url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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https://forums.wolfram.com/mathgroup/archive/1997/Mar/msg00117.html | 1,656,166,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00426.warc.gz | 321,682,828 | 7,472 | Re: Fourier Help!!
• To: mathgroup at smc.vnet.net
• Subject: [mg6438] Re: Fourier Help!!
• From: mightymouse at supertech.com (MightyMouse)
• Date: Fri, 21 Mar 1997 22:59:28 -0500 (EST)
• Organization: SuperTech
• Sender: owner-wri-mathgroup at wolfram.com
```In article <5gqklt\$94q at smc.vnet.net>, Woo Sanghyuk <jyun at planete.net> wrote:
> Why "Fourier" doesn't work??
>
> In = Fourier[{-1, -1, -1, 1, 1, 1}]
>
> Out = Fourier[{-1, -1, -1, 1, 1, 1}]
>
> I'm using Mathematica v2.2.2 on Macintosh performa 5200 with MacOS 7.6.
> I'm very urgent...
Hi, there....
My mma runs well....
I use mma 3.0 on pb5300cs..OS 7.5.5
In[1]:=Fourier[{-1, -1, -1, 1, 1, 1}]
Out[1]=
{0. + 0. I, -0.816497 - 1.41421 I,
-17 -16
-3.318 10 - 1.35974 10 I, -0.816497 + 0. I,
-17 -16
7.85046 10 - 1.35974 10 I, -0.816497 + 1.41421 I}
or
Out[1]=
{0.+0. I, -0.816497-1.414121 I,-3.318 10^-17 - 1.35974 10^-16 I,
-0.816497+0. I,7.85046 10^-17 - 1.35974 10^-16 I, -0.816497+1.41421 I}
If it never works, try this one...
Fourier[{-1, -1, -1, 1, 1, 1}] //N
It gives you a numerical result...
sungshin at mom.postech.ac.kr
```
• Prev by Date: Re: 3D Plots
• Next by Date: Re: Attributes
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• Next by thread: Re: Fourier Help!! | 555 | 1,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-27 | latest | en | 0.550101 |
https://developer.aliyun.com/article/536591 | 1,642,533,601,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00114.warc.gz | 271,695,012 | 13,765 | # Python多线程与多进程浅析之二
+关注继续查看
### Python 多线程 Step by Step
Python 在 CPU 密集运算的场景，多个线程并不能提高太多性能，而对于 I/O 阻塞的场景，可以使得运行效率获得几倍的提高。我们接下来会详细的分析一下。
# 计算一次 my_cal() 函数，获得耗用时间
>>> from time import time
>>> def my_cal(a):
... j = 0
... for i in range(a):
... j = j + i
... print(j)
... return j
>>> start = time()
>>> my_cal(1000000)
>>> end = time()
>>> print('cost time {:f} s'.format(end - start))
499999500000
cost time 0.106146 s
>>> from time import time
>>> def my_cal(a):
... j = 0
... for i in range(a):
... j = j + i
... print(j)
... return j
>>> start = time()
>>> list_01 = [1000000, 1500000, 2000000, 2500000, 3000000]
>>> for i in list_01:
... my_cal(i)
>>> end = time()
>>> print('cost time {:f} s'.format(end - start))
499999500000
1124999250000
1999999000000
3124998750000
4499998500000
cost time 0.906054 s
>>> from time import time
>>> def my_cal(a):
... j = 0
... for i in range(a):
... j = j + i
... print(j)
... return j
>>> start = time()
>>> list_01 = [1000000, 1500000, 2000000, 2500000, 3000000]
>>> list_02 = list(map(my_cal, list_01))
>>> end = time()
>>> print('cost time {:f} s'.format(end - start))
499999500000
1124999250000
1999999000000
3124998750000
4499998500000
cost time 0.927783 s
>>> from time import time
>>> def my_cal(a):
... j = 0
... for i in range(a):
... j = j + i
... print(j)
... return j
>>> start = time()
>>> list_01 = [1000000, 1500000, 2000000, 2500000, 3000000]
>>> for i in list_01:
... # 建立线程
... # 建立线程列表
# 等待线程结束
>>> end = time()
>>> print('\ncost time {:f} s'.format(end - start))
499999500000
1124999250000
1999999000000
3124998750000
4499998500000
cost time 1.074110 s
thread 的 join() 方法阻塞当前线程，等待子线程执行完毕后自己再继续执行。如果没有 join() 的阻塞，我们会先看到 cost time 的计算，而实际上上面的线程还没有完成自己的任务。
# 检查线程的状态
>>> from time import time
>>> def my_cal(a):
... j = 0
... for i in range(a):
... j = j + i
... print(j)
... return j
# 创建后的线程的状态
# 线程启动后的状态
4499998500000
### 多进程方式
# 多进程方式
>>> from time import time
>>> from concurrent.futures import *
>>> def my_cal(a):
... j = 0
... for i in range(a):
... j = j + i
... print(j)
... return j
>>> list_01 = [1000000, 2000000, 1500000, 2500000, 3000000]
>>> start = time()
>>> pool = ProcessPoolExecutor(max_workers=10)
>>> list_02 = list(pool.map(my_cal, list_01))
>>> end = time()
>>> print('cost time {:f} s'.format(end - start))
499999500000
1124999250000
1999999000000
3124998750000
4499998500000
cost time 0.374396 s
# 多进程方式，测试 workers
from concurrent.futures import *
def my_cal(a):
j = 0
for i in range(a):
j = j + i
return j
def my_cal_main(workers):
list_01 = [1000000, 2000000, 1500000, 2500000, 3000000]
pool = ProcessPoolExecutor(workers)
list_02 = list(pool.map(my_cal, list_01))
for i in range(2,11):
print(i)
%timeit -n6 my_cal_main(i)
# 输出结果
2
6 loops, best of 3: 491 ms per loop
3
6 loops, best of 3: 443 ms per loop
4
6 loops, best of 3: 437 ms per loop
5
6 loops, best of 3: 352 ms per loop
6
6 loops, best of 3: 336 ms per loop
7
6 loops, best of 3: 341 ms per loop
8
6 loops, best of 3: 346 ms per loop
9
6 loops, best of 3: 346 ms per loop
10
6 loops, best of 3: 347 ms per loop
# 获得 cpu 核心数量
>>> import multiprocessing
>>> pool = multiprocessing.Pool()
>>> print(pool._processes)
8
>>> print(multiprocessing.cpu_count())
8
>>> import psutil
>>> psutil.cpu_count()
8
java多线程 -- Condition 控制线程通信
Api文档如此定义： Condition 将 Object 监视器方法（wait、notify 和 notifyAll）分解成截然不同的对象，以便通过将这些对象与任意 Lock 实现组合使用，为每个对象提供多个等待 set（wait-set）。
692 0
python多线程，so easy
704 0
1318 0
python3,进程间的通信
866 0
+关注
11
0
《2021云上架构与运维峰会演讲合集》
《零基础CSS入门教程》
《零基础HTML入门教程》 | 1,447 | 3,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.273652 |
https://rdrr.io/cran/adventr/f/inst/tutorials/adventr_08/ais_08.Rmd | 1,642,926,867,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00522.warc.gz | 528,470,271 | 19,950 | # adventr: inferential statistics In adventr: Interactive R Tutorials to Accompany Field (2016), "An Adventure in Statistics"
```library(forcats)
library(learnr)
library(tidyverse)
library(boot)
knitr::opts_chunk\$set(echo = FALSE, warning = FALSE)
tutorial_options(exercise.cap = "Exercise")
#Read dat files needed for the tutorial
#setup objects for code blocks
```
# An Adventure in R: Inferential statistics and robust estimation
## Overview
This tutorial is one of a series that accompanies An Adventure in Statistics [@RN10163] by me, Andy Field. These tutorials contain abridged sections from the book so there are some copyright considerations but I offer them under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, ^[Basically you can use this tutorial for teaching and non-profit activities but do not meddle with it or claim it as your own work.]
• Who is the tutorial aimed at?
• What is covered?
• This tutorial looks at how to get (and plot) inferential statistics such as confidence intervals for the mean. We also look at robust confidence intervals. It would be a useful tutorial to run alongside teaching based on Chapters 8 and 9 of An Adventure in Statistics.
• This tutorial does not teach the background theory: it is assumed you have either attended my lecture or read the relevant chapter in the aforementioned books (or someone else's)
• The aim of this tutorial is to augment the theory that you already know by guiding you through fitting linear models using R and RStudio and asking you questions to test your knowledge along the way.
## Story précis
### Why a précis?
Because these tutorials accompany my book An adventure in statistics, which uses a fictional narrative to teach the statistics, some of the examples might not make sense unless you know something about the story. For those of you who don't have the book I begin each tutorial with a précis of the story. If you're not interested then fair enough - click past this section.
### General context for the story
It is the future. Zach, a rock musician and Alice, a geneticist, who have been together since high school live together in Elpis, the ‘City of Hope’.
Zach and Alice were born in the wake of the Reality Revolution which occurred after a Professor Milton Gray invented the Reality Prism – a transparent pyramid worn on the head – that brought honesty to the world. Propaganda and media spin became unsustainable, religions collapsed, advertising failed. Society could no longer be lied to. Everyone could know the truth about anything that they could look at. A gift, some said, to a previously self-interested, self-obsessed society in which the collective good had been eroded.
But also a curse. For, it soon became apparent that through this Reality Prism, people could no longer kid themselves about their own puffed-up selves as they could see what they were really like – by and large, pretty ordinary. And this caused mass depression. People lost faith in themselves. Artists abandoned their pursuits, believing they were untalented and worthless.
Zach and Alice have never worn a Reality Prism and have no concept of their limitations. They were born after the World Governance Agency (WGA) destroyed all Reality Prisms, along with many other pre-revolution technologies, with the aim of restoring community and well-being. However, this has not been straightforward and in this post-Prism world, society has split into pretty much two factions
• The Chippers who have had WiFi-enabled chips implanted into their brains, enabling them to record and broadcast what they see and think in real time; upload memories for future generations into a newly-created memoryBank and live-stream music and films directly into their brains.
• The Clocktarians, followers of the old pre-Prism ways who use steam punk style technologies, who have elected not to have chips in their brains, regarded by the Chippers as backward-looking stuck in a ‘clockwork, Victorian society’.
Everyone has a star, a limitless space on which to store their digital world.
Zach and Alice are Clocktarians. Their technology consists mainly of:
• A Proteus, a device made from programmable matter that can transform shape and function simply by the owners’ wishes. Zach calls his a diePad, in the shape of a tombstone in an ironic reference to an over-reliance on technology at the expense of memory.
• A Reality Checker, a clockwork mechanism that, at the point of critical velocity, projects an opaque human head that is linked to everything and can tell you anything. Every head has a personality and Zach’s is a handsome, laid back ‘dude’ who is like an electronic friend, who answers questions if he feels like it and often winds Zach up by giving him false information. And he often flirts with Alice.
### Main Protagonists
• Zach
• Rock musician in band called The Reality Enigma.
• Spellbinding performer, has huge fan-base.
• Only people living in Elpis get to see The Reality Enigma in the flesh. Otherwise all performances are done via an oculus riff, a multisensory headset for experiencing virtual gigs.
• Zach’s music has influenced and changed thousands of lives.
• Wishes he had lived pre-Revolutionary times, the turn of the 21st Century, a golden age for music when bands performed in reality at festivals.
• Kind, gentle and self-doubting.
• Believes science and maths are dull and uninspiring. Creates a problem between him and Alice as she thinks that because he isn’t interested in science, he isn’t interested in her. Leads to lots of misunderstandings between them.
• Alice
• Shy, lonely, academically-gifted – estranged from the social world until she met Zach in the college library.
• Serious scientist, works at the Beimeni Centre of Genetics.
• At 21, won the World Science Federation’s Einstein Medal for her genetics research
• Desperately wants Zach to get over his fear of science so he can open his mind to the beauty of it.
Alice has been acting strangely, on edge for weeks, disconnected and uncommunicative, as if she is hiding something and Zach can’t get through to her. Arriving home from band practice, unusually, she already home and listening to an old album that the two of them enjoyed together, back in a simpler, less complicated time in their relationship. During an increasingly testy evening, that involves a discussion with the Head about whether or not a Proteus causes brain cancer, Alice is interrupted by an urgent call which she takes in private. She returns looking worried and is once again, distracted. She tells Zach that she has ‘a big decision to make’. Before going to bed, Zach asks her if he can help with the decision but she says he ‘already has’, thanking him for making ‘everything easier.’ He has no idea what she means and goes to sleep, uneasy.
On waking, Zach senses that something is wrong. And he is right. Alice has disappeared. Her clothes, her possessions and every photo of them together have gone. He can’t get hold of any of her family or friends as their contact information is stored on her Proteus, not on his diePad. He manages to contact the Beimeni Centre but is told that no one by the name of Alice Nightingale has ever worked there. He logs into their constellation but her star has gone. He calls her but finds that her number never existed. She has, thinks Zach, been ‘wiped from the planet.’ He summons The Head but he can’t find her either. He tells Zach that there are three possibilities: Alice has doesn’t want to be found, someone else doesn’t want her to be found or she never existed.
Zach calls his friend Nick, fellow band member and fan of the WGA-installed Repositories, vast underground repositories of actual film, books, art and music. Nick is a Chipper – solely for the purpose of promoting the band using memoryBank – and he puts the word out to their fans about Alice missing.
Thinking as hard as he can, Zach recalls the lyrics of the song she’d been playing the previous evening. Maybe they are significant? It may well be a farewell message and the Head is right. In searching for clues, he comes across a ‘memory stone’ which tells him to read what’s on there. File 1 is a research paper that Zach can’t fathom. It’s written in the ‘language of science’ and the Head offers to help Zach translate it and tells him that it looks like the results of her current work were ‘gonna blow the world’. Zach resolves to do ‘something sensible’ with the report.
Zach doesn’t want to believe that Alice has simply just left him. Rather, that someone has taken her and tried to erase her from the world. He decides to find her therapist, Dr Murali Genari and get Alice’s file. As he breaks into his office, Dr Genari comes up behind him and demands to know what he is doing. He is shaking but not with rage – with fear of Zach. Dr Genari turns out to be friendly and invites Zach to talk to him. Together they explore the possibilities of where Alice might have gone and the likelihood, rating her relationship satisfaction, that she has left him. During their discussion Zach is interrupted by a message on his diePad from someone called Milton. Zach is baffled as to who he is and how he knows that he is currently discussing reverse scoring. Out of the corner of his eye, he spots a ginger cat jumping down from the window ledge outside. The counsellor has to go but suggests that Zach and ‘his new friend Milton’ could try and work things out.
## Packages and data
### Packages
This tutorial uses the following packages:
• `boot` to conduct bootstrapping [@RN11409]
• `Hmisc` to obtain confidence intervals [@RN11417]
• `tidyverse` for general data processing [@RN11407]
These packages are automatically loaded within this tutorial. If you are working outside of this tutorial (i.e. in RStudio) then you need to make sure that the package has been installed by executing `install.packages("package_name")`, where package_name is the name of the package. If the package is already installed, then you need to reference it in your current session by executing `library(package_name)`, where package_name is the name of the package.
Note that `boot` is installed by default in R, but you will need to execute `library(boot)` to use it.
### Data
This tutorial has the data files pre-loaded so you shouldn't need to do anything to access the data from within the tutorial. However, if you want to play around with what you have learnt in this tutorial outside of the tutorial environment (i.e. in a stand-alone RStudio session) you will need to download the data files and then read them into your R session. This tutorial uses the following file:
You can load the file in several ways:
• Assuming that you follow the workflow recommended in the tutorial adventr_02 (see also this online tutorial), you can load the data into an object called `jig_tib` by executing:
• `jig_tib <- readr::read_csv("../data/ais_c05_jigsaw.csv")`
• If you don't follow my suggested workflow, you will adjust the file location in the above command.
• Alternatively, if you have an internet connection (and my server has not exploded!) load the file direct from the URL by executing:
• `jig_tib <- readr::read_csv("http://www.discoveringstatistics.com/repository/ais_data/ais_c05_jigsaw.csv")`
## Confidence intervals
After breaking into the JIG:SAW complex and catching a glimpse of Alice, Zach is in shock. Returning home he is desolate: the evidence of his eyes suggests that Alice is not being held hostage at JIG:SAW and the reality kicks in that she may in fact have left him. He feels worthless. He meets up with Nick, the drummer in his band, at the Repository and they discuss what has happened. Nick is utterly convinced that Alice wouldn’t have left Zach out of choice and Zach returns home even more determined to get back into JIG:SAW to speak to her and find out for himself. Milton is reluctant and tries to convince Zach that he would need to step up his understanding of statistics to a level of which he is not capable. Based on data showing that performing under a different name in a maths test can free mental blocks to maths, it is decided that Zach needs to dress up as Florence Nightingale - a ‘gifted statistician’ as well as a nurse. Feel free to do the same.
Milton goes on to explain confidence intervals to Zach.
```quiz(
question("If a 95% confidence interval for the mean ranges from 45 to 51, what does this tell us?",
answer("If this confidence interval is one of the 95% that contains the population value then the population value of the mean lies between 45 and 51.", correct = TRUE),
answer("There is a 95% chance that the population value of the mean lies between 45 and 51", message = "You cannot make probability statements from a confidence interval. We don't know whether this particular CI is one of the 95% that contains the population value of the mean."),
answer("The probability of this confidence interval containing the population mean is 0.95.", message = "The probability of this confidence interval containing the population mean is either 0 (it doesn't) or 1 (it does) but it's impossible to know which."),
answer("I can be 95% confident that the population value of the mean lies between 45 and 51", message = "Confidence intervals do not quantify your subjective confidence."),
correct = "Correct - well done!",
allow_retry = T
)
)
```
In a previous tutorial we looked at how to summarize data to get summary statistics such as the mean, median, variance and so on. It is often useful to get the confidence interval too. This section looks at how to do that. We're going to use the data in the tibble called `jig_tib` again, which contains data about the strength, speed, and visual acuity of JIG:SAW employees and non-employees. Remember that this tibble contains 7 variables:
• id : employee ID
• employee: whether or not the employee works for JIG:SAW
• job_type: Categories of employee
• footspeed: Footspeed of the individual, in miles per hour
• strength: Maximal push force of the individual in newtons
• vision: Visual acuity
• sex: Biological sex of the individual
In the previous tutorial we looked at combining `group_by()` and `summarize()` in a pipe command to create a tibble containing summary statistics (split by group). For example, we used the following command to create a tibble called `speed_sum` that contained the mean of the variable footspeed grouped by combinations of the variables employee and sex from the tibble called `jig_tib`.
