | en | 0.923724 |
https://stats.stackexchange.com/questions/236194/sum-of-n-two-component-mixture-variates | 1,718,430,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00286.warc.gz | 498,175,332 | 39,175 | # Sum of N two-component mixture variates
I have a list of random variables, $X_1$, $X_2$, ..., $X_N$, associated with binary random variables $A_i$ such that $P(A_i) = \pi$ is known. I also know that, for all $i$ $$X_i|A_i\sim f(x)\\ X_i|\bar A_i\sim g(x)$$ where $f$ and $g$ are known, and thus the distribution of the $X_i$ is a mixture given by $$p(x) = \pi\cdot f(x) + (1-\pi)\cdot g(x)$$ Is there a general expression for the distribution of $Y=\sum_{i=1}^N X_i$?
If you denote by $$g^m\ast f^k$$ the convolution of $m$ $g$'s and $k$ $f$'s, meaning the density of the distribution of the sum of $m$ realisations of $X\sim g(x)$ and $k$ realisations of $X\sim f(x)$, with for instance $$g^0\ast f^1=f\,,\quad g^1\ast f^0=g\,,$$ \begin{align*}(g^m\ast f^k)(x)&=\int (g^{m-1}\ast f^k)(x-y)g(y)\,\text{d}y\\&=\int (g^{m}\ast f^{k-1})(x-y)f(y)\,\text{d}y\,,\end{align*}the density of the sum of $n$ variables from the mixture with density$$\pi g(x) + (1-\pi) f(x)$$is
$$\sum_{m=0}^n {n \choose m}\pi^m(1-\pi)^{n-m} (g^m\ast f^{n-m})(x)$$
For instance, if $g$ is the density of a N$(0,1)$ distribution and $f$ the density of a N$(\mu,\sigma^2)$ distribution, we have that $(g^m\ast f^{n-m})$ is the density of a $$\text{N}\{(n-m)\mu,(n-m)\sigma^2+m\}$$ distribution, hence the above sum has for distribution $$\sum_{m=0}^n {n \choose m}\pi^m(1-\pi)^{n-m} \text{N}\{(n-m)\mu,(n-m)\sigma^2+m\}$$ | 559 | 1,395 | {"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} | 3.453125 | 3 | CC-MAIN-2024-26 | latest | en | 0.731225 |
http://www.stat.nthu.edu.tw/~swcheng/Teaching/stat5410/index.php | 1,713,043,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00740.warc.gz | 54,109,228 | 6,145 | NTHU STAT 5410 - Linear Models
Sep 2022 ~ Jan 2023
Notes Jan 15 作業總成績, 期末考成績, 學期總成績, 及成績統計。 Jan 15 期末考解答。 Jan 03 期末考考古題及其解答。 Jan 03 期末考資訊及注意事項。 Nov 23 自下周(11/28)起，黃晨溦助教的office hour時間，將更改至每周三下午1:20~2:10，地點則維持不變(綜合三館840室)。更改後所有助教的office hour時間和地點，請見Syllabus。 Nov 23 期中考解答及成績統計。 Nov 07 期中考考古題及解答。 Nov 07 期中考資訊及注意事項。
Lecture Notes Lecture Notes with Hand-Written Notices Video 01 (1133 views) (465 views) (350 views) 02 (652 views) (398 views) (390 views) Sep 29 (443 views) 03 (572 views) (503 views) (499 views) Oct 06 (483 views) (420 views) (408 views) 04 (387 views) (341 views) (366 views) Oct 20 (362 views) (322 views) (339 views) 05 (316 views) (285 views) (267 views) Nov 03 (277 views) (229 views) (196 views) 06 (306 views) (251 views) (272 views) (273 views) Nov 17 (257 views) 07 (303 views) Nov 24 (273 views) (250 views) (235 views) Dec 01 (232 views) (211 views) (205 views) 08 (238 views) (219 views) (223 views) Dec 15 (217 views) (197 views) (205 views) Dec 22 (200 views) 09 (201 views) (171 views) Dec 29 (191 views) (179 views) (185 views) Jan 05 (193 views) 10 (197 views) (169 views)
• Lab
1 Introduction to R (332 views) 2 Initial Data Analysis (241 views) 3 Parameter Estimation (256 views) (179 views) 4 Hypotheses Testing (190 views) 5.01 Confidence Interval and Region (136 views) 5.02 Orthogonality (118 views) 5.03 Identifiability (88 views) 5.04 Interpretation of Parameter Estimates (87 views) 6.01 Generalized Least Squares (135 views) 6.02 Weighted Least Squares (134 views) 6.03 Testing for Lack of Fit (134 views) 7.01 Diagnostic 1 (residual, leverage, studentized residual, outlier, jacknife residual, influential observation, Cook's statistic) (157 views) 7.02 Diagnostic 2 (residual plots, non-constant variance, added variable plot, partial residual plot, Q-Q plot, half-normal plot, correlated errors) (141 views) 8.01 Mean Structure for Quantitative Predictors (no video) 8.02 Mean Structure for Qualitative Predictors (no video) 8.03 Box-Cox Transformation (no video) 9.01 Model (Variable) Selection (no video) 9.02 Error in Predictors (no video) 9.03 Collinearity, Principal Component, Ridge Regression (no video) 9.04 Analysis Strategy --- A Real Example: Chicago Insurance Redlining (no video) 10.01 Robust Regression (no video) 10.02 Incomplete Data (no video)
• Assignment and solution
Assignment Due Day Solution Homework 1 Oct 06 sol (revised 2) Homework 2 Oct 20 sol Homework 3 Nov 03 sol Homework 4 Nov 17 sol Homework 5 Dec 01 sol Homework 6 Dec 15 sol Homework 7 Jan 05 sol | 871 | 2,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.359889 |
https://en.academic.ru/dic.nsf/enwiki/3115497/Seismic_scale | 1,596,593,815,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735906.77/warc/CC-MAIN-20200805010001-20200805040001-00273.warc.gz | 251,026,728 | 15,112 | # Seismic scale
Seismic scale
A seismic scale is used to measure and compare the severity of earthquakes. (For a quick review, see the table of seismic scales at the end of this article.)
Two fundamentally different but equally important types of scales are commonly used by seismologists to describe earthquakes. The original force or energy of an earthquake is measured on a "magnitude scale", while the intensity of shaking occurring at any given point on the Earth's surface is measured on an "intensity scale".
Magnitude and Intensity
The severity of an earthquake is described by both magnitude and intensity. These two frequently-confused terms refer to different, but related, observations. "Magnitude," usually expressed as an Arabic numeral, characterizes the size of an earthquake by measuring indirectly the energy released. By contrast, "intensity" indicates the local effects and potential for damage produced by an earthquake on the Earth's surface as it affects humans, animals, structures, and natural objects such as bodies of water. Intensities are usually expressed in Roman numerals, and represent the severity of the shaking resulting from an earthquake. Ideally, any given earthquake can be described by only one "magnitude", but many "intensities" since the earthquake effects vary with circumstances such as distance from the epicenter and local soil conditions. In practise, the same earthquake might have magnitude estimates typically differing by few tenths of a unit, depending on which magnitude scale is used and which data are included in the analysis.
Charles Richter, the creator of the Richter magnitude scale, distinguished "intensity" and "magnitude" as follows: "I like to use the analogy with radio transmissions. It applies in seismology because seismographs, or the receivers, record the waves of elastic disturbance, or radio waves, that are radiated from the earthquake source, or the broadcasting station. Magnitude can be compared to the power output in kilowatts of a broadcasting station. Local intensity on the Mercalli scale is then comparable to the signal strength on a receiver at a given locality; in effect, the quality of the signal. Intensity, like signal strength, will generally fall off with distance from the source, although it also depends on the local conditions and the pathway from the source to the point."
eismic intensity scales
The first simple classification of earthquake intensity was devised by Domenico Pignataro in the 1780s. However, the first recognisable intensity scale in the modern sense of the word was drawn up by P.N.G. Egen in 1828; it was ahead of its time. The first widely adopted intensity scale, the Rossi-Forel scale, was introduced in the late 19th century. Since then numerous intensity scales have been developed and are used in different parts of the world: the scale currently used in the United States is the Modified Mercalli scale (MM), while the European Macroseismic Scale (EMS-94) is used in Europe, the Shindo scale is used in Japan, the MSK-64 scale is used in India, Israel, Russia and throughout the CIS, and the Liedu scale (GB/T 17742-1999) is used in mainland China; Hong Kong, on the other hand, uses the MM scale,cite web
url=http://www.weather.gov.hk/gts/equake/mag_and_int_e.htm
title=Magnitude and Intensity of an Earthquake
publisher=Hong Kong Observatory
accessdate=2008-09-15
] and Taiwan uses the Shindo scale. Most of these scales have twelve degrees of intensity, which are roughly equivalent to one another in values but vary in the degree of sophistication employed in their formulation.
Magnitude scales
The first attempt to qualitatively define a single, absolute value to describe the size of earthquakes was the magnitude scale (the name being taking from similarly formulated scales used to represent the brightness of stars).
Local magnitude scale and related scales
The local magnitude scale (ML), also popularly known as the Richter Scale, is a quantitative logarithmic scale. In the 1930s, California seismologist Charles F. Richter devised a simple numerical scale to describe the relative sizes of earthquakes in Southern California. The name "Richter Scale" was coined by journalists and is not generally used by seismologists in technical literature. ML is obtained by measuring the maximum amplitude of a recording on a Wood-Anderson torsion seismometer (or one calibrated to it) at a distance of 600 km from the earthquake. Other more recent magnitude measurements include: body wave magnitude (mb), surface wave magnitude (Ms), and duration magnitude (MD). Each of these is scaled to give values similar to those given by the local magnitude scale; but because each is based on a measurement of one aspect of the seismogram, they do not always capture the overall power of the source. Specifically, some can be affected by saturation at higher magnitude values—meaning that they systematically underestimate the magnitude of larger events. This problem sets in at around magnitude 6 for local magnitude; surface-wave magnitude saturates above 8. Despite the limitations of older magnitude scales, they are still in wide use, as they can be calculated rapidly, catalogues of them dating back many years are available, they are sufficient for the vast majority of observed events, and the public is familiar with them.
Moment magnitude scale
Because of the limitations of the magnitude scales, a new, more uniformly applicable extension of them, known as moment magnitude (MW) scale for representing the size of earthquakes, was introduced by Hiroo Kanamori in 1977. In particular, for very large earthquakes moment magnitude gives the most reliable estimate of earthquake size. This is because seismic moment is derived from the concept of moment in physics and therefore provides clues to the physical size of an earthquake—the size of fault rupture and accompanying slip displacement — as well as the amount of energy released. So while seismic moment, too, is calculated from seismograms, it can also be obtained by working backwards from geologic estimates of the size of the fault rupture and displacement. The values of moments for observed earthquakes range over more than 15 orders of magnitude, and because they are not influenced by variables such as local circumstances, the results obtained make it easy to objectively compare the sizes of different earthquakes.
Notes and References
ee also
*Earthquake
*Earthquake engineering
*Seismic performance
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• seismic scale — seisminė skalė statusas T sritis Standartizacija ir metrologija apibrėžtis Žemės drebėjimų intensyvumo vertinimo skalė. atitikmenys: angl. seismic scale vok. Intensitätsskala, f; seismische Skala, f rus. сейсмическая шкала, f pranc. échelle… … Penkiakalbis aiškinamasis metrologijos terminų žodynas
• Seismic moment — is a quantity used by earthquake seismologists to measure the size of an earthquake. The scalar seismic moment M 0 is defined by the equationM 0=mu Au, where *mu is the shear modulus of the rocks involved in the earthquake, typically 30… … Wikipedia
• scale of seismic intensity — seisminė skalė statusas T sritis ekologija ir aplinkotyra apibrėžtis Žemės drebėjimų stiprumo vertinimo skalė. Dažniausiai naudojamos seisminės skalės, vertinančios drebėjimų intensyvumą pagal matomus jų padarinius (pastatų sugriovimą), žemės… … Ekologijos terminų aiškinamasis žodynas
• scale of earthquake intensity — seisminė skalė statusas T sritis ekologija ir aplinkotyra apibrėžtis Žemės drebėjimų stiprumo vertinimo skalė. Dažniausiai naudojamos seisminės skalės, vertinančios drebėjimų intensyvumą pagal matomus jų padarinius (pastatų sugriovimą), žemės… … Ekologijos terminų aiškinamasis žodynas
• Richter magnitude scale — Part of a series on earthquakes Types Foreshock • Aftershock • Blind thrust Doublet • Interplate • … Wikipedia
• China Seismic Intensity Scale — The China Seismic Intensity Scale (CSIS) is a national standard in the People s Republic of China[1] used to measure seismic intensity. Similar to EMS 92 on which CSIS drew reference, seismic impacts are classified into 12 degrees of intensity,… … Wikipedia
• Mercalli intensity scale — Part of a series on earthquakes Types Foreshock • Aftershock • Blind thrust Doublet • Interplate • … Wikipedia
• Japan Meteorological Agency seismic intensity scale — The Japan Meteorological Agency seismic intensity scale is a measure used in Japan and Taiwan to indicate the strength of earthquakes. It is measured in units of . Unlike the Richter magnitude scale (which measures the total magnitude of the… … Wikipedia
• European Macroseismic Scale — The European Macroseismic Scale (EMS) is the basis for evaluation of seismic intensity in European countries and moreover in use on most other continents. Issued in 1998 as update of the test version from 1992, the scale is referred to as EMS… … Wikipedia
• Rossi-Forel scale — The Rossi Forel scale was one of the first seismic scales to reflect earthquake intensities. Developed by Michele Stefano Conte de Rossi of Italy and François Alphonse Forel of Switzerland in the 1800s, it was used for about two decades until the … Wikipedia | 2,122 | 9,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-34 | latest | en | 0.952 |
https://allresearchers.com/competitive-market-structure/ | 1,716,500,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00361.warc.gz | 75,034,579 | 40,673 | # Competitive Market Structure
Question
From the scenario, assuming Katrina’s Candies is operating in the monopolistically competitive market structure and faces the following weekly demand and short-run cost functions:
VC = 20Q+0.006665 Q2 with MC=20 + 0.01333Q and FC = \$5,000
P = 50-0.01Q and MR = 50-0.02Q
*Where price is in \$ and Q is in kilograms. All answers should be rounded to the nearest whole number.
· o Algebraically, determine what price Katrina’s Candies should charge in order for the company to maximize profit in the short run. Determine the quantity that would be produced at this price and the maximum profit possible.
Question
Find the two sets of raw data, AccRet.txt and AlbHomw.txt, attached under this assignment.
I. Compute (by hand) the N, Mean, Standard Deviation, Confidence Interval, and Z score for the variables “company market rate” and “accounting rate” respectively in the AccRet.txtdataset
Report your statistical results obtained from the AccRet.txt dataset.
II. Compute (by hand) the N, Mean, Standard Deviation, Confidence Interval, and Z score for the variables “Price”, SqFt”, Age, Feats, and “Tax” respectively in the AlbHome.txt dataset.
Report your statistical results obtained from the AlbHome.txt dataset.
Need help with this assignment or a similar one? Place your order and leave the rest to our experts! | 324 | 1,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.881516 |
http://www.chegg.com/homework-help/questions-and-answers/cubical-block-wood-107-cm-side-floats-interface-oil-water-lower-surface-210-cm-interface-f-q1352071 | 1,474,826,028,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660338.16/warc/CC-MAIN-20160924173740-00168-ip-10-143-35-109.ec2.internal.warc.gz | 395,602,401 | 13,820 | A cubical block of wood, 10.7 cm on a side, floats at the interface between oil and water with its lower surface 2.10 cm below the interface (the figure ). The density of the oil is 790 kg/m^3.
a) What is the gauge pressure at the upper face of the block?
? Pa
b)What is the gauge pressure at the lower face of the block?
? Pa | 89 | 328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.932758 |
http://www.heatinghelp.com/forum/profile/3148/Gordy/3 | 1,394,298,590,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999655239/warc/CC-MAIN-20140305060735-00079-ip-10-183-142-35.ec2.internal.warc.gz | 368,607,880 | 11,202 | ## Gordy
Joined on October 13, 2004
Last Post on March 8, 2014
### Recent Posts
« 1 2 3 4 5 6 7 ... 108 »
### Garbage in
@ February 11, 2014 12:03 PM in heat loss calculators
Garbage out. If the information requested is in accurate then the calculation will be.
How big is your sons room?
How many exterior walls?
How many windows? Size, type?
What kind of heat?
What inside temp, and outside design temp did you use?
Do you know the outdoor design temp for your area?
6000 btus sounds about right for an average size bedroom with one or two exterior walls.....depends.
### Mark
@ February 11, 2014 9:03 AM in delta T
Serious data logging, and controlling goin on there Mark!
Well being that I can't monitor 64 data points, and control 32 actions I'm only making some ass-umptions.
Initially my particular system had 21 loops all parallel piped. With the circ I have in place, and the calculated head I'm running 15 gpm @ 11' of head so IF the loops are equal length which they are most likely not the gpm to each loop would be .71 gpm. 15gpm/21=.71 gpm
Now I have added 5 more loops to the system so now I have 26 loops/15gpm=.58gpm....approximately.
Now what I have in place is thermometers at the main supply, and return at the boiler after mixing, and at the other end of the main supply, and return at the farthest point from the the boiler which supply the radiant ceilings. I also have thermometers on the main supply, and return of the floor loops I added which are parallel piped also. All other temperature monitoring is done with an IR thermometer at various points.
The main supply return temps for the radiant ceilings has always been a 15 delta before, and after. Adding floor loops. The delta should have increased because of the lower flow rates in the loops. The added floor loops delta is a 9.
What's stuck,in my head is the main supply/return delta should have increased due to a theoretically lowered flow rate due to the addition of extra loops to the parallel piping system. It has not changed, so what I'm thinking is now that I have 4 rooms with opposing radiant surfaces that those surfaces are giving up less heat due to double the radiant surface area. There for even though the flow rates decreased the delta t remained the same....pretty much by accident, unintended,and now wondering why.
@ February 10, 2014 9:36 PM in Radiant heat zone not letting water flow
Should only be about 12-15 psi. Bumping up the pressure was just to purge the line. If your pressure is to high you may pop the boilers relief valve which is 30psi.
Is the bladder tank properly charged? To 12-15 psi?
### The math
@ February 10, 2014 9:13 PM in propane line not buried
How many btus is the heater rated for?
How many hours a week is it on? Is it on a thermostat?
How big of a space is it heating? How well insulated is it?
15 x 91500 btus = 1372500 btus a week that's at 100% efficiency
That's 196071btus a day or 2.14 gal
Or 8170 btus an hour if running constantly.thats .09 gallons per hour.
Don't know if that's good or bad with out knowing the fore mentioned info.
I would guess you either have a huge heater, or a poorly insulated space with a high set point. Or both.
I highly doubt your using more gas because it's not buried unless you have a leak. If you do have a leak it would not make a difference buried or not.
Around here code is 12" deep.
### Just like the younger days
@ February 10, 2014 5:15 PM in Chicago Nearing Record For Days Below Zero
That. I remember ......62/63. 78/79. 81/82
We have been spoiled for some time some 50's winters were brutal to.
### Super emitterated
@ February 10, 2014 6:19 AM in delta T
Mark,
Did you ever keep track of those readings with just certain radiation surfaces in place?
I know you have ceilings, floors,walls, and windows radiant surfaces. But they were,not all in place at once were they? If not did you data log with only certain surfaces in place?
I know in my own home when I only had radiant ceilings the delta s in individual loops were 5-10 deltas with total system supply return delta of 15. It was always rock solid, and never varied with outdoor temps. Same delta always floats with supply temps.
Now that I have added radiant floors to some of the rooms which now have radiant floors, and ceilings the delta is always a 9 for the floors, and total system delta dropped to 13. Still always the same though.
What's interesting is my radiant is parallel piped. New loops were no longer than any existing loops. Circulator the same. So in theory my system flow rate gets divided up by additional loops added so flow rates in loops actually decreased.
In your case, and mine designing for a 20 delta, and experiencing a 10 is it possible a certain square feet of radiation decreases design delta by a factor?
### One last thing
@ February 9, 2014 2:43 PM in Electric heat, or infra-red
I should have emphasized this before. Make sure your base board is not blocked by furniture, or carpeting near the bottom opening. They depend on convection to operate efficiently. So if you block air from circulating through the bottom of the base board, and the top it will take longer to do its job.
### I'll make this easy
@ February 9, 2014 2:10 PM in Electric heat, or infra-red
Trust me on this.
Both heaters are electric right?
When both heaters are on they are using electricity right?
So the both cost the same to operate watt for watt.
The bigger heater(base board) is going to heat the space quicker, and be on probably less than the infra red space heater.
Use the base board. It will be more comfortable.
Now if we were comparing different types of heat sources gas, oil, kerosene, wood then it's worth doing some calculations. Your just robbing Peter to pay Paul.
### 18 cent KW
@ February 9, 2014 1:11 PM in Electric heat, or infra-red
Yikes!
### Chewy
@ February 9, 2014 1:05 PM in Electric heat, or infra-red
Do understand that the base boards do not run 24/7 so your ball park is off.
Unless not enough was installed for the heat load, or you dip way down below what the normal heat load was calculated for the sizing of the baseboard.
Also understand that 10 foot of base board is giving you more btus to heat with than your space heater. The base board if numbers are right is giving about 7500 btus compared to the space heater that gives about 5100 btus.
Like Harvey said evaluate a cheaper heat plant more than likely of an NG type of fuel source.
### Same info, and math
@ February 9, 2014 11:34 AM in Electric heat, or infra-red
Applies Chewy.
### Harvey's post
@ February 9, 2014 11:32 AM in Electric heat, or infra-red
Is probably due to an overloaded circuit, and the wiring got hot, and cooled multiple times creating a loose arcing connection. Or a sloppy outlet installation. Possibly aluminum wiring.
@ February 9, 2014 11:26 AM in Electric heat, or infra-red
For the cost to operate that heater. Is the comfort there compared to what you have in place?
That's 90 bucks a month!
Many people are getting caught up in trying to save money on their heating costs this year do to the weather. Spot heating in the end really reduces comfort, increases operating cost, and increases a potentially hazardous condition.
In the end a btu needed is a btu used. And a therm, KW, gal of fuel to make it is still needed to get you to a comfort level.
### Nice post Harvey
@ February 9, 2014 11:18 AM in Electric heat, or infra-red
### To find
@ February 9, 2014 11:15 AM in Electric heat, or infra-red
Total cost per KWH take your total monthly bill, and divide into total KWH used that month.
This will give you cost after all incurred fees. Usually the bill only states cost charge per KWH and then adds all taxes etc after.
Honestly if your base board zones are set up by certain zones, or areas ( multiple thermostats) you would be better served using what you have if the base boards are set up as to areas you are in can be run.
Otherwise an oil filled radiator type electric heater will give you far better comfort in a room than an infra red type heater for the same cost to operate. The radiator style heats the air not just objects or your person.
### It means
@ February 9, 2014 10:53 AM in Electric heat, or infra-red
For every hour that heater is on it uses 1.5 kW. So if your rate costs .10 cents a kilowatt. It costs 15 cents an hour to run.
So,if you use it for 24 hours it costs 3.60 a day.
If you use it 6 hours a day that's .90 cents or 27.00 a month.
If there is only an on off setting. If there are lower settings naturally it will use less. That rating may be maximum output.
### Mike
@ February 9, 2014 9:08 AM in Electric heat, or infra-red
Exactly Mike,
Not a huge portion of homeowners understand how their wiring is suppose to work in their home.
One scenario would be if your wiring is 12/3 or 12/2 with 20 amp breaker. Outlets are rated 15 amp (without the rotated slot) in using that adapter you could make the outlet fail causing a fire, or burn up the in this case kill-a-watt.
A worse scenario would be 14/3 or 14/2 circuit or in any portion of the circuit with 15 amp outlet, and 20 amp breaker that someone had enough knowledge to switch out but not the understanding of the implications. Then there are multiple points of failure in wiring, outlet etc.
I sure would not produce such an adapter, I'm sure there is do, and donts in the literature.
### Yes
@ February 9, 2014 1:27 AM in Electric heat, or infra-red
That one
### Where there is a will
@ February 9, 2014 1:22 AM in Electric heat, or infra-red
People find, or make a way.
All it takes from there is 12/3 wiring and a 20 amp breaker..
Amazing isn't it.
### Kill-a-watt
@ February 8, 2014 10:12 PM in Electric heat, or infra-red
Can be bought at most home improvement stores in the electrical section.
I'm only familiar with that brand it was the first device to do this,others followed. Cheaper is not always better.
Again if you know the heaters wattage the math is quite simple.
One other thing I hope you don't use this heater extensively, and unattended . They are not made to take the place of actual home heating means.
### Care should be given
@ February 8, 2014 7:13 PM in Modine HW Unit in Garage
Of course as to not forgetting to close garage door for the whole day in extreme conditions. Limiting supply return piping projection to, and from the unit helps. Mount unit on shared house garage wall.
### Power Purging
@ February 8, 2014 6:59 PM in Radiant heat zone not letting water flow
Did you after opening the system?
Is the zone valve oriented correctly?
Yes air in the line would be an issue. This would cause the loop not to flow, or be air bound.
You need to bump the pressure up with your water feed valve to about 25psi, and bleed the pipes.
If your piping allows you to isolate the boiler you can power purge under higher pressure.
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# Eat your heart out craigslist huckster p-3 system rundown Straight
Topic closed. 68 replies. Last post 9 months ago by lamaka10.
Page 1 of 5
United States
Member #164727
March 12, 2015
2675 Posts
Offline
Posted: April 4, 2016, 1:48 pm - IP Logged
I would never send money to some huckster. My gray matter does just fine, always coming up with new material, and this one isn't too bad. I will give a little credit to this huckster for giving me a rundown idea that I never did before.
First step is to make a vertical list of 0-9.
1
2
3
4
5
6
7
8
9
Then you take your last winning number and put the winning number in the 3 places it can attach to. For example, Florida, march 3 mid-258.
There is actually two rundowns to make this work for straight, see which one works better in your state.
Using the 258 below.
Now we always count up and stop and count up again starting with the new number. Star with the 0 in the middle position. But with the 5 in the middle section, we count back down to fill the two spots, which is 7 and 6.
And after the 8, we count up to fill in the next two spots, which is 9 and 0.
Now keep going up after the number two in the middle position, which is 3 and 4.
And lastly the final number after 5 which is 6.
Now do this for the last positions and you're done.
March 5 mid, two days later 063...straight hit.
6 3---straight hit..
7 4
2 8 5
3 9 6
4 0 7
5 2 8
6 3 9
7 4 0
8 5 1
9 6 3
Now you may have to do two versions to get a straight. Below is the other version which would bring a box hit. This depends on how you arrange the last winning number's orientation on the rundown.
0 3 6
1 4 7
2 8 5
3 6 9
4 7 0
5 8 2
6 9 3
7 0 4
8 1 5
9 3 6
Most of the time it takes more than two days, but this is not bad for a straight system.
IMPORTANT, SOMETIMES THE REPEATS WITHIN THE RUNDOWN IS ALL YOU NEED TO PLAY. THIS CAN REDUCE COST BY ONLY PLAYING RUNDOWNS THAT HAVE REPEATS.
March 5 eve-761
March 6 eve-409
0 6 5
1 7 6
2 8 7---repeats
3 9 8---repeats
4 0 9
5 1 0
6 1 7
7 6 1
8 7 2---repeats
9 8 3---repeats
March 4 eve-869
March 7 mid-790
0 2 3
1 3 4
2 4 5
3 5 6
4 6 7
5 7 8
6 8 9
7 9 0--straight hit
8 9 6
9 8 6
March 5 mid-063 ( this one took a little longer than I'd like )
March 11 eve-174
0 6 3
1 7 4
2 8 5---repeats
3 6 0
4 7 1--REPEATS
5 8 2---repeats
6 3 0
7 4 1--REPEATS
8 5 2---repeats
9 0 3
March 9 eve-572
March 13 mid-146
0 3 5
1 4 6- straight hit
2 5 7
3 6 8---repeats
4 7 9---repeats
5 7 2
6 8 3--repeats
7 2 5
8 3 6---repeats
9 4 7---repeats
March 11 eve-174
March 15 eve-063
0 6 3
1 7 4
2 8 5---REPEATS
3 9 6
4 7 1--REPEATS
5 8 2---REPEATS
6 9 3
7 1 4--REPEATS
8 2 5---REPEATS
9 3 6
This one took a whole 7 days to hit. But if you only play repeats it's 7 bucks a draw.
March 13 mid-146
March 20 mid-614
0 3 5
1 4 6---REPEATS
2 5 7---REPEATS
3 6 8---REPEATS
4 1 6---REPEATS
5 2 7---REPEATS
6 1 4
7 2 5---REPEATS
8 3 6---REPEATS
9 4 7
Only 5 repeats, so 5 bucks a draw.
March 16 eve-084
March 17 mid-951
0 8 4
1 9 5---REPEATS
2 0 6
3 1 7---REPEATS
4 8 0
5 9 1---REPEATS
6 0 2
7 1 3---REPEATS
8 4 0
9 5 1---REPEATS
At this point it takes a break till March 30. You may want to also take a break if it stops, then wait till it runs again.
Here, if you only play repeats, it would be two combos 436 and 643.
March 30 eve-235
April 2 mid-436
0 1 3
1 2 4
2 3 5
3 2 5
4 3 6---repeats
5 3 2
6 4 3---repeats
7 5 4
8 6 5
9 7 6
Here you would have won the very next day with only 4 combos if you only play repeats.
March 31 eve-906
April 1 mid-107
0 9 6
1 0 7
2 1 8---repeats
3 2 9
4 3 0
5 4 1
6 0 9
7 1 0---REPEATS
8 2 1---repeats
9 0 6
1
New Mexico
United States
Member #86099
January 29, 2010
11163 Posts
Offline
Posted: April 4, 2016, 2:00 pm - IP Logged
I would never send money to some huckster. My gray matter does just fine, always coming up with new material, and this one isn't too bad. I will give a little credit to this huckster for giving me a rundown idea that I never did before.
First step is to make a vertical list of 0-9.
1
2
3
4
5
6
7
8
9
Then you take your last winning number and put the winning number in the 3 places it can attach to. For example, Florida, march 3 mid-258.
There is actually two rundowns to make this work for straight, see which one works better in your state.
Using the 258 below.
Now we always count up and stop and count up again starting with the new number. Star with the 0 in the middle position. But with the 5 in the middle section, we count back down to fill the two spots, which is 7 and 6.
And after the 8, we count up to fill in the next two spots, which is 9 and 0.
Now keep going up after the number two in the middle position, which is 3 and 4.
And lastly the final number after 5 which is 6.
Now do this for the last positions and you're done.
March 5 mid, two days later 063...straight hit.
6 3---straight hit..
7 4
2 8 5
3 9 6
4 0 7
5 2 8
6 3 9
7 4 0
8 5 1
9 6 3
Now you may have to do two versions to get a straight. Below is the other version which would bring a box hit. This depends on how you arrange the last winning number's orientation on the rundown.
0 3 6
1 4 7
2 8 5
3 6 9
4 7 0
5 8 2
6 9 3
7 0 4
8 1 5
9 3 6
Most of the time it takes more than two days, but this is not bad for a straight system.
IMPORTANT, SOMETIMES THE REPEATS WITHIN THE RUNDOWN IS ALL YOU NEED TO PLAY. THIS CAN REDUCE COST BY ONLY PLAYING RUNDOWNS THAT HAVE REPEATS.
March 5 eve-761
March 6 eve-409
0 6 5
1 7 6
2 8 7---repeats
3 9 8---repeats
4 0 9
5 1 0
6 1 7
7 6 1
8 7 2---repeats
9 8 3---repeats
March 4 eve-869
March 7 mid-790
0 2 3
1 3 4
2 4 5
3 5 6
4 6 7
5 7 8
6 8 9
7 9 0--straight hit
8 9 6
9 8 6
March 5 mid-063 ( this one took a little longer than I'd like )
March 11 eve-174
0 6 3
1 7 4
2 8 5---repeats
3 6 0
4 7 1--REPEATS
5 8 2---repeats
6 3 0
7 4 1--REPEATS
8 5 2---repeats
9 0 3
March 9 eve-572
March 13 mid-146
0 3 5
1 4 6- straight hit
2 5 7
3 6 8---repeats
4 7 9---repeats
5 7 2
6 8 3--repeats
7 2 5
8 3 6---repeats
9 4 7---repeats
March 11 eve-174
March 15 eve-063
0 6 3
1 7 4
2 8 5---REPEATS
3 9 6
4 7 1--REPEATS
5 8 2---REPEATS
6 9 3
7 1 4--REPEATS
8 2 5---REPEATS
9 3 6
This one took a whole 7 days to hit. But if you only play repeats it's 7 bucks a draw.
March 13 mid-146
March 20 mid-614
0 3 5
1 4 6---REPEATS
2 5 7---REPEATS
3 6 8---REPEATS
4 1 6---REPEATS
5 2 7---REPEATS
6 1 4
7 2 5---REPEATS
8 3 6---REPEATS
9 4 7
Only 5 repeats, so 5 bucks a draw.
March 16 eve-084
March 17 mid-951
0 8 4
1 9 5---REPEATS
2 0 6
3 1 7---REPEATS
4 8 0
5 9 1---REPEATS
6 0 2
7 1 3---REPEATS
8 4 0
9 5 1---REPEATS
At this point it takes a break till March 30. You may want to also take a break if it stops, then wait till it runs again.
Here, if you only play repeats, it would be two combos 436 and 643.
March 30 eve-235
April 2 mid-436
0 1 3
1 2 4
2 3 5
3 2 5
4 3 6---repeats
5 3 2
6 4 3---repeats
7 5 4
8 6 5
9 7 6
Here you would have won the very next day with only 4 combos if you only play repeats.
March 31 eve-906
April 1 mid-107
0 9 6
1 0 7
2 1 8---repeats
3 2 9
4 3 0
5 4 1
6 0 9
7 1 0---REPEATS
8 2 1---repeats
9 0 6
1
Put this in excel. Adjacent numbers I see .
United States
Member #164727
March 12, 2015
2675 Posts
Offline
Posted: April 4, 2016, 2:10 pm - IP Logged
For tonight's prediction for Florida, I will be using the last 6 non double winning numbers that haven't produced a winner..Also, I will only play the repeats.
They are,
March 30 mid-065
April 1 mid-107
April 1 eve-167
April 2 mid-436
April 2 eve-495
April 3 eve-503
March 30 mid-065= 176 716 287 827 398 938
April 1 mid-107=218 812 329 923
April 1 eve-167= 278 827 389 938
April 2 mid-436=547 754
April 2 eve-495=(no repeats, so I will play all) 051 162 273 384 650 761 872 954
April 3 eve-503=164 641 275 752
jacksonville florida
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April 4, 2016
137 Posts
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Posted: April 4, 2016, 3:15 pm - IP Logged
Amber123
Could you break this down using last nights Florida April 3rd drawing for an exampIe am having a hard time understanding your process.
Thanks
Bke72\$
BOSTON
United States
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September 9, 2001
3611 Posts
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Posted: April 4, 2016, 3:31 pm - IP Logged
lakerben is right. excel would simplify the whole thing.
Ohio
United States
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July 31, 2014
1178 Posts
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Posted: April 4, 2016, 3:46 pm - IP Logged
An excel file would be great.
new jersey
United States
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December 31, 2013
513 Posts
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Posted: April 4, 2016, 3:50 pm - IP Logged
your explanation is terrible
United States
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March 12, 2015
2675 Posts
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Posted: April 4, 2016, 3:56 pm - IP Logged
Amber123
Could you break this down using last nights Florida April 3rd drawing for an exampIe am having a hard time understanding your process.
Thanks
Bke72\$
I know, after looking at my explanation, it does look confusing, it's very simple.
Last night was 503. You make a list below like this. Then you put 503 where it belongs.
0 5 3
1
2
3 0 5
4
5 0 3
6
7
8
9
Now do this below,
See the 5 in 053? count up...6-7
See the 3 in 053? count up...4-5
0 5 3
1 6 4
2 7 5
3 0 5
5 0 3
6
7
8
9
Then,
0 5 3
1 6 4
2 7 5
3 0 5
4 1 6 --------now we have to fill in the this space. after 0 count up to 1, and after 5 count up to 6
5 0 3
6
7
8
9
Then this,
0 5 3
1 6 4
2 7 5
3 0 5
4 1 6
5 0 3
6 1 ---fill in the remaining digits counting up, always count up with one exception. I'll show the exception later.
