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https://www.thecodingforums.com/threads/plotting-graph-functions-using-gnuplot.619881/
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# Plotting Graph Functions using Gnuplot
A
#### arslanburney
Hello. Needed some help again. Im trying to calculate the best fit
line here. Given a set of points in a list. However, wirte in the end
where i plot the line it tells me tht the variable is not defined.
Either try correcting this or tell me a subsitute that i could use.
Thnks. Heres the code:
#File name Bestfit.py
import Gnuplot
def bestfit(uinput):
if not isinstance(uinput, list):
return False
else:
sigmax = sigmay = sigmaxy = sigmaxwhl = sigmaxsq = 0
for i in range(len(uinput)):
n = len(uinput)
sigmax = uinput[0] + sigmax
sigmay = uinput[1] + sigmay
sigmaxy = uinput[0] * uinput [1] + sigmaxy
sigmaxwhl = sigmax * sigmax
sigmaxsq = uinput[0] * uinput[0] + sigmaxsq
sigmaxsigmay = sigmax * sigmay
num = sigmaxsigmay - (n * sigmaxy)
den = sigmaxwhl - (n* sigmaxsq)
num2 = (sigmax * sigmaxy) - (sigmay * sigmaxsq)
intercept = num2 / den
c = intercept
y = m*x + c
plot (y)
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https://www.indianuniversityquestionpapers.com/2016/10/ec2205-electronic-circuits-i-may-june-2012-question-paper.html
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# EC2205 Electronic Circuits I May June 2012 Question Paper
Anna University Chennai
Question Paper Code:
B.E./B.Tech. DEGREE EXAMINATION, May June 2012
Third Semester
Electronics and Communication Engineering
EC2205 Electronic Circuits I
(Common to Medical Electronics Engineering)
(Regulations 2008)
Time: Three Hours
Maximum: 100 Marks
Note: New Subject Code in R-2013 is EC6304
Part A (10 x 2 = 20 Marks)
1. Define Stability Factor
2. What are the different methods of biasing JFET?
3. Write the voltage gain equation for CE configuration including source resistance
4. What is the need of constant current circuit in differential amplifier?
5. Why are h parameters not used at high frequencies?
6. How is the high frequency gain of an amplifier limited?
7. What is IC biasing Current steering circuit using MOSFET?
8. Does this "enhancement load" resemble a resistor?
9. Why FET is called as a voltage controlled device?
10. Write the applications of MOSFET.
Part B (5 x 16 = 80 Marks)
11. (a) Explain fixed biasing in BJT and FET. Explain the procedure for locating operating point on the characteristic curves.
OR
(b) Draw a voltage divider bias BJT network. Derive expressions for ICQ and VCEQ and describe the method of drawing the dc load line on the output characteristics of transistor.
12. (a) (i) A CC amplifier is fed with the voltage source Vs of internal resistance Rs = 800Ω. The load resistance RL = 1600Ω. The CE hybrid parameters are hie = 1000Ω; hre = 2.2 x 10-4; hfe = 55; hoe = 23 µA/V. Compute voltage gain, current gain, input resistance, output resistance using approximate analysis. (8 Marks)
(ii) Explain the boot strapped Darlington emitter follower with circuit diagram. (8 Marks)
OR
(b) Draw the small signal hybrid model of CE amplifier and derive the expression for its AI, AV, RI, RO.
13. (a) (i) Define alpha cutoff frequency. (4 Marks)
(ii) Draw the high frequency hybrid π model for a transistor in the CE configuration and explain the significance of each component. (12 Marks)
OR
(b) Using hybrid π model of a CE amplifier derive the expression for its short circuit gain. (16 Marks)
14. (a) Derive the expression of the differential-mode voltage gain, common mode voltage gain and CMRR for a MOSFET differential amplifier.
OR
(b) Explain in detail the MOSFET amplifiers with enhancement load and depletion load
15. (a) With the neat sketch explain the principle of operation of cascode amplifier and also derive an expression for its performance measures?
OR
(b) A dc analysis of the source-follower network of the following figure will result in VGSQ= -2.86 V and IDQ = - 4.56 mA.
(a) Determining gm.
(b) Find rd.
(c) Determine Zi.
(d) Calculate Zo with and without rd. Compare results.
(e) Determine Av with and without rd. Compare results.
| 704
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CC-MAIN-2018-47
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longest
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| 0.809571
|
https://www.esaral.com/q/the-mean-of-31-results-is-60-if-the-mean-of-the-first-16-results-is-58-and-that-of-the-last-16-results-is-62-68986
| 1,723,612,743,000,000,000
|
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|
crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00195.warc.gz
| 586,423,504
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|
# The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62,
Question:
The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.
Solution:
Mean of 31 results = 60
Sum of 31 results $=31 \times 60=1860$
Mean of the first 16 results $=58$
Sum of the first 16 results $=58 \times 16=928$
Mean of the last 16 results = 62
Sum of the last 16 results $=62 \times 16=992$
Value of the 16th result $=$ (Sum of the first 16 results $+$ Sum of the last 16 results) $-$ Sum of 31 results
$=(928+992)-1860$
$=1920-1860$
$=60$
| 218
| 653
|
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| 4
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CC-MAIN-2024-33
|
latest
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en
| 0.840366
|
https://www.teacherspayteachers.com/Browse/Grade-Level/Tenth/PreK-12-Subject-Area/Physics?ref=filter/subject
| 1,571,094,098,000,000,000
|
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| 1,114,591,670
| 70,390
|
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I use this worksheet to help teach motion graphs. This distance-time graph worksheet includes six graphs to show constant speed, acceleration, deceleration, no motion, and combinations of these. Combine this worksheet with my Mo-7 worksheet on speed-time graphs to compare distance-time graphs and sp
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This resource is a compilation of editable daily quizzes or bell ringers to start off each class period for your an entire school year for a Physical Science course. It is an excellent way to check in with each student on a daily basis in a quick and efficient manner, and is my favorite way to star
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Newton’s 3 Laws of Motion Doodle Notes. 4 pages of doodle notes and 4 corresponding anchor charts covering:• an overview of Newton’s 3 Laws of Motion and Inertia• Newton’s First Law: The Law of Inertia with examples• Newton’s Second Law: The Law of Motion with an examination of the F=ma relationshi
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
| 2,100
| 8,866
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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
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CC-MAIN-2019-43
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https://www.bbc.co.uk/bitesize/guides/zw7xfcw/revision/6
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| 765,433,876
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# Using a calculator
## BODMAS/BIDMAS on a calculator
BODMAS or BIDMAS must also be used when using a calculator. Scientific calculators automatically apply the operations in the correct order, however extra brackets may be required.
### Example
Consider this calculation:
If was entered on a calculator, the calculator would use BODMAS/BIDMAS and calculate . The division would be completed before the addition or subtraction. This would give the wrong answer.
To make sure the calculation is correct, use the fraction button on the calculator or use brackets:
or
## Calculating indices on a calculator
On a calculator, brackets must be used when raising negative numbers to a power.
For example, to calculate -2 squared on a calculator, type in . This would give the correct answer of 4. If brackets were not used, BODMAS/BIDMAS would be applied and the answer given would be -4, as the power is applied before the subtraction.
### Example
Find the value of when .
Using BODMAS/BIDMAS, the answer is calculated as:
1. (power/indices first)
| 231
| 1,055
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https://admin.clutchprep.com/physics/practice-problems/139587/a-bass-guitar-string-is-89-cm-long-with-a-fundamental-frequency-of-30-hz-what-is
| 1,596,900,445,000,000,000
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crawl-data/CC-MAIN-2020-34/segments/1596439737883.59/warc/CC-MAIN-20200808135620-20200808165620-00186.warc.gz
| 200,040,700
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|
# Problem: A bass guitar string is 89 cm long with a fundamental frequency of 30 Hz. What is the wave speed on this string?
###### FREE Expert Solution
Fundamental frequency:
$\overline{){{\mathbf{f}}}_{{\mathbf{0}}}{\mathbf{=}}\frac{\mathbf{V}}{\mathbf{2}\mathbf{L}}}$
f0 = 30 Hz
L = 89 cm (1m/100cm) = 0.89 m
###### Problem Details
A bass guitar string is 89 cm long with a fundamental frequency of 30 Hz. What is the wave speed on this string?
| 145
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| 3.46875
| 3
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CC-MAIN-2020-34
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latest
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en
| 0.709772
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-10-radical-expressions-and-equations-10-5-graphing-square-root-functions-lesson-check-page-628/4
| 1,723,536,578,000,000,000
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crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00262.warc.gz
| 618,737,309
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|
## Algebra 1
$\sqrt 5$ is a constant (equal to 2.24). If we replace this in the equation, we have $y=2.24x$. This new equation is a linear function, not a square root function.
| 54
| 177
|
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| 3.109375
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CC-MAIN-2024-33
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http://gbhsweb.glenbrook225.org/gbs/science/phys/p163/audhelp/u5setb/prob13.html
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Physics 163
Unit 5: Work, Energy and Power
Problem Set B
Problem 13:
Esteemed Chinese ski jumper, Li Ping Phar, leaves the jump ramp with a kinetic energy of 16046 Joules and a potential energy (relative to the bottom of the ski hill) of 12836 Joules. She encounters -2052 Joules of work due to air resistance during her flight through the air. Determine the speed (in m/s) of her 49.2-kg body the instant prior to striking the ground at the bottom of the hill.
• Read the problem carefully and diagram the physical situation
• Identify known and unknown information in an organized manner
• Take the time to plot out a strategy prior to beginning the solution
• Identify an appropriate formula to use
• Perform step-by-step algebraic manipulations
Use the equation below to solve this problem. Cancel terms, substitute and solve.
KEi + PEi + Wnc = KEf + PEf
Work and energy principles and mathematical relationships can be reviewed on the Set B Overview page.
Further information about the work and energy relationship is available online at The Physics Classroom.
| 238
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https://www.teacherspayteachers.com/Product/EMOJI-Classify-Angles-Triangles-Acute-Right-Obtuse-3118876
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Total:
\$0.00
# EMOJI - Classify Angles & Triangles (Acute, Right, Obtuse)
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1.49 MB | 8 (including cover & answer key) pages
### PRODUCT DESCRIPTION
** This activity is available as a "GOOGLE INTERACTIVE" product COMING SOON ---
HERE IS WHATS AVAILABLE AT THIS TIME
**
✐ This product is a NO PREP- SELF CHECKING activity that engage students in "Classifying Angles & Triangles as Acute, Right, or Obtuse".
In this activity, students are asked to:
☑ (Questions 1-3) Determine if the given measure of an angle represents: (No images are provided)
✔ Acute Angle
✔ Right Angle
✔ Obtuse Angle
☑ (Questions 4-12) The measure of 2 angles of a triangle is provided, students would need to find the missing angle measure, then determine based on these measures if the triangle is:
✔ An Acute Triangle
✔ A Right Triangle
✔ An Obtuse Triangle
Students need to feel comfortable with:
☑ Combining Like Terms (integers in this case)
☑ Solving one step equation
☑ Understand that the sum of the measures of the angles in a triangle is equal to 180⁰
☑ The definition of Acute Angle, Right Angle, and Obtuse Angle
☑ The definition of Acute Triangle, Right Triangle, and Obtuse Triangle
Students would need to solve the mystery Emoji by following the color key.
Please take a look at the preview and see the questions of this EMOJI before purchasing
A suggested step-by-step answer key is provided in this activity
My students are fascinated with an additional coloring option that I have included where they could pick their own colors. They have become very very creative. I am including this option in this emoji and I truly hope that your kids will love it as much as mine have.
The purpose behind this activity is just to help students feel comfortable with the relationship between the measures and the naming the angle/triangle. If you would like to modify it in any way, please don't hesitate to contact me via Q and A. I will update and include a second version in this product JUST FOR YOU ☻
✶ ✶ ✶ My students truly were ENGAGED answering the questions and had a BLAST COLORING and finding the mystery picture much better than the textbook problems.✶ ✶ ✶
This RELAXING EMOJI could be used in: math centers, group work, a way to check for understanding, a review, recap of the lesson, pair-share, cooperative learning, exit ticket, entrance ticket, homework, individual practice, when you have time left at the end of a period, beginning of the period (as a warm up or bell work), before a quiz on the topic, substitute planning, and more.
☺Would love to hear your feedback☺. Please don't forget to come back and rate this product when you have a chance. You will also earn TPT credits. Enjoy and I ☺Thank You☺ for visiting my ☺Never Give Up On Math☺ store!!!
© Never Give Up On Math 2017
This product is intended for personal use in one classroom only. For use in multiple classrooms, please purchase additional licenses.
☺ HAVE A WONDERFUL DAY ☺
Total Pages
8 (including cover & answer key)
Included
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# 99.3lbs至kg99.3磅至千克
lbs
=
kg
## 如何轉換99.3磅至千克?
99.3 lbs * 0.45359237 kg = 45.041722341 kg 1 lbs
45041.722341 g
99.3 lbs
## 另一種拼寫
99.3 磅至千克, 99.3 磅成千克, 99.3 磅至kg, 99.3 磅成kg, 99.3 lb至千克, 99.3 lb成千克, 99.3 lbs至kg, 99.3 lbs成kg, 99.3 lb至kg, 99.3 lb成kg
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# STEREOX-RAY ANALYSIS
STEREOX-RAY ANALYSIS (Greek stereos firm, volume, space + a X-ray analysis) — the way of a X-ray analysis of a motionless object from two certain positions providing the subsequent visual reconstruction according to two roentgenograms of the volume (stereoscopic) image of the finished shooting object. The Page is the cornerstone the principle of the stereoscopic effect connected with properties solid vision (see).
The idea of use of stereoscopic effect in a radiology arose at the end of 19 century. E. Mach in 1896 offered S.'s idea, and after Levi-Dorn (M. of Levy-Dorn) developed S.'s technique and applied it for practical purposes.
The essence of a technique of S. is in what is made by two pictures (a so-called stereo pair) of the motionless studied object, displacing a X-ray tube in an interval between the first and second pictures in the plane parallel to the plane of the cartridge with a film. The distance between points of shooting (basis of shooting) is usually equal to 60 — 70 mm, i.e. visual basis (to distance between visual axes of eyes of the person). During the shooting of stereo pairs use special two-pipe stereographic X-ray apparatus or move the same X-ray tube at a size of basis of shooting. Sometimes for receiving a stereo pair of roentgenograms use the phenomenon of parallactical shift of the image (see Skialogiya) arising at rotation of an object. The size of basis of shooting is determined by a formula:
d = [N (N + E)] / 50E
where d — basis of shooting; N — distance from focus of a X-ray tube to the cartridge; E — thickness of the studied object.
For change of films at a X-ray analysis use an automatic prefix or the tunnel cartridge. The tunnel cartridge consists of three departments; average department of the cartridge misses x-ray emission, two other departments (extreme) are screened sheet lead. After receiving the first picture the exhibited x-ray film is stretched in a screened part of the cartridge, and in the unprotected part there is not exhibited part of a film again, on to-ruyu later shift of a X-ray tube and do the second picture, receiving thus a stereo pair.
For the purpose of obtaining stereoeffect received by the specified way of the roentgenogram consider on the stereonegatoscope.
In the analysis of roentgenograms on the special stereonegatoscope — the stereocomparator it is possible to make rentgenogrammetriya (see) structural elements of an object. For increase in accuracy and convenience of measurements create additional reference points, applying them on a body surface of the patient or entering X-ray contrast tags into hollow bodies.
In a crust, S.'s time apply seldom. Resort to it only in some cases at a contrast rentgenol. a research of a bronchial tree (see. Bronchography ), veins (see. Flebografiya ), at fistulografiya (see), in obstetric and gynecologic practice, at establishment of localization of metal foreign bodys (see), in the topometric analysis rentgenol. pictures of intracranial structures (e.g., Turkish saddle).
Bibliography: Bondarev I. M., etc. Studying of some indicators of function of bronchial tubes by method of a stereorentgenogrammet-riya, Vestn. rentgenol. and radio-gramophones., No. 4, page 20, 1975, bibliogr.; Poluektov JI. Century, etc. A clinical use of a stereox-ray analysis during the studying of venous system of the lower extremities, in the same place, No. 3, page 36, 1981, bibliogr.; Feoktistov V. I. The x-ray image, its metric properties and their use in clinic, L., 1966, bibliogr.; P e r N and y A. N. Use of a stereophotogrammetric method of a research in a medical radiology, Geodesy and aerial photography, century 5, page 119, 1971.
I. P. Korolyuk.
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Roman Numerals
# Roman Numerals: MMMCMLXVIII = 3968
## Convert Roman Numerals
Arabic numerals:
Roman numerals:
Arabicnumerals 0 1 M C X I 2 MM CC XX II 3 MMM CCC XXX III 4 CD XL IV 5 D L V 6 DC LX VI 7 DCC LXX VII 8 DCCC LXXX VIII 9 CM XC IX
The converter lets you go from arabic to roman numerals and vice versa. Simply type in the number you would like to convert in the field you would like to convert from, and the number in the other format will appear in the other field. Due to the limitations of the roman number system you can only convert numbers from 1 to 3999.
To easily convert between roman and arabic numerals you can use the table above. The key is to handle one arabic digit at a time, and translate it to the right roman number, where zeroes become empty. Go ahead and use the converter and observe how the table shows the solution in realtime!
## Current date and time in Roman Numerals
2021-11-27 20:53:41 MMXXI-XI-XXVII XX:LIII:XLI
Here is the current date and time written in roman numerals. Since the roman number system doesn't have a zero, the hour, minute, and second component of the timestamps sometimes become empty.
## The number 3968
The number 3968 is divisble by 2, 4, 8, 16, 31, 32, 62, 64, 124, 128, 248, 496, 992 and 1984 and can be prime factorized into 27×31.
3968 as a binary number: 111110000000
3968 as an octal number: 7600
3968 as a hexadecimal number: F80
## Numbers close to MMMCMLXVIII
Below are the numbers MMMCMLXV through MMMCMLXXI, which are close to MMMCMLXVIII. The right column shows how each roman numeral adds up to the total.
3965 = MMMCMLXV = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 5 3966 = MMMCMLXVI = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 5 + 1 3967 = MMMCMLXVII = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 5 + 1 + 1 3968 = MMMCMLXVIII = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 5 + 1 + 1 + 1 3969 = MMMCMLXIX = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 10 − 1 3970 = MMMCMLXX = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 10 3971 = MMMCMLXXI = 1000 + 1000 + 1000 + 1000 − 100 + 50 + 10 + 10 + 1
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Number 10691: mathematical and symbolic properties | Crazy Numbers
# Everything about number 10691
Discover a lot of information on the number 10691: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 10691
Is 10691 a prime number? Yes
Is 10691 a perfect number? No
Number of divisors 2
List of dividers
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1, 10691
Sum of divisors 10692
Prime factorization 10691
Prime factors
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10691
## How to write / spell 10691 in letters?
In letters, the number 10691 is written as: Ten thousand six hundred and ninety-one. And in other languages? how does it spell?
10691 in other languages
Write 10691 in english Ten thousand six hundred and ninety-one
Write 10691 in french Dix mille six cent quatre-vingt-onze
Write 10691 in spanish Diez mil seiscientos noventa y uno
Write 10691 in portuguese Dez mil seiscentos noventa e um
## Decomposition of the number 10691
The number 10691 is composed of:
2 iterations of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
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1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
## Mathematical representations and links
Other ways to write 10691
In letter Ten thousand six hundred and ninety-one
In roman numeral
In binary 10100111000011
In octal 24703
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# Composition question problem [closed]
So I have a line from composition:
'But balloons had the disadvantage of having to go wherever the wind blew them, so that one never knew where they would come down.'
And I was asked a question:
'What was the disadvantage of flying in a balloon?'
I was instructed to answer in my own words and be only one sentence long. I wrote:
'The disadvantage of flying in a balloon was that one was forced to go in the direction where the wind blew the balloon, and thus one had no idea where he would descend.'
My professor marked me zero. Why is this wrong?
• Zero is a bit harsh, in my opinion. It might be the use of "he" when you have previously used "one". (The best advice is probably to ask your teacher what is wrong.) Commented Nov 12, 2013 at 11:33
• @AndrewLeach thanks! So if I used 'he/she' or instead simply 'one', would I be still wrong? Commented Nov 12, 2013 at 11:37
• You're using a general impersonal "one", so consistency would be good. You could also use it, referring to the balloon. But I have no idea what was in your tutor's mind: only he has. Commented Nov 12, 2013 at 11:42
I could find some mistakes which your professor might have considered...
1. As @Andrew said: Use of 'he' which is inappropriate.
2. Use of 'one' for the second time is not necessary.
3. Descend means to Move downward and lower, but not necessarily all the way to the ground. So, 'where' he would descend doesn't sound that good. Because, the sentence in the paragraph speaks about the landing place.
The sentence may be like this: The disadvantage of flying in a baloon was that it moves with the direction of wind and thus the fliers didn't have the knowledge of where to land.
• thanks! But at the end, would not this be more appropriate: 'where they will land' instead of 'where to land.' Commented Nov 12, 2013 at 11:56
• Sure. You have the advantage of answering in your own words!! (But without deviating from the context. LOL!!) Commented Nov 12, 2013 at 12:00
Zero is harsh. Only the prof knows what he had in his mind. Apart from the "he-one" and "descend" issues already rightly pointed out, I would say you could have worded your sentence better without deviating from context.
The disadvantage of flying in a balloon is that there is little or no chance to navigate as flying direction is governed by the wind which makes landing at a specific location difficult.
This of course is a matter of taste.
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# direct sum
Also found in: Dictionary, Thesaurus, Wikipedia.
## direct sum
[də¦rekt ′səm]
(mathematics)
If each of the sets in a finite direct product of sets has a group structure, this structure may be imposed on the direct product by defining the composition “componentwise”; the resulting group is called the direct sum.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
Since A is g-Drazin invertible, X = R(I-P)[direct sum] N(I-P), A = [A.sub.1] [direct sum] [A.sub.2], where [A.sub.1] is closed and invertible and [A.sub.2] is bounded and quasinilpotent with respect to the direct sum. Therefore (3) has a solution if and only if each of the following two initial value problems has a solution on R(I - P) and R(P), respectively
Let [B.sub.i] be the direct sum of uniserial modules of length i and x, y [member of] [B.sub.i].
Proof: One can check these identities directly from the definitions of direct sum, restriction and deletion for matroids (see again the previous section).
This is an indefinite inner product space which has the structure of the direct sum of a Hilbert space and a negative Hilbert space.
1 and 2] give recurrences that reduce the computation of the Mobius function [mu]([sigma], [tau]) to Mobius function calculations of the form [mu]([sigma]', [tau]') where [tau]' is a single component of [tau] and [sigma]' is a direct sum of consecutive components of [sigma].
Faticoni (mathematics, Fordham U.) explores advanced topics in direct sum decompositions of abelian groups and their consequences.
The direct sum notation ([direct sum]) was introduced in the introduction.
The (canonical) isotypic decomposition of V is the direct sum of its kG-homogeneous components.
We define their direct sum M [direct sum] M' on the ground set E [??] E' by
Hereditary Noetherian prime rings may be the only non-commutative Noetherian rings whose projective modules, both finitely and infinitely generated, have nontrivial direct sum behavior and a structure theorem describing that behavior, say mathematicians Levy (U.
We may assume without loss of generality that the norm of the direct sum Y [direct sum] X is in fact equal to the maximum norm Y [[direct sum].sub.[infinity]] X.
We show that such a polytope is lattice equivalent to a direct sum of del Pezzo polytopes, pseudo del Pezzo polytopes, or a (possibly skew) bipyramid over (pseudo) del Pezzo polytopes.
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### THEORY OF DEMAND
DEFINITION OF DEMAND
Demand can be defined as the quantity of a commodity (goods and services) that consumers are willing and able to buy at a given price and at a particular place and time. Demand is quite different from wants, need or desire. Effective Demand’ in economics must meet three conditions which are:
• Ability to pay
• Willingness to pay
• Authority to buy a commodity
Demand must be related to price because to a great extent, price determines the quantity which consumers are willing to buy.
LAW OF DEMAND
The law of demand states that, all things being equal (Ceteris Paribus), The higher the price, the lower the quantity of goods that will be demanded, or the lower the price, the higher the quantity of goods that will be demanded. This law is often regarded as the first law of demand and supply. It simply means that when the price of a commodity, like yam for instance, is high in the market, very few quantity of it will be demanded by the consumers and vice-versa.
DEMAND SCHEDULE
It is a table of value showing the relationship between prices and quantity of that commodity demanded. This is a table, which shows the magnitude of demand at various prices. That is, the different quantities of a commodity, which would be bought at various prices, at a particular time.
DEMAND CURVE
The demand curve is the graphical representation of the information contained in the demand schedule. The price is plotted on the vertical axis and the quantity demanded is plotted on the horizontal axis. Normal demand curve slopes downwards from left to right. From Mr. Tunde's demand schedule, a demand curve is drawn as follows.
FACTORS INFLUENCING THE DEMAND OF A COMMODITY
• Price of the commodity: This is the most important factor influencing supply. The higher the price of the commodity, the higher the quantity supplied and vice versa.
• Cost of production: If the cost of producing a commodity falls, then more of that commodity could be supplied at the existing price. It therefore means that a producer will be able to produce more commodities with the existing raw materials, hence increase in supply.
• Technological development: An improvement in the level of technology will equate to improvement in the methods or techniques of production. This will encourage large scale production at lower costs which in turn increases supply e.g. the use of modern farming techniques and equipment
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# Cool Math Games for Kids: Online & Free!!
109 Flares 109 Flares
My kids are “cool” and I love math. It is fun to shake things up a bit and play some online cool math games! Even better, all of these sites offer free access so you’re not committed to buying something your child may only play a few times.
We use online math games as review work, & often as a reward. When my son completes his assignment, he can choose a game or two to play on the computer. I can’t say no to that…he’s having fun and learning. That’s a double bonus in my book.
::Multiplication.com This site is one of my son’s favorites! There are a variety of game choices such as Granny Prix (wheelchair races with Granny anyone?), Pumpkin Patch (grow pumpkins by answering multiplication problems & see how many happy customers you can get), and Cone Crazy (serve up some yummy ice cream to your customers as you answer multiplication questions). What I really like about Multiplication.com is that you can choose the level of difficulty for each game. Some games even let you focus on one particular number (such as all the factors of 9). This website is one we return to over and over again for fun practice.
::Interactive Decimal Games This is a new site to us, but there are some wonderful, easy review games here. Concepts as simple as identifying place value (tenths, hundreths, etc), all the way up to adding/subtracting/multiplying/dividing with decimals and figuring decimals based on percentages! My son’s favorite game for simple identification practice is Decimal Ships. The more questions you answer correctly, the more boats you launch across the bay. It has really helped him understand place value to the right of the decimal.
::Virtual Manipulatives This nifty website is useful for PreK-Middle School aged children…I love it when a website can serve so many needs! This site is perfect for home use, because you can practice math skills with manipulatives but you don’t have to spend a small fortune purchasing all of them! First you choose the grade level you want, then the other two drop down menus give you Backgrounds/Workmats & Manipulative choices. In the photo above, we created a word problem for a Kindergarten student, and obviously for older grades, the options go into much more depth.
::Arcademic Skill Builders This is another site with simple, fun games for children of various levels. They recreate the fun of an arcade game, and allow your children to practice important “quick thinking” math skills at the same time. I like using this site for a quick review because the games are short but target whatever skill I’m looking for at that time.
::Lemonade Stand This virtual lemonade stand offers plenty of opportunities to make choices! The player starts with a set amount of money, and then decides how to spend it on lemons, cups, ice, and sugar. The player sets the price per cup, and at the end of each “day”, you find out how many cups you sold and how much you earned. You also learn what supplies need to be restocked (all your ice melted!) before the next day. The customers that pass by provide feedback such as “Yum!” or “Too warm!” or the ever popular “Too expensive!” Your child has to make choices about how to please those customers, yet still keep some of their money for the next day. I have to admit, I even enjoy this simple little game!
Perhaps the next time your child needs to review math facts or concepts, you can incorporate some fun along the way! Do you know of any other fun free math websites for children? We’d love to know!
Heather spends time homeschooling, raising 3 great kids, & traveling with her family. She loves folding laundry but hates putting it away. She shares a love of Disney World with her Hubby, she's a cruise fanatic, & bird-watching hobbyist. You can follow her on Twitter as @CrazyZeus1.
From our partners:
1. Since you said you love your math curriculum and I am currently struggling to find one that I like, I am curious about what you use. Thanks.
• HappyCampers says:
We use Teaching Textbooks & my son loves it!
2. happycampers says:
We use Teaching Textbooks. For my son it is a perfect fit. Just beware, it is about a year behind, so bump your child up at least a year when ordering. Online, they have a placement test so you can find the best fit!
3. Great review! We also recently discovered Academic Skills Builders. Really, there are so many great sites out there and not enough time Thanks for sharing your choices with Afterschool.
4. Admiring the time and effort you put into your blog and detailed information you offer! I will bookmark your blog and have my children check up here often.
5. This is a wonderful site which provides the chance to learn through games. It is very helpful for children since they can learn through playing. The design and pattern is also very attractive. Try to include more like this. I would like to search more later since I am busy with my pets.
6. Nice collection! For Toddlers, I like http://www.pipogames.com there is a free section for the little ones. I hope you like it
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8. Some really good ones! I have tried the lemonade stand and it is pretty challenging, but very fun. Thanks for sharing at Mom’s Library! Pinned.
What may you suggest in regards to your put up that you just made some
days in the past? Any certain?
10. Can someone tell me how to discover more details about how to find deals on game for kids on the internet ?
my grammy want to buy me a present and she told me to find one online, , and i want to treasure a miracle time with her and see her satisfied with me play beside her.
i find this site,will it give me a value information , but it did not focus on game on the internet
game for kids
11. Excellent post and the information given in this blog is really knowledgeable.
12. Clearly you have been searching the net for games, as I have seen a lot of these before, tres bien.
I wonder if you have had a chance to stop in and check out , our math games. I hope you wont be disappointed by your time spent. And maybe you can let us know, which ones are your favorites., since we have almost 200 to choose from.
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This a great review! Kids would really learn fast if fun is being inserted to learning. My sons loves math, and would love it more with these cool math games.. They love sentence building too! i would recommend you try this app. My sons love this app too! https://play.google.com/store/apps/details?id=air.com.sierravistasoftware.SightWordsSB1
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Get a unique, high-quality and non-plagiarized paper from us today at the most affordable price
Email us : premieredtutorials@gmail.com
PREMIERED TUTORIALS
LED values
1. Write, compile, and load the program described in Example 1 above.
2. In the Data1: window, right click on num1, then left click on Show Location. The value of the initialized variables should appear in the Memory window. Cut and paste the Memory window.
3. Single step through the program using F11and verify the value of the result in Memory window. Cut and paste the Memory window to show the value of result.
4. What is the value displayed on the LEDs after you single step the line PTT = ~result?
5. Write, compile, and load the program described in Example 2 above.
6. Run the program on the board by pressing the Reset button on the Tower. What are the LED values?
7. Calculate the delay size.
1. Initialize an array with binary numbers 0 to 9. Use a “for loop” to display each array element on the Tower LEDs and have a one-second delay between each display. Define all of the variables as unsigned character. The delay should use inline assembly language.
2. Cut and paste the source code.
3. In the Data1: window, right click on the array name, then left click on Show Location. The initialized data array should appear in the Memory window. Cut and paste the Memory window.
4. Run the program on the board by pressing the Reset button on the Tower. Describe the values of LEDs. What is the last value that you see?
1. Modify the “for loop” in Problem 3 so that it displays numbers starting from 9 down to 0. In other words, the loop should become: for( i=9; i>= 0; i–). Run the program and observe the LED values. Does the loop end?
2. Cut and paste the source code.
3. Run the program by pressing the Reset button on the Tower. Describe the values of LEDs. What is the last value that you see? Does the loop end?
1. Recompile the code in Problem 4, but this time, do not declare the loop counter i as unsigned char. Declare i as char. Does the loop end?
2. Modify Problem 4 so that it displays the values from 0 to 9 continuously.
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Video Instruction
showmethephysics.com
Video
Show Me The Physics Website
This movie is part of the collection: Prelinger Archives
Production Company: ERPI Classroom Films, Inc. Distributed by Encyclopedia Britannica
Audio/Visual: sound, B&W
What is this speaker component called?
What kinds of waves does it produce?
High or low frequency?
Woofer, Outputs low frequency waves
II. Characteristics of Periodic Waves
A. Frequency(f) = #vibrations/sec
= # Cycles/# sec
= Hertz (Hz)
Source: Elroy M. Avery School Physics
(New York: Sheldon and Company, 1895) 202
Clipart ETC
Heinrich Hertz was the first to send
Frequency
- Determined by
Source
of Wave,
Not Medium
Cycle = (single vibration)
Ex)
How many cycles between the dots?
3 Cycles
Ex) 10 cycles pass a fixed point in a wave train in 5 seconds. What is the frequency of the wave?
f = 10 cycles = 2 cycles 5 seconds 1 sec
f = 2 Hertz (Hz)
Top View of a Wave
[ HTML5 ]
showmethephysics.com
Sound Waves
David Kirby - U. of Calif.
Top view of a periodic wave
(Each line is a crest)
Crest
Physics 2000
Review - Longitudinal or Transverse?
[ Flash ]
Ex) A wave generator operating for
4 seconds
produces the waves drawn below.
Top view of a periodic wave
(Each line is a crest)
How many waves are drawn between A and B? (Be careful!!)
What is the frequency of this
periodic wave train?
8 cycles
f = 8 cycles/4 sec
f = 2 cycles/sec
Ex) 2 cycles pass a fixed point in a wave train in 10 sec. What is the frequency ?
f = # Cycles/# sec
f = 2 Cycles/10 sec
f = .2 Hz
• Sound, frequency is pitch.
Source: Elroy M. Avery School Physics
(New York: Sheldon and Company, 1895) 247
• Light, frequency is color.
Human Ear:
Frequency Range
20 Hz - 20,000 Hz.
Did you know dogs are capable of hearing higher pitch sounds than humans?
The range of frequencies you
can hear changes with age.
How old are you?
How to get rid of teenagers
2006 McGraw-Hill Higher Education
Enrichment
How the Ear Works
How Sound Works
Sound of a bell in a vacuum
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/
# Hilbert Space¶
Hilbert spaces for quantum mechanics.
Authors: * Brian Granger * Matt Curry
class sympy.physics.quantum.hilbert.HilbertSpace[source]
An abstract Hilbert space for quantum mechanics.
In short, a Hilbert space is an abstract vector space that is complete with inner products defined [R232].