```speed_sum <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_speed = mean(footspeed)
)
```
```quiz(
question("In the above command, what does **dplyr::group_by(employee, sex)** do?",
answer("Groups the output by all combinations of the variables **employee** and **sex** (I.e. employee-male, employee-female, non-employee-male, non-employee-female.", correct = TRUE),
answer("Creates one output grouped by **employee** (i.e. employee, non-employee), and a separate output grouped by sex (male, female)."),
correct = "Correct - well done!",
allow_retry = T
),
question("What does the pipe operator (%>%) do?",
answer("Carries any instruction on the left of the operator forward into any instrucion on the right of the operator", correct = TRUE),
answer("Allows you to chain independent commands without them affecting each other.", message = "Sorry, that's incorrect. The pipe operator is does the opposite: it allows you to chain commands so that commands on the left are carried forwared in to commands on the right"),
answer("Multiplies anything on the left of the operator by anything on the right.", message = "Sorry, that's incorrect. The pipe operator is does the opposite: it allows you to chain commands so that commands on the left are carried forwared in to commands on the right"),
correct = "Correct - well done!",
allow_retry = T
)
)
```
To incorporate confidence intervals into this command isn't entirely straightforward. One way is to use the `mean_cl_normal()` function from the `ggplot2` package. (Under the hood this function is using the package `Hmisc` so if you're working outside of this tutorial then you'll need to install and load this package.) This function takes any model as its input and returns an object that contains estimates from the model and their confidence intervals. In general it takes the following form:
`ggplot2::mean_cl_normal(object, conf.int = 0.95, na.rm = TRUE)`
To keep things simple, imagine the 'object' we put into this function is a variable, like footspeed. We'd execute:
`ggplot2::mean_cl_normal(jig_tib\$footspeed, conf.int = 0.95, na.rm = TRUE)`
If we place only `jig_tib\$footspeed` into the function we'll get a 95% confidence interval and missing values will be removed, which is usually what we want. To get a different confidence interval specify a proportion in `conf.int =`. For example, `ggplot2::mean_cl_normal(jig_tib\$footspeed, conf.int = 0.9)` will give us a 90% confidence interval around the mean of footspeed. When you put a variable into the function it returns an object containing the mean (labelled y in the object), the lower boundary of the confidence interval (labelled ymin), and the upper boundary of the confidence interval (labelled ymax). To access these values simultaneously we execute the function (or create an object from it and execute that), but if we want to access them individually we can do that by appending their name after the `\$` symbol. For example:
`ggplot2::mean_cl_normal(jig_tib\$footspeed)\$ymin`
returns the value of the lower boundary of the (by default 95%) confidence interval. If we add this command within the `summarize()` statement in our pipe, we'll get the lower boundaries for the confidence intervals around the mean when the data are grouped by employee and sex. One final point is that we wouldn't include this line exactly as I've written it because we have specified the `jig_tib` tibble earlier in the pipe, which means it will be carried forward into the summarize statement. This means that we can specify the variable of interest as simply `footspeed` rather than `jig_tib\$footspeed` because the function will already know to look for footspeed within the `jig_tib` tibble.
Therefore, to create a tibble with means and confidence intervals grouped by employee and sex we can adapt the code that we've already used in previous tutorials to include the `mean_cl_normal()` function:
```speed_sum <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_speed = mean(footspeed),
ci_low = ggplot2::mean_cl_normal(footspeed)\$ymin,
ci_upp = ggplot2::mean_cl_normal(footspeed)\$ymax
)
```
Note that the code is the same as we used before except that we have added two lines (which should make sense based on the explanation above):
• `ci_low = ggplot2::mean_cl_normal(footspeed)\$ymin` creates a variable called ci_low by applying the function `mean_cl_normal()` to the variable footspeed and extracting the value labelled "CI lower", which is the value of the lower boundary of the 95% confidence interval.
• `ci_upp = ggplot2::mean_cl_normal(footspeed)\$ymax` creates a variable called ci_upp by applying the function `mean_cl_normal()` to the variable footspeed and extracting the value labelled "CI upper", which is the value of the upper boundary of the 95% confidence interval.
Copy the command into the code box and run it. Also, run the name of the `speed_sum` tibble that you create to see what it contains. Hopefully you'll see a tibble with the mean and 95% confidence interval boundaries split by employee and sex. Try to create tibbles called `strength_sum` and `vis_sum` that tabulate the mean and confidence intervals for the variables strength and vision respectively.
```
```
```speed_sum <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_speed = mean(footspeed),
ci_low = ggplot2::mean_cl_normal(footspeed)\$ymin,
ci_upp = ggplot2::mean_cl_normal(footspeed)\$ymax
)
speed_sum
```
## Plotting confidence intervals
We've seen in previous tutorials how to plot means. For example we created a plot of the strength data split by JIG:SAW employees and non-employees. We used this code:
```strength_plot <- ggplot2::ggplot(jig_tib, aes(employee, strength))
strength_plot +
stat_summary(fun = "mean", geom = "point", size = 4) +
labs(x = "Employee status", y = "Maximal push force (n)") +
coord_cartesian(ylim = c(1000, 1800)) +
scale_y_continuous(breaks = seq(1000, 1800, 100)) +
theme_bw()
```
Let's remind ourselves of what each part of this command is doing:
• `strength_plot <- ggplot(jig_tib, aes(employee, strength))` creates an object called `strength_plot` that contains the plot. The `ggplot()` function is then used to specify that the plot uses the `jig_tib` tibble and plots the variable employee on the x-axis and the variable strength on the y-axis.
• `strength_plot + stat_summary(fun = "mean", geom = "point", size = 4)` takes the object `strength_plot` and adds a statistics summary to it. The `stat_summary()` uses `fun = "mean"` to plot the mean using a dot (`geom = "point"`) of size 4 (`size = 4`).
• `labs(x = "Employee status", y = "Maximal push force (n)")` applies descriptive labels to the x and y axes.
• `coord_cartesian(ylim = c(1000, 1800))` sets the limits of the y-axis
• `scale_y_continuous(breaks = seq(1000, 1800, 100))` sets the breaks along the y-axis
• `theme_bw()` applies a black and white theme to the plot
If we want to plot confidence intervals around the means all we need to do is to edit the details of `stat_summary()`. Specifically there are three changes that we need to make:
1. Change `fun = "mean"` to `fun.data = "mean_cl_normal"`. This creates the data to be plotted for the mean and confidence intervals.
2. Change `geom = "point"` to `geom = "pointrange"`. This changes the geom to one that can display a confidence interval.
3. Delete `size = 4` because the default size for the 'pointrange' geom will look better than size 4.
The code box contains the code above and shows the resulting plot (that we created in previous tutorials). Change the two things listed above and run the code to see how the plot changes.
```strength_plot <- ggplot(jig_tib, aes(employee, strength))
strength_plot +
stat_summary(fun = "mean", geom = "point", size = 4) +
labs(x = "Employee status", y = "Maximal push force (n)") +
coord_cartesian(ylim = c(1000, 1800)) +
scale_y_continuous(breaks = seq(1000, 1800, 100)) +
theme_bw()
```
```strength_plot <- ggplot(jig_tib, aes(employee, strength))
strength_plot +
stat_summary(fun.data = "mean_cl_normal", geom = "pointrange") +
labs(x = "Employee status", y = "Maximal push force (n)") +
coord_cartesian(ylim = c(1000, 1800)) +
scale_y_continuous(breaks = seq(1000, 1800, 100)) +
theme_bw()
```
Hopefully your plot how looks like dots showing the means with vertical lines showing the confidence intervals. (Incidentally, you could layer a `stat_summary()` function underneath the confidence intervals that uses a bar geom to plot the mean. Something like `stat_summary(fun = "mean", geom = "bar", alpha = 0.6) +` placed on the line above the existing stat_summary() function would work. The result would be a bar graph of means with error bars layered on top. However, adding the bars does not add any new information, but does add a lot of unnecessary ink, so I wouldn't bother.)
What about more complex plots? In a previous tutorial we also created a similar plot but differentiated males and females by colours. The code box below replicates this code and shows the plot to remind you of how it looks. To add error bars we make exactly the same changes to `stat_summary()` as we did above:
1. Change `fun = "mean"` to `fun.data = "mean_cl_normal"`.
2. Change `geom = "point"` to `geom = "pointrange"`.
3. Delete `size = 4`.
Make these changes and run the code. You should see the same plot but with confidence intervals as well as means.
```strength_plot <- ggplot(jig_tib, aes(employee, strength, colour = sex))
strength_plot +
stat_summary(fun = "mean", geom = "point", size = 4, position = position_dodge(width = 0.9)) +
labs(x = "Employee status", y = "Maximal push force (n)", colour = "Biological sex") +
coord_cartesian(ylim = c(1000, 1800)) +
scale_y_continuous(breaks = seq(1000, 1800, 100)) +
scale_colour_manual(values = c("#56B4E9", "#E69F00")) +
theme_bw()
```
```strength_plot <- ggplot(jig_tib, aes(employee, strength, colour = sex))
strength_plot +
stat_summary(fun.data = "mean_cl_normal", geom = "pointrange", position = position_dodge(width = 0.9)) +
labs(x = "Employee status", y = "Maximal push force (n)", colour = "Biological sex") +
coord_cartesian(ylim = c(1000, 1800)) +
scale_y_continuous(breaks = seq(1000, 1800, 100)) +
scale_colour_manual(values = c("#56B4E9", "#E69F00")) +
theme_bw()
```
In the box below can you create a similar plot of the vision scores split by job_type on the x-axis and employees and non-employees in different colours.
```
```
```vis_plot <- ggplot(jig_tib, aes(job_type, vision, colour = employee))
vis_plot +
stat_summary(fun.data = "mean_cl_normal", geom = "pointrange", position = position_dodge(width = 0.9)) +
labs(x = "Type of job", y = "Visual acuity", colour = "Employee status") +
coord_cartesian(ylim = c(0, 1)) +
scale_y_continuous(breaks = seq(0, 1, 0.1)) +
scale_colour_manual(values = c("#56B4E9", "#E69F00")) +
theme_bw()
```
## Robust means
In Chapter 9, Zach steals the data we have been analysing that relates to JIG:SAW employees. Milton teaches him how the mean and confidence intervals can be biased by skew, kurtosis and outliers. Milton introduces the idea of the trimmed mean.
To get the trimmed mean for a variable we can adapt the code we have already used. For example, to get the mean footspeed split by employee and sex we used the code below. We discovered before that the mean has an option to specify a trim (`trim = 0`), so to produce a summary of trimmed means we can simply add this option to the command that gets the mean. For example, to get a 20% trimmed mean we might replace `mean_speed = mean(footspeed)` with:
`tr_mean_speed = mean(footspeed, trim = 0.2)`
Try adding this line to the `summarize()` function in the code box so that we store both the mean and the 20% trimmed mean (don't forget to include a comma after the first command within `summarize()`)
```speed_sum <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_speed = mean(footspeed)
)
speed_sum
```
```speed_sum <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_speed = mean(footspeed),
tr_mean_speed = mean(footspeed, trim = 0.2)
)
speed_sum
```
Try this for strength as well:
```
```
```strength_sum <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_strength = mean(strength),
tr_mean_strength = mean(strength, trim = 0.2)
)
strength_sum
```
## Robust confidence intervals
Milton also introduces Zach to the idea of bootstrapping as a way to create robust confidence intervals.
```quiz(
question("Bootstrapping is a technique from which the sampling distribution of a statistic is estimated by ...",
answer("Taking repeated samples (with replacement) from the data set.", correct = TRUE),
answer("Taking repeated samples from the population.", message = "Samples are not taken from the population because we don't have access to it."),
answer("Adjusting the standard error to compensate for heteroscedasticity.", message = "Bootstrapping is a *sampling* process."),
answer("Tying my shoelaces together so that I fall over.", message = "Now you're just being silly."),
correct = "Correct - well done!",
allow_retry = T
),
question("Which of these statements about bootstrap confidence intervals is **not** true?",
answer("Bootstrap confidence intervals have the same values each time you compute them.", correct = TRUE, message = "This stament *is* false. Because bootstrapping relies on repeated sampling, the results can vary slighty each time you impliment the process."),
answer("Bootstrap confidence intervals are robust to violations of homoscedasticity.", message = "This statement is true."),
answer("Bootstrap confidence intervals do not assume a normal sampling distribution.", message = "This statement is true: bootstrapping is a technique from which the sampling distribution of a statistic is estimated *empirically* from the data so no assumptions about its shape are made."),
answer("Bootstrap confidence intervals are most useful in small samples.", message = "Technically this statement is true because bootstrapping was designed for small sample situations (where the central limit theorem can't be invoked). However, there is evidence that the central limit theorem may not hold up in samples that are quite large so there is a case to be made that bootstrapping is still useful in larger samples."),
correct = "Correct - well done!",
allow_retry = T
)
)
```
We can get bootstrap confidence intervals in R by using the `boot` package, which is installed by default. However, you do need to load it before using it if you are working outside of this tutorial environment. The process of getting the bootstrap intervals is - and I'm not going to lie - complicated. You should not feel bad if this section makes no sense whatsoever (but by all means feel a sense of pride if you get through it in one piece).
The `boot()` function works in the following way: it takes a bootstrap sample from the data you specify, sends it out to a function (that the user defines) that returns the statistic of interest (in this case the mean), it stores this information and the repeats the process lots of times. The general format of the `boot()` function is:
`boot::boot(data, statistic, R)`
It has other options too, but the three key ones are `data` which specifies the data from which to take bootstrap samples, `statistic` which specifies a function that computes the information you want (e.g., the mean), and `R` which is the number of times you want to repeat the process. So, to get bootstrap samples of the strength data we could specify something like this:
`boot::boot(jig_tib\$strength, statistic, R = 1000)`
This specifies that the data from which to take bootstrap samples is `jig_tib\$strength` and that we want to repeat the process 1000 times (i.e. use 1000 bootstrap samples). The question is what do we replace `statistic` with. Given we're interested in the mean, you might reasonably think we can replace `statistic` with `mean()` or something. However, we can't do this because the function `mean()` will not return information about which bootstrap sample was used. Instead we need to write out own function that will include information about the particular bootstrap sample that is being used to compute the mean. We do this by executing:
`mean_boot <- function(data, index) mean(data[index], na.rm = TRUE)`
This creates a new function called `mean_boot()`. This function takes data and an index value as input `function(data, index)` and it returns the mean of the data with a particular index value after removing missing values `mean(data[index], na.rm = TRUE)`. In the main `boot()` function we would replace `statistic` with this function that we have created:
`boot::boot(jig_tib\$strength, mean_boot, R = 1000)`
To sum up, the `boot()` function is going to send data (the bootstrap samples) with an associated index value (from 1 to 1000) to the function called `mean_boot`. For each data set with each index value `mean_boot` will return the mean.
The code box contains these commands. Run the code.
```mean_boot <- function(data, index) mean(data[index], na.rm = TRUE)
boot::boot(jig_tib\$strength, mean_boot, R = 1000)
```
The output doesn't give us the confidence intervals. Instead we're told the original mean and an estimate of bias. To get the confidence intervals there is one more step, which is to use the `boot.mean_cl_normal()` function, which takes the general form:
`boot::boot.ci(boot.object, conf = 0.95, type = "all")`
There are other options but these ones are the main ones. First you input an object created with the `boot()` function, then specify the width of confidence interval that you want (the default is 0.95 so can leave this option out if that's what you want), and the type that you want. By default it will compute 5 different types of interval. For most purposes you don't need five different intervals so it would be best to change this option to specify a particular type. There's much to recommend Bias Corrected and Accelerated intervals (`type = "bca"`) or if those fail then a percentile interval (`type = "perc"`).
The code box below combines what we have learnt. First it creates the function to get the mean. The second line inserts our `boot()` function into the `boot.mean_cl_normal()` function and asked for BCa confidence intervals. Run the code.
```mean_boot <- function(data, index) mean(data[index], na.rm = TRUE)
boot::boot.ci(boot(jig_tib\$strength, mean_boot, R = 1000), type = "bca")
```
There is too much output to embed into a summarize function. We want to discard everything except for the numeric values that represent the lower and upper bound of the confidence interval. These values are stored in a variable called bca within the `boot.mean_cl_normal()` object. (Obviously if you chose a type of interval other than bca, the variable name will be different; for example, if you opted for `type = "Perc"` then the variable will be called perc). Like any variable, we can access it using `\$`. For example, to see the information relating to the BCa confidence intervals we could execute:
`boot::boot.ci(boot(jig_tib\$strength, mean_boot, R = 1000), type = "bca")\$bca`
Try it in the code box above and you'll see 5 values appear.Values 4 and 5 are the lower and upper limits of the interval. We can access these specific values using:
`boot::boot.ci(boot(jig_tib\$strength, mean_boot, R = 1000), type = "bca")\$bca[4]` `boot::boot.ci(boot(jig_tib\$strength, mean_boot, R = 1000), type = "bca")\$bca[5]`
Try these commands in the code box above. We can include these lines in the summarize function of the pipe that we used at the start of the tutorial. Remember that we can drop the `jig_tib\$` from the command because that tibble is specified earlier in the pipe. The code box below shows the commands.
```mean_boot <- function(data, index) mean(data[index], na.rm = TRUE)
strength_mean <-
jig_tib %>%
dplyr::group_by(employee, sex) %>%
dplyr::summarize(
mean_strength = mean(strength),
ci_low_boot = boot::boot.ci(boot(strength, mean_boot, R = 1000), type = "bca")\$bca[4],
ci_upp_boot = boot::boot.ci(boot(strength, mean_boot, R = 1000), type = "bca")\$bca[5]
)
strength_mean
```
Run this code and you should see that it creates a tibble with the means, the lower limit of the 95% bootstrap confidence interval for the mean (`ci_low_boot`) and the upper limit of the 95% bootstrap confidence interval for the mean (`ci_upp_boot`) split b all combinations of employee and sex.