7 2 after 0, count up to 1-2-3-4.
8 3
9 4
Then this,
0 5 3
1 6 4
2 7 5
3 0 5
4 1 6
5 0 3
6 1 4 ---and finally the last column. count up from 3, which is 4-5-6-7
7 2 5
8 3 6
9 4 7
Okay, here's the exception. Example, 297 last night.
0 ----since there is nothing to count up from, we have to count down from the 9 and 7 to fill in those two first slots.
1 Count down from 9 is 8-7 (LOOK BELOW)
2 9 7
3
4
5
6
7
8
9
0 7 5 ----since there is nothing to count up from, we have to count down from the 9 and 7 to fill in those
1 8 two first slots
2 9 7 Count down from 9 is 8-7....and then the 7, count down,6-5
3
4
5
6
7
8
9
toledo ohio
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Member #169147
October 4, 2015
70 Posts
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Posted: April 4, 2016, 4:09 pm - IP Logged
If the last draw was 617 how do you know how to place each version of 617 within the rundown?
United States
Member #164727
March 12, 2015
2675 Posts
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Posted: April 4, 2016, 4:21 pm - IP Logged
If the last draw was 617 how do you know how to place each version of 617 within the rundown?
That's a good question. Maybe do a reverse of 617. I have to look further to see which works better.
617
0
1 6 7---or 176
2
3
4
5
6 1 7---original number
7 1 6---reverse of 617
8
9
toledo ohio
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Posted: April 4, 2016, 4:25 pm - IP Logged
I notice that the middle digit was a 1 and you kept it that way is that something you do? i aslo notice it in other examples
United States
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Posted: April 4, 2016, 4:28 pm - IP Logged
I notice that the middle digit was a 1 and you kept it that way is that something you do? i aslo notice it in other examples
Nah, just a coincidence.
toledo ohio
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Posted: April 4, 2016, 4:32 pm - IP Logged
in the previous workouts the middle digit stays the same and it shifts to first position
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Posted: April 4, 2016, 5:37 pm - IP Logged
Amber123-
Is there any kind of reason that you think this would work?
toledo ohio
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Posted: April 4, 2016, 5:54 pm - IP Logged
Amber123-
Is there any kind of reason that you think this would work?
here we go
Page 1 of 5 | 4,785 | 12,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-04 | longest | en | 0.890844 |
https://bafybeiemxf5abjwjbikoz4mc3a3dla6ual3jsgpdr4cjr3oz3evfyavhwq.ipfs.dweb.link/wiki/Nagell%E2%80%93Lutz_theorem.html | 1,660,944,228,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573760.75/warc/CC-MAIN-20220819191655-20220819221655-00450.warc.gz | 142,868,669 | 3,980 | # Nagell–Lutz theorem
In mathematics, the Nagell–Lutz theorem is a result in the diophantine geometry of elliptic curves, which describes rational torsion points on elliptic curves over the integers.
## Definition of the terms
Suppose that the equation
defines a non-singular cubic curve with integer coefficients a, b, c, and let D be the discriminant of the cubic polynomial on the right side:
## Statement of the theorem
If P = (x,y) is a rational point of finite order on C, for the elliptic curve group law, then:
• 1) x and y are integers
• 2) either y = 0, in which case P has order two, or else y divides D, which immediately implies that y2 divides D.
## Generalizations
The Nagell–Lutz theorem generalizes to arbitrary number fields and more general cubic equations.[1] For curves over the rationals, the generalization says that, for a nonsingular cubic curve whose Weierstrass form
has integer coefficients, any rational point P=(x,y) of finite order must have integer coordinates, or else have order 2 and coordinates of the form x=m/4, y=n/8, for m and n integers.
## History
The result is named for its two independent discoverers, the Norwegian Trygve Nagell (1895–1988) who published it in 1935, and Élisabeth Lutz (1937). | 311 | 1,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-33 | latest | en | 0.887498 |
https://docs.sympy.org/1.1/modules/geometry/curves.html | 1,550,810,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247513222.88/warc/CC-MAIN-20190222033812-20190222055812-00135.warc.gz | 530,414,563 | 5,001 | # Curves¶
class sympy.geometry.curve.Curve[source]
A curve in space.
A curve is defined by parametric functions for the coordinates, a parameter and the lower and upper bounds for the parameter value.
Parameters: function : list of functions limits : 3-tuple Function parameter and lower and upper bounds. ValueError When $$functions$$ are specified incorrectly. When $$limits$$ are specified incorrectly.
Examples
>>> from sympy import sin, cos, Symbol, interpolate
>>> from sympy.abc import t, a
>>> from sympy.geometry import Curve
>>> C = Curve((sin(t), cos(t)), (t, 0, 2))
>>> C.functions
(sin(t), cos(t))
>>> C.limits
(t, 0, 2)
>>> C.parameter
t
>>> C = Curve((t, interpolate([1, 4, 9, 16], t)), (t, 0, 1)); C
Curve((t, t**2), (t, 0, 1))
>>> C.subs(t, 4)
Point2D(4, 16)
>>> C.arbitrary_point(a)
Point2D(a, a**2)
Attributes
functions parameter limits
arbitrary_point(parameter='t')[source]
A parameterized point on the curve.
Parameters: parameter : str or Symbol, optional Default value is ‘t’; the Curve’s parameter is selected with None or self.parameter otherwise the provided symbol is used. arbitrary_point : Point ValueError When $$parameter$$ already appears in the functions.
Examples
>>> from sympy import Symbol
>>> from sympy.abc import s
>>> from sympy.geometry import Curve
>>> C = Curve([2*s, s**2], (s, 0, 2))
>>> C.arbitrary_point()
Point2D(2*t, t**2)
>>> C.arbitrary_point(C.parameter)
Point2D(2*s, s**2)
>>> C.arbitrary_point(None)
Point2D(2*s, s**2)
>>> C.arbitrary_point(Symbol('a'))
Point2D(2*a, a**2)
free_symbols
Return a set of symbols other than the bound symbols used to parametrically define the Curve.
Examples
>>> from sympy.abc import t, a
>>> from sympy.geometry import Curve
>>> Curve((t, t**2), (t, 0, 2)).free_symbols
set()
>>> Curve((t, t**2), (t, a, 2)).free_symbols
{a}
functions
The functions specifying the curve.
Returns: functions : list of parameterized coordinate functions.
Examples
>>> from sympy.abc import t
>>> from sympy.geometry import Curve
>>> C = Curve((t, t**2), (t, 0, 2))
>>> C.functions
(t, t**2)
limits
The limits for the curve.
Returns: limits : tuple Contains parameter and lower and upper limits.
Examples
>>> from sympy.abc import t
>>> from sympy.geometry import Curve
>>> C = Curve([t, t**3], (t, -2, 2))
>>> C.limits
(t, -2, 2)
parameter
The curve function variable.
Returns: parameter : SymPy symbol
Examples
>>> from sympy.abc import t
>>> from sympy.geometry import Curve
>>> C = Curve([t, t**2], (t, 0, 2))
>>> C.parameter
t
plot_interval(parameter='t')[source]
The plot interval for the default geometric plot of the curve.
Parameters: parameter : str or Symbol, optional Default value is ‘t’; otherwise the provided symbol is used. plot_interval : list (plot interval) [parameter, lower_bound, upper_bound]
limits
Returns limits of the parameter interval
Examples
>>> from sympy import Curve, sin
>>> from sympy.abc import x, t, s
>>> Curve((x, sin(x)), (x, 1, 2)).plot_interval()
[t, 1, 2]
>>> Curve((x, sin(x)), (x, 1, 2)).plot_interval(s)
[s, 1, 2]
rotate(angle=0, pt=None)[source]
Rotate angle radians counterclockwise about Point pt.
The default pt is the origin, Point(0, 0).
Examples
>>> from sympy.geometry.curve import Curve
>>> from sympy.abc import x
>>> from sympy import pi
>>> Curve((x, x), (x, 0, 1)).rotate(pi/2)
Curve((-x, x), (x, 0, 1))
scale(x=1, y=1, pt=None)[source]
Override GeometryEntity.scale since Curve is not made up of Points.
Examples
>>> from sympy.geometry.curve import Curve
>>> from sympy import pi
>>> from sympy.abc import x
>>> Curve((x, x), (x, 0, 1)).scale(2)
Curve((2*x, x), (x, 0, 1))
translate(x=0, y=0)[source]
Translate the Curve by (x, y).
Examples
>>> from sympy.geometry.curve import Curve
>>> from sympy import pi
>>> from sympy.abc import x
>>> Curve((x, x), (x, 0, 1)).translate(1, 2)
Curve((x + 1, x + 2), (x, 0, 1)) | 1,162 | 3,896 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-09 | latest | en | 0.449813 |
http://www.fool.com/dripport/2000/dripport000224.htm | 1,496,117,982,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00243.warc.gz | 620,275,559 | 21,903 | Fool.com: Fun With Numbers [Drip] February 24, 2000
DRIP PORTFOLIO
Fun With Numbers
Back to Basics, Part 8
Format for printing Reuse/Reprint
By Vince Hanks (TMFElwood)
February 24, 2000
We've managed to plow our way through the entire balance sheet, defining and examining each element as it appears. It's been a long journey and now comes the reward: Math! There are a few formulaic tools we can use to assess a company's strength in asset management and measure its power of liquidity.
First up, the Current Ratio is a measure of a company's ability to use its current assets to pay off its current liabilities. It's simply the current assets divided by current liabilities. For instance, that hip dance club Fischer's Fandango Fortress has current assets of \$25 million and \$17 million in current liabilities, it's current ratio would be:
``` \$25 million
Current Ratio = ------------- = 1.47
\$17 million
```
Not bad, but it's about the cutoff of where we'd normally like to see it. Generally speaking, a current ratio of 1.5 or higher is reasonably sufficient to meet operational needs. Naturally, the higher the ratio, the greater the financial strength of the company. However, too high a current ratio may suggest that the company isn't optimizing its use of current assets to run and grow its business.
Look at the trend of the current ratio from quarter to quarter to determine if its financial position is maintaining or declining. It's also important to look at the ratios of the company's peers within an industry. Some sectors will have comparatively lower or higher norms than others.
The Quick Ratio is also a measure of a company's ability to cover its short-term liabilities. This ratio, however, is particularly focused on the liquidity of current assets. Since inventories are not always readily salable and are often sold at a deep discount to the amount recorded on the balance sheet, the quick ratio eliminates inventories from the equation. To determine a company's quick ratio, take current assets minus inventories, divided by current liabilities.
If we look back at Fischer's Fandango Fortress, we notice that a large percentage of its current assets are in the form of Citrus Zima, which has been piling up in the basement. We'll remove the snappy citrus beverage and other inventories from current assets in order to determine the quick ratio:
``` (Current Assets - Inventories)
Quick Ratio = ------------------------------ =
Current Liabilities
\$25 mm - \$10 mm
= -------------------------- = 0.88
\$17 mm
```
Fischer's may not be as ready to meet cash demands as we previously thought. We generally want to see a quick ratio of 1.0 or higher, indicating that a company has enough relatively liquid assets to pay off its current liabilities. Our lively Spanish dance club could run into trouble as it's forced to sell Zima for pennies a bottle and doesn't have enough liquid assets to pay the bills.
As with the current ratio (and pretty much any ratio), you'll want to compare the quick ratio with those of industry peers and also focus on its trend over time.
The final measure of liquidity is what is known as Working Capital. This is the fuel in the gastank of a company; what makes it zoom down the autobahn. Working capital is simply current assets minus current liabilities. This is the net amount the company has to funds its operations and provide growth.
Working capital is essential for the success of any business. If a company has abundant working capital, it has the ability to pay for everything it needs to and then some. If, however, its working capital is negative, it will lack the ability to spend as it desires, as well as meet obligations.
An interesting measure of value is a company's current working capital relative to its market capitalization. If you divide working capital by market capitalization, you can determine how much of the company is actually backed by working capital. Let's look at the working capital of Fischer's Fandango Fortress:
```Current Assets - Current Liabilities = Working Capital
\$25 mm - \$17 mm = \$8 mm
```
Now, if we divide that by Fischer's market cap of \$20 million, we get a working capital to market cap. ratio of 0.4 (8 / 20 = 0.4). 40% of the dance club's market capitalization is backed by working capital. Not terrible, but it could be better. Ratios of 50% or higher indicate a company that's in pretty good shape. Compare Fischer's to other dance clubs and see if it has more or less cash backing up its operations than its peers.
These are three useful tools to use when examining a company and comparing it with its peers. Keep those nuclear-powered calculators handy, as next week we'll look at a few more and close out the series on the balance sheet.
Drip on, Fools!
Drip Portfolio
Ticker Company Dly Pr Chg Price 2/24/00 Closing Numbers CPB Campbell Soup -1 13/32 \$27.47 INTC Intel Corp 5 3/16 \$114.25 JNJ Johnson & Johnson -3 31/32 \$72.16 MEL Mellon Financial -1 1/2 \$28.44
Day Week Month Year
To Date
Since
7/28/97
Annualized
Drip .30% 1.93% -.58% 4.94% 36.34% 12.77%
S&P 500 -.53% .55% -2.94% -7.88% 44.17% 15.24%
S&P 500(DA) -.53% .55% -2.94% -7.88% 46.79% 16.05%
S&P 500(DCA) n/a n/a n/a n/a 19.91% 7.29%
NASDAQ 1.48% 4.67% 17.19% 13.48% 194.20% 51.95%
Trade Date # Shares Ticker Cost/Share Price LT % Val Chg
9/8/9722.9799INTC45.635\$114.25150.36%
11/14/9711.811JNJ80.721\$72.16-10.61%
11/5/9831.3788MEL34.307\$28.44-17.11%
4/13/988.269CPB54.401\$27.47-49.51%
Trade Date # Shares Ticker Cost Value LT \$ Val Ch
9/8/9722.9799INTC\$1,048.68\$2,625.45\$1,576.77
11/14/9711.811JNJ\$953.39\$852.24(\$101.16)
11/5/9831.3788MEL\$1,076.51\$892.33(\$184.18)
4/13/988.269CPB\$449.84\$227.14(\$222.70)
Cash: \$24.41
Total: \$4,621.57
Key • S&P 500 (DA) = dividend adjusted. Dividends have been added to the total return of the index. Note Drip Port launched with \$500 on July 28, 1997, adds \$100 to invest every month, and the goal is to own \$150,000 in stock by August of the year 2017. Due to the slow nature of dollar-cost-averaging and our relatively significant starting costs, we do not expect to seriously challenge the S&P 500 for the first three to five years as we build an investment base. The long-term advantages of dollar-cost-averaging still overcome the short-term disadvantages, however. Final note: our investment in Campbell Soup is frozen due to fees instituted in its investment plan. Click here for a history of all Drip Port transactions. | 1,695 | 6,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-22 | latest | en | 0.967299 |
http://wiki.openstreetmap.org/wiki/Talk:Lanes | 1,513,442,477,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588294.67/warc/CC-MAIN-20171216162441-20171216184441-00481.warc.gz | 299,781,088 | 11,032 | # Talk:Lanes
## Node features
Can this be applied to tagged nodes on a highway, for example traffic lights which are only applicable to a subset of the lanes? --achadwick 13:31, 23 May 2012 (BST)
Hmm... I would say yes, if and only if the node is not an intersection of ways. However this was never discussed. --Imagic 15:17, 23 May 2012 (BST)
## Generalities-with-exceptions pattern?
It seems most natural to use the lanes scheme as a way of marking very occasional lane-specific exceptions to a general rule applicable to all of a highway. Hypothetically;
lanes:forward=3
maxspeed=50 mph
maxspeed:lanes:forward=40 mph||
In this case, the undefined lane-values should inherit the general maxspeed=50 mph, I think. Is that the intent? If so, the main page should clarify it. --achadwick 13:41, 23 May 2012 (BST)
Yes, this was the idea. The accepted proposal contained a section about default values, but obviously this information got lost.--Martinq 14:44, 23 May 2012 (BST)
It seems I have to update the article... this was indeed lost. I'll try to update the article within the next days. --Imagic 15:19, 23 May 2012 (BST)
Better now? --Imagic 10:09, 24 May 2012 (BST)
## Better explain how to add multiple values for turn lanes?
Could you please explain how adding multiple values for turn lanes work? Say I have two (forward) turn lanes, lane 1 with allowed direction: left, slight_left, straight; lane2 with straight, slight_right, right. how would i tag this? thanks, jose
You separate each lane-dependent value with | and if one lane-value needs to contain more than one value you separate them with ; . Your example would be; turn:lanes=through;left;slight_left|through;right;slight_right . Hope this helps. --Imagic 06:35, 29 November 2012 (UTC)
Thank you. Damn, I was always seeing two pairs of X|X, instead of one X, one X+Y and one Y. Also, please what is the difference between (turn) straight and through? Or are they synonymous (at least for one direction ways)?
You're welcome. Regarding straight vs. through: the value straight is not specified, only through is. So please only use through. I used the value straight right at the beginning when I started the :lanes-proposal, but dropped it in favour of through after some discussions. BTW: do you know the JOSM style Styles_Lane_features-style.mapcss? It's really great to verify your edits. It also started to support the proposed key change however only the positive values (like left, right) and not the proposed negative values (like not_right, not_left). Give it a try! If you intend to use the key change:lanes=* let me know and I can provide you with a patched version of the JOSM style which supports the proposed negative values. --Imagic 08:03, 29 November 2012 (UTC)
Thanks again. No need for allowed/forbidden values for now and that CSS for JOSM seems very useful!
## "Turn lanes as relation" proposal made obsolete?
Did I understand it correctly that now Relations/Proposed/turn_lanes proposal is obsolete?
I guess you have to ask the creator of that proposal. Info from taginfo: turn:lanes is today (29.11.2012) used about 7600 times and was proposed February 2012. The relation is used about 6600 times and was proposed March 2011. --Imagic 08:08, 29 November 2012 (UTC)
## More real-world examples
I think we need more real world examples, I started tagging a nice (but not too complex) junction with detailed bing coverage here -- could you possibly take a look at it and provide feedback? Also, I see you're very familiar with the subject so maybe you'd wanna join the discussion with MapFactor devs about the possibility of adding turn-lanes display to Navigator Free software. This could be the very first navigation software that could make use of the lanes info.
I started to create some motorway examples based on aerial images here. Currently I'm concentrating on motorway tagging, because 1) information about turning lanes is very important there (imagine you missed an exit on a motorway) and 2) tagging of junctions is still highly disputed.
But I also thought about junctions like yours and created some time ago a proposal that should make things a lot easier regarding turning restrictions. But as I can see you already mapped according to it - just without the highway=junction-area - so no news for you in there! Still may you have a look at the proposal of highway=junction and comment on it?
Regarding your tagging of the junction you provided: looks quite fine, but contains some errors. As this is pretty hard to explain here maybe you could contact me directly?
Regarding Navigator Free: another forum? Another login? I think I already have enough of this ;-) But regarding your last comment there: turn:lanes is not equal turn:lanes:forward - only if oneway=yes! And regarding the connectivity relation: I'm planning to write a proposal for it in the next weeks. Any feedback is very welcome. --Imagic 11:50, 29 November 2012 (UTC)
## Define 'forwards'
I'm having a bit of trouble with the tag lanes:forward and lanes:backward. With a two-way road, which way is "forwards"? For example, a set of traffic lights where the side approaching the intersection is divided into two lanes, one for turning left and the other for straight through/turn right. The side of the road leaving the intersection is a single lane. So do i put lanes=3 and turn:lanes=left|through;right|through with lanes:forward=2 and lanes:backward=1 ?
Thalass (talk) 05:08, 24 September 2013 (UTC)
The :forward and :backward key suffixes always refer to the direction of the osm data entity / way. It's explained on Forward & backward, left & right. The ways consist of an ordered list of nodes. If the way has been drawn towards the intersection, then your example would be lanes=3 lanes:forward=2 lanes:backward=1 turn:lanes:forward=left. If the way was draw in the other direction, you flip the :backward and :forward suffixes. Alv (talk) 06:06, 24 September 2013 (UTC)
Hi! It has been a little while since lanes (and turns) have been introduced. But I do wonder, is it adopted by any end-user application thus far, eg. navigation software? -- MrManny (talk) 09:43, 10 June 2014 (UTC)
OsmAnd uses the number of lanes to display the preferred lanes to drive on (see http://www.appsapk.com/wp-content/uploads/2013/09/OsmAnd-Maps-Navigation-1.jpg as an example). I don't think it uses turn:lanes due to its low adoption, but it counts the number of lanes from the way segment before and the segments after a highway split.--Sanderd17
you are right. The corresponding issue is http://code.google.com/p/osmand/issues/detail?id=1448 --Zuse (talk) 13:17, 29 August 2014 (UTC)
## Crossing with a designated lane for bicycles
Shouldn't this be
bicycle:lanes=yes|use_sidepath|designated|yes
It is not explicitly forbidden to use that lane.
I guess this depends on the country and other circumstances. Access tags should be discussed on the appropriate wiki page. Thanks. --Imagic (talk) 07:41, 22 December 2014 (UTC)
## Crossing with a designated lane for bicycles (second issue)
The page says
``` lanes=3
turn:lanes=left|through|through|right
vehicle:lanes=yes|yes|no|yes
bicycle:lanes=yes|no|designated|yes
```
This is in contradiction with the Vehicle types page. According to that page bicycle is a vehicle and automobile is the general term for all motorised vehicles. Hence this example should be:
...
``` automobile:lanes=yes|yes|no|yes
bicycle:lanes=yes|no|designated|yes
```
--voschix (talk) 17:44, 30 March 2015 (UTC)
That's not a contradiction. The tag bicycle is more specific than vehicle, so first we forbid all kind of vehicles on the third lane and afterwards - because "bicycle" is more specific - we specify the third lane as designated to bicycles. In fact you have to use "vehicle" and not "automobile", because otherwise all kind of non-motorized vehicles would be allowed on the third lane, which is not the case. And last but not least: I have never seen the tag automobile used anywhere and a quick look on taginfo shows that I am not alone ;-) --Imagic (talk) 07:34, 31 March 2015 (UTC)
### Another issue
Excluding bicycle lanes but not HOV, bus lanes or turn lanes breaks lane guidance. Bicycle facilities often have more than one lane and it's increasingly common to not have bicycle lanes as the rightmost or leftmost lane. Not being able to include these with lane tagging explicitly breaks accurate description of the lane layout. If width of the lane is a concern, tag for lane widths, don't exclude bicycle lanes. Paul Johnson (talk) 15:09, 25 September 2017 (UTC)
You are right, this is an unsolved issue, if the bicycle lane is not part of a vehicle lane and of significant width like shown in the example. Cause to change an existing tag is no prefered solution, I guess we need a new picture for small width bicycle lanes, as part of vehicle lanes, and a new example for separated bicycle lanes with full vehicle width. Have you ever thought about discussing this at tagging mailing list ? --Robybully (talk) 17:34, 25 September 2017 (UTC)
Ran into the chicken-egg problem there. Nobody wanted to work out the problem because nobody's tagging for completeness on lanes. Nobody's tagging for completeness on lanes because the wiki says specifically not to do so. Paul Johnson (talk) 18:25, 25 September 2017 (UTC)
## Time of day
How can we tag turn lanes and lanes counts for each direction which change based on the time of day? Aharvey (talk) 12:59, 26 June 2017 (UTC)
Using Conditional restrictions should work for that, i.e. using the lanes:[forward/backward:]conditional and turn:lanes:[forward/backward:]conditional keys. --Tordanik 15:47, 27 June 2017 (UTC)
## Where to split
Where do we split a way when a lane is added? Sometimes before an intersection a turn lane is added on the side, but it doesn't get lane markings right away (although there might be a sign on the side of the road indicating which direction each lane goes). With respect to the lanes suffix, do we add the lane as soon as it is full-width, or do we wait until there are road markings separating it? And if we add it right away, is there a way to mark that the lane markings do not start right away? Germyb (talk) 03:03, 30 September 2017 (UTC) | 2,522 | 10,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-51 | latest | en | 0.922837 |
https://www.convertunits.com/from/nail/to/sagene | 1,638,069,015,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00426.warc.gz | 791,810,402 | 12,811 | ## ››Convert nail to sagene
nail sagene
How many nail in 1 sagene? The answer is 37.333333333333.
We assume you are converting between nail and sagene.
You can view more details on each measurement unit:
nail or sagene
The SI base unit for length is the metre.
1 metre is equal to 17.497812773403 nail, or 0.4686914135733 sagene.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between nail and sagene.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of nail to sagene
1 nail to sagene = 0.02679 sagene
10 nail to sagene = 0.26786 sagene
20 nail to sagene = 0.53571 sagene
30 nail to sagene = 0.80357 sagene
40 nail to sagene = 1.07143 sagene
50 nail to sagene = 1.33929 sagene
100 nail to sagene = 2.67857 sagene
200 nail to sagene = 5.35714 sagene
## ››Want other units?
You can do the reverse unit conversion from sagene to nail, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Nail
old measure of two and a quarter inches
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 430 | 1,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-49 | latest | en | 0.808342 |
https://bmcmedresmethodol.biomedcentral.com/articles/10.1186/1471-2288-6-6/tables/4 | 1,582,480,760,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00124.warc.gz | 297,709,051 | 15,274 | Springer Nature is making SARS-CoV-2 and COVID-19 research free. View research | View latest news | Sign up for updates
# Table 4 Comparison of number of dialling, calling time and cost for an assumed electronic white pages sample to achieve the same precision of estimates from an assumed effective sample size of 1,000 (equivalent of a simple random sample size 1,000)
Average sample (design effect ratio = 1.19) High psychological distress (design effect ratio = 1.32)
LA-RDD n = 2,000 EWP n = 2,380 LA-RDD n = 1,960 EWP n = 2,600
Number of dialling 29,800 23,562 29,204 25,740
Costed dialling * 16,600 15,708 16,268 17,160
Calling time (in hours) 1,180 1,249.5 1,156.4 1,365
Cost† \$26,920 \$28,132 \$26,382 \$30,732
1. * excluding dialling of not connected numbers and dialling with no answer
2. † The costs is calculated as: A\$0.20 × number of costed dialling + A\$20 × number of calling hours.
3. Source: New South Wales Health Survey Program, Centre for Epidemiology and Research, New South Wales Department of Health. | 307 | 1,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-10 | latest | en | 0.825855 |
https://stackoverflow.com/questions/34913005/color-space-mapping-ycbcr-to-rgb/34913974 | 1,571,376,837,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00346.warc.gz | 716,814,825 | 30,040 | # Color Space Mapping YCbCr to RGB
I am experimenting with JPEG compression using python. I load in a tiff image and store it as numpy uint8 RGB array. I was doing this for color mapping.
``````def rgb2ycbcr(im):
cbcr = np.empty_like(im)
r = im[:,:,0]
g = im[:,:,1]
b = im[:,:,2]
# Y
cbcr[:,:,0] = .299 * r + .587 * g + .114 * b
# Cb
cbcr[:,:,1] = 128 - .169 * r - .331 * g + .5 * b
# Cr
cbcr[:,:,2] = 128 + .5 * r - .419 * g - .081 * b
return np.uint8(cbcr)
def ycbcr2rgb(im):
rgb = np.empty_like(im)
y = im[:,:,0]
cb = im[:,:,1] - 128
cr = im[:,:,2] - 128
# R
rgb[:,:,0] = y + 1.402 * cr
# G
rgb[:,:,1] = y - .34414 * cb - .71414 * cr
# B
rgb[:,:,2] = y + 1.772 * cb
return np.uint8(rgb)
``````
I did a simple RGB to YCbCr transformation followed with a inverse transformation.
``````img = rgb2ycbcr(img)
imshow(img)
img = ycbcr2rgb(img)
imshow(img)
``````
I got these two output image as YCbCr and RGB output after the color space transformation.
It seems that something is wrong with my color conversion and I cannot figure out what is wrong. I was using the JPEG color space conversion provided by Wikipedia. Thanks you for the help.
You have to do your intermediate calculations in floating point. The posterization should tip you off; you have a lot of "hot" (saturated) pixels.
``````def rgb2ycbcr(im):
xform = np.array([[.299, .587, .114], [-.1687, -.3313, .5], [.5, -.4187, -.0813]])
ycbcr = im.dot(xform.T)
ycbcr[:,:,[1,2]] += 128
return np.uint8(ycbcr)
def ycbcr2rgb(im):
xform = np.array([[1, 0, 1.402], [1, -0.34414, -.71414], [1, 1.772, 0]])
rgb = im.astype(np.float)
rgb[:,:,[1,2]] -= 128
rgb = rgb.dot(xform.T)
np.putmask(rgb, rgb > 255, 255)
np.putmask(rgb, rgb < 0, 0)
return np.uint8(rgb)
``````
• question i use the above to convert a color png image the format is on (1000, 800, 3). But then the rgb2ycbcr return shape [1000, 3 ] instead of [1000,800]. I am doubting what wrong I am doing because the answer is marked as correct. – partizanos May 25 at 13:20 | 685 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.728328 |
https://www.icserankers.com/2020/12/icse-solutions-for-chapter8-remainder-and-factor-theorem-class10-maths.html | 1,726,451,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00131.warc.gz | 756,201,466 | 46,864 | # ICSE Solutions for Chapter 8 Remainder and Factor Theorem Class 10 Mathematics
Question 1: Show that (x - 1) is a factor of x3 - 7x2 + 14x - 8. Hence, completely factorize the given expression.
Solution 1: Let f(x) = x3 - 7x+ 14x - 8
f(1) = (1)3 - 7(1)2 + 14(1) - 8 = 1 - 7 + 14 - 8 = 0
Hence, (x - 1) is a factor of f(x).
∴ x3 - 7x+ 14x - 8 = (x - 1)(x- 6x + 8)
= (x - 1)(x- 2x - 4x + 8)
= (x - 1)[x(x - 2) – 4(x - 2)]
= (x - 1)(x - 2)(x - 4)
Question 2: Using Remainder Theorem, factorise: x3 + 10x2 - 37x + 26 completely.
Solution 2: By Remainder Theorem,
For x = 1, the value of the given expression is the remainder.
x3 + 10x- 37x + 26
= (1)3 + 10(1)2 - 37(1) + 26
= 1 + 10 - 37 + 26
= 37 – 37
= 0
⇒ x - 1 is a factor of x3 + 10x– 37x + 26.
∴ x3 + 10x- 37x + 26 = (x - 1)(x+ 11x - 26)
= (x – 1)(x+ 13x - 2x - 26)
= (x - 1)[x(x + 13) - 2(x + 13)]
∴ x3 + 10x- 37x + 26 = (x - 1)(x + 13)(x - 2)
Question 3: Using the remainder Theorem, factorise the expression 3x3 + 10x2 + x - 6. Hence, solve the equation 3x3 + 10x2 + x - 6 = 0.
Solution 3: Let f(x) = 3x3 + 10x2 + x - 6
For x = -1
f(x) = f(-1) = 3(-1)3 + 10(-1)2 + (-1) - 6 = -3 + 10 – 1 - 6 = 0
Hence, (x + 1) is a factor of f(x).
∴ 3x3 + 10x2 + x – 6 = (x + 1)/(3x2 + 7x - 6)
= (x + 1)/(3x2 + 9x - 2x - 6)
= (x + 1)[3x(x + 3) – 2(x + 3)]
= (x + 1)(x + 3)(3x - 2)
Now, 3x3 + 10x2 + x - 6 = 0
⇒ (x + 1)(x + 3)(3x - 2) = 0
⇒ x = -1, -3, 2/3
Question 4: (i) If 2x + 1 is a factor of 2x2 + ax - 3, find the value of a
(ii) Find the value of k, if 3x - 4 is a factor of expression 3x2+ 2x - k.
Solution 4: (i) 2x + 1 is a factor of f(x) = 2x2 + ax - 3.
∴ f(-1/2) = 0
⇒ 2(-1/2)2 + a(-1/2) – 3 = 0
⇒1/2 – a/2 = 3
⇒ 1 – a = 6
⇒ a = -5
(ii) 3x – 4 is a factor of g(x) = 3x2 + 2x –k.
∴ f(4/3) = 0
⇒ 3(4/3)2 + 2(4/3) – k =0
⇒ 16/3 + 8/3 – k = 0
⇒ 24/3 = k
⇒ k = 8
Question 5: Given that x - 2 and x + 1 are factors of f(x) = x3 + 3x2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
Solution 5: f(x) = x3 + 3x2 + ax + b
Since, (x - 2) is a factor of f(x), f(2) = 0
⇒ (2)3 + 3(2)2 + a(2) + b = 0
⇒ 8 + 12 + 2a + b = 0
⇒ 2a + b + 20 = 0 ...(i)
Since, (x + 1) is a factor of f(x), f(-1) = 0
⇒ (-1)3 + 3(-1)2 + a(-1) + b = 0
⇒ -1 + 3 – a + b = 0
⇒ -a + b + 2 = 0 ...(ii)
Subtracting (ii) from (i), we get,
3a + 18 = 0
⇒ a = - 6
Substituting the value of a in (ii), we get,
b = a – 2 = - 6 - 2 = - 8
∴ f(x) = x3 + 3x2 – 6x – 8
Now, for x = - 1
f(x) = f(-1) = (-1)3 + 3(-1)2 – 6(-1) - 8 = -1 + 3 + 6 - 8 = 0
Hence, (x + 1) is a factor of f(x).
∴ x3 + 3x2 – 6x – 8 = (x + 1)(x2 + 2x – 8)
= (x + 1)(x2 + 4x – 2x – 8)
= (x + 1)[x(x + 4) – 2(x + 4)]
= (x + 1)(x + 4)(x – 2)
Question 6: If x - 2 is a factor of x2 + ax + b and a + b = 1, find the values of a and b.
Solution 6: Let f(x) = x2 + ax + b
Since. (x - 2) is a factor of f(x).
Remainder = f(2)= 0
(2)2 + a(2) + b = 0
4 + 2a + b = 0
2a + b = - 4 …(i)
It is given that:
a + b = 1 ...(ii)
Subtracting (ii) from (ii), we get,
a = -5
Substituting the value of a in (ii), we get,
b = 1 - (-5) = 6
Question 7: Factorise x3 + 6x2 + 11x + 6 completely using factor theorem
Solution 7: Let f(x) = x3 + 6x2 + 11x + 6
For x = -1
f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
= -1 + 6 - 11+ 6 = 12 - 12 = 0
Hence, (x + 1) is a factor of f(x).
∴ x3 + 6x2 + 11x + 6 = (x + 1)(x2 + 5x + 6)
= (x + 1)(x2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)]
= (x + 1)(x + 2)(x + 3)
Question 8: The polynomials ax3 + 3x2- 3 and 2x3 - 5x + a, when divided by x - 4, leave the same remainder in each case. Find the value of a.
Solution 8: Let f(x) = ax + 3x2 - 3
When f(x) is divided by (x - 4), remainder = f(4)
f(4) = a(4)3 + 3(4)2 - 3 = 64a + 45
Let g(x) = 2x3 - 5x + a
When g(x) is divided by (x - 4), remainder = g(4)
g(4) = 2(4)3 - 5(4) + a = a + 108
It is given that f(4) = g(4)
64a + 45 = a + 108
63a = 63
a = 1
Question 9: Find the value of 'a', if (x - a) is a factor of x3 - ax2 + x + 2.
Solution 9: Let f(x) = x2 - ax2 + x + 2
It is given that (x - a) is a factor of f(x).
Remainder = f(a) = 0
a3 - a3 + a + 2 = 0
a + 2 = 0
a = -2
Question 10: Find the value of a, if the division of ax3 + 9x2 + 4x - 10 by x + 3 leaves a remainder 5.
Solution 10: Let f(x) = ax3 + 9x2 + 4x - 10
x + 3 = 0 ⇒ x = - 3
On dividing f(x) by x + 3, it leaves a remainder 5.
∴ f(-3) = 5
a(-3)3 + 9(-3)2 + 4(-3) - 10 = 5
-27a + 81 – 12 - 10 = 5
54 = 27a
a = 2 | 2,453 | 4,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-38 | latest | en | 0.596662 |
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#1
03-24-2012, 11:25 PM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,478
Exercises and Problems
Please comment on the chapter problems in terms of difficulty, clarity, and time demands. This information will help us and other instructors in choosing problems to assign in our classes.