References
[R232] (1, 2) http://en.wikipedia.org/wiki/Hilbert_space
Examples
>>> from sympy.physics.quantum.hilbert import HilbertSpace
>>> hs = HilbertSpace()
>>> hs
H
dimension[source]
Return the Hilbert dimension of the space.
class sympy.physics.quantum.hilbert.ComplexSpace[source]
Finite dimensional Hilbert space of complex vectors.
The elements of this Hilbert space are n-dimensional complex valued vectors with the usual inner product that takes the complex conjugate of the vector on the right.
A classic example of this type of Hilbert space is spin-1/2, which is ComplexSpace(2). Generalizing to spin-s, the space is ComplexSpace(2*s+1). Quantum computing with N qubits is done with the direct product space ComplexSpace(2)**N.
Examples
>>> from sympy import symbols
>>> from sympy.physics.quantum.hilbert import ComplexSpace
>>> c1 = ComplexSpace(2)
>>> c1
C(2)
>>> c1.dimension
2
>>> n = symbols('n')
>>> c2 = ComplexSpace(n)
>>> c2
C(n)
>>> c2.dimension
n
class sympy.physics.quantum.hilbert.L2[source]
The Hilbert space of square integrable functions on an interval.
An L2 object takes in a single sympy Interval argument which represents the interval its functions (vectors) are defined on.
Examples
>>> from sympy import Interval, oo
>>> from sympy.physics.quantum.hilbert import L2
>>> hs = L2(Interval(0,oo))
>>> hs
L2([0, oo))
>>> hs.dimension
oo
>>> hs.interval
[0, oo)
class sympy.physics.quantum.hilbert.FockSpace[source]
The Hilbert space for second quantization.
Technically, this Hilbert space is a infinite direct sum of direct products of single particle Hilbert spaces [R233]. This is a mess, so we have a class to represent it directly.
References
[R233] (1, 2) http://en.wikipedia.org/wiki/Fock_space
Examples
>>> from sympy.physics.quantum.hilbert import FockSpace
>>> hs = FockSpace()
>>> hs
F
>>> hs.dimension
oo
#### Previous topic
Cartesian Operators and States
Operator
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > smupval Structured version Visualization version GIF version
Theorem smupval 15417
Description: Rewrite the elements of the partial sum sequence in terms of sequence multiplication. (Contributed by Mario Carneiro, 20-Sep-2016.)
Hypotheses
Ref Expression
smupval.a (𝜑𝐴 ⊆ ℕ0)
smupval.b (𝜑𝐵 ⊆ ℕ0)
smupval.p 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))
smupval.n (𝜑𝑁 ∈ ℕ0)
Assertion
Ref Expression
smupval (𝜑 → (𝑃𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))
Distinct variable groups: 𝑚,𝑛,𝑝,𝐴 𝐵,𝑚,𝑛,𝑝 𝑚,𝑁,𝑛,𝑝 𝜑,𝑛
Allowed substitution hints: 𝜑(𝑚,𝑝) 𝑃(𝑚,𝑛,𝑝)
Proof of Theorem smupval
Dummy variables 𝑘 𝑥 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 smupval.n . . . . 5 (𝜑𝑁 ∈ ℕ0)
2 nn0uz 11923 . . . . 5 0 = (ℤ‘0)
31, 2syl6eleq 2859 . . . 4 (𝜑𝑁 ∈ (ℤ‘0))
4 eluzfz2b 12556 . . . 4 (𝑁 ∈ (ℤ‘0) ↔ 𝑁 ∈ (0...𝑁))
53, 4sylib 208 . . 3 (𝜑𝑁 ∈ (0...𝑁))
6 fveq2 6332 . . . . . 6 (𝑥 = 0 → (𝑃𝑥) = (𝑃‘0))
7 fveq2 6332 . . . . . 6 (𝑥 = 0 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘0))
86, 7eqeq12d 2785 . . . . 5 (𝑥 = 0 → ((𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) ↔ (𝑃‘0) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘0)))
98imbi2d 329 . . . 4 (𝑥 = 0 → ((𝜑 → (𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥)) ↔ (𝜑 → (𝑃‘0) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘0))))
10 fveq2 6332 . . . . . 6 (𝑥 = 𝑘 → (𝑃𝑥) = (𝑃𝑘))
11 fveq2 6332 . . . . . 6 (𝑥 = 𝑘 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘))
1210, 11eqeq12d 2785 . . . . 5 (𝑥 = 𝑘 → ((𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) ↔ (𝑃𝑘) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘)))
1312imbi2d 329 . . . 4 (𝑥 = 𝑘 → ((𝜑 → (𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥)) ↔ (𝜑 → (𝑃𝑘) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘))))
14 fveq2 6332 . . . . . 6 (𝑥 = (𝑘 + 1) → (𝑃𝑥) = (𝑃‘(𝑘 + 1)))
15 fveq2 6332 . . . . . 6 (𝑥 = (𝑘 + 1) → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)))
1614, 15eqeq12d 2785 . . . . 5 (𝑥 = (𝑘 + 1) → ((𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) ↔ (𝑃‘(𝑘 + 1)) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1))))
1716imbi2d 329 . . . 4 (𝑥 = (𝑘 + 1) → ((𝜑 → (𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥)) ↔ (𝜑 → (𝑃‘(𝑘 + 1)) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)))))
18 fveq2 6332 . . . . . 6 (𝑥 = 𝑁 → (𝑃𝑥) = (𝑃𝑁))
19 fveq2 6332 . . . . . 6 (𝑥 = 𝑁 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁))
2018, 19eqeq12d 2785 . . . . 5 (𝑥 = 𝑁 → ((𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥) ↔ (𝑃𝑁) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁)))
2120imbi2d 329 . . . 4 (𝑥 = 𝑁 → ((𝜑 → (𝑃𝑥) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑥)) ↔ (𝜑 → (𝑃𝑁) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁))))
22 smupval.a . . . . . . 7 (𝜑𝐴 ⊆ ℕ0)
23 smupval.b . . . . . . 7 (𝜑𝐵 ⊆ ℕ0)
24 smupval.p . . . . . . 7 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))
2522, 23, 24smup0 15408 . . . . . 6 (𝜑 → (𝑃‘0) = ∅)
26 inss1 3979 . . . . . . . 8 (𝐴 ∩ (0..^𝑁)) ⊆ 𝐴
2726, 22syl5ss 3761 . . . . . . 7 (𝜑 → (𝐴 ∩ (0..^𝑁)) ⊆ ℕ0)
28 eqid 2770 . . . . . . 7 seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1)))) = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))
2927, 23, 28smup0 15408 . . . . . 6 (𝜑 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘0) = ∅)
3025, 29eqtr4d 2807 . . . . 5 (𝜑 → (𝑃‘0) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘0))
3130a1i 11 . . . 4 (𝑁 ∈ (ℤ‘0) → (𝜑 → (𝑃‘0) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘0)))
32 oveq1 6799 . . . . . . 7 ((𝑃𝑘) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) → ((𝑃𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)}) = ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)}))
3322adantr 466 . . . . . . . . 9 ((𝜑𝑘 ∈ (0..^𝑁)) → 𝐴 ⊆ ℕ0)
3423adantr 466 . . . . . . . . 9 ((𝜑𝑘 ∈ (0..^𝑁)) → 𝐵 ⊆ ℕ0)
35 elfzouz 12681 . . . . . . . . . . 11 (𝑘 ∈ (0..^𝑁) → 𝑘 ∈ (ℤ‘0))
3635adantl 467 . . . . . . . . . 10 ((𝜑𝑘 ∈ (0..^𝑁)) → 𝑘 ∈ (ℤ‘0))
3736, 2syl6eleqr 2860 . . . . . . . . 9 ((𝜑𝑘 ∈ (0..^𝑁)) → 𝑘 ∈ ℕ0)
3833, 34, 24, 37smupp1 15409 . . . . . . . 8 ((𝜑𝑘 ∈ (0..^𝑁)) → (𝑃‘(𝑘 + 1)) = ((𝑃𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)}))
3927adantr 466 . . . . . . . . . 10 ((𝜑𝑘 ∈ (0..^𝑁)) → (𝐴 ∩ (0..^𝑁)) ⊆ ℕ0)
4039, 34, 28, 37smupp1 15409 . . . . . . . . 9 ((𝜑𝑘 ∈ (0..^𝑁)) → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)) = ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑘) ∈ 𝐵)}))
41 elin 3945 . . . . . . . . . . . . . 14 (𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ↔ (𝑘𝐴𝑘 ∈ (0..^𝑁)))
4241rbaib 520 . . . . . . . . . . . . 13 (𝑘 ∈ (0..^𝑁) → (𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ↔ 𝑘𝐴))
4342adantl 467 . . . . . . . . . . . 12 ((𝜑𝑘 ∈ (0..^𝑁)) → (𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ↔ 𝑘𝐴))
4443anbi1d 607 . . . . . . . . . . 11 ((𝜑𝑘 ∈ (0..^𝑁)) → ((𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑘) ∈ 𝐵) ↔ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)))
4544rabbidv 3338 . . . . . . . . . 10 ((𝜑𝑘 ∈ (0..^𝑁)) → {𝑛 ∈ ℕ0 ∣ (𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑘) ∈ 𝐵)} = {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)})
4645oveq2d 6808 . . . . . . . . 9 ((𝜑𝑘 ∈ (0..^𝑁)) → ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑘) ∈ 𝐵)}) = ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)}))
4740, 46eqtrd 2804 . . . . . . . 8 ((𝜑𝑘 ∈ (0..^𝑁)) → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)) = ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)}))
4838, 47eqeq12d 2785 . . . . . . 7 ((𝜑𝑘 ∈ (0..^𝑁)) → ((𝑃‘(𝑘 + 1)) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)) ↔ ((𝑃𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)}) = ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) sadd {𝑛 ∈ ℕ0 ∣ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵)})))
4932, 48syl5ibr 236 . . . . . 6 ((𝜑𝑘 ∈ (0..^𝑁)) → ((𝑃𝑘) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) → (𝑃‘(𝑘 + 1)) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1))))
5049expcom 398 . . . . 5 (𝑘 ∈ (0..^𝑁) → (𝜑 → ((𝑃𝑘) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘) → (𝑃‘(𝑘 + 1)) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)))))
5150a2d 29 . . . 4 (𝑘 ∈ (0..^𝑁) → ((𝜑 → (𝑃𝑘) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘)) → (𝜑 → (𝑃‘(𝑘 + 1)) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1)))))
529, 13, 17, 21, 31, 51fzind2 12793 . . 3 (𝑁 ∈ (0...𝑁) → (𝜑 → (𝑃𝑁) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁)))
535, 52mpcom 38 . 2 (𝜑 → (𝑃𝑁) = (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁))
54 inss2 3980 . . . 4 (𝐴 ∩ (0..^𝑁)) ⊆ (0..^𝑁)
5554a1i 11 . . 3 (𝜑 → (𝐴 ∩ (0..^𝑁)) ⊆ (0..^𝑁))
561nn0zd 11681 . . . 4 (𝜑𝑁 ∈ ℤ)
57 uzid 11902 . . . 4 (𝑁 ∈ ℤ → 𝑁 ∈ (ℤ𝑁))
5856, 57syl 17 . . 3 (𝜑𝑁 ∈ (ℤ𝑁))
5927, 23, 28, 1, 55, 58smupvallem 15412 . 2 (𝜑 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ (𝐴 ∩ (0..^𝑁)) ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))
6053, 59eqtrd 2804 1 (𝜑 → (𝑃𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 382 = wceq 1630 ∈ wcel 2144 {crab 3064 ∩ cin 3720 ⊆ wss 3721 ∅c0 4061 ifcif 4223 𝒫 cpw 4295 ↦ cmpt 4861 ‘cfv 6031 (class class class)co 6792 ↦ cmpt2 6794 0cc0 10137 1c1 10138 + caddc 10140 − cmin 10467 ℕ0cn0 11493 ℤcz 11578 ℤ≥cuz 11887 ...cfz 12532 ..^cfzo 12672 seqcseq 13007 sadd csad 15349 smul csmu 15350 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1869 ax-4 1884 ax-5 1990 ax-6 2056 ax-7 2092 ax-8 2146 ax-9 2153 ax-10 2173 ax-11 2189 ax-12 2202 ax-13 2407 ax-ext 2750 ax-rep 4902 ax-sep 4912 ax-nul 4920 ax-pow 4971 ax-pr 5034 ax-un 7095 ax-inf2 8701 ax-cnex 10193 ax-resscn 10194 ax-1cn 10195 ax-icn 10196 ax-addcl 10197 ax-addrcl 10198 ax-mulcl 10199 ax-mulrcl 10200 ax-mulcom 10201 ax-addass 10202 ax-mulass 10203 ax-distr 10204 ax-i2m1 10205 ax-1ne0 10206 ax-1rid 10207 ax-rnegex 10208 ax-rrecex 10209 ax-cnre 10210 ax-pre-lttri 10211 ax-pre-lttrn 10212 ax-pre-ltadd 10213 ax-pre-mulgt0 10214 ax-pre-sup 10215 This theorem depends on definitions: df-bi 197 df-an 383 df-or 827 df-3or 1071 df-3an 1072 df-xor 1612 df-tru 1633 df-fal 1636 df-had 1680 df-cad 1693 df-ex 1852 df-nf 1857 df-sb 2049 df-eu 2621 df-mo 2622 df-clab 2757 df-cleq 2763 df-clel 2766 df-nfc 2901 df-ne 2943 df-nel 3046 df-ral 3065 df-rex 3066 df-reu 3067 df-rmo 3068 df-rab 3069 df-v 3351 df-sbc 3586 df-csb 3681 df-dif 3724 df-un 3726 df-in 3728 df-ss 3735 df-pss 3737 df-nul 4062 df-if 4224 df-pw 4297 df-sn 4315 df-pr 4317 df-tp 4319 df-op 4321 df-uni 4573 df-int 4610 df-iun 4654 df-disj 4753 df-br 4785 df-opab 4845 df-mpt 4862 df-tr 4885 df-id 5157 df-eprel 5162 df-po 5170 df-so 5171 df-fr 5208 df-se 5209 df-we 5210 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-pred 5823 df-ord 5869 df-on 5870 df-lim 5871 df-suc 5872 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-isom 6040 df-riota 6753 df-ov 6795 df-oprab 6796 df-mpt2 6797 df-om 7212 df-1st 7314 df-2nd 7315 df-wrecs 7558 df-recs 7620 df-rdg 7658 df-1o 7712 df-2o 7713 df-oadd 7716 df-er 7895 df-map 8010 df-pm 8011 df-en 8109 df-dom 8110 df-sdom 8111 df-fin 8112 df-sup 8503 df-inf 8504 df-oi 8570 df-card 8964 df-cda 9191 df-pnf 10277 df-mnf 10278 df-xr 10279 df-ltxr 10280 df-le 10281 df-sub 10469 df-neg 10470 df-div 10886 df-nn 11222 df-2 11280 df-3 11281 df-n0 11494 df-xnn0 11565 df-z 11579 df-uz 11888 df-rp 12035 df-fz 12533 df-fzo 12673 df-fl 12800 df-mod 12876 df-seq 13008 df-exp 13067 df-hash 13321 df-cj 14046 df-re 14047 df-im 14048 df-sqrt 14182 df-abs 14183 df-clim 14426 df-sum 14624 df-dvds 15189 df-bits 15351 df-sad 15380 df-smu 15405 This theorem is referenced by: smup1 15418 smueqlem 15419
Copyright terms: Public domain W3C validator
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## What is Calculus? (Mathematics)
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What is calculus? In this video, we give you a quick overview of calculus and introduce the limit, derivative and integral.
We begin with the question “Who invented Calculus?” Next, we talk about the two main tools you’ll study: derivatives and integrals. To understand both of these you’ll first learn about limits. After you learn how to compute the derivative and integral for basic functions and apply them to real-world problems, you’ll move up to higher dimensions and study things like “partial derivatives” and “multiple integrals.”
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If you would like to help us make new videos, you can support us through Patreon at https://www.patreon.com/socratica
We also welcome Bitcoin donations! Our Bitcoin address is: 1EttYyGwJmpy9bLY2UcmEqMJuBfaZ1HdG9
Thank you!!
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Written and Produced by Michael Harrison
Michael Harrison received his BS in math from Caltech, and his MS from the University of Washington where he studied algebraic number theory. After teaching math for a few years, Michael worked in finance both as a developer and a quantitative analyst (quant). He then worked at Google for over 5 years before leaving to found Socratica.
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Jonny quest says:
Wish I watched this while still in high-school but I will learn maybe I will be of help. Kudos
Jonny quest says:
Wish I watched this while still in high-school but I will learn maybe I will be of help. Kudos
Venkatesh babu says:
The force of push and pull of matter.
melodie gauthier says:
Very very good video!
Saurin S says:
very good explanation. thanks..
Safa Ashraf says:
This is really awesome..!!
Doug R. says:
Actually, Archimedes used a summation technique to find the volume and area of the sphere, cylinder and cone, anticipating the same developments in calculus by 2000 years. His Codex A and C manuscripts were lost, and his Codex B was almost erased and written over by monks. Codex B was re-discovered in 1906 in Istanbul (in the Hagia Sophia Library)and portions were analyzed, translated and published. Archimedes proofs use the same techniques for integration taught today.
Jake Hallam says:
Those graphs pleased me massively. I can't say why, I can only say that they do.
Arbaaz mir says:
Thanks! it gave me a good insight of calculus.
mike johnson says:
Some people are just good at math period & other people suck at it I'm not sure if it's because of the brain not getting it or they simply find it boring, people are not created equal who ever said that is the biggest ass.
Sanjay Shinde says:
Excellent. Thank you very much.
Sania Maroof says:
whats a function and what do those lines mean in the graph?
Witch Returns says:
I understood nothing…
Raimo Kohonen says:
Too easy xd
Pain Peace says:
I need to re study about calculus..
νεκρός says:
I didn’t know Alberto from wwe is math legend.
Vivek Ramanaidoo says:
Justin Epperly says:
Excellent intro.
Khagendra Kumar says:
Khagendra Kumar says:
Awesome thanks
Hamid safi says:
charming…and i have easily understand
Sera K says:
Your lectures are so easy to understand. Please post more videos for calculus and vectors.
husam botros says:
Actually a new analysis of a set of ancient clay tablets has revealed that ancient astronomers of Babylonia used advanced geometrical methods to calculate the position of Jupiter ie calculus – a conceptual leap that was previously thought to have occurred in 14 th century Europe. Ha ha ha …u need to update your information m8..
Sickage [GD] says:
Before: Didn’t even know what Calculus was.
After: Full book of Calculus.
That’s me when I watched this video…
smit shah says:
how does he finds the slope
mark fullilove says:
I wish you were my calculus teacher.
Sandra Hawkins says:
Calculus looks really interesting!
Einar Ekeberg says:
This is an absolutely fantastic video. Thank you so much
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# Seeing Improper Fractions
Today I want to put in a good word for improper fractions. You know, those fractions where the numerator is larger than the denominator, like 5/3 or 289/18.
We don’t usually tell kids about improper fractions until kind of late in their fraction learning trajectory, after they’re comfortable with “normal” fractions like 1/2 and 3/4 and 2/3 and 5/8. So children naturally get used to thinking that a fraction is like a partially filled pie. 3/4, for example, means we have a pie with 4 pieces but there are only 3 left. And then once they’re really good at this idea, we spring 5/4 on them, and the kids think, “Huh? If there are only 4 pieces in the pie to begin with, how is it possible to have 5?”
Even adults can have a hard time with improper fractions.
The conceptual root of this problem (if you’ll allow me to go into math educator mode for a moment) is that when kids see only fractions less than one, they start to think that 3/4 means “3 out of 4 things” which is not quite right, because 5/4 is a totally legitimate fraction but “5 out of 4 things” doesn’t make any sense. A better way to think about 3/4 is as three 1/4’s, where it takes four 1/4’s to make a whole. 5/4 then means five 1/4’s, where it takes four 1/4’s to make a whole. When we think of a 1/4 as a unit, we can have as many of them as we want.
So moving back to parent mode, when we give our child chances to see improper fractions, like 5/4, in real life, and when we do this early in their fraction learning trajectory, we’re not only making improper fractions themselves easier, we’re helping them develop a strong and solid understanding of what a fraction is in the first place.
Here are a few ideas for seeing improper fractions in everyday situations:
Graham Crackers
Graham crackers are great for introducing fractions to young learners because they break naturally into halves and into fourths, and because those halves and fourths are an identifiable unit. A graham cracker square can be called a half. A small graham cracker rectangle can be called a fourth. Four small rectangles make a whole cracker – that’s why they’re fourths. And it’s not at all inconceivable that you could have 5 or 6 or 7 or more small rectangles: 5/4, 6/4, 7/4, and so on.
Measuring Cups
Measuring cups are also great for thinking of fractions as units. It takes three 1/3-cup measuring cups to fill up a 1-cup measuring cup; that’s why it’s called 1/3. 1/2-cup measures and 1/4-cup measures are similar. You could experiment and see how many 1/3-cup measures would fill a glass measuring cup up to the 2 cup line. That’s six thirds (6/3)! How many thirds would it take to fill it all the way to the top, above the line? 7 thirds? 8?*
Pizza
Or quesadillas, pies, mini bread loaves… The important thing is that a) you can cut the food into equal-sized servings, and b) you have more than one whole (whole pizza, whole quesadilla, whole pie, whole loaf). If a child can identify a piece of pizza as 1/8 of a pizza, and can count pieces as eights (one pieces is 1/8, three pieces are 3/8, etc.), they can also tell how many eights 10 pieces would be, or 15, or how many eights there are in 2 pizzas.
Ruler Measurements
It’s common to use fractions in measurements – a quarter inch, a half centimeter. We have to have a way of naming measurements that are in between whole number measurements. If you’re working on measurement with your child, you’re probably using mixed numbers (e.g., 2 1/2 inches) rather than improper fractions. But go ahead and try to make the leap. If something measures 2 1/2 inches, ask: “How many half inches is that?” or “How many quarter inches is it?” This is more for older children – this is more challenging than pizza where you can see and hold and count an eighth. But it never hurts to ask something that’s beyond the child, and come back to it later if you find it’s beyond the child’s current capacity.
* As an aside, a colorful set of plastic measuring cups and spoons (maybe even one with a 1/8-cup measure) is a great Christmas or birthday gift for a child. It’s not terribly expensive, and opens up opens up all sorts of opportunities for experiences, creativity, and one-on-one time with parents or older siblings and (bonus!) they have their very own tool for thinking about and reasoning with fractions in a completely natural context.
# Measure! Early Measurement in Everyday Interactions
A child’s very early math experience is made up in large part of learning how to quantify, or to use numbers to describe characteristics of their world. Usually we think of this skill in terms of counting. But another form of quantifying is measurement. Whereas counting answers the question “How many?”, measuring assigns a number to attributes such as height, weight, length, and temperature.
Measurement allows kids access to all sorts of interesting questions about their environment. It’s also useful—we use measurement in chores, cooking, art, sports, sewing, building, etc. And measurement provides fertile ground for starting to think about fractions: what, for instance, do we call a measurement that’s somewhere between 4 inches and 5 inches?
Formal measurement skills (such as using a ruler) develop out of informal measurement experiences. Here ere are some easy, natural ways to introduce early measurement ideas in conversations and play with your toddler or preschooler.
Comparison Questions
Ask your child, “Which is bigger, ___ or ___?” This is a simple question with infinite variations, and easily tailored to your child’s ability and interests.
• Use different measurement words for different attributes. “Which is bigger?” is appropriate for some comparisons, but you might also use, “Which is taller?” or “Which is longer?” or “Which is more full?” Use comparisons questions to compare length, height, weight, volume, width, area, or even coldness or loudness.
• Don’t shy away from silly questions! “Which is bigger, the real car or the toy car?” “Which is heavier, the piano or the pencil?” These silly questions can be very fun for a young child, and still get them to think about, notice, and talk about measurable attributes.
• When your child is ready, make comparisons of objects that have very similar measurements. Your child might just guess, and that’s okay. When they begin to show interest in actually knowing, you have a great gateway to introducing formal measurement. If it’s not clear whether the pink cup or the glass tumbler has more water in it, pull out the measuring cups and find out!
Find Something Bigger/Smaller
This is a game that can be played in the home, outdoors, at the grocery store, in the car. It is better for preschoolers than toddlers, but older children can join in. Start with an object, and then ask your child to find something bigger. Proceed from there, taking turns finding bigger and bigger objects.
• You can also, of course, find smaller and smaller objects – just be sure to start with something very big.
• You can also use different measurable attributes – find something longer, or heavier. The term “bigger” is actually vague, because big could mean lots of things. Is a bookshelf bigger than a couch or smaller? It depends on what you’re measuring! But don’t avoid the term “bigger” – using the word can give you insight into how your child is thinking about “big”, and can open up conversations that lead to noticing other size attributes.
Fill It Up
We usually measure things with tools, like rulers, measuring cups, or scales. But informal measuring techniques can help children understand what formal tools are actually doing. Most informal techniques involve “filling up”:
• Fill different-size containers (cups, bowls, tupperware, etc.) with “scoops” of water or Cheerios or flour (whatever you’re willing to clean up!). Use a single measuring cup (it doesn’t matter what size) to scoop with. Notice that some containers take lots of scoops and some take very few.
• Draw outlines of shapes on paper and fill up the outline with objects of the same size, like small blocks, cotton balls, dry pasta. Again, notice that some shapes take more and some take fewer. Or you can fill the same shape with different-sized objects – goldfish crackers first, and then wheat squares. Notice that the size of the objects relates to how many fit.
• Line silverware up end-to-end and see how many pieces of silverware it takes to go from one end to the other, or all the way around. Then use silverware to measure other lengths as well. You can also use pens, crayons, pretzel sticks – anything long and thin and approximately the same length.
As with all activities on this blog, the most important thing is to follow your child’s interest, and to have fun! Your child naturally wants to learn about his/her world, and you as a parent are there to give them more tools to do so.
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# Chem 1 - Amount
### Amount of a Substance
Relative Atomic Mass The weighted Average of a mass of an atom of an element, taking into account its naturally occurring isotopes, relative to 1/12th the mass of Carbon-12.
Relative Atomic Mass Formula = Ar = (Average mass of an atom of an element) / (1/12 of mass of atom of Carbon-12)
Relative Molecular Mass The relative molecular mass of a molecule is the mass of that molecule relative the mass of an atom of Carbon-12. Add the Ar of the Atoms in the molecule.
Relative Molecular Mass Formula = Mr = (Average Mass of a Molecule) / (1/12th mass of an atom of Carbon-12).
Avagardo Constant: The Number of molecules in 12 grams of Carbon-12. 6.022 x10^3
Number of Moles = Mass of what you have / Mass of 1 mole (Mr).
Pressure = Force / Area
Describe the effects on pressure for the following variables: Temp, Volume, Number of Moles: Temp: increasing temp increases pressure, the inverse is true. Volume: Increasing Volume decreases pressure, inverse also true. No of Moles, Increasing the no of moles increases pressure, inverse true.
Ideal Gas Constant: PV=nRT (Pressure x Vol = No of Moles x Constant x Temp).
Units of the Ideal Gas Constant Vol = M^3, Temp = Kelvin (K), Pressure = Pa, R = J K-1 Mol-1.
Empirical Formula: A formula showing the simplest whole number ratio of atoms of each element in a compound.
Molecular Formula: A formula showing the actual number of each atom in a compound.
What method is used to find Empirical Formula? Combustion Analysis. Compound is burnt in Oxygen and then the products by mass are analysed.
How to Calculate Empirical Formula: Workout Number of moles of each element, divide all numbers by lowest amount, find ratio.
Moles in a Solution: (Concentration (M) x Volume (V)) / 1000
Concentration = Number of moles / Volume
When can we use the ideal gas equation? To calculate the amount of gas produced in a Reaction
How can we find concentration, experimentally? Using a Titration. So long as we know the concentration of the acid and equation between them.
Write a common Ionic equation: H+ + OH- ---> H20
Titration explain the steps: Fill burette with Acid (known concentration), measure out Alkali with Pipette and add to Flask with indicator, slowly add in acid until neutral, repeat until 2 results are the same.
Atom Economy (percentage): (Mass of desired produce / Amount of reactants) x 100
Reaction Yield (%): (Mass (or moles) of desired product / theoretical maximum amount of desired product) x 100
Created by: mjwilson1988
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# Chance News 84
## Quotations
"There's a saying in the San Francisco Bay Area: There are lies, damned lies, and next bus."
reported in Get onboard: It's time to stop hating the bus, Talk of the Nation, NPR, 29 March 2012
Submitted by Jerry Grossman
"Skepticism enjoins scientists — in fact all of us — to suspend belief until strong evidence is forthcoming, but this tentativeness is no match for the certainty of ideologues and seems to suggest to many the absurd idea that all opinions are equally valid. ...What else explains the seemingly equal weight accorded to the statements of entertainers and biological researchers on childhood vaccines?..."
--John Allen Paulos, in Why don’t Americans elect scientists?, New York Times, 13 February 2012
Submitted by Bill Peterson
From Significance magazine, February 2012:
“Diaconis and Mosteller … introduced an adage that they called the law of truly large numbers: with a large enough sample, almost anything outrageous will happen.”
“What a Coincidence? It’s not as unlikely as you think”
“[After a drug trial] no one in the world wants to know what the chance is that, for this experimental group, A was better than B …. We know exactly what the chance is, because the event has already happened. …. The probability that A was better, given this evidence, is either 1 or 0. …. What the drug company actually wants to know is, given the evidence from the trial, and possibly given other evidence about the two drugs, what are the chances that a greater proportion of future patients will get better if they take A instead of B? …. [After a hypothesis test] the civilian assumes that his original question has been answered and that the probability that A is better is high. But he should not believe this, because of course it might not be true. The p-value can be small, as small as you like, and the probability that A is better could still be low.”
“Why do statisticians answer silly questions that no one ever asks”
Submitted by Margaret Cibes
## Forsooth
“Let's start with T. Rowe Price's U.S. stock fund lineup. I have plugged in 15 of its largest actively managed U.S. equity funds. Let's start at the top with T. Rowe Price Blue Chip Growth. [Note] that T. Rowe Price Growth Stock is a tight fit with a 1.00 correlation--the highest it can get. So, we know that owning those two large-growth funds is rather redundant. [Note] that Small-Cap Value has the lowest correlation at 0.91--thus, it's a good choice for diversification purposes.”
“Fun With Mutual Correlation Matrices”
Morningstar, April 2, 2012
Submitted by Margaret Cibes
“This past week, Obama sports a 53-44 fav[orable] / unfav[orable] rating among women, but 43-54 among men. That's a whopping 20-point gender gap. .... In 2008, President Barack Obama won women 56-43, while narrowly edging out John McCain among men 49-48. That 12-point gender gap appeared massive at the time, but it appears that we're headed toward an even bigger margin in 2012.”
“Republicans face massive gender deficit”
Daily KOS, March 23, 2012
Submitted by Margaret Cibes
## Chocolate hype
“Association Between More Frequent Chocolate Consumption and Lower Body Mass Index”
Archives of Internal Medicine, March 26, 2012
This report (full text not yet available online) was published as a “research letter.” It involved about 1000 Southern Californians, who were surveyed about their eating and drinking habits and whose BMIs were computed. Funded by the National Institutes of Health, the study found that, among this group, people who ate chocolate more frequently had lower BMIs than less frequent chocolate eaters. The authors, in the “research letter” at least, apparently did not claim any cause-and-effect result and indicated that a controlled study would be needed before jumping to any such conclusion.
The Knight Science Journalism Tracker (subtitled “Peer review within science journalism”) presented a nice critique of press accounts of the “research letter” in “Eat Chocolate. Lose Weight. Yeah, right.”.
Before I became aware of the Knight project, I had tracked down a number of press accounts of the study and was encouraged to find it appropriately described in the articles, if not in the headlines, as non-definitive. (This is not to say that all reports were accurate or complete with respect to other aspects of the study.) Comparing media reports to original study reports might be a good exercise for a stats class, when a similarly enticing topic presents itself in the news, and when the original study report, or even an abstract, is available for comparison.
(a) The New York Times[1]: “Dietary studies can be unreliable, ... and it is difficult to pinpoint cause and effect.”
(b) BBC News[2]: “But the findings only suggest a link - not proof that one factor causes the other.”
(c) TODAY [3]: “The researchers only found an association, not a direct cause-effect link.”
(d) The Times of India[4]: “[T]he reasons behind this link between chocolate consumption and weight loss remain unclear.”
(e) Reuters[5]: “Researchers said the findings … don't prove that adding a candy bar to your daily diet will help you shed pounds.”
(f) Poughkeepsie Journal[6]: “The study is limited. It was observational, ... rather than a controlled trial ....”
(g) The Wall Street Journal[7]: “[B]efore people hoping to lose weight indulge in an extra scoop of chocolate fudge swirl, the researchers caution that the study doesn't prove a link between frequent chocolate munching and weight loss…..”
### Discussion
1. According to Knight, the original purpose of the project was to study “non-cardiac effects on statin drugs,” not chocolate consumption. Should this information have been reported in the articles? Why or why not?
2. Do you think that the size of the chocolate-consuming group was the same as the size of the entire group of respondents? Why would you want to know the “sample” size behind the chocolate results?
3. How would you monitor a controlled study involving a group of people who were instructed to eat chocolate occasionally and another group who were instructed to eat it more frequently?
Submitted by Margaret Cibes
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### Create an Account
Home / Questions / Consider Pacific Energy Company and US Bluechips Inc both of which reported earnings of \$9...
# Consider Pacific Energy Company and US Bluechips Inc both of which reported earnings of \$957000 Without new projects both firms will continue to generate earnings
Consider Pacific Energy Company and U.S. Bluechips, Inc., both of which reported earnings of \$957,000. Without new projects, both firms will continue to generate earnings of \$957,000 in perpetuity. Assume that all earnings are paid as dividends and that both firms require a return of 14 percent.
a.
What is the current PE ratio for each company? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
PE ratio
7.14
times
b.
Pacific Energy Company has a new project that will generate additional earnings of \$107,000 each year in perpetuity. Calculate the new PE ratio of the company. (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
PE ratio
times
c.
U.S. Bluechips has a new project that will increase earnings by \$207,000 in perpetuity. Calculate the new PE ratio of the firm. (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.)
PE ratio
times
I need the answer to b and c.
Apr 11 2020 View more View Less
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# BOLZANO WEIERSTRASS THEOREM PROOF PDF
Two other proofs of the Bolzano-Weierstrass Theorem. We prove the result: If \$ \ mathbb{X} = \{x_n: n \in \mathbb is a sequence of real numbers. Theorem. (Bolzano-Weierstrass). Every bounded sequence has a convergent subsequence. proof: Let be a bounded sequence. Then, there exists an interval. The proof doesn’t assume that one of the half-intervals has infinitely many terms while the other has finitely many terms; it only says that at least one of the halves .