## Plotting bootstrap confidence intervals
The last section was, I imagine, not the most fun that you have ever had. I expect you are dreading the added complexity of plotting these bootstrap intervals. Well, fear not. Earlier on we plotted this graph:
```strength_plot <- ggplot(jig_tib, aes(employee, strength, colour = sex))
strength_plot +
stat_summary(fun.data = "mean_cl_normal", geom = "pointrange", position = position_dodge(width = 0.9)) +
labs(x = "Employee status", y = "Maximal push force (n)", colour = "Biological sex") +
coord_cartesian(ylim = c(1000, 1800)) +
scale_y_continuous(breaks = seq(1000, 1800, 100)) +
scale_colour_manual(values = c("#56B4E9", "#E69F00")) +
theme_bw()
```
To convert these confidence intervals to bootstrap confidence intervals requires only one change to the command:
• In the `stat_summary()` function change `mean_cl_normal` to `mean_cl_boot`.
It's a simple as that. Make the change in the code box above and re-run the code to observe the change on the plot.
## Other resources
### Statistics
• The tutorials typically follow examples described in detail in @RN10163, so for most of them there's a thorough account in there. You might also find @RN4832 useful for the R stuff.
• There are free lectures and screen casts on my YouTube channel
• There are free statistical resources on my website www.discoveringstatistics.com | 8,615 | 36,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-05 | latest | en | 0.885409 |
http://www.askphysics.com/tag/florida/ | 1,632,474,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057508.83/warc/CC-MAIN-20210924080328-20210924110328-00397.warc.gz | 69,928,974 | 23,606 | Home » Posts tagged 'Florida'
# Tag Archives: Florida
## Question from image formation by a plane mirror
if a ball is located 4 meters in front of a mirror, what distance separates the object from the image? please Use your diagram to aid your answer.
## Resistance In Circuit
Please solve this problem. And please keep in mind that i have just passed 10th standard guyz.
## Coulomb’s law
at each of the four corners of a sqare of side A, a charge +q is placed freely. what charge should be placed at the centre of the sqare so that whole system be in equilibrium
## Mechanics Questions from Mohammed
a. Determine the work done by the motor when the wheelchair starts at rest and speeds up to its normal speed.
b. Determine the maximum distance that the wheelchair can travel on a horizontal surface at its normal speed, using its stored energy. (Ignore the energy needed for it to speed up when it starts.)
c. Suppose that 0.023 percent of the power required for driving is expended against drag due to the flexing of the wheelchair’s soft rubber tires. Calculate the magnitude of the drag force.
### Hits so far @ AskPhysics
• 2,274,296 hits | 256 | 1,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-39 | latest | en | 0.909038 |
https://www.enotes.com/homework-help/determine-polynomial-function-400457 | 1,511,390,040,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806676.57/warc/CC-MAIN-20171122213945-20171122233945-00610.warc.gz | 786,716,519 | 8,868 | # Determine the polynomial function The zeroes of polynomial are 1-√3,1 + √3 - 2
giorgiana1976 | Student
Since the polynomial has 3 zeroes, that means that the polynomial has 3 roots, then the polynomial is of 3rd degree.
f(x) = ax^3 + bx^2 + cx + d
According to the rule, a polynomial could be written as product of linear factors, also.
Since the roots of polynomial are:
x1 = 1 - sqrt3
x2 = 1 + sqrt3
x3 = -2
f(x) = (x - 1 + sqrt3)(x - 1 - sqrt3)(x + 2)
f(x) = [(x-1)^2 - 3](x+2)
f(x) = (x^2 - 2x + 2 - 3)(x+2)
f(x) = (x^2 - 2x - 1)(x+2)
We'll remove the brackets and we'll get:
f(x) = x^3 + 2x^2 - 2x^2 - 4x - x - 2
We'll eliminate and combine like terms:
f(x) = x^3 - 5x - 2
The function that has the 3 given roots is:
f(x) = x^3 - 5x - 2 | 300 | 761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2017-47 | latest | en | 0.881613 |
https://math.stackexchange.com/questions/313632/why-must-a-finite-symmetry-group-be-discrete?noredirect=1 | 1,566,349,182,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00387.warc.gz | 549,655,157 | 31,896 | # Why must a finite symmetry group be discrete?
I'm having trouble justifying why a finite symmetry group is discrete. Can someone help?
• What do you think? What would one that isn't discrete look like? – vonbrand Feb 25 '13 at 4:37
• Where is the symmetry group getting its topology from exactly? In a somewhat recent question, it was seen that any finite topological group is discrete modulo a normal subgroup N (which is in particular the connected component of the identity), meaning the cosets in G/N form a topological basis. Thus there can be topologies on finite groups that are not quite discrete. – anon Feb 25 '13 at 4:39
• I know an infinite symmetry group would describe a figure like a circle since there are infinitely many rotations but I don't quite understand how a finite symmetry group has to be discrete. – Peter12 Feb 25 '13 at 4:43
• It doesn't have to be discrete (only discrete modulo the identity component), unless perhaps the finite symmetry group is getting its topology in some particular manner from how it acts or there are extra restrictions on the topology. – anon Feb 25 '13 at 4:45
• Peter, what does it mean to you to say a group is discrete? What does it mean to say a group is not discrete? – Gerry Myerson Feb 25 '13 at 4:55
Consider a finite symmetry group, $G=S_n$ (I'm assuming that you meant the full symmetry group, otherwise, as was pointed out in the comments, the question seems to be not well-defined).
A topology $\mathcal T$ on $G$ would be a subset of $2^G$, satisfying the axioms of a topological space. A topology is discrete if any point is open, or equivalently $\mathcal T = 2^G$. A topology is trivial if $\mathcal T = \{\emptyset, G\}$. Since any set can have the trivial topology, we will need to assume that $\mathcal T$ is not trivial.
In a topological group a translation by a group element is a homeomorphism, so if U is open then so is $gU$ for all $g \in G$.
Lets verify that $\mathcal T$ is indeed discrete. Because it is not trivial there is some open set $U \in \mathcal T$, such that $U\neq G$ and $U \neq \emptyset$. Lets say that $|U|=k$ with $0<k<n$. Observe that as $g$ runs over $G$, $\{gU\}$ consists of all possible subsets of $\{1...n\}$ of size $k$, and since the union of open sets is open, we get that any $V \subset G$ with $k\le |V| \le n$ is open.
To see that every set with less than $k$ elements is open as well, we use the fact that the intersection of two open sets is open, and prove by downward induction on the size of smallest non-trivial open set. Namely, given $x\in U, y \notin U$ define $U' = U \setminus \{x\} \cup \{y\}$ which is also open, and note that $|U \cap U'| = |U| - 1$. | 719 | 2,681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-35 | latest | en | 0.945849 |
https://opensourcedigest.org/geometric-shapes-worksheet-2nd-grade/ | 1,618,155,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00139.warc.gz | 551,203,034 | 8,247 | # 30 Check Out Geometric Shapes Worksheet 2nd Grade
worksheet 2nd Grade Geometry Worksheets Image Inspirations from geometric shapes worksheet 2nd grade, image source: bigmetalcoal.com
2nd grade geometry worksheets grade 2 geometry worksheets our grade 2 geometry worksheets focus on deepening students understanding of the basic properties of two dimensional shapes as well as introducing the concepts of congruency symmetry area and perimeter our final worksheets introduce 3d shapes 2nd grade geometry worksheets & free printables take your students geometry skills to the next level with our second grade geometry worksheets and printables begin by reviewing 2d shapes and advance to introducing more plex 3d shapes and rare polygons explore concepts of angles lines and symmetry and use visual guides to practice fractions 2nd grade geometry & shapes worksheets teachervision in this geometry worksheet students write the name of each 2 dimensional shape naming 2 dimensional shapes ii review 2 dimensional shapes with this worksheet students use the names in the word box to name each of the shapes then they draw a pentagon and hexagon naming 3 dimensional shapes i identify 2d shapes worksheets geometry worksheets the basic two dimensional shapes in the first worksheet students are asked to identify squares rectangles triangles and circles the second worksheet covers rectangles pentagons and hexagons rectangle square triangle circle rectangle pentagons hexagons similar making rectangles from squares edges & vertices worksheets | 284 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-17 | latest | en | 0.863847 |
http://msgroups.net/excel.charting/help-with-setting-up-spreadsheet-and-charts-pl/231710 | 1,575,861,826,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00163.warc.gz | 104,668,305 | 14,671 | #### Help with setting up spreadsheet and charts please
```Hi, I would be extremely grateful for some help with my spreadsheet.
I have the following data which I need to use to produce charts for a
hospital report fairly urgently.
The rows are the names of patients. This could be up to 200 for each month,
but this varies from month to month. I will have each month on a different
worksheet.
Column F - Ward
Column G - Hospital
Column K - I have a number 1 in the row if the patient meets the criteria
I need to produce a chart(s) to show and compare the following:
How many patients from each ward met the criteria (as indicated by 1 in
column K) and how many did not.
How many patients from each hospital met the criteria and how many did not.
I would really like some help!!!
```
0
11/6/2008 11:52:02 AM
excel.charting 18370 articles. 0 followers.
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```Matthew,
Classical homework, isn’t it? It’s an example for using Excel function SUMIF
(find out the Help for this function). Make somewhere the column list of you
Wards. Next to the first Ward insert the function SUMIF. As its Range
argument select the range of your Wards from the patients list, as Criteria
select the link to the cell with Ward name, as Sum all of yours 1’s and
nothings or zeros. The ranges should be anchored properly (\$’s). Copy the
SUMIF down for all Ward names. You will obtain the correct count in each
category. For graphical representation create bar chart.
For negative counting you have to know the totals of patients in Wards. You
obtain them by means of COUNTIF function. So, for each Ward from your list
insert this function to the next column with quite the same range and
criteria arguments as with SUMIF. The count of the not-meetings is mere a
difference.
Do the same with Hospitals.
--
Petr Bezucha
"matthewluck1" wrote:
> Hi, I would be extremely grateful for some help with my spreadsheet.
>
> I have the following data which I need to use to produce charts for a
> hospital report fairly urgently.
>
> The rows are the names of patients. This could be up to 200 for each month,
> but this varies from month to month. I will have each month on a different
> worksheet.
>
> Column F - Ward
> Column G - Hospital
> Column K - I have a number 1 in the row if the patient meets the criteria
>
> I need to produce a chart(s) to show and compare the following:
>
> How many patients from each ward met the criteria (as indicated by 1 in
> column K) and how many did not.
>
> How many patients from each hospital met the criteria and how many did not.
>
> I would really like some help!!!
>
>
```
0
PBezucha (46)
11/6/2008 3:48:01 PM
```Thanks, I have got as far as counting the number of 1's in the patient list
using the formula: =SUMIF(\$F1:F89,L1:L26,J1:J89)
I am stuck using the COUNTIF function to count which cells are have 0's. I
have my summary of all wards mentioned in column L (L1:L26). My 0's and 1's
are found in column J (J1:J89). The ward on the main part of the sheet is on
each row in column G (G1:G89).
I would like the spreadsheet to look at the ward name in my summary in
column L, then if this matches in column G and there is an 0 in column J
alongside it, then this should be counted to add up to a cumulative total of
0's per ward.
Sorry for my lack of understanding I've not used excel for ages!!!
"PBezucha" wrote:
> Matthew,
> Classical homework, isn’t it? It’s an example for using Excel function SUMIF
> (find out the Help for this function). Make somewhere the column list of you
> Wards. Next to the first Ward insert the function SUMIF. As its Range
> argument select the range of your Wards from the patients list, as Criteria
> select the link to the cell with Ward name, as Sum all of yours 1’s and
> nothings or zeros. The ranges should be anchored properly (\$’s). Copy the
> SUMIF down for all Ward names. You will obtain the correct count in each
> category. For graphical representation create bar chart.
> For negative counting you have to know the totals of patients in Wards. You
> obtain them by means of COUNTIF function. So, for each Ward from your list
> insert this function to the next column with quite the same range and
> criteria arguments as with SUMIF. The count of the not-meetings is mere a
> difference.
> Do the same with Hospitals.
>
> --
> Petr Bezucha
>
>
> "matthewluck1" wrote:
>
> > Hi, I would be extremely grateful for some help with my spreadsheet.
> >
> > I have the following data which I need to use to produce charts for a
> > hospital report fairly urgently.
> >
> > The rows are the names of patients. This could be up to 200 for each month,
> > but this varies from month to month. I will have each month on a different
> > worksheet.
> >
> > Column F - Ward
> > Column G - Hospital
> > Column K - I have a number 1 in the row if the patient meets the criteria
> >
> > I need to produce a chart(s) to show and compare the following:
> >
> > How many patients from each ward met the criteria (as indicated by 1 in
> > column K) and how many did not.
> >
> > How many patients from each hospital met the criteria and how many did not.
> >
> > I would really like some help!!!
> >
> >
```
0
11/11/2008 4:34:07 PM
```Put
=SUMIF(\$G\$1:\$G\$100;L1;\$J1:\$J100) into M-column
=COUNTIF(\$G1:\$G100;L1)-M1 into N-column.
I was overly didactic, take my apologies. Countif has also by one argument
fewer, I failed to mention, and that was what may have confused you.
Regards
--
Petr Bezucha
"matthewluck1" wrote:
> Thanks, I have got as far as counting the number of 1's in the patient list
> using the formula: =SUMIF(\$F1:F89,L1:L26,J1:J89)
>
> I am stuck using the COUNTIF function to count which cells are have 0's. I
> have my summary of all wards mentioned in column L (L1:L26). My 0's and 1's
> are found in column J (J1:J89). The ward on the main part of the sheet is on
> each row in column G (G1:G89).
>
> I would like the spreadsheet to look at the ward name in my summary in
> column L, then if this matches in column G and there is an 0 in column J
> alongside it, then this should be counted to add up to a cumulative total of
> 0's per ward.
>
> Sorry for my lack of understanding I've not used excel for ages!!!
>
> "PBezucha" wrote:
>
> > Matthew,
> > Classical homework, isn’t it? It’s an example for using Excel function SUMIF
> > (find out the Help for this function). Make somewhere the column list of you
> > Wards. Next to the first Ward insert the function SUMIF. As its Range
> > argument select the range of your Wards from the patients list, as Criteria
> > select the link to the cell with Ward name, as Sum all of yours 1’s and
> > nothings or zeros. The ranges should be anchored properly (\$’s). Copy the
> > SUMIF down for all Ward names. You will obtain the correct count in each
> > category. For graphical representation create bar chart.
> > For negative counting you have to know the totals of patients in Wards. You
> > obtain them by means of COUNTIF function. So, for each Ward from your list
> > insert this function to the next column with quite the same range and
> > criteria arguments as with SUMIF. The count of the not-meetings is mere a
> > difference.
> > Do the same with Hospitals.
> >
> > --
> > Petr Bezucha
> >
> >
> > "matthewluck1" wrote:
> >
> > > Hi, I would be extremely grateful for some help with my spreadsheet.
> > >
> > > I have the following data which I need to use to produce charts for a
> > > hospital report fairly urgently.
> > >
> > > The rows are the names of patients. This could be up to 200 for each month,
> > > but this varies from month to month. I will have each month on a different
> > > worksheet.
> > >
> > > Column F - Ward
> > > Column G - Hospital
> > > Column K - I have a number 1 in the row if the patient meets the criteria
> > >
> > > I need to produce a chart(s) to show and compare the following:
> > >
> > > How many patients from each ward met the criteria (as indicated by 1 in
> > > column K) and how many did not.
> > >
> > > How many patients from each hospital met the criteria and how many did not.
> > >
> > > I would really like some help!!!
> > >
> > >
```
0
PBezucha (46)
11/12/2008 7:36:01 AM
```Thank you so much for your help. It is working fine now. :D
"PBezucha" wrote:
> Put
>
> =SUMIF(\$G\$1:\$G\$100;L1;\$J1:\$J100) into M-column
>
> =COUNTIF(\$G1:\$G100;L1)-M1 into N-column.
>
> I was overly didactic, take my apologies. Countif has also by one argument
> fewer, I failed to mention, and that was what may have confused you.
>
> Regards
> --
> Petr Bezucha
>
>
> "matthewluck1" wrote:
>
> > Thanks, I have got as far as counting the number of 1's in the patient list
> > using the formula: =SUMIF(\$F1:F89,L1:L26,J1:J89)
> >
> > I am stuck using the COUNTIF function to count which cells are have 0's. I
> > have my summary of all wards mentioned in column L (L1:L26). My 0's and 1's
> > are found in column J (J1:J89). The ward on the main part of the sheet is on
> > each row in column G (G1:G89).
> >
> > I would like the spreadsheet to look at the ward name in my summary in
> > column L, then if this matches in column G and there is an 0 in column J
> > alongside it, then this should be counted to add up to a cumulative total of
> > 0's per ward.
> >
> > Sorry for my lack of understanding I've not used excel for ages!!!
> >
> > "PBezucha" wrote:
> >
> > > Matthew,
> > > Classical homework, isn’t it? It’s an example for using Excel function SUMIF
> > > (find out the Help for this function). Make somewhere the column list of you
> > > Wards. Next to the first Ward insert the function SUMIF. As its Range
> > > argument select the range of your Wards from the patients list, as Criteria
> > > select the link to the cell with Ward name, as Sum all of yours 1’s and
> > > nothings or zeros. The ranges should be anchored properly (\$’s). Copy the
> > > SUMIF down for all Ward names. You will obtain the correct count in each
> > > category. For graphical representation create bar chart.
> > > For negative counting you have to know the totals of patients in Wards. You
> > > obtain them by means of COUNTIF function. So, for each Ward from your list
> > > insert this function to the next column with quite the same range and
> > > criteria arguments as with SUMIF. The count of the not-meetings is mere a
> > > difference.
> > > Do the same with Hospitals.
> > >
> > > --
> > > Petr Bezucha
> > >
> > >
> > > "matthewluck1" wrote:
> > >
> > > > Hi, I would be extremely grateful for some help with my spreadsheet.