Also, please comment on the exercises in terms of how useful they are in understanding the material.
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Where everyone thinks alike, no one thinks very much
#2
07-08-2014, 01:56 PM
jhmiller@tricity.wsu.edu Junior Member Join Date: Jun 2014 Location: Richland WA Posts: 3
Re: Exercises and Problems
How is values of Eout calculated in tables on page 121 that relate to Figure 4.1?
For the table referring to Figure 4.1a, I can see using the formula in Exercise 3.4e on page 88 if the value of sigma were known since d can be inferred from the degree of polynomial fit. I don't see that formula applying to values in the table that relate to Figure 4.1b because sigma is zero.
I can see using the formula for Eout in Exercise 3.4e on page 88 in Exercise 4.2 on page 123. Is this correct?
#3
07-09-2014, 01:58 AM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,478
Re: Exercises and Problems
Quote:
Originally Posted by jhmiller@tricity.wsu.edu How is values of Eout calculated in tables on page 121 that relate to Figure 4.1? For the table referring to Figure 4.1a, I can see using the formula in Exercise 3.4e on page 88 if the value of sigma were known since d can be inferred from the degree of polynomial fit. I don't see that formula applying to values in the table that relate to Figure 4.1b because sigma is zero. I can see using the formula for Eout in Exercise 3.4e on page 88 in Exercise 4.2 on page 123. Is this correct?
Each part of Figure 4.1 illustrates a specific target and a specific training set. The value of in the table is calculated directly as the mean-squared error between this specific and the that resulted from fitting the training data shown. Later, we compute averages over targets and training sets in the more elaborate overfitting experiment.
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Where everyone thinks alike, no one thinks very much
#4
07-09-2014, 05:36 PM
jhmiller@tricity.wsu.edu Junior Member Join Date: Jun 2014 Location: Richland WA Posts: 3
Re: Exercises and Problems
Thanks a million. I assume the mean square error between f and g is restricted to the range of x-values where training data exist. Is this correct?
#5
07-09-2014, 11:26 PM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,478
Re: Exercises and Problems
Quote:
Originally Posted by jhmiller@tricity.wsu.edu Thanks a million. I assume the mean square error between f and g is restricted to the range of x-values where training data exist. Is this correct?
Correct, but the cause and effect is the other way around. The training data lies within the range of x-values where the mean-squared error is computed, namely the input space which is a finite range in this case.
__________________
Where everyone thinks alike, no one thinks very much
#6
08-15-2015, 10:30 AM
prithagupta.nsit Junior Member Join Date: Jun 2015 Posts: 7
Re: Exercises and Problems
For Exercise 4.4
I am not able to understand that in this exercise 4.4, what is actually w and do we have to consider the regularizatio also and if the formulae:
Ein(w) =1/N (Zwlin − y)T * (Zwlin − y)
how can w-wlin come in picture.
by this formula I am able to get the second term but not the second term?
Can anyone help me derive this expression or can anyone share his/her solution with me.
http://book.caltech.edu/bookforum/showthread.php?t=4512
same is the Ein used here??? is it out of sample error... I am not able to understand the conflict. What my understanding is that we have derived wlin but for a variable vector we first check how much does it vary from the wlin and multiplied by Z vector gives us how much function vary from average hypothesis and the second term gives us the error of Wlin predicting outputs. Is my understanding right??
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The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. Abu-Mostafa, Malik Magdon-Ismail, and Hsuan-Tien Lin, and participants in the Learning From Data MOOC by Yaser S. Abu-Mostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity. | 1,454 | 5,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.883758 |
https://www.math-only-math.com/number-system.html | 1,726,578,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00327.warc.gz | 813,747,081 | 13,238 | Number System
In number system modern method of representing numbers symbolically is based on positional notations.
In this method, each number is represented by a string of symbols where each symbol is associated with a specific weight depending upon its positions. The total number of different symbols which are used in a particular number system is called the base or radix of the system and the weight of each position of a particular number is expressed as a power of the base. When a number is formed with the combination of the symbols, each symbol is then called a digit and the position of each symbol is referred to as the digit position.
Thus if a number system has symbols starting from 0, and the digits of the system are 0, 1, 2, ….. (r - 1) then the base or radix is r. If a number D of this system be represented by
D = d₀ d₀ ……. d₀…….. d₁ d
then the magnitude of this number is given by
|D| = dn-1 rn-1 + dn-2 rn-2 + …… di ri + …… d1 r1 + d0 r0
Where each d₀ ranges from 0 to r - 1, such that
0 ≤ d₀ ≤ r - 1, i = 0, 1, 2 ...... (n - 1).
The digit at the extreme left has the highest positional value and is generally called the Most Significant Digit, or in short MSD; similarly, the digit occupying the extreme right position has the least positional value and is referred to as the Least Significant Digit or LSD.
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We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma… | 669 | 2,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.895629 |
http://hotmath.com/help/gt/genericalg1/section_9_1.html | 1,412,106,702,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663135.34/warc/CC-MAIN-20140930004103-00118-ip-10-234-18-248.ec2.internal.warc.gz | 151,848,423 | 8,142 | # Hotmath Practice Problems
Title:
Hotmath Algebra 1
Author:
Hotmath Team
Chapter:Relations and FunctionsSection:Definitions: Relations and Functions
Problem: 1
Draw a mapping diagram and a graph of the following relation.
{(6, 1), (3, 1), (6, –5), (2, 7), (–8, 3)}
Problem: 3
Draw a mapping diagram and a graph of the following relation.
{(5, 0), (5, –2), (2, –1), (4, –1)}
Problem: 5
For the given relation, find the domain and range. Determine whether the relation is a function.
{(a, e), (c, b), (a,d)}
Problem: 7
For the given relation, find the domain and range. Determine whether the relation is a function.
Problem: 9
Is the relation a function? If so, give the domain and the range.
Problem: 11
Is the relation a function? If so, give the domain and the range.
Problem: 13
Is the relation a function? If so, give the domain and the range.
Problem: 15
Check whether the relation is a function. If yes, then give the domain and range.
Problem: 17
Check whether the relation is a function. If yes, then give the domain and range.
Problem: 19
Check whether the relation is a function. If yes, then give the domain and range.
Problem: 21
Does the graph represent y as a function of x? Explain.
Problem: 23
Does the graph represent y as a function of x? Explain.
Problem: 25
Check whether the graph represents a function. Explain your reasoning.
Problem: 27
Check whether the graph in the figure represents a function. Explain your reasoning.
Problem: 29
Check whether the graph in the figure represents a function. Explain your reasoning.
Problem: 31
Check whether the graph in the figure represents a function. Explain your reasoning.
Problem: 33
Check whether the graph in the figure represents a function. Explain your reasoning.
Problem: 35
The relation y = –3x is a set of ordered pairs (x, y) that satisfy the equation y = –3x. Identify whether this relation is a function.
Problem: 37
The relation x = |y| – 3 is a set of ordered pairs (x, y) that satisfy the equation x = |y| – 3. Identify whether this relation is a function.
Problem: 39
Find f(2),f(6), and f(–8) for the given function machine.
Problem: 41
Find h(–1),h(7), and h(12) for the given function machine.
Problem: 43
Evaluate the function at the given points.
h(y) = |y|, find h(–3),h(3), and h(–1).
Problem: 45
Find the indicated function values.
f(x) = |x| – 4, find f(4),f(90), and f(–151).
Problem: 47
Find the indicated function values.
h(x) = x3 – 12, find h(1), h(–2), and h(4).
Problem: 49
Find the value of f(–2) if f(x) = 5x + 3.
Problem: 51
Find the value of g(1/3) if g(x) = x2 – 5x.
Problem: 53
Find the value of g(2.5) if g(x) = x2 – 5x.
Problem: 55
Find the value of 4[ f(–3)] if f(x) = 5x + 3.
Problem: 57
Find the value of g(2b) if g(x) = x2 – 5x.
Problem: 59
Find the value of –4[g(2)] if g(x) = x2 – 5x.
Problem: 61
Find the value of 4[ g(a2)] if g(x) = x2 – 5x.
Problem: 63
A large tank is filled with a certain liquid. The function
gives the pressure in atmospheres of the liquid as a function of d in feet.
Find the pressure exerted on a object submerged (a) 20 feet, (b) 30 feet, (c) 50 feet in the tank. | 959 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | longest | en | 0.780568 |
https://www.techylib.com/el/view/pancakesnightmute/singular_value_decomposition | 1,548,186,801,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583867214.54/warc/CC-MAIN-20190122182019-20190122204019-00196.warc.gz | 949,941,281 | 12,868 | # Singular Value Decomposition
Τεχνίτη Νοημοσύνη και Ρομποτική
5 Νοε 2013 (πριν από 5 χρόνια και 3 μήνες)
293 εμφανίσεις
Singular Value Decomposition
in Digital Signal Processing
By Tin Sheen
Signals
Flow of information
Measured quantity that varies with time (or
position)
Electrical signal received from a transducer
(microphone, thermometer, accelerometer,
antenna, etc.)
Electrical signal that controls a process
SVD background
The Singular Value Decomposition (SVD) of a rectangular matrix
A
is a decomposition of the form
A = U S V
T
U
and
V
are orthogonal matrices
and
S
is a diagonal matrix
The singular vectors form orthonormal bases, and the important
relation
A v
i
= s
i
u
i
shows that each right singular vectors is mapped onto the
corresponding left singular vector, and the "magnification factor" is
the corresponding singular value.
What is Digital Signal Processing?
Digital signal processing
(
DSP
) is the study
of signals in a digital representation and the
processing methods of these signals.
DSP includes subfields like: audio signal
processing, control engineering, digital image
processing and speech processing.
Noise reduction
The SVD has also applications in digital signal processing, e.g., as a
method for noise reduction. The central idea is to let a matrix
A
represent the noisy signal, compute the SVD, and then discard small
singular values of
A
. It can be shown that the small singular values
mainly represent the noise, and thus the rank
-
k
matrix
A
k
represents
a filtered signal with less noise.
SVD’s usefulness
SVD is sufficient and is the most optimal in
given image.
It is packed with energy in a given number of
transformation coefficients is maximized.
Easy to calculate.
Normal vs Noised Image
Block SVD filtering algorithm
Let the original, non
-
corrupted image F be represented as KxL matrix. Adding the noise to the
original F image will produces the noised image G of the same size
G = F + N
Where N = random KxL noise field
G will be divided into bxb matrix image
G
ij
= U
ij
S
ij
V
ij
T
where i = 1,2,3,4,…k; j =1,2,3,4,…,l
Where U
ij
is the left singular vector, S
ij
is the diagonal, and V
ij
is the right singular vector
From here we can calculate the mean value of the rank (rank obtained from G
ij
* S
ij
)
SVD’s calculation
The proposed algorithm has two basic steps: first, the noise variance
(
variance
of a random variable (or somewhat more precisely, of a
probability distribution) is a measure of its statistical dispersion, indicating
how its possible values are spread around the expected value) is
estimated,
And then filtering is performed on singular values and vectors. Noised
image is divided into square blocks of size
bxb
. For each block the singular
value decomposition is performed. In the consequent step, the average sum
of the last
t
singular values is calculated over all image
Noised calculation
Previously calculated SVD of image blocks will now be used for filtering.
First step in filtering is decreasing of noised singular value
s
i jr
for every image block :
ŝ
i jr
= s
i jr
p
1
* n
s
*w(r)
where ŝ
i jr
is filtered singular value,
p
1
is image dependent parameter and
w(r)
is a
weighting function that determines percentage of estimated noise variance to be
subtracted from noised
singular value
s
i jr
. The weighting function used in this work is chosen to be:
Implementation of DSP
Conclusion
Digital signal processing is a stealth technology. It is
the core enabling technology in everything from your
cell phone to the Mars Rover. It goes much further
than just enabling a one
-
time breakthrough product.
It provides ever
-
increasing capability; compare the
performance gains made by dial
-
up modems with
the recent performance gains of DSL and cable
modems. Remarkably, digital signal processing has
become ubiquitous with little fanfare, and most of its
users are not even aware of what it is. | 976 | 3,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.842394 |
http://math.stackexchange.com/questions/266411/lhospitals-rule-question/266419 | 1,462,399,773,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860124045.24/warc/CC-MAIN-20160428161524-00119-ip-10-239-7-51.ec2.internal.warc.gz | 183,196,160 | 20,383 | # L'Hospital's Rule Question.
show that if $x$ is an element of $\mathbb R$ then $$\lim_{n\to\infty} \left(1 + \frac xn\right)^n = e^x$$
(HINT: Take logs and use L'Hospital's Rule)
i'm not too sure how to go about answer this or putting it in the form $\frac{f'(x)}{g'(x)}$ in order to apply L'Hospitals Rule.
so far i've simply taken logs and brought the power in front leaving me with $$n\log \left(1+ \frac xn\right) = x$$
-
I think, I saw something like this before at the site. – Babak S. Dec 28 '12 at 11:08
@BabakSorouh. Yes definitley – Amr Dec 28 '12 at 11:08
@BabakSorouh do either of you remember the question title by any chance? – jill Dec 28 '12 at 11:10
Possible duplicate: math.stackexchange.com/questions/39170/… – Julian Kuelshammer Dec 28 '12 at 11:19
Sometimes this is the definition of $e^x$. What is your definition of $e^x$? – Henning Makholm Dec 28 '12 at 12:05
The ‘$=x$’ is getting ahead of yourself a bit. Let $$L=\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\;,$$ and take the logarithm to get
\begin{align*} \ln L&=\ln\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\\ &=\lim_{n\to\infty}\ln\left(1+\frac{x}n\right)^n\\ &=\lim_{n\to\infty}n\ln\left(1+\frac{x}n\right)\;, \end{align*}
where the interchange of the log and the limit is justified by the fact that the logarithm function is continuous.
This limit is now a so-called $\infty\cdot 0$ indeterminate form, and there is a standard approach to those: move one of the factors into the denominator. In this case we have
$$\ln L=\lim_{n\to\infty}\frac{\ln\left(1+\frac{x}n\right)}{1/n}\;,$$
a limit in which both numerator and denominator approach $0$ as $n\to\infty$. Now you can apply l’Hospital’s rule.
Don’t forget that at this point you’re actually finding $\ln L$, not $L$, so you’ll have to exponentiate to get $L$.
-
Taking $\,n\,$ as a continuous variable:
$$\lim_{n\to\infty}\log\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}n\log\left(1+\frac{x}{n}\right)=\lim_{n\to\infty}\frac{\log\left(1+\frac{x}{n}\right)}{\frac{1}{n}}\stackrel{\text{L'Hospital}}=$$
$$=\lim_{n\to\infty}-\frac{x}{n^2}\frac{\frac{1}{1+\frac{x}{n}}}{-\frac{1}{n^2}}=x\lim_{n\to\infty}\frac{1}{1+\frac{x}{n}}=x$$
Now just apply the exponential function at both ends of the above, remembering this function is a continuous one on the whole real line.
-
@jill: Note that to ... this function is continuous one... above. It is very important for getting the desire result. – Babak S. Dec 28 '12 at 11:17
$$\lim_{n\to\infty} (1 + \frac xn)^n =\lim_{n\to\infty} e^{n\ln(1 + \frac xn)}$$ The limit $$\lim_{n\to\infty} n\ln(1 + \frac xn)=\lim_{n\to\infty} \frac{\ln(1 + \frac xn)}{\frac1n}=\lim_{n\to\infty} \frac{\frac{1}{1 + \frac xn}\frac{-x}{n^2}}{-\frac1{n^2}}=\lim_{n\to\infty} \frac{x}{1 + \frac xn}=x$$ By continuity of $e^x$, $$\lim_{n\to\infty} (1 + \frac xn)^n =\lim_{n\to\infty} e^{n\ln(1 + \frac xn)}=e^x$$
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If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ be as $1^{+\infty}$, which is an indeterminate form, then we have this fact that: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$ Try to verify and then prove it. :)
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Write the limit in the following form $$\lim_{n\to\infty}\frac{\log(1+x/n)}{1/n}=\lim_{n\to\infty}\frac{f(n)}{g(n)}$$ where $f(n)=\log (1+x/n)$ and $g(n)=1/n$ and the limit is in $0/0$ form, so applying L'Hospitals rule we have the above limit is same as $$\lim_{n\to\infty}\frac{f^\prime(n)}{g^\prime(n)}=\lim_{n\to\infty}\frac{1/(1+x/n).(-x/n^2)}{-1/n^2}=\lim_{n\to\infty}\frac{x}{1+x/n}=x$$
- | 1,347 | 3,543 | {"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} | 4.03125 | 4 | CC-MAIN-2016-18 | longest | en | 0.736308 |
https://la.mathworks.com/matlabcentral/cody/problems/65-word-counting-and-indexing/solutions/1812771 | 1,590,947,635,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00377.warc.gz | 422,120,643 | 16,379 | Cody
Problem 65. Word Counting and Indexing
Solution 1812771
Submitted on 11 May 2019 by Gregory
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
sl = {'one two three','two one four zero'}; wtc = {'four','one','three','two','zero'}; silc = {[2 4 3],[4 2 1 5]}; [wt,sil] = word_index(sl); assert(isequal(wt,wtc))
str_index_list = 1×2 cell array {1×3 double} {1×4 double}
2 Pass
sl = {'one two three'}; wtc = {'one','three','two'}; silc = {[1 3 2]}; [wt,sil] = word_index(sl); assert(isequal(wt,wtc))
str_index_list = 1×1 cell array {1×3 double}
3 Pass
sl = {'this little piggy went to market', ... 'and this little piggy stayed home', ... 'this little piggy had roast beef', ... 'and this one studied computer science at piggy university'}; wtc = { ... 'and','at','beef','computer','had','home', ... 'little','market','one','piggy','roast','science', ... 'stayed','studied','this','to','university','went'}; silc = { ... [15 7 10 18 16 8], ... [1 15 7 10 13 6], ... [15 7 10 5 11 3], ... [ 1 15 9 14 4 12 2 10 17]}; [wt,sil] = word_index(sl); assert(isequal(wt,wtc))
str_index_list = 1×4 cell array {1×6 double} {1×6 double} {1×6 double} {1×9 double} | 452 | 1,276 | {"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} | 3.875 | 4 | CC-MAIN-2020-24 | latest | en | 0.54816 |
http://www.numbersaplenty.com/401131 | 1,591,292,008,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00077.warc.gz | 188,196,583 | 3,388 | Search a number
401131 = 3671093
BaseRepresentation
bin1100001111011101011
3202101020201
41201323223
5100314011
612333031
73260323
oct1417353
9671221
10401131
11254415
12174177
13110773
14a6283
157dcc1
hex61eeb
401131 has 4 divisors (see below), whose sum is σ = 402592. Its totient is φ = 399672.
The previous prime is 401119. The next prime is 401161. The reversal of 401131 is 131104.
Adding to 401131 its reverse (131104), we get a palindrome (532235).
It is a happy number.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 401131 - 29 = 400619 is a prime.
It is a super-2 number, since 2×4011312 = 321812158322, which contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (401101) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 180 + ... + 913.
It is an arithmetic number, because the mean of its divisors is an integer number (100648).
2401131 is an apocalyptic number.
401131 is a deficient number, since it is larger than the sum of its proper divisors (1461).
401131 is a wasteful number, since it uses less digits than its factorization.
401131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1460.
The product of its (nonzero) digits is 12, while the sum is 10.
The square root of 401131 is about 633.3490348931. The cubic root of 401131 is about 73.7500086182.
The spelling of 401131 in words is "four hundred one thousand, one hundred thirty-one".
Divisors: 1 367 1093 401131 | 501 | 1,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.888741 |
https://utzx.com/units/convert-63-cm-to-inches/ | 1,720,922,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00425.warc.gz | 518,358,345 | 21,492 | Convert 63 CM to Inches
1 centimeter is how many inches?
Do you are seeking to convert 63 centimeters into the equivalent of inches? first, you obviously should be aware of how many inches 1 cm is equal to.
You may use this conversion from cm to inches to calculate the conversion.
Cm Implication
Centimeters, or Cm are the unit of length measurement in the metric system. Its symbol is cm. The length unit meter has been defined internationally to an “International System of Units”, but the unit cm is not. However, one cm is equals 100 meters. It is also 39.37 in.
Inch
The unit “inch” or “In” is an Anglo-American length unit. Its symbol is in. In bulk of European local languages, “inch” can be used interchangeably with , or is derived from “thumb”. Because a man’s thumb is around an inch long.
• Electronic components like the dimensions of the PC screen.
• The size of the tires of a car or a truck.
What is 63 centimetres Converted to inches?
The centimeters to inches converter is a tool that lets me to convert cm into inches. This simple principle is used to translate cm into inches.
You have fully grasped of cm to inches by the above. You can use the formula to answer related questions:
• What’s the formula to convert inches from 63 cm?
• How can you convert cm into inches?
• How can you change cm into inches?
• How do you measure cm to inches?
• How tall are 63 cm to inches?
cm inches 62.6 cm 24.64562 inches 62.65 cm 24.665305 inches 62.7 cm 24.68499 inches 62.75 cm 24.704675 inches 62.8 cm 24.72436 inches 62.85 cm 24.744045 inches 62.9 cm 24.76373 inches 62.95 cm 24.783415 inches 63 cm 24.8031 inches 63.05 cm 24.822785 inches 63.1 cm 24.84247 inches 63.15 cm 24.862155 inches 63.2 cm 24.88184 inches 63.25 cm 24.901525 inches 63.3 cm 24.92121 inches 63.35 cm 24.940895 inches 63.4 cm 24.96058 inches | 516 | 1,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.863545 |
https://rpg.stackexchange.com/questions/130971/rolling-a-d10-0-9-for-hitpoints-what-do-i-get-with-a-zero/130977#130977 | 1,638,650,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00527.warc.gz | 554,164,495 | 34,502 | Rolling a D10 (0-9) for hitpoints. What do I get with a zero?
My character leveled up and I rolled a D10 to determine the rise in hitpoints. The dice that I used goes from 0 to 9 and I rolled a zero. The DM told me that this means I only add my modifier (0 + 3 = 3).
I didn't agree with this and think it counts as 10. The 0 is probably there for cosmetic reasons.
Who is right in this situation?
• Sep 2 '18 at 14:05
• Does your DM always treat 10 as "0" for d10 (for damage, etc)? Or was it only about hit points? Sep 2 '18 at 14:27
A zero on a d10 counts as 10.
This isn't clearly specified in the 5th edition rules, but is clearer in earlier editions of the game, and widely understood among long-time RPG players. However, we can still surmise from the D&D 5th edition rules that d10 is 1-10 like all other dice, from the following:
• Hit Dice by Size (Monster Manual p.7) confirms that the average of 1d10 is 5½, which is consistent with a range of 1-10. (The average of 0-9 would be 4½.)
• The fighter receives 10 hit points at 1st level. This wouldn't make sense if a d10 couldn't actually reach 10. All other classes receive the highest value of their roll at 1st level (e.g. the monk receives 1d8 per level, and 8 at first level).
• The fighter may take 6 hit points instead of rolling 1d10. This is consistent with the average result of 1-10 (5.5) rounded up. All other classes work the same way (e.g. the monk can take 5 hit points, equivalent to the average of 1-8 (4.5) rounded up). But the average of 0-9 is 4.5, so if 1d10 really meant 0-9, we would not expect to see the fighter allowed to take 6.
• The D&D Beyond character builder can confirm that if you create a fighter above level 1 with rolled hit points, they can indeed gain a roll of 10 on their hit points.
The reason for the zero is that in many games including D&D, two d10s can be rolled together to generate a two-digit percentile number between 00 and 99. As per PHB p.6, "Two 0s represent 100," another precedent which suggests that zero on the dice doesn't necessarily mean zero.
• Nice proof you've collected here, but imho you should also point out the Defining Event table of the Folk Hero background (PHB, p. 131): It lists all the possible outcomes of the d10 from 1 to 10. Sep 4 '18 at 8:49
• It will always bug me that it's 00-99 instead of 01-100. Considering the 0 on a d10 being 10 by itself you would think rolling 00 would be 1-10, 10 would be 11-20... 90 would be 91-100. But nooooo, it magically turns into a 0. Sep 5 '18 at 15:54
You are correct. Many old school d10s are marked 0-9 but you count the 0 as 10.
If he doesn’t listen to reason, next level use one of the d10s marked 00-90 instead and demand your 45 or so hit points from that roll.
• Let's make D6 and D10 dices with 99.'s on all sides.
– Cœur
Sep 9 '18 at 11:02
The Player's Handbook discusses dice on p. 6, in the subsection titled "Game Dice":
In these rules, the different dice are referred to by the letter d followed by the number of sides: d4, d6, d8, d10, d12, and d20. For instance, a d6 is a six-sided die (the typical cube that many games use).
Percentile dice, or d100, work a little differently. You generate a number between 1 and 100 by rolling two different ten-sided dice numbered from 0 to 9. One die (designated before you roll) gives the tens digit, and the other gives the ones digit. If you roll a 7 and a 1, for example, the number rolled is 71. Two 0s represent 100. Some ten-sided dice are numbered in tens (00, 10, 20, and so on), making it easier to distinguish the tens digit from the ones digit. In this case, a roll of 70 and 1 is 71, and 00 and 0 is 100.
It does not actually say that a d10 generates a number between 1 and 10, but that is how it is interpreted. The zero means 10.
• The 0 does mean ten on a d10, but percentile dice aren't a good way of showing it. If you roll e.g. 40 and 0, that 0 is zero.
– Ray
Sep 6 '18 at 1:30
A 0 counts as a 10
The Player's Handbook talks about dice referring to each dX as a die with X sides implying that the number of sides is the maximum roll through context.
There is also some circumstantial evidence in the spell confusion, which states:
An affected target ... must roll a d10 at the start of each of its turns to determine its behavior for that turn.
$$\begin{array} {|c|l|} \hline 1 & \text{The creature uses all its movement to move in a random direction...} \\ \hline 2-6 & \text{The creature doesn't move or take actions...} \\ \hline 3-7 & \text{The creature uses its action to make a melee attack ... randomly} \\ \hline 9-10 & \text{The creature can act and move normally.} \\ \hline \end{array}$$
Since there is no 0 in the table but a 10 is included, the most reasonable assumption is that a 0 is a 10.
A d6 goes from 1 to 6. A d12 goes from 1 to 12. A d10 goes from 1 to 10. That zero is the ten.
And in case your GM starts wondering...
Why is it a zero then?
While rarer in modern DnD, it's quite a common use of d10 to roll two at the same time to simulate a roll of d100. This is why these dice are often sold in sets with one die having "tens": 00, 10, 20 and so on, and the other having just "ones": 0, 1, 2, 3 and so on. One'd roll two dice and combine the tens and ones to get a result, eg. 80 and 7 yield an 87, with the special case of 00 and 0 typically but not universally meaning 100. Using d10's as d100 is the use where zero means a zero (except in the 00+0 case).
Even when intended for percentage rolls, the "ones" die could of course include a ten instead of a zero, at risk of being suspectible to be confused with the "tens" die (both'd have a 10), being somewhat uglier (at least to my taste) and causing a fair amount of "sixty, uh, tens" which might be jarring to those of us who aren't French. Indeed, d10s with numbers 1-10 do exist, they just don't seem to be as common.
When used as a single die, the aforementioned purpose of the 0 doesn't really exist, so a roll 0 on a d10 is a ten.
• Ah, how I love to work in a number-intensive role and knowing just enough french that people expect me to follow along as they read up 5 or 6-digit numbers. I hear somewhere between 8 and 19 digits. Quattre-vingt-seize - did you mean 96 and more numbers will follow or was that 802016? You guys... ;-) Sep 4 '18 at 7:26
• @zakinster Nay, 00+1 for instance is 1, not 101. Sep 5 '18 at 10:12
• @kviiri ah yes, didn't think this through, I'll delete my confusing comment. Sep 5 '18 at 11:32
• Also may be worth mentioning, all dice used single digits in the very early days of D&D. D20s were marked 0-9 twice, with one set inked in a different color. I assume there's some reason tied to how the dice were manufactured. Sep 5 '18 at 17:22 | 1,940 | 6,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.965596 |
https://collegephysicsanswers.com/openstax-solutions/what-force-meter-lightning-bolt-equator-carries-20000-perpendicular-earths | 1,721,244,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00184.warc.gz | 159,195,915 | 26,119 | Question
(a) What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to the Earth’s $3.00\times 10^{-5}\textrm{ T}$ field? (b) What is the direction of the force if the current is straight up and the Earth’s field direction is due north, parallel to the ground?
1. $0.600\textrm{ N/m}$
2. West
# OpenStax College Physics, Chapter 22, Problem 34 (Problems & Exercises)
In order to watch this solution you need to have a subscription. | 132 | 477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.828823 |
https://ete-online.biomedcentral.com/articles/10.1186/s12982-020-00091-z/tables/3 | 1,619,180,343,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00603.warc.gz | 339,014,610 | 32,766 | # Table 3 Estimates of vaccine effectiveness ($${\varvec{V}}{\varvec{E}}$$) under non-differentially imperfect sensitivity and high cumulative risk of infection in absence of false-positive events
True Estimation adjusted for $${se}_{0}={se}_{1}=0.04$$ Naïve estimation
$$VE$$ $$\widehat{VE}$$ $$\widehat{SE}$$ $${SE}_{\widehat{VE}}$$ $${\sqrt{MSE}}_{\widehat{VE}}$$ Bias Cov $$\widehat{VE}$$ $$\widehat{SE}$$ $${SE}_{\widehat{VE}}$$ $${\sqrt{MSE}}_{\widehat{VE}}$$ Bias Cov
Cohort of 50,000 individuals (30% vaccinated at season onset)
10% 10% 0.05 0.10 0.10 ± 0 67% 4% 0.05 0.05 0.08 − 6 82%
30% 30% 0.04 0.08 0.08 ± 0 70% 15% 0.05 0.05 0.16 − 15 11%
50% 50% 0.03 0.05 0.05 ± 0 75% 30% 0.04 0.04 0.20 − 20 0%
70% 70% 0.02 0.03 0.03 ± 0 80% 51% 0.03 0.03 0.19 − 19 0%
90% 90% 0.01 0.01 0.01 ± 0 88% 81% 0.02 0.02 0.09 − 9 0%
Cohort of 1,000,000 individuals (50% vaccinated at season onset)
10% 10% 0.01 0.02 0.02 ± 0 66% 4% 0.01 0.01 0.06 − 6 0%
30% 30% 0.01 0.02 0.02 ± 0 69% 15% 0.01 0.01 0.15 − 15 0%
50% 50% 0.01 0.01 0.01 ± 0 71% 30% 0.01 0.01 0.20 − 20 0%
70% 70% 0.00 0.01 0.01 ± 0 75% 52% 0.01 0.01 0.18 − 18 0%
90% 90% 0.00 0.00 0.00 ± 0 83% 81% 0.00 0.00 0.09 − 9 0%
1. Mean of the vaccine effectiveness estimates ($$\widehat{VE}$$), mean of the standard error estimates ($$\widehat{SE}$$), standard error of the vaccine effectiveness estimates ($${SE}_{\widehat{VE}}$$), root-mean-squared error of the vaccine effectiveness estimates ($${\sqrt{MSE}}_{\widehat{VE}}$$), bias in percentage points, and empirical coverage probability (Cov) of the 95% confidence intervals when estimating vaccine effectiveness from 104 repeated data sets under non-differential sensitivity ($${se}_{0}={se}_{1}$$) of 0.04 and a cumulative risk of 0.81 in the unvaccinated in absence of false-positive events. Naïve estimation was conducted under the incorrect assumption of perfect sensitivity ($${se}_{0}={se}_{1}=1$$) | 840 | 1,922 | {"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} | 2.609375 | 3 | CC-MAIN-2021-17 | latest | en | 0.575934 |
https://kmmiles.com/569-22-miles-in-km | 1,674,963,349,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00131.warc.gz | 365,976,507 | 6,248 | kmmiles.com
Search
# 569.22 miles in km
## Result
569.22 miles equals 915.875 km
You can also convert 569.22 mph to km.
## Conversion formula
Multiply the amount of miles by the conversion factor to get the result in km:
569.22 mi × 1.609 = 915.875 km
## How to convert 569.22 miles to km?
The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:
1 mi = 1.609 km
To convert 569.22 miles into km we have to multiply 569.22 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:
1 mi → 1.609 km
569.22 mi → L(km)
Solve the above proportion to obtain the length L in km:
L(km) = 569.22 mi × 1.609 km
L(km) = 915.875 km
The final result is:
569.22 mi → 915.875 km
We conclude that 569.22 miles is equivalent to 915.875 km:
569.22 miles = 915.875 km
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case five hundred sixty-nine point two two miles is approximately nine hundred fifteen point eight seven five km:
569.22 miles ≅ 915.875 km
## Conversion table
For quick reference purposes, below is the miles to kilometers conversion table:
miles (mi) kilometers (km)
570.22 miles 917.48398 km
571.22 miles 919.09298 km
572.22 miles 920.70198 km
573.22 miles 922.31098 km
574.22 miles 923.91998 km
575.22 miles 925.52898 km
576.22 miles 927.13798 km
577.22 miles 928.74698 km
578.22 miles 930.35598 km
579.22 miles 931.96498 km
## Units definitions
The units involved in this conversion are miles and kilometers. This is how they are defined:
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world. | 645 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-06 | latest | en | 0.835647 |
https://platformmnlaibnw.netlify.app/enriques29290bak/calculate-present-value-of-future-cash-flows-formula-397.html | 1,723,301,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00871.warc.gz | 364,796,024 | 11,874 | ## Calculate present value of future cash flows formula
In this module, you will focus on how to estimate number of periods, (annual) That is, firm value is present value of cash flows a firm generates in the future. Learn how to calculate net present value with our simple guide. the time value of money (TVM), translating future cash flows into the value of today's dollars. cash flow, you can use the following net present value formula to calculate NPV:.
Using the Excel FV Function to Calculate the Future Value of a Single Cash Flow. Instead of using the above formula, the future value of a single cash flow can be calculated using the built-in Excel FV function (which is generally used for a series of cash flows). We calculate that the present value of the free cash flows is \$326. Thus, if you were to sell this business based on its expected cash flows and a 10% discount rate, \$326.00 would be a very fair Compute the net present value of a series of annual net cash flows. To determine the present value of these cash flows, use time value of money computations with the established interest rate to convert each year’s net cash flow from its future value back to its present value. Using the Excel PV Function to Calculate the Present Value of a Single Cash Flow. Instead of using the above formula, the present value of a single cash flow can be calculated using the built-in Excel PV function (which is generally used for a series of cash flows).
## The Formula for Calculating Present Value of an Even Cash Flow the annuity formula discounts a series of future payments to calculate their present value.
23 Dec 2016 You understand, of course, that projections about the future are To calculate the present value of any cash flow, you need the formula below:. Review the calculation. The formula for finding the present value of future cash flows (PV) = C * [(1 - (1+i)^-n)/i], where C = the cash flow each period, i = the Most capital projects are expected to provide a series of cash flows over a period of time. necessary for calculating NPV when you have a series of future cash flows: According to this figure, the total present value of these future cash flows Use Excel Formulas to Calculate the Present Value of a Single Cash Flow or a fv is the future value of the investment;; rate is the interest rate per period (as a The Formula for Calculating Present Value of an Even Cash Flow the annuity formula discounts a series of future payments to calculate their present value. Use NPVs to evaluate future cash-flows in today's time value of money. By calculating risk-adjusted NPVs, you can quantitatively compare different investments.
### Use Excel Formulas to Calculate the Present Value of a Single Cash Flow or a fv is the future value of the investment;; rate is the interest rate per period (as a
21 Jun 2019 Future cash flows are discounted at the discount rate, and the higher the Calculating present value involves making an assumption that a rate Calculate the present value of uneven, or even, cash flows. Finds the present value (PV) of future cash flows that start at the end or beginning We start with the formula for PV of a future value ( FV ) single lump sum at time n and interest rate i,. Here is the mathematical formula for calculating the present value of an individual cash flow. NPV = F / [ (1 + i)^n ]. Where,. PV = Present Value. F = Future PV calculation a. Constant Annuity b. Future cash flows are discounted at the discount rate, and the higher the discount The difference between the present value of cash inflows and the present value of Formulas Summary. • Constant 23 Dec 2016 You understand, of course, that projections about the future are To calculate the present value of any cash flow, you need the formula below:. Review the calculation. The formula for finding the present value of future cash flows (PV) = C * [(1 - (1+i)^-n)/i], where C = the cash flow each period, i = the
### 18 Feb 2013 Another example using discounted cash flows, to value an annuity who want to start calculating discounted cash flows in their own life. to use the discounted cash flows formula Present Value = Future Value/ (1+Yield/p)N.