Author: Tur Akinokus Country: Mexico Language: English (Spanish) Genre: History Published (Last): 10 October 2008 Pages: 367 PDF File Size: 15.82 Mb ePub File Size: 17.80 Mb ISBN: 783-4-54901-167-5 Downloads: 99732 Price: Free* [*Free Regsitration Required] Uploader: Baramar
It has since become an essential theorem of analysis.
### Bolzano–Weierstrass theorem – Wikipedia
The Bolzano—Weierstrass theorem allows one to prove that if the set of allocations is compact weirstrass non-empty, then the system has a Pareto-efficient allocation. The theorem states that each bounded sequence in R n has a convergent subsequence.
By using this site, you agree to the Terms of Use and Privacy Policy. Home Questions Tags Users Unanswered. Theorems in weiestrass analysis Compactness theorems. Bolzno it makes sense now! Since you can choose either one in this case, why not always just choose the left hand one? The proof doesn’t assume that one of the half-intervals has infinitely many terms while the other has finitely many terms; it only says that at least one of the halves has infinitely many terms, and one can be chosen arbitrarily.
PAULA ZUBIAUR PDF
I just can’t convince myself to accept this part.
Retrieved from ” https: I am now satisfied and convinced, thank you so much for the explanation! Indeed, we have the following result. Views Read Edit View history.
Now, to answer your question, as others have said and you have said yourselfit’s entirely possible weirrstrass both intervals have infinitely many elements from the sequence in them.
I know because otherwise you wouldn’t have thought to ask this question. Does that mean this proof only proves that there is only one subsequence that is convergent? It doesn’t matter, but it’s a neater proof to say “choose the left hand thforem. To show existence, you just have to show you can find one.
ISO 14887 PDF
Some fifty years later weierstrass result was identified as significant in its own right, and proved again by Weierstrass. In fact, general topology tells us that a metrizable space is compact if and only if it is sequentially compact, so that the Bolzano—Weierstrass and Heine—Borel theorems are essentially the same.
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# Search by Topic
#### Resources tagged with Interactivities similar to Ribbon Squares:
Filter by: Content type:
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### There are 190 results
Broad Topics > Information and Communications Technology > Interactivities
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Fault-free Rectangles
##### Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Countdown
##### Age 7 to 14 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### More Transformations on a Pegboard
##### Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Teddy Town
##### Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Difference
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### A Square of Numbers
##### Age 7 to 11 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
##### Age 5 to 11 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### First Connect Three for Two
##### Age 7 to 11 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Combining Cuisenaire
##### Age 7 to 11 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
##### Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Domino Numbers
##### Age 7 to 11 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### More Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Square Tangram
##### Age 7 to 11 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### Got it Article
##### Age 7 to 14
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Number Differences
##### Age 7 to 11 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Triangles All Around
##### Age 7 to 11 Challenge Level:
Can you find all the different triangles on these peg boards, and find their angles?
### Four Triangles Puzzle
##### Age 5 to 11 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Tetrafit
##### Age 7 to 11 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Have You Got It?
##### Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### One to Fifteen
##### Age 7 to 11 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
### Junior Frogs
##### Age 5 to 11 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Winning the Lottery
##### Age 7 to 11 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### World of Tan 26 - Old Chestnut
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### Rabbit Run
##### Age 7 to 11 Challenge Level:
Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths?
### Round Peg Board
##### Age 5 to 11 Challenge Level:
A generic circular pegboard resource.
### Coordinate Tan
##### Age 7 to 11 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### World of Tan 1 - Granma T
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T?
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Sorting Symmetries
##### Age 7 to 11 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Number Pyramids
##### Age 11 to 14 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### World of Tan 27 - Sharing
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
##### Age 7 to 11 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### World of Tan 29 - the Telephone
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### World of Tan 20 - Fractions
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### World of Tan 30 - Logical Thinking
##### Age 7 to 11 Challenge Level:
Can you logically construct these silhouettes using the tangram pieces?
### World of Tan 25 - Pentominoes
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 21 - Almost There Now
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 22 - an Appealing Stroll
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### Triangle Pin-down
##### Age 7 to 11 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### World of Tan 24 - Clocks
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### World of Tan 16 - Time Flies
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
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Chat with us, powered by LiveChat I need help with my discussions | WriteMoh
• # Sampling with a Pair of Dice
• Roll the 2 virtual dice and calculate the sum of the pair of virtual dice. Do this 10 times.
• Then after you have rolled the virtual pair dice 10 times and calculated 10 sums calculate the average of these 10 sums.
• Conduct this experiment again but this time roll the virtual pair of dice 20 times and calculate the 20 sums and then find the average of these 20 sums.
• Discuss the Central Limit Theorem. What is it? What does it say? Summarize it in your own words.
• Post your results and discuss how the dice activity relates to this week’s lesson, particularly the Central Limit Theorem.
• # Constructing Confidence Intervals
• What is a confidence interval? What information do confidence intervals give you?
• What advantages do confidence intervals give over a single number?
• How do you compute a confidence interval?
• Why do confidence intervals have two numbers? What does each represent?
• In the discussion for week 4, you rolled a pair of dice 10 times and calculated the average sum of your rolls. Then you did the same thing with 20 rolls. Use your results from the week 4 discussion for the average of 10 rolls and for the average of 20 rolls to construct a confidence interval for the true mean of the sum of a pair of dice (assume σ = 2.41, and you are doing a 95% confidence interval).
• What do you notice about the length of the interval for the mean of 10 rolls versus the mean of 20 rolls? Did you expect this? Why or why not?
• What would happen to the length of the interval if the confidence level was 99% instead of 95%? Why? What if it were a 90% confidence interval?
• # Errors in Hypothesis Testing
• What is hypothesis testing? What are you trying to find out?
• What are the steps in hypothesis testing?
• What is a null hypothesis, and why is it important to hypothesis testing?
• Now consider the situation where a husband and wife go to the doctor’s office to each get some tests run and the doctor accidently mixes up their charts. The doctor comes into the exam room with the results of the tests and declares that the wife is NOT pregnant but her husband IS indeed pregnant with a baby. In section 9.2, the concepts of Type I and Type II errors are introduced.
• How does this scenario illustrate the concepts behind Type I and Type II errors?
• With this situation in mind, what type of error (Type I or Type II) is worse? Explain your reasoning.
• Remember to focus on the statistical concepts (rather than biological concepts) in this posting. The post's focus is on the error types and not whether or not men can have babies.
• # Relationship of Height and Weight
• What is regression analysis?
• In every-day language, what is a trendline, and what is it telling us?
• What does it mean to interpolate? What does it mean to extrapolate?
• Using the given Height and Weight data set, follow the steps in the weekly video or on pages 584-585 of the textbook for performing a regression analysis using Excel to analyze the Height and Weight Data set (assume height is the input variable x and weight is the output variable y).
• Once you have performed the analysis in Excel, state the correct simple linear regression equation and use the regression equation to predict the weight (in pounds) of a person who is 65 inches tall and the weight (in pounds) of a person who is 100 inches tall.
• Why might the regression equation you have found not be a good predication of the weight of someone who is 100 inches tall? Are you interpolating or extrapolating when you use the trendline to predict the weight?
Do you need an answer to this or any other questions?
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Home / Power Conversion / Convert Foot Pound-force/second to Btu (th)/minute
# Convert Foot Pound-force/second to Btu (th)/minute
Please provide values below to convert foot pound-force/second to Btu (th)/minute, or vice versa.
From: foot pound-force/second To: Btu (th)/minute
### Foot Pound-force/second to Btu (th)/minute Conversion Table
Foot Pound-force/secondBtu (th)/minute
0.01 foot pound-force/second0.0007715567 Btu (th)/minute
0.1 foot pound-force/second0.0077155666 Btu (th)/minute
1 foot pound-force/second0.0771556664 Btu (th)/minute
2 foot pound-force/second0.1543113329 Btu (th)/minute
3 foot pound-force/second0.2314669993 Btu (th)/minute
5 foot pound-force/second0.3857783322 Btu (th)/minute
10 foot pound-force/second0.7715566643 Btu (th)/minute
20 foot pound-force/second1.5431133286 Btu (th)/minute
50 foot pound-force/second3.8577833216 Btu (th)/minute
100 foot pound-force/second7.7155666431 Btu (th)/minute
1000 foot pound-force/second77.1556664312 Btu (th)/minute
### How to Convert Foot Pound-force/second to Btu (th)/minute
1 foot pound-force/second = 0.0771556664 Btu (th)/minute
1 Btu (th)/minute = 12.9608108679 foot pound-force/second
Example: convert 15 foot pound-force/second to Btu (th)/minute:
15 foot pound-force/second = 15 × 0.0771556664 Btu (th)/minute = 1.1573349965 Btu (th)/minute
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 7.6: Measure Spaces. More on Outer Measures
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I. In §5, we considered premeasure spaces, stressing mainly the idea of $$\sigma$$-subadditivity (Note 5 in §5). Now we shall emphasize $$\sigma$$-additivity.
## Definition 1
A premeasure
$m : \mathcal{M} \rightarrow[0, \infty]$
is called a measure (in $$S$$) iff $$\mathcal{M}$$ is a $$\sigma$$-ring (in $$S$$), and $$m$$ is $$\sigma$$-additive on $$\mathcal{M}.$$
If so, the system
$(S, \mathcal{M}, m)$
is called a measure space; $$m X$$ is called the measure of $$X \in \mathcal{M}$$; $$\mathcal{M}$$-sets are called $$m$$-measurable sets.
Note that $$m$$ is nonnegative and $$m \emptyset=0,$$ as $$m$$ is a premeasure (Definition 2 in §5).
## Corollary $$\PageIndex{1}$$
Measures are $$\sigma$$-additive, $$\sigma$$-subadditive, monotone, and continuous.
Proof
Use Corollary 2 in §5 and Theorem 2 in §4, noting that $$\mathcal{M}$$ is a $$\sigma$$-ring.$$\quad \square$$
## Corollary $$\PageIndex{2}$$
In any measure space $$(S, \mathcal{M}, m),$$ the union and intersection of any sequence of $$m$$-measurable sets is $$m$$-measurable itself. So also is $$X-Y$$ if $$X, Y \in \mathcal{M}.$$
This is obvious since $$\mathcal{M}$$ is a $$\sigma$$-ring.
As measures and other premeasures are understood to be $$\geq 0,$$ we often write
$m : \mathcal{M} \rightarrow E^{*}$
for
$m : \mathcal{M} \rightarrow[0, \infty].$
We also briefly say "measurable" for "$$m$$-measurable."
Note that $$\emptyset \in \mathcal{M},$$ but not always $$S \in \mathcal{M}$$.
## Examples
(a) The volume of intervals in $$E^{n}$$ is a $$\sigma$$-additive premeasure, but not a measure since its domain (the intervals) is not a $$\sigma$$-ring.
(b) Let $$\mathcal{M}=2^{S}.$$ Define
$(\forall X \subseteq S) \quad m X=0.$
Then $$m$$ is trivially a measure (the zero-measure). Here each set $$X \subseteq S$$ is measurable, with $$m X=0$$.
(c) Let again $$\mathcal{M}=2^{S}.$$ Let $$m X$$ be the number of elements in $$X,$$ if finite, and $$m X=\infty$$ otherwise.
Then $$m$$ is a measure ("counting measure"). Verify!
(d) Let $$\mathcal{M}=2^{S}.$$ Fix some $$p \in S.$$ Let
$m X=\left\{\begin{array}{ll}{1} & {\text { if } p \in X}, \\ {0} & {\text { otherwise }}.\end{array}\right.$
Then $$m$$ is a measure (it describes a "unit mass" concentrated at $$p$$).
(e) A probability space is a measure space $$(S, \mathcal{M}, m$$), with
$S \in \mathcal{M} \text { and } m S=1.$
In probability theory, measurable sets are called events; $$m X$$ is called the probability of $$X,$$ often denoted by $$p X$$ or similar symbols.
In Examples (b), (c), and (d),
$\mathcal{M}=2^{S} \text { (all subsets of } S \text{).}$
More often, however,
$\mathcal{M} \neq 2^{S},$
i.e., there are nonmeasurable sets $$X \subseteq S$$ for which $$m X$$ is not defined.
Of special interest are sets $$X \in \mathcal{M},$$ with $$m X=0,$$ and their subsets. We call them $$m$$-null or null sets. One would like them to be measurable, but this is not always the case for subsets of $$X.$$
This leads us to the following definition.
## Definition 2
A measure $$m : \mathcal{M} \rightarrow E^{*}$$ is called complete iff all null sets (subsets of sets of measure zero) are measurable.
We now develop a general method for constructing complete measures.
II. From §5 (Note 5) recall that an outer measure in $$S$$ is a $$\sigma$$-subadditive premeasure defined on all of $$2^{S}$$ (even if it is not derived via Definition 3 in §5). In Examples (b), (c), and (d), $$m$$ is both a measure and an outer measure. (Why?)
An outer measure
$m^{*} : 2^{S} \rightarrow E^{*}$
need not be additive; but consider this fact:
$\text { Any set } A \subseteq S \text { splits } S \text { into two parts: } A \text { itself and }-A.$
It also splits any other set $$X$$ into $$X \cap A$$ and $$X-A;$$ indeed,
$X=(X \cap A) \cup(X-A) \text { (disjoint).}$
We want to single out those sets $$A$$ for which $$m^{*}$$ behaves "additively," i.e., so that
$m^{*} X=m^{*}(X \cap A)+m^{*}(X-A).$
This motivates our next definition.
## Definition 3
Given an outer measure $$m^{*} : 2^{S} \rightarrow E^{*}$$ and a set $$A \subseteq S,$$ we say that $$A$$ is $$m^{*}$$-measurable iff all sets $$X \subseteq S$$ are split "additively" by $$A;$$ that is,
$(\forall X \subseteq S) \quad m^{*} X=m^{*}(X \cap A)+m^{*}(X-A).$
As is easily seen (see Problem 1), this is equivalent to
$(\forall X \subseteq A)(\forall Y \subseteq-A) \quad m^{*}(X \cup Y)=m^{*} X+m^{*} Y.$
The family of all $$m^{*}$$-measurable sets is usually denoted by $$\mathcal{M}^{*}.$$ The system $$\left(S, \mathcal{M}^{*}, m^{*}\right)$$ is called an outer measure space.
Note 1. Definition 3 applies to outer measures only. For measures, "$$m$$-measurable" means simply "member of the domain of $$m$$" (Definition 1).
Note 2. In (1) and (2), we may equivalently replace the equality sign $$(=)$$ by $$(\geq).$$ Indeed, $$X$$ is covered by
$\{X \cap A, X-A\},$
and $$X \cup Y$$ is covered by $$\{X, Y\};$$ so the reverse inequality $$(\leq)$$ anyway holds, by subadditivity.
Our main objective is to prove the following fundamental theorem.
## Theorem $$\PageIndex{1}$$
In any outer measure space
$\left(S, \mathcal{M}^{*}, m^{*}\right),$
the family $$\mathcal{M}^{*}$$ of all $$m^{*}$$-measurable sets is a $$\sigma$$-field in $$S,$$ and $$m^{*},$$ when restricted to $$\mathcal{M}^{*},$$ is a complete measure (denoted by $$m$$ and called the $$m^{*}$$-induced measure; so $$m^{*}=m$$ on $$\mathcal{M}^{*}$$).
Proof
We split the proof into several steps (lemmas).
## lemma 1
$$\mathcal{M}^{*}$$ is closed under complementation:
$\left(\forall A \in \mathcal{M}^{*}\right) \quad-A \in \mathcal{M}^{*}.$
Indeed, the measurability criterion (2) is same for $$A$$ and $$-A$$ alike.
## lemma 2
$$\emptyset$$ and $$S$$ are $$\mathcal{M}^{*}$$ sets. So are all sets of outer measure 0.
Proof
Let $$m^{*} A=0.$$ To prove $$A \in \mathcal{M}^{*},$$ use (2) and Note 2.
Thus take any $$X \subseteq A$$ and $$Y \subseteq-A.$$ Then by monotonicity,
$m^{*} X \leq m^{*} A=0$
and
$m^{*} Y \leq m^{*}(X \cup Y).$
Thus
$m^{*} X+m^{*} Y=0+m^{*} Y \leq m^{*}(X \cup Y),$
as required.
In particular, as $$m^{*} \emptyset=0, \emptyset$$ is $$m^{*}$$-measurable $$\left(\emptyset \in \mathcal{M}^{*}\right)$$.
So is $$S$$ (the complement of $$\emptyset)$$ by Lemma 1.$$\quad \square$$
## lemma 3
$$\mathcal{M}^{*}$$ is closed under finite unions:
$\left(\forall A, B \in \mathcal{M}^{*}\right) \quad A \cup B \in \mathcal{M}^{*}.$
Proof
This time we shall use formula (1). By Note 2, it suffices to show that
$(\forall X \subseteq S) \quad m^{*} X \geq m^{*}(X \cap(A \cup B))+m^{*}(X-(A \cup B)).$
Fix any $$X \subseteq S;$$ as $$A \in \mathcal{M}^{*},$$ we have
$m^{*} X=m^{*}(X \cap A)+m^{*}(X-A).$
Similarly, as $$B \in \mathcal{M}^{*},$$ we have (replacing $$X$$ by $$X-A$$ in (1))
\begin{aligned} m^{*}(X-A) &=m^{*}((X-A) \cap B)+m^{*}(X-A-B) \\ &=m^{*}(X \cap-A \cap B)+m^{*}(X-(A \cup B)), \end{aligned}
since
$X-A=X \cap-A$
and
$X-A-B=X-(A \cup B).$
Combining (4) with (3), we get
$m^{*} X=m^{*}(X \cap A)+m^{*}(X \cap-A \cap B)+m^{*}(X-(A \cup B)).$
Now verify that
$(X \cap A) \cup(X \cap-A \cap B) \supseteq X \cap(A \cup B).$
As $$m$$ is subadditive, this yields
$m^{*}(X \cap A)+m^{*}(X \cap-A \cap B) \geq m^{*}(X \cap(A \cup B)).$
Combining with (5), we get
$m^{*} X \geq m^{*}(X \cap(A \cup B))+m^{*}(X-(A \cup B)),$
so that $$A \cup B \in \mathcal{M}^{*},$$ indeed.$$\quad \square$$
Induction extends Lemma 3 to all finite unions of $$\mathcal{M}^{*}$$-sets.
Note that by Problem 3 in §3, $$\mathcal{M}^{*}$$ is a set field, hence surely a ring. Thus Corollary 1 in §1 applies to it. (We use it below.)
## lemma 4
Let
$X_{k} \subseteq A_{k} \subseteq S, \quad k=0,1,2, \ldots,$
with all $$A_{k}$$ pairwise disjoint.
Let $$A_{k} \in \mathcal{M}^{*}$$ for $$k \geq 1.$$ ($$A_{0}$$ and the $$X_{k}$$ need not be $$\mathcal{M}^{*}$$-sets.) Then
$m^{*}\left(\bigcup_{k=0}^{\infty} X_{k}\right)=\sum_{k=0}^{\infty} m^{*} X_{k}.$
Proof
We start with two sets, $$A_{0}$$ and $$A_{1};$$ so
$A_{1} \in \mathcal{M}^{*}, A_{0} \cap A_{1}=\emptyset, X_{0} \subseteq A_{0}, \text { and } X_{1} \subseteq A_{1}.$
As $$A_{0} \cap A_{1}=\emptyset,$$ we have $$A_{0} \subseteq-A_{1};$$ hence also $$X_{0} \subseteq-A_{1}$$.
since $$A_{1} \in \mathcal{M}^{*},$$ we use formula (2), with
$X=X_{1} \subseteq A_{1} \text { and } Y=X_{0} \subseteq-A,$
to obtain
$m^{*}\left(X_{0} \cup X_{1}\right)=m^{*} X_{0}+m^{*} X_{1}.$
Thus (6) holds for two sets.
Induction now easily yields
$(\forall n) \sum_{k=0}^{n} m^{*} X_{k}=m^{*}\left(\bigcup_{k=0}^{n} X_{k}\right) \leq m^{*}\left(\bigcup_{k=0}^{\infty} X_{k}\right)$
by monotonicity of $$m^{*}.$$ Now let $$n \rightarrow \infty$$ and pass to the limit to get
$\sum_{k=0}^{\infty} m^{*} X_{k} \leq m^{*}\left(\bigcup_{k=0}^{\infty} X_{k}\right).$
As $$\bigcup X_{k}$$ is covered by the $$X_{k},$$ the $$\sigma$$-subadditivity of $$m^{*}$$ yields the reverse inequality as well. Thus (6) is proved.$$\quad \square$$
Proof of Theorem 1. As we noted, $$\mathcal{M}^{*}$$ is a field. To show that it is also closed under countable unions (a $$\sigma$$-field), let
$U=\bigcup_{k=1}^{\infty} A_{k}, \quad A_{k} \in \mathcal{M}^{*}.$
We have to prove that $$U \in \mathcal{M}^{*};$$ or by (2) and Note 2,
$(\forall X \subseteq U)(\forall Y \subseteq-U) \quad m^{*}(X \cup Y) \geq m^{*} X+m^{*} Y.$
We may safely assume that the $$A_{k}$$ are disjoint. (If not, replace them by disjoint sets $$B_{k} \in \mathcal{M}^{*},$$ as in Corollary 1 §1.)
To prove (7), fix any $$X \subseteq U$$ and $$Y \subseteq-U,$$ and let
$X_{k}=X \cap A_{k} \subseteq A_{k},$
$$A_{0}=-U,$$ and $$X_{0}=Y,$$ satisfying all assumptions of Lemma 4. Thus by (6), writing the first term separately, we have
$m^{*}\left(Y \cup \bigcup_{k=1}^{\infty} X_{k}\right)=m^{*} Y+\sum_{k=1}^{\infty} m^{*} X_{k}.$
But
$\bigcup_{k=1}^{\infty} X_{k}=\bigcup_{k=1}^{\infty}\left(X \cap A_{k}\right)=X \cap \bigcup_{k=1}^{\infty} A_{k}=X \cap U=X$
(as $$X \subseteq U).$$ Also, by $$\sigma$$-subadditivity,
$\sum m^{*} X_{k} \geq m^{*} \bigcup X_{k}=m^{*} X.$
Therefore, (8) implies (7); so $$\mathcal{M}^{*}$$ is a $$\sigma$$-field.
Moreover, $$m^{*}$$ is $$\sigma$$-additive on $$\mathcal{M}^{*},$$ as follows from Lemma 4 by taking
$X_{k}=A_{k} \in \mathcal{M}^{*}, A_{0}=\emptyset.$
Thus $$m^{*}$$ acts as a measure on $$\mathcal{M}^{*}$$.
By Lemma 2, $$m^{*}$$ is complete; for if $$X$$ is "null" ($$X \subseteq A$$ and $$m^{*} A=0$$), then $$m^{*} X=0;$$ so $$X \in \mathcal{M}^{*},$$ as required.
Thus all is proved.$$\quad \square$$
We thus have a standard method for constructing measures: From a premeasure
$\mu : \mathcal{C} \rightarrow E^{*}$
in $$S,$$ we obtain the $$\mu$$-induced outer measure
$m^{*} : 2^{S} \rightarrow E^{*} \text{ (§5);}$
this, in turn, induces a complete measure
$m : \mathcal{M}^{*} \rightarrow E^{*}.$
But we need more: We want $$m$$ to be an extension of $$\mu,$$ i.e.,
$m=\mu \text { on } \mathcal{C},$
with $$\mathcal{C} \subseteq \mathcal{M}^{*}$$ (meaning that all $$\mathcal{C}$$-sets are $$m^{*}$$-measurable). We now explore this question.
## lemma 5
Let $$(S, \mathcal{C}, \mu)$$ and $$m^{*}$$ be as in Definition 3 of §5. Then for a set $$A \subseteq S$$ to be $$m^{*}$$-measurable, it suffices that
$m^{*} X \geq m^{*}(X \cap A)+m^{*}(x-A) \quad \text {for all } X \in \mathcal{C}.$
Proof
We must show that (9) holds for any $$X \subseteq S,$$ even not a $$\mathcal{C}$$-set.
This is trivial if $$m^{*} X=\infty.$$ Thus assume $$m^{*} X<\infty$$ and fix any $$\varepsilon>0$$.
By Note 3 in §5, $$X$$ must have a basic covering $$\left\{B_{n}\right\} \subseteq \mathcal{C}$$ so that
$X \subseteq \bigcup_{n} B_{n}$
and
$m^{*} X+\varepsilon>\sum \mu B_{n} \geq \sum m^{*} B_{n}.$
(Explain!)
Now, as $$X \subseteq \cup B_{n},$$ we have
$X \cap A \subseteq \bigcup B_{n} \cap A=\bigcup\left(B_{n} \cap A\right).$
Similarly,
$X-A=X \cap-A \subseteq \bigcup\left(B_{n}-A\right).$
Hence, as $$m^{*}$$ is $$\sigma$$-subadditive and monotone, we get
\begin{aligned} m^{*}(X \cap A)+m^{*}(X-A) & \leq m^{*}\left(\bigcup\left(B_{n} \cap A\right)\right)+m^{*}\left(\bigcup\left(B_{n}-A\right)\right) \\ & \leq \sum\left[m^{*}\left(B_{n} \cap A\right)+m^{*}\left(B_{n}-A\right)\right]. \end{aligned}
But by assumption, (9) holds for any $$\mathcal{C}$$-set, hence for each $$B_{n}.$$ Thus
$m^{*}\left(B_{n} \cap A\right)+m^{*}\left(B_{n}-A\right) \leq m^{*} B_{n},$
and (11) yields
$m^{*}(X \cap A)+m^{*}(X-A) \leq \sum\left[m^{*}\left(B_{n} \cap A\right)+m^{*}\left(B_{n}-A\right)\right] \leq \sum m^{*} B_{n}.$
Therefore, by (10),
$m^{*}(X \cap A)+m^{*}(X-A) \leq m^{*} X+\varepsilon.$
Making $$\varepsilon \rightarrow 0,$$ we prove (10) for any $$X \subseteq S,$$ so that $$A \in \mathcal{M}^{*},$$ as required.$$\quad \square$$
## Theorem $$\PageIndex{2}$$
Let the premeasure
$\mu : \mathcal{C} \rightarrow E^{*}$
be $$\sigma$$-additive on $$\mathcal{C}, a$$ semiring in $$S.$$ Let $$m^{*}$$ be the $$\mu$$-induced outer measure, and
$m : \mathcal{M}^{*} \rightarrow E^{*}$
be the $$m^{*}$$-induced measure. Then
(i) $$\mathcal{C} \subseteq \mathcal{M}^{*}$$ and
(ii) $$\mu=m^{*}=m$$ on $$\mathcal{C}$$.
Thus $$m$$ is a $$\sigma$$-additive extension of $$\mu$$ (called its Lebesgue extension) to $$\mathcal{M}^{*}$$.
Proof
By Corollary 2 in §5, $$\mu$$ is also $$\sigma$$-subadditive on the semiring $$\mathcal{C}.$$ Thus by Theorem 2 in §5, $$\mu=m^{*}$$ on $$\mathcal{C}.$$
To prove that $$\mathcal{C} \subseteq \mathcal{M}^{*},$$ we fix $$A \in \mathcal{C}$$ and show that $$A$$ satisfies (9), so that $$A \in \mathcal{M}^{*}.$$
Thus take any $$X \in \mathcal{C}.$$ As $$\mathcal{C}$$ is a semiring, $$X \cap A \in \mathcal{C}$$ and
$X-A=\bigcup_{k=1}^{n} A_{k} \text { (disjoint)}$
for some sets $$A_{k} \in \mathcal{C}.$$ Hence
\begin{aligned} m^{*}(X \cap A)+m^{*}(X-A) &=m^{*}(X \cap A)+m^{*} \bigcup_{k=1}^{n} A_{k} \\ & \leq m^{*}(X \cap A)+\sum_{k=1}^{n} m^{*} A_{k}. \end{aligned}
As
$X=(X \cap A) \cup(X-A)=(X \cap A) \cup \bigcup A_{k} \text { (disjoint),}$
the additivity of $$\mu$$ and the equality $$\mu=m^{*}$$ on $$\mathcal{C}$$ yield
$m^{*} X=m^{*}(X \cap A)+\sum_{k=1}^{n} m^{*} A_{k}.$
Hence by (12),
$m^{*} X \geq m^{*}(X \cap A)+m^{*}(X-A);$
so by Lemma 5, $$A \in \mathcal{M}^{*},$$ as required.
Also, by definition, $$m=m^{*}$$ on $$\mathcal{M}^{*},$$ hence on $$\mathcal{C}.$$ Thus
$\mu=m^{*}=m \text { on } \mathcal{C},$
as claimed.$$\quad \square$$
Note 3. In particular, Theorem 2 applies if
$\mu : \mathcal{M} \rightarrow E^{*}$
is a measure (so that $$\mathcal{C}=\mathcal{M}$$ is even a $$\sigma$$-ring).
Thus any such $$\mu$$ can be extended to a complete measure $$m$$ (its Lebesgue extension) on a $$\sigma$$-field
$\mathcal{M}^{*} \supseteq \mathcal{M}$
via the $$\mu$$-induced outer measure (call it $$\mu^{*}$$ this time), with
$\mu^{*}=m=\mu \text { on } \mathcal{M}.$
Moreover,
$\mathcal{M}^{*} \supseteq \mathcal{M} \supseteq \mathcal{M}_{\sigma}$
(see Note 2 in §3); so $$\mu^{*}$$ is $$\mathcal{M}$$-regular and $$\mathcal{M}^{*}$$-regular (Theorem 3 of §5).
Note 4. A reapplication of this process to $$m$$ does not change $$m$$ (Problem 16).
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# Data Mining: Finding Similar Items and Users
Because we want to give kick-ass product recommendations.
I'm showing you how to find related items based on a really simple formula. If you pay attention, this technique is used all over the web (like on Amazon) to personalize the user experience and increase conversion rates.
To get one question out of the way: there are already many available libraries that do this, but as you'll see there are multiple ways of skinning the cat and you won't be able to pick the right one without understanding the process, at least intuitively.
## Defining the Problem
To find similar items to a certain item, you've got to first define what it means for 2 items to be similar and this depends on the problem you're trying to solve:
• on a blog, you may want to suggest similar articles that share the same tags, or that have been viewed by the same people viewing the item you want to compare with
• Amazon has this section called "customers that bought this item also bought", which is self-explanatory
• a service like IMDB, based on your ratings, could find users similar to you, users that liked or hated approximately the same movies you did, thus giving you suggestions on movies you'd like to watch in the future
In each case you need a way to classify these items you're comparing, whether it is tags, or items purchased, or movies reviewed. We'll be using tags, as it is simpler, but the formula holds for more complicated instances.
## Redefining the Problem in Terms of Geometry
We'll be using my blog as sample. Let's take some tags:
``````["API", "Algorithms", "Amazon", "Android", "Books", "Browser"]
``````
That's 6 tags. Well, what if we considered these tags as dimensions in a 6-dimensional Euclidean space? Then each item you want to sort or compare becomes a point in this space, in which a coordinate (representing a tag) is either one (tagged) or zero (not tagged).
So let's say we've got one article tagged with API and Browser. Then its associated point will be:
``````[ 1, 0, 0, 0, 0, 1 ]
``````
Now these coordinates could represent something else. For instance they could represent users. If say you've got a total of 6 users in your system, 2 of them rating an item with 3 and 5 stars respectively, you could have for the article in question this associated point (do note the order is very important):
``````[ 0, 3, 0, 0, 5, 0 ]
``````
So now you can go ahead and calculate distances between these points. For instance you could calculate the angle between the associated vectors, or the actual euclidean distance between the 2 points. For a 2-dimensional Euclidean space, here's how it would look like:
## Euclidean Distance
The mathematical formula for the Euclidean distance is really simple. Considering 2 points, A and B, with their associated coordinates, the distance is defined as:
The lower the distance between 2 points, then the higher the similarity. Here's some Ruby code:
``````# Returns the Euclidean distance between 2 points
#
# Params:
# - a, b: list of coordinates (float or integer)
#
def euclidean_distance(a, b)
sq = a.zip(b).map{|a,b| (a - b) ** 2}
Math.sqrt(sq.inject(0) {|s,c| s + c})
end
# Returns the associated point of our tags_set, relative to our
# tags_space.
#
# Params:
# - tags_set: list of tags
# - tags_space: _ordered_ list of tags
def tags_to_point(tags_set, tags_space)
tags_space.map{|c| tags_set.member?(c) ? 1 : 0}
end
# Returns other_items sorted by similarity to this_item
# (most relevant are first in the returned list)
#
# Params:
# - items: list of hashes that have [:tags]
# - by_these_tags: list of tags to compare with
def sort_by_similarity(items, by_these_tags)
tags_space = by_these_tags + items.map{|x| x[:tags]}
tags_space.flatten!.sort!.uniq!
this_point = tags_to_point(by_these_tags, tags_space)
other_points = items.map{|i|
[i, tags_to_point(i[:tags], tags_space)]
}
similarities = other_points.map{|item, that_point|
[item, euclidean_distance(this_point, that_point)]
}
sorted = similarities.sort {|a,b| a[1] <=> b[1]}
return sorted.map{|point,s| point}
end
``````
And here is the test you could do, and btw you can copy the above and the bellow script and run it directly:
``````# SAMPLE DATA
all_articles = [
{
:article => "Data Mining: Finding Similar Items",
:tags => ["Algorithms", "Programming", "Mining",
"Python", "Ruby"]
},
{
:article => "Blogging Platform for Hackers",
:tags => ["Publishing", "Server", "Cloud", "Heroku",
"Jekyll", "GAE"]
},
{
:article => "UX Tip: Don't Hurt Me On Sign-Up",
:tags => ["Web", "Design", "UX"]
},
{
:article => "Crawling the Android Marketplace",
:tags => ["Python", "Android", "Mining",
"Web", "API"]
}
]
# SORTING these articles by similarity with an article
# tagged with Publishing + Web + API
#
#
# The list is returned in this order:
#
# 1. article: Crawling the Android Marketplace
# similarity: 2.0
#
# 2. article: "UX Tip: Don't Hurt Me On Sign-Up"
# similarity: 2.0
#
# 3. article: Blogging Platform for Hackers
# similarity: 2.645751
#
# 4. article: "Data Mining: Finding Similar Items"
# similarity: 2.828427
#
sorted = sort_by_similarity(
all_articles, ['Publishing', 'Web', 'API'])
require 'yaml'
puts YAML.dump(sorted)
``````
### The Problem (or Strength) of Euclidean Distance
Can you see one flaw with it for our chosen data-set and intention? I think you can - the first 2 articles have the same Euclidean distance to ["Publishing", "Web", "API"], even though the first article shares 2 tags with our chosen item, instead of just 1 tag as the rest.