> > > >
> > > > I have the following data which I need to use to produce charts for a
> > > > hospital report fairly urgently.
> > > >
> > > > The rows are the names of patients. This could be up to 200 for each month,
> > > > but this varies from month to month. I will have each month on a different
> > > > worksheet.
> > > >
> > > > Column F - Ward
> > > > Column G - Hospital
> > > > Column K - I have a number 1 in the row if the patient meets the criteria
> > > >
> > > > I need to produce a chart(s) to show and compare the following:
> > > >
> > > > How many patients from each ward met the criteria (as indicated by 1 in
> > > > column K) and how many did not.
> > > >
> > > > How many patients from each hospital met the criteria and how many did not.
> > > >
> > > > I would really like some help!!!
> > > >
> > > >
```
0
11/12/2008 10:41:01 AM
```Thanks again. Sorry to be a pain, but I have one more question / formula to
work out and have tried to change the ones already used but to no avail.
In cell O3 and O4 I have the total number of patients (one cell per
hospital). In cell P3 and P4, I need a formula to display the number of
patients who me the criteria.
Back to the main part of my spreadsheet, I have the name of the hospital in
the F row from F3 to F91. In the same rows but in the G column, I have a 1 if
the criteria was met and 0 if it was not.
I need some sort of formula to look at the number of patients who met the
criteria by hospital name.
Matthew
"PBezucha" wrote:
> Put
>
> =SUMIF(\$G\$1:\$G\$100;L1;\$J1:\$J100) into M-column
>
> =COUNTIF(\$G1:\$G100;L1)-M1 into N-column.
>
> I was overly didactic, take my apologies. Countif has also by one argument
> fewer, I failed to mention, and that was what may have confused you.
>
> Regards
> --
> Petr Bezucha
>
>
> "matthewluck1" wrote:
>
> > Thanks, I have got as far as counting the number of 1's in the patient list
> > using the formula: =SUMIF(\$F1:F89,L1:L26,J1:J89)
> >
> > I am stuck using the COUNTIF function to count which cells are have 0's. I
> > have my summary of all wards mentioned in column L (L1:L26). My 0's and 1's
> > are found in column J (J1:J89). The ward on the main part of the sheet is on
> > each row in column G (G1:G89).
> >
> > I would like the spreadsheet to look at the ward name in my summary in
> > column L, then if this matches in column G and there is an 0 in column J
> > alongside it, then this should be counted to add up to a cumulative total of
> > 0's per ward.
> >
> > Sorry for my lack of understanding I've not used excel for ages!!!
> >
> > "PBezucha" wrote:
> >
> > > Matthew,
> > > Classical homework, isn’t it? It’s an example for using Excel function SUMIF
> > > (find out the Help for this function). Make somewhere the column list of you
> > > Wards. Next to the first Ward insert the function SUMIF. As its Range
> > > argument select the range of your Wards from the patients list, as Criteria
> > > select the link to the cell with Ward name, as Sum all of yours 1’s and
> > > nothings or zeros. The ranges should be anchored properly (\$’s). Copy the
> > > SUMIF down for all Ward names. You will obtain the correct count in each
> > > category. For graphical representation create bar chart.
> > > For negative counting you have to know the totals of patients in Wards. You
> > > obtain them by means of COUNTIF function. So, for each Ward from your list
> > > insert this function to the next column with quite the same range and
> > > criteria arguments as with SUMIF. The count of the not-meetings is mere a
> > > difference.
> > > Do the same with Hospitals.
> > >
> > > --
> > > Petr Bezucha
> > >
> > >
> > > "matthewluck1" wrote:
> > >
> > > > Hi, I would be extremely grateful for some help with my spreadsheet.
> > > >
> > > > I have the following data which I need to use to produce charts for a
> > > > hospital report fairly urgently.
> > > >
> > > > The rows are the names of patients. This could be up to 200 for each month,
> > > > but this varies from month to month. I will have each month on a different
> > > > worksheet.
> > > >
> > > > Column F - Ward
> > > > Column G - Hospital
> > > > Column K - I have a number 1 in the row if the patient meets the criteria
> > > >
> > > > I need to produce a chart(s) to show and compare the following:
> > > >
> > > > How many patients from each ward met the criteria (as indicated by 1 in
> > > > column K) and how many did not.
> > > >
> > > > How many patients from each hospital met the criteria and how many did not.
> > > >
> > > > I would really like some help!!!
> > > >
> > > >
```
0
11/12/2008 11:39:01 AM
```Matthew,
Just by the strike of fortune I looked, after ages, into my post, and found
that after the acknowledgment an additional post came from you.
I had taken it that the wards and hospitals were classifying variables of
the same position towards meeting criteria. I don’t see any difference
whether there are wards in the column F or G or hospitals in the column F or
G, suppose there is still the same column (G or K??) with 0 or 1’s. Exactly
the same function COUNTIF and SUMIF should have been applied, as I had
Where have you lost your head since?
Regards
Petr
--
Petr Bezucha
"matthewluck1" wrote:
> Thanks again. Sorry to be a pain, but I have one more question / formula to
> work out and have tried to change the ones already used but to no avail.
>
> In cell O3 and O4 I have the total number of patients (one cell per
> hospital). In cell P3 and P4, I need a formula to display the number of
> patients who me the criteria.
>
> Back to the main part of my spreadsheet, I have the name of the hospital in
> the F row from F3 to F91. In the same rows but in the G column, I have a 1 if
> the criteria was met and 0 if it was not.
>
> I need some sort of formula to look at the number of patients who met the
> criteria by hospital name.
>
> Matthew
>
>
>
> "PBezucha" wrote:
>
> > Put
> >
> > =SUMIF(\$G\$1:\$G\$100;L1;\$J1:\$J100) into M-column
> >
> > =COUNTIF(\$G1:\$G100;L1)-M1 into N-column.
> >
> > I was overly didactic, take my apologies. Countif has also by one argument
> > fewer, I failed to mention, and that was what may have confused you.
> >
> > Regards
> > --
> > Petr Bezucha
> >
> >
> > "matthewluck1" wrote:
> >
> > > Thanks, I have got as far as counting the number of 1's in the patient list
> > > using the formula: =SUMIF(\$F1:F89,L1:L26,J1:J89)
> > >
> > > I am stuck using the COUNTIF function to count which cells are have 0's. I
> > > have my summary of all wards mentioned in column L (L1:L26). My 0's and 1's
> > > are found in column J (J1:J89). The ward on the main part of the sheet is on
> > > each row in column G (G1:G89).
> > >
> > > I would like the spreadsheet to look at the ward name in my summary in
> > > column L, then if this matches in column G and there is an 0 in column J
> > > alongside it, then this should be counted to add up to a cumulative total of
> > > 0's per ward.
> > >
> > > Sorry for my lack of understanding I've not used excel for ages!!!
> > >
> > > "PBezucha" wrote:
> > >
> > > > Matthew,
> > > > Classical homework, isn’t it? It’s an example for using Excel function SUMIF
> > > > (find out the Help for this function). Make somewhere the column list of you
> > > > Wards. Next to the first Ward insert the function SUMIF. As its Range
> > > > argument select the range of your Wards from the patients list, as Criteria
> > > > select the link to the cell with Ward name, as Sum all of yours 1’s and
> > > > nothings or zeros. The ranges should be anchored properly (\$’s). Copy the
> > > > SUMIF down for all Ward names. You will obtain the correct count in each
> > > > category. For graphical representation create bar chart.
> > > > For negative counting you have to know the totals of patients in Wards. You
> > > > obtain them by means of COUNTIF function. So, for each Ward from your list
> > > > insert this function to the next column with quite the same range and
> > > > criteria arguments as with SUMIF. The count of the not-meetings is mere a
> > > > difference.
> > > > Do the same with Hospitals.
> > > >
> > > > --
> > > > Petr Bezucha
> > > >
> > > >
> > > > "matthewluck1" wrote:
> > > >
> > > > > Hi, I would be extremely grateful for some help with my spreadsheet.
> > > > >
> > > > > I have the following data which I need to use to produce charts for a
> > > > > hospital report fairly urgently.
> > > > >
> > > > > The rows are the names of patients. This could be up to 200 for each month,
> > > > > but this varies from month to month. I will have each month on a different
> > > > > worksheet.
> > > > >
> > > > > Column F - Ward
> > > > > Column G - Hospital
> > > > > Column K - I have a number 1 in the row if the patient meets the criteria
> > > > >
> > > > > I need to produce a chart(s) to show and compare the following:
> > > > >
> > > > > How many patients from each ward met the criteria (as indicated by 1 in
> > > > > column K) and how many did not.
> > > > >
> > > > > How many patients from each hospital met the criteria and how many did not.
> > > > >
> > > > > I would really like some help!!!
> > > > >
> > > > >
```
0
PBezucha (46)
11/24/2008 2:20:10 PM
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https://ecampusontario.pressbooks.pub/fundamentalsofbusinessmath/chapter/section-4-2-markup/ | 1,659,944,144,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00138.warc.gz | 228,251,988 | 31,000 | # 4.2 Markup
As you wait in line to purchase your Iced Caramel Macchiato at Starbucks, you look at the pricing menu and think that $4.99 seems like an awful lot of money for a frozen coffee beverage. Clearly, the coffee itself doesn’t cost anywhere near that much. But then gazing around the café, you notice the carefully applied colour scheme, the comfortable seating, the high-end machinery behind the counter, and a seemingly well-trained barista who answers customer questions knowledgeably. Where did the money to pay for all of this come from? You smile as you realize your$4.99 pays not just for the macchiato, but for everything else that comes with it.
The process of taking a product’s cost and increasing it by some amount to arrive at a selling price is called markup. This process is critical to business success because every business must ensure that it does not lose money when it makes a sale. From the consumer perspective, the concept of markup helps you make sense of the prices that businesses charge for their products or services. This in turn helps you to judge how reasonable some prices are (and hopefully to find better deals).
## The Components in a Selling Price
Before you learn to calculate markup, you first have to understand the various components of a selling price. Then, in the next section, markup and its various methods of calculation will become much clearer.
When your business acquires merchandise for resale, this is a monetary outlay representing a cost. When you then resell the product, the price you charge must recover more than just the product cost. You must also recover all the selling and operating expenses associated with the product. Ultimately, you also need to make some money, or profit, as a result of the whole process.
Most people think that marking up a product must be a fairly complex process. It is not. Below is an illustration of the relationship between cost, expenses, and profits in calculating the selling price.
#### The selling price of a product
\begin{align*} \text{selling price}&=\text{cost}+\overbrace{\text{expenses}+\text{profit}}^{\text{markup amount}}\\ &=\text{cost}+\text{markup amount}\\ \\ \\ S&=C+\overbrace{E+P}^{M}\\ &=C+{M} \end{align*}
$S$ is selling price: Once you calculate what the business paid for the product (cost), the bills it needs to cover (expenses), and how much money it needs to earn (profit), you arrive at a selling price by summing the three components.
$C$ is cost: The cost is the amount of money that the business must pay to purchase or manufacture the product. If manufactured, the cost represents all costs incurred to make the product. If purchased, this number results from applying an appropriate discount formula from Section 4.1. There is a list price from which the business will deduct discounts to arrive at the net price. The net price paid for the product equals the cost of the product. If a business purchases or manufactures a product for $10 then it must sell the product for at least$10. Otherwise, it fails to recover what was paid to acquire or make the product in the first place—a path to sheer disaster!
$E$ is expenses: Expenses are the financial outlays involved in selling the product. Beyond just purchasing the product, the business has many more bills to pay, including wages, taxes, leases, equipment, electronics, insurance, utilities, fixtures, décor, and many more. These expenses must be recovered and may be calculated as:
• A fixed dollar amount per unit.
• A percentage of the product cost. For example, if a business forecasts total merchandise costs of $100,000 for the coming year and total business expenses of$50,000, then it may set a general guideline of adding 50% ($50,000 ÷$100,000) to the cost of a product to cover expenses.
• A percentage of the product selling price based on a forecast of future sales. For example, if a business forecasts total sales of $250,000 and total business expenses of$50,000, then it may set a general guideline of adding 20% ($50,000 ÷$250,000) of the selling price to the cost of a product to cover expenses.
$P$ is profit: Profit is the amount of money that remains after a business pays all of its costs and expenses. A business needs to add an amount above its costs and expenses to allow it to grow. If it adds too much profit, though, the product’s price will be too high, in which case the customer may refuse to purchase it. If it adds too little profit, the product’s price may be too low, in which case the customer may perceive the product as shoddy and once again refuse to purchase it. Many businesses set general guidelines on how much profit to add to various products. As with expenses, this profit may be expressed as:
• A fixed dollar amount per unit.
• A percentage of the product cost.
• A percentage of the selling price.
$M$ is markup amount: Markup is the amount of money that has to be added to the cost of the item to cover for expenses and produce profit from sale of the item.
For example, assume a business pays a net price of $75 to acquire a product. Through analyzing its finances, the business estimates expenses at$25 per unit, and it figures it can add $50 in profit. Calculate the selling price. We have to calculate the selling price and we are given the cost $C = 75$, the expenses $E = 25$ and the profit $P = 50$. Therefore, the unit selling price is: $\text{selling price} =C+E+P= 75+ 25+ 50= 150$ When building the equation, you must adhere to the basic rule of linear equations requiring all terms to be in the same unit. That is, you could use the formula above to solve for the selling price of an individual product, where the three components are the unit cost, unit expenses, and unit profit. When you add these, you calculate the unit selling price. Alternatively, you could use the same formula in an aggregate form where the three components are total cost, total expenses, and total profit. In this case, the selling price is a total selling price, which is more commonly known as total revenue. But you cannot mix individual components with aggregate components. The most common mistake in working with pricing components occurs in understanding the context, the task, and the conditions given. It is critical to identify and label information correctly. You have to pay attention to details such as whether you are expressing the expenses in dollar format or as a percentage of either cost or selling price. Systematically work your way through the information provided piece by piece to ensure that you do not miss an important detail. Give It Some Thought: 1. What three components make up a selling price? In what units are these components commonly expressed? 2. In what three ways are expenses and profits expressed? 3. What is the relationship between net price and cost? #### Example 4.2A – Setting a Price on Fashion in Dollars Mary’s Boutique purchases a dress for resale at a cost of$23.67. The owner determines that each dress must contribute $5.42 to the expenses of the store. The owner also wants this dress to earn$6.90 toward profit. What is the regular selling price for the dress?
Answer: regular selling price for the dress? $\rightarrow S$.
Conditions: unit cost of the dress $C = \ 23.67$, unit expense $E = \ 5.42$,
unit profit $P = \ 6.90$
$\text{selling price} =C+E+P= 23.67+5.42+6.90=\ 35.99$
Therefore, Mary’s Boutique will set the regular price of the dress at $35.99. #### Example 4.2B – Setting the Price Using Percentage of Cost John’s Discount Store just completed a financial analysis. The company determined that expenses average 20% of the product cost and profit averages 15% of the product cost. John’s Discount Store purchases Chia Pets from its supplier for an MSRP of$19.99 less a trade discount of 45%. What will be the regular selling price for the Chia Pets?
Answer: the regular selling price for the Chia pets?
\begin{align*} \text{reg. selling price}&=\overset{?}{\text{cost}}+\overset{?}{\text{expenses}}+\overset{?}{\text{profit}}\\ \\ &=\overset{?}{\text{cost}}+0.2(\overset{?}{\text{cost}})+0.15(\overset{?}{\text{cost}})\\ \\ &=1.35\cdot(\overset{?}{\text{cost}})\\ \\ &=1.35\cdot\text{MSRP}(1-(\text{trade disc. rate}))\\ \\ &=1.35\cdot 19.99\cdot (1-0.45)\\ \\ &=\14.84 \end{align*}
Therefore, John’s Discount Store will sell the Chia Pet for $14.84. #### Example 4.2C – Setting the Price Using Percentage of Selling Price Based on last year’s results, Benthal Appliance learned that its expenses average 30% of the regular selling price. It wants a 25% profit based on the selling price. If Benthal Appliance purchases a fridge for$1,200, what is the regular unit selling price?
Plan:
You are looking for the regular unit selling price for the fridge, or $S$.
Understand:
Step 1: The cost, expenses, and profit for the fridge are known:
\begin{align*} E &= 30\%\textrm{ of S, or }0.3S\\ P &= 25\%\textrm{ of S, or }0.25S\\ C &= 1,200.00\\ \end{align*}
Step 2: Apply Formula 4.5.
Perform:
Step 2:
\begin{align*} S&= 1,200.00+0.3S+0.25S\\ S&= 1,200+0.55S\\ S-0.55S&= 1,200.00\\ 0.45S&= 1,200.00\\ S&= 2,666.67\\ \end{align*}
Present
Benthal Appliance should set the regular selling price of the fridge at $2,666.67. #### Example 4.2D – Using Selling Price to Figure Out the Cost If a company knows that its profits are 15% of the selling price and expenses are 30% of cost, what is the cost of an MP3 player that has a regular selling price of$39.99?
You are looking for the cost of the MP3 player, or $C$.