6 Jun 2019 Click here to understand the formula and concept of present value. present value are time, expected rate of return, and the size of the future cash flow. In the stock world, calculating present value can be a complex, inexact 29 Apr 2019 Investors use the NPV to determine the value of future deposits and payouts The formula for calculating net present value in bold: cash flow 20 Mar 2019 So how do you determine today's value of the future cash flows that we Besides calculating the net present value in the period 2017 – 2021, 18 Feb 2013 Another example using discounted cash flows, to value an annuity who want to start calculating discounted cash flows in their own life. to use the discounted cash flows formula Present Value = Future Value/ (1+Yield/p)N. 14 Jul 2015 To calculate present value from a cash flow stream, you must use the present value This can be written as Where PV is present value, C is future value, i is we calculate each yearly cash flow individually using this formula. The company may divide the remaining capital among shareholders as dividends or invest it in future ventures. Future debts and obligations cause capital's value
## PV calculation a. Constant Annuity b. Future cash flows are discounted at the discount rate, and the higher the discount The difference between the present value of cash inflows and the present value of Formulas Summary. • Constant
10 Jul 2019 Net present value discounts the cash flows expected in the future back to the present to show their today's worth. Microsoft Excel has a special 6 Aug 2018 Once you get the value from the Discounted Cash Flow formula, you long-term asset, you would have to first estimate its future cash flows. 9 Mar 2020 The cash flows in the future will be of lesser value than the cash flows of today. And hence the further the cash flows, lesser will the value. This is The value of money in the future can be calculated to Present Value or Present Worth with the "discount rate" as. P = F / (1 + i)n (1). where. F = future cash flow 3 Sep 2019 Calculating the sum of future discounted cash flows is the gold standard to determine how much an investment is worth. This guide show you Discounted Cash Flow is a term used to describe what your future cash flow is worth in today's value. This is also known as the present value (PV) of a future cash
10 Jul 2019 Net present value discounts the cash flows expected in the future back to the present to show their today's worth. Microsoft Excel has a special 6 Aug 2018 Once you get the value from the Discounted Cash Flow formula, you long-term asset, you would have to first estimate its future cash flows. 9 Mar 2020 The cash flows in the future will be of lesser value than the cash flows of today. And hence the further the cash flows, lesser will the value. This is The value of money in the future can be calculated to Present Value or Present Worth with the "discount rate" as. P = F / (1 + i)n (1). where. F = future cash flow 3 Sep 2019 Calculating the sum of future discounted cash flows is the gold standard to determine how much an investment is worth. This guide show you | 1,616 | 7,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-33 | latest | en | 0.909663 |
https://discourse.shapr3d.com/t/create-a-substration-with-a-gap-on-the-subtracted-one/15105 | 1,701,169,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00860.warc.gz | 256,133,455 | 7,316 | # Create a substration with a gap on the subtracted one
Hi, i wanna make stands for my resin vat, however i want to make a little gap to make easy to insert the « foots » on the vat. How can i make this ?
Hi! Are you looking to create a small tolerance gap between the two bodies? In this case, just create a copy of the purple body, scale it up by a few percent, then subtract it. The result will be a tiny (0.1-0.2 mm) gap between the two bodies which will make it easier to assemble them after manufacturing.
using scaling for this task has many problems:
• the direction of the scaling is tied to the current xyz axes. at least i couldn’t find a way to rotate the direction, i need to rotate my body instead.
• can’t set the size of the gap properly. i figured out that i can play with the distance between a given point on my body and the origin of scaling and the percentage value to get the desired result. like i can move my gizmo 100mm away from an edge and set the value to 0.99 to get a 1mm gap. but only at that point.
• scaling distorts the surface because of the percentage value so the gap is not consistent. in theory i can move my gizmo very far away (100 000mm) and set the difference to a very small number (0.99999) but what’s the lowest number i can use? the ui shows hundredths by default.
two years passed, is there a better tool for this job by any chance?
Hi @gex,
The pivot point of the scale can be adjusted to faces and edges where it will automatically align to the snapped elements, there is no need to rotate the body. However, it is indeed not the most accurate solution.
With the recent release of the history-based beta, offsetting the faces with a given distance will allow you to set a clearance with a defined gap.
The pivot point of the scale can be adjusted to faces and edges where it will automatically align to the snapped elements, there is no need to rotate the body. However, it is indeed not the most accurate solution.
sorry i haven’t checked this. i used a very far point to reduce the size difference and there was nothing to snap to.
With the recent release of the history-based beta, offsetting the faces with a given distance will allow you to set a clearance with a defined gap.
i tried this many times in stable but did not in beta. it works as i expected!
Gif with offset face in action
I select all mating surfaces on dovetails and use Offset Face to add gaps to make sure they’re mateable on a 3D print. Having the history to be able to change the gap widths easily will be useful, especially after named parameters are added.
With the recent release of the history-based beta, offsetting the faces with a given distance will allow you to set a clearance with a defined gap.
what are the limitations of this tool? i tried to use it on more complex bodies and it doesn’t work all the time, especially on merged and/or intersected bodies.
here is a simplified example:
first body is union of a cylinder and a spline swept around it. i get an error. second one is just a swept body. now this shape was easy to replace with something else but in my original project i couldn’t come up with a solution like this.
offset-face-example.shapr (12.4 KB)
There are no limitations set, but offsetting the faces could end up in non-manifold bodies if the geometry is complex or has some tiny details.
In your example, the shape on the left has a small remaining face from the cylinder. Please select that face too and the Offset Face tool will work fine. | 788 | 3,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.916151 |
https://spiral.ac/sharing/avudjjc/area-of-rectangles-using-sums-and-differences-kristakingmath | 1,621,037,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00374.warc.gz | 551,943,662 | 10,384 | area-of-rectangles-using-sums-and-differences-kristakingmath
# Interactive video lesson plan for: area of rectangles using sums and differences (KristaKingMath)
#### Activity overview:
► My Geometry course: https://www.kristakingmath.com/geometry-course
In this video we'll learn how to find the area of a figure that's the compilation of multiple rectangles, by finding the area of the larger rectangle and subtracting out part of the area, or by finding the areas of the smaller rectangles and then adding them together.
● ● ● GET EXTRA HELP ● ● ●
If you could use some extra help with your math class, then check out Krista’s website // http://www.kristakingmath.com
● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: http://www.kristakingmath.com
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Tagged under: differences,rectangular area sums,rectangles,expert,rectangular area differences,educational,calculus,Krista King, ,sums,area
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Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks! | 753 | 3,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-21 | latest | en | 0.877033 |
https://eric.ed.gov/?id=EJ892005 | 1,544,850,389,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00197.warc.gz | 591,819,076 | 3,709 | Collection
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Peer reviewed
ERIC Number: EJ892005
Record Type: Journal
Publication Date: 2007-May
Pages: 4
Abstractor: As Provided
Reference Count: 0
ISBN: N/A
ISSN: ISSN-0031-921X
Thermodynamics of a Block Sliding across a Frictional Surface
Mungan, Carl E.
Physics Teacher, v45 n5 p288-291 May 2007
The following idealized problem is intended to illustrate some basic thermodynamic concepts involved in kinetic friction. A block of mass m is sliding on top of a frictional, flat-topped table of mass M. The table is magnetically levitated, so that it can move without thermal contact and friction across a horizontal floor. The table is initially stationary, while the block has initial speed v[subscript i] and slides to rest relative to the table. The block and table are inside a large vacuum tank, so there is no air resistance, buoyancy, nor thermal losses to the atmosphere. Furthermore the inner surface of the vacuum tank is a perfect mirror so that the tank does not radiatively exchange heat with the block and table. The block and table are homogeneous, both initially have temperature T[subscript i], and they each have large thermal conductivities so that they rapidly attain a common final temperature T[subscript f] after the block has come to rest. The specific heat capacity of the block is c[subscript b] and that of the table is c[subscript t], and these heat capacities are assumed to be temperature independent over the range of temperatures that arises in this problem. (a) Find the common final speed v[subscript f] and temperature T[subscript f] of the block and table. (b) Find the changes in the bulk kinetic energies K (in the center-of-mass frame of the isolated block-table system), the internal energies U, and the entropies S of the block and table. (c) Discuss the first and second laws of thermodynamics in connection with these results.
American Association of Physics Teachers. One Physics Ellipse, College Park, MD 20740. Tel: 301-209-3300; Fax: 301-209-0845; e-mail: pubs@aapt.org; Web site: http://scitation.aip.org/tpt
Publication Type: Journal Articles; Reports - Descriptive
Education Level: N/A
Audience: N/A
Language: English | 511 | 2,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-51 | latest | en | 0.869774 |
http://es.wikidoc.org/index.php/Robust_statistics | 1,590,545,435,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00030.warc.gz | 43,364,953 | 24,900 | # Robust statistics
Robust statistics provides an alternative approach to classical statistical methods. The motivation is to produce estimators that are not unduly affected by small departures from model assumptions.
## Introduction
In statistics, classical methods rely heavily on assumptions which are often not met in practice. In particular, it is often assumed that the data are normally distributed, at least approximately, or that the central limit theorem can be relied on to produce normally distributed estimates. Unfortunately, when there are outliers in the data, classical methods often have very poor performance. Robust statistics seeks to provide methods that emulate classical methods, but which are not unduly affected by outliers or other small departures from model assumptions.
In order to quantify the robustness of a method, it is necessary to define some measures of robustness. Perhaps the most common of these are the breakdown point and the influence function, described below.
Good books on robust statistics include those by Huber (1981), Hampel et al (1986) and Rousseeuw and Leroy (1987). A modern treatment is given by Maronna et al (2006). Huber's book is quite theoretical, whereas the book by Rousseew and Leroy is very practical (although the sections discussing software are rather out of date, the bulk of the book is still very relevant). Hampel et al (1987) and Maronna et al (2006) fall somewhere in the middle ground. All four of these are recommended reading, though Maronna et al is the most up to date.
Robust parametric statistics tends to rely on replacing the normal distribution in classical methods with the t-distribution with low degrees of freedom (high kurtosis; degrees of freedom between 4 and 6 have often been found to be useful in practice) or with a mixture of two or more distributions.
## Example: speed of light data
Gelman et al. in Bayesian Data Analysis (2004) consider a data set relating to speed of light measurements made by Simon Newcomb. The data sets for that book can be found via the Classic data sets page, and the book's website contains more information on the data.
Although the bulk of the data look to be more or less normally distributed, there are two obvious outliers. These outliers have a large effect on the mean, dragging it towards them, and away from the center of the bulk of the data. Thus, if the mean is intended as a measure of the location of the center of the data, it is, in a sense, biased when outliers are present.
Also, the distribution of the mean is known to be asymptotically normal due to the central limit theorem. However, outliers can make the distribution of the mean non-normal even for fairly large data sets. Besides this non-normality, the mean is also inefficient in the presence of outliers and less variable measures of location are available.
### Estimation of location
The plot below shows a density plot of the speed of light data, together with a rug plot (panel (a)). Also shown is a normal QQ-plot (panel (b)). The outliers are clearly visible in these plots.
Panels (c) and (d) of the plot show the bootstrap distribution of the mean (c) and the 10% trimmed mean (d). The trimmed mean is a simple robust estimator of location that deletes a certain percentage of observations (10% here) from each end of the data, then computes the mean in the usual way. The analysis was performed in R and 10 000 bootstrap samples were used for each of the raw and trimmed means.
The distribution of the mean is clearly much wider than that of the 10% trimmed mean (the plots are on the same scale). Also note that whereas the distribution of the trimmed mean appears to be close to normal, the distribution of the raw mean is quite skewed to the left. So, in this sample of 66 observations, only 2 outliers cause the central limit theorem to be inapplicable.
Robust statistical methods, of which the trimmed mean is a simple example, seek to outperform classical statistical methods in the presence of outliers, or, more generally, when underlying parametric assumptions are not quite correct.
Whilst the trimmed mean performs well relative to the mean in this example, better robust estimates are available. In fact, the mean, median and trimmed mean are all special cases of M-estimators. Details appear in the sections below.
### Estimation of scale
The outliers in the speed of light data do not just have an adverse effect on the mean. The usual estimate of scale is the standard deviation, and this quantity is even more badly affected by outliers because the squares of the deviations from the mean go into the calcuation, so the outliers' effects are exacerbated.
The plots below show the bootstrap distributions of the standard deviation, median absolute deviation (MAD) and Qn estimator of scale (Rousseeuw and Croux, 1993). The plots are based on 10000 bootstrap samples for each estimator, and some normal random noise was added to the resampled data (smoothed bootstrap). Panel (a) shows the distribution of the standard deviation, (b) of the MAD and (c) of Qn.
The distribution of standard deviation is erratic and wide, a result of the outliers. The MAD is better behaved, and Qn is a little bit more efficient than MAD. This simple example demonstrates that when outliers are present, the standard deviation cannot be recommended as an estimate of scale.
### Manual screening for outliers
Traditionally, statisticians would manually screen data for outliers, and remove them, usually checking the source of the data to see if the outliers were erroneously recorded. Indeed, in the speed of light example above, it is easy to see and remove the two outliers prior to proceeding with any further analysis. However, in modern times, data sets often consist of large numbers of variables being measured on large numbers of experimental units. As such, manual screening for outliers is impractical.
Outliers can often interact in such a way that they mask each other. As a simple example, consider a small univariate data set containing one modest and one large outlier. The estimated standard deviation will be grossly inflated by the large outlier. The result is that the modest outlier looks relatively normal. As soon as the large outlier is removed, the estimated standard deviation shrinks, and the modest outlier now looks unusual.
This problem of masking gets worse as the complexity of the data increases. For example, in regression problems, diagnostic plots are used to identify outliers. However, it is common that once a few outliers have been removed, others become visible. The problem is even worse in higher dimensions.
Robust methods provide automatic ways of detecting, downweighting (or removing), and flagging outliers, largely removing the need for manual screening.
### Variety of applications
Although this article deals with general principles for univariate statistical methods, robust methods also exist for regression problems, generalized linear models, and parameter estimation of various distributions.
## Measures of robustness
The basic tools used to describe and measure robustness are, the breakdown point', the 'influence function and the sensitivity curve.
### Breakdown point
Intuitively, the breakdown point of an estimator is the proportion of incorrect observations (i.e. arbitrarily large observations) an estimator can handle before giving an arbitrarily large result. For example, given ${\displaystyle n}$ independent random variables ${\displaystyle (X_{1},\cdots ,X_{n})\sim {\mathcal {N}}(0,1)}$ and the corresponding realizations ${\displaystyle x_{1},\cdots ,x_{n}}$, we can use ${\displaystyle {\overline {X_{n}}}:={\frac {X_{1}+\cdots +X_{n}}{n}}}$ to estimate the mean. Such an estimator has a breakdown point of 0 because we can make ${\displaystyle {\overline {x}}}$ arbitrarily large just by changing any of ${\displaystyle x_{1},\cdots ,x_{n}}$.
The higher the breakdown point of an estimator, the more robust it is. Intuitively, we can understand that a breakdown point cannot exceed 50% because if more than half of the observations are contaminated, it is not possible to distinguish between the underlying distribution and the contaminating distribution. Therefore, the maximum breakdown point is 0.5 and there are estimators which achieve such a breakdown point. For example, the median has a breakdown point of 0.5. The X% trimmed mean has breakdown point of X%, for the chosen level of X. Huber (1981) and Maronna et al (2006) contain more details.
#### Example: speed of light data
In the speed of light example, removing the two lowest observations causes the mean to change from 26.2 to 27.75, a change of 1.55. The estimate of scale produced by the Qn method is 6.3. Intuitively, we can divide this by the square root of the sample size to get a robust standard error, and we find this quantity to be 0.78. Thus, the change in the mean resulting from removing two outliers is approximately twice the robust standard error.
The 10% trimmed mean for the speed of light data is 27.43. Removing the two lowest observations and recomputing gives 27.67. Clearly, the trimmed mean is less affected by the outliers and has a higher breakdown point.
Notice that if we replace the lowest observation, -44, by -1000, the mean becomes 11.73, whereas the 10% trimmed mean is still 27.43. In many areas of applied statistics, it is common for data to be log-transformed to make them near symmetrical. Very small values become large negative when log-transformed, and zeroes become negatively infinite. Therefore, this example is of practical interest.
### Empirical influence function
The empirical influence function gives us an idea of how an estimator behaves when we change one point in the sample and relies on the data (i.e. no model assumptions). The following picture is Tukey's biweight function, which, as we will later see, is an example of what a "good" (in a sense defined later on) empirical influence function should look like:
File:Biweight.png
The context is the following:
1. ${\displaystyle (\Omega ,{\mathcal {A}},P)}$ is a probability space,
2. ${\displaystyle ({\mathcal {X}},\Sigma )}$ is a measure space (state space),
3. ${\displaystyle \Theta }$ is a parameter space of dimension ${\displaystyle p\in \mathbb {N} ^{*}}$,
4. ${\displaystyle (\Gamma ,S)}$ is a measure space,
5. ${\displaystyle \gamma :\Theta \rightarrow \Gamma }$ is a projection,
6. ${\displaystyle {\mathcal {F}}(\Sigma )}$ is the set of all possible distributions on ${\displaystyle \Sigma }$
For example,
1. ${\displaystyle (\Omega ,{\mathcal {A}},P)}$ is any probability space,
2. ${\displaystyle ({\mathcal {X}},\Sigma )=(\mathbb {R} ,{\mathcal {B}})}$,
3. ${\displaystyle \Theta =\mathbb {R} \times \mathbb {R} ^{+}}$
4. ${\displaystyle (\Gamma ,S)=(\mathbb {R} ,{\mathcal {B}})}$,
5. ${\displaystyle \gamma :\mathbb {R} \times \mathbb {R} ^{+}\rightarrow \mathbb {R} }$ is defined by ${\displaystyle \gamma (x,y)=x}$.
The definition of an empirical influence function is: Let ${\displaystyle n\in \mathbb {N} ^{*}}$ and ${\displaystyle X_{1},\cdots ,X_{n}:(\Omega ,{\mathcal {A}})\rightarrow ({\mathcal {X}},\Sigma )}$ are iid and ${\displaystyle (x_{1},\cdots ,x_{n})}$ is a sample from these variables. ${\displaystyle T_{n}:({\mathcal {X}}^{n},\Sigma ^{n})\rightarrow (\Gamma ,S)}$ is an estimator. Let ${\displaystyle i\in \{1,\cdots ,n\}}$. The empirical influence function ${\displaystyle EIF_{i}}$ at observation ${\displaystyle i}$ is defined by:
${\displaystyle EIF_{i}:x\in {\mathcal {X}}\mapsto T_{n}(x_{1},\cdots ,x_{i-1},x,x_{i+1},\cdots ,x_{n})\in \Gamma }$
What this actually means is that we are replacing the i-th value in the sample by an arbitrary value and looking at the output of the estimator.
### Influence function and sensitivity curve
Instead of relying solely on the data, we could use the distribution of the random variables. The approach is quite different from that of the previous paragraph. What we are now trying to do is to see what happens to an estimator when we change the distribution of the data slightly.
Let ${\displaystyle A}$ be a convex subset of the set of all finite signed measures on ${\displaystyle {\mathcal {X}}}$. We want to estimate the parameter ${\displaystyle \theta \in \Theta }$ of a distribution ${\displaystyle F}$ in ${\displaystyle A}$. Let the functional ${\displaystyle T:A\rightarrow \Gamma }$ be the asymptotic value of some estimator sequence ${\displaystyle (T_{n})_{n\in \mathbb {N} }}$. We will suppose that this functional is Fisher consistent, i.e. ${\displaystyle \forall \theta \in \Theta ,T(F_{\theta })=\theta }$. this means that at the model ${\displaystyle F}$, the estimator sequence asymptotically measures the right quantity.
Let ${\displaystyle G}$ be some distribution in ${\displaystyle A}$. What happens when the data doesn't follow the model ${\displaystyle F}$ exactly but another, slightly different, "going towards" ${\displaystyle G}$?
We're looking at: ${\displaystyle dF_{G-F}(F)=\lim _{t\rightarrow 0^{+}}{\frac {T(tG+(1-t)F)-T(F)}{t}}}$,
which is the directional derivative of ${\displaystyle T}$ at ${\displaystyle F}$, in the direction of ${\displaystyle G}$.
Let ${\displaystyle x\in {\mathcal {X}}}$. ${\displaystyle \Delta _{x}}$ is the probability measure which gives mass 1 to ${\displaystyle x}$. We chose ${\displaystyle G=\Delta _{x}}$. The influence function is then defined by:
${\displaystyle IF(x;T;F):=\lim _{t\rightarrow 0^{+}}{\frac {T(t\Delta _{x}+(1-t)F)-T(F)}{t}}}$
It describes the effect of an infinitesimal contamination at the point ${\displaystyle x}$ on the estimate we are seeking, standardized by the mass ${\displaystyle t}$ of the contamination (the asymptotic bias caused by contamination in the observations).
## Desirable properties
Properties of an influence function which bestow it with desirable performance are:
1. Finite rejection point ${\displaystyle \rho ^{*}}$,
2. Small gross-error sensitivity ${\displaystyle \gamma ^{*}}$,
3. Small local-shift sensitivity ${\displaystyle \lambda ^{*}}$.
### Rejection point
${\displaystyle \rho ^{*}:=\inf _{r>0}\{r:IF(x;T;F)=0,|x|>r\}}$
### Gross-error sensitivity
${\displaystyle \gamma ^{*}(T;F):=\sup _{x\in {\mathcal {X}}}|IF(x;T;F)|}$
### Local-shift sensitivity
${\displaystyle \lambda ^{*}(T;F):=\sup _{(x,y)\in {\mathcal {X}}^{2},x\neq y}\left\|{\frac {IF(y;T;F)-IF(x;T;F)}{y-x}}\right\|}$
This value, which looks a lot like a Lipschitz constant, represents the effect of shifting an observation slightly from ${\displaystyle x}$ to a neighbouring point ${\displaystyle y}$, i.e. add an observation at ${\displaystyle y}$ and remove one at ${\displaystyle x}$.
## M-estimators
(The mathematical context of this paragraph is given in the section on empirical influence functions.)
Historically, several approaches to robust estimation were proposed, including R-estimators and L-estimators. However, M-estimators now appear to dominate the field as a result of their generality, high breakdown point, and their efficiency. See Huber (1981).
M-estimators are a generalization of maximum likelihood estimators (MLEs). What we try to do with MLE's is to maximize ${\displaystyle \prod _{i=1}^{n}f(x_{i})}$ or, equivalently, minimize ${\displaystyle \sum _{i=1}^{n}-\log f(x_{i})}$. In 1964, Huber proposed to generalize this to the minimization of ${\displaystyle \sum _{i=1}^{n}\rho (x_{i})}$, where ${\displaystyle \rho }$ is some function. MLE are therefore a special case of M-estimators (hence the name: "Maximum likelihood type" estimators).
Minimizing ${\displaystyle \sum _{i=1}^{n}\rho (x_{i})}$ can often be done by differentiating ${\displaystyle \rho }$ and solving ${\displaystyle \sum _{i=1}^{n}\psi (x_{i})=0}$, where ${\displaystyle \psi (x)={\frac {d\rho (x)}{dx}}}$ if ${\displaystyle \rho }$ has a derivative, that is.
Several choices of ${\displaystyle \rho }$ and ${\displaystyle \psi }$ have been proposed. The two figures below show four ${\displaystyle \rho }$ functions and their corresponding ${\displaystyle \psi }$ functions.
For squared errors, ${\displaystyle \rho (x)}$ increases at an accelerating rate, whilst for absolute errors, it increases at a constant rate. When Windsorizing is used, a mixture of these two effects is introduced: for small values of x, ${\displaystyle \rho }$ increases at the squared rate, but once the chosen threshold is reached (1.5 in this example), the rate of increase becomes constant.
Tukey's biweight (also known as bisquare) function behaves in a similar way to the squared error function at first, but for larger errors, the function tapers off.
### Properties of M-estimators
Notice that M-estimators do not necessarily relate to a probability density function. As such, off-the-shelf approaches to inference that arise from likelihood theory can not, in general, be used.
It can be shown that M-estimators are asymptotically normally distributed, so that as long as their standard errors can be computed, an approximate approach to inference is available.
Since M-estimators are normal only asymptotically, for small sample sizes it might be appropriate to use an alternative approach to inference, such as the bootstrap. However, M-estimates are not necessarily unique (i.e. there might be more than one solution that satisfies the equations). Also, it is possible that any particular bootstrap sample can contain more outliers than the estimator's breakdown point. Therefore, some care is needed when designing bootstrap schemes.
Of course, as we saw with the speed of light example, the mean is only normally distributed asymptotically and when outliers are present the approximation can be very poor even for quite large samples. However, classical statistical tests, including those based on the mean, are typically bounded above by the nominal size of the test. The same is not true of M-estimators and the type I error rate can be substantially above the nominal level.
These considerations do not "invalidate" M-estimation in any way. They merely make clear that some care is needed in their use, as is true of any other method of estimation.
### Influence function of an M-estimator
It can be shown that the influence function of an M-estimator ${\displaystyle T}$ is proportional to ${\displaystyle \psi }$ (see Huber, 1981 (and 2004), page 45), which means we can derive the properties of such an estimator (such as its rejection point, gross-error sensitivity or local-shift sensitivity) when we know its ${\displaystyle \psi }$ function.
${\displaystyle IF(x;T,F)=M^{-1}\psi (x,T(F))}$ with the ${\displaystyle p\times p}$ given by: ${\displaystyle M=-\int _{\mathcal {X}}\left({\frac {\partial \psi (x,\theta )}{\partial \theta }}\right)_{T(F)}dF(x)}$.
### Choice of ${\displaystyle \psi }$ and ${\displaystyle \rho }$
In many practical situations, the choice of the ${\displaystyle \psi }$ function is not critical to gaining a good robust estimate, and many choices will give similar results that offer great improvements, in terms of efficiency and bias, over classical estimates in the presence of outliers (Huber, 1981).
Theoretically, redescending ${\displaystyle \psi }$ functions are to be preferred, and Tukey's biweight (also known as bisquare) function is a popular choice. Maronna et al (2006) recommend the biweight function with efficiency at the normal set to 85%.
## Robust parametric approaches
M-estimators do not necessarily relate to a density function and so are not fully parametric. Fully parametric approaches to robust modelling and inference, both Bayesian and likelihood approaches, usually deal with heavy tailed distributions such as Student's t-distribution.
For the t-distribution with ${\displaystyle \nu }$ degrees of freedom, it can be shown that
${\displaystyle \psi (x)={\frac {x}{x^{2}+\nu }}}$.
For ${\displaystyle \nu =1}$, the t-distribution is equivalent to the Cauchy distribution. Notice that the degrees of freedom is sometimes known as the kurtosis parameter. It is the parameter that controls how heavy the tails are. In principle, ${\displaystyle \nu }$ can be estimated from the data in the same way as any other parameter. In practice, it is common for there to be mulitple local maxima when ${\displaystyle \nu }$ is allowed to vary. As such, it is common to fix ${\displaystyle \nu }$ at a value around 4 or 6. The figure below displays the ${\displaystyle \psi }$-function for 4 different values of ${\displaystyle \nu }$.
### Example: speed of light data
For the speed of light data, allowing the kurtosis parameter to vary and maximizing the likelihood, we get
${\displaystyle {\hat {\mu }}=27.40,{\hat {\sigma }}=3.81,{\hat {\nu }}=2.13.}$
Fixing ${\displaystyle \nu =4}$ and maximizing the likelihood gives
${\displaystyle {\hat {\mu }}=27.49,{\hat {\sigma }}=4.51.}$
## Key contributors
Key contributors to the field of robust statistics include Frank Hampel, Peter J. Huber, Peter J. Rousseeuw, and John Tukey.
## References
Robust Statistics - The Approach Based on Influence Functions, Frank R. Hampel, Elvezio M. Ronchetti, Peter J. Rousseeuw and Werner A. Stahel, Wiley, 1986 (republished in paperback, 2005)
Robust Statistics, Peter. J. Huber, Wiley, 1981 (republished in paperback, 2004)
Robust Regression and Outlier Detection, Peter J. Rousseeuw and Annick M. Leroy, Wiley, 1987 (republished in paperback, 2003)
Robust Statistics - Theory and Methods, Ricardo Maronna, Doug Martin and Victor Yohai, Wiley, 2006
Bayesian Data Analysis, Andrew Gelman, John B. Carlin, Hal S. Stern and Donald B. Rubin, Chapman & Hall/CRC, 2004
Alternatives to the Median Absolute Deviation, P. J. Rousseeuw and C. Croux, C., Journal of the American Statistical Association, 88, 1993 | 5,316 | 21,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 98, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.931789 |
https://astarmathsandphysics.com/gcse-physics-notes/803-pulleys.html?tmpl=component&print=1&page= | 1,534,407,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00583.warc.gz | 614,451,532 | 3,093 | ## Pulleys
Pulleys aresimilar to levers. Each allows a force to be magnified to a point many times over so that an object can be moved or a force overcome. Pulleys are especially useful for lifting objects vertically. It is important to realise that, ignoring friction, that tension is the same at each point in a rope, so that if you pull the end of the rope bottom left with a force T, the weight will experience a force 3T and will rise as long as T>W/3.
Similarly the weight in the centre experiences a force 3T upwards and will rise as long as T>W/3. The weight at the right hand side experiences an upwards force of 4T and will rise as long as T>W/4. In general you can reduce the force required to lift a weight by adding more pulleys. Adding another pulley at the bottom results in an extra force T acting upwards and this reduces the force required to lift the weight W.
Since Work=Force times Distance, moving a weight by applying a smaller force means you must pull the string further. If you raise the weight by a height h, each string attached to the bottom pulley must shorten by h so you must apply the force over a distance 3h. | 269 | 1,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.915356 |
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Statistics exercise one | Statistics homework help
Assignment 4
Statistics Exercise I
These weekly exercises provide the opportunity for you to understand and apply statistical methods and analysis.
All assignments MUST be typed, double-spaced, in APA style and must be written at graduate level English, citing the text in APA format.
#1. Given the following values of the mean and median, state the likely shape of the distribution and which measure should be used to summarize it.
(a) mean = 4, median = 4 _______________
(b) mean = 12, median = 2 _______________
(c) mean = 8, median = 18 _______________
(d) mean = 6, median = 14 _______________
(e) mean = 10, median = 3 _______________
(f) mean = 8, median = 8 _______________
#2. What are two steps to locate proportions under the normal curve?
#3. The sample mean is an unbiased estimator of the population mean. Explain this statement.
Use SPSS and the provided data to answer the following questions. Round your answers to the nearest dollar, percentage point, or whole number.
#4. What is the mean annual income (INC1) of the participants? A.\$43,282 B.\$72,133 C.\$47,113 D.\$34,282
#5. What percent of the participants are married (RELAT)? A.28% B.33% C.51% D.59%
#6. What is the modal level of relationship happiness (HAPPY)? A.Mixed B.Happy C.Very Happy D.Cannot be determined
#7. What is the median income of the participants’ partners (INC2)? A.\$24,212 B.\$28,945 C.\$32,000 D.\$48,975
#8. What percent of the participants are age 51 or older? A.4% B.5% C.7% D.10%
Assignment Outcomes:
Assess the concepts underlying appropriate use of various research methodologies
Analyze how to recognize the inappropriate or deceptive use of research methodology
Compare/contrast the basic assumptions underlying various statistical operations Summarize the consequences of using various methodological approaches Differentiate between the appropriate and inappropriate application and interpretation of research methods and statistics
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https://boffyblog.blogspot.com/2013/01/capital-i-chapter-20-part-2.html | 1,591,238,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00330.warc.gz | 266,096,871 | 34,479 | Tuesday, 22 January 2013
Capital I, Chapter 20 - Part 2
In much of what has gone before, Marx has described the dire consequences for workers from overwork. Now he describes the dire consequences of not enough work. We have set out that the annual cost of reproducing the workers is £10,000, which on the basis of a 50 week year, 40 hour week, and 8 hour day amounts to £5 per hour. But, this cost is the cost of reproducing the workers' labour power during this period, not the value created by that labour in the same period. For example, in a year, with a 100% rate of surplus value, the worker will create £20,000 of new value. Deducting the £10,000 cost of reproducing their labour-power leaves a surplus value of £10,000. By the same token, in a week they will produce £400 of new value, £200 as wages, £200 surplus value. In a day, £80 of new value, £40 wages, £40 surplus value. Finally, in an hour, £10 of new value, £5 wages, £5 surplus value.
If the worker is paid by the hour, then if the number of hours worked in the day is reduced, the amount of wages, and of surplus value will fall equally. For example, if the day is cut from 8 hours to 6 hours, only £60 of new value will be created, £30 as wages, and £30 as surplus value. But, whether the worker is working these 8 hours or not the cost of reproducing their labour-power does not fall. The worker does not stop living during these 2 hours.
The £10,000 cost for the year, was the minimum amount required for their reproduction from one year to the next. But, if the worker now only works 6 hours a day, receiving £30 in wages, that amounts to only £7,500 for the year, which means they are unable to reproduce their labour-power.
If the hour’s wage is fixed so that the capitalist does not bind himself to pay a day’s or a week’s wage, but only to pay wages for the hours during which he chooses to employ the labourer, he can employ him for a shorter time than that which is originally the basis of the calculation of the hour-wage, or the unit-measure of the price of labour. Since this unit is determined by the ratio
daily value of labour-power working-day of a given number of hours’
it, of course, loses all meaning as soon as the working-day ceases to contain a definite number of hours. The connection between the paid and the unpaid labour is destroyed. The capitalist can now wring from the labour a certain quantity of surplus-labour without allowing him the labour-time necessary for his own subsistence. He can annihilate all regularity of employment, and according to his own convenience, caprice, and the interest of the moment, make the most enormous overwork alternate with relative or absolute cessation of work. He can, under the pretense of paying “the normal price of labour,” abnormally lengthen the working-day without any corresponding compensation to the labourer.” (p 510-11)
Marx then turns to overtime. He then shows that even where the price of labour i.e. the hourly wage, rises, this can be consistent with falling real wages. If the daily rate remains constant, but the number of hours in the day rises, then the hourly rate clearly falls. But, it also falls if the increase in hours is proportionately more than the rise in the daily wage. For example, £10 per day with a 5 hour day = £2 per hour. But, £12 per day, with an 8 hour day is only £1.50 per hour.
But, even where the hourly rate itself increases this can still represent a fall in real wages. As described previously, the value of labour power is calculated on the cost of reproducing labour-power over a given period, and part of that cost is to cover the average wear and tear of the worker. But, that assumes that the worker works for the normal working day, at the normal level of intensity. If the worker works, for any consistent period of time, for longer than the normal working day, or at a higher than normal level of intensity, they will suffer greater wear and tear, and this wear and tear rises by a proportionately greater amount than the additional work.
As a consequence, the value of labour-power rises proportionately more, the more the worker is over worked. This is not only manifest in the need for more food etc. but also in a shorter life, greater medical costs, and so on. Suppose then that the worker goes from working an 8 hour day to a 12 hour day. But, this 50% increase in hours worked causes the value of their labour-power to rise from £5 per hour to £10 per hour. Their hourly rate rises from £5 to say £8. Their wages for the year rise to £24,000, and yet their real wage has fallen because the value of their labour-power is now £30,000, required to meet the now much higher cost of reproducing their labour power.
In many branches of industry where time-wage is the general rule without legal limits to the working-time, the habit has, therefore, spontaneously grown up of regarding the working day as normal only up to a certain point, e.g., up to the expiration of the tenth hour (“normal working-day,” “the day’s work,” “the regular hours of work”). Beyond this limit the working-time is over-time, and is, taking the hour as unit-measure, paid better (“extra pay”), although often in a proportion ridiculously small. The normal working-day exists here as a fraction of the actual working-day, and the latter, often during the whole year, lasts longer than the former. The increase in the price of labour with the extension of the working-day beyond a certain normal limit, takes such a shape in various British industries that the low price of labour during the so-called normal time compels the labourer to work during the better paid over-time, if he wishes to obtain a sufficient wage at all. Legal limitation of the working-day puts an end to these amenities.” (p 512-3)
A low price of labour begets longer working hours, because the worker is forced to work longer to obtain enough wages to live on.