To visualize why, look at the points used in calculating the distance for the first article:
``````[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1]
[1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1]
``````
So 4 coordinates are different. Now look at the points used for the second article:
``````[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1]
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1]
``````
Again, 4 coordinates are different. So here's the deal with Euclidean distance: it measures dissimilarity. The coordinates that are the same are less important than the coordinates that are different. For my purpose here, this is not good - because articles with more tags (or less) tags than the average are going to be disadvantaged.
## Cosine Similarity
This method is very similar to the one above, but does tend to give slightly different results, because this one actually measures similarity instead of dissimilarity. Here's the formula:
If you look at the visual with the 2 axis and 2 points, we need the cosine of the angle theta that's between the vectors associated with our 2 points. And for our sample it does give better results.
The values will range between -1 and 1. -1 means that 2 items are total opposites, 0 means that the 2 items are independent of each other and 1 means that the 2 items are very similar (btw, because we are only doing zeros and ones for coordinates here, this score will never get negative for our sample).
Here's the Ruby code (leaving out the wiring to our sample data, do that as an exercise):
``````def dot_product(a, b)
products = a.zip(b).map{|a, b| a * b}
products.inject(0) {|s,p| s + p}
end
def magnitude(point)
squares = point.map{|x| x ** 2}
Math.sqrt(squares.inject(0) {|s, c| s + c})
end
# Returns the cosine of the angle between the vectors
#associated with 2 points
#
# Params:
# - a, b: list of coordinates (float or integer)
#
def cosine_similarity(a, b)
dot_product(a, b) / (magnitude(a) * magnitude(b))
end
``````
Also, sorting the articles in the above sample gives me the following:
``````- article: Crawling the Android Marketplace
similarity: 0.5163977794943222
- article: "UX Tip: Don't Hurt Me On Sign-Up"
similarity: 0.33333333333333337
- article: Blogging Platform for Hackers
similarity: 0.23570226039551587
- article: "Data Mining: Finding Similar Items"
similarity: 0.0
``````
Right, so much better for this chosen sample and usage. Ain't this fun? BUT, you guessed it, there's a problem with this too …
### The Problem with Our Sample; The Tf-Idf Weight
Our data sample is so simple that we could have simply counted the number of common tags and use that as a metric. The result would be the same without getting fancy with Cosine Similarity :-)
Clearly a tag such as "Heroku" is more specific than a general purpose tag such as "Web". Also, just because Jekyll was mentioned in an article, that doesn't make the article about Jekyll. Also an article tagged with "Android" may be twice as Android-related as another article also tagged with "Android".
So here's a solution to this: the Tf-Idf weight, a statistical measure used to evaluate how important a word is to a document in a collection or corpus. With it you can give values to your coordinates that are much more specific than simple ones and zeros. But I'll leave that for another day.
Also, related to our simple data-set here, perhaps an even simpler metric, like the Jaccard index would be better.
## Pearson Correlation Coefficient
The Pearson Correlation Coefficient for finding the similarity of 2 items is slightly more sophisticated and doesn't really apply to my chosen data-set. This coefficient measures how well two samples are linearly related.
For example, on IMDB we may have 2 users. One of them, lets call him John, has given the following ratings to 5 movies: [1, 2, 3, 4, 5]. The other one, Mary, has given the following ratings to the same 5 movies: [4, 5, 6, 7, 8]. The 2 users are very similar, as there is a perfect linear correlation between them, since Mary just gives the same rankings as John plus 3. The formula itself or the theory is not very intuitive though. But it is simple to calculate:
Here's the code:
``````def pearson_score(a, b)
n = a.length
return 0 unless n > 0
# summing the preferences
sum1 = a.inject(0) {|sum, c| sum + c}
sum2 = b.inject(0) {|sum, c| sum + c}
# summing up the squares
sum1_sq = a.inject(0) {|sum, c| sum + c ** 2}
sum2_sq = b.inject(0) {|sum, c| sum + c ** 2}
# summing up the product
prod_sum = a.zip(b).inject(0) {|sum, ab| sum + ab[0] * ab[1]}
# calculating the Pearson score
num = prod_sum - (sum1 *sum2 / n)
den = Math.sqrt((sum1_sq - (sum1 ** 2) / n) * (sum2_sq - (sum2 ** 2) / n))
return 0 if den == 0
return num / den
end
puts pearson_score([1,2,3,4,5], [4,5,6,7,8])
# => 1.0
puts pearson_score([1,2,3,4,5], [4,5,0,7,8])
# => 0.5063696835418333
puts pearson_score([1,2,3,4,5], [4,5,0,7,7])
# => 0.4338609156373132
puts pearson_score([1,2,3,4,5], [8,7,6,5,4])
# => -1
``````
## Manhattan Distance
There is no one size fits all and the formula you're going to use depends on your data and what you want out of it.
For instance the Manhattan Distance computes the distance that would be traveled to get from one data point to the other if a grid-like path is followed. I like this graphic from Wikipedia that perfectly illustrates the difference with Euclidean distance:
Red, yellow and blue lines all have the same length and the distance is bigger than the corresponding green diagonal, which is the normal Euclidean distance.
Personally I haven't found a usage for it, as it is more related to path-finding algorithms, but it's a good thing to keep in mind that it exists and may prove useful. Since it measures how many changes you have to do to your origin location to get to your destination while being limited to taking small steps in a grid-like system, it is very similar in spirit to the Levenshtein distance, which measures the minimum number of changes required to transform some text into another.
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# B PEMDAS and Physicists
#### John3509
In case you have not seen the equation that has gone viral recently https://www.popularmechanics.com/science/math/a28569610/viral-math-problem-2019-solved/
Similarly, there can be ambiguity in the use of the slash symbol / in expressions such as 1/2x.[5] If one rewrites this expression as 1 ÷ 2x and then interprets the division symbol as indicating multiplication by the reciprocal, this becomes:
1 ÷ 2 × x = 1 × 1/2 × x = 1/2 × x.
With this interpretation 1 ÷ 2x is equal to (1 ÷ 2)x.[1][6] However, in some of the academic literature, multiplication denoted by juxtaposition (also known as implied multiplication) is interpreted as having higher precedence than division, so that 1 ÷ 2x equals 1 ÷ (2x), not (1 ÷ 2)x. This higher precedence itself implies the need for an updated mnemonic PEIMDAS, with I = Implied multiplication.
For example, the manuscript submission instructions for the Physical Review journals state that multiplication is of higher precedence than division with a slash,[7] and this is also the convention observed in prominent physics textbooks such as the Course of Theoretical Physics by Landau and Lifshitz and the Feynman Lectures on Physics.[a]
Why are physicists allowed to break the rules? What reasoning does the Physics Review and Feynman have for making multiplication of higher precedence than division with a slash? And doesn't this cause problems for consistency within Physics?
#### Ibix
In Britain an 18 year old may drink alcohol in a bar. In America they'd risk arrest. Why do Brits get to break the rules?
Answer: we don't. Not everybody agrees on the same set of rules. As long as everyone is clear what the rules are in their circumstances there's no problem. The "puzzle" you linked (what is 8/2(2+2)?) boils down to whether you believe you should apply strict left-to-right evaluation or not. It's just an excuse for a fight. The correct answer, of course, is that it depends what convention you use.
Also many physics equations use quantities with units, which frequently resolve any ambiguity. For example, deliberately being sloppy, kinetic energy is $1/2mv^2$. Knowing that's an energy, interpreting it as $1/(2mv^2)$ is clearly wrong since it has the wrong units so I must, in this case, mean $(1/2)mv^2$. That means that an editorial instruction to never write $1/2mv^2$ is largely a style choice.
Edit: just to add, multiplication having a higher precedence than addition is also a convention. In this case it's a natural one, since everyone interprets "I've got three twenty pence pieces and two ten pence pieces" as me having 80p. It makes sense for 3×20+2×10 to be interpreted the same way as the verbal version. But systems that don't respect that rule are easy to write and (as long as I make clear that I'm writing in such a convention) they aren't wrong, just different.
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#### fresh_42
Mentor
2018 Award
This entire discussion about the order is ridiculous. It only shows that division signs lack context in a linear word. I liked @Ibix's comparison. If I write "either", is it eether or ither? Is it is-sue or ish-you? There is no way to figure it out.
My personally preferred answer to such questions is: Divisions and subtractions are multiplications and additions. There is no such thing as a division or a subtraction, ergo no problem. We only have two operations, not four, and what we really do is $a \div b = a \cdot b^{-1}$ and $a-b = a + b^{-1}$. As the latter leads to confusion if we used both operations and the convention to write inverse elements as ${*}^{-1}$, it is acceptable to write $b^{-1}=-b$ in the additive case, whence $a-b=a+ -b$.
To debate a convention which is wrong by nature is ridiculous.
#### ZapperZ
Staff Emeritus
2018 Award
In case you have not seen the equation that has gone viral recently https://www.popularmechanics.com/science/math/a28569610/viral-math-problem-2019-solved/
Why are physicists allowed to break the rules? What reasoning does the Physics Review and Feynman have for making multiplication of higher precedence than division with a slash? And doesn't this cause problems for consistency within Physics?
This is nothing to do with mathematical rules. It has everything to do with typesetting style!
Many publications have their own set of styles, and more often, it is based on what the common readers of that publication are familiar with. The Physical Review editor isn't trying to rewrite the math rules, but rather to clarify that if you type it that way, it will be interpreted by their editor and copywriters as such, and will be typeset as such!
Sometime people make a mountain out of a molehill.
Zz.
#### Mark44
Mentor
This is nothing to do with mathematical rules. It has everything to do with typesetting style!
I agree completely.
This expression --
$$\frac 1 {2x}$$
-- is unambiguous, but this one --
$1/2x$
-- is ambiguous.
In the first expression, the vinculum clearly indicates that $2x$ is the denominator, thereby playing the same role as parentheses in the equivalent expression $1/(2x)$.
#### John3509
In Britain an 18 year old may drink alcohol in a bar. In America they'd risk arrest. Why do Brits get to break the rules?
Answer: we don't. Not everybody agrees on the same set of rules. As long as everyone is clear what the rules are in their circumstances there's no problem. The "puzzle" you linked (what is 8/2(2+2)?) boils down to whether you believe you should apply strict left-to-right evaluation or not. It's just an excuse for a fight. The correct answer, of course, is that it depends what convention you use.
Also many physics equations use quantities with units, which frequently resolve any ambiguity. For example, deliberately being sloppy, kinetic energy is $1/2mv^2$. Knowing that's an energy, interpreting it as $1/(2mv^2)$ is clearly wrong since it has the wrong units so I must, in this case, mean $(1/2)mv^2$. That means that an editorial instruction to never write $1/2mv^2$ is largely a style choice.
Edit: just to add, multiplication having a higher precedence than addition is also a convention. In this case it's a natural one, since everyone interprets "I've got three twenty pence pieces and two ten pence pieces" as me having 80p. It makes sense for 3×20+2×10 to be interpreted the same way as the verbal version. But systems that don't respect that rule are easy to write and (as long as I make clear that I'm writing in such a convention) they aren't wrong, just different.
Right, but PEMDAS/BEDMAS is a standardized set of rules for math. There are no different "countries" in math, there is just...math. I do agree that multiplication before addition is more intuitive, that's probably why PEMDAS is the way it is, but what is the reason for Physicists choosing multiplication to take priority over division with a slash when it goes against PEMDAS. I know you said you can chose what ever convention you want but even when all technology and calculators they use are programed to strictly follow PEMDAS, this decision seems weird and pointless to me.
#### fresh_42
Mentor
2018 Award
Right, but PEMDAS/BEDMAS is a standardized set of rules for math. There are no different "countries" in math, there is just...math. I do agree that multiplication before addition is more intuitive, that's probably why PEMDAS is the way it is, but what is the reason for Physicists choosing multiplication to take priority over division with a slash when it goes against PEMDAS. I know you said you can chose what ever convention you want but even when all technology and calculators they use are programed to strictly follow PEMDAS, this decision seems weird and pointless to me.
Read this thread. The notation is not uniquely defined, hence worthless. No mathematician or physicist would ever write down such a literal nonsense. It is a character string which needs further interpretation rules. Same as you cannot write $(123)(321)$ without telling us whether this has to be read from left to right or right to left. Language isn't unique, that's why scientists prefer mathematics. The character string in your example isn't mathematics, it's crap.
Edit: "A road crosses a chicken." is a syntactically correct sentence, but without semantics it is useless. The example in post #1 is also syntactically correct, but readers apply their own semantic when they read it. However, nobody has defined a semantic for $\div$.
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#### John3509
Read this thread. The notation is not uniquely defined, hence worthless. No mathematician or physicist would ever write down such a literal nonsense. It is a character string which needs further interpretation rules. Same as you cannot write $(123)(321)$ without telling us whether this has to be read from left to right or right to left. Language isn't unique, that's why scientists prefer mathematics. The character string in your example isn't mathematics, it's crap.
But that is what i am trying to understand. Why do physicists think it is not uniquely defined or ambiguous? If you plug it in to a ti84 it will 16 as the solution, programing has it clearly defined, / = (division symbol with 2 dots). When written as 1/2x why would anyone be unsure if that both 2 and x are in the denominator if there are no paranthesis around them?
#### PeroK
Homework Helper
Gold Member
2018 Award
But that is what i am trying to understand. Why do physicists think it is not uniquely defined or ambiguous? If you plug it in to a ti84 it will 16 as the solution, programing has it clearly defined, / = (division symbol with 2 dots). When written as 1/2x why would anyone be unsure if that both 2 and x are in the denominator if there are no paranthesis around them?
A computer is a computer. Mathematicians and physicist's are not mere computers, parsing strings of symbols mechanically according to a fixed set of rules.
For example, I would write $e^{i\pi}$, literally as you see it on the screen. But, to communicate via a computer I have to render that as a Latex string. Which is another arbitrary convention.
Also, regarding your assertion that mathematics has no national boundaries, the Chinese might have something to say about that.
#### Ibix
Right, but PEMDAS/BEDMAS is a standardized set of rules for math. There are no different "countries" in math
You yourself cited a specific example of a journal using different rules from the ones you want to use, but you are here trying to tell me no-one uses different rules. Your position seems somewhat self contradictory.
what is the reason for Physicists choosing multiplication to take priority over division with a slash
Saves on typing, I imagine. With your rules, there's no way to write $1/(2x)$ without using brackets. With the journal's rules you can write $1/2x$ and with either set of rules you can write $x/2$ if you meant $(1/2)x$.
when all technology and calculators they use are programed to strictly follow PEMDAS, this decision seems weird and pointless to me.
So your value judgement differs from that of the editors of the journal. But note that your only argument is "follow the majority", not anything technical.
#### Mark44
Mentor
...but even when all technology and calculators they use are programed to strictly follow PEMDAS...
This is not true of calculators in general. In the same wiki article that you cited (https://en.wikipedia.org/wiki/Order_of_operations#Calculators), it mentions several calculators, including Texas Instrument models and Casio models, that don't follow this convention. It also mentions that the Windows calculator evaluates $1 + 2 \times 3$ in two different ways, depending on whether the calculator is in Standard mode or Scientific mode.
Also, the same wiki article talks about an ambiguous representation of $1/2\sqrt N$ on a page in the Feynman lectures. On the page cited, this expression is written as
$$\frac 1 {2\sqrt N}$$
In a later paragraph it is written as $1/2\sqrt N$, so there is some context for interpreting the latter expression to mean what is written the first time.
As already mentioned, a major reason for writing this as $1/2\sqrt N$ is as a space-saving feature in typesetting.
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#### russ_watters
Mentor
Windows calculator evaluates $1 + 2 \times 3$ in two different ways, depending on whether the calculator is in Standard mode or Scientific mode.
This threw me just the other day!
And can I please get a 1/x in scientific mode?
#### Ibix
And can I please get a 1/x in scientific mode?
You just type 1/1x.
This is not true of calculators in general.
Nor programming languages. I've written a program in postscript, which is stack based and effectively uses reverse Polish notation. Like Yoda maths doing it is. Perfectly understandable once you spot how it works, but a completely distinct notation.
#### fresh_42
Mentor
2018 Award
There is a simple rule which always leads to success, whether in programming or in writing formulas:
If in doubt, use a bracket!
#### Mark44
Mentor
If in doubt, use a bracket!
And even better, use a pair of them...
#### John3509
You yourself cited a specific example of a journal using different rules from the ones you want to use, but you are here trying to tell me no-one uses different rules. Your position seems somewhat self contradictory.
Well then if many people break the rules in a consistent way I suppose you can call that "using different rules"
I think that is how dielects work too.
#### John3509
This is not true of calculators in general. In the same wiki article that you cited (https://en.wikipedia.org/wiki/Order_of_operations#Calculators), it mentions several calculators, including Texas Instrument models and Casio models, that don't follow this convention. It also mentions that the Windows calculator evaluates $1 + 2 \times 3$ in two different ways, depending on whether the calculator is in Standard mode or Scientific mode.
that's interesting, I did not notice that. I can understand why / would be interpreted as everything to the right is denominator, it saves the need to type parentheses, but why this?
#### PeroK
Homework Helper
Gold Member
2018 Award
Well then if many people break the rules in a consistent way I suppose you can call that "using different rules"
I think that is how dielects work too.
Well, you are breaking the rules by spelling "dialect" your own way.
#### Klystron
Gold Member
Well then if many people break the rules in a consistent way I suppose you can call that "using different rules"
I think that is how dielects work too.
Well, you are breaking the rules by spelling "dialect" your own way.
Perhaps, but I read it as dialectic, more consistent with content but still a reach.
[Edit: It was dialect. See below.]
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#### John3509
Well, you are breaking the rules by spelling "dialect" your own way.
Not sure it that's supposed to mock me or not but I did not know I was spelling it wrong,
But if some English professor or language expert published everything with, and required all papers submitted in class as well to have dialect misspelled it would be expected to raise questions among the students.
"PEMDAS and Physicists"
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colorNumeric
0th
Percentile
Color mapping
Conveniently maps data values (numeric or factor/character) to colors according to a given palette, which can be provided in a variety of formats.
Usage
colorNumeric(palette, domain, na.color = "#808080", alpha = FALSE)colorBin(palette, domain, bins = 7, pretty = TRUE, na.color = "#808080", alpha = FALSE)colorQuantile(palette, domain, n = 4, probs = seq(0, 1, length.out = n + 1),
na.color = "#808080", alpha = FALSE)colorFactor(palette, domain, levels = NULL, ordered = FALSE, na.color = "#808080",
alpha = FALSE)
Arguments
palette
The colors or color function that values will be mapped to
domain
The possible values that can be mapped.
For colorNumeric and colorBin, this can be a simple numeric range (e.g. c(0, 100)); colorQuantile needs representative numeric data; and colorFactor
na.color The color to return for NA values. Note that na.color=NA is valid. alpha Whether alpha channels should be respected or ignored. If TRUE then colors without explicit alpha information will be treated as fully opaque. bins Either a numeric vector of two or more unique cut points or a single number (greater than or equal to 2) giving the number of intervals into which the domain values are to be cut. pretty Whether to use the function pretty() to generate the bins when the argument bins is a single number. When pretty = TRUE, the actual number of bins may not be the number of bins you n Number of equal-size quantiles desired. For more precise control, use the probs argument instead. probs See quantile. If provided, the n argument is ignored. levels An alternate way of specifying levels; if specified, domain is ignored ordered If TRUE and domain needs to be coerced to a factor, treat it as already in the correct order
Details colorNumeric is a simple linear mapping from continuous numeric data to an interpolated palette.colorBin also maps continuous numeric data, but performs binning based on value (see the cut function).colorQuantile similarly bins numeric data, but via the quantile function.colorFactor maps factors to colors. If the palette is discrete and has a different number of colors than the number of factors, interpolation is used.The palette argument can be any of the following: A character vector of RGB or named colors. Examples:palette(),c("#000000", "#0000FF", "#FFFFFF"),topo.colors(10) The name of an RColorBrewer palette, e.g. "BuPu" or "Greens". A function that receives a single value between 0 and 1 and returns a color. Examples: colorRamp(c("#000000", "#FFFFFF"), interpolate="spline"). Value A function that takes a single parameter x; when called with a vector of numbers (except for colorFactor, which expects factors/characters), #RRGGBB color strings are returned (unless alpha=TRUE in which case #RRGGBBAA may also be possible). Aliases colorBin colorFactor colorNumeric colorQuantile Examples pal = colorBin("Greens", domain = 0:100) pal(runif(10, 60, 100)) # Exponential distribution, mapped continuously previewColors(colorNumeric("Blues", domain = NULL), sort(rexp(16))) # Exponential distribution, mapped by interval previewColors(colorBin("Blues", domain = NULL, bins = 4), sort(rexp(16))) # Exponential distribution, mapped by quantile previewColors(colorQuantile("Blues", domain = NULL), sort(rexp(16))) # Categorical data; by default, the values being colored span the gamut... previewColors(colorFactor("RdYlBu", domain = NULL), LETTERS[1:5]) # ...unless the data is a factor, without droplevels... previewColors(colorFactor("RdYlBu", domain = NULL), factor(LETTERS[1:5], levels = LETTERS)) # ...or the domain is stated explicitly. previewColors(colorFactor("RdYlBu", levels = LETTERS), LETTERS[1:5]) Documentation reproduced from package leaflet, version 1.0.0, License: GPL-3 | file LICENSE Community examples Looks like there are no examples yet. Post a new example:
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# SBI Clerk 2018: Reasoning Ability Quiz – 25
Hello and welcome to exampundit. Here is a set of Reasoning Quiz on Puzzle problems for SBI Clerk 2018 Prelims.
Directions (Q. 1-5) : Study the following information carefully and answer the questions given below:
M, N, O, P, Q, R, S and T are eight friends sitting around a circular table. Six of them are facing the centre. All of them have a different hobbies viz— Dancing, Singing, Cooking, Novel Writing, Watching Netflix, Blogging, Playing Carrom and Reading Novel, but not necessarily in the same order. O is an immediate neighbour of P. The person who is opposite N is facing outside and N is the immediate neighbour of the person his hobby is Novel Writing. T sits third to the right of Q and his hobby is Playing Carrom. M and T are not the neighbours of the person whom hobby is Blogging. R’s hobby is not cooking or Novel Writing. The person whom hobby is Reading Novel is sitting opposite Q. The one whom hobby is Novel Writing, sits immediate right of N. P and S are immediate neighbours of N, whom hobby is Watching Netflix. S sits third to the right of M and his hobby is Dancing. P sits third to the left of M.
1. Whom hobby is Blogging ?
1) M
2) Q
3) O
4) N
5) None of these
1. How many persons sit between O and the person his hobby is Singing, when counted in clockwise direction, starting from O ?
1) None
2) One
3) Two
4) Three
5) None of these
1. The person who is sitting second to the right of N, what is his hobby ?
2) Singing
3) Cooking
4) Dancing
5) None of these
1. What is the position of Q with respect to P ?
1) Third to the left
2) Second to the right
3) Immediate right
4) Third to the right
5) None of these
1. Which of the following statements is true ?
1) Q’s hobby is Singing
2) P sits immediate left of N.
3) N sits second to the left of Q.
4) T’s hobby is Cooking.
5) None of these
Directions (Q. 6-10) : Study the following information carefully and answer the questions given below: Eight candidates B, H, M, O, Q, R, S and T have applied for banking exam and interview for different international banks viz— AXIS, HDFC, STANDARD CHARTERED, Union Bank, Punjab Bank, Laxmi Vilas Bank, Bank of Baroda and Kotak Mahindra Bank but not necessarily in same order. There are five male and three females in the group. Each male and each female has applied from his/her city viz— Kolkata, Delhi, Pune, Mumbai, Patna, Agartala, Bengaluru and Gurugram. No male has applied from Agartala and Mumbai. Q has applied for Laxmi Vilas Bank from Gurugram. The one who has applied for Bank of Baroda is neither from Pune nor from Mumbai. H has applied for Punjab Bank. His sister M has applied for STANDARD CHARTERED Bank from Kolkata. The one who belongs to Delhi has applied for AXIS Bank. The one who has applied from Pune is not a female. O has applied for Union Bank and her friend has applied for AXIS. R is from Bengaluru and has not applied either for Bank of Baroda or for Kotak Mahindra Bank. The one who has applied for Kotak Mahindra Bank has applied neither from Patna nor from Pune. T has applied from Mumbai. B does not apply from Patna.
1. Which of the following groups is a group of female applicants ?
1) H, O, M
2) O, R, T
3) M, O, T
4) Q, R, S
5) None of these
1. Who among the following has applied from Agartala ?
1) Q
2) O
3) T
4) M
5) None of these
1. Who among the following has applied for AXIS ?
1) B
2) T
3) O
4) R
5) None of these
1. H belongs to which of the following cities ?
1) Delhi
2) Agartala
3) Pune
4) Bengaluru
5) None of these
1. Four of the five are alike in a certain way and hence from a group. Which is the one that does not belong to that group?
1) Q
2) S
3) B
4) M
5) R
Show Solution
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# A used cars salesman receives an annual bonus if he meets a certain
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07 Sep 2016, 04:30
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A used cars salesman receives an annual bonus if he meets a certain quota. This year, the salesman has so far sold 1/5 of last year's quota. If this year's quota is reduced by 25 percent from last year, the quantity that he still needs to sell this year in order to receive the annual bonus is what fraction of the preceding year's car quota?
A. 4/15
B. 4/20
C. 3/4
D. 11/20
E. 9/20
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A used cars salesman receives an annual bonus if he meets a certain [#permalink]
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07 Sep 2016, 04:45
4
A used cars salesman receives an annual bonus if he meets a certain quota. This year, the salesman has so far sold 1/5 of last year's quota. If this year's quota is reduced by 25 percent from last year, the quantity that he still needs to sell this year in order to receive the annual bonus is what fraction of the preceding year's car quota?
Let Last year Quota be = 100x
This year Quota = 100x - 25%of100x = 75x
Salesman has so far sold = 1/5(100x) = 20x
Quantity still need to be sold = 75x - 20x = 55x
Question :: What fraction of 100x is 55x = 55x/100x = 11/20.
Ans:: D
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Re: A used cars salesman receives an annual bonus if he meets a certain [#permalink]
### Show Tags
07 Sep 2016, 18:56
1
1
We have 1/5 and 25% which is 1/4. Let's pick some smart numbers.
Assume last year's quota to be 20.
This year's quota = 25% reduction from 20 = 15
Sold so far = 1/5 * 20 = 4
Remaining = 15 - 4 = 11
Required = 11/20
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Re: A used cars salesman receives an annual bonus if he meets a certain [#permalink]
### Show Tags
23 Jun 2017, 18:16
Bunuel wrote:
A used cars salesman receives an annual bonus if he meets a certain quota. This year, the salesman has so far sold 1/5 of last year's quota. If this year's quota is reduced by 25 percent from last year, the quantity that he still needs to sell this year in order to receive the annual bonus is what fraction of the preceding year's car quota?
A. 4/15
B. 4/20
C. 3/4
D. 11/20
E. 9/20
All this is really asking is that if we have say 500 cars for last years quota, then we have 100 cars sold so far and that 500 has been fortunately reduced to 350 (500 x .75) - so 275 cars need to be sold and those 275 cars are what percent of 500 in terms of a fraction?
275/500= 11/20
Thus
"D"
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Re: A used cars salesman receives an annual bonus if he meets a certain [#permalink]
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03 Jan 2018, 02:52
1
1/5 * 100% = 20% sold
Target reduced by 25% = 100% - 25% = 75%
He still has to sell 75% - 20% = 55%
55/100 = 11/20
Option D
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A used cars salesman receives an annual bonus if he meets a certain [#permalink]
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05 Jan 2018, 15:17
Bunuel wrote:
A used cars salesman receives an annual bonus if he meets a certain quota. This year, the salesman has so far sold 1/5 of last year's quota. If this year's quota is reduced by 25 percent from last year, the quantity that he still needs to sell this year in order to receive the annual bonus is what fraction of the preceding year's car quota?
A. 4/15
B. 4/20
C. 3/4
D. 11/20
E. 9/20
Last year's quota: 100
He has sold $$\frac{1}{5}$$ of that: 20
This year's quota down by 25% (100*.75): 75
Must still sell: (75-20) = 55
55 as fraction of last year's quota? $$\frac{55}{100}=\frac{11}{20}$$
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## Solution
The focal length of the concave mirror, f = 10 cm. So, radius of curvature, r = 2f = 20 cm. Since, the object is placed at a distance of 16 cm from the pole, hence it is situated in between the centre of curvature and the focus. Hence, the image formed will be magnified, real and inverted.
Chapter: Light
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# Book Review: Statistics Done Wrong
Many times we data scientists, not being statisticians in the strictest sense, have concern that we may commit some kind of statistical faux pas. Fear no more! With the release of a probing new book “Statistics Done Wrong” by Alex Reinhart, we now have a curious road map for avoiding statistical fallacies. As a Ph.D. student and statistics instructor at Carnegie Mellon University, Reinhart shows how scientific progress depends on good research, and good research needs good statistics. But statistical analysis is tricky to get right, even for the best data scientists. You’ll be surprised how many practicing data scientists are doing it wrong.
Although written for a broad audience of scientific researchers, I found the book compelling for me personally as someone working daily in data science. Many of the same principles I use regularly, such as linear regression, overfitting, confounding variables, cross validation, feature selection, p-values, confidence intervals, etc., are familiar concepts covered in the book.
The best part of the book is all the examples of statistical blunders in modern science. Reinhart provides ample cases of embarrassing errors and omissions in recent research. You’ll learn about the misconceptions and scientific politics that allow these mistakes to happen, and lead you to a path of reform in the way you do statistics.
Here is a list of chapters:
Chapter 1 – An Introduction to Statistical Signficance
Chapter 2 – Statistical Power and Underpowered Statistics
Chapter 3 – Pseudoreplication: Choose Your Data Wisely
Chapter 4 – The P Value and The Base Rate Fallacy
Chapter 5 – Bad Judges of Signficance
Chapter 6 – Double-Dipping in the Data
Chapter 7 – Continuity Errors
Chapter 8 – Model Abuse
Chapter 9 – Researcher Freedom: Good Vibrations
Chapter 10 – Everybody Makes Mistakes
Chapter 11 – Hiding the Data
Chapter 12 – What Can Be Done?
I think “Statistics Done Wrong” is an important addition to any data scientists library. In addition, the pithy writing style will keep your interest and fuel your creativity for future projects. Highly recommended.
Daniel D. Gutierrez – Managing Editor
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Te Kete Ipurangi
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Level Six > Geometry and Measurement
Hinea's Watch's Hands
Achievement Objectives:
Specific Learning Outcomes:
find an angle greater than 360 degrees that relates to a given position
solve a linear equation
devise and use problem solving strategies to explore situations mathematically (be systematic, draw a diagram, guess and check, use algebra)
Description of mathematics:
This problem involves taking a problem about time, looking at it geometrically, and solving it algebraically. The problem is about when hands on a watch coincide. This is easily seen as a geometry problem as a diagram can be drawn to show the situation. The difficulty is to extract the algebra. This requires knowing about the relative speed of the hands of a watch and the significance of angle.
A similar problem can be found under Hinea’s Other Watch.
Required Resource Materials:
Copymaster of the problem (English).
Copymaster of the problem (Māori).
Activity:
The Problem
Hinea took a second look at her watch! There was only one hand! Then she realised that one was on top of the other one. This made her wonder how many times a day the hour hand and the minute hand on her watch were in the same position.
How many would you think?
Teaching sequence
1. Focus the class on the clock on the wall or their own watch and pose the problem.
2. Encourage the students to estimate the number of possible answers, using questions that require them to justify their guesses.
How many answers will there be altogether? Why?
What position will the hands be in?
3. While the students are working on the problem ask questions that enable them to clarify the variables involved in the time situation.
What starting strategy did you use?
What changes in this problem?
What variables do you need to consider.
4. Questions that ensure students have knowledge of angles greater than 360 degrees may be needed:
About when are the angles the same?
How far has the minute hand moved?
Through what angle is this?
How can these angles be the same?
What angle is at the same position as 390 degrees? 450 degrees?
5. Encourage the students to write down any connections they have found in words and link this to a possible algebraic form.
6. Connections need to be made between the angle and the time it represents.
How does the angle relate to the time that has passed?
7. Get the students to share their approaches and explain the links between the speed, the angles and the time.
8. Compare the number of answers with the predicted total from the starting discussion.
Why are there only 22 answers and not 24?
Solution
One way to do this is to turn the hands of a real clock around and count the number of times the hands overlap. Then ask how you can be sure that you have the right answer.
Hinea’s (and everyone else’s for that matter) minute hand, goes through 360° in 1 hour. Her hour hand goes through 360° in 12 hours, or 30° an hour. So the minute hand moves 12 times as quickly as her hour hand. While the hour hand is moving through an angle , the minute hand is moving through an angle of 12.
On the other hand, because they are on top of each other, 12 - 360° = . So 11 = 360° or = 360/11 = 32.72. This represents (32.72 / 360) x 60 minutes. This is approximately 5 minutes and 27.3 seconds.
The hands are therefore on top of each other at 0:00:00; 1:05:27; 2:10:55; 3:16:22; 4:21:49; 5:27:16; 6:32:44; 7:38:11; 8:43:38; 9:49:05; 10:54:33; and every 12 hours afterwards.
This therefore happens 22 times in a day.
AttachmentSize
HineasWatch.pdf40.3 KB
HineasWatchMaori.pdf46.77 KB
Hinea's Other Watch
Solve problems using the relationships between angles in a circle.
Devise and use problem solving strategies to explore situations mathematically (be systematic, guess and check, think, use algebra).
Working Backwards
Use their mathematical knowledge to invent problems
Devise and use problem solving strategiesto explore situations mathematically.
Moana's Watch
Convert seconds to minutes.
Subtract minutes and seconds using a 24-hour clock
Devise and use problem solving strategies to explore situations mathematically (be systematic).
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1. ## Expectation
The time interval Y between successive feedings of a certain type of insect has an exponential distribution with mean proportional to a positive measured characteristic x of the insect. Suppose we observe n independent insects, observing time intervals Y1, . . . , Yn with associated characteristics x1, . . . , xn.
Let E(Y)=a.
I have need to calculate $E(Y{_1}/X{_1})$. I supposed they were independent and then using the information i have E(Y)=cX so I get a/(c/a)=a^2/c. I think however that it is wrong.
2. ## Re: Expectation
It doesnt sound like Y and X are indep at all
Furthermore that would be E(YlX) not E(Y).