Condition(s):
selling price = cost + expenses + profit
So,
\begin{align*} &\overset{\checkmark}{\text{selling price}}=\overset{?}{\text{cost}}+\overset{?}{\text{expenses}}+\overset{?}{\text{profit}}\\\\ &\overset{\checkmark}{\text{selling price}}=\overset{?}{\text{cost}}+0.3(\overset{?}{\text{cost}})+0.15(\overset{\checkmark}{\text{selling price}})\\\\ \Rightarrow &\text{selling price}-0.15(\text{selling price})=\text{cost}+0.3(\text{cost})\\\\ \Rightarrow &0.85(\text{selling price})=1.3(\text{cost})\\\\ \Rightarrow &\text{cost}=\frac{0.85}{1.3}(\text{selling price})\\\\ \Rightarrow &\text{cost}=\frac{0.85}{1.3}\cdot 39.99 = \26.15 \end{align*}
Therefore, the cost of the MP3 Player is $26.15. #### Example 4.2E – Determining Profitability Peak of the Market considers setting the regular unit selling price of its strawberries at$3.99 per kilogram. If it purchases these strawberries from the farmer for 2.99 per kilogram and expenses average 40% of product cost, does Peak of the Market make any money? Answer: profit = ? ($P =?$) \begin{align*} \overset{S}{\text{selling price}} &= \overset{C}{\text{cost}}+\overset{E}{\text{expenses}}+\overset{P}{\text{profit}}\\ P&=\overset{\checkmark}{S}-\overset{\checkmark}{C}-\overset{?}{E}\\ P&=S-C-0.4C\\ &=3.99-2.99-0.4(3.99)\\ &=- 0.20\\ \end{align*} The negative sign on the profit means that Peak of the Market would take a loss of0.20 per kilogram if it sells the strawberries at 3.99. Unless Peak of the Market has a marketing reason or sound business strategy for doing this, the company should reconsider its pricing. ## Calculating the Markup Dollars Most companies sell more than one product, each of which has different price components with varying costs, expenses, and profits. Can you imagine trying to compare 50 different products, each with three different components? You would have to juggle 150 numbers! To make merchandising decisions more manageable and comparable, many companies combine expenses and profit together into a single quantity, either as a dollar amount or a percentage. This section focuses on the markup as a dollar amount. One of the most basic ways a business simplifies its merchandising is by adding the dollar amounts of its expenses and profits, as expressed in the basic selling price formula below: #### Markup amount \begin{align*} \text{markup amount}&=\text{expenses}+\text{profit}\\ \\ M& =E+P \end{align*} $M$ is Markup amount: Markup is taking the cost of a product and converting it into a selling price. The markup amount represents the dollar amount difference between the cost and the selling price. $E$ is Expenses: The expenses associated with the product. $P$ is Profit: The profit earned when the product sells. Note that since the markup amount ($M$) represents the expenses ($E$) and profit ($P$) combined, you can substitute the variable for markup amount instead of expenses plus profit to calculate the regular selling price. #### Selling price using markup $S=C+M$ $S$ is Selling price: The regular selling price of the product. $C$ is Cost: The amount of money needed to acquire or manufacture the product. If the product is being acquired, the cost is the same amount as the net price paid. $M$ is Markup Amount: This is the single value that represents the total of the expenses and profit. Recall from Example 4.2D that the MP3 player’s expenses are7.84, the profit is $6.00, and the cost is$26.15. Here is how we would calculate the markup amount and the selling price.
markup:
$\text{markup}=\text{expenses}+\text{profit}=7.84+6.00=\ 13.84$
selling price:
$\text{selling price}=\text{cost}+\text{markup}= 26.15+ 13.84=\ 39.99$
You might have already noticed that many of the formulas in this chapter are interrelated. The same variables appear numerous times but in different ways. To help visualize the relationship between the various formulas and variables, many students have found it helpful to create a markup chart, as shown below.
This chart demonstrates the relationships between selling price, cost, expenses, profit and markup amount. It is evident that the selling price (the green line) consists of cost, expenses, and profit (the red line); or it can consist of cost and the markup amount (the blue line). The markup amount on the blue line consists of the expenses and profit on the red line.
#### Example 4.2F – Markup as a Dollar Amount
A cellular retail store purchases an iPhone with an MSRP of 779 less a trade discount of 35% and volume discount of 8%. The store sells the phone at the MSRP. a. What is the markup amount? b. If the store knows that its expenses are 20% of the cost, what is the store’s profit? Answer: a. markup amount =? Conditions: list price (MSRP), trade discount, volume discount, selling price (MSRP) \begin{align*} &\text{selling price}=\text{cost}+\text{markup}\\ \Rightarrow \text{markup}&=\overset{\checkmark}{\text{selling price}}-\overset{?}{\text{cost}}\\ \\ &=\text{selling price}\\ &\ \ \ -(\text{list price})(1-\text{trade disc. rate})(1-\text{volume disc. rate})\\ \\ &= 779-779(1-0.35)(1-0.08)\\ \\ &=\313.16 \end{align*} b. profit = ? Conditions: expenses = 20% of cost \begin{align*} \text{markup}&=\text{expenses}+\text{profit}\\ \\ \Rightarrow \text{profit}&=\overset{\checkmark}{\text{markup}}-\overset{?}{\text{expenses}}\\ \\ &=\text{markup}-0.2(\text{cost})\\ \\ &=\text{markup}-0.2(\text{list price})(1-\text{trade disc.rate})(1-\text{vol. disc. rate})\\ \\ &=313.16-0.2(779)(1-0.35)(1-0.08)\\ \\ &=\219.99\\ \end{align*} The markup amount for the iPhone is313.16. When the store sells the phone for $779.00, its profit is$219.99.
## Calculating the Markup Percent
It is important to understand markup in terms of the actual dollar amount; however, it is more common in business practice to calculate the markup as a percentage. There are three benefits to converting the markup dollar amount into a percentage:
1. Easy comparison of different products having vastly different price levels and costs. Markup rate can help you see how each product contributes toward the financial success of the company. For example, if a chocolate bar has a 50¢ markup included in a selling price of $1, while a car has a$1,000 markup included in a selling price of $20,000, it is difficult to compare the profitability of these items. If these numbers were expressed as a percentage of the selling price such that the chocolate bar has a 50% markup and the car has a 5% markup, it is clear that more of every dollar sold for chocolate bars goes toward list profitability. 2. Simplified translation of costs into a regular selling price—a task that must be done for each product, making it helpful to have an easy formula, especially when a company carries hundreds, thousands, or even tens of thousands of products. For example, if all products are to be marked up by 50% of cost, an item with a$100 cost can be quickly converted into a selling price of $150. 3. An increased understanding of the relationship between costs, selling prices, and the list profitability for any given product. For example, if an item selling for$25 includes a markup on selling price of 40% (which is $10), then you can determine that the cost is 60% of the selling price ($15) and that $10 of every$25 item sold goes toward list profits.
You can translate the markup dollars into a percentage using two methods, which express the amount either as a percentage of cost or as a percentage of selling price:
• Method 1: Markup as a Percentage of Cost.
This method expresses the markup rate using cost as the base. Many companies use this technique internally because most accounting is based on cost information. The result, known as the markup on cost percentage, allows a reseller to convert easily from a product’s cost to its regular unit selling price.
• Method 2: Markup as a Percentage of Selling Price.
This method expresses the markup rate using the regular selling price as the base. Many other companies use this method, known as the markup on selling price percentage, since it allows for quick understanding of the portion of the selling price that remains after the cost of the product has been recovered. This percentage represents the list profits before the deduction of expenses and therefore is also referred to as the list profit margin.
The markup rate on cost and markup rate on selling price are calculated using the relationships below:
#### Markup rate on cost
\begin{align*} \text{markup rate on cost}&=\frac{\text{markup amount}}{\text{cost}}\\ \\ m_C&=\frac{M}{C}\\ \\ &=\frac{M}{C}\cdot 100\% \end{align*}
$m_C$ is the markup rate on cost (or markup with respect to cost): This is the percentage rate by which the cost of the product needs to be increased to arrive at the selling price for the product.
$M$ is the markup amount: The total dollars of the expenses and the profits; this total is the difference between the cost and the selling price.
$C$ is the cost: The amount of money needed to acquire or manufacture the product. If the product is being acquired, the cost is the same amount as the net price paid.
× 100 is Percent Conversion: The markup on cost is always a percentage.
#### Markup rate on selling price
\begin{align*} \text{markup rate on selling price}&=\frac{\text{markup amount}}{\text{selling price}}\\ \\ m_S&=\frac{M}{S}\\ \\ &=\frac{M}{S}\cdot 100\% \end{align*}
$m_S$ is the markup rate on selling price (or markup with respect to price): This is the percentage of the selling price that remains available as list profits after the cost of the product is recovered.
$M$ is the markup amount: The total dollars of the expenses and the profits; this total is the difference between the cost and the selling price.
$S$ is the selling price: The regular selling price of the product.
× 100 is the percent conversion: The markup on selling price is always a percentage.
Continuing to work with Example 4.2D, recall that the cost of the MP3 player is $26.15, the markup amount is$13.84, and the selling price is 39.99. Calculate both markup percentages. Step 1: The known variables are \begin{align*} C &=26.15\\
M &= $13.84\\ S &=$39.99\\
\end{align*}
Step 2: To calculate the markup rate on cost:
\begin{align*}m_C\%=\frac{$13.84}{$ 26.15}=52.9254\%\end{align*}
In other words, you must add 52.9254% of the cost on top of the unit cost to arrive at the regular unit selling price of 39.99. To calculate the markup rate on selling price: \begin{align*}m_S\%=\frac{ 13.84}{39.99}=34.6087\%\end{align*} In other words, 34.6087% of the selling price represents list profits after the business recovers the26.15 cost of the MP3 player.
Merchandising involves many variables. Nine formulas have been established so far, and a few more are yet to be introduced. Though you may feel bogged down by all of these formulas, just remember that you have encountered most of these merchandising concepts since you were very young and that you interact with retailers and pricing every day. This chapter merely formalizes calculations you already perform on a daily basis, whether at work or at home. The calculation of discounts is no different than going to Walmart and finding your favourite CD on sale. You know that when a business sells a product, it has to recoup the cost of the product, pay its bills, and make some money. And you have worked with percentages since elementary school.
Do not get stuck in the formulas. Think about the concept presented in the question. Change the scenario of the question and put it in the context of something more familiar. Ultimately, if you really have difficulties then look at the variables provided and cross-reference them to the merchandising formulas. Your goal is to find formulas in which only one variable is unknown. These formulas are solvable. Then ask yourself, “How does knowing that new variable help solve any other formula?”
You do not need to get frustrated. Just be systematic and relate the question to what you already know.
Sometimes you need to convert the markup on cost percentage to a markup on selling price percentage, or vice versa. Two shortcuts allow you to convert easily from one to the other:
\begin{align*} m_C&=\frac{m_S}{1-m_S}\\ \\ m_S&=\frac{m_C}{1+m_C}\\ \end{align*}
Notice that these formulas are very similar. How do you remember whether to add or subtract in the denominator? In normal business situations, the $m_C$ is always larger than the $m_S$. Therefore, if you are converting one to the other you need to identify whether you want the percentage to become larger or smaller.
• To calculate $m_C$, you want a larger percentage. Therefore, make the denominator smaller by subtracting $m_S$ from 1.
• To calculate $m_S$, you want a smaller percentage. Therefore, make the denominator larger by adding $m_C$ to 1.
#### Give It Some Thought:
4. The markup on selling price percentage can be higher than 100%.
5. The markup dollar amount can be more than the selling price.
6. The markup on cost percentage can be higher than 100%.
7. The markup on cost percentage in most business situations is higher than the markup on selling price percentage.
8. If you know the markup on cost percentage and the cost, you can calculate a selling price.
9. If you know the markup on selling price percentage and the cost, you can calculate a selling price.
#### Example 4.2G – Markup as a rate
A large national retailer wants to price a Texas Instruments BAII Plus calculator at the MSRP of $39.99. The retailer can acquire the calculator for$17.23.
a. What is the markup rate on cost?
b. What is the markup rate on selling price?
a. markup rate on cost =?
Conditions: selling price = $39.99, cost =$17.23
\begin{align*} \text{markup rate on cost}&=\frac{\overset{?}{\text{markup amount}} }{\underset{\checkmark}{\text{cost}}}\\ \\ &=\frac{\text{selling price}-\text{cost}}{\text{cost}}\\ \\ &=\frac{39.99-17.23}{17.23}\\ \\ &=1.320952=132.0952\% \end{align*}
b. markup rate on selling price =?
Conditions: selling price = $39.99, cost =$17.23
\begin{align*} \text{markup rate on selling price}&=\frac{\overset{?}{\text{markup amount}} }{\underset{\checkmark}{\text{selling price}}}\\ \\ &=\frac{\text{selling price}-\text{cost}}{\text{selling price}}\\ \\ &=\frac{39.99-17.23}{39.99}\\ \\ &=0.569142=56.9142\% \end{align*}
The markup on cost percentage is 132.0952%. The markup on selling price percentage is 56.9142%.
## Break-Even Pricing
In running a business, you must never forget the “bottom line.” In other words, if you fully understand how your products are priced, you will know when you are making or losing money. Remember, if you keep losing money you will not stay in business for long! 2% of new businesses will not make it past their first year, and 37% fail in their first five years. This number becomes even more staggering with a 57% failure rate within the first decade.[1] Do not be one of these statistics! With your understanding of markup, you now know what it takes to break even in your business. Break-even means that you are earning no profit, but you are not losing money either. Your profit is zero.
If the regular unit selling price must cover three elements—cost, expenses, and profit—then the regular unit selling price must exactly cover your costs and expenses when the profit is zero. In other words,:
$\text{break-even selling price per unit}=\text{cost per unit}+\text{expenses per unit}$
This is not a new formula. It just summarizes that at break-even there is no profit or loss, so the profit ( $P$) is eliminated from the formula.
Recall Example 4.2D that the cost of the MP3 player is $26.15 and expenses are$7.84.
Therefore
$\text{break-even price}=26.15+ 7.84=\ 33.99$
This means that if the MP3 player is sold for anything more than $33.99, it is profitable; if it is sold for less, then the business does not cover its costs and expenses and takes a loss on the sale. #### Example 4.2H – Knowing Your Break-Even Price John is trying to run an eBay business. His strategy has been to shop at local garage sales and find items of interest at a great price. He then resells these items on eBay. On John’s last garage sale shopping spree, he only found one item—a Nintendo Wii that was sold to him for$100. John’s vehicle expenses (for gas, oil, wear/tear, and time) amounted to $40. eBay charges a$2.00 insertion fee, a flat fee of $2.19, and a commission of 3.5% based on the selling price less$25. What is John’s minimum list price for his Nintendo Wii to ensure that he at least covers his expenses?
Answer: break-even selling price = ?
Conditions: selling price = cost + expenses
cost = $100.00, vehicle expenses =$40.00, insertion expenses = $2.00, flat fee expense=$2.19, commission expense = 3.5% on selling price less 25.00 \begin{align*} \overset{S_{BE}}{\text{b-e selling price}}&=\overset{C\checkmark}{\text{cost}}+\overset{E_{vehicle}\checkmark}{\text{vehicle expenses}}+ \overset{E_{insertion}\checkmark}{\text{insertion expenses}}\\ &\ \ \ \ +\underset{E_{flat}\checkmark}{\text{flat fee expense}}+\underset{?}{\text{commission expense}}\\ \\ \Rightarrow S_{BE}&={C}+{E_{vehicle}}+{E_{insertion}}+{E_{flat}}+0.035(S_{BE}-25)\\ \\ \Rightarrow S_{BE}&=100+40+2+2.19+0.035S_{BE}-0.875 \\ &\Rightarrow S_{BE}-0.035S_{BE}=100+40+2+2.19+0.035-0.875\\ \\ &\Rightarrow 0.965S_{BE}=143.315\\ \\ &\Rightarrow S_{BE}=148.51 \end{align*} Therefore, at a price of148.51, John would cover all of his costs and expenses but realize no profit or loss. Therefore, $148.51 is his minimum price. ## Give It Some Thought Answers 1. Cost, expenses, and profit. They are expressed either per unit or as a total. 2. A specific dollar amount, a percentage of cost, or a percentage of the selling price. 3. The net price paid for a product is the same as the cost of the product. 4. False. The markup amount is a portion of the selling price and therefore is less than 100%. 5. False. The markup amount plus the cost equals the selling price. It must be less than the selling price. 6. True. A cost can be doubled or tripled (or increased even more) to reach the price. 7. True. The base for markup on cost percentage is smaller, which produces a larger percentage. 8. True. You could combine formulas to arrive at the selling price. 9. True. You could convert the $m_S$ to a $m_C$ and solve as in the previous question. ## Exercises Round all money to two decimals and percentages to four decimals for each of the following exercises. ### Mechanics For questions 1–8, solve for the unknown variables (identified with a ?) based on the information provided. Regular Unit Selling Price Cost Expenses Profit Markup Amount Break-Even Price Markup on Cost Markup on Selling Price 1. ?$188.42 $48.53$85.00 ? ? ? ? 2. $999.99 ? 30% of $C$ 23% of $C$ ? ? ? ? 3. ? ? ? 10% of $S$$183.28 ? 155% ? 4. $274.99 ? 20% of $S$ ? ? ? ? 35% 5. ? ? 45% of $C$ ?$540.00 $1,080.00 ? ? 6. ?$200 less 40% ? 15% of $S$ ? ? 68% ? 7. ? ? $100.00 ?$275.00 ? ? 19% 8. ? ? 15% of $C$ 12% of $S$ ? $253.00 ? ? ### Applications 9. If a pair of sunglasses sells at a regular unit selling price of$249.99 and the markup is always 55% of the regular unit selling price, what is the cost of the sunglasses?
10. A transit company wants to establish an easy way to calculate its transit fares. It has determined that the cost of a transit ride is $1.00, with expenses of 50% of cost. It requires$0.75 profit per ride. What is its markup on cost percentage?
11. Daisy is trying to figure out how much negotiating room she has in purchasing a new car. The car has an MSRP of $34,995.99. She has learned from an industry insider that most car dealerships have a 20% markup on selling price. What does she estimate the dealership paid for the car? 12. The markup amount on an eMachines desktop computer is$131.64. If the machine regularly retails for $497.25 and expenses average 15% of the selling price, what profit will be earned? 13. Manitoba Telecom Services (MTS) purchases an iPhone for$749.99 less discounts of 25% and 15%. MTS’s expenses are known to average 30% of the regular unit selling price.
a. What is the regular unit selling price if a profit of $35 per iPhone is required? b. What are the expenses? c. What is the markup on cost percentage? d. What is the break-even selling price? 14. A snowboard has a cost of$79.10, expenses of $22.85, and profit of$18.00.
a. What is the regular unit selling price?
b. What is the markup amount?
c. What is the markup on cost percentage?
d. What is the markup on selling price percentage?
e. What is the break-even selling price? What is the markup on cost percentage at this break-even price?
### Challenge, Critical Thinking, & Other Applications
15. A waterpark wants to understand its pricing better. If the regular price of admission is $49.95, expenses are 20% of cost, and the profit is 30% of the regular unit selling price, what is the markup amount? 16. Sally works for a skateboard shop. The company just purchased a skateboard for$89.00 less discounts of 22%, 15%, and 5%. The company has standard expenses of 37% of cost and desires a profit of 25% of the regular unit selling price. What regular unit selling price should Sally set for the skateboard?