On the other hand, the extension of the working-time produces, in its turn, a fall in the price of labour, and with this a fall in the day’s or week’s wages.” (p 513)
Competition between large numbers of small firms led them to cut prices by things such as using low wages, but also adulterating food etc. When workers created their own co-ops in the 19th Century, it meant they could provide themselves with quality products, as well as better working conditions. The Co-op Consumer Research Department also did valuable work in identifying substandard products being sold by private companies.
If one man does the work of 1½ or 2 men, the supply of labour increases, although the supply of labour-power on the market remains constant. The competition thus created between the labourers allows the capitalist to beat down the price of labour, whilst the falling price of labour allows him, on the other hand, to screw up still further the working-time. Soon, however, this command over abnormal quantities of unpaid labour, i.e., quantities in excess of the average social amount, becomes a source of competition amongst the capitalists themselves. A part of the price of the commodity consists of the price of labour. The unpaid part of the labour-price need not be reckoned in the price of the commodity. It may be presented to the buyer. This is the first step to which competition leads. The second step to which it drives is to exclude also from the selling price of the commodity at least a part of the abnormal surplus-value created by the extension of the working-day. In this way, an abnormally low selling price of the commodity arises, at first sporadically, and becomes fixed by degrees; a lower selling price which henceforward becomes the constant basis of a miserable wage for an excessive working-time, as originally it was the product of these very circumstances.” (p 513-4)
Marx does not expand on this further here, because he says it would require an analysis of competition, which he intended to deal with later. However, as is the case later with his analysis of the Falling Rate of Profit, and as was the case with the establishment of the normal working day, its possible to identify countervailing tendencies. For example, for so long as the abnormal surplus value is accumulated it creates an abnormal demand for labour, which will tend to push the price of labour higher. When it is discounted from selling prices, it will cause an abnormal increase in demand for this commodity (because its price does not fully reflect the value of the labour-power consumed), which again will cause an increase in demand for labour-power raising the price of labour. Finally, if this particular labour is paid a price, which does not cover the value of the labour-power, for any considerable period, the supply of labour power of the appropriate quality will fall, thereby again raising its price. However, the periods involved could be considerable.
Marx again cites the example of the competition between the “full priced” and “under priced” bakers, when the latter reduced the price of bread not only by gross adulteration of the product, but by the overwork and underpaying of their workers. Such competition and consequences between small capitalists continues today, but capital today is dominated by Big Capital, which long ago, as Engels described, found these penny-pinching measures to be counter-productive.
Marx quotes the Reports of the Children's Employment Commission.
““In Birmingham there is so much competition of masters one against another that many are obliged to do things as employers that they would otherwise be ashamed of; and yet no more money is made, but only the public gets the benefit.”” (p 514)
That is certainly a lesson that oligopolies have learned. As Paul Sweezy demonstrated, an oligopoly that reduces its prices, will be followed by others, creating a price war that is destructive of profits. One that raises prices, is not likely to be followed by others. This is why oligopolies try to avoid reducing prices (as Kliman points out they will cut output rather than prices when demand falls) and why central banks were introduced to increase money supply to prevent falls in nominal price levels. Its why oligopolies began to compete not on lower prices, but on better quality, and sought to increase profits, via increased market share from higher quality, new products, and via continual innovation to raise productivity and reduce costs.
For much of the 20th. Century the Tory Party was just as much an element of Social Democracy as the Labour Party. It was reflected in "Buttskillism" reflecting the similarity of position of Tory R.A. Butler, and Hugh Gaitskill. That Social Democracy was the form of Bourgeois Democracy applicable to a period of Fordist Regulation, which met the needs of Big Capital. It broke down in the 1970's, and led to the rise of Thatcher who personified the small capitalist, shopkeeper mentality, that has dominated the Tories since.
As Marx points out elsewhere, what the capitalists demand above all else is a level playing field, which requires regulation. The more Big Capital dominates, the more those regulations meet its needs and undermine those of the small capitalists. Where regulations have been introduced by social democracy (meaning the modern form of bourgeois democracy, and therefore including the ideas that have dominated most bourgeois parties be they of the Right or the Left) that has been to meet the needs of Big Capital. The fact that governments, like that of Cameron, talk about removing such regulation is an indication of how much these parties are beholden to small capital, and its attendant social layers.
So, for example, Marx cites the testimony of the “full priced” bakers.
The “full-priced” denounced their rivals before the Parliamentary Committee of Inquiry: “They only exist now by first defrauding the public, and next getting 18 hours’ work out of their men for 12 hours’ wages.... The unpaid labour of the men was made ... the source whereby the competition was carried on, and continues so to this day.... The competition among the master bakers is the cause of the difficulty in getting rid of night-work. An underseller, who sells his bread below the cost-price according to the price of flour, must make it up by getting more out of the labour of the men.... If I got only 12 hours’ work out of my men, and my neighbour got 18 or 20, he must beat me in the selling price. If the men could insist on payment for over-work, this would be set right.... A large number of those employed by the undersellers are foreigners and youths, who are obliged to accept almost any wages they can obtain.” (p 514)
A hundred and fifty years later, Cameron's Government represents the interests of the “under priced” producers of goods, not the “full priced” producers. In so doing, it is a far cry from the position of the Liberal Winston Churchill, who more than 100 years ago introduced the first Minimum Wage, precisely to undermine the kind of race to the bottom that the policies of today's Liberal-Tories lead to. As President of the Board of Trade in 1909, Churchill introduced the Minimum Wage, saying,
It is a national evil that any class of Her Majesty’s subjects should receive less than a living wage in return for their utmost exertions… where you have what we call sweated trades, you have no organisation, no parity of bargaining, the good employer is undercut by the bad and the bad by the worst; the worker, whose whole livelihood depends upon the industry, is undersold by the worker who only takes up the trade as a second string… where these conditions prevail you have not a condition of progress, but a condition of progressive degeneration.”
Thatcherism typified a period of support for the "undersellers", the "bad employers" as Churchill had described them. It set the course for the de-industrialisation of the UK economy, reliant on inefficient firms that could only survive on the back of low wages and poor conditions. They are today's "zombie" firms that survive by those means, and on unsustainably low interest rates.
But, today in Britain, a large minority of workers are in jobs that do not pay a living wage. As a consequence, the State has to make up the wages of these workers with various Welfarist measures, such as Housing Benefit, Child Benefit, Tax Credits and so on, which subsidise these low-paying, “under-selling” employers that often also provide poor conditions of employment. These “bad” employers are subsidised as a result of taxes taken from better paid workers, used as benefits for worse paid workers.
One of the main influences on undermining this, of course, was the development by workers themselves of the co-operatives. Their own shops were able to provide quality produce at reasonable prices without the kind of exploitation of workers producing those goods, that occurred amongst capitalist producers. The co-ops were able to introduce decent pay and working conditions, as well as introducing welfare benefits for their employees and members. Yet, they were still able to out compete the private producers and retailers, and to grow until by the end of the 19th Century, the Co-op dominated the retail and wholesale sector. The extension of the Co-op into Consumer Research, also meant it was able to highlight poor and injurious products being produced and sold by other retailers. The Co-op was also able to work with the rest of the Labour Movement to highlight things like profiteering. For example, after WWI, the London Co-ops worked with the London Trades Councils in the Food Vigilance Committee.
Back To Part 1
Forward To Chapter 21
Back To Index | 3,358 | 15,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.965407 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-2-linear-equations-and-functions-2-7-use-absolute-value-functions-and-transformations-2-7-exercises-skill-practice-page-127/4 | 1,721,410,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00108.warc.gz | 686,758,885 | 16,304 | ## Algebra 2 (1st Edition)
The is a transformation of the parent function $|x|$ The transformation applied was a horizontal shift of negative 2.
The is a transformation of the parent function $|x|$ The transformation applied was a horizontal shift of negative 2. | 58 | 263 | {"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} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.784425 |
https://math.stackexchange.com/questions/2505495/singularity-of-the-function-fz-sin-left-cos-left-frac1z-right-right | 1,620,674,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00180.warc.gz | 406,422,717 | 39,099 | Singularity of the function $f(z)=\sin\left(\cos\left(\frac{1}{z}\right)\right)$ at the point $z=0$ is …
For the function $f(z)=\sin(\cos(\frac{1}{z}))$, the point $z=0$ is
(a) a Removable singularity
(b) a pole
(c) an essential singularity
(d) Non-isolated singularity
I have written Laurent's series expansion $f(z)=\sin(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+...)= (1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+...)-\frac{(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+...)^3}{3!}+...$ all the negative powers coming in the expansion. so singularity with respect to zero is an essential singularity. Am I correct? Please help me to judge.
No, it is not correct. For two reasons:
1. the Laurent series of $\cos\left(\frac1z\right)$ at $0$ is$$1-\frac1{2!z^2}+\frac1{4!z^4}-\cdots;$$
2. you did not prove that the negative powers do not cancel each other after a certain point.
You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $\cos\left(\frac1z\right)$, if $V$ is a neighborhood of $0$, then the set $W=\left\{\cos\left(\frac1z\right)\,\middle|\,z\in V\setminus\{0\}\right\}$ is a dense subset of $\mathbb C$ and, since $\sin$ is non-constant entire function, $\sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.
• Thank you for pointing mistake. – user464147 Nov 5 '17 at 10:04
• Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance. – Aditya Prasad Jan 22 '19 at 16:30
• What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $f\bigl(U\setminus\{0\}\bigr)$ would not be dense in $\mathbb C$. Since it is… – José Carlos Santos Jan 22 '19 at 16:44
• ...it is a isolated singularity. Am I right? – Aditya Prasad Jan 22 '19 at 17:21
• That should be obvious. No: it is an essential singularity. – José Carlos Santos Jan 22 '19 at 17:24
Notice that the limit $\lim_{z \to 0} \cos(1/z)$ does not exist. Therefore the singularity is of essential type.
• consider $\frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right? – user464147 Nov 5 '17 at 10:09
• @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z)$, then $\lim_{z \to z_0} f (z) = \infty$. – user371838 Nov 5 '17 at 10:37
• Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity? – user464147 Nov 5 '17 at 10:42
• math.stackexchange.com/questions/1806582/… – user464147 Nov 5 '17 at 10:46
• this one is essential right? – user464147 Nov 5 '17 at 10:47 | 974 | 2,855 | {"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} | 3.84375 | 4 | CC-MAIN-2021-21 | latest | en | 0.798768 |
https://saintmungo.org/what-factors-determine-alveolar-partial-pressures/ | 1,653,516,837,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00738.warc.gz | 555,089,962 | 12,388 | What factors determine alveolar partial pressures?
Patm is the atmospheric pressure (at sea level 760 mm Hg), PH2O is partial pressure of water (approximately 45 mm Hg). FiO2 is the fraction of inspired oxygen. PaCO2 is partial pressure of carbon dioxide in alveoli (in normal physiological conditions around 40 to 45 mmHg).
Patm is the atmospheric pressure (at sea level 760 mm Hg), PH2O is partial pressure of water (approximately 45 mm Hg). FiO2 is the fraction of inspired oxygen. PaCO2 is partial pressure of carbon dioxide in alveoli (in normal physiological conditions around 40 to 45 mmHg).
Likewise, where is the partial pressure of oxygen the lowest? The partial pressure of oxygen is lower in the blood than in alveoli, so it diffuses into the blood. It’s important to note that, for each gas, the partial pressures equilibrate, or balance out, across the respiratory membrane, and they do so as the blood flows through the lungs.
Likewise, what is normal alveolar po2?
1) PO2 in alveoli is 104 mmHg vs. 40 mmHg for the deoxygenated blood of the pulmonary arteries. That means that PO2 in the pulmonary capillary blood = 104 mmHg. 2) PCO2 in alveoli is at 40 mmHg vs. 45 mmHg in blood returning from tissues.
How do you find the partial pressure of alveolar oxygen?
The alveolar gas equation is a formula used to approximate the partial pressure of oxygen in the alveolus (PAO2):PAO2=(PB−PH2O)FiO2−(PaCO2÷R)where PB is the barometric pressure, PH2O is the water vapor pressure (usually 47mmHg), FiO2 is the fractional concentration of inspired oxygen, and R is the gas exchange ratio.
Where is partial pressure of oxygen the highest?
It is at this point, in the pulmonary veins that carry blood away from the lungs and back to the heart, that the partial pressure of oxygen is highest, typically 100 millimeters of mercury.
Where is partial pressure of co2 the highest?
The partial pressure of CO2 will be the highest in the pulmonary artery because it brings the deoxygenated blood from the rest of the body to the lungs. Therefore, it will have the most CO2. It is then oxygenated in the lungs so after it leaves the lungs partial pressure of CO2 will be low.
What is partial pressure of co2?
The partial pressure of carbon dioxide in the blood of the capillary is about 45 mm Hg, whereas its partial pressure in the alveoli is about 40 mm Hg.
How is partial pressure calculated?
The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: Ptotal=P1+P2+… +Pn. + P n . The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
What does PaO2 mean?
The partial pressure of oxygen, also known as PaO2, is a measurement of oxygen pressure in arterial blood.
What is Dalton’s partial pressure?
In chemistry and physics, Dalton’s law (also called Dalton’s law of partial pressures) states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases. This empirical law was observed by John Dalton in 1801 and published in 1802.
What is PaO2 normal range?
Most healthy adults have a PaO2 within the normal range of 80–100 mmHg. If a PaO2 level is lower than 80 mmHg, it means that a person is not getting enough oxygen. A low PaO2 level can point to an underlying health condition, such as: emphysema.
What is the partial pressure of carbon dioxide in the veins?
Carbon dioxide tension PvCO2 – Partial pressure of carbon dioxide at sea level in venous blood is between 40 mmHg and 50 mmHg.
What affects PaO2?
PaO2, the partial pressure of oxygen in the arterial blood, is determined solely by the pressure of inhaled oxygen (the PIO2), the PaCO2, and the architecture of the lungs. The O2 dissociation curve (and hence the SaO2 for a given PaO2) is affected by PaCO2, body temperature, pH and other factors.
What is the normal po2?
As an example, the normal PO2 (partial pressure of oxygen) is 80? 100 mmhg. All this should really mean to us is that in arterial blood, 80 to 100 mmHg represents the “amount” of oxygen that is dissolved in each 100 ml of the arterial blood.
What determines the direction of gas movement?
Movement of Oxygen and carbon dioxide: 1) The direction of gas movement is determined by partial pressure differences. 2) At the arterial end of the pulmonary capillaries, O2 diffuses from the alveoli into the blood, while CO2 diffuses from the blood into the alveoli.
What happens when pO2 is high?
It primarily measures the effectiveness of the lungs in pulling oxygen into the blood stream from the atmosphere. Elevated pO2 levels are associated with: Increased oxygen levels in the inhaled air. Polycythemia.
How do you measure PaO2?
PaO2 is directly measured by a Clark electrode and can be used to assess oxygen exchange through a few relationships. The PaO2 rises with increasing FiO2. Inadequate or decreased oxygen exchange decreases the ratio.
What should PaO2 be on 100 oxygen?
A patient’s PaO2 (at sea level) should be 5 x the inspired oxygen percentage (FIO2). For example, a patient on room air is breathing 21% oxygen and so the PaO2 should be ~ 105 mmHg. A patient on 100% oxygen should have a PaO2 of ~500 mmHg. A patient on 40% FIO2 should have a PaO2 of ~200 mmHg. | 1,323 | 5,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-21 | latest | en | 0.890207 |
http://www.matholympiad.org.bd/forum/search.php?author_id=4265&sr=posts | 1,563,720,102,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527048.80/warc/CC-MAIN-20190721144008-20190721170008-00547.warc.gz | 238,113,014 | 6,032 | Search found 5 matches
Sun May 27, 2018 4:15 pm
Forum: Divisional Math Olympiad
Topic: Chittagong Secondary 2017#4
Replies: 7
Views: 828
Re: Chittagong Secondary 2017#4
No,you can't Check practically in a real chessboard.
Wed May 02, 2018 8:03 pm
Forum: Divisional Math Olympiad
Topic: Chittagong Secondary 2017#4
Replies: 7
Views: 828
Re: Chittagong Secondary 2017#4
Hint There are \$2015\$ black squares in that chess board Answer \$2015\$ Sorry brother,your solution is not right :oops: :!: .It doesn't mean that if there are 2015 black squares then there can be placed 2015 horses at the maximum rate.First think about 5*2 chessboard.You can put at most 4 horses in 5...
Tue Mar 27, 2018 6:10 pm
Forum: Primary Level
Topic: Combinatorics
Replies: 6
Views: 942
Combinatorics
In how many ways can COMPUTER be spelled by moving either down or diagonally to the right showed in the figure below??
Tue Mar 27, 2018 4:11 pm
Forum: Introductions
Topic: Combinatorics
Replies: 4
Views: 812
Re: Combinatorics
The problem is solved for ordered pairs in your suggested book.But I want the solution for the unordered pairs.Detailed help will be much appreciated.
Thu Mar 15, 2018 10:48 am
Forum: Introductions
Topic: Combinatorics
Replies: 4
Views: 812
Combinatorics
How many pairs of natural number can be formed whose LCM will be 7000?This problem has been taken from bdmo regional contest. As I am a new learner,please help me solving the problem in detail. | 445 | 1,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-30 | latest | en | 0.884411 |
https://puzzling.stackexchange.com/questions/15059/another-puzzling-alien-signal/15155 | 1,563,334,129,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00015.warc.gz | 516,769,286 | 36,558 | # Another Puzzling Alien Signal
There is great excitement in the astronomy community. The search for extraterrestrial intelligence is finally bearing fruit. A strong clear repeating signal is being received from a not so distant star. The pattern being transmitted is as follows.
A AB AACABB ABA AABCABBB AAACABBA ABAB AABBCABBBB
A represents a high pitch tone, B a low pitch tone, and C a tone that oscillates rapidly between the two.
The message lasts for roughly 14 seconds, then stops for about half an hour before repeating again. The time between each repetition is decreasing by about 35 seconds each time.
After the expected duration the message changes.
The original message is still present but when it ends a new message starts after a few second pause.
AAAAAAAAAAAAAAAAAAAAAABACADAAAAAAAAAAAAAAAAAAAAAAAAAAAABEACADDAAAAAAAAAAAAAAAAAAAAAAAAAABEEACADDDAAAAAAAAAAAAAAAAAAAAAAAAABEBACADDDDAAAAAAAAAAAAAAAAAAAAAAABEEEACADDDDDAAAAAAAAAAAAAAAAAAAAAABEEBACADDDDDDAAAAAAAAAAAAAAAAAAAAABEBEACADDDDDDDAAAAAAAAAAAAAAAAAAABEEEEACADDDDDDDD
This new message is much longer than the first and with different tones, each a a different frequency.
• I can't believe that another astronomy question was posted right before my own one! What a coincidence! – leoll2 May 16 '15 at 14:40
• @Bob Some other puzzler has bugged your computer! ;-) – Rand al'Thor May 16 '15 at 14:46
• Is it possible to solve this puzzle with no knowledge of music? – Rand al'Thor May 16 '15 at 15:50
• What exactly is the message is not clear to me but it is someone that will hit earth (if the time stays exactly 35 seconds between messages it means that the move is directly to the earth) in about 30*60/35 repetitions. You may estimate the speed (the speed of light multiplied by 35). The message could be: "I can not stop (please?) move away" – Moti May 16 '15 at 18:20
So the aliens are counting in binary. Here's the signal with A replaced with 1, B replaced with 0, and C replaced with ':'
1 10 11:100 101 110:1000 111:1001 1010 1100:10000
They're telling us they have three fingers! Here's a commentary:
One, two, three:ten, eleven, twelve:one hundred, thirteen:one hundred and one, one hundred and ten, one hundred and a hundred:thousand
Everytime they run out of fingers, they add C[base increment] to the number. This should be our response:
1 10 11 101 111 1000 1001 1010 1011 1100C10000
Hopefully that will be enough to let them know we understand their message, and we have ten fingers!
Edit: If they "count one, two, ten" then they have three fingers, not two.
• I don't understand how you converted the binary into a message. Also, what's the meaning of ":" ? – leoll2 May 18 '15 at 16:17
• : is just what I replaced C with. I've replaced the characters A, B, and C with 0, 1, and ":". It's not a message, it's a pattern, it's showing that everytime they want to count three, they say ten. Everytime they want to count thirteen, they say one hundred, etc. – Mark May 18 '15 at 16:19
• Hmmm, then we know that they know the meaning of three. We can speculate that they have 3 fingers when born, then amputee one for a some kind of ritual – leoll2 May 18 '15 at 16:23
• We can't make that deduction, they know the meaning of three exactly as much as we know the meaning of eleven, they're just telling us what base their counting system is in – Mark May 18 '15 at 16:25
• That said, they'd have no reason to skip 111 if they were actually counting in binary. Base 2 usage doesn't care what your "native" base is, any more than we'd skip 11, 12, 13, 14, and 15 when counting in hex. I suspect this answer is close, though. – Bobson May 18 '15 at 18:31 | 975 | 3,650 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-30 | longest | en | 0.951646 |
http://research.stlouisfed.org/fred2/series/M10097USM486SNBR | 1,427,938,338,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131309986.49/warc/CC-MAIN-20150323172149-00119-ip-10-168-14-71.ec2.internal.warc.gz | 224,442,165 | 19,241 | Net Change In Number of Operating Businesses for United States
1961-08: 5,100 Number Of Operating Businesses (+ see more)
Monthly, Seasonally Adjusted, M10097USM486SNBR, Updated: 2012-08-17 3:20 PM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Series Is Presented Here As Two Variables--(1)--Seasonally Adjusted Data, 1945-1962 (2)--Seasonally Adjusted Data, 1948-1961. Source: Unpublished Material From The U.S. Bureau Of The Census
This NBER data series m10097 appears on the NBER website in Chapter 10 at http://www.nber.org/databases/macrohistory/contents/chapter10.html.
NBER Indicator: m10097
Release: NBER Macrohistory Database
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(a) Net Change In Number of Operating Businesses for United States, Number Of Operating Businesses, Seasonally Adjusted (M10097USM486SNBR)
Series Is Presented Here As Two Variables--(1)--Seasonally Adjusted Data, 1945-1962 (2)--Seasonally Adjusted Data, 1948-1961. Source: Unpublished Material From The U.S. Bureau Of The Census
This NBER data series m10097 appears on the NBER website in Chapter 10 at http://www.nber.org/databases/macrohistory/contents/chapter10.html.
NBER Indicator: m10097
Net Change In Number of Operating Businesses for United States
Integer Period Range: to copy to all
Create your own data transformation: [+]
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
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``` National Bureau of Economic Research, Net Change In Number of Operating Businesses for United States [M10097USM486SNBR], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/M10097USM486SNBR/, April 1, 2015. ```
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Name: Email: | 679 | 2,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2015-14 | latest | en | 0.774577 |
http://www.talkstats.com/threads/paired-t-test-h-e-l-p.17514/ | 1,652,820,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00270.warc.gz | 118,532,125 | 9,817 | # Paired t-test. H-E-L-P.
#### Malcolm
##### New Member
Hi!
I'm really, really hoping someone can help me with this, because I've been trying to figure it out on my own, and I'm getting NOWHERE.
Here goes: I'm a student working on a paper regarding the use of cholesterol- lowering meds and cholesterol-leves in blood. My paper is basically based upon several published articles. I'm doing no research of my own (I'm a freshman pharmacy- student), and therefore the data I have, is restricted to the data given in the articles. Anyway, what I'm doing, is that I am trying to calculate if the reduction of blood- cholesterol is statistically significant after starting to use cholesterol-lowering meds, in different groups of patients (age, medical history, gender, etc). I have the mean pre- and posttreatment values of blood-cholesterol of these patient, and the mean pre- and post standard deviation of this. I figured I had to use a paired t-test to calculate p. The problem is, though, that I only have the mean values of the groups, I do not have the values of each patient, so that I cannot calculate "s" in the formula for the paired t-test. The more I read about this, the more confused I get, so I would be eternally grateful if someone could point me in the right direction. How do i calculate p from these mean- values?
I hope I'm making sense, English is not my native language.
Thank you!
#### duskstar
##### New Member
Are you trying to calculate a p-value for each study or an overall one (in some form of meta-analysis)? Do you know n (the number of observations) for each study?
#### Malcolm
##### New Member
Hello.
I'm trying to calculate p for each study. I know the number of observations for each study, but I don't have every patients individual value. I only have the mean pre- and postoperative value, and the standard deviation for these. | 425 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.941814 |
https://becalculator.com/81-5-inches-to-feet-converter-2.html | 1,675,703,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00657.warc.gz | 139,249,719 | 17,495 | 81.5 inches to feet converter
How much feet in an inch?
Let’s talk about some methods to determine between units of length, such as to convert 81.5 inches to feet. How tall is 81.5 inches in feet?
You are able to calculate 81.5 inches in feet when you know the conversion factor of an inch in feet.
1 inch equals 0.083333 feet.
The answers to these questions are about 1 inch in feet:
• What is the number of an inch to feet?
• 1 inch is how much feet?
• What is conversion inches to ft?
• How to convert 1 inch to feet?
Implication of Inch
An inch (symbol in) is an Anglo-American unit of length measurement.. The symbol is in. In many European languages, “inch” can be used interchangeably with or derived from “thumb”. Since a person’s thumb is around an inch long.
Use:
• Electronic components, like the size of the screen.
• Size of car/truck tires.
About Feet
Feet or foot (symbol: ft) is a measure of length used in the Anglo-American customary system of measurement It equals a third of a yard and 12 inches.
Use:
• For measuring heights, and shorter distances, field lengths.
• People foot size.
How to Transfer 81.5 Inches to Feet?
There are a variety of measurement methods that can be employed worldwide. Every conversion system is widely used across different regions and countries.
To convert a number in inches into the equivalent value in feet, Simply multiply the amount in inches by 0.083333.
81.5 inches to feet = 81.5 inches × 0.083333 = 6.7916395 feet
Frequently Asked Questions About Inches to Feet
• How many in in feet?
1 inch is equals to 0.083333 feet. To turn others, use cminchesconverter.
• connection between inches and feet?
1 foot = 12 inches
1 inch = 0.08333 feet
• What is formula for inches to feet?
The conversion factor to convert inches to feet is 0.083333. To calculate feet, simply multiply the inches by 0.083333.
• How to convert in in ft?
feet = in × 0.083333
For example:
81.5 inches to ft = 0.083333 × 81.5 = 6.7916395 ft
Inches to Feet Formula
Value in ft = value in in × 0.083333
Final Thought
Up to now, do you know how much are 81.5 in to ft?
Our homepage provides more information about inches in feet.
Popular Inches into Feet Conversions Table
6 inches to feet 0.5 71 inches to feet 5.92 72 inches to feet 6 67 inches to feet 5.58 60 inches to feet 5 36 inches to feet 3 48 inches to feet 4 80 inches to feet 6.67
Common Inches to Feet Conversion Table
inches feet 81.1 inches 6.7583063 feet 81.15 inches 6.76247295 feet 81.2 inches 6.7666396 feet 81.25 inches 6.77080625 feet 81.3 inches 6.7749729 feet 81.35 inches 6.77913955 feet 81.4 inches 6.7833062 feet 81.45 inches 6.78747285 feet 81.5 inches 6.7916395 feet 81.55 inches 6.79580615 feet 81.6 inches 6.7999728 feet 81.65 inches 6.80413945 feet 81.7 inches 6.8083061 feet 81.75 inches 6.81247275 feet 81.8 inches 6.8166394 feet 81.85 inches 6.82080605 feet 81.9 inches 6.8249727 feet | 846 | 2,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.908292 |
https://converterin.com/area/manzana-nicaragua-to-acre-ac.html | 1,582,032,754,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143695.67/warc/CC-MAIN-20200218120100-20200218150100-00123.warc.gz | 338,336,881 | 9,634 | # MANZANA [NICARAGUA] TO ACRE CONVERTER
FROM
TO
The result of your conversion between manzana [Nicaragua] and acre appears here
## MANZANA [NICARAGUA] TO ACRE (manzana TO ac) FORMULA
To convert between Manzana [Nicaragua] and Acre you have to do the following:
First divide 70.44*100 / 0.09290304*43560 = 1.74061031
Then multiply the amount of Manzana [Nicaragua] you want to convert to Acre, use the chart below to guide you.
## MANZANA [NICARAGUA] TO ACRE (manzana TO ac) CHART
• 1 manzana [Nicaragua] in acre = 1.74061031 manzana
• 10 manzana [Nicaragua] in acre = 17.40610307 manzana
• 50 manzana [Nicaragua] in acre = 87.03051535 manzana
• 100 manzana [Nicaragua] in acre = 174.06103071 manzana
• 250 manzana [Nicaragua] in acre = 435.15257676 manzana
• 500 manzana [Nicaragua] in acre = 870.30515353 manzana
• 1,000 manzana [Nicaragua] in acre = 1740.61030705 manzana
• 10,000 manzana [Nicaragua] in acre = 17406.10307055 manzana
Symbol: manzana
No description
Symbol: ac
No description | 375 | 1,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.584502 |
http://au.metamath.org/mpegif/df-rocatset.html | 1,513,217,448,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00079.warc.gz | 25,067,294 | 5,552 | Mathbox for Frédéric Liné < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > df-rocatset Unicode version
Definition df-rocatset 26059
Description: Composition of two morphisms in the category Set. Experimental. (Contributed by FL, 6-Nov-2013.)
Assertion
Ref Expression
df-rocatset
Distinct variable group: ,,,
Detailed syntax breakdown of Definition df-rocatset
StepHypRef Expression
1 crocase 26058 . 2
2 vx . . 3
3 cgru 8428 . . 3
4 va . . . . . . . 8
54cv 1631 . . . . . . 7
62cv 1631 . . . . . . . 8
7 ccmrcase 26013 . . . . . . . 8
86, 7cfv 5271 . . . . . . 7
95, 8wcel 1696 . . . . . 6
10 vb . . . . . . . 8
1110cv 1631 . . . . . . 7
1211, 8wcel 1696 . . . . . 6
13 c1st 6136 . . . . . . . . 9
1413, 13ccom 4709 . . . . . . . 8
155, 14cfv 5271 . . . . . . 7
16 c2nd 6137 . . . . . . . . 9
1716, 13ccom 4709 . . . . . . . 8
1811, 17cfv 5271 . . . . . . 7
1915, 18wceq 1632 . . . . . 6
209, 12, 19w3a 934 . . . . 5
21 vc . . . . . . 7
2221cv 1631 . . . . . 6
2311, 14cfv 5271 . . . . . . . 8
245, 17cfv 5271 . . . . . . . 8
2523, 24cop 3656 . . . . . . 7
265, 16cfv 5271 . . . . . . . 8
2711, 16cfv 5271 . . . . . . . 8
2826, 27ccom 4709 . . . . . . 7
2925, 28cop 3656 . . . . . 6
3022, 29wceq 1632 . . . . 5
3120, 30wa 358 . . . 4
3231, 4, 10, 21coprab 5875 . . 3
332, 3, 32cmpt 4093 . 2
341, 33wceq 1632 1
Colors of variables: wff set class This definition is referenced by: isrocatset 26060
Copyright terms: Public domain W3C validator | 721 | 1,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.114423 |
http://yutsumura.com/row-equivalent-matrix-bases-for-the-null-space-range-and-row-space-of-a-matrix/ | 1,513,326,006,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00406.warc.gz | 496,660,262 | 28,071 | # Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix
## Problem 260
Let $A=\begin{bmatrix} 1 & 1 & 2 \\ 2 &2 &4 \\ 2 & 3 & 5 \end{bmatrix}.$
(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.
(b) Find a basis for the null space of $A$.
(c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.
(d) Exhibit a basis for the row space of $A$.
## Hint.
In part (c), you may use the following theorem.
Theorem (leading-1 method): If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.
In part (d), you may use the following theorem.
Theorem (Row-space method): If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.
## Solution.
### (a) Find a matrix $B$ in reduced roe echelon form such that $B$ is row equivalent to the matrix $A$.
We apply the elementary row operations to the matrix $A$ and obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}
\xrightarrow{\substack{R_2-2R_1\\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 2 \\
0 &0 &0 \\
0 & 1 & 1
\end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}\\
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form that is row equivalent to $A$.
Thus we set
$B=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 &0 &0 \end{bmatrix}.$
### (b) Find a basis for the null space of $A$.
The null space $\calN(A)$ of the matrix is the set of solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$.
By part (a), the augmented matrix $[A\mid \mathbf{0}]$ is row equivalent to $[B \mid \mathbf{0}]$.
Thus, the solution $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ must satisfy
$x_1=-x_3 \text{ and } x_2=-x_3,$ where $x_3$ is a free variable.
Thus the solutions are given by
$\mathbf{x}=\begin{bmatrix} -x_3 \\ -x_3 \\ x_3 \end{bmatrix}=x_3\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}$ for any number $x_3$.
Hence we have
\begin{align*}
\calN(A)&=\{ \mathbf{x}\in \R^3 \mid \mathbf{x}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \} \\
-1 \\
-1 \\
1
\end{align*}
From this, we deduce that the set
$\left\{ \quad\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}\quad \right \}$ is a basis for the null space of $A$.
### (c) Find a basis for the range of $A$ that consists of columns of $A$. (Longer version)
Let us write
$A=\begin{bmatrix} 1 & 1 & 2 \\ 2 &2 &4 \\ 2 & 3 & 5 \end{bmatrix}=[A_1, A_2, A_3],$ where $A_1, A_2, A_3$ are the column vectors of the matrix $A$.
The range $\calR(A)$ of $A$ is the same as the column space of $A$.
Thus, it suffices to find the maximal number of linearly independent vectors among the column vectors $A_1, A_2, A_3$.
Consider the linear combination
$x_1A_1+x_2A_2+x_3A_3=\mathbf{0}. \tag{*}$ Then this is equivalent to the homogeneous system
$A\mathbf{x}=\mathbf{0}$ and we already found the solutions in part (b):
$x_1=-x_3 \text{ and } x_2=-x_3.$ This tells us the vectors $A_1, A_2, A_3$ are linearly dependent.
For example, $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of the system (*).
Thus, we have
$\calR(A)=\Span(A_1, A_2, A_3)=\Span(A_1, A_2).$
On the other hand, if we consider only $A_1$ and $A_2$, they are linearly independent.
(To see this, you just need to repeat the above argument without $A_3$. This amounts to just ignoring the third columns in the computations.)
Therefore, the set $\{A_1, A_2\}$ is a linearly independent spanning set for the range.
Hence a basis of the range consisting column vectors of $A$ is
$\{A_1, A_2\}.$
The only column vector which is not a basis vector is $A_3$.
We already found that $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of (*).
Thus we have
$-A_1-A_2+A_3=\mathbf{0}.$ Solving this for $A_3$, we obtain the linear combination for $A_3$ of the basis vectors:
$A_3=A_1+A_2.$
#### (c) A shorter solution using the leading 1 method
Here is a shorter solution which uses the following theorem.
Theorem (leading-1 method): If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.
Looking at the matrix $B$, we see that the first and the second columns of $B$ have leading 1’s. Thus the first and the second column vectors of $A$ form a basis for the range of $A$.
### (d) Exhibit a basis for the row space of $A$.
We use the following theorem.
Theorem (Row-space method): If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.
We have already found such $B$ in part (a), and the first and the second row vectors are nonzero. Thus they form a basis for the row space of $A$.
Hence a basis for the row space of $A$ is
$\left\{\,\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \, \right \}.$
## Comment.
The longer solution of part (c) is essentially the proof of Theorem (leading 1 method).
It is good to know where the theorem came from, but when you solve a problem you may forget and just use the theorem like the shorter solution of (c).
### 2 Responses
1. 03/16/2017
[…] that these vectors are linearly independent, thus a basis for the range. (Basically, this is the leading 1 method.) Hence we have [calR(T)=calR(A)=Span left{begin{bmatrix} 1 \ 1 \ 0 end{bmatrix}, […]
2. 06/25/2017
[…] Since the both columns contain the leading $1$’s, we conclude that [left{, begin{bmatrix} 1 \ 0 \ 1 end{bmatrix}, begin{bmatrix} -1 \ 1 \ 1 end{bmatrix} ,right}] is a basis of the range of $A$ by the leading $1$ method. […]
##### Determine a Matrix From Its Eigenvalue
Let $A=\begin{bmatrix} a & -1\\ 1& 4 \end{bmatrix}$ be a $2\times 2$ matrix, where $a$ is some real number. Suppose...