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https://grandunifiedtheory.org.il/grav/grav6.htm
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Dr. Chaim H. Tejman's Grand Unified Theory: Wave Theory Introduction Summary Book Wave Formation Photons Time Atoms Life Cancer Fundamental Force Gender/Why Sex? Sexual Reproduction Schrodinger & Heisenberg Creation Supernova Speed of Light Cloud Formations Global Warming Thermodynamics Backward Time Quantum Mechanics Compton Effect Equations Predictions Mysteries
Gravitation — a Pushing Force
An active, pushing unidirectional force
(Continued — Page 6)
In his famous experiment involving magnetic bars and wires, Michael Faraday clearly proved that magnetic matter and energetic matter corroborate and that both entities advance in the same exclusive direction.
Changing the direction of the magnet alters the flow of electricity. Matter moves in but one direction and its functionality is strictly dependent on the interaction between its two loops. Faraday’s experiment, which assumes an even greater significance according to wave theory, proves that only one form of matter exists and that its flow produces both an electric and magnetic form of this same energetic matter. Moreover, Faraday showed that only a change in the configuration of energetic matter can alter its behavior. However, the electric and magnetic flows are always connected perpendicularly regardless of the direction of either current.
Field lines connecting positively and negatively charged points. There are only pushing forces; consequently, the positive charges are dispersed and the negative charges contract.
Nature thus points to but one unidirectional force. Although gravitation is widely perceived to be a pulling force, there is absolutely no evidence backing this claim. On the other hand, many observations have verified the existence of a pushing force that results from the advancing, swirling movement of energetic matter. The configuration of the energetic path in a magnetic swirl enables the magnetic swirl to push the energetic matter into the swirl, and the latter then pushes in everything in its vicinity by dint of its spinning and swirling movement. Moreover, the spinning and swirling regulates the pace in which the energy is propagated so that the energy does not jettison straight out into space. This process, then, maintains the energy and preserves the structure of the wave formation. It constitutes the only feasible framework (as exhibited in Faraday’s experiments) for the behavior of energetic matter.
Click to enlarge; NASA Propagation of energetic matter in a closed formation.
This evidence shows that, since the Big Bang, a single form of energetic matter has expanded in a pushing, spinning and swirling mode and creates vortices.
Next Page 8 9 10 11 12 13 Printer-Friendly Version
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http://gmatclub.com/forum/a-woman-has-11-close-friends-find-the-number-of-ways-she-17181-20.html?sort_by_oldest=true
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# A woman has 11 close friends. Find the number of ways she
Author Message
Intern
Joined: 30 Apr 2005
Posts: 11
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### Show Tags
14 Jul 2005, 01:44
In my opinion answer is 504
We have 10c2 or 252 options if we ignore 1 out of the two fighting friends, and 252 ways for the other. so it makes 504. Though i m not that good in probabilty, but this is the solution i can think of. :~
_________________
i believe that all the problem solving calculations in Quantitative section have simple calculations. No Long Mahcine requiring calculations r required :D
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# Can you try to solve this question ( I am stuck )
2 views (last 30 days)
A recursive implementation of reverberation is given by (2.103) which is given below y[n] = x[n] + ay[n − D], where D = τFs is the delay in sampling interval given the delay τ in seconds and sampling rate Fs and a is an attenuation factor. To generate digital reverberation we will use the sound file handel which is recorded at Fs = 8192 samples per second. (See Problem 6 for using this file.)
(a) For τ = 50 ms and a = 0.7, obtain a difference equation for the digital reverberation and process the sound in handel. Comment on its audio quality
Hello i have this question and i am stuck in a part where i can't form the equation i am looking for someone to help me out
Fs=8192;
tao=milliseconds(50);
alpha=0.7;
filename = 'handel.wav';
audiowrite(filename,y,Fs);
d=tao*Fs;
x=transpose(y)
---then??
sound(y,Fs)
Walter Roberson on 26 Mar 2019
You do not appear to have constructed the difference equation ?
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# Is the Hurwitz zeta function in $l^1(\mathbb{N})$?
Let $$Re(s) >1, n = 1,2,\dots$$ and let
$$\zeta(s,n) = \sum_{m=0}^{\infty}\frac{1}{{(n+m)}^{s}}$$ be the Hurwitz zeta function. I want to prove that there is some $$p \geq 1$$ such that for all $$Re(s) >1$$ we have that $$\sum_{n=1}^\infty\lvert \zeta(s,n) \rvert^p < \infty.$$ I have the intuition that $$p = 2$$ works but I am unable to prove this. Is there any literature regarding this question?
I also think this can be proven using formula $$\zeta(s,q) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.$$
• For $a >0$ and $\Re(s)> 1$ and by analytic continuation for $\Re(s) >0$ $\zeta(s,a) - \frac{a^{1-s}}{s-1}= \sum_{m=0}^\infty (a+m)^{-s}-\int_m^{m+1} (a+x)^{-s}dx$ $= \sum_{m=0}^\infty \int_m^{m+1} \int_m^x s(a+t)^{-s-1}dtdx = \sum_{m=0}^\infty O(\int_m^{m+1} |s| (a+x)^{-\Re(s)-1}dx)=O(|s| \frac{ a^{-\Re(s)}}{\Re(s)})$ Sep 4, 2019 at 3:30
Note that for $$k \ge 1, a>1, \frac{1}{k^a}+\frac{1}{(k+1)^a}+...> \int_{k}^{\infty}\frac{1}{x^a}dx=\frac{1}{(a-1)k^{a-1}}$$, so $$\zeta(a,k) >\frac{1}{(a-1)k^{a-1}}$$. In particular unless you impose a condition $$\Re s \ge a_0 >1$$ there is no such $$p$$, while for $$\Re s \ge a_0 >1$$, you need to take $$p > \frac{1}{a_0-1}$$
(which works since for $$k \ge 2, a>1, \frac{1}{k^a}+\frac{1}{(k+1)^a}+...< \int_{k-1}^{\infty}\frac{1}{x^a}dx=\frac{1}{(a-1)(k-1)^{a-1}}$$, hence for all $$k \ge 2$$, we have $$|\zeta(s,k)| \le \zeta(\Re{s},k) \le \frac{1}{(a_0-1)(k-1)^{a_0-1}}$$ , so $$p(a_0-1) >1$$ insures convergence for the required series as the first term doesn't matter for that)
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https://socratic.org/questions/what-is-the-graph-of-f-x-x-2-1
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# What is the graph of f(x) = -x^2?
This is a parabola that opens downwards. It has a vertex at the origin. And it is symmetrical about the vertical line $x = 0$. It has a maximum value of $0$.
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http://www.physicsforums.com/showthread.php?t=372116
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# Sequence of real numbers | Proof of convergence
by kingwinner
Tags: convergence, numbers, proof, real, sequence
P: 1,633 Since you seem to have spent quite some time on the problem i will try to give you some hints, i hope i don't get another warning from pf moderators for offering too much help(solving >90% of the problem for the op) :( This might not be the nicest proof in the world, but i think it works. As you have figured out the main problem is when 0a_0+a_1[/tex] If we continue in this fashion, after n-2 steps we would get something like: $$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_n}>a_0+a_1+...+a_n>n*min\{a_0,a_1,...,a_ n\}=n*a$$ So, now you see that if we let n>N=1/a we get our result. where a=min{a_o,...,a_n} cheers!
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https://citizenmaths.com/frequency/degrees-per-millisecond-to-yoctohertz
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# Degree per Millisecond to Yoctohertz Conversions
From
Degree per Millisecond
• Action per Minute
• Attohertz
• Centihertz
• Cycle per Day
• Cycle per Hour
• Cycle per Microsecond
• Cycle per Millisecond
• Cycle per Minute
• Cycle per Month
• Cycle per Nanosecond
• Cycle per Picosecond
• Cycle per Second
• Cycle per Year
• Decahertz
• Decihertz
• Degree per Hour
• Degree per Millisecond
• Degree per Minute
• Degree per Second
• Exahertz
• Femtohertz
• Frame per Second
• Fresnel
• Gigahertz
• Hectohertz
• Hertz
• Kilohertz
• Megahertz
• Microhertz
• Millihertz
• Nanohertz
• Petahertz
• Picohertz
• Revolution per Minute
• Terahertz
• Yoctohertz
• Yottahertz
• Zeptohertz
• Zettahertz
To
Yoctohertz
• Action per Minute
• Attohertz
• Centihertz
• Cycle per Day
• Cycle per Hour
• Cycle per Microsecond
• Cycle per Millisecond
• Cycle per Minute
• Cycle per Month
• Cycle per Nanosecond
• Cycle per Picosecond
• Cycle per Second
• Cycle per Year
• Decahertz
• Decihertz
• Degree per Hour
• Degree per Millisecond
• Degree per Minute
• Degree per Second
• Exahertz
• Femtohertz
• Frame per Second
• Fresnel
• Gigahertz
• Hectohertz
• Hertz
• Kilohertz
• Megahertz
• Microhertz
• Millihertz
• Nanohertz
• Petahertz
• Picohertz
• Revolution per Minute
• Terahertz
• Yoctohertz
• Yottahertz
• Zeptohertz
• Zettahertz
Formula 7,468 deg/ms = 7468 x 2777777777777779878658048 yHz = 2.1e+28 yHz
## How To Convert From Degree per Millisecond to Yoctohertz
1 Degree per Millisecond is equivalent to 2.8e+24 Yoctohertz:
1 deg/ms = 2.8e+24 yHz
For example, if the Degree per Millisecond number is (3.6), then its equivalent Yoctohertz number would be (1.0e+25).
Formula:
3.6 deg/ms = 3.6 x 2777777777777779878658048 yHz = 1.0e+25 yHz
## Degree per Millisecond to Yoctohertz conversion table
Degree per Millisecond (deg/ms) Yoctohertz (yHz)
0.1 deg/ms 2.8e+23 yHz
0.2 deg/ms 5.6e+23 yHz
0.3 deg/ms 8.3e+23 yHz
0.4 deg/ms 1.1e+24 yHz
0.5 deg/ms 1.4e+24 yHz
0.6 deg/ms 1.7e+24 yHz
0.7 deg/ms 1.9e+24 yHz
0.8 deg/ms 2.2e+24 yHz
0.9 deg/ms 2.5e+24 yHz
1 deg/ms 2.8e+24 yHz
1.1 deg/ms 3.1e+24 yHz
1.2 deg/ms 3.3e+24 yHz
1.3 deg/ms 3.6e+24 yHz
1.4 deg/ms 3.9e+24 yHz
1.5 deg/ms 4.2e+24 yHz
1.6 deg/ms 4.4e+24 yHz
1.7 deg/ms 4.7e+24 yHz
1.8 deg/ms 5.0e+24 yHz
1.9 deg/ms 5.3e+24 yHz
2 deg/ms 5.6e+24 yHz
2.1 deg/ms 5.8e+24 yHz
2.2 deg/ms 6.1e+24 yHz
2.3 deg/ms 6.4e+24 yHz
2.4 deg/ms 6.7e+24 yHz
2.5 deg/ms 6.9e+24 yHz
2.6 deg/ms 7.2e+24 yHz
2.7 deg/ms 7.5e+24 yHz
2.8 deg/ms 7.8e+24 yHz
2.9 deg/ms 8.1e+24 yHz
3 deg/ms 8.3e+24 yHz
3.1 deg/ms 8.6e+24 yHz
3.2 deg/ms 8.9e+24 yHz
3.3 deg/ms 9.2e+24 yHz
3.4 deg/ms 9.4e+24 yHz
3.5 deg/ms 9.7e+24 yHz
3.6 deg/ms 1.0e+25 yHz
3.7 deg/ms 1.0e+25 yHz
3.8 deg/ms 1.1e+25 yHz
3.9 deg/ms 1.1e+25 yHz
4 deg/ms 1.1e+25 yHz
4.1 deg/ms 1.1e+25 yHz
4.2 deg/ms 1.2e+25 yHz
4.3 deg/ms 1.2e+25 yHz
4.4 deg/ms 1.2e+25 yHz
4.5 deg/ms 1.3e+25 yHz
4.6 deg/ms 1.3e+25 yHz
4.7 deg/ms 1.3e+25 yHz
4.8 deg/ms 1.3e+25 yHz
4.9 deg/ms 1.4e+25 yHz
5 deg/ms 1.4e+25 yHz
5.1 deg/ms 1.4e+25 yHz
5.2 deg/ms 1.4e+25 yHz
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #5001 2012-10-03 23:17:02
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
No, just be pleasant in conversation. Then when you ask her she will not want to lie and blurt everything out.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5002 2012-10-03 23:21:57
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
I think I was pleasant in conversation when walking with her...
The problem is getting her to confess the truth (about how she feels toward me) -- it took H a long time to finally say.
## #5003 2012-10-03 23:24:13
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
How did you get H to do that?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
zetafunc.
Guest
## #5005 2012-10-03 23:37:15
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
That wasn't very tough. Why not try the same approach?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5006 2012-10-03 23:38:50
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
I will end up with egg on my face if she doesn't.
## #5007 2012-10-03 23:51:55
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
You do not ask until you are sure of her reply. Best done face to face.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5008 2012-10-04 00:15:45
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Yes, I want to be more sure first. With H I was almost certain. With PJ, there is not a lot to work with but I can find out more.
## #5009 2012-10-04 00:26:58
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
If you get the answer that it was not a coincidence then you will not need to ask.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5010 2012-10-04 00:28:59
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Yes, I know. I think she would be reluctant to say that though since that does seem like the kind of thing a stalker would say.
## #5011 2012-10-04 00:38:03
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
And if you find out she is interested what do you do?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5012 2012-10-04 00:40:13
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Date her and find out more. I prefer not to have a 1652-email conversation about it like I did with H (that is the exact number).
## #5013 2012-10-04 00:43:11
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
It just might happen.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5014 2012-10-04 00:45:39
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
I have a feeling that won't happen. PJ is more social than H.
## #5015 2012-10-04 00:46:32
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
No, I meant dating her.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5016 2012-10-04 00:49:26
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Who knows, it is difficult to say at this point. I need some signs to work with. H opened up to me emotionally almost immediately... if PJ does that then the probability of a date increases dramatically, I would say.
## #5017 2012-10-04 01:04:16
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
Same thing might happen with PJ. She might be a little less likely than H but you should be able to detect that.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5018 2012-10-04 01:07:03
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Well, I have *something*, but it's still not enough for me. PJ sent me fairly lengthy e-mails which went in detail about her holidays, what she was doing for her work experience, etc., and she also did not reply when I said I wanted to tell IY something 'personal'. She also started shying away from me a bit (e.g. when knowing I was close by, she wouldn't make an attempt to talk to me) in the month of September since I did not reply to her last e-mail, which H did too. So there is something, but I need more than this.
## #5019 2012-10-04 01:17:17
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
Wait for the answer from the email you sent. You will know what to do when you get some answer.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5020 2012-10-04 01:18:56
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
Okay. I am also heading to school now (to talk to a few teachers about my uni application) so I may see PJ there.
## #5021 2012-10-04 01:24:03
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
Okay, good luck. See you tomorrow.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
## #5022 2012-10-04 05:42:41
zetafunc.
Guest
### Re: Linear Interpolation FP1 Formula
"Hi,
Nah, I've always lived in the same house in _____.
PJ"
What to do from here...?
## #5023 2012-10-04 05:51:42
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
Is that house close to your route to school?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
zetafunc.
Guest
Not really, no.
## #5025 2012-10-04 06:02:29
bobbym
From: Bumpkinland
Registered: 2009-04-12
Posts: 84,363
### Re: Linear Interpolation FP1 Formula
When you saw her first were you close to school?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Offline
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# Excel Char Function
Character Encoding
The character encoding used by your computer depends on the operating system used.
Windows uses the ANSI character set, whereas Macintosh uses the Macintosh character set.
Therefore, the characters returned by the Excel Char Function for specific number codes may be different on different operating environments.
## Basic Description
The Excel Char function returns the character relating to a supplied character set number (from 1 to 255).
Note that the character set may vary across different operating systems and so, for a specific number, the Char function may return different results on different computers.
The syntax of the Char function is:
CHAR( number )
Where the number argument is an integer from 1 to 255.
## Char Function Examples
### Example 1
The following spreadsheet shows the Excel Char function, used to return the characters associated with four different character set numbers.
Note that these results are from the Ansi character set (used on the Windows operating system).
Formulas:
AB
1 =CHAR( 65 )
2 =CHAR( 97 )
363=CHAR( A3 )
451=CHAR( A4 )
Results:
AB
1 A
2 a
363?
4513
### Example 2
One use of the CHAR function is when inserting line breaks into text. This is shown in the example below (Note that, in the Ansi character set, the line break is associated to the character set number 10):
Formula:
A
1="This line contains a" & CHAR( 10 ) & "line break"
Result:
A
1This line contains a
line break
Note that, in the example above, in order to display the result with the line break, you will need to ensure that the cell text wrapping is enabled.
#### Enabling Text Wrapping
The easiest way to enable text wrapping in a cell is to use the Wrap Text button, which is located in the 'Alignment' group of the Home tab of the Excel ribbon (see below).
Alternatively, if you have an older version of Excel, that does not have the ribbon, you can wrap text in a cell as follows:
• Right click on the cell and select Format Cells ...
• Select the Alignment tab of the dialog box that pops up;
• Check the box next to the Wrap text option and click OK.
For further information and examples of the Excel Char Function, see the Microsoft Office website.
## Char Function Error
If you get an error from the Excel Char function, this is likely to be the #VALUE! error:
Common Error
#VALUE! - Occurs if either:The supplied number argument is not recognised as a numeric value;The supplied number argument is < 1 or > 255.
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### Home > GB8I > Chapter 11 Unit 12 > Lesson INT1: 11.2.2 > Problem11-62
11-62.
A share of ABC stock was worth $60$ in 2005 and only worth $45$ in 2010.
1. Find the multiplier and the percent decrease.
Solve for $m: 60 · m^{5} = 45$
Use the general form of an exponential equation: $y = a · m^{x}$, where m is the multiplier found in part (a), and $a$ is the initial value.
$25.28$
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Boundedness of an Oscillating Integral
Let $g(x):\mathbb{R}_{\geq0}\rightarrow\mathbb{R}$ be real analytic s.t. $g(0)\neq 0$ and $g(x)=O(x^{-2})$ as $x\rightarrow\infty$. I think the following integral should be bounded as $\lambda\rightarrow0$, but would like to have a proof: $$J(\lambda) = \int_0^{\infty}dy \sin\left(y\right)\log(y)g(\lambda y)$$
I know how to deal with similar asymptotic expansions when there is no log term (integration by parts, method of steepest descent), but these seem to fail here. Can somebody advise how to approach this?
• Do you have similar decay information on the derivatives of $g$? Then I think it's OK. Apr 28, 2013 at 19:08
• Yes, you may assume $g^{(n)}(x)=O(x^{-2-n})$. Apr 29, 2013 at 17:17
• On a second thought (after quite a bit more time of thinking) I'm not at all sure whether your $J(\lambda)$ is bounded in $\lambda$. I keep getting an upper bound of the type $\log(1/\lambda)$ though I cannot prove yet that this is also a lower bound. Apr 30, 2013 at 12:19
Yes it is bounded. Assume throughout that $\lambda<1$ and split $$J(\lambda)=\int_0 ^1 \sin(y) \log(y) g(\lambda y)dy + \int_1 ^\infty \sin(y) \log(y) g(\lambda y)dy=:J_0(\lambda)+J_1(\lambda) .$$ In $[0,1]$ we have that $0<\lambda y <\lambda <1$ so that $|g(\lambda y)|\lesssim_g 1$ and this is best possible since $g(0)\neq 0$. We estimate $J_0$ by brute force $$|J_0(\lambda)|\lesssim_g \int_0 ^1 |\sin(y)\log(y)|dy\leq\int_0 ^1 y \log(1/y)dy\lesssim 1.$$ For $J_1$ we first integrate by parts once using that $\lim_{y\to\infty}g(y)=0$: $$J_1(\lambda)=-\cos(y)\log(y)g(\lambda y)\bigg|_{1} ^\infty +\int_1 ^\infty \cos(y)\frac{d}{dy}\big(\log (y) g(\lambda y)\big)dy$$ $$= \int_1 ^\infty \cos(y)\frac{g(\lambda y)}{y}dy +\lambda \int_1 ^\infty \cos(y)\log(y) g'(\lambda y)dy=:A+B.$$ For $A,B$ it is convenient to split the interval of integration to $[1,1/\lambda]$ and $[1/\lambda,\infty]$ where different bounds for $g(\lambda \cdot)$ apply and integrate by parts as many times as necessary. $$A_1:=\int_1 ^\frac{1}{\lambda} \cos(y)\frac{g(\lambda y)}{y}dy= \sin(y) \frac{g(\lambda y)}{y}\bigg|_{1} ^{1/\lambda}-\int_1 ^\frac{1}{\lambda} \sin(y) \big(\frac{\lambda g'(\lambda y)}{y}-\frac{g(\lambda y)}{y^2} \big)dy$$ $$=\lambda\sin(1/\lambda)g(1)-\sin(1)g(\lambda)-\lambda\int_1 ^\frac{1}{\lambda}\sin(y)\frac{g'(\lambda y)}{y}dy +\int_1 ^\frac{1}{\lambda}\sin(y)\frac{g(\lambda y)}{y^2}dy.$$ Since $|\lambda y|< 1$ and $g$ is real analytic we have $|g'(\lambda y)|+|g(\lambda y)|\lesssim_g 1$ for $y\in[1,1/\lambda]$. Using these estimates we get $|A_1|\lesssim_g 1$ (one can be more careful here in order to get the main term).
Similarly $$A_2:=\sin(y)\frac{g(\lambda y)}{y}\bigg|_{\frac{1}{\lambda}}^\infty - \int _ {\frac{1}{\lambda}} ^\infty \sin(y)\frac{d}{dy}\big( \frac{g(\lambda y)}{y}\big)dy$$ so that $|A_2|\lesssim \lambda |g(1)|\lesssim_g \lambda.$ Now for $B$ we split again $$B_1:= \lambda \sin(y) \log(y) g'(\lambda y) \bigg|_{1} ^{1/\lambda}-\lambda\int_1 ^\frac{1}{\lambda}\sin(y)\frac{d}{dy}\big(\log(y)g'(\lambda y)\big)dy$$ which shows that $|B_1|\lesssim \lambda \sin(1/\lambda)\log(1/\lambda)|g'(1)|\lesssim_g \lambda \log(1/\lambda).$
Finally $$B_2:= \lambda \sin(y)\log(y)g'(\lambda y)\big|_\frac{1}{\lambda} ^\infty-\lambda \int _\ frac{1}{\lambda} ^\infty \sin(y)\bigg(\frac{g'(\lambda y)}{y}+\log(y)\lambda g''(\lambda y)\bigg)dy.$$
Using the asmyptotic decay for the derivatives of $g(y)$, $g^{(n)}(x)=O(x^{-2-n})$ for large $y$, and the boundedness for small $y$ we can estimate $$|B_2|\leq |g'(1)|\lambda \log(1/\lambda)+\lambda \int_{\frac{1}{\lambda}} ^\infty \frac{1}{\lambda^3 y^4}dy+\lambda^2 \int_{\frac{1}{\lambda}} ^\infty \frac{\log(y)}{\lambda^4 y^4}dy$$ $$\lesssim_g \lambda \log(1/\lambda)+\lambda+\lambda\log(1/\lambda).$$ So $|A|\lesssim_g 1$ and $|B|\lesssim_g \lambda \log(1/\lambda)$ as $\lambda \to 0^+$ which implies that $|J(\lambda)|\lesssim_g 1$.
If the function $g$ is something like $g(x)=1/(1+x^2)$ (which satisfies all the hypotheses), I am pretty sure that you also have $|J(\lambda)|\gtrsim 1$ but I don't have the patience to check the details.
• Great! Numerical experiments suggest that for $g(x)=1/(1+x^2)$, the integral tends to a nonzero constant, so the limit is optimal. Thanks very much. May 2, 2013 at 1:00
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# Create a Pathway
#### Main Core Tie
Science - 5th Grade
Standard 4 Objective 2
#### Time Frame
1 class periods of 45 minutes each
Individual
Teresa Hislop
KIRSTIN REED
### Summary
Students build and experiment with electric circuits using easily available materials.
### Materials
For the Student: (Individual or Pairs)
• 1 battery (size D)
• 1 Christmas tree light with wires
• 2 strips of aluminum foil
• 2 paper clips
### Background for Teachers
Electricity can be made to flow along materials that will conduct the flow. Wire is a good example of a material that will allow electricity to flow. A circuit includes a source for electrical energy, a conducting path, and something that will use the electrical energy. The devices that use the electrical energy are placed on separate paths or connecting paths. When two or more dry cells are connected, a battery is created.
### Intended Learning Outcomes
• Observe simple objects, patterns, and events and report their observations.
• Compare things, processes, and events.
• Plan and conduct simple experiments.
• Predict results of investigations based on prior data.
• Use data to construct a reasonable conclusion.
### Instructional Procedures
Lab Preparation:
1. Prepare an aluminum foil ribbon by doing the following.
2. Lay parallel strips of 1/2 inch strip of clear tape on the dull side of the aluminum foil.
3. Use scissors to cut beside and between the strips. You should end up with two 1/2 inch wide strips of tape coated aluminum ribbon.
4. This ribbon will act as the wire in these activities. Paper clips can be used to connect the wires.
5. To prepare the Christmas tree bulbs, cut the wire between each bulb and scrape off enough of the plastic covering to expose about one inch of the wires. (These lights are better than flashlight bulbs because they do not burn out.)
Lab:
1. Give each student (or pairs) the supplies you have prepared.
2. Challenge them to use the battery, light bulb, 2 strips of aluminum ribbon, and 2 paper clips and find three ways to make the bulb light.
3. Challenge them to also find three ways that will not work to make the bulb light.
4. After they have had sufficient time to experiment and "play," students should draw pictures of each of the six ways in their science journals.
5. Have the students compare and contrast their findings.
6. Draw a picture of a circuit. Label the electrical source, the electrical path, and the user of the electrical energy.
7. Randomly select one component of the circuit and have students predict what will happen if that part is changed.
8. Test their hypothesis and discuss the results.
9. Select additional components to change and have students hypothesize how it will affect the circuit.
10. Continue this process until students understand the components of a complete electrical circuit.
### Extensions
Once students understand how to make a complete electrical circuit, give them additional materials (insulators and conductors) and see how they affect their ability to make the bulb light.
### Assessment Plan
Use the Science Lab Report Rubric to evaluate student science journal entries for their experiments.
### Rubrics
Created: 09/21/2002
Updated: 02/05/2018
17293
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You are on page 1of 2
NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY
Section : A-H
Last Date of Submission:
Assignment : 3
QUIZ QUESTIONS:
1. The rms value of the resultant current in a wire which carries a dc current of 10 A and a
sinusoidal alternating current of peak value 20 A is
a. 14.1 A
b. 17.3 A
c. 22.4 A
d. 30 A
2. A capacitor that stores a charge of 0.5 C at 10 volts has a capacitance of farad.
a. 5
b. 20
c. 10
d. 0.05
3. An a.c. current given by i= 14.14 sin (t+ ) has an r.m.s value of amperes.
a.
10
b. 14.14
c. 1.96
d. 7.07
4. The r.m.s. value of sinusoidal a.c. current is equal to its value at an angle of .degrees.
a.
60
b. 45
c. 30
d.90
5. A sine wave has a frequency of 50Hz. Its angular frequency is .radian/second.
a. 50/
b. 50/2
c. 50
d. 100
6. If e1= A sin t and e2= B sin (t-), then
a. e1 lags e2 by b. e2 lags e1 by
d. e1 is in phase with e2
7. A complex current wave is given by i= 5+ 5 sin 100 t ampere. Its average value is ampere.
a. 10
b. 0
c.
d. 5
8. The voltage applied across an R-L circuit is equal to .. of VR and VL.
a. Arithmetic sum b. algebraic sum
c. phasor sum
d. sum of the squares
9. The p.f. of an R-C circuit is
a. often zero
b. between zero and 1 c. always unity d. between zero and -1
10. The power in an a.c. circuit is given by
a. VI cos
b. VI sin
c. I2Z
d. I2XL
1. In a series R-L circuit the potential difference across resistance R is 12 V and potential
difference across inductance L is 5 V. Find the supply voltage and phase angle between current
and voltage.
2. Define form factor and peak factor of an alternating quantity.
3. A resistor of 25 in series with a 0.75 H inductor is connected across a supply at 250V, 50Hz.
Find the current through the inductor.
4. A circuit consists of a resistor of 10 in series with an inductor of 4.5 henries. The frequency is
70 Hz. Calculate the conductance and the susceptance of the circuit.
5. Two impedances of value (2+j6) and (8-j12) are connected in series. What would be the
resulting impedance in polar form?
6. Use the definition of complex power to calculate real and reactive power of the load taking a
current of (3+j4) amp when connected to a source of (8+j6) V.
7. In a circuit the voltage and current equations are given by V=15 sin (t+30 0) and I=15 sin (t300). Find the power consumed in the circuit.
8. A direct voltage 0f 200V is applied to a coil of resistance 20 and inductance of 2000 mH. Find
the time taken for the current through the coil to reach one half of its final value.
9. What is the time constant of a RC circuit having R=10 and C=10F?
10. A DC voltage V is switched on to a series R-L circuit. Write down the differential equation for
the circuit and also the expression for instantaneous current.
1. What is Complex Power, Apparent Power, Real Power and Reactive Power in a single phase AC
circuit? Explain by drawing the power triangle.
2. A series circuit has R=10, L=50mH, and C=100F and is supplied with 200V, 50Hz AC
supply. Find (i) Current and Power Factor (ii) Active Power (iii) Voltage drop across each
element.
3. Derive the mathematical expression for the charge stored in the capacitor of an RC series circuit
connected across a dc voltage source.
4. A 10F capacitor is connected in series with a resistor of 10M. This combination is connected
across a dc voltage of 200V which is switched on at t=0. Find (i) Time constant (ii) Steady
State Voltage across capacitor (iii) time taken for the capacitor to charge for 50% of supply
voltage and (iv) Voltage across capacitor after 50seconds.
5. In a single phase series circuit consisting of a non-inductive resistor of 10, an inductance of
0.159H and a capacitance of 106 F. The circuit is energized from a 230V, 50Hz supply.
Calculate (i) Impedance and Current (ii) Powe factor and power absorbed in the circuit.
6. In an AC single phase circuit three impedances of value 5<30 , 10<60 and 4-8j are
connected in series with a 230V, 50Hz supply. Find (i) Total combined impedance in rectangular
form (ii) Magnitude of current flowing in the circuit.
7. A resistance of 100 and an inductor of 0.5H are connected in series to a 100V dc supply.
Assuming that the steady state is already achieved, the battery is suddenly removed and
replaced by a shorted link at t=0. Calculate (i) time constant of the circuit and (ii) initial and
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http://stackoverflow.com/questions/13114010/max-min-and-sum-random-number-php
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# Max, Min and sum Random Number php
Here's the code
<?php
echo "<table border=\"0\">";
for ($d = 1;$d <= $times;$d++ )
{
echo "<tr><td>";
echo rand(1,6), "\n";
echo "</td></tr>";
}
echo "</table>"; ?>
I'm trying to get the max,min and sum of the rand(1,6), "\n"; But i can't figure it out. And it's killing me.
-
"I'm trying" --- how exactly? – zerkms Oct 28 '12 at 23:17
put the results in an array, from there its easy – Dagon Oct 28 '12 at 23:18
You need to collect the random numbers in an array, too:
echo "<table border=\"0\">";
$rands = array(); ################# for ($d = 1; $d <=$times; $d++) { echo "<tr><td>"; echo$rands[] = rand(1,6), "\n";
###########
echo "</td></tr>";
}
echo "</table>"; ?>
Afterwards you can make use of max, min and array_sum (all these links come with nice examples).
As your code already shows you should start to differ between code that does data-processing and code that does the HTML output:
// handle the data
$randomNumbers = array(); foreach (range(1,$times) as $d) {$randomNumbers[$d] = rand(1,6); } // output the data echo '<table border="0">'; foreach ($randomNumbers as $number) { printf("<tr><td>%d</tr></td>",$number);
}
echo "</table>";
-
$rands = array(); // rand() storage for($d = 1; $d <=$times; $d++){$rands[$d] = rand(1, 6); // store rands } var_dump($min = min($rands)); // min() of rands var_dump($min = max($rands)); // max() of rands ^ see sample code. (PS: I use [$d] as he has a 1-based increment and it may be needed for his further logic. This way the rands keys match his $d and can be easily accessed later on.) - Why not $rands[]? – zerkms Oct 28 '12 at 23:19
$rands[] = rand(1, 6); //$d not needed – Dagon Oct 28 '12 at 23:19
@zerkms To stay within his logic. He uses 1-based increment. So if he will walk the array, the key will be 1-off. I'm playing by his rules. – CodeAngry Oct 28 '12 at 23:19
@Claudrian: there is a foreach to iterate over arrays – zerkms Oct 28 '12 at 23:20
@Claudrian: omg, for the user it doesn't matter what underlying container is used. For 99% users to be array - it should implement all array related interfaces and behave like array. Anyway, 1-based arrays are more harm than good usually. – zerkms Oct 28 '12 at 23:38
<?php
$min = 10;$max = -1;
$sum = 0; for ($d = 1; $d <=$times; $d++) {$n = rand(1, 6);
if ($n <$min) $min =$n;
if ($n >$max) $max =$n;
$sum +=$n;
}
echo $min . ' ' .$max . ' ' . $sum . '<br/>'; ?> - $sum=0;
for ($d = 1;$d <= $times;$d++ ) {
echo "<tr><td>";
$r=rand(1,6);$sum +=$r; echo "$r, \n";
echo "</td></tr>";
if ($d==1) {$min=$r;$max=$r; } if ($r>$max)$max=$r; if ($r<$min)$min=$r; } // do something with$min, $max and$sum;
-
Without $min and $max initialization - the code in this answer is terrible – zerkms Oct 28 '12 at 23:23
<?php
$total = 0; echo "<table border=\"0\">"; for ($d = 1; $d <=$times;$d++ ) {$rand = rand(1,6);
$total +=$rand;
$array[] =$rand;
echo "<tr><td>";
echo $rand, "\n"; echo "</td></tr>"; } echo "</table>"; ?> echo$total;
echo min($array); echo max($array);
-
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http://math.stackexchange.com/questions/222221/problem-on-entire-function
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# problem on entire function
Let $f$ be an entire function. For which of the following cases $f$ is not necessarily a constant
1. $\operatorname{im}(f'(z))>0$ for all $z$
2. $f'(0)=0$ and $|f'(z)|\leq3$ for all $z$
3. $f(n)=3$ for all integer $n$
4. $f(z) =i$ when $z=(1+\frac{k}{n})$ for every positive integer $k$
I think 1 is true since $f'(z)=$constant so $f(z)=cz$ for some $c$
For 2, $f=0$
For 3, $f$ is not constant since $f(z)=3 \cos2\pi z$
I have no idea for 4
am i right for other three options
-
In your statement number $4$: what's $n$? – Kevin Carlson Oct 27 '12 at 19:19
You're right for $1$, where the $c$ you mention should have positive imaginary part. For example $f(z)=iz$ does the job.