17. If an item has a 75% markup on cost, what is its markup on selling price percentage?
18. A product received discounts of 33%, 25%, and 5%. A markup on cost of 50% was then applied to arrive at the regular unit selling price of $349.50. What was the original list price for the product? 19. Mountain Equipment Co-op (MEC) wants to price a new backpack. The backpack can be purchased for a list price of$59.95 less a trade discount of 25% and a quantity discount of 10%. MEC estimates expenses to be 18% of cost and it must maintain a markup on selling price of 35%.
a. What is the cost of backpack?
b. What is the markup amount?
c. What is the regular unit selling price for the backpack?
d. What profit will Mountain Equipment Co-op realize?
e. What happens to the profits if it sells the backpack at the MSRP instead?
20. Costco can purchase a bag of Starbucks coffee for $20.00 less discounts of 20%, 15%, and 7%. It then adds a 40% markup on cost. Expenses are known to be 25% of the regular unit selling price. a. What is the cost of the coffee? b. What is the regular unit selling price? c. How much profit will Costco make on a bag of Starbucks coffee? d. What markup on selling price percentage does this represent? e. Repeat questions (a) through (d) if the list price changes to$24.00.
1. Innovation, Science and Economic Development Canada. Canadian New Firms: Birth and Survival Rates over the Period 2002–2014, May 2018. https://www.ic.gc.ca/eic/site/061.nsf/eng/h_03075.html#section-3.1 | 8,247 | 32,313 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | longest | en | 0.962562 |
http://www.enki-village.com/binary-to-octal.html | 1,488,113,955,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172000.93/warc/CC-MAIN-20170219104612-00014-ip-10-171-10-108.ec2.internal.warc.gz | 403,221,889 | 8,336 | Consider any binary number. A binary number consists of only 0's and 1's.
## Part 1
1
Group all the bits in the binary number, as a set of 3 bits.
Start from the right: approaching from least significant bit to the most significant bit.
• If any bit remains ungrouped in a set of 3 bits then you can add a leading '0' to it on the left, to make it a perfect set.
2
Find the octal numbers and their equivalent binary numbers, as given in the picture.
3
Replace each 3-bit binary number set to its equivalent octal number.
Hence the number obtained after replacing is the equivalent octal number. | 147 | 604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-09 | longest | en | 0.920626 |
https://www.mathallstar.org/Problem/Details/364 | 1,611,681,262,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00319.warc.gz | 887,141,490 | 2,568 | LogicalAndReasoning AMC10/12
2015
Problem - 364
A league with $12$ teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores $2$ points for every game it wins and $1$ point for every game it draws. Which of the following is NOT a true statement about the list of $12$ scores? | 97 | 389 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-04 | latest | en | 0.913063 |
https://www.convertunits.com/from/ell/to/point+%5BTeX%5D | 1,638,781,282,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00422.warc.gz | 776,123,311 | 12,996 | ## ››Convert ell [English] to point [TeX]
ell point [TeX]
Did you mean to convert ell [English] ell [Scotland] to point [Adobe] point [Britain, US] point [Didot] point [TeX]
How many ell in 1 point [TeX]? The answer is 0.00030748889195101.
We assume you are converting between ell [English] and point [TeX].
You can view more details on each measurement unit:
ell or point [TeX]
The SI base unit for length is the metre.
1 metre is equal to 0.87489063867017 ell, or 2845.2755906694 point [TeX].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between ell [English] and points.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of ell to point [TeX]
1 ell to point [TeX] = 3252.15 point [TeX]
2 ell to point [TeX] = 6504.3 point [TeX]
3 ell to point [TeX] = 9756.45 point [TeX]
4 ell to point [TeX] = 13008.6 point [TeX]
5 ell to point [TeX] = 16260.75 point [TeX]
6 ell to point [TeX] = 19512.9 point [TeX]
7 ell to point [TeX] = 22765.05 point [TeX]
8 ell to point [TeX] = 26017.2 point [TeX]
9 ell to point [TeX] = 29269.35 point [TeX]
10 ell to point [TeX] = 32521.5 point [TeX]
## ››Want other units?
You can do the reverse unit conversion from point [TeX] to ell, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 524 | 1,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.752539 |
https://codedump.io/share/rNX2tu2dG7dW/1/calculate-weekly-mean-from-time-series-with-missing-data-in-r | 1,529,585,823,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00080.warc.gz | 589,058,386 | 9,222 | ulima2_ - 1 year ago 99
R Question
# Calculate weekly mean from time series with missing data in R
I have a time series object with daily values that starts in the 19th century and reaches into the 20th century. There are a lot of missing values in there.
I'm trying to calculate weekly means and here's a minimal example:
``````library(zoo)
library(xts)
# Create time series that starts in 19th century
T <- 100 # number of days
myTS <- xts(rnorm(T), as.Date(1:T, origin="1899-11-05"))
# Insert some missing values
myTS[4:7] <- NA
myTS[33:34] <- NA
myTS[67:87] <- NA
# Try calculating weekly means
weekData <- apply.weekly(myTS, colMeans, na.rm = TRUE)
``````
which only returns the weekly mean for the last week:
1900-02-13 [some value]
I use
`colMeans`
`mean`
because I'm operating on a larger dataset with several variables.
I would like the mean for all the weeks. Does somebody have an idea what I'm doing wrong?
Updated based on your comment to use week-year combination:
``````library(zoo)
library(xts)
# Create time series that starts in 19th century
T <- 100 # number of days
myTS <- xts(rnorm(T), as.Date(1:T, origin="1899-11-05"))
# Insert some missing values
myTS[4:7] <- NA
myTS[33:34] <- NA
myTS[67:87] <- NA
# Let's use a flexible class
myTS <- data.frame(dates=index(myTS),v1=myTS[,1])
# Here's an easy way to transform dates to weeks
require(lubridate)
week_num <- week(myTS[,1])
year_num <- year(myTS[,1])
week_yr <- paste(week_num, year_num)
# Weekly means
aggregate(myTS\$v1,by=list(week_yr),mean,na.rm=T)
``````
`````` Group.1 x
1 1 1900 0.05405322
2 2 1900 0.31981319
3 3 1900 NaN
4 4 1900 NaN
5 45 1899 0.85081053
6 46 1899 0.34064255
7 47 1899 0.02880424
8 48 1899 -0.34408119
9 49 1899 -0.38089026
10 5 1900 0.62292188
11 50 1899 -0.59666955
12 51 1899 0.57756987
13 52 1899 -0.41325485
14 53 1899 0.88013634
15 6 1900 0.01514668
16 7 1900 -0.50863942
``````
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 720 | 2,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-26 | latest | en | 0.812616 |
https://math.libretexts.org/Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/05%3A__Algebra | 1,719,039,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00425.warc.gz | 335,765,833 | 30,117 | # 5: Algebra
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Thumbnail: The red and blue lines on this graph have the same slope (gradient); the red and green lines have the same y-intercept (cross the y-axis at the same place). (CC BY-SA 1.0; ElectroKid via Wikipedia)
This page titled 5: Algebra is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 770 | 2,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.364288 |
www.ctech.com | 1,695,791,634,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00095.warc.gz | 798,200,587 | 15,322 | Precision Issues in EVS
This is an important subject that is long overdue to be discussed in detail. There are several aspects of numeric precision that are relevant to EVS and our users, but two in particular warrant a detailed description.
1. Precision of calculations and deliverable results
2. Precision of graphical displays
Both topics above are driven by the way that computers represent numbers, therefore we will address first how computers represent Float and Double numbers. For those of you not familiar with the terms “Float” and “Double” as used in computer jargon, computer programming languages represent numbers in several different ways (in order of increasing precision) including “Short”, “Integer”, “Half Float”, “Float” and “Double”.
In EVS, Float and Double are most used, and the limitations of Float are the most important to understand. The table below lists the Float and Double Precision for Ranges of numbers from 1.0 to 1 trillion.
Let me explain what this table is telling us. For Float Precision, it tells us:
• For numbers in the range of 1 to 2, we can expect precision of 1 part in ~10 million.
• For numbers between 4 to 8 million, our precision goes down to 1 part in 0.5.
• For numbers between 1 to 2 billion, our precision is only 1 part in 128!
What this shows us is that regardless of the magnitude of the numbers, Floats have only 6-7 digits of precision and Doubles have 15-16 digits of precision.
When I look at this table from a non-programmer’s perspective, the obvious question is “why don’t you just do everything in Double Precision?”. Our answer is that we do everything we can in Double Precision, but all computer graphic cards render their displays with Floats.
We do use Double Precision for all internal computations including all of our estimation methods (e.g. kriging), and volumetric computations. However, when it is time to send this “double precise” model to the viewer, we’re subject to the limitations of graphics cards which do everything in Float Precision.
The consequences of this are:
1. You don’t need to worry about the precision of analytical results in EVS.
1. Volumes and masses in volumetrics will be accurate.
2. Sites with large X and/or Y coordinates and small X-Y extents can have some problems in the viewer.
What do we mean by large X-Y coordinate and small extents? With the table above, you can see if your site might have issues. Here is how:
Here’s an example.
• Take the larger of your Maximum X or Y coordinate and find the “Float Precision” in the table above.
• If your Y coordinates are bigger than X coordinates and Ymax = 10,315,442, then the Float Precision is 1.0
• Use file_statistics to get the X-Y extents. Let’s assume Xext = 65.2 and Yext = 62.8. Take the lesser extent = 62.8
• Compute the Precision Ratio by taking the lesser extent divided by the Float Precision
• If this is less than ~400, you’ll likely see some viewer issues. In the above case it would be < 63, so the problems would be very obvious.
• If you zoom into a small region of your site, a Precision Ratio of 400 may not be high enough.
What do we mean by “problems in the viewer”?
• The good news is that your model will look exactly the same (in EVS’s viewer) when your Precision Ratio is 63 or 1,000.
• Problems are only obvious when you try rotating or zooming on the model
• Rotation problems are the most obvious. The model will be jerky when rotating.
• Zooming is only obvious if you use Shift-Middle-button with your mouse. This (normally) gives smooth zooming and it too will be jerky.
• If you tend to use the Az-El controls, you might never see the above issues.
• Making a bitmap animation with rotations and or zooming will likely be a big problem.
• Your bitmap animation (e.g. AVI file) will clearly exhibit these problems to the point that it will likely be unacceptable.
What other problems might you encounter?
• Exporting models to other software (e.g. CAD) may be problematic.
• Tests in Autodesk’s AutoCAD 2020 showed no problems importing DWG files created in EVS with low Precision Ratios. (however, it does not have an option to perform smooth rotations)
• Tests in ZWCAD 2020 Pro showed some problems importing DWG files created in EVS with low Precision Ratios.
• When displaying the model in Wireframe model it showed no issues
• When rendering as surfaces (Gouraud or Flat), there were severe rendering problems.
• Tests in ESRI’s ArcScene 10.7.1 had no major problems (but rotation is slow for modest sized models).
How can you fix this issue for your project?
• If you discover that this is happening to you, there is a relatively straightforward solution:
• Remember, this only happens when you have big coordinates and small coordinate extents!
• Translate the coordinates of your data (all of it) to move the range of your coordinates
• For example, if your:
• X coordinates range from 211,400 to 211,470 (extent 70), and
• Y coordinates range from 6,133,200 to 6,133,260 (extent 60), then
• Your precision is 0.5 with a Precision ratio of 120 (60 / 0.5)
• If you translate the Y coordinates by -6,000,000, then
• Your precision drops to 0.015625
• Your Precision Ratio increases to 3,840
• What are the consequences and limitations in EVS?
• Your model’s coordinates displayed in the viewer will be offset by 6 million.
• Probing will report Y coordinates that are off by 6 million.
• Normally the axes module would label the translated Y coordinates (offset by 6 million)
• However, the axes module has a “Set Axes Origin” toggle. Turn it on and set Y to be -6,000,000.
• This will correct the coordinates displayed by axes.
• For non-ASCII data, use this approach:
• For CAD and Shapefiles, use transform_field to move the coordinates
• For georeferenced images (for use in overlay_aerial), save/convert your image to a format with a world file (e.g. PNG with .PGW or .JPG with .JGW).
• Translate the appropriate coordinate(s) in the world file.
C Tech’s Long-Term Plans:
We know how to solve this problem completely, but it will take a major restructuring of EVS and changing many of the 3rd party libraries upon which our software is built. We are moving in this direction. | 1,460 | 6,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-40 | longest | en | 0.894666 |
https://www.lmfdb.org/EllipticCurve/Q/135240bs6/ | 1,638,597,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00258.warc.gz | 896,461,992 | 33,088 | # Properties
Label 135240bs6 Conductor $135240$ Discriminant $-2.972\times 10^{22}$ j-invariant $$-\frac{957928673903042}{123339801817575}$$ CM no Rank $1$ Torsion structure $$\Z/{2}\Z$$
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, -1, 0, -1277936, 8313136140])
gp: E = ellinit([0, -1, 0, -1277936, 8313136140])
magma: E := EllipticCurve([0, -1, 0, -1277936, 8313136140]);
$$y^2=x^3-x^2-1277936x+8313136140$$
## Mordell-Weil group structure
$\Z\times \Z/{2}\Z$
### Infinite order Mordell-Weil generator and height
sage: E.gens()
magma: Generators(E);
$P$ = $$\left(\frac{66001}{16}, \frac{17317553}{64}\right)$$ $\hat{h}(P)$ ≈ $11.391198642557390873245303271$
## Torsion generators
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
$$\left(-2235, 0\right)$$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$\left(-2235, 0\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$135240$$ = $2^{3} \cdot 3 \cdot 5 \cdot 7^{2} \cdot 23$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $-29718127296585484646400$ = $-1 \cdot 2^{11} \cdot 3^{2} \cdot 5^{2} \cdot 7^{7} \cdot 23^{8}$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$-\frac{957928673903042}{123339801817575}$$ = $-1 \cdot 2 \cdot 3^{-2} \cdot 5^{-2} \cdot 7^{-1} \cdot 23^{-8} \cdot 78241^{3}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $2.9916103966809121293800111604\dots$ Stable Faltings height: $1.3832704066399722765282053440\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $1$ sage: E.regulator() magma: Regulator(E); Regulator: $11.391198642557390873245303271\dots$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.096479895531833524700093484459\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $16$ = $1\cdot2\cdot2\cdot2\cdot2$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $2$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L'(E,1)$ ≈ $4.3960866200652037102195307161382827050$
## Modular invariants
Modular form 135240.2.a.e
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - q^{3} - q^{5} + q^{9} - 4 q^{11} + 2 q^{13} + q^{15} - 2 q^{17} + 4 q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 11010048 $\Gamma_0(N)$-optimal: no Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 5 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $1$ $II^{*}$ Additive -1 3 11 0
$3$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
$5$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
$7$ $2$ $I_{1}^{*}$ Additive -1 2 7 1
$23$ $2$ $I_{8}$ Non-split multiplicative 1 1 8 8
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2B 8.24.0.96
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 add nonsplit nonsplit add ordinary ordinary ordinary ordinary nonsplit ordinary ordinary ordinary ordinary ordinary ordinary - 3 1 - 1 1 1 1 1 1 1 1 1 1 1 - 0 0 - 0 0 0 0 0 0 0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2, 4 and 8.
Its isogeny class 135240bs consists of 4 curves linked by isogenies of degrees dividing 8.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $2$ $$\Q(\sqrt{-14})$$ $$\Z/2\Z \times \Z/2\Z$$ Not in database $2$ $$\Q(\sqrt{7})$$ $$\Z/4\Z$$ Not in database $2$ $$\Q(\sqrt{-2})$$ $$\Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{-2}, \sqrt{7})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{5}, \sqrt{7})$$ $$\Z/8\Z$$ Not in database $4$ $$\Q(\sqrt{7}, \sqrt{-10})$$ $$\Z/8\Z$$ Not in database $8$ Deg 8 $$\Z/2\Z \times \Z/4\Z$$ Not in database $8$ 8.0.39969909374976.50 $$\Z/8\Z$$ Not in database $8$ 8.0.98344960000.6 $$\Z/2\Z \times \Z/8\Z$$ Not in database $8$ Deg 8 $$\Z/6\Z$$ Not in database $16$ Deg 16 $$\Z/4\Z \times \Z/4\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database $16$ Deg 16 $$\Z/16\Z$$ Not in database $16$ Deg 16 $$\Z/16\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/6\Z$$ Not in database $16$ Deg 16 $$\Z/12\Z$$ Not in database $16$ Deg 16 $$\Z/12\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 2,336 | 6,030 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.229239 |
https://practice.geeksforgeeks.org/problems/minimum-cost-to-fill-given-weight-in-a-bag1956/1 | 1,623,705,138,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00095.warc.gz | 425,453,798 | 25,633 | Showing:
Handle Score
@Ibrahim Nash 6420
@mb1973 5704
@Quandray 5245
@akhayrutdinov 5111
@saiujwal13083 5046
@sanjay05 3762
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@marius_valentin_dragoi 3523
@mantu_singh 3510
@sushant_a 3459
Minimum cost to fill given weight in a bag
Medium Accuracy: 40.64% Submissions: 9686 Points: 4
Given an array cost[] of positive integers of size N and an integer W, cost[i] represents the cost of ‘i’ kg packet of oranges, the task is to find the minimum cost to buy W kgs of oranges. If it is not possible to buy exactly W kg oranges then the output will be -1
Note:
1. cost[i] = -1 means that ‘i’ kg packet of orange is unavailable
2.
It may be assumed that there is infinite supply of all available packet types.
Example 1:
Input: N = 5, arr[] = {20, 10, 4, 50, 100}
W = 5
Output: 14
Explanation: choose two oranges to minimize
cost. First orange of 2Kg and cost 10.
Second orange of 3Kg and cost 4.
Example 2:
Input: N = 5, arr[] = {-1, -1, 4, 3, -1}
W = 5
Output: -1
Explanation: It is not possible to buy 5
kgs of oranges
You don't need to read input or print anything. Complete the function minimumCost() which takes N, W, and array cost as input parameters and returns the minimum value.