Close | 2,139 | 6,295 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2017-51 | latest | en | 0.708533 |
https://www.hackmath.net/en/math-problem/8413 | 1,708,741,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00297.warc.gz | 838,326,532 | 9,242 | # Chester
Chester has a piece of wood measuring 1 2/3 cm. He needs to cut it into pieces measuring 3/4 cm long. How many pieces of wood did Chester cut?
n = 20/9 = 2
r = 0 cm
### Step-by-step explanation:
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Do you want to round the number? | 138 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.914478 |
http://oeis.org/A153229 | 1,500,561,374,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423222.65/warc/CC-MAIN-20170720141821-20170720161821-00707.warc.gz | 229,804,826 | 4,852 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A153229 a(0) = 0, a(1) = 1, and for n>=2, a(n) = (n-1) * a(n-2) + (n-2) * a(n-1). 9
0, 1, 0, 2, 4, 20, 100, 620, 4420, 35900, 326980, 3301820, 36614980, 442386620, 5784634180, 81393657020, 1226280710980, 19696509177020, 335990918918980, 6066382786809020, 115578717622022980, 2317323290554617020, 48773618881154822980, 1075227108896452857020 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS Previous name was: Weighted Fibonacci numbers. From Peter Bala, Aug 18 2013: (Start) The sequence occurs in the evaluation of the integral I(n) := int {u = 0..inf} exp(-u)*u^n/(1 + u) du. The result is I(n) = A153229(n) + (-1)^n*I(0), where I(0) = int {0..inf} exp(-u)/(1 + u) du = 0.5963473623... is known as Gompertz's constant. See A073003. Note also that I(n) = n!*int {u = 0..inf} exp(-u)/(1 + u)^(n+1) du. (End) ((-1)^(n+1))*a(n) = p(n,-1), where the polynomials p are defined at A248664. - Clark Kimberling, Oct 11 2014 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..200 FORMULA a(0) = 0, a(1) = 1, and for n>=2, a(n) = (n-1) * a(n-2) + (n-2) * a(n-1). For n>=1, a(n) = A058006(n-1) * (-1)^(n-1). G.f.: G(0)*x/(1+x)/2, where G(k)= 1 + 1/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013 G.f.: 2*x/(1+x)/G(0), where G(k)= 1 + 1/(1 - 1/(1 - 1/(2*x*(k+1)) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013 G.f.: W(0)*x/(1+sqrt(x))/(1+x), where W(k) = 1 + sqrt(x)/( 1 - sqrt(x)*(k+1)/(sqrt(x)*(k+1) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 17 2013 a(n) ~ (n-1)! * (1 - 1/n + 1/n^3 + 1/n^4 - 2/n^5 - 9/n^6 - 9/n^7 + 50/n^8 + 267/n^9 + 413/n^10), where numerators are Rao Uppuluri-Carpenter numbers, see A000587. - Vaclav Kotesovec, Mar 16 2015 EXAMPLE a(20) = 19 * a(18) + 18 * a(19) = 19 * 335990918918980 + 18 * 6066382786809020 = 6383827459460620 + 109194890162562360 = 115578717622022980 MAPLE t1 := sum(n!*x^n, n=0..100): F := series(t1/(1+x), x, 100): for i from 0 to 40 do printf(`%d, `, i!-coeff(F, x, i)) od: # Zerinvary Lajos, Mar 22 2009 # second Maple program: a:= proc(n) a(n):= `if`(n<2, n, (n-1)*a(n-2) +(n-2)*a(n-1)) end: seq(a(n), n=0..25); # Alois P. Heinz, May 24 2013 MATHEMATICA Join[{a = 0}, Table[b = n! - a; a = b, {n, 0, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *) PROG (C) unsigned long a(unsigned int n) { if (n == 0) return 0; if (n == 1) return 1; return (n - 1) * a(n - 2) + (n - 2) * a(n - 1); } (PARI) a(n)=if(n, my(t=(-1)^n); -t-sum(i=1, n-1, t*=-i), 0); \\ Charles R Greathouse IV, Jun 28 2011 CROSSREFS Cf. A000045, A000587, A058006. Sequence in context: A158094 A108879 A058006 * A013329 A102087 A052573 Adjacent sequences: A153226 A153227 A153228 * A153230 A153231 A153232 KEYWORD nonn AUTHOR Shaojun Ying (dolphinysj(AT)gmail.com), Dec 21 2008 EXTENSIONS Edited by Max Alekseyev, Jul 05 2010 Better name by Joerg Arndt, Aug 17 2013 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 1,397 | 3,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-30 | latest | en | 0.482642 |
https://derangedphysiology.com/main/cicm-primary-exam/viva-stations/non-canonical-vivas-practice/viva-biii | 1,653,103,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00539.warc.gz | 259,166,213 | 6,670 | # Viva B(iii)
This viva is relevant to Section B(iii) of the 2017 CICM Primary Syllabus, which expects the exam candidate to "Describe factors influencing the distribution of drugs".
###### Define the term "volume of distribution".
Volume of distribution is ".not a "real volume"... It is the parameter relating the concentration of a drug in the plasma to the total amount of the drug in the body" (Birkett, 2009)
Alternatively:
"(Vd) is defined as the apparent volume into which a drug disperses in order to produce the observed plasma concentration" (Peck & Hill)
###### How does the timing of measurement influence the volume of distribution?
• Drug concentration in the body will vary over time due to clearance and redistribution
• Taking measurements at different points in time will yield different volumes of distribution, which will depend on the dominant influences on drug distribution at the measurement time.
###### What is the initial volume of distribution (Vinitial)?
• This is the Vd of the drug in the central compartment
• It is possible to calculate this soon after a drug is administered intravenously, by extrapolating an imaginary line from plasma concentration measurements, extended to time zero.
• You can use this relationship to estimate the volume of the central compartment
• For therapeutic purposes, it can be used to estimate high peak plasma concentrations so that - if need be- you can divide your loading dose to avoid toxicity.
###### What is the extrapolated volume of distribution?
• Extrapolated volume of distribution (Vextrap) is the Vd of tissue distribution.
• This method .extrapolates a line of best fit from the terminal elimination phase.
• This will cause an overestimation of Vd for drugs which distribute extensively into the tissues.
###### What is meant by the term "non-compartmenal volume of distribution"?
• Non-compartmental Vd (Varea) is an extrapolated Vd which uses the area under the concentration/time curve.
###### Define the term "steady state volume of distribution"
• Vss describes the volume of distribution during steady state conditions, i.e. when there is a stable drug concentration.
• Of all the volumes of distribution, Vss is probably the most useful for calculating the loading dose
###### Which factors influence the volume of distribution?
Measurement and pharmacokinetic modelling of Vd Timing of measurements Depending on when the measurements are taken, the Vd will be different (i.e. it will correspond to Vinitial if the measurements are taken too early, and Vextrap if they are taken during the elimination phase). Pharmacokinetic model Vinitial, Vextrap, Varea and Vss are various ways to estimate the Vd of a drug from empirical measurements. All of these methods will yield slightly different results - or, occasionally completely different results. Free vs. total drug levels In highly protein bound drugs, the calculated volume of distribution for the "total" drug levels will be totally different to the Vd calculated for the free drug. Total Vd will correspond to the Vd of the binding protein rather than the drug itself. Properties of the drug Molecule size The larger the molecule, the harder it will be for it to passively diffuse out of the central compartment, and therefore the smaller the Vd. Molecule charge Highly ionised charged molecules will have higher water solubility, and may even be trapped in the central compartment by electrostatic factors which keep them bound to proteins with corresponding charge. pKa pKa determines the degree of ionisation and therefore influences lipid solubility Lipid solubility Lipid solubility is one of the major determinants of Vd; highly lipid-soluble drugs will have the highest Vd values because of the low fat content of the bloodstream. Water solubility Highly water-soluble drugs will have difficulty penetrating lipid bilayer membranes and generally tent do have smaller volumes of distribution, essentially being limited to extracellular water. Properties of the patient's body fluids pH pH interactes with the drug's pKa to influence the degree of lipid solubility. pH also influences the degree of protein binding (a good exmaple of this is ionised calcium) Body water volume Dehydrated patients will have drug levels concentrated in the plasma just as all dissolved substances are concentrated by loss of water. Protein levels For highly protein-bound drugs, lower serum protein levels will result in a higher free (unbound) drug fraction. This may have little effect on the Vd as calculated from total drug concentration, but if you are measuring free drug levels it will make the Vd appear smaller. Displacement Drugs may be displaced from their protein and tissue binding sites by the effects of pH or by competition from other drugs/substances (eg. urea). Displaced drugs mayl redistribute into plasma, decreasing the calculated Vd. Effects of physiology and pathological states Age As an old professor of mine had put it, babies are grapes and the elderly are raisins. As you age, body water content decreases, shrinking the Vd of water-soluble drugs. Muscle mass also decreases, and so tissue binding diminishes. Gender Female Vds tend to be higher than male Vds due to the generally higher body water content Pregnancy Both the body water and the body fat content increases, and therefore the Vd increases for most drugs. Not to speak of the possible distribution into amniotic fluid and foetus. Oedema Oedema represents increased body water and this influences water-soluble substances; Vd for these will increase Ascites / effusions Just as in oedema, large fluid collections may sequester water soluble drugs and act as reservoirs. Effects of apparatus Adsorption on to apparatus Dialysis filters and ECMO circuits tend to adsorb drugs in an unpredictable fashion, resulting in an apparent increase in the volume of distribution. Volume expansion In the context of bypass circuits and other large extracorporeal machinery, there may be 2000-2500ml of additional extracorporeal fluid, which will change the volume of distribution (particularly for drugs which are largely confined to the central compartment)
## References
Gibaldi, M., and P. J. McNamara. "Apparent volumes of distribution and drug binding to plasma proteins and tissues." European journal of clinical pharmacology 13.5 (1978): 373-378.
Toutain, Pierre-Louis, and Alain BOUSQUET‐MÉLOU. "Volumes of distribution." Journal of veterinary pharmacology and therapeutics 27.6 (2004): 441-453.
Riegelman, S., J. Loo, and M. Rowland. "Concept of a volume of distribution and possible errors in evaluation of this parameter." Journal of Pharmaceutical Sciences 57.1 (1968): 128-133.
Krishna, Sanjeev, and Nicholas J. White. "Pharmacokinetics of quinine, chloroquine and amodiaquine." Clinical pharmacokinetics30.4 (1996): 263-299.
Wagner, John G. "Significance of ratios of different volumes of distribution in pharmacokinetics." Biopharmaceutics & drug disposition 4.3 (1983): 263-270. | 1,524 | 7,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-21 | latest | en | 0.863933 |
http://library.thinkquest.org/06aug/01704/properties%20practice.htm | 1,371,535,187,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706961352/warc/CC-MAIN-20130516122241-00018-ip-10-60-113-184.ec2.internal.warc.gz | 122,183,889 | 2,940 | Properties Practice Multiplication with Jungle JoePracticing Properties of Multiplication with Froodle Click the answer you think is correct. What property of multiplication is this an example of? 6 x 1= 6 A: commutative property B: identity property C: distributive property What property of multiplication is this an example of?4 x 6= 24 6 x 4= 24 A: commutative property B: zero property C: associative property What property of multiplication is this an example of?7 x (4 x 2) = 56 4 x (7 x 2) = 56 A: identity property B: distributive property C: associative property What property of multiplication is this an example of?6 x 0 = 0 A: zero property B: commutative property C: distributive property What property of multiplication is this an example of?3 x (8 + 3) = 33 (3 x 8) + (3 x 3) = 33 A: associative property B: identity property C: zero property Front Page Froodle's Page | 242 | 913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.761135 |
http://www.mathworks.com/matlabcentral/cody/problems/1-times-2-start-here/solutions/150003 | 1,485,278,593,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00044-ip-10-171-10-70.ec2.internal.warc.gz | 569,006,599 | 11,596 | Cody
# Problem 1. Times 2 - START HERE
Solution 150003
Submitted on 17 Oct 2012 by Daniel Boland
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
%% assert(isequal(times2(1),2));
``` y = 2 ```
2 Pass
%% assert(isequal(times2(11),22));
``` y = 22 ```
3 Pass
%% assert(isequal(times2(-3),-6));
``` y = -6 ```
4 Pass
%% assert(isequal(times2(29),58));
``` y = 58 ``` | 160 | 498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-04 | latest | en | 0.672531 |
http://www.purplemath.com/learning/viewtopic.php?t=2570 | 1,527,046,444,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00369.warc.gz | 442,231,734 | 6,382 | ## can't remember how to solve (x+5)-2(4x-1)=0
Simple patterns, variables, the order of operations, simplification, evaluation, linear equations and graphs, etc.
angelajoyner31
Posts: 2
Joined: Tue Apr 24, 2012 1:40 pm
Contact:
### can't remember how to solve (x+5)-2(4x-1)=0
I cant remember how to solve for x!!!
buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
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### Re: can't remember how to solve (x+5)-2(4x-1)=0
I cant remember how to solve for x!!!
you can see how here. if your having trouble with the brackets, try this article on "simplifying w/ parentheses". the 1st step will be x+5-8x+2=0
MrAlgebra
Posts: 11
Joined: Tue May 24, 2011 7:36 pm
Location: NC
Contact:
### Re: can't remember how to solve (x+5)-2(4x-1)=0
Start off by simplifying everything that you can. Maybe try multiplying out the -2(4x-1) part?
(x + 5) - 2(4x - 1) = 0
x + 5 - 8x + 2 = 0
Don't forget that you have to multiply out by -2 instead of just 2, so you'll get -8x + 2 instead of 8x - 2.
From there, you can simplify some more and try to get x by itself so you have something that looks like x = ___ that tells you the answer. | 386 | 1,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-22 | latest | en | 0.926828 |
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The solution of a 6x6 and a 9x9 puzzles with "strange cages"
Author Message
Posted on: Thu Nov 17, 2011 12:33 am
Posts: 694
Joined: Fri May 13, 2011 6:51 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
pnm wrote:
That last one ("version e") had 9 solutions, computed in 0.03 seconds.
You see how hard it is to create puzzles with a single solution: especially
for the larger puzzles I like to add more large cages, but they
(1) increase solving time, and (2) make it more likely there is more than one
solution. Also, generally, puzzles with more symmetry are more likely
to have more than one solution (can't prove this, only intuition confirmed).
Another cheap way of reducing the number of solutions is by adding single cell cages
(e.g. split a double cell, or remove one from a larger cell).
version a (8 cages): 42672 solutions
version b (9 cages): 4280
version c (10 cages): 512
version d (11 cages): 66
version e (12 cages): 9
Looking to the curve I think that one more restriction would finally give the unique solution, maybe we can try it tomorrow, each number is obtained dividing the previous one by 8, more or less.
Yes, it is hard to create puzzles with the unique solution. When we add single cell cages we are limiting the "liberty" of those numbers to fly over the grid, this is what apparently happens when we have many cages, we are reducing the degrees of "freedom" of the numbers to move elsewhere (in the extreme case 36 cages in a 6x6... solved!!!), so a big proportion of cages... easy puzze, less cages... difficult puzzle... your intuition is good. It looks like to create a very difficult puzzle it is necessary to find the best compromise between the size of the puzzle and the size of the wider cage. This "crossing" point seems to be in 5 (in the very difficult 9x9's on tuesdays, which curiously are symmetric with respect to the horizontal and to the vertical axis). Sometimes we see 7-cell or 6-cell (big cages) in the 10x10's or the 12x12's but by increasing the total number of cages the difficulty of the puzzle decreases, so a big number of cages looks to compensate for the presence of a few "big cages" (I do not know if the "solver rating" would be related in some way with this "proportionality" but intuitively I would say that yes).
Posted on: Thu Nov 17, 2011 7:48 pm
Posts: 694
Joined: Fri May 13, 2011 6:51 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
pnm wrote:
That last one ("version e") had 9 solutions, computed in 0.03 seconds.
... Another cheap way of reducing the number of solutions is by adding single cell cages
(e.g. split a double cell, or remove one from a larger cell).
Here goes a new split (but for two cells).
Version f:
6
11,+,a1b1c1
2,+,b2
16,+,a2a3b3
4,+,c2c3
150,x,d1d2d3e3f3
4,+,e2
2,-,e1f1f2
450,x,a4a5a6b6c6
1,-,b4c4c5
3,+,b5
19,+,d4e4f4f5f6
3,-,d5d6e6
5,+,e5
Posted on: Thu Nov 17, 2011 8:57 pm
Posts: 2215
Joined: Thu May 12, 2011 11:58 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
version f: 4 solutions, computed in 0.0029 seconds.
now we have only four, here they are (will help in determining
how to make it have a unique solution):
2 4 5 3 6 1
6 2 1 5 4 3
4 6 3 1 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
---
solution 2:
2 4 5 3 6 1
6 2 1 5 4 3
4 6 3 2 1 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 1 2 4
---
solution 3:
2 4 5 1 3 6
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 6 3
1 3 4 6 5 2
3 5 6 2 1 4
---
solution 4:
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Posted on: Fri Nov 18, 2011 11:23 pm
Posts: 694
Joined: Fri May 13, 2011 6:51 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
pnm wrote:
version f: 4 solutions, computed in 0.0029 seconds.
now we have only four, here they are (will help in determining
how to make it have a unique solution):
...
solution 4:
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Hi, Patrick, the solution 4 is the one we have being working with, let's try a new split (14 cages in total) version g, and see this time, thanks:
6
11,+,a1b1c1
2,+,b2
16,+,a2a3b3
4,+,c2c3
180,x,d1d2d3e1e3
4,+,e2
3,+,f1
5,x,f2f3
450,x,a4a5a6b6c6
1,-,b4c4c5
3,+,b5
19,+,d4e4f4f5f6
3,-,d5d6e6
5,+,e5
Posted on: Sat Nov 19, 2011 12:31 am
Posts: 2215
Joined: Thu May 12, 2011 11:58 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
2 solutions this time (0.0007 sec):
2 4 5 1 6 3
6 2 1 3 4 5
4 6 3 5 2 1
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
---
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Posted on: Sat Nov 19, 2011 2:46 pm
Posts: 694
Joined: Fri May 13, 2011 6:51 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
pnm wrote:
2 solutions this time (0.0007 sec):
2 4 5 1 6 3
6 2 1 3 4 5
4 6 3 5 2 1
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
---
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Hi, Patrick, I have moved the "cage limitting line" from the left of cell d3 to the top of this cell d3, then keeping fourteen cages for this version h:
6
11,+,a1b1c1
2,+,b2
16,+,a2a3b3
9,+,c2c3d3e3
30,x,d1d2e1
4,+,e2
3,+,f1
5,x,f2f3
450,x,a4a5a6b6c6
1,-,b4c4c5
3,+,b5
19,+,d4e4f4f5f6
3,-,d5d6e6
5,+,e5
Posted on: Sat Nov 19, 2011 3:03 pm
Posts: 2215
Joined: Thu May 12, 2011 11:58 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
Still two solutions (version g)
2 4 5 1 6 3
6 2 1 3 4 5
4 6 3 5 2 1
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
---
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Posted on: Sat Nov 19, 2011 3:33 pm
Posts: 694
Joined: Fri May 13, 2011 6:51 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
pnm wrote:
Still two solutions (version g)
2 4 5 1 6 3
6 2 1 3 4 5
4 6 3 5 2 1
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
---
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
No, something is wrong, the version h is the one to be tested, not the g (already tested), see that the first of those two solutions is not valid since d1 x e1 x d2 = 1 x 6 x 3 = 18 and not 30, which is the code I sent (graphic), thanks
Posted on: Sat Nov 19, 2011 3:48 pm
Posts: 2215
Joined: Thu May 12, 2011 11:58 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
Ah, yes, sorry: 1 solution (0.00036 seconds):
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Difficulty factor for this puzzle would be 21 (so it would rate somewhere
between the medium and difficult puzzles of the site)
Patrick
Posted on: Sat Nov 19, 2011 4:12 pm
Posts: 694
Joined: Fri May 13, 2011 6:51 pm
Re: The solution of a 6x6 and a 9x9 puzzles with "strange ca
pnm wrote:
Ah, yes, sorry: 1 solution (0.00036 seconds):
2 4 5 1 6 3
6 2 3 5 4 1
4 6 1 3 2 5
5 1 2 4 3 6
1 3 4 6 5 2
3 5 6 2 1 4
Difficulty factor for this puzzle would be 21 (so it would rate somewhere
between the medium and difficult puzzles of the site)
Patrick
Yes, it has finally become really easy, but it has been very instructive the full process to insure the unique solution starting from 42672 solutions. To do something similar with the 9x9 is an utopia due to the huge number of initial solutions in case of finding all (probably many hours!... or days!). The conclusion is that it's better not using so big cages or "strange cages" and to find always that compromise we were talking about between the size of the puzzle and the maximum size of the cages.
Also, your intuition with respect to the introduction of single-cell cages has resulted good. In fact, it can be observed that those single cells many times are the key for the solution of the big puzzles (10x10's and 12x12's) and probably are, as you say, not only the cheap but the only way to guarantee a unique solution.
Thank you very much (I will take soon the liberty of "bombarding" your solver with new challenges...).
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https://homeworkontime.com/2022/07/08/american-university-r-studio-programing-data-science-task-1-use-relative-paths-to-load-these-data-frames-into-r-r-evaltrue2-these-data-ar/ | 1,675,349,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00749.warc.gz | 305,809,547 | 13,265 | # American University R Studio Programing Data Science Task 1. Use relative paths to load these data frames into R. “`{r, eval=TRUE} “` 2. These data ar
American University R Studio Programing Data Science Task 1. Use relative paths to load these data frames into R.
“`{r, eval=TRUE}
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2. These data are messy. The observational units in `fert`, `life`, and `pop` are locations in space-time (e.g. Aruba in 2017). Recall that tidy data should have one observational unit per row.
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+ Include only the years 1960, 1970, 1980, 1990, 2000, and 2010. Facet by these years.
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6. Make a bar plot of population vs region for the year 2017.
+ Order the bars on the \$y\$-axis in **decreasing** order of population.
+ Your final plot should look like this:
“`{r, eval = TRUE}
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https://www.belgeci.com/6-0-the-structure-mapping-theory-of-analogy.html | 1,721,299,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00405.warc.gz | 583,638,975 | 17,237 | # 6.0 The Structure-Mapping Theory of Analogy
Induction, as we have analyzed it, requires a store of patterns on which to operate. We have
not said how these patterns are to be obtained. Any general global optimization algorithm could
be applied to the problem of recognizing patterns in an environment. But pattern recognition is a
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One might propose to determine, by induction, an effective pattern recognition algorithm. But
although this might be possible, it would be very slow. Such a process probably occurred during
the evolution of intelligent species, but within the lifespan of one organism there is simply not
time.
In Chapter 9, I will propose that intelligent entities solve pattern recognition problems with a
"perceptual hierarchy" that applies the multilevel philosophy of global optimization sketched in
Chapter 2. Among other things, this perceptual hierarchy makes continual use of two processes:
analogy and deduction. And deduction, it will be argued in Chapter 8, is also essentially
dependent on analogy. Hence analogical reasoning is an essential part of the picture of
intelligence drawn at the end of the previous chapter.
WHAT IS ANALOGY
What I mean by analogy is, roughly, reasoning of the form "A is similar to B in respect X,
therefore A is also similar to B in respect Y." As with induction, it is difficult to say exactly why
analogy works as well as it does. But there is no disputing its effectiveness. Bronowski (1956)
has driven the point home so forcefully that an extended quote seems appropriate:
Man has only one means to discovery, and that is to find likeness between things. To
him, two trees are like two shouts and like two parents, and on this likeness he has built all
mathematics. A lizard is like a bat and like a man, and on such likenesses he has built the theory
of evolution and all biology. A gas behaves like a jostle of billiard balls,and on this and kindred
likenesses rests much of our atomic picture of matter.
In looking for intelligibility in the world, we look for unity; and we find this (in the arts
as well as in science) in its unexpected likenesses. This indeed is man’s creative gift, to find or
make a likeness where none was seen before — a likeness between mass and energy, a link
between time and space, an echo of all our fears in the passion of Othello.
So, when we say that we can explain a process, we mean that we have mapped it in the likeness of another process which we know to work. We say that a metal crystal stretchesbecause its layers slide over one another like cards in a pack, and then that some polyester yarns stretch and harden like a metal crystal. That is, we take from the world round us a few models of structure and process (the particle, the wave, and so on), and when we research into nature, we try to fit her with these models.
Even more intriguingly, Bronowski goes on to relate analogy with structure:
Yet one powerful procedure in research, we know, is to break down complex events into
simpler parts. Are we not looking for the understanding of nature in these? When we probe
below the surface of things, are we not trying, step by step, to reach her ultimate and
fundamental constituents?
We do indeed find it helpful to work piecemeal. We take a sequence of events or an
assembly to pieces: we look for the steps in a chemical reaction, we carve up the study of an
animal into organs and cells and smaller units within a cell. This is our atomic approach, which
tries always to see in the variety of nature different assemblies from a few basic units. Our search
is for simplicity, in that the distinct units shall be few, and all units of one kind identical.
And what distinguishes one assembly of these units from another? the elephant from the
giraffe, or the right-handed molecule of sugar from the left-handed? The difference is in the
organization of the units into the whole; the difference is in the structure. And the likenesses for
which we look are also likenesses of structure.
This is the true purpose of the analytic method in science: to shift our gaze from the thing
or event to its structure. We understand a process, we explain it, when we lay bare in it a
structure which is like one we have met elsewhere.
What Bronowski observed in the history and psychology of science, Gentner and Gentner
(1983) have phrased in a more precise and general way. They speak of the "Generative Analogy
Hypothesis" — the hypothesis that analogies are used in generating inferences. And in order to
test this hypothesis, they setforth a specific theoretical framework for analogical processing,
called "structure-mapping." According to this framework, analogical reasoning is concerned with
deriving statements about a target domain T from statements about a base domain B. Each
domain is understood to consist of a number of "nodes" and a collection of relations between
these nodes. Essentially, a node may be any sort of entity — an object, a color, etc. A structure-
mapping begins with a relation which takes certain base nodes into certain target nodes: if the
source nodes are (b1,…,bn) and the target nodes are (t1,…,tn), it is a map M(bi)=tj, where i ranges
over some subset of (1,…,n). Analogy occurs when it is assumed that a relation which holds
between bi and bk also holds between M(bi) and M(bk).
The theory of structure-mapping analogy states that reasoning of this form is both common
and useful. This hypothesis has been verified empirically — e.g. by studying the way people
reason about electricity by analogy to water and other familiar "base" domains. Furthermore, the
evidence indicates that, as Gentner and Gentner put it, relations "are more likely to be imported
into the target if they belong to a system of coherent, mutually constraining relationships, the others of which map into the target." If a relation is part of a larger pattern of relationships which have led to useful analogies, people estimate that it is likely to lead to a useful analogy.
The structure-mapping theory of analogy — sketched by Bronowski and many others and
formalized by Gentner and Gentner — clearly captures the essence of analogical reasoning. But it
is not sufficiently explanatory — it does not tell us, except in a very sketchy way, why certain
relations are better candidates for analogy than others. One may approach this difficult problem
by augmenting the structure-mapping concept with a more finely-grained pattern-theoretic
approach.
INDUCTION, DEDUCTION, ANALOGY
Peirce proclaimed the tendency to take habits to be the "one law of mind", and he divided this
law into three parts: deduction, induction, and abduction or analogy. The approach of computer
science and mathematical logic, on the other hand, is to take deductive reasoning as primary, and
then analyze induction and analogy as deviations from deduction. The subtext is that deduction,
being infallible, is the best of the three. The present approach is closer to Peirce than it is to the
standard contemporary point of view. When speaking in terms of pattern, induction and analogy
are more elementary than deduction. And I will argue that deduction cannot be understood
except in the context of a comprehensive understanding of induction and analogy.
STRUCTURAL SIMILARITY
As in Chapter 3, define the distance between two sequences f and g as d#(f,g)=%(P(f)-
P(g))U(P(g)-P(f)%#. And define the approximation to d#(f,g) with respect to a given set of
functions S as
dS(f,g)=%[(S%P(f))-(S%P(g))]U[(S%P(g))-(S%P(f))]%#. This definition is the key to our
analysis of analogy.
A metric is conventionally defined as a function d which satisfies the following axioms:
1) d(f,g) % d(g,h) + d(f,h)
2) d(f,g) = d(g,f)
3) d(f,g) % 0
4) d(f,g)=0 if and only if f=g.
Note that d# is not a metric, because it would be possible for P(f) to equal P(g) even if f and g
were nonidentical. And it would not be wise to consider equivalence classes such that f and g are
in the same class if and only if d#(f,g)=0, because even if d#(f,g)=0, there might exist some h
such that d#(Em(f,h),Em(g,h)) is nonzero. That is, just because d#(f,g)=0, f and g are not for all
practical purposes equivalent. And the same argument holds for dS — for dS(f,g)=0 does not in
general imply dS(Em(f,h),Em(g,h)), and hence there is little sense in identifying f and g. A
function d which satisfies the first three axioms given above might be called a pseudometric; that
is how d# and dS should be considered.
TRANSITIVE REASONING
To understand the significance of this pseudometric, let us pause to consider a "toy version" of
analogy that might be called transitive reasoning. Namely, if we reason that "f is similar to g, and
g is similar to h, so f is similar to h," then we are reasoning that "d(f,g) is small, and d(g,h) is
small, so d(f,h) is small." Obviously, the accuracy of this reasoning is circumstance-dependent.
Speaking intuitively, in the following situation it works very well:
gf
h
But, of course, one may also concoct an opposite case:
f gh
Since our measure of distance, d#, satisfies the triangle inequality, it is always the case that
d#(f,h) % d#(g,h) + d#(f,g). This puts a theoretical bound on thepossible error of the associated
form of transitive reasoning. In actual circumstances where transitive reasoning is utilized, some
approximation to d# will usually be assumed, and thus the relevant observation is that dS(f,h) %
dS(g,h) + dS(f,g) for any S. The fact that dS(f,h) is small, however, may say as much about S as
about the relation between f and h. The triangle inequality is merely the final phase of transitive
reasoning; equally essential is to the process is the pattern recognition involved in approximating
d#.
Kaynak: A New Mathematical Model of Mind
belgesi-944 | 2,324 | 9,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.969219 |
https://www.code4example.com/tag/c-programs-with-output/ | 1,718,642,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00504.warc.gz | 618,731,604 | 26,799 | # Tag - C++ Programs with Output
C++
## C++ Relational Operators with Examples
In C++, Relational Operators are useful to check the relation between two operands like we can...
C++
## C++ Programs Examples with Output
C++ stands out as a robust and advanced programming language, offering high-level capabilities...
C++
## C++ Program to Calculate the Power of a Number Using pow Function
In this program, we use C++ Math.pow() function to calculate the power of the given base. C++...
C++
## C++ Program to Calculate the Power of a Number Using For Loop
In this example, i’ll show you How to Calculate power of a number using a for loop in C++...
C++
## C++ Program to Calculate the Power of a Number Using While Loop
In this program, base and exponent are assigned values 3 and 4 respectively. Using the while loop...
C++
## C++ Program to Find Average of Marks
Formula: Average=Obtain Marks * 100 / Total Marks will be used in this program to find Average of...
C++
## C++ Program to Convert Minutes into a number of Years and Days
Program to calculate Number of Years and Days according to given minutes. There is total 525600...
C++
## Reverse an Integer in C++
C++ program to reverse the number entered by the user, then displays the reversed number on the...
C++
## BMI Calculator Program in C++
The Body Mass Index (BMI) is a quick way to assess your body size simply with your weight and...
C++
## C++ Program to Calculate Perimeter of Triangle
In this Program, we will calculate the Perimeter of the Triangle using Formula Perimeter=(a+b+c) in... | 362 | 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} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.704361 |
https://circlecoder.com/random-pick-with-weight/ | 1,680,247,376,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00169.warc.gz | 199,527,669 | 9,336 | # Random Pick With Weight Problem
## Description
LeetCode Problem 528.
You are given a 0-indexed array of positive integers w where w[i] describes the weight of the i^th index.
You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w).
• For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%).
Example 1:
``````1
2
3
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8
Input
["Solution","pickIndex"]
[[[1]],[]]
Output
[null,0]
Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.
``````
Example 2:
``````1
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Input
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output
[null,1,1,1,1,0]
Explanation
Solution solution = new Solution([1, 3]);
solution.pickIndex(); // return 1. It is returning the second element (index = 1) that has a probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.
Since this is a randomization problem, multiple answers are allowed.
All of the following outputs can be considered correct:
[null,1,1,1,1,0]
[null,1,1,1,1,1]
[null,1,1,1,0,0]
[null,1,1,1,0,1]
[null,1,0,1,0,0]
......
and so on.
``````
Constraints:
• 1 <= w.length <= 10^4
• 1 <= w[i] <= 10^5
• pickIndex will be called at most 10^4 times.
``````1
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class Solution {
public:
vector<int> s;
Solution(vector<int> w) {
for (int i : w) s.push_back(s.empty() ? i : (i + s.back()));
}
int pickIndex() {
int m = s.back();
int r = rand() % m;
auto it = upper_bound(s.begin(), s.end(), r);
return it - s.begin();
}
};
`````` | 699 | 2,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.736328 |
https://www.sarthaks.com/331108/the-density-material-in-si-units-is-128-kg-in-certain-units-in-which-the-unit-of-length-is-25-cm?show=331111 | 1,555,629,233,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00162.warc.gz | 814,380,586 | 12,375 | # The density of a material in SI units is 128 kg m^–3. In certain units in which the unit of length is 25 cm
12 views
The density of a material in SI units is 128 kg m–3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is :
(1) 40
(2) 16
(3) 410
(4) 640
+1 vote
selected ago
Correct option: (1) 40
Explanation: | 128 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-18 | longest | en | 0.841421 |
https://codeforces.com/blog/entry/64444 | 1,560,831,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998605.33/warc/CC-MAIN-20190618023245-20190618045245-00179.warc.gz | 393,450,624 | 19,607 | Apptica's blog
By Apptica, history, 5 months ago, ,
Hello everyone!
I would like to invite you to participate in HackerEarth HourStorm #7. It’s the seventh version of the short contest, that runs for 1 hour! The problem set consists of 3 traditional algorithmic tasks of various difficulties. The contest starts on the 11th of January, 9.30- PM IST
For traditional algorithmic tasks, you will receive points for every test case your solution passes — so you can get some points with partial solutions as well. Check contest page for more details about in-contest schedule and rules.
Great thanks to raghavkapoor1997 for preparing the tasks and isaf27 for testing them. As usual, there will be some prizes for the top three competitors:
$75 Amazon gift card$50 Amazon gift card
In addition, top 5 on the scoreboard with rating less than 1600 will win HackerEarth t-shirts. Good luck to everyone, and let's discuss the problems after the contest!
• +41
» 5 months ago, # | 0 Auto comment: topic has been updated by Apptica (previous revision, new revision, compare).
» 5 months ago, # | +16 I didn't participate in, but when I look at "Number Game": the statement definitely is wrong.
• » » 5 months ago, # ^ | 0 And what's wrong with it?
• » » » 5 months ago, # ^ | +1 The statement might miss condition {Ai} must be pairwise distinct.
• » » » » 5 months ago, # ^ | 0 What theorem does it use anyway?
• » » » » » 5 months ago, # ^ | +10 If a polynomial of degree ≤ N is zero at N + 1 distinct points then it is zero polynomial.
» 5 months ago, # | +24 The problem statement for Orthogonal Pairs was not clear at all.The problem says to compute "the number of ordered quadruples" (P1, P2, P3, P4) such that the angle between the lines P1P2 and P3P4 is exactly . This was not actually what the problem was asking:First, P1, P2, P3, P4 do not have to be pairwise distinct; in particular, you are allowed to form a angle with just three points. This was not specified anywhere. Moreover, all of the following quadruples are considered the same as (P1, P2, P3, P4): (P2, P1, P3, P4), (P1, P2, P4, P3), (P3, P4, P1, P2), etc.In other words the quadruples aren't ordered at all. (And they aren't quadruples at all, but rather pairs of pairs.) A better formulation of the problem would have been to first generate all of the line segments formed by an (unordered) pair of the N points. Then count the number of (unordered) pairs of line segments that form perpendicular lines to each other.The sample explanation was also broken during the entirety of the contest, which made things even worse. But having seen the actual explanation, it did not clear up any of the above issues anyway. The only way I was able to eventually guess what the problem was asking was by reverse engineering the sample output.