For $2$, $f$ need not be identically $0$. $f(z)=2$ satisfies the requirements, for instance. $f$ does have to be constant, though, because if $f$ is entire, $f'$ is too, and so by Liouville's theorem since $f'$ is bounded it's constant, and the other condition forces it to be $0$ everywhere.
You're right on $3$.
For $4$, I'm still not sure what $n$ is. My guess is that this will be an identity theorem application, and that $f$ will be $i$ on a set with an accumulation point, which will force it to be $i$ everywhere.
But as it's written now, for $n$ is a fixed integer you could get $f$ non-constant in the same way as you did in number $3$: set $$f(z)=i\cos(2n\pi(z-1))$$
| 456
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http://energy.gov/energysaver/articles/electric-kettle-takes-down-microwave-final-round-energyfaceoff
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## Electric Kettle Takes Down Microwave in Final Round of #EnergyFaceoff
November 24, 2014 - 12:13pm
The electric kettle wins the final round of #EnergyFaceoff. | Graphic by Stacy Buchanan, National Renewable Energy Laboratory
Residential microwaves vary in their wattages; the higher the wattage, the faster your food or liquid generally cooks. For this Energy War, we used two products we had available to compare the energy use: a 1,000 watt (W) microwave and a 1,500 W electric kettle.
Looking at just the wattages, you might think that the microwave would be more efficient. But what we found is that the time to heat the water makes all the difference.
For our test, we heated an eight ounce cup of water and timed how long it took each. For the microwave, we used the "Beverage" setting, and it took about four minutes to heat the cup of water. The kettle took 1:30 to heat the same amount of water. So not only are you saving precious minutes, you're also saving energy!
Assuming someone might do this three times a day, five days a week, we followed the steps for calculating annual energy use and cost on Energy Saver (note that we used weekly energy consumption instead of daily):
##### Microwave
Time used: 4:00, 3 times per day, 5 days a week. (1 hour per week)
Wattage: 1000 W
(1,000 W x 1 hour) / 1000 = 1 kWh/week (weekly energy consumption)
1 kWh/week x 52 weeks = 52 kWh (annual energy consumption)
52 kWh x \$0.11 = \$5.72 per year
##### Kettle
Time used: 1:30, 3 times per day, 5 days a week (.375 hours per week)
Wattage: 1500 W
(1,500 W x .375) / 1000 = .56 kWh/week (weekly energy consumption)
.56 kWh/week x 52 weeks = 29.12 kWh (annual energy consumption)
29.12 kWh x \$0.11 = \$3.20 per year
Those aren't enormous savings, but if you find yourself boiling water a lot (or if you're in an office with a lot of tea drinkers), they can add up. One thing to also consider is that you are more likely to heat the exact amount of water you need in the microwave, whereas you might overfill the electric kettle and negate your savings with a longer time to boil (so just heat what you need!).
Do you use an electric kettle? Try this test with your own kettle and microwave and let us know your results (be sure to share the wattages and the time it took each to heat the water).
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https://cassno.com/casino/learn-the-hands-of-poker/
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# Learn The Hands Of Poker
If you have played poker and are familiar with all the bits and bobs of the game, then you are, more than likely familiar with poker hands. If you are not here is your opportunity to pick up some necessary information. A poker hand is each time somebody has had five cards dealt in their hands. As with any regular card game, there is a precise ranking for poker hands. Understanding these rankings and becoming familiar with them is important if you plan on having any success in the poker arena. Listed below are some examples of poker hand rankings.
HIGH CARD – The high card is ordered by the value of the highest card. If your poker hand
Was seven, six, five, then the seven would be the highest card. If {, however ,} two of the cards had the same worth, then it’s the next highest card.
PAIRS – This is kind of self explanatory. If 2 of your 5 cards have the same worth, you have got a pair.
TWO PAIRS – This, as in the example, would be a hand consisting of two different pairs.
THREE OF A KIND – In this hand, 3 of the five cards are the same.
STRAIGHT – The straight is composed of five consecutive valued cards which are ranked according to the rule of the highest card.
FLUSH – A flush occurs when all 5 of the cards in the hand are of the same suit or classification.
FULL HOUSE – The full house has three of the five cards with the same worth while the. Remaining two cards are a pair.
FOUR OF A KIND – Having four of the five cards in your hand with the same worth and is ordered by the value of the 4 cards.
STRAIGHT FLUSH – This hand mixes a straight and a flush. Or put in other terms, the cards. Are the same suit with successive values.
ROYAL FLUSH – In this hand you have all of the best cards.
Alright, now you have read this perhaps we will be able to showwhat’s explained above in some easier terms : High Card – 3,4,5,6,7
Pair – 4, 4 & 7, 7
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Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic: complex numbers
Replies: 8 Last Post: Jan 31, 2007 7:33 PM
Messages: [ Previous | Next ]
Faton Berisha Posts: 39 Registered: 1/9/07
Re: complex numbers
Posted: Jan 31, 2007 7:03 PM
On Jan 31, 7:52 pm, "Stephen J. Herschkorn" <sjhersc...@netscape.net>
wrote:
> jennifer wrote:
> [snip]
> >Also how do i describe a family of curves in the complex plane:
>
> >1. Re(1/z)= C
>
> >The answer is a family of circles centered at 1/(2C)+0i of radius 1/
> >modulus 2C but with the origin excluded.
>
> That should be 1 / (2 |C|).
>
> >My question is how did they get the center as 1/(2C)+0i and radius 1/
> >modulus 2C.
>
> >I got the Re(1/z)= x/(x^2+y^2) = C but how do i obtain the radius and
> >the center as indicated above?
>
> So x^2 - x / C + y^2 = 0. Complete the square.
>
> >I know the equation of a circle is: x^2 +y^2= r^2
>
> That is for a circle centerer at the origin. Do you know the equation
> for an arbitrary center in the plane?
>
Or if you like it better, without switching to coordinates,
C = Re(1/z) = 1/2 (1/z + 1/ bar z)
= 1/2 (z + bar z)/(z bar z);
hence,
z + bar z = 2Cz bar z,
or
(z - 1/(2C)) (bar z - 1/(2C)) = (1/(2C))^2.
Since 1/(2C) is real,
(z - 1/(2C)) bar (z - 1/(2C)) = (1/(2C))^2,
hence
|z - 1/(2C)|^2 = (1/(2C))^2,
or
|z - 1/(2C)| = (1/(2|C|)).
Regards,
Faton Berisha
Date Subject Author
1/30/07 nonton
1/30/07 Bart Goddard
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# What Measures Out Time Until in Time All is Smashed To It? Riddle - Take a Look At The Answer With Detailed Explanation of This Innovative Riddle Here
by Shalini K | Updated December 09, 2020
What Measures Out Time Until in Time All is Smashed To It? Riddle - What Measures Out Time Until in Time All is Smashed To It? Riddle is a very popular and interesting riddle these days that is trending on social media platforms like Twitter, WhatsApp Groups, Facebook, etc. These days Riddles are an integral part of our daily routine and at least we should practice one riddle per day. This interesting riddle is a great way to start with. Read on to know about the answer and detailed explanation of What Measures Out Time Until in Time All is Smashed To It? Riddle.
## Why We Should Solve Riddles on a Daily Basis?
Solving Riddles us to focus on each and every Word. During the process of solving riddles, It is very much important to focus on each detail of the Riddles in order to find the answer to the riddle. This ability to focus on each and every detail helps us in our day to day life routine, especially at our workspace. Focusing on Small Details drastically improves our quality of Work. Take a look at What Measures Out Time Until in Time All is Smashed To It? Riddle.
## Know About this What Measures Out Time Until in Time All is Smashed To It? Riddle
This Amazing What Measures Out Time Until in Time All is Smashed To It? Riddle is stated as follows:
What Measures Out Time Until in Time All is Smashed To It?
## Answer to this amazing What Measures Out Time Until in Time All is Smashed To It? Riddle
Answer to this amazing What Measures Out Time Until in Time All is Smashed To It? Riddle is Sand.
## Explanation:
Hour clock is a type of clock that contains sand in it. Sand inside the Hour Clock is used to measure the time and sand at one end smashes into the other end when a limited time-period is completed.
## What Measures Out Time Until in Time All is Smashed To It? Riddle - FAQs
1. You see me once in the Bear, twice in Rabbit, and not at all in Cow. What am I?
The answer to this amazing You see me once in the Bear riddle is the letter ‘B’.
2. Two in a Sheep, one in a Hare, zero in a Goat, but one in a Zebra. What is it?
The answer to this mind-blowing Two in a Sheep riddle is the letter ‘E’.
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# 34642 - Electromechanical systems modelling M
• Docente: Angelo Tani
• Credits: 6
• SSD: ING-IND/32
• Language: Italian
• Campus: Bologna
• Corso: Second cycle degree programme (LM) in Electrical Energy Engineering (cod. 9066)
• from Sep 18, 2023 to Dec 18, 2023
## Learning outcomes
The aim of the course is to provide instruments useful in order to define mathematical models suitable for studying, designing and controlling electromechanical systems.
## Course contents
Prerequisites
Basic skills in electrical engineering.
Course contents
Electromechanical systems dynamics
Introduction to mathematical modelling of electromechanical systems, Input-Output and Input-State-Output differential equations, stability analysis, small signal analysis, numerical solution of differential equations. Analysis of electrodynamic and electromagnetic levitation systems.
Space vectors in three-phase systems
Definition of space vector and zero sequence component, differential equations of three-phase systems in terms of space vectors, Fourier expansion of space vectors, mathematical representation of quantities depending on space and time, multiple space vectors for multiphase systems.
Analysis of electric machines by space vectors
Modelling principles for rotating electrical machines, dynamic model of the induction machine in terms of space vectors and zero-sequence components, machine parameter estimation, direct torque and flux control (DTC) of induction machines.
Discrete-time systems
Introduction to mathematical modelling of discrete-time systems, Input-Output and Input-State-Output discrete equations, stability analysis, discretization of the differential equations of continuous-time systems.
Parameters and state estimation of an electromechanical system
On line parameters estimation of an electromechanical system by MRAS method, full order state observer, adaptive state observer.
Fuzzy controllers
Introduction to fuzzy logic, linguistic variables, fuzzy sets, membership functions, inference process, fuzzy logic for electromechanical system modelling and control.
Artificial neural network
Artificial neurons and activation functions, multilayer neural networks, learning process by back propagation algorithm, neural networks for electromechanical system modelling and control.
The lessons are supported by exercises with Personal Computer.
It is not necessary to buy specific books. The pdf files of the slides utilized during the lessons are indispensable and sufficient for the preparation for the exam, and are available on VIRTUALE.
## Teaching methods
The frontal lessons are supported by exercises with Personal Computer (MATLAB-Simulink).
## Assessment methods
The exam consists of an oral examination, which is based on three questions.
## Teaching tools
Lessons are carried out with the help of a personal computer and a computer projector (Power Point). The pdf files of the slides utilized during the lessons are available on VIRTUALE. Exercises are carried out with the help of a personal computer (MATLAB-Simulink).
## Office hours
See the website of Angelo Tani
### SDGs
This teaching activity contributes to the achievement of the Sustainable Development Goals of the UN 2030 Agenda.
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EMISIVITY OF CARBON DIOXIDE AND STEAM (WATER VAPOR)
Author: Nasif Nahle. Published: ©14 March 2007 by Biology Cabinet.
Web www.biocab.org
CLICK ON THE DIAGRAM TO SEE AN OPTIMIZED IMAGE
DENSITY-EMISSIVITY COMPARISON IF DENSITY IS MULTIPLIED BY 10
In the table inserted above we observe a very tiny red bar that represents the present concentration of atmospheric CO2, which is 0.032% or a density of 0.000614 Kg/m^3. At the moment, the emissivity of CO2 at such concentrations is 0.00092 (with no units), for that reason their thermal capability to increase the tropospheric temperature is hardly 0.02 K (1K=1C). (Pitts & Sissom, 1997)
However, the blue bar that represents the water vapor (I will call it “steam” from this line ahead) fills 15% from the total amount of atmospheric gases, but it is not taken into account for filling the atmosphere’s composition because its concentration fluctuates enormously depending of the incoming solar radiation and interstellar cosmic rays (Shaviv. 2005). The capability of the steam to store energy is 0.4 (without units). The increase of the tropospheric temperature by steam is of 0.15 K. 5.7 times higher than the increase by carbon dioxide density.
The emissivity of Oxygen (green bar) and Nitrogen (yellow bar) are inconsequential; hence, they are not taken into account for the estimate of the oscillation of the Earth’s tropospheric temperature.
What would happen if we increase 10 times the densities of CO2 and steam in the atmosphere?
If we increase the density of CO2 from 381 ppmv to 3810 ppmv it would not happen anything, absolutely nothing, because the emissivity of CO2 would increase barely to 0.19 (without units). (Pitts & Sissom, 1997)
I will consider a concentration of 5% of steam to make the calculi. If we had 595311 ppmv (concentration = 50%) of steam in the whole atmosphere, its emissivity would increase to 0.7 (no units). Some investigators of the US Navy have reported a steam emissivity of 1.2, which is blatantly flawed because steam is not a black body, but a grey body, and no known system behaves like a black body. The maximum emissivity for a perfect Black body is 1.
The fluctuation of the temperature provoked by the emission of heat by 50% of steam in the atmosphere would be 0.4 K; 0.12 K lower than the observed fluctuation. However, if there is not more incoming Interstellar Cosmic Rays and more heat incoming from the Sun, the steam in the atmosphere would not increase. An interesting discovery of Dr. Nir Shaviv has been that if the incoming Interstellar Cosmic Rays increase, then the steam would form bigger droplets that would form clouds. Those clouds would reflect the Solar Radiation back to space, cooling so the Earth. But if the intensity of Solar Activity increases, then the interstellar cosmic particles would be blown off by the Solar Wind and the cloudiness in the Earth would decrease, increasing thus the tropospheric temperatures.
As we can see, THERE IS NOT a single scientific reason or a single factual reason to assure that the human beings are causing the global warming on the Solar System. THE WHOLE THING IS NATURAL, and as well as it began, it must finish, unless the stupid greens commit the foolishness of “fertilizing” the oceans with Iron, causing thus an eutrophication that will be lethal to the marine animals; or that NASA think to create its artificial volcano, filling the atmosphere with dust, aggravating thus the glaciations in progress; or that the same NASA commit the stupidity of filling the atmosphere with sulphates, which would cause the collapse of life on the Earth by global acid rains.
Let us leave that the Earth fixes the problem by itself; it could be that, instead of fixing the “problem”, we could make it worsen to the point of exterminating all the life on the planet only for earning 25 million dollars.
TOP OF PAGE ^^
BIBLIOGRAPHY
Pitts, Donald and Sissom, Leighton. Heat Transfer. 1997. McGraw Hill. New York, NY.
Documentary "The Great Global WArming Swindle". Revisado por última vez el 14 de Marzo de 2007.
TOP OF PAGE ^^
This Website created and kept up by Nasif Nahle et al.
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# ART - Carmen
About the thimble you described, please look at the maker's mark on the right:
http://home.comcast.net/~brl/thimbles/stern.jpg
Sra
1. 👍
2. 👎
3. 👁
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## Hello I'm class12maths!
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# 09ma6cHW5 - S ‘ A for any wff A(30 2 Prove Proposition...
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Ma/CS 6c Assignment #5 Due Thursday, May 6 at 1 p.m. IN ALL PROBLEMS BELOW YOU CANNOT USE THEOREM 1.11.5 IN THE NOTES (since the exercises below form parts of the proof of that theorem). ALSO “FORMULA” OR “WFF” MEANS “WELL-FORMED FORMULA IN PROPOSITIONAL LOGIC BUILT USING PROPOSITIONAL VARIABLES, PARENTHESES, AND ONLY ¬ , .” (40%) 1. * (i) Show that for any formula A , ( ¬¬ A A ) . (ii) Prove Proposition 1.11.8 in the notes: Let S be any set of wffs and A any wff. If S ∪ { A } is formally inconsistent, then S ‘ ¬ A . (iii) Show that if S is formally inconsistent, then
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Unformatted text preview: S ‘ A for any wff A . (30%) 2. Prove Proposition 1.11.9 in the notes: Let S be any set of wffs and A,B any wffs. Then: S ∪ { A } ‘ ¬ B iff S ∪ { B } ‘ ¬ A. (40%) 3. (i) Show that ¬ ( A ⇒ B ) ‘ A and ¬ ( A ⇒ B ) ‘ ¬ B (ii) Prove Lemma 1.11.12 in the notes: If ¯ S is a formally consistent and complete set of formulas and ν is the valuation defined by ν ( p i ) = ( 1 , if p i ∈ ¯ S , if p i 6∈ ¯ S, then for any formula A , ν ( A ) = 1 iff A ∈ ¯ S. 1...
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Python program to convert int to exponential
• Last Updated : 23 Dec, 2020
Given a number of int type, the task is to write a Python program to convert it to exponential.
Examples:
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```Input: 19
Output: 1.900000e+01
Input: 2002
Output: 2.002000e+03
Input: 110102
Output: 1.101020e+05```
Approach:
• We will first declare and initialise an integer number
• Then we will use formate method to convert the number from integer to exponential type.
• Then we will print the converted value.
Syntax:
`String {field_name:conversion} Example.format(value)`
Errors and Exceptions:
ValueError: Error occurs during type conversion in this method.
More parameters can be included within the curly braces of our syntax. Use the format code syntax {field_name: conversion}, where field_name specifies the index number of the argument to the str.format() method, and conversion refers to the conversion code of the data type.
Example:
Python3
`# Python program to convert int to exponential ` ` ` `# Declaring the integer number ` `int_number ``=` `110102` ` ` `# Converting the integer number to exponential number ` `exp_number ``=` `"{:e}"``.``format``(int_number) ` ` ` `# Printing the converted number ` `print``(``"Integer Number:"``,int_number) ` `print``(``"Exponent Number:"``,exp_number)`
Output:
```Integer Number: 110102
Exponent Number: 1.101020e+05```
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# Video: Determining the Degree of a Polynomial Function
Find the degree of the function 𝑓(𝑥) = −3𝑥(−6𝑥 + 10)³.
02:26
### Video Transcript
Find the degree of the function 𝑓 𝑥 is equal to negative three 𝑥 multiplied by negative six 𝑥 plus 10 to the power of three.
To solve this problem, first of all, we need to know what is the degree of a function. Well, actually, the degree of a function in this case a polynomial — cause that’s what we’re looking at — is the highest degree of its monomials with nonzero coefficients.
Well, is this useful? What’s a monomial? Well, actually, a monomial is actually an individual term. So actually, this is really useful because it tells us that if we want to find out the degree of our function, what we need to do is find out the highest degree of one of the individual terms.
So if we take a look at our function, so we got 𝑓𝑥 is equal to negative three 𝑥 multiplied by negative six 𝑥 plus 10 all cubed, then we can actually see that if we want to find the highest degree of this function, then we’re actually only concerned with the negative six 𝑥 cubed from the terms that are in the parenthesis. And the reason this is is actually because it’s gonna give us the highest degree from any of those terms.
So therefore, if we’re gonna try and find the highest degree of our function, we just need to multiply negative three 𝑥 by negative six 𝑥 all cubed. So this gonna be equal to negative three 𝑥 multiplied by negative 216 𝑥 cubed. And it’s still negative 216 𝑥 cubed because you’ve got negative six cubed, which is negative 216, and then 𝑥 cubed, which is 𝑥 cubed. And this will give us an answer of 648 𝑥 to the power of four.
We’re not really interested in this 648, which is the coefficient of our 𝑥 to the power of four. However, what it does tell us is that because it’s nonzero, then, okay, we can look at this monomial as the one that we’re looking for for the highest degree.
What we’re really interested in is the power of 𝑥. And we can see in this case that it’s four. So therefore, we can say that the degree of 𝑓𝑥 is equal to negative three 𝑥 multiplied by negative six 𝑥 plus 10 all cubed is equal to four. And that’s because four is the highest degree of any individual term within the function.
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Essay Help Services
# CSC1401| Chinese Zodiac Specification Assignment | Science
Home Recent Questions CSC1401| Chinese Zodiac Specification Assignment | Science
Your task will be to write a program to accept user inputs for the date of birth, calculate the number of days the user has been alive, and give the corresponding Chinese Zodiac sign.
Inputs
The following inputs should be declared as variables and initialised at the beginning of your pro-gram:
- The year of birth (a valid 4 digit year);
- The month of birth (a valid integer number from 1 to 12, representing January to December);
- The date (in the month) of birth (an integer number from 1 to 31, representing a valid date of the month).
Code has been provided to handle the following tasks1:
- to receive inputs from the user;
- to convert the user inputs (String) to the proper type (Number).
//Accept user inputs for DOB and convert them into Number. The default value is set as 31/12/2016.
parseInt(prompt(’Enter the year of birth as a 4 digit integer’, ’2016’)); parseInt(prompt(’Enter the month of birth as an integer, ranging from 1 to 12’, ’12’)); parseInt(prompt(’Enter the date of birth as an integer, ranging from 1 to 31’, ’31’));
Such statements will return the user inputs in proper Number type. You need to create some variables to store such returned results for further calculations in the program. In this assignment, we assume that the user inputs are always valid, e.g., the value entered for month is an integer number within the range of 1 – 12, etc. You are not required to validate the values input by the user.
Functional Requirements
Calculation
1. Create a Date object for the date of birth based on the user input.
- Send the Date constructor the year, month number and date. Please notice that, in the Date constructor, the number for month is starting from 0, rather than 1.
2. Create a Date object for the current time;
- No arguments supplied to the constructor
3. Calculate the user’s age in milliseconds
- Dates are stored in milliseconds since 1 Jan 1970;
- Use the getTime() function to get a Date object’s time in milliseconds
- Deduct the user’s date of birth time from the current time
4. Calculate the user’s age in days
- Divide the user’s age in milliseconds by the number of milliseconds in a day.
- use the Math.floor() function to reduce the number to an integer.
5. Calculate the Chinese Zodiac
(a) Initialise an array object to host the String values of 12 Chinese Zodiac animal signs in the correct order;
(b) In 1024 the Chinese Zodiac started a new cycle with the animal “Rat”. Create a constant for the year of 1024;
(c) Use the minutes operator to calculate the number of years between the user’s year of birth and the constant created in (b);
(d) Use the mod operator to find the remainder after dividing the result of (c) by the constant created for the number of years in a Chinese Zodiac cycle;
(e) Use the reminder as an index to find the correct animal sign in the Chinese Zodiac sign array created in (a).
Output
1. An alert statement to display the user’s birthday by using the toDateString() function for Date object created for the user’s date of birth. For example, “Your date of birth is Tue Apr 01 1980 ”;
2. An alert statement to display how many days the user has been alive. For example, “You have been alive for 13489 days”;
3. An alert statement to display the corresponding Chinese Zodiac animal sign. For example, “Your Chinese Zodiac sign is Monkey”.
The following figures illustrate the results displayed regarding inputs of a user whose birthday is April 01, 1980, tested on March 07, 2017. Note that the layout of alert window and the font of text on di↵erent web browsers could be di↵erent, and they are not a concern in this assignment.
Testing
• Enter the user birthday into the “Your Age in Days” system on http://jalu.ch/coding/days/en and compare the result with that calculated by your program to test the days alive calculation;
• Compare the Chinese Zodiac result calculated in your program with the chart in http://en.wikipedia.org/wiki/Chinese zodiac to perform the logic test for your program.
Non-functional Requirements
Structure of the Source Code
- All code should appear in the script section in the head of the HTML document.
- Do not write any code in the HTML body. Deliver all functionality by JavaScript.
- In the script order your code as follows:
(a) Constants;
(b) Variables and objects (declared and initialised);
(c) Other statements.
- You are required to add at least three comments to the source code.
- Do not comment every single line, instead, comment on blocks of code with a common purpose.
- Do not simply translate the syntax into English for comments, instead, describe the purpose of the code.
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# pound (metric) to mark conversion
Conversion number between pound (metric) and mark is 2.0094216605392. This means, that pound (metric) is bigger unit than mark.
### Contents [show][hide]
Switch to reverse conversion:
from mark to pound (metric) conversion
### Enter the number in pound (metric):
Decimal Fraction Exponential Expression
pound (metric)
eg.: 10.12345 or 1.123e5
Result in mark
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 pound (metric) = (exactly) (0.5) / (0.2488278144) = 2.0094216605392 mark
• 1 mark = (exactly) (0.2488278144) / (0.5) = 0.4976556288 pound (metric)
• ? pound (metric) × (0.5 ("kg"/"pound (metric)")) / (0.2488278144 ("kg"/"mark")) = ? mark
### High precision conversion
If conversion between pound (metric) to kilogram and kilogram to mark is exactly definied, high precision conversion from pound (metric) to mark is enabled.
Decimal places: (0-800)
pound (metric)
Result in mark:
?
### pound (metric) to mark conversion chart
Start value: [pound (metric)] Step size [pound (metric)] How many lines? (max 100)
visual:
pound (metric)mark
00
1020.094216605392
2040.188433210785
3060.282649816177
4080.37686642157
50100.47108302696
60120.56529963235
70140.65951623775
80160.75373284314
90180.84794944853
100200.94216605392
110221.03638265932
Copy to Excel
## Multiple conversion
Enter numbers in pound (metric) and click convert button.
One number per line.
Converted numbers in mark:
Click to select all
## Details about pound (metric) and mark units:
Convert Pound (metric) to other unit:
### pound (metric)
Definition of pound (metric) unit: ≡ 500 g . = 500 g = 0.5 kg
Convert Mark to other unit:
### mark
Definition of mark unit: ≡ 8 oz t . The Mark is originally a medieval weight or mass unit, which supplanted the pound weight as a precious metals and coinage weight from the 11th century = 248.8278144 g
← Back to Mass units
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Difficulty level: Beginner
Duration:
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Lesson title:
Tutorial describing the basic search and navigation features of the Allen Mouse Brain Atlas
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Duration: 6:40
Speaker: : Unknown
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Difficulty level: Beginner
Duration: 6:35
Speaker: : Unknown
This tutorial demonstrates how to use the differential search feature of the Allen Mouse Brain Atlas to find gene markers for different regions of the brain and to visualize this gene expression in three-dimensional space. Differential search is also available for the Allen Developing Mouse Brain Atlas and the Allen Human Brain Atlas.
Difficulty level: Beginner
Duration: 6:31
Speaker: : Unknown
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Duration: 7:57
Speaker: : Barton Poulson
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Speaker: : Barton Poulson
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Difficulty level: Beginner
Duration: 4:01
Speaker: : Barton Poulson
Lesson title:
Statistics is exploring data.
Difficulty level: Beginner
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Speaker: : Barton Poulson
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Graphical data exploration
Difficulty level: Beginner
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Speaker: : Barton Poulson
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Numerical data exploration
Difficulty level: Beginner
Duration: 5:05
Speaker: : Barton Poulson
Lesson title:
Simple description of statistical data.
Difficulty level: Beginner
Duration: 10:16
Speaker: : Barton Poulson
Lesson title:
Basics of hypothesis testing.
Difficulty level: Beginner
Duration: 06:04
Speaker: : Barton Poulson
In this lecture, the speaker demonstrates Neurokernel's module interfacing feature by using it to integrate independently developed models of olfactory and vision LPUs based upon experimentally obtained connectivity information.
Difficulty level: Intermediate
Duration: 29:56
Speaker: : Aurel A. Lazar
Lesson title:
This lecture covers modeling the neuron in silicon, modeling vision and audition and sensory fusion using a deep network.
Difficulty level: Beginner
Duration: 1:32:17
Speaker: : Shih-Chii Liu
Course Name:
Lesson title:
Presentation of a simulation software for spatial model neurons and their networks designed primarily for GPUs.
Difficulty level: Beginner
Duration: 21:15
Lesson title:
Presentation of past and present neurocomputing approaches and hybrid analog/digital circuits that directly emulate the properties of neurons and synapses.
Difficulty level: Beginner
Duration: 41:57
Speaker: : Giacomo Indiveri
Lesson title:
Presentation of the Brian neural simulator, where models are defined directly by their mathematical equations and code is automatically generated for each specific target.
Difficulty level: Beginner
Duration: 20:39
Speaker: : Giacomo Indiveri
Lesson title:
The lecture covers a brief introduction to neuromorphic engineering, some of the neuromorphic networks that the speaker has developed, and their potential applications, particularly in machine learning.
Difficulty level: Intermediate
Duration: 19:57
This lecture will highlight our current understanding and recent developments in the field of neurodegenerative disease research, as well as the future of diagnostics and treatment of neurodegenerative diseases
Difficulty level: Beginner
Duration: 1:02:29
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# Module 3 Interpreting Scatterplots Form: linear, curved, clusters, no pattern Direction: positive, negative, no direction Strength: How well
points fit form Outliers that are deviations from pattern Correlation Coefficient is the measure of the direction and strength of a linear relationship. r = (1-(n-1)) Between -1 and 1 Values of r close to 0 imply weak or no linear association. X and Y are independent, knowing X tells nothing about Y. WATCH out for Outliers Rule of thumb Values larger than .8 or smaller than -.8 represent VERY strong correlation. Anything between -.3 and .3 indicated weak or no association. Least Squares Regression Analysis Describes the linear relationship and defines how one variable changes the other. Y = a + bX Interpreting Slope Slope b represent the increase or decrease in Y for any one-unit increase in X. 2 Coefficient of determination R 2 R varies between 0 and 1. If close to 1, then there is almost a perfect linear association between Y and X. Values of R^2 larger than .6 show that x is a strong predictor of y, and that the regression line on x provides a good explanation of changes in y. If R^2 values less than .25, indicate x is not a good predictor for y, and that there is a large fraction of variation in y that is not explained by changes in X.
90% Example residual explanation: The residual plot of least squares regression line for CPU usage time on LINET shows no pattern in residuals. Residuals are randomly scattered around the zero line in a horizontal band. The regression line appears to be a good representation between TIME and LINET. MODULE 4 Sampling approaches use statistical techniques to produce representative samples from a large population. Sampling plan states objectives of study, target population, method of sampling, set of characteristics and variables of interest, and more. Probabilistic sampling chooses individuals or units in the sample using random procedure. Eliminates bias. Following are types: Simple random sample a set of individuals chosen in such a way that everyone has the same change of being chose. Stratified Sampling population is divided into groups of similar units (or strata). Convenience Sampling individuals are chosen based on ease with which the data can be collected. Conclusions can t be generalized to entire population. Voluntary Response Sampling individuals that select themselves to respond to data collection appeal. Observational Study vs. Experiment Observational is when research has no control on the factors, can t assign subjects to variables of interest. Just observe. (Hard to prove causation) Experiment researched is able to control factors and assign individuals to diff. treatments. (Can prove cause and effect). Experimental Design: Controlled Experiments comparing several treatments. One is many times a placebo. Randomization: Subjects are selected randomly to each treatment. Reduces bias. Replication Each treatment is done on many units to reduce chance of variation. Double-blind process Both the researches and subjects know what treatment they are in reduces unconscious bias. Module 6 Point estimate single value computed from a sample to estimate the value of a population parameter. Point estimates alone provide no information on accuracy of estimate. Confidence Interval range of values that are believes to contain the value of the population parameter. Sample mean denoted by Population mean denoted by Central Limit Theorem given the size, the population mean and standard deviation, and the size is large, the sampling distribution of the sample mean is approx. normal with the center in the population mean and the standard deviation equal to (st. dev / (sqrrot of n)) To get more accurate estimates, we should increase the sample size. Confidence Interval 95% 99% Reduce Margin of error: Reduce confidence level Increase Sample size Reduce variation in S (Sample size should be over 30) Example: analysis shows with 95% confidence that the user s average mean time between strokes is within 0.225 and 0.375 seconds. The observed attempt has average time of 0.39 seconds that is significantly higher than the 95% confidence range. We conclude that the recent attempt cannot be explained by chance variation and therefore is an unauthorized intrusion. Module 7 Sampling Distribution of Sampling Proportions We use proportions when dealing with categorical data when there are two possible outcomes for variables. For large samples, the sampling distribution of the sample proportion is approximately normal with mean and
Standard deviation In large samples, the std. deviation of the sample proportion is computed by replacing the population proportion p with the sample estimate
Large Sample Confidence Interval for population mean p: 95%
90% 99%
Residuals Vertical deviations difference between the observed y and the y predicted by the least squares regression line. Average of residuals is equal to zero. st 1 Linear randomly scattered around the zero line nd 2 Problem non-linear association curved pattern indicates the relationship between y and x is curved and not linear. rd 3 Problem non-constant variance variation in y increases as x increase (fan shaped)
Module 8 Test of significance a probability that measures how well the data support the hypothesis. Null and Alternative Hypotheses Null hypothesits no difference Alternative hypothesis must be true if null hypothesis is false. Ex. Verify mean response time of call center is less than 15.
Null and Alternative Hypotheses on population average
Calculating a Test Statistic How far away is the sample mean from general mean? If Z is 2 or 3 standard errors away from general mean, then it is Far away.
P-Value probability that test statistic is equal to or more extreme than the value obtained from data when null hypothesis is true. The smaller the p-value, the stronger evidence against the null hypothesis. - If p-value is LESS than =0.05, the null hypothesis IS REJECTED at 5% significance level. The test result is called statistically significant . -If p-value is LESS than =0.01, the null hypothesis IS REJECTED at 1% significance level. The test result is called highly significant . - If p-value is LARGER than 0.05, the null hypothesis CANNOT BE REJECTED. The test is not significant . For One sided test Ha < mean Find the entry in standard normal table corresponding to z*. For One sided test Ha > mean Find the entry in standard normal table corresponding to z* and do 1 that value. For Two sided test Ha != mean Find entry in table corresponding to absolute value of -|z*| and compute 2 * that value Example answers: - The p-value is extremely small, so we can conclude that data provide very strong evidence against the null hypothesis. The test is highly significant indicating that the mean response time is significantly lower than 15 minutes. - Note that the p-value is 0.053 that is slightly larger than 0.05. So we cannot reject the null hypothesis at 5% significance level. However, since the p-value < 0.10, we could reject it at 10% significance level. Thus the test does not provide enough evidence at 5% level to support the claim that there is a significant change in the mean percentage of visits from search engines. More data are necessary to evaluate the test hypotheses. Confidence Interval based decision rule: Construct a 95% interval. Use the sample mean, not general mean to compute it. IF null hypothesis is NOT contained in 95% C.I. then reject the null hypothesis at 5% significance level in favor of Ha. IF null hypothesis IS contained then you cann reject the null hypothesis at 5% significance level. WHY? The sample data provide a range of plausible values for the general mean. If is not contained in the C.I. there is strong evidence that the population value is not equal to , and therefore we can reject HO.