Expected Time Complexity: O(N*W)
Expected Auxiliary Space: O(N*W)
Constraints:
1 ≤ N, W ≤ 103
-1 ≤ cost[i] ≤ 105
cost[i] ≠ 0 | 440 | 1,335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-25 | longest | en | 0.766276 |
https://www.shiksha.com/online-courses/articles/how-to-add-two-numbers-in-java/ | 1,713,390,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00763.warc.gz | 894,757,693 | 83,793 | How to Add Two Numbers in Java
# How to Add Two Numbers in Java
clickHere
Updated on Mar 13, 2024 14:23 IST
The addition is the fundamental aspect we use to learn programming in any language. In this article, we will learn different methods to add two numbers in java. So, without further delay let’s dive deep to learn how to add numbers.
In our earlier education days, we must learn addition, subtraction, division, and multiplication. We use the ‘+’ operator to add two integers. Similarly, we use the ‘+’ operator to add two numbers in Java. We say operands to these numbers because we do operations on operands. When we add two numbers, always remember that the size of the result should not extend the size of the data type.
Addition uses ‘+’ operator on two numbers
## a + b
Here, a and b are two operands.
Must Check: Variables in Java
Let’s look at the different ways by which we can add two numbers in Java.
### Addition of two numbers using ‘+’ operator
In the below code, we take two integers a =10 and b =20 and add these two numbers using ‘+’ operator and store the result in sum which is the integer type.
Code:
`public class Main{ public static void main(String args[]) { int a = 10; int b = 20; int sum = a + b; System.out.println("Addition of a and b is:"+ sum); }}Copy code`
Output
Must Check: Java Online Courses and Certification
### Addition of two numbers in Java with User input
In the below code, we use nextInt() method of Scanner class to take input from user.
Code
```import java.util.Scanner; public class Main{ public static void main(String args[]) { Scanner sc=new Scanner(System.in); System.out.println("Enter a first number:"); int a = sc.nextInt(); System.out.println("Enter a second number:"); int b = sc.nextInt(); int sum = a + b; System.out.println("Addition of a and b is:"+ sum); }}Copy code```
Output:
Must Read: Understanding Java Scanner Class
### The sum of two numbers in Java using a method
#### By using sum() method
Integer class in Java has a method sum(), which we use two add two numbers by passing arguments in the sum() method. Observe the code for better understanding.
Code:
```import java.util.Scanner; public class Main{ public static void main(String args[]) { Scanner sc=new Scanner(System.in); System.out.println("Enter a first number:"); // Taking input of integer int a = sc.nextInt(); System.out.println("Enter a second number:"); // Taking input of integer int b = sc.nextInt(); // calling sum method by passing integer a and b int sum = Integer.sum(a, b); System.out.println("Addition of a and b is:"+ sum); }}Copy code```
Output:
#### By using a user-defined method
In the below code, we make a static method sum that takes two integers as arguments and has return type int. This method adds two integers passed as arguments and returns the result. Observe the code.
Code:
```import java.util.Scanner; public class Main{ public static int sum( int a, int b) { int ans=a+b; return ans; } public static void main(String args[]) { Scanner sc=new Scanner(System.in); System.out.println("Enter a first number:"); // Taking input of integer int a = sc.nextInt(); System.out.println("Enter a second number:"); // Taking input from integer int b = sc.nextInt(); // calling sum method int sum = Main.sum(a, b); System.out.println("Addition of a and b is:"+ sum); }}Copy code```
Output:
### By using for loop
We can add numbers using for loop. In the below code, we first take input from the user on how many numbers to be added; suppose the user enters n numbers to be added, then we enter into for loop n times and take number input one by one and add the number to sum, which is initialized to zero. After the addition of all the numbers, we will return the sum. Observe the code for better understanding.
Code:
```import java.util.Scanner; public class Main{ public static void main(String args[]) { Scanner sc=new Scanner(System.in); System.out.println("Enter total numbers to be added:"); int n=sc.nextInt(); int sum=0; for(int i=0;i<n;i++) { System.out.println("Enter a number:"); int num=sc.nextInt(); sum+=num; } System.out.println("Sum of numbers:"+ sum); }}Copy code```
Output
Must Read: Mastering For Loop in Java
## Special Case: When Integers are in String Form
Let’s take a case when integers are in String form and how we can add these integers. So, for this type of scenario in Java, we have some inbuilt methods in the Integer class. We can use Integer.parseInt() method to Integer, and then we typecast the Integer to int. Let’s take a program to understand this clearly. For example, we have String a and String b as
String a=”400″;
String b=”100″;
Code:
```public class Main{ public static void main(String args[]) { String a="400"; String b="100"; //we use parseInt() method of Integer to convert a String value to Integer // Converted Integer is type cast into int; int val1=(int)Integer.parseInt(a); int val2=(int)Integer.parseInt(b); int sum=val1+val2; System.out.println("Sum of a and b is:"+sum); }}Copy code```
Output
## Conclusion
In this article, we have discussed different methods to add two numbers in Java with the help of examples.
Hope you will like the article.
Keep Learning!!
Keep Sharing!!
Contributed By: Shubham Kumar | 1,234 | 5,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.814491 |
https://www.isnt.org.in/java-2d-array-loop-with-code-examples.html | 1,716,902,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00283.warc.gz | 716,218,809 | 51,195 | Java 2D Array Loop With Code Examples
In this article, we will look at how to get the solution for the problem, Java 2D Array Loop With Code Examples
What type of loop is the most appropriate for iterating a 2D array?
Now let's jump into nested for loops as a method for iterating through 2D arrays. A nested for loop is one for loop inside another. Take a look below to see what this means. The first for loop loops through each row of the 2D array one by one.
``````// Program start method.
public static void main(String[] args) {
String[][] array2D = new String[10][10]; // Create 2D array.
for (int row = 0; row < array2D.length; row++) { // Loop through the rows in the 2D array.
for (int col = 0; col < array2D[row].length; col++) { // Loop through the columns in the 2D array.
array2D[row][col] = "row: " + row + ", column: " + col; // Set the data in the right row and column.
}
}
}```
```
``````public static void main(String[] args) {
int[][] array = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
for(int[] count : array) {
for(int elem : count) {
System.out.print(elem + " "); //output: 1 2 3 4 5 6 7 8 9
}
}
}```
```
How do you declare and print a 2D array in Java?
public class Print2DArray { public static void main(String[] args) { final int[][] matrix = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; for (int i = 0; i < matrix. length; i++) { //this equals to the row in our matrix. for (int j = 0; j < matrix[i]. length; j++) { //this equals to the column in each row.
What is 2D array explain with syntax?
Declaration of Two-Dimensional Arrays The syntax of two-dimensional arrays is: Data_type name_of the array[rows][index]; Here is one example: int multi_dim[2][3]; In the above example, the name of the 2d array is multi_dim consisting of 2 rows and three columns of integer data types.
How do you do two for loops in Java?
If a loop exists inside the body of another loop, it's called a nested loop. Here's an example of the nested for loop. // outer loop for (int i = 1; i <= 5; ++i) { // codes // inner loop for(int j = 1; j <=2; ++j) { // codes } .. } Here, we are using a for loop inside another for loop.
How do you print a 2D array for loop?
In other words Preferences[0][2] is out of the bounds of the array because the length is 2. Inner loop should be j < 2, not 3.Using a for loop to Print and Two-Dimensional Array in Java
• Declare a 2d String Array,
• Assign Values to the array of two people and their favourite drink.
• Output using a for loop.
Why do we use two for loops with two-dimensional arrays?
For a two-dimensional array, in order to reference every element, we must use two nested loops. This gives us a counter variable for every column and every row in the matrix.
Can you use a For-Each loop on a 2D array Java?
Since 2D arrays are really arrays of arrays you can also use a nested enhanced for-each loop to loop through all elements in an array.
How do you fill a 2D array in Java?
how to fill a 2d array in java
• int rows = 5, column = 7; int[][] arr = new int[rows][column];
• for (int row = 0; row < arr. length; row++)
• { for (int col = 0; col < arr[row]. length; col++)
• { arr[row][col] = 5; //Whatever value you want to set them to.
How do you loop a 2D array?
In order to loop over a 2D array, we first go through each row, and then again we go through each column in every row. That's why we need two loops, nested in each other. Anytime, if you want to come out of the nested loop, you can use the break statement.
How does a nested loop work?
When a loop is nested inside another loop, the inner loop runs many times inside the outer loop. In each iteration of the outer loop, the inner loop will be re-started. The inner loop must finish all of its iterations before the outer loop can continue to its next iteration.
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## Ordinal Numbers 1 5 Worksheets
Ordinal Numbers 1 5 Worksheets – A limitless number of sets can easily be enumerated with ordinal numerals as a tool. They also can help generalize ordinal quantities. 1st The basic concept of math is the ordinal. It is a number that indicates the position of an object in a set of objects. In general, … Read more
## Ordinal Numbers 1st To 5th Worksheet
Ordinal Numbers 1st To 5th Worksheet – A limitless number of sets are easily counted using ordinal numbers as a tool. They also can help generalize ordinal quantities. 1st The ordinal numbers are one of the most fundamental ideas in math. It is a numerical value that represents the position of an object within an … Read more | 236 | 1,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-18 | latest | en | 0.900379 |
https://www.scribd.com/document/88507916/Example-1 | 1,511,131,581,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00361.warc.gz | 875,857,051 | 29,932 | # Example 1 A firm is currently selling a product @ Rs 10 per unit.
The most recent annual sales (all credit) were 30,000 units. The variable cost per unit is Rs .6 and the average cost per unit, given a sales volume of 30,000 units, is Rs 8. The total fixed cost is Rs 60,000. The average collection period may be assumed to be 30 days. The firm is contemplating a relaxation of credit standards that is expected to result in a 15 per cent increase in units sales; the average collection period would increase to 45 days with no change in bad debt expenses. It is also expected that increased sales will result in additional net working capital to the extent of Rs 10,000. The increase in collection expenses may be assumed to be negligible. The required return on investment is 15 per cent. Should the firm relax the credit standard? Example 2 Assume that the firm in our Example 1 is contemplating to allow 2 per cent discount for payment within 10 days after a credit purchase. It is expected that if discounts are offered, sales will increase by 15 per cent and the average collection period will drop to 15 days. Assume bad debt expenses will not be affected; return on investment expected by the firm is 15 per cent; 60 per cent of the total sales will be on discount. Should the firm implement the proposal? Example 3 Suppose, a firm is contemplating an increase in the credit period from 30 to 60 days. The average collection period which is at present 45 days is expected to increase to 75 days. It is also likely that the bad debt expenses will increase from the current level of 1 per cent to 3 per cent of sales. Total credit sales are expected to increase from the level of 30,000 units to 34,500 units. The present average cost per unit is Rs 8, the variable cost and sales per unit is Rs 6 and Rs 10 per unit respectively. Assume the firm expects a rate of return of 15 per cent. Should the firm extend the credit period? Example 4 A firm is contemplating stricter collection policies. The following details are available: 1. 2. At present, the firm is selling 36,000 units on credit at a price of Rs 32 each; the variable cost per unit is Rs 25 while the Rs 29; average collection period is 58 days; and collection expenses amount to Rs 10,000; bad debts are 3 per cent.
If the collection procedures are tightened, additional collection charges amounting to Rs 20,000 would be required, bad cent; the collection period will be 40 days; sales volume is likely to decline by 500 units.
Assuming a 20 per cent rate of return on investments, what would be your recommendation? Should the firm implement the Example 5
Rs 6 lakh per annum Required (pre-tax) return on investment: 20 per cent Credit policy A B C Average collection period (days) 45 60 75 Annual sales (Rs lakh) 56 60 62 63 D 90 Determine which policy the company should adopt. The company’s variable costs are 70 per cent of the selling price.00.50.50.4 .Super Sports. Fixed costs.000 3.00.000 4. 80 per cent of sales. but the dealers have difficulty in financing their therefore. Currently. Variable costs.000 4 Policy option I Rs 60. has an annual sale of Rs 50 lakh and currently extending 30 days’ credit to the dealers pick up considerably if the dealers are willing to carry increased stocks.000 receivable Present policy Rs 50. dealing in sports goods.50. which is a better option? Particulars Annual credit sales Accounts turnover ratio Bad debt losses 1. The following information is available: The average collection period now is 30 days. The current level of loss due to bad debts is Rs 1.000 3 Policy option II Rs 67. The firm is required to give a return of 25 per cent on the investment in new accounts receivable. the firm has annual credit sales of Rs 50 lakh and accounts receivable turnover ratio of 4 times a year. Given the following information. considering shifts in credit policy.000.50.000 2.00. Example 6 XYZ Corporation is considering relaxing its present credit policy and is in the process of evaluating two alternative policies. | 931 | 4,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-47 | latest | en | 0.938037 |
http://mathhelpforum.com/differential-geometry/89235-limit-print.html | 1,500,909,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424884.51/warc/CC-MAIN-20170724142232-20170724162232-00312.warc.gz | 208,917,463 | 2,990 | # limit?
• May 16th 2009, 10:54 AM
Showcase_22
limit?
Quote:
Find the limit (if it exists) of $\lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}$.
It's apparent that $\sin \left( \frac{1}{x} \right)$ is discontinuous (I have a proof for that involving the sequence $a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $b_n=\frac{1}{2n \pi -\frac{\pi}{2}}$).
Define $-\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}$
$\lim_{x \rightarrow 0} \frac{x^2}{\sin x}=\lim_{x \rightarrow} \frac{2x}{\cos x}=0$
Hence $\lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}=0$
Is this the correct way of doing this type of question or is there an easier way?
• May 16th 2009, 11:10 AM
HallsofIvy
Quote:
Originally Posted by Showcase_22
It's apparent that $\sin \left( \frac{1}{x} \right)$ is discontinuous (I have a proof for that involving the sequence $a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $b_n=\frac{1}{2n \pi -\frac{\pi}{2}}$).
Trying this as $\left(\frac{x}{sin x}\right)\left(\frac{sin(1/x)}{(1/x)}\right)$ should make it much easier! As x goes to 0, 1/x goes to infinity while sin(1/x) is always between -1 and 1.
Quote:
Define $-\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}$
$\lim_{x \rightarrow 0} \frac{x^2}{\sin x}=\lim_{x \rightarrow} \frac{2x}{\cos x}=0$
Hence $\lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}=0$
Is this the correct way of doing this type of question or is there an easier way? | 668 | 1,719 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-30 | longest | en | 0.585834 |
https://studylib.net/doc/10844879/close-please-your-laptops--and-get-out-your-note- | 1,653,555,158,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00311.warc.gz | 641,055,170 | 11,956 | ```Please
CLOSE
and turn off and put away your
cell phones,
and get out your notetaking materials.
Sections 1.6-1.8, 2.1
Order of Operations, Part 2
Properties of Real Numbers
Simplifying Algebraic Expressions.
You should work the homework
problems in this assignment
WITHOUT A CALCULATOR
(after this assignment, a calculator can
be used and will be available as a
button in each online HW assignment)
Working with positive and negative signs
A number line is a line on which each point is
associated with a number.
–5 –4–3 –2 –1
Negative
numbers
0
1
2
3
4
Positive
numbers
5
To add two numbers, especially if one (or both) of them is
negative, it often helps to picture them on a number line.
Example 1 : 3 + (-5)
•Locate the first number on the number line.
• Starting from that number, if the second number is positive, move to the
right by that many units. If it’s negative, move to the left that many units.
(-5)
3
(Count 5 units to the left from 3)
–5
–4 –3
–2
–1
0
1
2
3
4
5
Example 2: -2 + (-3)
•Locate the first number on the number line.
• Start from that first number. The second number is negative, so move to
the left by that many units.
(-3)
2
(Count 3 units to the left from -2)
–5
–4 –3
–2
–1
0
1
2
3
4
5
Sample homework problem:
Compare to this problem:
6
Absolute Value:
•
The absolute value of a number is the distance of that
number away from 0 on a number line.
a 0 always, since distances are non-negative.
(We say non-negative to include positive numbers and zero.)
a= a, if a is 0 or a positive number.
Example: |0| = 0, 10= 10
a= -a, if a is a negative number
Example: -10= -(-10) = 10
Examples:
1.
5 = 5
2.
-5= 5
3.
-5= -5
4.
0= 0
5. --5= -5
Note the difference the placement
of the negative sign makes!!!
It may help to think of this as -1 · |5|.
Sample homework problems
using absolute value:
9
10
Properties of Real Numbers:
• Commutative property
• of addition: a + b = b + a
• of multiplication: a · b = b · a
• Associative property
• of addition: (a + b) + c = a + (b + c)
• of multiplication: (a · b) · c = a · (b · c)
• Distributive property of multiplication over
• a(b + c) = ab + ac
Sample problem from today’s homework:
Sample problem from today’s homework:
Sample problems from today’s homework:
Simplifying Algebraic Expressions:
The terms of an expression are the addends of
the expression (parts separated by + or –
signs).
Like terms contain the same variables raised
to the same powers.
To combine like terms, add or subtract their
numerical coefficients, then multiply the result by
the common variable factors.
(The coefficient is the number in front of a term.)
Examples of Combining Terms
Terms Before Combining
6x2 + 7x2
19xy – 30xy
13xy2 – 7x2y
After Combining Terms
13x2
-11xy
Can’t be combined
(since the terms are not like
terms)
Sample problems from today’s homework:
“help me solve this” function, because you won’t
If you’re having a lot of trouble understanding
how to do a set of problems, come to the open lab
and our TAs will work with you one-on-one to help
you learn the easiest way to work the problems on
Open lab hours:
M - Th, 8 a.m. – 6:30 p.m.
The assignment on this material (HW 1.6/7&2.1) is due
at the start of class tomorrow. (Remember: You should
do these problems by hand, without a calculator.)
You also have the Practice Gateway Test
due at the start of class tomorrow. This is a
required assignment worth 3 points.
Tomorrow at the end of class you will be taking the
25-point Gateway Test, which is similar to the quiz
you did on the first day of class.
The practice test has similar questions and can be done as many
times as you want. Only your best score on that counts for
points. No calculators can be used on this test, so do the
practice test without one, too.