• » » 5 months ago, # ^ | +18 The second line in the problem statement about the quadruples is ambiguous and we have updated the problem statement. We apologize for the same and thank you for the suggestion. | 795 | 3,036 | {"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} | 3.015625 | 3 | CC-MAIN-2019-26 | latest | en | 0.925492 |
https://www.mathnasium.com/uk/blog/20151208-mental-math-from-fahrenheit-to-celsius-and-back | 1,718,829,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00171.warc.gz | 766,153,736 | 11,779 | Get Started by Finding a Local Centre
# Mental Math: From Fahrenheit to Celsius... and Back!
Dec 8, 2015
(Image: Ilya via Flickr)
Winter break is almost here! If you're spending the holiday season visiting the United States, you'll probably have to make some adjustments on how you view temperature. This is because the United States remains the only country in the world that measures temperature in degrees Fahrenheit (ËšF). Everywhere else in the world, temperature is measured in degrees Celsius (ËšC). So, if you travel to the United States, you’d have to convert the temperature expressed in ËšF to its equivalent in ËšC to understand how hot or cold it is... unless you get lucky and find a thermometer that uses both units of measure!
Converting temperatures from ËšC to ËšF involves multiplying the temperature in Celsius by 1.8 and adding 32 to your answer. This can take a while to do in your head!
However, we can make a quick estimate by rounding the numbers we use when converting between the two units. Rounding makes it easier to work with the numbers and do the conversion mentally. 1.8 rounded to the nearest whole number is 2, and 32 rounded to the nearest ten is 30. Now here’s what to do:
1. Take the temperature in degrees Celsius and double it (multiply by 2).
2. Add 30 to your answer. The sum will be approximately the same temperature in degrees Fahrenheit.
Looking to convert from Fahrenheit to Celsius? Estimate by working backwards using inverse operations! Take the temperature in Fahrenheit, subtract 30, and then divide the answer by 2.
This winter break, where will your travels take you?
## OUR METHOD WORKS
Mathnasium meets your child where they are and helps them with the customised program they need, for any level of mathematics. | 395 | 1,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.906991 |
http://www.public-domain-content.com/encyclopedia/Family/Function.shtml | 1,369,511,712,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706153698/warc/CC-MAIN-20130516120913-00032-ip-10-60-113-184.ec2.internal.warc.gz | 668,257,499 | 7,359 | #### Webmasters, increase productivity, download the whole site in zip files.Database size Public: 874.98 Megs.Premium Members: 4.584 Gig.Message Boards
Function
The word "function" refers to basic concepts in several disciplines:
* In sociology, social functions are the basis of functionalism.
* In computer science, a function is a subprogram or subroutine. See also
functional programming.
Introduction
The concept of function is a generalization of the common notion of a
"mathematical formula". Functions describe special mathematical
relationships between two objects, x and y=f(x). The object x is called the
argument of the function f, and y is said to "depend functionally" on x.
Intuitively, a function is a way to assign to each value of the argument x a
unique value of the function f(x). This could be specified by a formula, a
relationship, and/or a rule. This concept is deterministic, always producing
the same result from the same input (the generalization to random values is
called a stochastic process). A function may be thought of as a "machine" or
"black box" converting valid input into a unique output.
The most familiar kind of function is that where the argument and the
function's value are both numbers, and the functional relationship is
expressed by a formula, and the value of the function is obtained from the
arguments by direct substitution. Consider for example
f(x) = x2
which assigns to any number x its square.
A straightforward generalization is to allow functions depending not on a
single number, but on several. For instance,
g(x,y) = xy
which takes two numbers x and y and assigns to them their product, xy.
According to how a function is specified, it is said to be an explicit
function (as above) or an implicit function, as in
xf(x) = 1
which implicitly specifies the function
f(x) = 1 / x
We have seen that the intuitive notion of function is not limited to
computations using single numbers, but the mathematical notion of function
is not limited to computations, or even to situations involving numbers at
all. Because of this generality, functions appear in a wide variety of
mathematical contexts, and several mathematical fields are based on the
study of functions.
It should be noted that the words "function", "mapping", "map" and
"transformation" are usually used synonymously. Furthermore, functions may
occasionally referred to as well-defined function or total function (See the
section "Formal Definition" below).
History
As a mathematical term, "function" was coined by Leibniz in 1694, to
describe a quantity related to a curve; such as a curve's slope or a
specific point of said curve. Functions related to curves are nowaday called
differentiable functions and are still the most frequently type of functions
encounted by non-mathematicians. For such kind of functions, one can talk
about limits and derivatives; both are measurements of the change of output
values associated to a change of input values, and they are the basics of
calculus.
The word function was later used by Euler during the mid-18th Century to
describe an expression involving various arguments; ie: y = F(x). By
broadening the definition of functions, mathematicians were then able to
study "strange" mathematical objects such as functions which are nowhere
differentiable. Those functions, first thought as purely imaginary and
called collectively "monsters" as late as the turn of the 20th century, were
later found to be important in the modelling of physical phenomena such as
Brownian motion.
During the 19th Century, mathematicians started to formalize all the
different branches of mathematics. Weierstrass advocated building calculus
on arithmetic rather than on geometry, which favoured Euler's definition
over Leibniz's (see arithmetization of analysis). Towards the end of the
century, mathematicians started trying to formalize all of mathematics using
set theory and they sought definitions of every mathematical object as a
set. It was Dirichlet that gave the modern "formal" definition of function.
In Dirichlet's definition, a function is a special case of a relation, which
is a set. In most cases of practical interest, however, the differences
between the modern definition and Euler's definition is negligible.
Formal Definition
Formally, a function f from a set X of input values to a set Y of possibly
output values (written as f: X → Y) is a relation between X and Y which
satisfies:
1. f is functional: if x f y (x is f-related to y) and x f z, then y = z.
i.e., for each input value, there should only be one possible output
value.
2. f is total: for all x in X, there exists a y in Y such that x f y. i.e.
for each input value, the formula should produce at least one output
value within Y.
For each input value x in the domain, the corresponding unique output value
y in the codomain is denoted by f(x).
Consider the following three examples:
[NotMap1.png]This is not a "well-defined" function; because, the element 3, in X, is associated with two elements b and c in Y
(Condition 1 is violated). This is a multivalued function.
[NotMap2.png]This is not a "well-defined" function; because, the element 1, in XÊ, is associated with nothing (Condition 2 is
violated). This is a partial function.
This is a function, called a discrete function (or rarely piecewise function); of which the range is {a,c,d}. It can
be stated explicitly as
[Mathmap.png]
[f(x)=\left\{\begin{matrix} a, & \mbox{if }x=1 \\ d, & \mbox{if }x=2 \\ c, & \mbox{if }x=3. \end{matrix}\right.]
Occasionally, all three relations above are called functions. In this case,
the function satisfies Conditions (1) and (2) is said to be a "well-defined
function" or "total function". In this encyclopedia, the terms "well-defined
function", "total function" and "function" are synonymous.
Domains, Codomains, and Ranges
X, the set of input values, is called the domain of f and Y, the set of
possible output values, is called the codomain. The range of f is the set of
all actual outputs {f(x) : x in the domain}. Beware that sometimes the
codomain is wrongly called the range because of a failure to distinguish
between possible and actual values.
In computer science, specifying the datatypes of the arguments and return
values sets the domain and codomain (respectively) of a subprogram. So the
domain and codomain are constraints imposed initially on a function; on the
other hand the range has to do with how things turn out in practice.
Graph of a functions
The graph of a function f is the collection of all points(x, f(x)), for all
x in set X. In the example of the discrete function, the graph of f is
{(1,a),(2,d),(3,c)}. There are theorems formulated or proved most easily in
terms of the graph, such as the closed graph theorem.
If X and Y are real lines, then this definition coincides with the familiar
sense of graph.
Note that since a relation on the two sets X and Y is usually formalized as
a subset of X×Y, the formal definition of function actually identifies
the function f with its graph.
Images and preimages
The image of an element x∈X under f is the output f(x).
The image (or direct image) of A⊂X under f is the subset of Y defined by
f(A)Ê:= {f(x)Ê: x in A}.
Notice that the range of f is the image f(X) of its domain. In our example
of discrete function, the image of {2,3} under f is f({2,3})={c,d} and the
range of f is {a,c,d}.
The preimage of y∈Y is the set f−1(y)={x∈X : f(x)=y}. If the
set is a singleton {x}, then we simply say that x=f−1(y) is the
preimage of y.
The preimage (or inverse image) of B ⊂ Y under f is the subset of X
defined by
fÊ−1(B)Ê:= {x in XÊ: f(x)∈B}.
In our example of discrete function, the preimage of {a,b} is
fÊ−1({a,b})={1}.
Note that with this definiton, fÊ-1 becomes a function whose domain is the
set of all subsets of Y (also known as the power set of Y) and whose
codomain is the power set of X.
Some consequences that follow immediately from these definitions are:
* f(A1Ê∪ÊA2)Ê= f(A1)Ê∪Êf(A2).
* f(A1Ê∩ÊA2)Ê⊆ f(A1)Ê∩Êf(A2).
* fÊ−1(B1Ê∪ÊB2)Ê= fÊ−1(B1)Ê∪ÊfÊ−1(B2).
* fÊ−1(B1Ê∩ÊB2)Ê= fÊ−1(B1)Ê∩ÊfÊ−1(B2).
* f(fÊ−1(B))Ê⊆ÊB.
* fÊ−1(f(A))Ê⊇ÊA.
The results relating images and preimages to the algebra of intersection and
union work for any number of sets, not just for 2.
Injective, surjective and bijective functions
Several types of functions are very useful, deserve special names:
* injective (one-to-one) functions send different arguments to different
values; in other words, if x and y are members of the domain of f, then
f(x) = f(y) if and only if x = y. Our example is an injective function.
* surjective (onto) functions have their range equal to their codomain;
in other words, if y is any member of the codomain of f, then there
exists at least one x such that f(x) = y.
* bijective functions are both injective and surjective; they are often
used to show that the sets X and Y are "the same" in some sense.
Examples of functions
(More can be found at List of functions.)
* The relation wght between persons in the United States and their
weights.
* The relation between nations and their capitals.
* The relation sqr between natural numbers n and their squares n2.
* The relation nlog between positive real numbers x and their natural
logarithms ln(x). Note that the relation between real numbers and their
natural logarithms is not a function because not every real number has
a natural logarithm; that is, this relation is not total and is
therefore only a partial function.
* The relation dist between points in the plane R2 and their distances
from the origin (0,0).
* The relation grav between a point in the punctured plane R2Ê\ {(0,0)}
and the vector describing the gravitational force that a certain mass
at that point would experience from a certain other mass at the origin
(0,0).
Most commonly used types of mathematical functions involving addition,
division, exponents, logarithms, multiplication, polynomials, radicals,
rationals, subtraction, and trigonometric expressions. They are sometimes
collectively referred as Elementary functions -- but the meaning of this
term varies among different branches of mathematics. Example of
non-elementary functions are Bessel functions and gamma functions.
n-ary function: function of several variables
Functions in applications are often functions of several variables: the
values they take depend on a number of different factors. From a
mathematical point of view all the variables must be made explicit in order
to have a functional relationship - no 'hidden' factors are allowed. Then,
again from the mathematical point of view, there is no qualitative
difference between functions of one and of several variables. A function of
three real variables is just a function that applies to triples of real
numbers. The following paragraph says this in more formal language.
If the domain of a function is a subset of the Cartesian product of n sets
then the function is called an n-ary function. For example, the relation
dist has the domain RÊ×ÊR and is therefore a binary function. In that
case dist((x,y)) is simply written as dist(x,y).
Another name applied to some types of functions of several variables is
operation. In abstract algebra, operators such as "*" are defined as binary
functions; when we write a formula such as x*y in this context, we are
implicitly invoking the function *(x,y), but writing it in a convenient
infix notation.
An important theoretical paradigm, functional programming, takes the
function concept as central. In that setting, the handling of functions of
several variables becomes an operational matter, for which the lambda
calculus provides the basic syntax. The composition of functions (see under
composing functions immediately below) becomes a question of explicit forms
of substitution, as used in the substitution rule of calculus. In
particular, a formalism called currying can be used to reduce n-ary
functions to functions of a single variable.
Composing functions
The functions f:ÊXÊ→ÊY and g:ÊYÊ→ÊZ can be composed by first
applying f to an argument x and then applying g to the result. Thus one
obtains a function gÊoÊf: XÊ→ÊZ defined by (gÊoÊf)(x)Ê:= g(f(x)) for
all x in X. As an example, suppose that an airplane's height at time t is
given by the function h(t) and that the oxygen concentration at height x is
given by the function c(x). Then (cÊoÊh)(t) describes the oxygen
concentration around the plane at time t.
If [Y\sub X] then f may compose with itself; this is sometimes denoted
f². (Do not confuse it with the notation commonly seen in
trigonometry.) The functional powers [f\circ f^n=f^n\circ f=f^{n+1}] for
natural n follow immediately. On their heels comes the idea of functional
root; given f and n, find a g such that gn=f. (Feynman illustrated practical
use of functional roots in one of his anecdotal books. Tasked with
building an analogue arctan computer and finding its parts overstressed, he
instead designed a machine for a functional root of arctan and
chained enough copies to make the arctan machine.)
Inverse function
If a function f:X→Y is bijective then preimages of any element y in the
codomain Y is a singleton. A function taking y∈Y to its preimage
f−1(y) is a well-defined function called the inverse of f and is
denoted by f−1.
An example of an inverse function, for f(x) = x2, is f(x)−1 =
√x. Likewise, the inverse of 2x is x/2. The inverse function is the
function that "undoes" its original.
Pointwise operations
If f:ÊXÊ→ÊR and g:ÊXÊ→ÊR are functions with common domain X and
codomain is a ring R, then one can define the sum function fÊ+Êg: XÊ→ÊR
and the product function fÊ×Êg: XÊ→ÊR as follows:
(fÊ+Êg)(x)Ê:= f(x)Ê+Êg(x);
(fÊ×Êg)(x)Ê:= f(x)Ê×Êg(x);
for all x in X.
This turns the set of all such functions into a ring. The binary operations
in that ring have as domain ordered pairs of functions, and as codomain
functions. This is an example of climbing up in abstraction, to functions of
more complex types.
By taking some other algebraic structure A in the place of R, we can turn
the set of all functions from X to A into an algebraic structure of the same
type in an analogous way.
Enumeration
The number of computable functions (aka calculable functions) from integers
to integers is countable and its size is the transfinite number aleph-null,
which is written [\aleph_0]. The number of all functions from integers to
integers is higher: the same as the cardinality of the real numbers. This
counting argument shows that there are functions from integers to integers
that are not computable. For examples of noncomputable functions, see the
articles on the halting problem and Rice's theorem.
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Alibabacloud.com offers a wide variety of articles about sudoku solver 3x3, easily find your sudoku solver 3x3 information here online.
### [Leetcode 36&37] Valid Sudoku & SudokuSolver (Sudoku problem)
Title Link: Valid-sudokuImport java.util.arrays;/** * Determine if a Sudoku is valid, according To:sudoku puzzles-the rules.the Sudoku Board Co Uld be partially filled, where empty cells is filled with the character '. A partially filled sudoku which is valid. Note:a Valid Sudoku board (partially filled) is not necessa
### [Leetcode] SudokuSolver Solving Sudoku
numbers are legal, which can make the program more efficient, as shown in the code:classSolution { Public: voidSolvesudoku (vectorChar> > Board) { if(Board.empty () | | board.size ()! =9|| board[0].size ()! =9)return; Solvesudokudfs (board,0,0); } BOOLSolvesudokudfs (vectorChar> > board,intIintj) {if(i = =9)return true; if(J >=9)returnSolvesudokudfs (board, i +1,0); if(Board[i][j] = ='.') { for(intK =1; K 9; ++k) {Board[i][j]= (Char) (k +'0'); if(isValid (board, I, J))
### [Leetcode] [Python]37:sudokuSolver
#-*-Coding:utf8-*-‘‘‘__author__ = ' [email protected] '37:sudoku Solverhttps://oj.leetcode.com/problems/sudoku-solver/Write a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character '.Assume that there would be is only one unique solution.===comments by dabay===Progressive
### [Leetcode] SudokuSolver
Write a program to solve a Sudoku puzzle by filling the empty cells. Empty cells are indicated by the character‘.‘. You may assume that there will be only one unique solution. A Sudoku puzzle... ... And its solution numbers marked in red. The sudoku solver was finally reached. With the basis of valid
### The implementation method of C + + for programming beauty SudokuSolver
Programming the beauty of the first chapter of the 15th section, talking about the construction of Sudoku, a start to get this problem really no idea, but read the book in the introduction, found that the original solution of the idea and the N queen problem is consistent, but do not know why, anyway, at first did not think of this backtracking method, know that is solved by backtracking, The problem becomes much easier.Here we do not intend to implem
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### Leetcode Notoginseng SudokuSolver (C,c++,java,python)
Solvesudoku (vectorPython source code (636MS):Class Solution: # @param {character[][]} board # @return {void} do not return anything, modify board In-place Instea D. def solvesudoku (self, Board): self. SudoKu (board,0,0) def SudoKu (self,board,i,j): If I==8 and J==9:return True if j==9:i+=1;j=0 if boa Rd[i][j]!= '. ': if self. SudoKu (board,i,j+1
### [Leetcode] SudokuSolver @ Python
Original title address: https://oj.leetcode.com/problems/sudoku-solver/Test instructionsWrite a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character ‘.‘ .Assume that there would be is only one unique solution.A Sudoku Puzzle ...... and its solution numbers marked in red.
### LeetCode: SudokuSolver, leetcodesudoku
LeetCode: Sudoku Solver, leetcodesudoku Sudoku Solver Total Accepted: 13937 Total Submissions: 66832My Submissions Write a program to solve a Sudoku puzzle by filling the empty cells. Empty cells are indicated by the character'.'. You may assume that there will be only one
### Leetcode DFS SudokuSolver
Sudoku Solver Total accepted:11799 Total submissions:56732 My submissions Write a program to solve a Sudoku puzzle by filling the empty cells. Empty cells are indicated by the character'.'. You may assume that there will be only one unique solution. A Sudoku puzzle... ... And its solution numbers marked in
### Leetcode---37. SudokuSolver
complexity: O (n2)1 class Solution2 {3 int used1[9][9], Used2[9][9], used3[9][9];4 5 Public:6 void Solvesudoku(VectorVectorChar> > Board)7 {8 VectorVectorint>> Place;9 for(int I = 0; I Board.size(); ++ I)Ten for(int J = 0; J Board[I].size(); ++ J) One { A if(Board[I][J] == '. ') - Place.push_back({I, J}); - Else the { - int Num = Board[I][J] - '
### Leetcode SudokuSolver
{ - return true; - } theBoard[i][j]= '. '; - } - return false; - } + } - } + return true; A } at Private BooleanIsvalidsudoku (Char[] board,intRowintColum) { - for(intj=0;j) - if(J! = Colum Board[row][j] = =Board[row][colum]) - return false; - -
### [Leetcode] [Java] SudokuSolver
Title:Write a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character ‘.‘ .Assume that there would be is only one unique solution.A Sudoku Puzzle ...... and its solution numbers marked in red.Test Instructions:Write a program to solve Sudoku problems by filling in whitespace.These spaces are characters‘.‘填充。You can as
### Leetcode SudokuSolver
Topic Connectionhttps://leetcode.com/problems/sudoku-solver/Sudoku solverdescriptionWrite a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character ‘.‘ .Assume that there would be is only one unique solution.A Sudoku Puzzle .....
### [Leetcode] Algorithm topic-SudokuSolver
the termination of the condition of recursion, to reach the termination conditions, the inspection results are qualified, qualified records down. And the essence of backtracking is in the following for loop, a total of 5 steps, the last two steps is just the first two steps of the anti-operation, and therefore become ' backtracking '. We can argue that backtracking is an optimization of brute force search (brute).A little gossip, to see Sudoku
### [Leetcode] SudokuSolver
Sudoku SolverWrite a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character ‘.‘ .Assume that there would be is only one unique solution.A Sudoku Puzzle ...... and its solution numbers marked in red.Problem Solving Ideas:I didn't play Sudoku before, so I went to the interne
### "Leetcode" SudokuSolver
} - - - for(intj=0;j9; j + +) + { - if(j==j0)Continue; + if(board[i0][j]==target) A { at return false; - } - } - - for(inti=i0/3*3; i3*3+3; i++) - { in - for(intj=j0/3*3; j3*3+3; j + +) to { + if(i==i0j==j0)Continue; - if(board[i][j]==target) the { * return false; \$
### [Leetcode] (python): 037-sudokuSolver
Source of the topic https://leetcode.com/problems/sudoku-solver/Write a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character ‘.‘ .Assume that there would be is only one unique solution. Test instructions Analysis Input:a Unsolved SudokuOutput:a solved SudokuConditions: Meet the
### Panax SudokuSolver
/** 37. Sudoku Solver * 2015.12.13 by Mingyang * 1. Length standard: None * 2. Optional range: All the wood is worth the point, pick a number from 1 to 9 * 3. Take a step forward: if put in is validate, then Put * 4. Take a step back: if you put it in and the back is false, change the point back to * 5. Special case: No * 6. On repetition: none * The IsValid of this topic is difficult * row = I/3 * 3; Row *
### Panax SudokuSolver (Graph; WFS)
Write a program to solve a Sudoku puzzle by filling the empty cells.Empty cells is indicated by the character ‘.‘ .Assume that there would be is only one unique solution.A Sudoku Puzzle ...... and its solution numbers marked in red.classSolution { Public: voidSolvesudoku (vectorChar>> Board) {Backtracking (board,0); } BOOLBacktracking (vectorChar>> board,intLine ) { //Find first empty cell
### 037. SudokuSolver
1 classSolution {2 Public:3 voidSolvesudoku (vectorChar>>Board) {4 Sudoku (board);5 }6 Private:7 BOOLSudoku (vectorChar>>Board)8 {9 for(inti =0; I 9; ++i) {Ten for(intj =0; J 9; ++j) { One if(Board[i][j] = ='.') { A for(intK =0; K 9; ++k) { -BOARD[I][J] ='1'+K; - if(IsValid (board, I, J) Sudoku (board))retur
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How to calculate marginal costs and benefits (from total costs and benefits), and how to use that information to calculate equilibrium
This example problem goes over the degree of comfort experienced at different levels of clean air. The different dollar value amounts are shown for every 10% increase in clean air. We want to find the optimum amount of clean air that we should have in this area, and it is important to remember that there is an optimal amount of pollution. Having 100% clean air is probably never going to be the solution.
So the first step is to recognize the type of data we are working with. In this problem we have a table of information showing us what the benefits and costs are for different levels of clean air.
As we would expect, the more clean air we have in our economy, the higher the benefit we receive (we prefer clean air over dirty air). However, as we produce (or clean) more of the clean air we also incur a cost. As more clean air is present, the higher our costs. Note that benefits here is the same as utility,
% of Clean Air Total Benefits Total Cost 0% 0 0 10% 50 45 20% 130 50 30% 205 58 40% 269 68 50% 319 81 60% 351 96 70% 371 115 80% 386 150 90% 398 200 100% 406 280
The next logical step is to calculate the marginal benefits (marginal utility), and marginal costs. In order to do this we should begin at 0% clean air. When we move to 10% clean air, we see that benefits go up by 50, and costs go up by 45. This means that our marginal benefit from 10% clean air is 50, and our marginal cost of 10% clean air is 45. When we move from 10% to 20% we see total benefit change from 50 to 130. This means that marginal benefit from another 10% increase in clean air (from 10% to 20%) is 80 (130-50). The increase in cost is 5 (50-45) which is also our marginal cost. We can do these calculations all the way through and we will end up with the following table:
% of Clean Air Total Benefits Total Cost Marginal Benefits Marginal Cost 0% 0 0 10% 50 45 50 45 20% 130 50 80 5 30% 205 58 75 8 40% 269 68 64 10 50% 319 81 50 13 60% 351 96 32 15 70% 371 115 20 19 80% 386 150 15 35 90% 398 200 12 50 100% 406 280 8 80
You can see that marginal costs are rising as more clean air is produced, and that marginal benefits are decreasing as more clean air is produced. This result is consistent with the theory of diminishing marginal utility (for marginal benefits), and diminishing marginal returns (for increasing marginal costs).
If we graph out the marginal costs and marginal benefits, we will get our typical looking supply and demand graph where marginal costs represent supply (supply of c lean air) and marginal benefits represent demand (demand for clean air).
Now there are two ways to find the optimum amount of clean air. The most basic economic way is to figure out where marginal benefits equal marginal costs. In our table above, they are never perfectly equal, but at 70% clean air, we see that marginal benefit is 20, and marginal cost is 19. This is pretty close, and since MB are greater than MC then this is a valid solution. The true answer will probably be around 71% but 70\$ is pretty close.
Another way to do it is to find the difference between total benefits and total costs and choose the biggest number. Theoretically either method will give us the same result so let’s find out:
% of Clean Air Total Benefits Total Cost Net Benefit 0% 0 0 0 10% 50 45 5 20% 130 50 80 30% 205 58 147 40% 269 68 201 50% 319 81 238 60% 351 96 255 70% 371 115 256 80% 386 150 236 90% 398 200 198 100% 406 280 126
As you can see, our 70% clean air level gives us the highest net benefit. Although it is only 1 higher than the 60% clean air level, it is higher. This result is consistent with our MB = MC analysis so we should be confident that 70% clean air is the correct amount for this economy.
Essay Writer on February 19, 2016 at 1:53 AM said...
Very well-written article! Thaks for sharing this informative post! Keep posting your articles!
assignment writer on August 9, 2016 at 4:10 AM said...
Reading this article was interesting for me! It is also quite surprising that 100% clean air is never going to be the solution. It is certain that the clearer the air is, the higher the costs are. However, the clearer the air is, the more benefits we get. assignment writer
Rory on August 9, 2016 at 11:29 PM said...
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https://justaaa.com/accounting/117886-find-the-future-value-of-an-ordinary-annuity-of | 1,686,215,804,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00682.warc.gz | 364,860,462 | 9,398 | Question
# Find the future value of an ordinary annuity of \$1,000 paid quarterly for 9 years, if...
Find the future value of an ordinary annuity of \$1,000 paid quarterly for 9 years, if the interest rate is 9%, compounded quarterly. (Round your answer to the nearest cent.)
Future value is present value of cash flow+ interest earned.
When there is uniform series of cash flow occuring at the end of each period , we shall use ordinary annuity.
Here \$1,000 is paid at the end of each quarter for 9 years.
We will use the following formula:
Where,
FV = ?
A =\$1,000
m=number of period (compounded per year) 4 [quarterly]
t= number of years = 9
r=0.09 (9%)
=\$54,569.6
Thus, the future value would be \$54,569.6
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#### Earn Coins
Coins can be redeemed for fabulous gifts. | 246 | 950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.954447 |
https://math.stackexchange.com/questions/2257558/deriving-inverse-of-laplace-transform | 1,566,105,990,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313617.6/warc/CC-MAIN-20190818042813-20190818064813-00351.warc.gz | 554,862,773 | 29,639 | # deriving inverse of Laplace transform
I am trying to derive the inverse of the laplace transform myself, but I'm immediately stuck.
The laplace transform is: $$\mathcal L[f(t)](s)=\int_0^\infty e^{-st}f(t)dt$$
Intuitively, my first attempt would be to get rid of the integral with respect to $t$, by taking the derivative: $$\frac{d \mathcal L[f(t)](s)}{dt}=e^{-st}f(t), \text{or something...}$$
But this doesn't work, because the integral is not a function of $t$, since it is definite.
Can someone give me a hint on how to solve it, or a possible misconception that I may have?
• Are you try to get the inverse result for a particular function by the inverse theorem, or you want to prove the inverse theorem? As the proof of the inverse theorem is never about algebraic operations, it's very involving. – Yujie Zha Apr 29 '17 at 13:19
• I'm trying to derive the general formula for the inverse of the laplace transform. – user56834 Apr 29 '17 at 13:28
• Maybe you could check out this doc – Yujie Zha Apr 29 '17 at 13:32 | 289 | 1,031 | {"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} | 2.9375 | 3 | CC-MAIN-2019-35 | latest | en | 0.920089 |
https://videocide.com/glossary/binary-coded-decimal/ | 1,585,403,746,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491998.11/warc/CC-MAIN-20200328134227-20200328164227-00521.warc.gz | 762,913,523 | 13,942 | # Binary-coded decimal
A coding system in which each decimal digit from 0 to 9 is represented by four binary (0 or 1) digits.
In computing and electronic systems, binary-coded decimal (BCD) is a class of binary encodings of decimal numbers where each decimal digit is represented by a fixed number of bits, usually four or eight. Special bit patterns are sometimes used for a sign or for other indications (e.g., error or overflow).
In byte-oriented systems (i.e. most modern computers), the term unpacked BCD usually implies a full byte for each digit (often including a sign), whereas packed BCD typically encodes two decimal digits within a single byte by taking advantage of the fact that four bits are enough to represent the range 0 to 9. The precise 4-bit encoding may vary, however, for technical reasons, see Excess-3 for instance. The ten states representing a BCD decimal digit are sometimes called tetrades (for the nibble typically needed to hold them also known as tetrade) with those don't care-states unused named pseudo-tetrad(e)s or pseudo-decimal digit).
BCD's main virtue is its more accurate representation and rounding of decimal quantities as well as an ease of conversion into human-readable representations, in comparison to binary positional systems. BCD's principal drawbacks are a small increase in the complexity of the circuits needed to implement basic arithmetics and slightly less dense storage.
BCD was used in many early decimal computers and is implemented in the instruction set of machines such as the IBM System/360 series and its descendants, Digital Equipment Corporation's VAX, the Burroughs B1700, and the Motorola 68000-series processors. Although BCD per se is not as widely used as in the past and is no longer implemented in newer computers' instruction sets (such as ARM; x86 does not support its BCD instructions in long mode anymore), decimal fixed-point and floating-point formats are still important and continue to be used in financial, commercial, and industrial computing, where subtle conversion and fractional rounding errors that are inherent in floating-point binary representations cannot be tolerated.
### Key Terms
bcd
binary coded decimal
class binary encodings
decimal digit
decimal numbers
electronic systems
fixed number bits
indications
sign
special bit patterns
No ressources found.
BCD
### Synonymns
Binary-coded decimal
(none found) | 499 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | longest | en | 0.940344 |
http://www.mathworks.com/help/physmod/simscape/ref/rotarypneumaticpistonchamber.html?requestedDomain=www.mathworks.com&nocookie=true | 1,500,685,886,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423839.97/warc/CC-MAIN-20170722002507-20170722022507-00466.warc.gz | 499,665,417 | 13,724 | # Documentation
### This is machine translation
Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.
# Rotary Pneumatic Piston Chamber
Rotational pneumatic piston chamber based on ideal gas law
## Library
Pneumatic Elements
## Description
The Rotary Pneumatic Piston Chamber block models a pneumatic rotary piston chamber based on the ideal gas law and assuming constant specific heats. Use this model as a building block for pneumatic rotational actuators. The piston can generate torque in one direction only, and the direction is set by the Chamber orientation parameter.
The continuity equation for the network representation of the piston chamber is
`$G=\frac{{V}_{0}+D·\theta }{RT}\left(\frac{dp}{dt}-\frac{p}{T}\frac{dT}{dt}\right)+\frac{D}{RT}·p·\frac{d\theta }{dt}$`
where
G Mass flow rate at input port V0 Initial chamber volume D Piston displacement (volume per unit angle) Θ Piston angle p Absolute pressure in the chamber R Specific gas constant T Absolute gas temperature t Time
The energy equation is
`$q=\frac{{c}_{v}}{R}\left({V}_{0}+D·\theta \right)\frac{dp}{dt}+\frac{{c}_{p}·D}{R}p\frac{d\theta }{dt}-{q}_{w}$`
where
q Heat flow due to gas inflow in the chamber (through the pneumatic port) qw Heat flow through the chamber walls (through the thermal port) cv Specific heat at constant volume cp Specific heat at constant pressure
The torque equation is
`$\tau =p·D$`
Port A is the pneumatic conserving port associated with the chamber inlet. Port H is a thermal conserving port through which heat exchange with the environment takes place. Ports C and R are mechanical rotational conserving ports associated with the piston case and rod, respectively. The gas flow and the heat flow are considered positive if they flow into the chamber.
### Variables
Use the Variables tab in the block dialog box (or the Variables section in the block Property Inspector) to set the priority and initial target values for the block variables prior to simulation. For more information, see Set Priority and Initial Target for Block Variables.
## Basic Assumptions and Limitations
• The gas is ideal.
• Specific heats at constant pressure and constant volume, cp and cv, are constant.
## Parameters
Displacement
Specify the effective piston displacement, as volume per unit angle. The default value is `.001` m^3/rad.
Initial angle
Specify the initial piston angle. The default value is `0`.
Specify the volume of gas in the chamber at zero piston position. The default value is `1e-5` m^3.
Chamber orientation
Specify the direction of torque generation. The piston generates torque in a positive direction if this parameter is set to `1` (the default). If you set this parameter to `2`, the piston generates torque in a negative direction.
## Ports
The block has the following ports:
`A`
Pneumatic conserving port associated with the chamber inlet.
`H`
Thermal conserving port through which heat exchange with the environment takes place.
`R`
Mechanical rotational conserving port associated with the piston (rod).
`C`
Mechanical rotational conserving port associated with the reference (case). | 731 | 3,205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.809673 |
https://rdrr.io/cran/HyperbolicDist/man/momIntegrated.html | 1,642,954,433,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00150.warc.gz | 530,454,665 | 9,353 | # momIntegrated: Moments Using Integration In HyperbolicDist: The hyperbolic distribution
## Description
Calculates moments and absolute moments about a given location for the generalized hyperbolic and related distributions.
## Usage
`1` ```momIntegrated(densFn, order, param = NULL, about = 0, absolute = FALSE) ```
## Arguments
`densFn` Character. The name of the density function whose moments are to be calculated. See Details. `order` Numeric. The order of the moment or absolute moment to be calculated. `param` Numeric. A vector giving the parameter values for the distribution specified by `densFn`. If no `param` values are specified, then the default parameter values of each distribution are used instead. `about` Numeric. The point about which the moment is to be calculated. `absolute` Logical. Whether absolute moments or ordinary moments are to be calculated. Default is `FALSE`.
## Details
Denote the density function by f. Then if `order`=k and `about`=a, `momIntegrated` calculates
integral_{-infinity}^infinity (x - a)^k f(x) dx
when `absolute = FALSE` and
integral_{-infinity}^infinity |x - a|^k f(x) dx
when `absolute = TRUE`.
Only certain density functions are permitted.
When `densFn="ghyp"` or `"generalized hyperbolic"` the density used is `dghyp`. The default value for `param` is `c(1,1,0,1,0)`.
When `densFn="hyperb"` or `"hyperbolic"` the density used is `dhyperb`. The default value for `param` is `c(0,1,1,0)`.
When `densFn="gig"` or `"generalized inverse gaussian"` the density used is `dgig`. The default value for `param` is `c(1,1,1)`.
When `densFn="gamma"` the density used is `dgamma`. The default value for `param` is `c(1,1)`.
When `densFn="invgamma"` or `"inverse gamma"` the density used is the density of the inverse gamma distribution given by
f(x) = u^alpha exp(-u)/(x Gamma(alpha)), u = theta/x
for x > 0, alpha > 0 and theta > 0. The parameter vector `param = c(shape,rate)` where `shape` =alpha and `rate`=1/theta. The default value for `param` is `c(-1,1)`.
When `densFn="vg"` or `"variance gamma"` the density used is `dvg` from the package VarianceGamma. In this case, the package VarianceGamma must be loaded or an error will result. The default value for `param` is `c(0,1,0,1)`.
## Value
The value of the integral as specified in Details.