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# Fourier transform f(x)=sinax/x, a>0
• catcherintherye
In summary, the question involves showing the transform of f(x)=(sinax)/x, where a>0, is 0 for |k|>a and (pi/2)^1/2 for |k|<a. The attempt at a solution involves using l'hopital's rule to evaluate the quotient at x=0 and then finding the integral (-inf, +inf) of exp[-ikx](sinax)/x. The suggestion of using contour integration is mentioned, but it is easier to show the answer by writing sin(ax)/x as (exp(iax)-exp(-iax))/2ix and comparing it to the Fourier transform of a step function.
## Homework Statement
I am trying to show given f(x)=(sinax)/x, a>0
that the transform is 0, |k|>a
(pi/2)^1/2, |k|<a
## The Attempt at a Solution
so far i have f transform =1/(2pi)^1/2.[integral from -inf to +inf]exp[-ikx](sinax)/x.dk, i am concerned about the singularity at x =0, does this compel me to use contour integration?
There's no singularity at 0 for sin(x)/x. (Use l'hopital's rule to give Lim x->0 =1)
okay so I've used l'hopital to evaluate the quotient, so this tells me my rational function q->1 as x-> 0 right ? but i don't see how this helps me in the evaluation of the transform? I suppose I have to findintegral (-inf, +inf) of exp[-ikx](sinax)/x. First I use the fact that my integrand is an even function, I've ended up with lim R->inf of the integral(0,R) of (sinaxcoskx)/xdx. not really sure if this right and the form of the answer seems to suggest that contour integration was used??
Last edited:
You could do contour integration, but since you know the answer, it's easier to show it's right. Write sin(ax)/x as (exp(iax)-exp(-iax))/2ix. Now compare this expression with $$\int^a_{-a} e^{i k x} dk$$. Do you see the relation between your function and the Fourier transform of a step function?
## 1. What is a Fourier transform?
A Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It is often used in signal processing and image analysis to analyze the frequency components of a given function.
## 2. What is the significance of "a" in the given function?
The "a" in the given function represents the frequency of the sine wave. A higher value of "a" will result in a higher frequency and a lower value of "a" will result in a lower frequency. The value of "a" is typically given in units of cycles per unit distance.
## 3. Why is "a" required to be greater than 0?
The function given, f(x)=sinax/x, is known as the sinc function and it has a singularity at x=0. Therefore, in order to avoid any mathematical complications, "a" must be greater than 0 so that the function is well-behaved and continuous.
## 4. What is the domain and range of the given function?
The domain of the given function is all real numbers except for x=0. The range of the function will depend on the value of "a" and will be between -1 and 1.
## 5. How is the Fourier transform of this function calculated?
The Fourier transform of f(x)=sinax/x can be calculated using the Fourier transform formula. However, since the function is not integrable, the Fourier transform cannot be directly calculated. Instead, it can be approximated using numerical methods such as the Fast Fourier Transform algorithm.
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Python: Get the items from a given list with specific condition - w3resource
Python: Get the items from a given list with specific condition
Python List: Exercise - 164 with Solution
Write a Python program to get the items from a given list with specific condition.
Number of Items of the following list which are even and greater than 73
[12,45,23,67,78,90,45,32,100,76,38,62,73,29,83]
Output:
5
Sample Solution:
Python Code:
``````def first_index(l1):
return sum(1 for i in l1 if (i> 45 and i % 2 == 0))
nums = [12,45,23,67,78,90,45,32,100,76,38,62,73,29,83]
print("Original list:")
print(nums)
n = 45
print("\nNumber of Items of the said list which are even and greater than",n)
print(first_index(nums))
```
```
Sample Output:
```Original list:
[12, 45, 23, 67, 78, 90, 45, 32, 100, 76, 38, 62, 73, 29, 83]
Number of Items of the said list which are even and greater than 45
5
```
Pictorial Presentation:
Flowchart:
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Python Code Editor:
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Test your Python skills with w3resource's quiz
Python: Tips of the Day
Floor Division:
When we speak of division we normally mean (/) float division operator, this will give a precise result in float format with decimals.
For a rounded integer result there is (//) floor division operator in Python. Floor division will only give integer results that are round numbers.
```print(1000 // 300)
print(1000 / 300)```
Output:
```3
3.3333333333333335```
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$t=-\dfrac{1}{5}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|6-5t|=|5t+8| ,$ use the definition of an absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $|x|=|y|$ implies $x=y \text{ or } x=-y,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 6-5t=5t+8 \\\\\text{OR}\\\\ 6-5t=-(5t+8) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 6-5t=5t+8 \\\\ -5t-5t=8-6 \\\\ -10t=2 \\\\ t=\dfrac{2}{-10} \\\\ t=-\dfrac{1}{5} \\\\\text{OR}\\\\ 6-5t=-(5t+8) \\\\ 6-5t=-5t-8 \\\\ -5t+5t=-8-6 \\\\ 0=-14 \text{ (FALSE)} .\end{array} Hence, $t=-\dfrac{1}{5} .$
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Physics Help Forum What precisely are "inertial" forces?
General Physics General Physics Help Forum
Aug 22nd 2017, 07:08 PM #1 Senior Member Join Date: Nov 2013 Location: New Zealand Posts: 510 What precisely are "inertial" forces? This concept seems to be hanging me. It's used in a lot of D'Alembert problems and even on various texts on the web. The wiki page on "D'Alembert's principle" points to a page called "fictitious forces". https://en.wikipedia.org/wiki/Fictitious_force Except the wiki page on this starts talking about accelerated frame's of references which aren't the sort of problems I have been working on. So some questions...Are D'Alembert forces real or fictitious forces ? What makes these forces special and why do we need them? What advantage is D'Alembert's method over just applying Newton's second law? Does it make sense to use D'Alembert's method without employing generalised co-ordinates?
Aug 22nd 2017, 07:24 PM #2
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Originally Posted by kiwiheretic This concept seems to be hanging me. It's used in a lot of D'Alembert problems and even on various texts on the web. The wiki page on "D'Alembert's principle" points to a page called "fictitious forces". https://en.wikipedia.org/wiki/Fictitious_force
WARNING! Whomever wrote that page made a mistake. The inertial forces from D'Alembert's principle are not the same thing as inertial forces in the Wikipedia page it links to.
Originally Posted by kiwiheretic Except the wiki page on this starts talking about accelerated frame's of references which aren't the sort of problems I have been working on.
Exactly. Didn't I mention this to you before? The term inertial force is an overloaded term which means it has two different meanings. Context determines which one you want.
Originally Posted by kiwiheretic Are D'Alembert forces real or fictitious forces ?
Real
Originally Posted by kiwiheretic What makes these forces special and why do we need them?
One inertial force which is referred to as the Coriolis force is because its what causes certain weather phenomena. See: https://www.nationalgeographic.org/e...riolis-effect/
Another inertial force of renown is the centrifugal force. Its what gives the spaceship inhabitants in the movie "Martian" the effect of gravitational field so they can walk around that part of the ship. It's for reasons like these that they were of great importance to Einstein when he was working on GR
Originally Posted by kiwiheretic What advantage is D'Alembert's method over just applying Newton's second law?
In certain instances it can make the formulation of the mathematical problem easier to formulate.
Originally Posted by kiwiheretic Does it make sense to use D'Alembert's method without employing generalised co-ordinates?
Not really. You can certainly do it. You just make it less generalizable and thus harder to apply.
Aug 22nd 2017, 07:36 PM #3 Senior Member Join Date: Nov 2013 Location: New Zealand Posts: 510 So then does D'Alembert's principle have anything useful to say about problems involving inclined planes, rolling balls, frictionless sliding blocks and pulley's where no accelerated frames are involved?
Aug 22nd 2017, 07:43 PM #4
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Originally Posted by kiwiheretic So then does D'Alembert's principle have anything useful to say about problems involving inclined planes, rolling balls, frictionless sliding blocks and pulley's where no accelerated frames are involved?
I don't know what you consider useful. As I said, sometimes it just makes solving a problem easier. Is that useful?
Aug 22nd 2017, 08:49 PM #5 Senior Member Join Date: Nov 2013 Location: New Zealand Posts: 510 By making a dynamics problem a static one by putting the problem in an accelerated frame of reference?
Aug 22nd 2017, 09:08 PM #6
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Do you recall the book I mentioned by Lanczos on the variational principles of mechanics? Its a great book on D'Alembert's principle and addresses what you wish to know.
I should state that while I have read and studied D'Alembert's principle of virtual work in graduate school it was a very long time ago and I don't recall much of it. I did reread that part of Lanczos and recommend you find it online and read that section. I cannot stress that enough. Let me quote part of it here so you get how great that text is. From page 93
D'Alembert's principle is fundamental in still another respect; it makes possible the use of ,moving reference systems, and is thus a forerunner of Einstein's revolutionary ideas concerning the relativity of motion, explaining - within the scope of Newtonian physics - the nature of those "apparent forces" which are present in moving systems.
The author also points out that Hamilton's principle can be derived from D'Alembert's principle by a mathematical transformation. Hamilton's principle is restricted to holonomic systems whereas D'Alembert's principle applies to both holonomic and non-holonomic systems. My back is killing me now so I'll let you look those terms up yourself.
By the way. When I suggest taking a look at a particular text/book its always for a very good reason. Some books explain certain things exceptionally well. I'd quote them but it takes time and its painful for me to post lately since my HDTV bit the dust. I used it as a monitor which made things easy to read and thus post. Now I'm restricted to smaller monitor and have to sit on the floor making it very painful for me to post right now. The idea here is that reading that text is better all around rather than having me quote it or try to recall things from that dusty part of my memory.
On another note - I have three cats. Two are brothers. Right now one is licking the butt of the other. Now that is what I call brotherly love. Lol!
Last edited by Pmb; Aug 22nd 2017 at 09:27 PM.
Aug 22nd 2017, 09:33 PM #7 Senior Member Join Date: Nov 2013 Location: New Zealand Posts: 510 Thanks, I'll try the Taylor's book first and work through some of the problems in Chaper 6 Calculus of Variations, Chapter 7 Lagrange Equations and possibly Chapter 8 Two Body Central Force problems and then decide where to from there. I am hoping this pursuing the "holy grail" of "not relying on co-ordinate systems" is really worth it and the advantages of generalised co-ordinates hasn't been oversold.
Aug 22nd 2017, 10:16 PM #8
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Originally Posted by kiwiheretic Thanks, I'll try the Taylor's book first and work through some of the problems in Chaper 6 Calculus of Variations, Chapter 7 Lagrange Equations and possibly Chapter 8 Two Body Central Force problems and then decide where to from there. I am hoping this pursuing the "holy grail" of "not relying on co-ordinate systems" is really worth it and the advantages of generalised co-ordinates hasn't been oversold.
Generalized coordinates is extremely important.
Aug 22nd 2017, 10:25 PM #9
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Originally Posted by kiwiheretic Are D'Alembert forces real or fictitious forces ?
It's not a good idea to refer to inertial forces as "fictitious". There's significant caution in the physics literature against doing that, although this is a point of disagreement among physicists.
The point on thinking of inertial forces as not being "real" is addressed in that text I mentioned, i.e. The Variational Principles of Mechanics - 4th Ed., Cornelius Lanczos, Dover Pub., page 98.
Whenever the motion of the reference system generates a force which has to be added to the relative force of inertia I’, measured in that system, we call that force an “apparent force.” The name is well chosen, inasmuch as that force does not exist in the absolute system. The name is misleading, however, if it is interpreted as a force which is not as “real” as any given physical force. In the moving reference system the apparent force is a perfectly real force, which is not distinguishable in its nature from any other impressed force. Let us suppose that the observer is not aware of the fact that his reference system is in accelerated motion. Then purely mechanical observations cannot reveal to him that fact.
From Introducing Einstein's Relativity, by Ray D'Inverno, Oxord/Clarendon Press, (1992) page 122
Notice that all inertial forces have the mass as a constant of proportionality in them. The status of inertial forces is again a controversial one. One school of thought describes them as apparent or fictitious which arise in non-inertial frames of reference (and which can be eliminated mathematically by putting the terms back on the right hand side). We shall adopt the attitude that if you judge them by their effects then they are very real forces. [Author gives examples]
Last edited by Pmb; Aug 22nd 2017 at 10:27 PM.
Tags forces, inertial, precisely
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# 66636
## 66,636 is an even composite number composed of three prime numbers multiplied together.
What does the number 66636 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 24 divisors.
66636 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twenty-four divisors.
## Prime factorization of 66636:
### 22 × 33 × 617
(2 × 2 × 3 × 3 × 3 × 617)
See below for interesting mathematical facts about the number 66636 from the Numbermatics database.
### Names of 66636
• Cardinal: 66636 can be written as Sixty-six thousand, six hundred thirty-six.
### Scientific notation
• Scientific notation: 6.6636 × 104
### Factors of 66636
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 622
### Divisors of 66636
• Number of divisors d(n): 24
• Complete list of divisors:
• Sum of all divisors σ(n): 173040
• Sum of proper divisors (its aliquot sum) s(n): 106404
• 66636 is an abundant number, because the sum of its proper divisors (106404) is greater than itself. Its abundance is 39768
### Bases of 66636
• Binary: 100000100010011002
• Base-36: 1FF0
### Squares and roots of 66636
• 66636 squared (666362) is 4440356496
• 66636 cubed (666363) is 295887595467456
• The square root of 66636 is 258.1394971715
• The cube root of 66636 is 40.5417949883
### Scales and comparisons
How big is 66636?
• 66,636 seconds is equal to 18 hours, 30 minutes, 36 seconds.
• To count from 1 to 66,636 would take you about eighteen hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 66636 cubic inches would be around 3.4 feet tall.
### Recreational maths with 66636
• 66636 backwards is 63666
• 66636 is a Harshad number.
• The number of decimal digits it has is: 5
• The sum of 66636's digits is 27
• More coming soon!
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# The relation between the co-efficient of discharge (Cd), co-efficient of contraction (Cc), and co-efficient of velocity (Cv) for an orifice-meter is given by equation:
This question was previously asked in
JSSC JE Mechanical Re-Exam 23 Oct 2022 Official Paper-II
View all JSSC JE Papers >
1. Cd = CvCc
2. C= 2 CvCc
3. $$C_d=\frac{C_c}{C_v}$$
4. $$C_d=\frac{C_v}{C_c}$$
Option 1 : Cd = CvCc
Free
JSSC JE Full Test 1 (Paper 1)
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120 Questions 360 Marks 120 Mins
## Detailed Solution
Coefficient of Velocity(Cv) is defined as the ratio between the actual velocity of a jet of liquid at vena-contracta and the theoretical velocity of the jet.
$${C_v} = \frac{V}{{\sqrt {2gH} }}$$
Co-efficient of contraction(Cc) is defined as the ratio of the area of the jet at vena-contracta to the area of the orifice.
$${C_c} = \frac{{{a_c}}}{a}$$
Co-efficient of Discharge (Cd) is defined as the ratio of the actual discharge from an orifice to the theoretical discharge from the orifice.
By the relation
Cd = Cc × Cv
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Question
# The intercept cut-off by a line from y-axis is twice than from x-axis and the line passes through the point (1, 2). The equation of the line is (a) 2x + y = 4 (b) 2x + y + 4 = 0 (c) 2x – y = 4 (d) 2x – y + 4 = 0
Open in App
Solution
## Let us suppose line make intercept "a" an x-axis. Then according to given condition, it makes intercept 2a an y-axis i.e. equation of the line is given by $\frac{x}{a}+\frac{y}{2a}=1\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.2x+y=2a$ also given line passes through (1, 2) i.e. 2(1) + 2 = 2a i.e. 4 = 2a i.e. a = 2 ∴ equation of line is 2x + y = 4 Hence, the correct answer is option A.
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# How many moles are there in a certain volume of a gaseous substance ?
To calculate the number of moles of a gaseous substance in a certain volume, we use the relationship that one mole of any gaseous substance under NTP conditions occupies 22.4 liter (L), or 22400 ml of volume. Let,
Volume of the gaseous substances at NTP= VL
Molar volume under NTP= 22.4 L/mol
So, No. of moles of the substances in V litre
= Volume (in litres) of the substance at NTP/ 22.4 litre/mol
= V litre / 22.4 litre/ mol =V/22.4 mol
And No. of molecules in Very litre of a gas
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# Fight Finance
#### CoursesTagsRandomAllRecentScores
A 180-day Bank Accepted Bill has a face value of $1,000,000. The interest rate is 8% pa and there are 365 days in the year. What is its price now? A 90-day Bank Accepted Bill (BAB) has a face value of$1,000,000. The simple interest rate is 10% pa and there are 365 days in the year. What is its price now?
A 30-day Bank Accepted Bill has a face value of $1,000,000. The interest rate is 8% pa and there are 365 days in the year. What is its price now? A 90-day Bank Accepted Bill has a face value of$1,000,000. The interest rate is 6% pa and there are 365 days in the year. What is its price?
A 60-day Bank Accepted Bill has a face value of $1,000,000. The interest rate is 8% pa and there are 365 days in the year. What is its price now? A 30-day Bank Accepted Bill has a face value of$1,000,000. The interest rate is 2.5% pa and there are 365 days in the year. What is its price now?
On 27/09/13, three month Swiss government bills traded at a yield of -0.2%, given as a simple annual yield. That is, interest rates were negative.
If the face value of one of these 90 day bills is CHF1,000,000 (CHF represents Swiss Francs, the Swiss currency), what is the price of one of these bills?
Refer to the below graph when answering the questions.
Which of the following statements is NOT correct?
A 90 day bank bill has a face value of $100,000. Investor A bought the bill when it was first issued at a simple yield to maturity of 3% pa and sold it 20 days later to Investor B who expected to earn a simple yield to maturity of 5% pa. Investor B held it until maturity. Which of the following statements is NOT correct? A bank bill was bought for$99,000 and sold for \$100,000 thirty (30) days later. There are 365 days in the year. Which of the following formulas gives the simple interest rate per annum over those 30 days?
Note: To help you identify which is the correct answer without doing any calculations yourself, the formulas used to calculate the numbers are given.
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##### Differences
This shows you the differences between two versions of the page.
— nlp-private:dan [2015/04/23 13:32] (current)ryancha created 2015/04/23 13:32 ryancha created 2015/04/23 13:32 ryancha created Line 1: Line 1: + == Other == + + == Items from Qualifying Paper == + * Initialize Gibbs with noisy marginal technique + * Tom Griffiths: measure convergence using Inter-chain metrics (that are immune to label switching) + * Evolution (or lack thereof) over time between samples within the same chain + ** Auto-correlation + *** likelihood within chains + *** favorite clustering metrics (e.g., ARI, unnormalized K-L divergence of each cell in diagonal) + ** Plot these over time (like the divergence movie in a single graph) + * Question about negative correlation between MAP sample and metrics on "comb" + * EM as refinement of Gibbs to climb to mode of local (?) maxima + * Gelman: converged when inter-chain variance is same as intra-chain variance + ** on likelihood + ** on metrics + * Another chain summary idea from Kevin Seppi: most frequent label occurring in last 100 samples + + == Near Term == + * Get comparable likelihood measures for Gibbs and EM + * Implement and run Variational EM + * Do a Tech Report with the full derivation of the collapsed sampler + * Start EM with Gibbs + * Do Tech-Paper with a full derivation of the collapsed sampler + * Develop the "comb" idea + * Implement versions of the partition comparison metrics that can be run on samples (inter-chain and cross-chain) + * Look at the mean entropy metric. Can this be adapted for + * Experiment with feature selectors/dimensionality reducers + * Split out held-out dataset to compute held-out likelihood on Enron + * Auto stop-word detection / feature selection + * Complete bibliography of clustering techniques in prep + + == Longer term: == + + * Reproduce a result from one of the papers (LDA) + * Identify something in the model that can be improved + * Implement differences and write a paper + + + ==CS 601R== + + * Fix held-out set handling for CS 601R + + == Done: == + * 9/15: Present hierachical bisecting k-means clustering algorithm at NLP Lab Meeting + * 9/25: Finish LogCounter (or set it aside for near-term experiments) + * 9/21: Label with name for every PC + * 9/25: Get a copy of Hal's evalutation script + * 9/30: Figure out the profiling situation - JProfiler + * 10/2: Send me your 598R PowerPoint + * 10/6: Subscribe to topic-models at Princeton [https://lists.cs.princeton.edu/mailman/listinfo/topic-models] + * 10/6: Factor clustering away from classification + * 10/6: Hoist computation out of "foreach document" loop and getProbabilities() for anything not specific to the current document. e.g., logDistOfC + * 10/13: Mine Hal's script for good metrics, etc. + * 10/11: Fix Adjusted Rand index calculation + * 11/1: Prepare 10-15 minute presentation detailing your current activities for a CS + Machine Learning audience at the UofU for 11/2. + * 11/3 : Implement P, R, and F_1 as metrics. + * 11/10 : Implement Variation of information metric + + == Brainstorming == + + [[Dan/Brainstorming|Brainstorming List]]
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# Measuring Angles Worksheet Works Answer Key
Subtracting the known angle from 360 provides the measurement of missing angle in the figure. Here is a graphic preview for all of the angles worksheets you can select different variables to customize these angles worksheets for your needs.
Geometry Worksheets Angles Worksheets For Practice And Study In 2020 Geometry Worksheets Angles Worksheet Free Math Worksheets
### Measuring angles and finding x get kids to measure the angle first and then equate the expression to the size of the angle followed by isolating x making it the subject and solving the expression in these pdfs on measuring angles and finding x.
Measuring angles worksheet works answer key. This is a math pdf printable activity sheet with several exercises. Naming angles using three points. Our premium worksheet bundles contain 10 activities and answer key to challenge your students and help them understand each and every topic within their grade level.
View premium worksheets often when they are investigating geometry students are shown angles of different sizes and given various tools to draw them and measure them without. It has an answer key attached on the second page. This worksheet provides the student with a set of angles.
Geometry worksheets angles worksheets for practice and study. This worksheet is a supplementary fourth grade resource to help teachers parents and children at home and in school. The actual angle is provided on the answer key so in addition to identifying whether an angle is obtuse or acute you can suggest that students mesaure the angle with a protractor and supply the measurement as part of their work.
Each angle depicted here can be labeled in two ways keeping the vertex in the middle. Perfect for fourth graders who are learning about lines and angles it asks students to make simple measurements in degrees. His or her job is to use a standard protractor to measure the angles in degrees extending the lines with a straight edge if necessary.
Practice using a protractor and measuring angles in this fill in the blank math worksheet. The options below let you specify the range of angles used on the worksheet. The angles worksheets are randomly created and will never repeat so you have an endless supply of quality angles worksheets to use in the classroom or at home.
Measure angles worksheet for 4th grade children. Each angles worksheet in this section also provide great practice for measuring angles with a protractor. Develop protractor usage skills.
Teachers tutors parents or students can check or validate the solved questions by using the corresponding answers key which comprises the step by step work on how to find the missing angle in the figure by without using the protractor. For a bigger geometry challenge they can see if they are able to find the supplementary angles as well. Practice with these 4th grade pdf worksheets and learn to name an angle using the vertex and two points on its arms.
Finding Missing Angle Worksheet Angles Worksheet Geometry Worksheets Angles Math
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Printable Worksheets Drawing And Measuring Angles Maths Worksheet Angles Worksheet Year 7 Maths Worksheets Measuring Angles Worksheet
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# Fiat Shamir transformation of zero knowledge proof
Consider the following three stage interactive zero knowledge proof
1. The prover sends some information $$a$$ to the verifier.
2. The verifier picks a challenge $$c\in \{0 ,1\}$$
3. Depending on the challenge, the prover responds with $$r(c)$$ that convinces the verifier of his knowledge.
Each round can be won with probability $$\frac{1}{2}$$ by a malicious prover and so we must play this for $$k$$ rounds till $$\frac{1}{2^k}$$ is sufficiently small.
The Fiat Shamir transformation replaces the challenge generation with a random oracle that generates $$c$$ instead. Let $$i$$ be the round number (we play $$k$$ rounds). Typically, one can use a hash function $$c_i = H(p_{i-1}, a_{i-1})$$ for this purpose, where $$p_{i-1}$$ is some input generated from prior rounds and $$a_{i-1}$$ is the previous round's commitment. The prover then generates a transcript of the following form after many rounds
$$T = \{a_i, c_i = H(p_{i-1}, a_{i-1}), r_i\}$$
It is not clear to me why this is secure. In each round, the dishonest prover can simply keep trying different $$a_{i-1}$$ until $$H(p_{i-1}, a_{i-1})$$ is the challenge he is prepared to meet. Since $$c\in\{0,1\}$$ (i.e. the range of $$H$$ is just $$\{0,1\}$$), this does not take him many tries. The prover will only include such challanges (which he is prepared to meet) in the transcript.
What exactly is the correct Fiat Shamir transformation that prevents this attack by a dishonest prover?
• en.wikipedia.org/wiki/Fiat%E2%80%93Shamir_heuristic Oct 7 '19 at 17:17
• @kelalaka, thank you for the link! Unfortunately, I'm still not sure why the cheating strategy I have elaborated in the question fails (or I have misunderstood the Fiat Shamir transform). In particular, the challenge space being so small (two choices only) seems to make it very weak? Oct 7 '19 at 20:31
You are correct about the attack, given the way you have defined the Fiat-Shamir-transformed protocol.
Since Fiat-Shamir is subtle for multi-round protocols, the standard way to resolve this is to use parallel instead of sequential composition. Start with a protocol that has soundness error 1/2, and run it $$k$$ times in parallel (not in sequence) to amplify its soundness error to $$1/2^k$$.
Then apply Fiat-Shamir to the resulting protocol, which just has one round. All the challenge bits are generated from a single call to $$H$$ as $$c_1 \cdots c_k = H(a_1 \cdots a_k)$$. Note that if the prover is proving a false statement, then for any first protocol message $$a_1 \cdots a_k$$ there is at most one $$c_1 \cdots c_k$$ for which she can generate a totally convincing response. With $$q$$ queries to $$H$$, an adversary can find such a valid first message with probability only $$q/2^k$$.
• In short, OP's method proves bit-by-bit (linear time to break), secure method proves all-bits-together, (exponential time to break). Oct 8 '19 at 2:16
The main idea behind the Fiat-Shamir heuristic is to eliminate the interaction in public coin protocols. In the interactive model, the randomly selected challenges by the verifier force a malicious prover to provide a wrong proof. As you mention, it is negligible for a malicious prover to convince the verifier after $$k$$ round.
To make this scheme non-interactive we have to provide a hash function as a Random Oracle model (the output is not distinguishable from a truly random string) to be unpredictable. It means the prover is not able to predict the output of this hash function in each iteration. Since the input challenge for each round is the output of the previous round so the malicious prover can not cheat. To the best of my knowledge, there is no Hash function to be as a ROM.
• Thank you for the answer! However, I'm not sure I see why this statement "Since the input challenge for each round is the output of the previous round so the malicious prover can not cheat" is true. A dishonest prover that can only answer one of the challenges but not both is easy to construct. Assume he can always answer $c= 0$ but not $c = 1$. Now, if $c_i = 1$, he simply tries different $a_{i-1}$ until one of them satisfies $c_i = H(p_{i-1}, a_{i-1}) = 0$. He will find one quite fast since the range of $H$ is just $\{0,1\}$. Oct 7 '19 at 22:40
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Holidays
# of Days
Seasons
Trivia
Math
100
This holiday is the last Thursday in November.
What is Thanksgiving
100
Number of weeks in a year.
What is 52?
100
The number of seasons
What is 4
100
The month school starts
What is August?
100
Two weeks from today
What is the May 24th?
200
A week long festival of lights around Christmastime
What is Hannuka
200
Number of days in September, April, June, and November
What is 30
200
Change of fall to winter, and spring to summer
What is the solstice
200
The month school ends.
What is May?
200
If the date is the 21st, 14 days earlier
What is the 7th
300
Third Monday in January
What is Martin Luther King Day
300
Every fourth year (February 29th)
What is Leap Year
300
Change of winter to spring, and summer to fall
What is an equinox?
300
The length of winter break.
What is two weeks?
300
If today is Sunday, 4 days later
What is Thursday
400
Third Sunday in June
What is Father's Day
400
January, March, May, July, August, October, December all have this in common.
What are months with 31 days?
400
Longest day, shortest night of the year.
What is first day of summer?
400
The months with NO SCHOOL.
What is June and July? (unless you have summer school)
400
Months that have 28 days.
What is every month?
500
A spring day of jokes and pranks
What is April Fool's Day
500
Days in a year
What is 365
500
Longest night, shortest day of the year.
What is First day of winter?
500
Months with the most letters.
What is November and December?
500
When we change the clocks forward.
What is spring?
Click to zoom
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# Weight of Ammonium chloride
## substance volume to weight conversions
### calculate weight of compounds and materials per volume
#### Weight of 1 cubic centimeter of Ammonium chloride
carat 7.64 ounce 0.05 gram 1.53 pound 0 kilogram 0 tonne 1.53 × 10-6 milligram 1 527.4
#### How many moles in 1 cubic centimeter of Ammonium chloride?
There are 28.56 millimoles in 1 cubic centimeter of Ammonium chloride
#### The entered volume of Ammonium chloride in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
• Reference (ID: 110)
• 1. National Center for Biotechnology Information; U.S. National Library of Medicine; 8600 Rockville Pike; Bethesda, MD 20894 USA.
• 2. Handbook of Inorganic Chemicals by Pradyot Patnaik (Author). Published by McGraw-Hill Professional; 1 edition (November 20, 2002).
• Disclaimer of Liability: Technical data present herein is furnished without charge or obligation and is given and accepted at recipient’s sole risk. Because conditions of measurements may vary and are beyond our control, this site or its parent company AVCalc LLC makes no representation about, and is not responsible or liable for the accuracy or reliability of data and calculation results associated with any particular property of any product, substance or material described herein.
#### Foods, Nutrients and Calories
HENNING'S, AGED CHEDDAR, UPC: 018482020003 contain(s) 393 calories per 100 grams or ≈3.527 ounces [ price ]
PUMPKIN SPICE, UPC: 852965001123 contain(s) 394 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Zinc, Zn and Recommended Dietary Allowances (RDAs) for Zinc
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Florida Crushed Coral weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Monosodium carbonate [CH2NaO3] weighs 2 540 kg/m³ (158.56702 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-407C, liquid (R407C) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
Gram per square nanometer (kg/nm²) is a metric measurement unit of surface or areal density [ kilogram per square nanometer ]
The surface density of a two-dimensional object, also known as area or areal density, is defined as the mass of the object per unit of area.
ft² to mm² conversion table, ft² to mm² unit converter or convert between all units of area measurement.
#### Calculators
Calculate gravel and sand coverage in a cylindrical aquarium
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# 1998 AJHSME Problems/Problem 1
## Problem
For $x=7$, which of the following is the smallest?
$\text{(A)}\ \dfrac{6}{x} \qquad \text{(B)}\ \dfrac{6}{x+1} \qquad \text{(C)}\ \dfrac{6}{x-1} \qquad \text{(D)}\ \dfrac{x}{6} \qquad \text{(E)}\ \dfrac{x+1}{6}$
## Solution
### Solution 1
Plugging $x$ in for every answer choice would give
$\text{(A)}\ \dfrac{6}{7} \qquad \text{(B)}\ \dfrac{6}{8} \qquad \text{(C)}\ \dfrac{6}{6} \qquad \text{(D)}\ \dfrac{7}{6} \qquad \text{(E)}\ \dfrac{8}{6}$
From here, we can see that the smallest is answer choice $\boxed{B}$
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https://math.answers.com/Q/What_is_the_average_of_4_and_6_and_11
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0
# What is the average of 4 and 6 and 11?
Wiki User
2012-05-24 20:49:42
Best Answer
( 4 + 6 + 11 )/3
= 7
===
Wiki User
2012-05-24 20:49:42
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https://www.gurufocus.com/term/Total%20Liabilities/MSFT/Total%252BLiabilities/Microsoft+Corporation
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Switch to:
Microsoft Corp (NAS:MSFT) Total Liabilities: \$168,692 Mil (As of Jun. 2017)
Microsoft Corp's total liabilities for the quarter that ended in Jun. 2017 was \$168,692 Mil.
Microsoft Corp's quarterly total liabilities declined from Dec. 2016 (\$155,801.00 Mil) to Mar. 2017 (\$155,288.00 Mil) but then increased from Mar. 2017 (\$155,288.00 Mil) to Jun. 2017 (\$168,692.00 Mil).
Microsoft Corp's annual total liabilities increased from Jun. 2015 (\$94,389.00 Mil) to Jun. 2016 (\$121,471.00 Mil) and increased from Jun. 2016 (\$121,471.00 Mil) to Jun. 2017 (\$168,692.00 Mil).
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Microsoft Corp Annual Data
Jun08 Jun09 Jun10 Jun11 Jun12 Jun13 Jun14 Jun15 Jun16 Jun17 Total Liabilities 63,487.00 82,600.00 94,389.00 121,471.00 168,692.00
Microsoft Corp Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Total Liabilities 121,471.00 142,152.00 155,801.00 155,288.00 168,692.00
Calculation
Total liabilities are the liabilities that the company has to pay others. It is a part of the balance sheet of a company that shareholders do not own, and would be obligated to pay back if the company liquidated.
Microsoft Corp's Total Liabilities for the fiscal year that ended in Jun. 2017 is calculated as
Total Liabilities = Total Current Liabilities + Total Long Term Liabilities = Total Current Liabilities + (Long-Term Debt & Capital Lease Obligation + Other Long Term Assets = 64527 + (76073 + 17184 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 10908 + 0 + 0) = 168,692
Total Liabilities = Total Assets (A: Jun. 2017 ) - Total Equity (A: Jun. 2017 ) = 241086 - 72394 = 168,692
Microsoft Corp's Total Liabilities for the quarter that ended in Jun. 2017 is calculated as
Total Liabilities = Total Current Liabilities + (Total Long Term Liabilities) = Total Current Liabilities + (Long-Term Debt & Capital Lease Obligation + Long-Term Debt & Capital Lease Obligation = 64527 + (76073 + 17184 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 10908 + 0 + 0) = 168,692
Total Liabilities = Total Assets (Q: Jun. 2017 ) - Total Equity (Q: Jun. 2017 ) = 241086 - 72394 = 168,692
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
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insurance claims: evaluating period of enrollment prior to diagnosis date
Occasional Contributor
Posts: 5
insurance claims: evaluating period of enrollment prior to diagnosis date
hello - I'm stuck on something I was hoping to get some help with.