## Author(s)
David Scott d.scott@auckland.ac.nz, Christine Yang Dong c.dong@auckland.ac.nz
`dghyp`, `dhyperb`, `dgamma`, `dgig`, `VarianceGamma`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17``` ```### Calculate the mean of a generalized hyperbolic distribution ### Compare the use of integration and the formula for the mean m1 <- momIntegrated("ghyp", param = c(1/2,3,1,1,0), order = 1, about = 0) m1 ghypMean(c(1/2,3,1,1,0)) ### The first moment about the mean should be zero momIntegrated("ghyp", order = 1, param = c(1/2,3,1,1,0), about = m1) ### The variance can be calculated from the raw moments m2 <- momIntegrated("ghyp", order = 2, param = c(1/2,3,1,1,0), about = 0) m2 m2 - m1^2 ### Compare with direct calculation using integration momIntegrated("ghyp", order = 2, param = c(1/2,3,1,1,0), about = m1) momIntegrated("generalized hyperbolic", param = c(1/2,3,1,1,0), order = 2, about = m1) ### Compare with use of the formula for the variance ghypVar(c(1/2,3,1,1,0)) ``` | 985 | 3,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-05 | longest | en | 0.565113 |
https://philoid.com/question/35154-suppose-that-for-a-particular-economy-investment-is-equal-to-200-government-purchases-are-150-net-taxes-that-is-lump-sum-taxes-m | 1,696,405,590,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511361.38/warc/CC-MAIN-20231004052258-20231004082258-00691.warc.gz | 500,972,788 | 10,923 | ## Book: Introductory Macroeconomics
### Chapter: 5. The Government: Budget And The Economy
#### Subject: Social Science - Class 12th
##### Q. No. 5 of Exercises
Listen NCERT Audio Books - Kitabein Ab Bolengi
5
##### Suppose that for a particular economy, investment is equal to 200, government purchases are 150, net taxes (that is lump-sum taxes minus transfers) is 100 and consumption is given by C = 100 + 0.75Y (a) What is the level of equilibrium income? (b) Calculate the value of the government expenditure multiplier and the tax multiplier. (c) If government expenditure increases by 200, find the change in equilibrium income.
Given is –
I = 200
G = 150
T = 100
C = 100 + 0.75 Y
So, C (Autonomous consumption) = 100
And, MPC (c) = 0.75
(a) What is the level of equilibrium income?
Level of equilibrium income = 1/(1-c) [C - cT + I + G]
= 1/ (1 – 0.75) [100+ (0.75 x 100) + 200 + 150]
= 1/0.25 [375]
= 1500
(b) Calculate the value of the government expenditure multiplier and the tax multiplier.
Government expenditure multiplier ∆Y/ ∆G = 1/1-c
= 1/ 1- 0.75
= 1/0.25
= 4
Tax multiplier ∆Y/ ∆T = -c/1-c
= -0.75/0.25
= - 3
(c) If government expenditure increases by 200, find the change in equilibrium income.
New level of equilibrium income = 1/(1-c) [C - cT + I + G + ∆G], Where ∆G = 200
= 2300
Change in equilibrium income = 2300 – 1500 = 800
5
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14 | 473 | 1,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.844389 |
http://stackoverflow.com/questions/8812073/ray-and-square-rectangle-intersection-in-3d | 1,430,011,427,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246652114.13/warc/CC-MAIN-20150417045732-00088-ip-10-235-10-82.ec2.internal.warc.gz | 255,689,876 | 19,914 | # Ray and square/rectangle intersection in 3D
Hei. Are making a game and are looking for a ray intersection onto a square or a rectangle only in 3D space. Have search the web and found many solutions but nothing i can understand have a line and line segment intersection script in 2D but i cant figure out have to make it 3D. It is not important from what side it intersect the square or rectangle but it must be able to retrive the point of intersection vector so that later can be tested for distance to se if it occurred before or after other intersections on the same ray intersection.
Any examples in python or other similar scripting languages will be greatly appreciated
Edit: Dont know have to modify the 2D to show an exaple but made a new and posting both.
``````//this is the exaple it test a ray onto a plane then look to se if that point is in the rectangle and saves it to test for distanse later
list Faces; //triangle faces
list Points; //
vector FindPoint(){
//calcute the point of intersection onto the plane and returns it
//if it can intersect
//else return ZERO_VECTOR
}
//return 1 if the point is in the rectangular on the plane
//else return 0
}
default{
state_entry(){
integer n = (Faces != []); //return number of elements
integer x = 0;
while(x < n){
vector intersection = FindPoint( FromList(Faces, x) ); //take out a element and runs it trough the function
if(intersection != ZERO_VECTOR){
integer test = point-in-quadrilateral( FromList(Faces, x) ); //find out if the point is in rectangular
if(test == 1){ //if so
Points += intersection; //save the point
}
}
++x;
}
float first; //the distanse to the box intersection
integer l = (Points != []);
integer d;
while(d < l){
if(Dist( FromList(Points, d) ) < first) //if the new distanse is less then first
return 0; //then end script
++d;
}
}
}
//this is the 2D version
vector lineIntersection(vector one, vector two, vector three, vector four){
float bx = two.x - one.x;
float by = two.y - one.y;
float dx = four.x - three.x;
float dy = four.y - three.y;
float b_dot_d_perp = bx*dy - by*dx;
if(b_dot_d_perp == 0.0) {
return ZERO_VECTOR;
}
float cx = three.x-one.x;
float cy = three.y-one.y;
float t = (cx*dy - cy*dx) / b_dot_d_perp;
if(LineSeg){ //if true tests for line segment
if((t < 0.0) || (t > 1.0)){
return ZERO_VECTOR;
}
float u = (cx * by - cy * bx) / b_dot_d_perp;
if((u < 0.0) || (u > 1.0)) {
return ZERO_VECTOR;
}
}
return <one.x+t*bx, one.y+t*by, 0.0>;
``````
}
-
Create a vector equation for a line in R3, then solve for the intersection of that line in the plane of the rectangle that you are testing it against. After that, it's simple enough to test if that point of solution lies within the bounds.
the parameter t of the solution can be found with:
``````t = (a * (x0 - rx) + b * (y0 - ry) + c * (x0 - rz)) / (a * vx + b * vy + c * vz)
``````
where:
``````a(x - x0) + b(y - y0) + c(z - z0) = 0
``````
is the equation of the plane that your rectangle lies on
and:
``````<x, y, z> = <rx + vx * t, ry + vy * t, rz + vz * t>
``````
is the vector equation of the line in question.
note that:
``````<rx, ry, rz>
``````
is the initial point of the vector equation, and
``````<vx, vy, vz>
``````
is the direction vector of the above equation
After that, plugging the parameter t into your vector equation will give you the point to test for distance.
-
Thank for the respons. Can you explane some part of it? Are the "x0 y0 z0" the equalizer to its equal x, y and z to make the second equation equal to zero? if so how to calculate the value for each of them? – TeaWave Jan 11 '12 at 18:02
Essentially, yes. x0, y0, and z0 are the values of SOME point (it can be any point) that lies on the same plane as your rectangle. As for x, y, and z within that equation, you would plug in the values for the line equation listed above (r_ + v_ * t). Can you be more specific about how you plan on writing the code? Maybe an example of the 2d version you said you were able to work out? I might be able to explain better that way. – Joel Cornett Jan 13 '12 at 2:27
Have edited the main post for an exaple and the 2D version. Have done some more founding out and are woundring if the a, b and c are the unit vector of the plane? and are the x0, y0, and z0 the line equation? have run a multiple combinations but can geat it to equal zero. – TeaWave Jan 14 '12 at 12:58
I edited the post to include an image for clarification. Anyway, the second equation I gave is not important. What you really want is to plug the values into the first equation to get the parameter "t". After that you would plug the "t" into the vector equation for the line (the ray you are tracing -- that's the third equation btw). – Joel Cornett Jan 14 '12 at 21:57
Thanks for the good and persistent help. Have it figured out now : ) – TeaWave Jan 15 '12 at 21:53
The solution is very easy when you define a ray with a point(= vector) and a direction vector, and the rectangle with a point(= vector) and two vectors representing the sides.
Suppose the ray is defined as `R0 + t * D`, where `R0` is the origin of the ray, `D` is an unit vector representing its direction and `t` is its length.
The rectangle can be represented with a corner point `P0`, and two vectors `S1` and `S2` which should represent the sides (their length being equal to the length of the sides). You will need another vector `N` normal to its surface, which is equal to the unit vector along the cross product of `S1` and `S2`.
Now, assume the ray intersects the rect at `P`. Then, the direction of the ray, `D` must make a nonzero angle with the normal `N`. This can be verified by checking `D.N < 0`.
To find the intersection point, assume `P = R0 + a * D` (the point must be on the ray). You need to find the value of `a` now.Find the vector `P0P`. This must be perpendicular to `N`, which means `P0P.N = 0` which reduces to `a = ((P0 - R0).N) / (D.N)`.
Now you need to check if the point is inside the rect or not. To do this, take projection `Q1` of `P0P` along `S1` and `Q2` of `P0P` along `S2`. The condition for the point being inside is then `0 <= length(Q1) <= length(S1)` and `0 <= length(Q2) <= length(S2)`.
This method is appropriate for any type of parallelograms, not only for rectangles.
-
Thanks for the good post. Your post helped triangulate the problem and solve it – TeaWave Jan 15 '12 at 21:56
You don't say whether the square/rectangle in 3D is aligned with the coordinate axes or not. Assuming the 3D rectangle R is oriented arbitrarily in space, here is one method. First interesect your ray r with the plane containing R. This can be accomplished by requiring a scale factor s to multiply r and place it on the plane of R, and solving for s. This gives you a point p on the plane. Now project the plane, and R and p, on to one of the coordinate planes {xy, yz, zx}. You only have to avoid projecting perpendicular to the normal vector to the plane, which is always possible. And then solve the point-in-quadrilateral problem in the plane of projection.
Before beginning, check if your line segment lies in the 3D plane of R, and if so, handle that separately.
-
Thanks for respons. The squares and rectangles can be rotated any way in 3D. So if i understand it correctly the part you write 'First interesect your ray r with the plane containing R. This can be accomplished by requiring a scale factor s to multiply r and place it on the plane of R, and solving for s' is the solution given by Joel Carnett then i use the point of intersection vector given from that and solve it with point-in-quadrilateral to find if it is inside the boundaries. – TeaWave Jan 11 '12 at 17:42
Yes, you understand it. I apparently answered at too high a level, but Joel is essentially articulating in equations what I described in prose. – Joseph O'Rourke Jan 13 '12 at 2:13 | 2,134 | 7,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2015-18 | latest | en | 0.82928 |
http://everything.explained.today/Principle_of_relativity/ | 1,566,640,310,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00117.warc.gz | 73,133,294 | 8,680 | # Principle of relativity explained
In physics, the principle of relativity is the requirement that the equations describing the laws of physics have the same form in all admissible frames of reference.
For example, in the framework of special relativity the Maxwell equations have the same form in all inertial frames of reference. In the framework of general relativity the Maxwell equations or the Einstein field equations have the same form in arbitrary frames of reference.
Several principles of relativity have been successfully applied throughout science, whether implicitly (as in Newtonian mechanics) or explicitly (as in Albert Einstein's special relativity and general relativity).
## Basic concepts
See main article: Galilean invariance and History of special relativity. Certain principles of relativity have been widely assumed in most scientific disciplines. One of the most widespread is the belief that any law of nature should be the same at all times; and scientific investigations generally assume that laws of nature are the same regardless of the person measuring them. These sorts of principles have been incorporated into scientific inquiry at the most fundamental of levels.
Any principle of relativity prescribes a symmetry in natural law: that is, the laws must look the same to one observer as they do to another. According to a theoretical result called Noether's theorem, any such symmetry will also imply a conservation law alongside.[1] [2] For example, if two observers at different times see the same laws, then a quantity called energy will be conserved. In this light, relativity principles make testable predictions about how nature behaves, and are not just statements about how scientists should write laws.
## Special principle of relativity
According to the first postulate of the special theory of relativity:[3]
This postulate defines an inertial frame of reference.
The special principle of relativity states that physical laws should be the same in every inertial frame of reference, but that they may vary across non-inertial ones. This principle is used in both Newtonian mechanics and the theory of special relativity. Its influence in the latter is so strong that Max Planck named the theory after the principle.[4]
The principle requires physical laws to be the same for any body moving at constant velocity as they are for a body at rest. A consequence is that an observer in an inertial reference frame cannot determine an absolute speed or direction of travel in space, and may only speak of speed or direction relative to some other object.
The principle does not extend to non-inertial reference frames because those frames do not, in general experience, seem to abide by the same laws of physics. In classical physics, fictitious forces are used to describe acceleration in non-inertial reference frames.
### In Newtonian mechanics
See main article: Galilean invariance. The special principle of relativity was first explicitly enunciated by Galileo Galilei in 1632 in his Dialogue Concerning the Two Chief World Systems, using the metaphor of Galileo's ship.
Newtonian mechanics added to the special principle several other concepts, including laws of motion, gravitation, and an assertion of an absolute time. When formulated in the context of these laws, the special principle of relativity states that the laws of mechanics are invariant under a Galilean transformation.
### In special relativity
See main article: Special relativity.
Joseph Larmor and Hendrik Lorentz discovered that Maxwell's equations, the cornerstone of electromagnetism, were invariant only by a certain change of time and length units. This left some confusion among physicists, many of whom thought that a luminiferous aether was incompatible with the relativity principle, in the way it was defined by Henri Poincaré:
In their 1905 papers on electrodynamics, Henri Poincaré and Albert Einstein explained that with the Lorentz transformations the relativity principle holds perfectly. Einstein elevated the (special) principle of relativity to a postulate of the theory and derived the Lorentz transformations from this principle combined with the principle of the independence of the speed of light (in vacuum) from the motion of the source. These two principles were reconciled with each other (in Einstein's treatment, though not in Poincaré's) by a re-examination of the fundamental meanings of space and time intervals.
The strength of special relativity lies in its derivation from simple, basic principles, including the invariance of the laws of physics under a shift of inertial reference frames and the invariance of the speed of light in a vacuum. (See also: Lorentz covariance.)
It is, in fact, possible to derive the Lorentz transformations from the principle of relativity alone and obtain the constancy of the speed of light as a consequence. Using only the isotropy of space and the symmetry implied by the principle of special relativity, one can show that the space-time transformations between inertial frames are either Galilean or Lorentzian. In the Lorentzian case, one can then obtain relativistic interval conservation and the constancy of the speed of light.[5]
## General principle of relativity
The general principle of relativity states:[6] That is, physical laws are the same in all reference frames—inertial or non-inertial. An accelerated charged particle might emit synchrotron radiation, though a particle at rest doesn't. If we consider now the same accelerated charged particle in its non-inertial rest frame, it emits radiation at rest.
Physics in non-inertial reference frames was historically treated by a coordinate transformation, first, to an inertial reference frame, performing the necessary calculations therein, and using another to return to the non-inertial reference frame. In most such situations, the same laws of physics can be used if certain predictable fictitious forces are added into consideration; an example is a uniformly rotating reference frame, which can be treated as an inertial reference frame if one adds a fictitious centrifugal force and Coriolis force into consideration.
The problems involved are not always so trivial. Special relativity predicts that an observer in an inertial reference frame doesn't see objects he would describe as moving faster than the speed of light. However, in the non-inertial reference frame of Earth, treating a spot on the Earth as a fixed point, the stars are observed to move in the sky, circling once about the Earth per day. Since the stars are light years away, this observation means that, in the non-inertial reference frame of the Earth, anybody who looks at the stars is seeing objects which appear, to them, to be moving faster than the speed of light.
Since non-inertial reference frames do not abide by the special principle of relativity, such situations are not self-contradictory.
### General relativity
See main article: General relativity.
General relativity was developed by Einstein in the years 1907 - 1915. General relativity postulates that the global Lorentz covariance of special relativity becomes a local Lorentz covariance in the presence of matter. The presence of matter "curves" spacetime, and this curvature affects the path of free particles (and even the path of light). General relativity uses the mathematics of differential geometry and tensors in order to describe gravitation as an effect of the geometry of spacetime. Einstein based this new theory on the general principle of relativity, and he named the theory after the underlying principle.
## Notes and references
1. Book: Classical Mechanics: Hamiltonian and Lagrangian Formalism . Alexei . Deriglazov . Springer . 2010 . 978-3-642-14037-2 . 111 . Extract of page 111
2. Book: The Noether Theorems: Invariance and Conservation Laws in the Twentieth Century . Bertram E. . Schwarzbach . Yvette . Kosmann-Schwarzbach . Yvette Kosmann-Schwarzbach . Springer . 2010 . 0-387-87868-8 . 174 . Extract of page 174
3. Book: The Principle of Relativity: A Collection of Original Memoirs on the Special and General Theory of Relativity . Einstein, A., Lorentz, H. A., Minkowski, H., and Weyl, H.. 111. Dover Publications. Mineola, NY. 1952. 1923. Arnold Sommerfeld. 0-486-60081-5.
4. Book: Einstein's Pathway to the Special Theory of Relativity . Galina . Weistein . Cambridge Scholars Publishing . 2015 . 978-1-4438-7889-0 . 272 . Extract of page 272
5. Yaakov Friedman, Physical Applications of Homogeneous Balls, Progress in Mathematical Physics 40 Birkhäuser, Boston, 2004, pages 1-21.
6. Book: The Theory of Relativity . C. Møller . Oxford University Press . 2nd . Delhi . 0-19-560539-X . 220 . 1952. | 1,873 | 8,755 | {"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} | 3.109375 | 3 | CC-MAIN-2019-35 | latest | en | 0.914857 |
https://stats.stackexchange.com/questions/211092/if-textvarx-infty-is-textvarxy-infty-for-0-le-y-le-1/211138 | 1,713,812,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00571.warc.gz | 472,770,695 | 42,833 | # If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$?
I have a variable $X$ that I know has finite variance (and therefore also finite mean). Is it always true that its variance remains finite after scaling by $0 \le Y \le 1$?
Note that $X$ and $Y$ are not necessarily independent.
Edit: I believe the "worst-case" $Y$ is $0$ whenever $X < c$ and $1$ whenever $X \ge c$, for some $c$ (and the mirrored case)?
• Is $Y$ a random variable? May 5, 2016 at 19:02
• Yes but it may depend on $X$. May 5, 2016 at 19:08
• An utterly trivial inequality sometimes is quite useful in such situations: $\mathbb{E}(X^2Y^2) \le \mathbb{E}(X^2)\sup(Y^2)$. (This is perhaps the simplest special case of Hölder's inequality for $p=1,q=\infty$ applied to $X^2$ and $Y^2$.)
– whuber
May 5, 2016 at 19:35
• Thanks whuber. I believe this leads to the correct solution (see the answer I made)! May 6, 2016 at 0:03
I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.
The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:
\begin{align} \text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\ &\le E(X^2Y^2) \\ &\le E(X^2)\sup(Y^2) \\ &= E(X^2) \\ &= \text{Var}(X) + E(X)^2 \\ &< \infty \end{align}
Note that the result also holds for any bounded $Y$ (since $\sup(Y^2)$ will be finite).
• Also note that we can conclude $|E(XY)| \le \sqrt{\text{Var}(X) + E(X)^2}$, since $\text{Var}(XY) \ge 0 \implies E(XY)^2 \le E((XY)^2)$. May 6, 2016 at 0:41
• I don't believe kjetil assumed independence between $X$ and $Y$. The law of total variation holds in general, and not assuming independence. So I can find nothing in his statement that assumes independence. Also notice that your conclusion is exactly the same as my conclusion that is based off of kjetil's answer. May 6, 2016 at 2:12
• Independence must have been used somewhere (I guess when factoring out the expectation), otherwise the first equation from your answer (as shown in my comment) is stating that $\text{Var}(XY)$ is the same whether or not $X$ and $Y$ are independent, which is a contradiction. The fact that we came to the same conclusion is kind of a "coincidence", because we are both giving upper bounds. Mine comes from dropping the $E(XY)^2$ term and $Y^2 \le \sup(Y^2)$, and yours comes from $\text{Var}(Y), E(Y^2) \le \sup(Y^2)$. May 6, 2016 at 2:30
• I think I figured out where kjetil is using independence. By the law of total variance $Var(XY) = Var(E(XY|Y)) + E(Var(XY|Y))$. If we just look at the first term $Var(E(XY \mid Y)) = Var(Y^2E(X \mid Y))$ which is not the same as $Var(Y^2E(X))$. It is the same only when $X$ and $Y$ are independent. May 6, 2016 at 2:37
• I changed my answer to reflect the changes. May 6, 2016 at 2:41
You need to use the formula $$\DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}} \Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y))$$ where $\Var$ is the variance operator. Take it term for term, write $\mu=\E X, \sigma^2=\Var X$, $\E (XY | Y= y) = \E (yX) = y \E (X) =\mu y$ with variance (over $Y$) $\Var (\mu Y)$ which is finite since $Y$ is bounded.
Then the other term, $\Var (XY | Y=y) = \Var (yX) = y^2 \Var (X) = \sigma^2 y^2$ which again has a finite expectation since $Y$ is bounded. So the answer is yes.
• Nice. To check my understanding this is using the law of total variance? Also this seems to prove something more general: that the variance $XY$ is finite as long as the variance of both $X$ and $Y$ are finite? May 5, 2016 at 19:23
• @Aaron Voelker: There is no need for independence in the calculations. May 6, 2016 at 7:59
• @kjetilbhalvorsen $\mathbb{E}(XY \mid Y=y) = \mathbb{E}(yX)$ does not hold without some assumptions (such as independence). May 6, 2016 at 11:15
• @Juho $\mathbb{E}(1 X) = \mathbb{E}(X)=0.5$, too. The relation $\mathbb{E}(XY\mid Y=y)=\mathbb{E}(X)y$ is an example of a very general theorem called "taking out what is known." It does not require independence of $(X,Y)$. See en.wikipedia.org/wiki/Conditional_expectation#Basic_properties.
– whuber
Nov 9, 2017 at 15:07
• @Juho Sorry, my comment was stupid. Of course I needed to write the conditional expectation in $\mathbb{E}(XY\mid Y=y)=\mathbb{E}(X\mid Y=y) y$. For some reason I just automatically understood these expectations as being conditional even when they weren't... .
– whuber
Nov 9, 2017 at 19:55 | 1,464 | 4,443 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-18 | longest | en | 0.915422 |
https://www.thestudentroom.co.uk/showthread.php?t=102882&page=12 | 1,544,787,361,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825512.37/warc/CC-MAIN-20181214092734-20181214114234-00166.warc.gz | 1,064,792,836 | 47,859 | You are Here: Home >< Maths
1. i wish i could remember my answers
2. Can anyone remember what they got for standard deviation?
15 or 64?
3. (Original post by sequence123)
Can anyone remember what they got for standard deviation?
15 or 64?
it was aorund 15
4. for some strange reason i wrote quartiles, instead od quadrants when justifying which pmcc represented which diagram, in question 1 I them. will I get at least one mark each for identifying correctly.
yeah I did that too.. even if everything else we wrote (that was right0 would have got us the 6 marks do you reckon we will loose marks for writing a bit of rubbish at the end?
5. Anyone remember how they did the interpolation?? I put everyhing into the formula but still ended up with reali high values
6. I'm a bit pissed off about making silly mistakes on a paper that easy...
But I'll tell you what I got wrong:
1. Question 3 I think (with the histogram and frequency density). I answered parts a to d and I *think* I got them right. Couldn't do e and f because I spent half a ****in hour trying to remember the stupid interpolation formula.
2. Buggered up a bit on the box plot I think - I didn't make the line go till 68 or what ever it was - I put it as an outlier - I put Q3+(1.5*IQR) as my max.
3. Their dodgey use of words made me think that there were 6 art students overall... I dunno what overall effect that had on the question.
Apart from that I think I was ok, so anyone up for guessing what grade I might get?
I'm also slightly pissed off about OCR putting new stuff into the Computing papers... cheap way to make us drop marks.
7. i'm sure you'll do fine ...prolly an A im sure :P, i thought there were 6 students for art overall for the last ques too, dammit, and i got the reges. ques wrong too, it depends how u do on your other modules as well but either way theres no pint in worrying, whats done is done, you can't change anything, wait and see till the 18th, until then..positive thinking or not thinking about it at all usually works for me!!
8. To the whisker thing: the text book said if there is an outlier in the upper part, then the upper boundary will be Q3+1.5(IQR) etc...mark a cross as outlier.....still i m not sure abt that.......
9. (Original post by PennyRoyalTea)
i'm sure you'll do fine ...prolly an A im sure :P, i thought there were 6 students for art overall for the last ques too, dammit, and i got the reges. ques wrong too, it depends how u do on your other modules as well but either way theres no pint in worrying, whats done is done, you can't change anything, wait and see till the 18th, until then..positive thinking or not thinking about it at all usually works for me!!
i did that at first then changed it. When i saw "corresponding values" i looked at it for a good 10 mins and realised.
As for the boxplot, there was 1 value about 45mins => this was an outlier therefore it should go to 45mins + a cross at 60 something [the outlier].
10. Do you guys know about the dates Edexcel will release results this year? (Sorry if I'm missing a thread that answers this question :/) Because Maths is my only Edexcel subject, and I'd like to know, but maybe no-one knows when the results will be out yet.
11. (Original post by Paperdoll)
Do you guys know about the dates Edexcel will release results this year? (Sorry if I'm missing a thread that answers this question :/) Because Maths is my only Edexcel subject, and I'd like to know, but maybe no-one knows when the results will be out yet.
All A-level results come out on the same day regardless of the exam board. I think its the third Thursday in August but I'd wait for someone else to confirm that.
12. I think it is always the second to last Thursday in August (GCSE results are on the last Thrusday).
This year I think it is the 18th.
13. (Original post by Alexdel)
I used outlier 52 but didnt know what to do with the 63....I put 45 as highest
52 was the upper extreme of the distribution.
63 was the outlier.
Newton.
14. (Original post by ClaireC)
i wish i could remember my answers
lol same, i have a really bad memory. however i did find S1 pretty ok!
15. (Original post by Chris87)
All A-level results come out on the same day regardless of the exam board. I think its the third Thursday in August but I'd wait for someone else to confirm that.
Thanks, just wondering. Thing is, I'm in Northern Ireland, and last year I got my results on the 24th, but the 3 AQA GCSEs I did didn't make their way over 'til the 26th. I think. Or maybe that's just when they were posted out to me. Woah. I've just realised I have, like, no memory of last year.
(Original post by Newton)
52 was the upper extreme of the distribution.
63 was the outlier.
Are you absolutely sure? I can't remember what I put anyway, but there seems to be huge debate in this thread over the 52/45 thing.
16. (Original post by Paperdoll)
Are you absolutely sure? I can't remember what I put anyway, but there seems to be huge debate in this thread over the 52/45 thing.
Yeah there is a large debate. But what i want to know is, how many marks would you actually lose?
I mean i think the question was 6 or 7 marks?
You definatly got 1/2marks for finding any outliers above/below Q1/Q3 (i think)
You got 1 mark for a correct scale possibley joint with a mark for the scale being labeled
3-4 Marks for drawing the box-plot - with Q1, Q2, Q3, and the lower and upper limits?
Then theres indicating the outlier - a mark for that maybe?
That comes to 6-8 marks already, so i think you would only lose 1 or max 2 if u put the upper limit wrong (as long as you got everything else right)
And also, the 45/52 dilemma on here is most probably the same with everyone else who sat the exam. So its probably something in the examiners report which will say "confusion" or something.
17. (Original post by Paperdoll)
Thanks, just wondering. Thing is, I'm in Northern Ireland, and last year I got my results on the 24th, but the 3 AQA GCSEs I did didn't make their way over 'til the 26th. I think. Or maybe that's just when they were posted out to me. Woah. I've just realised I have, like, no memory of last year.
Are you absolutely sure? I can't remember what I put anyway, but there seems to be huge debate in this thread over the 52/45 thing.
everyone ive spoken to has said thats its 45. THe question said only one value was higher than 45. Fair enough it didnt say there was one at 45, but I did a paper for revision and the question was very similar and in that case 45 would have been the upper limit.
Anyone know how much the m=0.300 question was out of? Thats hopefully the only marks ive dropped.
18. the m=0.300 bit was worth 4 marks. i messed it up took < sign the wrong way lol. but i still standardised so hopefully i should get a mark there (says my maths teacher).
19. (Original post by spikeyguru)
the m=0.300 bit was worth 4 marks. i messed it up took < sign the wrong way lol. but i still standardised so hopefully i should get a mark there (says my maths teacher).
do you know what the answer was. what was the actual value of 'm' ?
20. (Original post by cherc2005)
do you know what the answer was. what was the actual value of 'm' ?
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Question:How do I combine my vectors to get a big matrix?
Hi, Here is what i have
with(LinearAlgebra):
nodes:=9:
Xfull:=RandomMatrix(nodes):
for i from 1 to nodes do
newXfull(i):=zip(`/`,Column(Xfull,i),Row(Xfull,1)(i)):
end do:
the answers are correct, but it gives me 9 sets of 9x1 matrix. How do I write so that it would give me a 9x9 output? As in, all the answers represented in one big matrix.
OR
If I join it one by one, I wrote << newXfull[1]>|< newXfull[2]>|<newXfull[3]>|< newXfull[4]>|< newXfull[5]>|< newXfull[6]>|< newXfull[7]>|< newXfull[8]>|< newXfull[9]>>
But my number for nodes changes, how do I write it so that it changes accordingly? | 247 | 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} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.792806 |
https://origin.geeksforgeeks.org/how-to-make-pca-plot-with-r/ | 1,652,890,210,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00509.warc.gz | 476,209,183 | 28,335 | How To Make PCA Plot with R
• Last Updated : 23 Sep, 2021
Principal component analysis(PCA) in R programming is the analysis of the linear components of all existing attributes. Principal components are linear combinations (orthogonal transformation) of the original predictor in the dataset. It is a useful technique for EDA(Exploratory data analysis) and allowing you to better visualize the variations present in a dataset with many variables.
It works best with numerical data values. In this process the principal components of data are calculated and are used for performing a change of basis on the data, sometimes using only the first few principal components while ignoring the rest. One can take PCA as a kind of linear transformation of the data on the basis of certain data spaces. This transformation fits the data into a coordinate system where the most significant variance is found on the first coordinate, and each subsequent coordinate is orthogonal to the last and has a lesser variance than the previous.
PCA Plot in R
We are gonna work on the ‘Iris’ dataset, which is built into R. It is a multivariate dataset that consists of data on 50 samples from each of three species of Iris (Iris setosa, Iris virginica, and Iris versicolor).
R
`# structure of the iris ` `# dataset ` `str``(iris) ` ` ` `# print the iris dataset ` `head``(iris)`
Output:
As mentioned PCA works best with numerical data we will neglect the categorical variable Species. We are now left with a matrix of 4 columns and 150 rows which we will pass through prcomp( ) function for the principal component analysis. This function returns the results as an object of class ‘prcomp’. We will assign the output to a variable named iris.pca.
R
`iris.pca <- ``prcomp``(iris[,``c``(1:4)], ` ` ``center = ``TRUE``, ` ` ``scale. = ``TRUE``) ` ` ` `# summary of the ` `# prcomp object ` `summary``(iris.pca)`
Output:
Here we get four principal components named PC1-4. Each of these explains a percentage of the total variation in the dataset. For example, PC1 explains nearly 72% of the total variance i.e. around three-fourth of the information of the dataset can be encapsulated by just that one Principal Component. PC2 explains 22% and so on.
Let us take a glance at the structure of the PCA object so formed.
<
class="noIdeBtnDiv">
R
`# structure of the pca object ` `str``(iris.pca) `
Output:
Plotting PCA
While talking about plotting a PCA we generally refer to a scatterplot of the first two principal components PC1 and PC2. These plots reveal the features of data such as non-linearity and departure from normality. PC1 and PC2 are evaluated for each sample vector and plotted.
The autoplot( ) function of the ‘ggfortify package’ gives ease in plotting PCA’s in R.
R
`# loading library ` `library``(ggfortify) ` `iris.pca.plot <- ``autoplot``(iris.pca, ` ` ``data = iris, ` ` ``colour = ``'Species'``) ` ` ` `iris.pca.plot`
Output:
For a better understanding of the linear transformation of features, biplot( ) function also be used to plot PCA.
R
`biplot.iris.pca <- ``biplot``(iris.pca) ` `biplot.iris.pca`
[tabb
ing]
Output:
The X-axis of the biplot represents the first principal component where the petal length and petal width are combined and transformed into PC1 with some parts of sepal length and sepal width. Whereas the vertical part of the sepal length and sepal width forms the second principal component.
For determining the ideal features which can be justified after performing PCA, the plot( ) function can be used to plot the precomp object.
R
`plot.iris.pca <- ``plot``(iris.pca, type=``"l"``) ` `plot.iris.pca`
Output:
In a screeplot the ‘arm-bend’ represents a decrease in cumulative contribution. The above plot shows the bend at the second principal component.
My Personal Notes arrow_drop_up
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# what comes after googolplex
Posted by: | Posted on: November 27, 2020
There’d be ten-duotrigintillion of them! Even though we see a million and a billion as large numbers, there are 1 x 1094 ”millions” or 1 x 1091 “billions” in a googol, which shows you how much larger a googol is than these numbers. [citation needed]. i was joking One googol is presumed to be greater than the number of atoms in the observable universe, which has been estimated to be approximately 1078. our complete guide to Vygotsky scaffolding. In this guide, we’ll give you googolplex and googol definitions, show how you can write them out, explain how they’re useful, and give examples on how you can start to understand huge numbers like these. SAT® is a registered trademark of the College Entrance Examination BoardTM. How Many Zeros in a Googol? 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| 6,001 | 18,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.756373 |
https://whoatwherewhy.com/how-do-you-wire-two-light-sockets-together/ | 1,632,030,623,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00097.warc.gz | 649,816,647 | 12,996 | 1. Step 1: Shut Off the Power to the Circuit.
2. Step 2: Install a Double Switch Wall Box and Run the Feed Cable.
3. Step 3: Run Cables from Wall Box to Light Fixture Locations.
4. Step 4: Attach Pigtails to the Switches.
5. Step 5: Join the Ground Wires.
6. Step 6: Connect the Hot Feed Wires.
How do you wire in parallel? Method 2 Building a Parallel Circuit with Wires and a Switch Choose this method for a slightly advanced project. Gather the main components of a parallel circuit. Prepare your wires. Connect the first lightbulb to the battery. Begin to connect the switch to the battery. Connect the switch to the first lightbulb.
## how do you wire multiple light sockets?
You run another 12/2 or 14/2 cable from the top of the switch box to the first light. You connect the black wire to the switch. You connect the 2 white wires together, and then connect the other terminal on the switch to the black wire of the cable leaving this junction box towards the first light.
Should outlets be wired in series or parallel? It's common to describe household wall receptacles that are wired together using the device terminals as wired in series. But, in fact, all household receptacles are always wired in parallel, and never in series. In a series circuit, current must pass through a load at each device.
## how many lights can I daisy chain?
Unscrew the plate cover on the wall and look at the switch. The wattage is marked on the switch. Then divide that number by the wattage of the bulbs you're using to determine how many bulbs you can connect to this chain. Common switch wattages are 300, 600, and 1,000.
How many outlets can be on a breaker? By rule of thumb you would stick with 8 to 10 outlets and or lights per breaker. Now this is of course they are 120 volts 60 Hz (USA or Canada).
## how do you wire two light switches together?
If you've mounted two switches in the same electrical box, prepare two black wires. Connect one end of the 6-inch wire to the top terminal of the first switch. Twist the other end together with the black wire from the incoming circuit cable and the black wire from the cable going to the second switch to form a pigtail.
Can I wire lights and outlets on the same circuit? An option for you if you don't want two switches is just to use a blank face plate (check this question). Basic answer to your question of can a mixture of lights and receptacles be installed on a single circuit is yes. The neutral will be white but some switches are wired up with a white wire that is not a neutral.
### How many lights can I put on one switch?
A maximum of 96 lights you can connect in parallel but to prevent the risk of unnecessary sparks and arc when operating the switch with having moisture on the switch or you hand, you can connect less than the maximum or you can use several switches instead.
How many outlets and lights can be on a 15 amp circuit? for a 15 amp circuit this allowed for 10 outlets and on a 20 amp circuit 13 outlets were allowed.
### Can two separate circuits be in the same junction box?
2 Answers. The answer is yes you can have 2 separate circuits in the same box (they can have a splice also but not needed in your case). The only concern would be the total box fill. | 738 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-39 | latest | en | 0.895504 |
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