I am working with insurance claims data, in this case, the enrollment file. I have 48 months worth of data, and am trying to do two things:
1. for each id, determine whether the person had a continuous 12 or more month period of enrollment at any time during the 48 months, and if so, store the first and last months of enrollment (there may be gaps in enrollment, so a given id may have more than one 12+ month continuous period).
2. for each id, evaluate the dx_month, and see if the person had 12 months of continuous enrollment for the 12 months prior. So for a person diagnosed in month 13, was he continuously enrolled between months 1 and 12.
• dx_month = month during the 4-yr period that the person was diagnosed with hypertension
• e1-e26 = binary variables indicating enrollment during each month (1=yes). My dataset actually has e1-e48 but I didn't include all here.
data temp;
input id dx_month e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15 e16
e17 e18 e19 e20 e21 e22 e23 e24 e25 e26;
cards;
101 13 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1
102 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
103 15 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0
104 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0
105 22 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0
106 11 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0
107 20 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1
Any help would be greatly appreciated. I've tried lags, macros, etc. but am not getting it to work.. Thank you.
Super Contributor
Posts: 3,176
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Using an ARRAY for your Enn variables, have a DO/END loop that tests the prior 12 variable-values in the array to check for a 1 -- if any are 0, then you have failed to match a qualification.
Scott Barry
SBBWorks, Inc.
data step array site:sas.com
Occasional Contributor
Posts: 5
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
thank you for the reply Scott. l used arrays to set the e1-e48 indicators, but do not have much experience with syntax for them...specifically, I am getting an 'array subscript out of range' error message when trying to evaluate the specific values (i.e., if arrayname{i} = 0 then DO...). I wrote my 'DO i=' statement as
do i = (dx_month-12) to (dx_month-1);
Can you (or anyone else), help further with this?
thank you..
Super Contributor
Posts: 3,176
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Did you by opportunity review the SAS support website technical document references - those revealed with the Google advanced search argument provided?
Scott Barry
SBBWorks, Inc.
http://www2.sas.com/proceedings/sugi30/242-30.pdf
Occasional Contributor
Posts: 5
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
yes
Super Contributor
Posts: 578
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
How is your enrollment data structured? I would recommend restructuring it to a vertical format like:
id month_start month_end
101 1 12
101 15 48
102 1 48
103 9 22
etc
proc sql;
create table work.dx_history as
select t1.id, t1.dx_month, intck('month',t2.month_start,t1.dx_month) as Months_Continuous_Enrollment_Prior_to_dx
from work.temp t1 inner join work.enrollment_vertical t2
on t1.id=t2.id and t2.dx_month between t2.month_start and t2.month_end;
quit;
Not quite where you were going...but it shouldn't be too hard to restructure your enrollment data.
Super User
Posts: 10,787
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
I give you answer code for Question One.
I will give code for Question Two tomorrow.I have to leave now;
I make a variable e_end to flag the end for lines;
[pre]
data temp;
input id dx_month e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15 e16
e17 e18 e19 e20 e21 e22 e23 e24 e25 e26 e_end;
cards;
101 13 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0
102 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0
103 15 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0
104 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0
105 22 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0
106 11 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0
107 20 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0
;
run;
data QuestionOne;
set temp;
array e{*} e:;
do i=1 to dim(e)-12;
if e{i} = 1 then do;
j=i+1;count=0;count+1;
do while(e{j}=1);
j+1;count+1;
end;
if count ge 12 then do;
range=catx('-',vname(e{i}),vname(e{j-1}));
output;
end;
end;
end;
keep id range;
run;
[/pre]
Ksharp
Message was edited by: Ksharp
Message was edited by: Ksharp
PROC Star
Posts: 8,167
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
The following attempts to correct for one oversight in ksharp's suggestion and also offers a possible suggestion for your other question:
data QuestionOne;
set temp;
array e{*} e:;
do i=1 to dim(e)-12;
if e{i} = 1 then do;
j=i+1;count=0;count+1;
do while(j lt dim(e) and e{j}=1);
j+1;count+1;
end;
if count ge 12 then do;
range=catx('-',vname(e{i}),vname(e{j-1}));
output;
end;
end;
end;
keep id range;
run;
%let nmonths=26;
data QuestionTwo (drop=i);
set temp;
array enroll(*) e1-e&nmonths.;
if dx_month le 12 then months=0;
else do;
do i=1 to &nmonths.;
if not(i ge dx_month-12 and i le (dx_month-1))
then enroll(i)=0;
end;
months=sum(of enroll(*));
if months eq 12 then Met_Criterion=1;
end;
run;
Super User
Posts: 10,787
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Hi.
Art.T ,that is not to need to add it.Because I have make a variable e_end which is always to equal zero,So the condition in loop will be false finally.
[pre]
do while(j lt dim(e) and e{j}=1);
[/pre]
And For the Second Question,I think it will be easy after yielding my dataset, So you can do it by yourself.
Ksharp Message was edited by: Ksharp
Occasional Contributor
Posts: 5
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Art - thanks...I didn't think the calculation through completely before I posted that..I understand what the code is doing now. thank you..
KSharp - thanks for your help also.
Regular Contributor
Posts: 241
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Adding a dummy element at both (or either) ends of the array -- as Ksharp does -- is a handy coding technique worth learning. Below finds all the enrollment spans. HTH.
/* test data */
data temp;
input id dx_month e1-e26;
cards;
101 13 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1
102 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
103 15 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0
104 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0
105 22 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0
106 11 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0
107 20 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1
108 99 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
;
run;
/* find all the countinous enrollment "spans" */
data spans;
set temp;
keep id spanId start finish length;
call missing(sentinel, spanId, start, finish, length);
array enrolled[1:27] e1-e26 sentinel;
do m = lbound(enrolled) to hbound(enrolled);
if start & ^enrolled then do;
length = m - start;
finish = m - 1;
output;
start = .;
endelse if ^start & enrolled then do;
spanId + 1;
start = m;
end;
end;
run;
/* check */
proc print data=spans;
where length >= 12;
run;
/* on lst
span
Obs id Id start finish length
1 101 1 1 12 12
2 101 2 15 26 12
3 102 1 1 26 26
4 103 1 9 22 14
5 104 1 1 15 15
6 106 1 3 19 17
7 107 1 1 12 12
8 107 2 15 26 12
*/
Occasional Contributor
Posts: 5
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Chung - thank you. I saw also that this is picking up the second enrollment spans for the first and last records.
Art297...for the QuestionTwo code, 'months' for ID #107 is 10, but should be 5 months of continuous enrollment (there was a 2-month gap in between). I see why it returned 10 (because of the sum statement). How can I sum starting with e{dx_month-1} back to e{dx_month-12}, but stopping if a '0' is encountered?
Thanks very much to all who have responded...shows the variety of ways to approach it.
Judy
PROC Star
Posts: 8,167
Re: insurance claims: evaluating period of enrollment prior to diagnosis date
Judy,
Since your specs indicated that you were only interested in finding those records with 12 consecutive months, the only relevant number from the count (for Question 2) was whether it was 12 or not.
Art
----------
> Chung - thank you. I saw also that this is picking
> up the second enrollment spans for the first and last
> records.
>
> Art297...for the QuestionTwo code, 'months' for ID
> #107 is 10, but should be 5 months of continuous
> enrollment (there was a 2-month gap in between). I
> see why it returned 10 (because of the sum
> statement). How can I sum starting with
> e{dx_month-1} back to e{dx_month-12}, but stopping if
> a '0' is encountered?
>
> Thanks very much to all who have responded...shows
> the variety of ways to approach it.
>
> Judy
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# Inverse projection mapping
Hello,
I have the following situation. I have a projector. Also I have calculate a view and a projection matrix. Now I can do a projaction mapping. But I would like the inverse way. I have the texture (maybe 512 x 512) and each pixel would describe by a texture coordinate. For example the pixel on 256 x 256 has the u = 0.5 and v = 0.5. That is what I known. Can I calculate the point in world space with the uv-coordinate, projection matrix and the view matrix? Is it possible?
Thanks, Martin
My take:
Let P = projection matrix , V = view matrix and t = tex coord.
Let t=(0,0) for actual projected point p(0,0)
t(0,1) = p(0,512) [because your image is 512*512]
t(1,0) = p(512,0) and t(1,1) = p(512,512)
Now apply P inverse on all points to get points in view space.
Then apply V inverse to get them in world space.
p(x,y) in world space --> p(m,n) in view space --> p(a,b) on Image after projection. Now p(a,b) is assigned texel coordinate
t(s,t).Thus you have the correspondence.
Problem occurs when you do not know the transformation matrices.In that case you need to get correspondence data of of more than one texel to find out a parameter in the matrix and use equations.
Hope this helps.
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# A company has the task of producing spheres of volume `288pi pm 5` cm^3. Use integration to find the radius of the spheres, writing this in the form `a pm b` cm with b rounded to three decimal...
A company has the task of producing spheres of volume `288pi pm 5` cm^3.
Use integration to find the radius of the spheres, writing this in the form `a pm b` cm
with b rounded to three decimal places.
### 1 Answer |Add Yours
mathsworkmusic | (Level 2) Educator
Posted on
The formula for the volume of a sphere is
`V = 4/3pir^3`
Think of this as adding two hemispheres together, where each of those hemispheres is the sum of many many circle slices/discs from `x=0` ro `r` where the radius of the disc at `x ` is (using Pythagoras) `sqrt(r^2-x^2)` . Then, since the area of each disc is `pi(r^2-x^2)` and integrating over `x` we get
`V = 2 times int_0^r pi (r^2-x^2) dx`
`= 2 pi times (xr^2 - 1/3 x^3|_0^r = 2pi times (r^3 - 1/3r^3) = 4/3pir^3`
Now, we have that the volume ` `of the spheres in question is
`V = 288pi pm 5` cm^3
`= 4/3pi(288(3/4) pm 5/pi(3/4)) = 4/3pi(216 pm 15/(4pi))`
Therefore
`r^3 = 216 pm 15/4pi`
The minimum `r` can be is ` ``root(3)(216-15/(4pi)) =5.989`
The maximum `r` can be is `root(3)(216+15/(4pi)) = 6.011`
Therefore
`r = 6 pm 0.011` cm
a = 6 and b = 0.011 to 3dp
We’ve answered 317,406 questions. We can answer yours, too.
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A143980 Binomial transform of A079260. 1
0, 0, 0, 0, 1, 6, 21, 56, 126, 252, 462, 792, 1288, 2016, 3108, 4928, 8569, 17154, 38931, 94164, 229824, 550088, 1278662, 2884752, 6335005, 13590930, 28575315, 59014620, 119878606, 239662236, 471605976, 913302656, 1740247806, 3262146492, 6015853242 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,6 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..1000 FORMULA a(n) = Sum_{k=5..n} C(n,k) * A079260(k). EXAMPLE a(13) = [1287,1716,1716,1287,715,286,78,13,1] * [1,0,0,0,0,0,0,0,1] = 1287+1 = 1288. MAPLE bintrans:= proc(p) proc (n) add (p(k) *binomial(n, k), k=0..n) end end: f:= proc(n) if isprime(n) and modp(n, 4)=1 then 1 else 0 fi end: a:= bintrans(f): seq (a(n), n=1..40); CROSSREFS Cf. A007318, A079260. Sequence in context: A341203 A090581 A000389 * A140228 A264926 A006090 Adjacent sequences: A143977 A143978 A143979 * A143981 A143982 A143983 KEYWORD nonn AUTHOR Alois P. Heinz, Sep 06 2008 STATUS approved
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## molar masses on the test
Kassidy Tran 1E
Posts: 77
Joined: Fri Sep 28, 2018 12:15 am
### molar masses on the test
Will we be given the molar mass of elements on the exam? or will we have to memorize the molar masses of the elements? I noticed that in the module assessments we were given the molar mass for some of the elements, but not all.
Iris Bai 2K
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
### Re: molar masses on the test
We're given a periodic table, so all the molar masses should be on there.
Noh_Jasmine_1J
Posts: 71
Joined: Fri Sep 28, 2018 12:15 am
### Re: molar masses on the test
You can solve for the molar mass if the question gives you the chemical formula because there is a periodic table given for the test
Albert_Luu3K
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am
### Re: molar masses on the test
Even though we'll be given molar masses from the periodic table, it doesn't hurt to memorize some of the more common elements! After doing a whole bunch of practice problems, molar masses of common elements used will just be ingrained in your memory. Oxygen, Carbon, Hydrogen, and Nitrogen are nice ones to remember, so you won't have to go back and forth between the periodic table and your test.
Kyither Min 2K
Posts: 60
Joined: Wed Oct 03, 2018 12:15 am
### Re: molar masses on the test
You'd typically will have to solve for molar mass which you can do by analyzing the molecule and seeing how many of each element there is, and referencing the periodic table, you can solve for molar mass.
whitney_2C
Posts: 74
Joined: Fri Sep 28, 2018 12:28 am
### Re: molar masses on the test
You'll also find that after doing chemistry for a while you'll just naturally memorize the molar masses of elements and compounds. After doing IB chemistry I had the molar masses of carbon, nitrogen, water, hydrogen, sodium hydroxide, and many more memorized without having to even study them because we used them so much, so don't fret over memorization.
Andrewwiner4D
Posts: 32
Joined: Fri Sep 28, 2018 12:28 am
### Re: molar masses on the test
You will be provided a periodic table but will have to calculate molar masses on molecules.
Andrewwiner4D
Posts: 32
Joined: Fri Sep 28, 2018 12:28 am
### Re: molar masses on the test
That reply is by Andrew Winer (4D).^^^
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# Polynomials
## Objective
Identify features of polynomial functions including end behavior, intervals where the function is positive or negative, and domain and range of function.
## Common Core Standards
### Core Standards
?
• F.BF.B.3 — Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.
• F.IF.B.4 — For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. Modeling is best interpreted not as a collection of isolated topics but in relation to other standards. Making mathematical models is a Standard for Mathematical Practice, and specific modeling standards appear throughout the high school standards indicated by a star symbol (★). The star symbol sometimes appears on the heading for a group of standards; in that case, it should be understood to apply to all standards in that group.
• F.IF.B.6 — Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. Modeling is best interpreted not as a collection of isolated topics but in relation to other standards. Making mathematical models is a Standard for Mathematical Practice, and specific modeling standards appear throughout the high school standards indicated by a star symbol (★). The star symbol sometimes appears on the heading for a group of standards; in that case, it should be understood to apply to all standards in that group.
?
• F.IF.B.4
## Criteria for Success
?
1. Describe end behavior as the value of y the function approaches as the value of x either increases to infinity or decreases to negative infinity.
2. Describe how the shape of the function, domain of the function, and end behavior change based on whether the degree of the function is odd or even.
3. Describe the effect the leading coefficient has on the shape of the function and the end behavior of the function.
4. Explain that when all coefficients and terms are unchanged except the degree of the polynomial, the rate of change over a fixed interval will increase with an increasing degree.
## Anchor Problems
?
### Problem 1
What do all of these functions have in common? What's different?
${h(x)=x^2}$ ${k(x)=x^4}$ ${l(x)=x^6}$ ${r(x)=x^8}$
What do all of these functions have in common? What's different?
${t(x)=x}$ ${j(x)=x^3}$ ${m(x)=x^5}$ ${w(x)=x^7}$
### Problem 2
Write a function and draw a sketch that meets the constraints below.
• Your function tends toward negative infinity when $x$ is very small and tends toward positive infinity when $x$ is very large.
• Your function has the marked points as $x$-intercepts.
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
• Include problems where an “always sometimes never” statement is given, such as “the end behavior is the same for even functions.”
1 a. ${y=\frac{1}{2}x^2}$ 2 b. ${y=x^3-8}$ 3 c. ${y=-x^4-x^3+4x+2}$ 4 d. ${-3x^3}$ 5 e. ${3x^5-x^3+4x+2}$
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# Gas concentration from ppm to mg/m³ and vice versa
This online calculator converts the gas concentration value from ppm to mass concentration and back taking into account temperature and pressure
### This page exists due to the efforts of the following people:
#### Timur
Created: 2021-08-20 06:26:17, Last updated: 2021-08-20 06:26:57
The calculator below converts between volume concentration of gas, i.e. the ratio of the gas volume to the total volume, and mass concentration of gas, i.e. the ratio of the mass of the gas to the total volume. Most of such calculators perform calculations for standard temperature and pressure (0°C, 100.00 kPa), but this calculator allows you to set your own temperature and pressure values. A detailed description of the calculation can be found under the calculator.
#### Parts per million (ppm) to milligram per cubic meter (mg/m³) converter for gases
Concentration, ppm
Concentration, mg/m³
Molar mass
Molar volume
Digits after the decimal point: 2
### Conversion from mass concentration to volume concentration and vice versa
As mentioned above, the volume concentration of gas is defined as the ratio of the gas volume to the total volume. It is a dimensionless value, for convenience expressed in percent (%), permille (parts per thousand, ‰), or ppm (parts per million). Sometimes, to emphasize that we are talking about volume concentration, you can see %vol. or ppmv (parts per million by volume), although IUPAC recommends avoiding such notation1.
Ppm is often used for low concentration values, typical for air pollution. By definition, it is the number of gas particles per million air particles, 1 ppm = 1/1000000 = 0.0001% = 0.001‰.
Sometimes, however, the gas concentration is expressed as the mass concentration in milligrams per cubic meter (mg/m3). For example, permissible exposure limits (PEL or OSHA PEL) are expressed both in ppm and in mg/m³.
The conversion from one unit to another is based on the Law of Avogadro's, according to which equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. According to the first corollary from Avogadro's law, one mole of any gas under the same - isobaric and isothermal - conditions occupies the same volume.
You can find the molar volume for a given temperature and pressure using the Clapeyron equation:
$V_m = \frac {RT} {p}$,
where the universal gas constant R ≈ 8.314 J / (mol K)
#### Conversion from ppm to mass concentration
Knowing ppm, we can find how many liters of gas are contained in one cubic meter of total volume. By dividing this number by the molar volume of the gas (for a given temperature and pressure), we can get the number of moles of gas in one cubic meter of total volume. Finally, by multiplying the number of moles of gas by the molar mass of gas, we can obtain the mass of gas in one cubic meter of total volume or mass concentration.
#### Conversion from mass concentration to ppm
By dividing the mass of gas in a cubic meter by the molar mass of gas, we can get the number of moles of gas in a cubic meter. Multiplying the number of moles of gas by the molar volume of gas (for a given temperature and pressure), we get the volume of gas in liters per cubic meter of total volume. Now we only need to express this volume in parts per million, thus obtaining the ppm value.
1. Quantities, Units and Symbols in Physical Chemistry. INTERNATIONAL UNION OF PURE AND APPLIED CHEMISTRY, PHYSICAL CHEMISTRY DIVISION
URL copied to clipboard
PLANETCALC, Gas concentration from ppm to mg/m³ and vice versa
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# QlikView App Dev
Discussion Board for collaboration related to QlikView App Development.
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Not applicable
## Time in SET
Hi friends,
It's been to long since I used time in SET, try to avoid that, but now I have to build my time limits in a chart.
I need to see last period - always previous month, this year and for last year. Trying this but can't seem to get it to work:
Sum({\$<Year = {\$(=Year(Today()))}, Month = {\$(=Month(Today())-1)},Year=, Month=>} Amount))
and
Sum({\$<Year = {\$(=Year(Today())-1)}, Month = {\$(=Month(Today())-1)},Year=, Month=>} Amount))
I also need the acumulated amount from yearstart to monthend previous month...
Best regards
Torbjörn Ungvall (@Ungvall)
1 Solution
Accepted Solutions
Partner
Sum of amount from january to july (previous month):
Sum of amount from august last year to july (previous month):
Cheers!!
Jagan
11 Replies
Master II
I think you don't have to "clear" it with ",Year=, Month=" in the end of your set expression.
Sum({\$<Year = {\$(=Year(Today()))}, Month = {\$(=Month(Today())-1)}>} Amount))
Not applicable
Author
This can be done by putting the periods into date ranges.
Declare two variables
Then use an expression such as this to find the amount
sum({\$< DateMonth = {">=\$(#vDate12)<=\$(#vDateNow)"} >} Amount)
Hope that helps
Partner
Hello,
Please check this file. My data looks like this:
Customer DateField Month Amount
A 01/05/2011 May 200
B 02/06/2011 Jun 300
C 03/07/2011 Jul 400
D 04/08/2011 Aug 500
A 01/05/2012 May 600
B 02/06/2012 Jun 700
C 03/07/2012 Jul 800
D 04/08/2012 Aug 900
Dimension as Customer
Current Year- Previous Month:=Sum({}Amount)Previous Year-Previous month:=Sum({}Amount)
Cheers !!
Jagan
Not applicable
Author
Works just fine - could I use the same for acumulated amount (yearstart to monthend previous month) and rolling 12 month, also to previous monthend?
Best regards
Torbjörn Ungvall (@Ungvall)
Partner
Hi,
I didn't get you what you want. Can you explain with small example what you want exactly. I mean How is your data and what you need the output. Did you got any help from my previous post ? So that i can move forward.
Cheers!!
Jagan
Not applicable
Author
It works just fine!
Can't give you example for various reasons.
I also need the acumulated amount this year:
Sum of amount from january to july (previous month)
and
The amount for rolling 12 month back:
Sum of amount from august last year to july (previous month)
Hope this helps!
Thanks/Toby
Partner
What I do for all these time period comparisons is this...
Sum({\$<\$(='[' & Concat({1<\$Table = {"MasterCalendar"}>}\$Field,'] = , [' ) & '] = '), MasterDate = {"\$(='>=' & MonthStart(Max(MasterDate),-1) & '<=' & AddMonths(Max(MasterDate),-1))"}>} SalesAmount)
So let's break this into parts
• \$(='[' & Concat({1<\$Table = {"MasterCalendar"}>}\$Field,'] = , [' ) & '] = ')
This part overrides all selections on the MasterCalendar table. The problem I always run into with this stuff is that YTD, MTD, Prior MTD and all other pre selected time slices you can think of need to be sensitive to the "current" selections of the user, if they are not handled properly you end up with charts/columns that go blank when the selections are not compatible. So overriding the selections on the calendar are the first part of preventing incompatible selections.
• MasterDate = {"\$(='>=' & MonthStart(Max(MasterDate),-1) & '<=' & AddMonths(Max(MasterDate),-1))"}
This part makes the selection on the primary date field for your data to match whatever range of dates you want. This example gives prior month to date. so if the user has selected July of 2012 you will see ALL of June if the user selects July 2012 up to July 15 then only June1 through June 15. There are lots of variations you can use based on adding and subtracting months or years from the min max or whatever dates are available.
I hope this helps
Chris
Partner
Sum of amount from january to july (previous month):
Sum of amount from august last year to july (previous month):
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# ©Silberschatz, Korth and Sudarshan2.1Database System Concepts Huiswerk Lees delen 3.2, 3.3 van hoofdstuk 3. opgaven voor hoofdstuk 2: modelleeropgave 5.
## Presentation on theme: "©Silberschatz, Korth and Sudarshan2.1Database System Concepts Huiswerk Lees delen 3.2, 3.3 van hoofdstuk 3. opgaven voor hoofdstuk 2: modelleeropgave 5."— Presentation transcript:
©Silberschatz, Korth and Sudarshan2.1Database System Concepts Huiswerk Lees delen 3.2, 3.3 van hoofdstuk 3. opgaven voor hoofdstuk 2: modelleeropgave 5 opgaven voor hoofdstuk 3: maak de queries voor de vragen uit 3.5 in relationele algebra; maak de queries 1-6 voor de bierdrinkerdatabase in tuppel calculus EN in relationele algebra.
©Silberschatz, Korth and Sudarshan2.2Database System Concepts Silberschats 3.5 a employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who work for FBC { t | w works ( t[person-name] =w[person-name] Λ w[company-name]=“FBC” )}
©Silberschatz, Korth and Sudarshan2.3Database System Concepts Silberschats 3.5 b employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names and the cities of residence of all employees who work for FBC { t | e employee (t[person-name] = e[person-name] Λ t[city] = e[city] Λ w works ( w[person-name] =e[person-name] Λ w[company-name]=“FBC”))}
©Silberschatz, Korth and Sudarshan2.4Database System Concepts Silberschats 3.5 c employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names, street address, and the cities of residence of all employees who work for First Bank Corporation and earn more that \$10000 per annum. { t | e employee (t[person-name] = e[person-name] Λ t[city] = e[city] Λ t[street] = e[street] Λ w works ( w[person-name] =e[person-name] Λ w[company-name]=“FBC” Λ w[salary] >10000))} { t | t employee Λ w works ( w[person-name] =t[person-name] Λ w[company-name]=“FBC” Λ w[salary] >10000))}
©Silberschatz, Korth and Sudarshan2.5Database System Concepts Silberschats 3.5 d employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who live in the same city as the company for which they work. { t | w works ( t[person-name] =w[person-name] Λ c company ( c[company-name=w[company-name] Λ e employee ( e[person-name] = w[person-name] Λ e[city] = c[city] )))}
©Silberschatz, Korth and Sudarshan2.6Database System Concepts Silberschats 3.5 e employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who live in the same city and on the same street as do their managers. { t | e employee (t[person-name] = e[person-name] Λ m manages ( m[person-name] = e[person-name] Λ em employee (em[person-name] = m[person-name] Λ e[city] = em[city] Λ e[street] = em[street] )))}
©Silberschatz, Korth and Sudarshan2.7Database System Concepts Silberschats 3.5 f employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees in this database who do not work for FBC. single company assumption { t | w works (t[person-name]=w[person-name] Λ [company-name]≠“FBC”)} Multiple company assumption { t | w works (t[person-name]=w[person-name] Λ w1 works (w[person-name] = w1[person-name] w1[company-name]≠“FBC”))} Which assumption holds according to the definition of the database?
©Silberschatz, Korth and Sudarshan2.8Database System Concepts Silberschats 3.5 g employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Find the names of all employees who earn more than every employee of SBC. { t | w works (t[person-name]=w[person-name] Λ w1 works (w1[company-name] = “SBC” w[salary] > w1[salary] ))}
©Silberschatz, Korth and Sudarshan2.9Database System Concepts Silberschats 3.5 h employee (person-name, street, city) works (person-name, company-name, salary) company (company-name, city) manages (person-name, manager_name) Assume the companies may be located in several cities. Find all companies located in every city in which SBC is located. { t | c company (t[company-name]=c[company-name] Λ c1 company (c1[company-name] = “SBC” c2 company (c2[company-name]=c[company-name] Λ c2[city] = c1[city] )))}
©Silberschatz, Korth and Sudarshan2.10Database System Concepts Reduction of an E-R Schema to Tables Primary keys allow entity sets and relationship sets to be expressed uniformly as tables which represent the contents of the database. A database which conforms to an E-R diagram can be represented by a collection of tables. For each entity set and relationship set there is a unique table which is assigned the name of the corresponding entity set or relationship set. Each table has a number of columns (generally corresponding to attributes), which have unique names. Converting an E-R diagram to a table format is the basis for deriving a relational database design from an E-R diagram.
©Silberschatz, Korth and Sudarshan2.11Database System Concepts Representing Entity Sets as Tables A strong entity set reduces to a table with the same attributes.
©Silberschatz, Korth and Sudarshan2.12Database System Concepts Composite and Multivalued Attributes Composite attributes are flattened out by creating a separate attribute for each component attribute E.g. given entity set customer with composite attribute name with component attributes first-name and last-name the table corresponding to the entity set has two attributes name.first-name and name.last-name A multivalued attribute M of an entity E is represented by a separate table EM Table EM has attributes corresponding to the primary key of E and an attribute corresponding to multivalued attribute M E.g. Multivalued attribute dependent-names of employee is represented by a table employee-dependent-names( employee-id, dname) Each value of the multivalued attribute maps to a separate row of the table EM E.g., an employee entity with primary key John and dependents Johnson and Johndotir maps to two rows: (John, Johnson) and (John, Johndotir)
©Silberschatz, Korth and Sudarshan2.13Database System Concepts Representing Weak Entity Sets A weak entity set becomes a table that includes a column for the primary key of the identifying strong entity set
©Silberschatz, Korth and Sudarshan2.14Database System Concepts Representing Relationship Sets as Tables A many-to-many relationship set is represented as a table with columns for the primary keys of the two participating entity sets, and any descriptive attributes of the relationship set. E.g.: table for relationship set borrower
©Silberschatz, Korth and Sudarshan2.15Database System Concepts Redundancy of Tables Many-to-one and one-to-many relationship sets that are total on the many-side can be represented by adding an extra attribute to the many side, containing the primary key of the one side E.g.: Instead of creating a table for relationship account- branch, add an attribute branch to the entity set account
©Silberschatz, Korth and Sudarshan2.16Database System Concepts Redundancy of Tables (Cont.) For one-to-one relationship sets, either side can be chosen to act as the “many” side That is, extra attribute can be added to either of the tables corresponding to the two entity sets If participation is partial on the many side, replacing a table by an extra attribute in the relation corresponding to the “many” side could result in null values The table corresponding to a relationship set linking a weak entity set to its identifying strong entity set is redundant. E.g. The payment table already contains the information that would appear in the loan-payment table (i.e., the columns loan-number and payment-number).
©Silberschatz, Korth and Sudarshan2.17Database System Concepts Representing Specialization as Tables Method 1: Form a table for the higher level entity Form a table for each lower level entity set, include primary key of higher level entity set and local attributes table table attributes personname, street, city customername, credit-rating employeename, salary Drawback: getting information about, e.g., employee requires accessing two tables
©Silberschatz, Korth and Sudarshan2.18Database System Concepts Representing Specialization as Tables (Cont.) Method 2: Form a table for each entity set with all local and inherited attributes table table attributes personname, street, city customername, street, city, credit-rating employee name, street, city, salary If specialization is total, table for generalized entity (person) not required to store information Can be defined as a “view” relation containing union of specialization tables But explicit table may still be needed for foreign key constraints Drawback: street and city may be stored redundantly for persons who are both customers and employees
©Silberschatz, Korth and Sudarshan2.19Database System Concepts Relations Corresponding to Aggregation To represent aggregation, create a table containing primary key of the aggregated relationship, the primary key of the associated entity set Any descriptive attributes
©Silberschatz, Korth and Sudarshan2.20Database System Concepts Relations Corresponding to Aggregation (Cont.) E.g. to represent aggregation manages between relationship works-on and entity set manager, create a table manages(employee-id, branch-name, title, manager-name) Table works-on is redundant provided we are willing to store null values for attribute manager-name in table manages
Example
©Silberschatz, Korth and Sudarshan2.22Database System Concepts Vertaal naar het relationeel model
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https://mathoverflow.net/questions/303521/is-there-a-connection-nabla-for-which-this-particular-non-geodesible-vector-f?noredirect=1
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# Is there a connection $\nabla$ for which this particular non geodesible vector field $X$ satisfy $\nabla_X X=0$?
Let $X$ be the following vector field on $\mathbb{R}^2\setminus \{0\}$ \begin{align} x' &= x\,(1-x^2-y^2)(x^2+y^2-3) - y\,(2-x^2-y^2)\\ y' &= y\,(1-x^2-y^2)(x^2+y^2-3) + x\,(2-x^2-y^2). \end{align}
It is well known that there is no a Riemannian metric on the punctured plane such that the orbits of the above vector field would be unparametrized geodesics. The proof is based on existence of opposite orientation of two consecutive nested limit cycles in the phase portrait of the vector field.
Now we reduce the above geodesibility requirement to the following question;(Is this an obvious reduced question)?
Is there a (torsion free or metric or arbitrary) connection $\nabla$ on the punctured plane with $\nabla_X X=0$?
• Are you asking for $X$ to satisfy $\nabla X = 0$ (as in the title) or $\nabla_X X = 0$ (as in the body---this is a weaker condition)? – Travis Jun 24 '18 at 16:48
• @Travis Thank you I revise the title. – Ali Taghavi Jun 24 '18 at 19:29
Your vector field can be written in polar coordinates as $$X=X^r\partial_r+X^{\varphi}\partial_{\varphi}=r(4r^2-r^4-3)\partial_r+(2-r^2)\partial_{\varphi},$$ which exhibits the rotational symmetry. The condition $\nabla_X X=0$ implies that \begin{align} X^r\partial_rX^r+\Gamma^r_{rr}(X^r)^2+(\Gamma^r_{r\varphi}+\Gamma^r_{\varphi r}) X^rX^{\varphi}+\Gamma^r_{\varphi\varphi}(X^{\varphi})^2=0, \\ X^r\partial_rX^{\varphi}+\Gamma^{\varphi}_{rr}(X^r)^2+(\Gamma^{\varphi}_{r\varphi}+\Gamma^{\varphi}_{\varphi r}) X^rX^{\varphi}+\Gamma^{\varphi}_{\varphi\varphi}(X^{\varphi})^2=0. \end{align} For a connection with vanishing torsion, we have $\Gamma^r_{r\varphi}=\Gamma^r_{\varphi r}$ and $\Gamma^{\varphi}_{r\varphi}=\Gamma^{\varphi}_{\varphi r}$, which leads to the simplification \begin{align} X^r\partial_rX^r+\Gamma^r_{rr}(X^r)^2+2\Gamma^r_{r\varphi} X^rX^{\varphi}+\Gamma^r_{\varphi\varphi}(X^{\varphi})^2=0, \\ X^r\partial_rX^{\varphi}+\Gamma^{\varphi}_{rr}(X^r)^2+2\Gamma^{\varphi}_{r\varphi} X^rX^{\varphi}+\Gamma^{\varphi}_{\varphi\varphi}(X^{\varphi})^2=0. \end{align} Since the vector field has rotational symmetry, I will look for a connection with coefficients that depend only on $r$. By making a fourth-order polynomial-in-$r$ ansatz for all Christoffel symbols, $$\Gamma^i_{jk}=\sum_{m=0}^4\Gamma^i_{jkm}r^m,$$ one can find the following solution: $$\Gamma^r_{rr}=-\frac{r}{2}, \quad \Gamma^r_{r\varphi}=\Gamma^r_{\varphi r}=\frac{1}{4}(3-12r^2+r^4), \quad \Gamma^r_{\varphi\varphi}=0,$$ $$\Gamma^{\varphi}_{rr}=2, \quad \Gamma^{\varphi}_{r\varphi}=\Gamma^{\varphi}_{\varphi r}=r(2-r^2), \quad \Gamma^{\varphi}_{\varphi\varphi}=0.$$
• I don't think it is a stupid question. When you try to solve the equations $\nabla_{\rho}g_{\mu\nu}=0$ for the unknown metric, using the connection given above, a quick calculation seems to suggest (but I may have made a mistake, so I would suggest to try for yourself) that you can find a nontrivial solution in some finite disk, but not the entire punctured plane (the metric seems to become singular on the boundary of said disk). I did not study the proof of the non-geodesibility that you linked to, so I cannot vouch for its accuracy. – S.Surace Jun 25 '18 at 12:25
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