You may now OPEN | 1,144 | 3,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-21 | latest | en | 0.880378 |
https://www.alltexcommercial.com/2017/08/19/30-x-60-table/ | 1,524,768,536,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948464.28/warc/CC-MAIN-20180426183626-20180426203626-00533.warc.gz | 749,210,455 | 9,226 | » » 30 X 60 Table
# 30 X 60 Table
This image about 30 X 60 Table was posted at August 19, 2017 at 6:29 am. This blog post is published at the Table category. 30 X 60 Table is tagged with 30 X 60 Table, 30, X, 60, Table..
### X
Roman numerals,
• the numerals in the ancient Roman system of notation, still used for certain limited purposes, as in some pagination, dates on buildings, etc. The common basic symbols are I (=1), V (=5), X (=10), L (=50), C (=100), D (=500), and M (=1000). The Roman numerals for one to nine are: I, II, III, IV, V, VI, VII, VIII, IX. A bar over a letter multiplies it by 1000;
thus, X̄ equals 10,000. Integers are written according to these two rules: If a letter is immediately followed by one of equal or lesser value, the two values are added;
thus, XX equals 20, XV equals 15, VI equals 6. If a letter is immediately followed by one of greater value, the first is subtracted from the second;
thus, IV equals 4, XL equals 40, CM equals 900. Examples: XLVII(=47), CXVI(=116), MCXX(=1120), MCMXIV(=1914). Roman numerals may be written in lowercase letters, though they appear more commonly in capitals.
• ### Table
ta•ble (tābəl),USA pronunciation n., v., -bled, -bling, adj.
n.
1. an article of furniture consisting of a flat, slablike top supported on one or more legs or other supports: a kitchen table; an operating table; a pool table.
2. such a piece of furniture specifically used for serving food to those seated at it.
3. the food placed on a table to be eaten: She sets a good table.
4. a group of persons at a table, as for a meal, game, or business transaction.
5. a gaming table.
6. a flat or plane surface;
a level area.
7. a tableland or plateau.
9. an arrangement of words, numbers, or signs, or combinations of them, as in parallel columns, to exhibit a set of facts or relations in a definite, compact, and comprehensive form;
a synopsis or scheme.
10. (cap.) the constellation Mensa.
11. a flat and relatively thin piece of wood, stone, metal, or other hard substance, esp. one artificially shaped for a particular purpose.
• a course or band, esp. of masonry, having a distinctive form or position.
• a distinctively treated surface on a wall.
12. a smooth, flat board or slab on which inscriptions may be put.
13. tables:
• the tablets on which certain collections of laws were anciently inscribed: the tables of the Decalogue.
• the laws themselves.
14. the inner or outer hard layer or any of the flat bones of the skull.
15. a sounding board.
16. [Jewelry.]
• the upper horizontal surface of a faceted gem.
• a gem with such a surface.
17. on the table, [Parl. Proc.]
• [U.S.]postponed.
• [Brit.]submitted for consideration.
18. turn the tables, to cause a reversal of an existing situation, esp. with regard to gaining the upper hand over a competitor, rival, antagonist, etc.: Fortune turned the tables and we won. We turned the tables on them and undersold them by 50 percent.
19. under the table:
• drunk.
• as a bribe;
secretly: She gave money under the table to get the apartment.
20. wait (on) table, to work as a waiter or waitress: He worked his way through college by waiting table.Also, wait tables.
v.t.
1. to place (a card, money, etc.) on a table.
2. to enter in or form into a table or list.
3. [Parl. Proc.]
• [Chiefly U.S.]to lay aside (a proposal, resolution, etc.) for future discussion, usually with a view to postponing or shelving the matter indefinitely.
• to present (a proposal, resolution, etc.) for discussion.
1. of, pertaining to, or for use on a table: a table lamp.
2. suitable for serving at a table or for eating or drinking: table grapes.
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May 28, 2017 | 1,699 | 6,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-17 | longest | en | 0.846977 |
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solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
3 years ago
mycroft holmes
272 Points
a cos A = b cos B $\Rightarrow$ 2R sin A cos A = 2R sin B cos B $\Rightarrow$ sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180o – 2B so that A+B = 90o.(Right-angled)
3 years ago
mycroft holmes
272 Points
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H.The following can be proved easily:1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends equal angles at C and D, so that $\angle HCE = \ang HDE = 90^{\circ} - A$. Similarly $\angle HDF = 90^{\circ} - A$. 2. Now look at $\triangle FBD$. You have $\angle FDB = \angle ADB - \angle HDF = A$ So, $\angle BFD = C$. In other words $\triangle DBF \sim \triangle ABC$ 3. From right $\triangle ABD$, we get BD = c cos B. From $\triangle DBF \sim \triangle ABC$, we have $\frac{DF}{AC} = \frac{DB}{AB}$ so that $DF = \frac{b c \cos B}{c} = b \cos B$
3 years ago
mycroft holmes
272 Points
To continue the prev post, we have $DF = b \cos B = 2R \sin B \cos B = R \sin 2B$
Hence the sides of the orthic triangle are in the ration$\sin 2A: \sin 2B : \sin 2C$
3 years ago
mycroft holmes
272 Points
Draw $\triangle OBC$ which is Isoceles as OB = OC. Now $\angle BOC = 2A$ which means $\angle OBC = 90^{\circ} - A$. Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required ratio is cos A: cos B : cos C
3 years ago
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• Previous Year Exam Questions | 741 | 2,245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-10 | latest | en | 0.610466 |
https://medium.com/@nabanita.sarkar/simulating-amplitude-modulation-using-python-6ed03eb4e712 | 1,713,507,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00059.warc.gz | 347,648,247 | 29,352 | # Simulating Amplitude Modulation using Python
In communication labs in our colleges we all generate amplitude modulated signals using CROs. But the same can be performed by using Python and few of its additional libraries and the end result can be equally dope.
At first let’s revise the formulas needed to generate amplitude modulated signal
Now to simulate the signals in python first we have to import 2 python libraries: numpy and matplotlib
Then we have to take carrier amplitude, carrier frequency, message amplitude, message frequency and modulation index as inputs.
The built-in input function in python returns string value. So we have to convert them to integer or float. Now float is preferred because amplitude or frequency can be in decimal too.
The time function in analog communication is continuous function. To replicate its behavior we will use linspace function which will provide large number of discrete points which will act almost similar to continuous function. The linspace function will generate evenly spaced numbers within a given interval. The first argument of linspace is starting point, second argument is ending point and third argument is number of breakpoints between the given interval.
Now we will create our carrier, modulator(message), and product function. Here we will use sin , cos and pi from numpy.
Here comes the plotting part. We will utilize matplotlib library functions. The subplot function creates more than one plot in the canvas. The plot function plots the given function. In plot function we can pass color names or their acronyms to plot the graph in any color we wish. The ‘g’ , ‘r’ stands for green and red respectively. The xlabel and ylabel prints the x-axis and y-axis variable names. And the title function prints the title of the over all plot.
Now we can customize the plot as much we want. We can change the space between plots, font, size etc. And if we want to save the picture we can easily do so by savefig function where dpi means dots per inch which in today’s world, can be inter-changeably used with ppi(pixel per inch). More of such functions can be found in official matplotlib documentation.
Thus we can create simple amplitude modulation plot in python just using two external libraries only.
The full code is here: | 448 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-18 | latest | en | 0.838982 |
dearsicilia.com | 1,685,741,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648858.14/warc/CC-MAIN-20230602204755-20230602234755-00548.warc.gz | 233,734,787 | 30,244 | # How to Measure Copperplate Guidelines (Part 1 – 3:2:3 ratio)
calligraphy
What is the 3:2:3 ratio?
Hey. I made this video to (hopefully) answer some questions submitted by a fellow learner to my blog. How to measure the guidelines? Is the 3:2:3 ratio different from the 1/4″ height? What does the numbers represent? Why is it expressed in mm and inches? So this is basically the blog post/written verison of the video.
In this video, I will assume that you’re familiar with the terms x-height, ascenders, and descenders. Maybe you don’t enjoy math, it’s okay, I will try to explain without sounding like your math teacher from school. I do enjoy math to an extent – I like measuring and being precise in things. But I really want to try and make this easy to understand, and keep the calculating to minimum.
But first, pound cake.
Have you ever had a pound cake? Do you know why it’s called that? Well, it’s traditionally made with a pound of each for ingredients: flour, butter, sugar, and eggs. We can say it’s a 1:1:1:1 ratio – equal parts of each ingredients. Mix ‘em, bake’ em, and you’ll get a standard sized pound cake.
Well, what if we want to make a smaller pound cake? Easy. Say we want to half the size. The ratio stays the same – so you’ll need half a pound of flour, half a pound of butter, half a pound of sugar, and (you guessed it) half a pound of eggs. Mix, put in a smaller pan, and bake. Still a pound cake, but half the size.
How about a bigger pound cake? You want to double the size? That’s right – double the ingredients. Two pounds of flour, two pounds of butter, two pounds of sugar, and two pounds of eggs. We use the same ratio. We get bigger cake. It’s still a pound cake.
Back to our standard, normal sized pound cake. What happens if you put 1 pound of each flour, butter, and sugar – but then 2 pounds of eggs? That would mean you changed the ratio. It’s no longer a 1:1:1:1 ratio. No longer a pound cake… not a traditional one anyway.
In summary, you can get whatever size of pound cake by sticking to the ratio. The ratio numbers represent part(s) of something. In pound cake’s case, the weight of the ingredients, in this case measured in unit of pound. (Which incidentally, is equal to 16 ounces, or 453.592 grams).
And now, to make our ‘Copperplate’ cake.
Traditional Copperplate ratio is 3:2:3. Not equal parts of everything, like the pound cake. Copperplate ratio is 3 parts of ascender space, 2 parts of x-height, and 3 parts of descender space. The ratio numbers in this case represent the height of the letters. Measured in what unit? It’s really up to you. Inches, milimetres, number of lines in your legal pad…
To measure your guideline, we can start with the x-height. First, choose how big your x-height is going to be. So if your x-height is 2 parts of something, the ascender space will be 3 parts of something, as will the descender space. The parts are however tall or short and in whatever units you choose – as long as they are the same height.
For example, if a part is 1” tall, the x-height would be 2” tall, and the ascender and descender space 3” tall each. 3:2:3.
Question. Imagine writing Copperplate calligraphy with x-height 2 mugs tall. How tall will the ascenders space be?……… 3 mugs tall. Very good.
More difficult question. If 1/4” is our x-height, how tall will the Copperplate writing be from ascender to descender line?
I know I promised you won’t need to get your calculator out, so if you want to find out the answer to that question, get my free printable Copperplate calligraphy guide lines through the link in the description.
Thank you for submitting your questions, and please, please keep doing that, so we can keep learning together.
To summarise:
– The ratio is used to calculate each space so that we can maintain Copperplate proportions no matter how big or small we write
– The quarter inch is like the “standard” or shall we say the “learning” x-height.
S.
# My Copperplate 5-minutes-each Practice Videos
calligraphy
I’ve recorded and uploaded my copperplate practice videos. They’re imperfect to boot, and not necessarily instructional, but it’s great if you can glean something from them. Or- you might just find them relaxing. Enjoy, and as usual – let me know if you have any questions!
I learn from the book: Mastering Copperplate Calligraphy by Eleanor Winters. These guides are for learning from that book! If you haven’t got it, read my review here. All with 1/4 inch x-height perfect for first-timers, like me. I like to divide ascender and descenders practice to save paper, ink, and space. Read somewhere that the green colour is good for tired eyes – plus: the black ink stands out more. I use 100 GSM A4 paper. Do not scale when you print, use actual size. Enjoy!
Copperplate complete
Copperplate ascender
Copperplate descender
Copperplate flat-x
Writing smaller:
Copperplate at 87 percent
Copperplate at 75 percent
Copperplate at 66 percent
Copperplate at 50 percent
Writing bigger:
Copperplate at 112 percent
Copperplate at 125 percent
A free option is the interactive Script in the Copperplate Style (Pointed Pen Calligraphy) by Dr. Joseph M. Vitolo on Apple Books, but you’d need different guidelines from this one.
Check out my 5-minutes-each letter practice videos on Youtube
P.S. Swipe some A to Z: Word Lists and Pangrams for Practising Calligraphy here! PLUS: discover search bar and categories by clicking the + plus sign at the very bottom of the page! ↓
S.
# Questions?
What are the things that will make your life easier on your calligraphy journey? What you are wondering about, unsure about, want to know more about, whatever you wish someone could explain to you about learning calligraphy – ask away. Write me personally now!
P.S. Hi fellow calligraphers! I’m crafting a basic Copperplate calligraphy course. It will be an online video course, but I might open private additional live classes if there is interest. When I have the beginning prepared, I’m planning to open the class for free preview to get your feedback. Please fill in this form to get notified! S.
# Why Calligraphy
thoughts
Hullo! My name is Sicilia, and I’d like to start this blog by answering the (kinda) big question – the foundation of all my endeavours here: why calligraphy?
Why learn it, why use it, why look at it at all? The number one reason, of course, is because it’s beautiful. Calligraphy is just the epitome of hand writing. But you don’t have to have good hand writing to be able to excel at calligraphy, because in calligraphy, you’re basically drawing letters. Anything you can write, you can enhance by drawing it in calligraphy.
Reason number two: truth is, it just makes people feel special if they received something hand made, or in this case, handwritten. When many things in this world are computerised, automated, and tagged, it’s always something else when you receive an old wordly handwritten letter addressed to you. It displays care and attention, and for a moment time actually slows down as you give attention to the human made detail.
Lastly, imperfections are beautiful. Calligraphy created by hand has an earthy, organic feel that never fails to surprise. When pen glides on paper with endless possibilities, that’s one thing a computer font cannot duplicate.
S. | 1,750 | 7,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-23 | latest | en | 0.938096 |
https://jjj.de/arctan/arctanpage.html | 1,696,356,021,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00697.warc.gz | 363,685,212 | 3,425 | # Arctan relations for Pi
### Description of the files
In a relation
` M1*arctan(1/A1)+M2*arctan(1/A2)+...+Mj*arctan(1/Aj) == k*Pi/4 `
the left hand side is abbreviated as
` M1[A1]+M2[A2]+...+Mj[Aj] `
The term of least convergence is listed first. Relations of n arctan terms are in one file. The files are ordered according to the arguments, the "best" relation is first. When the first arguments coincide the next is used for ordering. An example (6-term relations):
```+322[577] +76[682] +139[1393] +156[12943] +132[32807] +44[1049433] == 1 * Pi/4
+122[319] +61[378] +115[557] +29[1068] +22[3458] +44[27493] == 1 * Pi/4
+100[319] +127[378] +71[557] -15[1068] +66[2943] +44[478707] == 1 * Pi/4
+337[307] -193[463] +151[4193] +305[4246] -122[39307] -83[390112] == 1 * Pi/4
+183[268] +32[682] +95[1568] +44[4662] -166[12943] -51[32807] == 1 * Pi/4
+183[268] +32[682] +95[1483] -7[9932] -122[12943] +51[29718] == 1 * Pi/4
+29[268] +269[463] +154[2059] +122[2943] -186[9193] +71[390112] == 1 * Pi/4
```
Each relation is followed by a list of primes of the form 4*k+1. These are obtained by factoring Ai^2+1 for each (inverse) argument Ai. An example (a 5-term relation):
```+88[192] +39[239] +100[515] -32[1068] -56[173932] == 1 * Pi/4
{5, 13, 73, 101}
```
We have
```192^2+1 == 36865 == 5 73 101
239^2+1 == 57122 == 2 13 13 13 13
515^2+1 == 265226 == 2 13 101 101
1068^2+1 == 1140625 == 5 5 5 5 5 5 73
173932^2+1 == 30252340625 == 5 5 5 5 5 13 73 101 101
```
The search is described in the fxtbook. Here are the slides of my talk "Search for the best arctan relation" given 2006 in Berlin (gzip compressed): dvi (8kB), ps (75kB), or pdf (80kB).
All relations were computed April-2006. Some of them improve on my April-1993 computation. The files near the bottom of the hfloat page contain the (now obsolete) 1993 data.
### The relations
Note added 2010: Amrik Singh Nimbran communicated (23-July-2010) a 25-term identity (arctan-25term-nimbran.gp) which is 'better' than my best 25-term relation: the first argument is 218,123,852,367 (my identity has 160,422,360,532). The prime 941 involved shows that my search could not have found this relation.
He also gave (27-October-2010) a 20-term relation (arctan-20term-nimbran.gp) with first argument 3,739,944,528 (my identity has 2,674,664,693). The primes 977 and 1409 were not used in my 2006 search.
Also (2-June-2011) a 24 term relation (arctan-24term-nimbran.gp) with first argument 121,409,547,033 (improving on my 102,416,588,812). The prime 941 was not included in my search.
### The state of art
``` n-terms min-arg
2 5 Machin (1706)
3 18 Gauss (YY?)
4 57 Stormer (1896)
5 192 JJ (1993), prev: Stormer (1896) 172
6 577 JJ (1993)
7 2,852 JJ (1993)
8 5,357 JJ (2006), prev: JJ (1993) 4,246
9 34,208 JJ (2006), prev: JJ (1993) 12,943, prev: Gauss (Y?) 5,257
10 54,193 JJ (2006), prev: JJ (1993) 51,387
11 390,112 JJ (1993)
12 1,049,433 JJ (2006), prev: JJ (1993) 683,982
13 3,449,051 JJ (2006), prev: JJ (1993) 1,984,933
14 6,826,318 JJ (2006)
15 20,942,043 HCL (1997), prev: MRW (1997) 18.975,991
16 53,141,564 JJ (2006)
17 201,229,582 JJ (2006)
18 299,252,491 JJ (2006)
19 778,401,733 JJ (2006)
20 2,674,664,693 JJ (2006) [Note: superseded by Nimbran: 3,739,944,528 (2010)]
21 5,513,160,193 JJ (2006)
22 17,249,711,432 JJ (2006), prev: 16,077,395,443 MRW (27-Jan-2003)
23 58,482,499,557 JJ (2006)
24 102,416,588,812 JJ (2006) [Note: superseded by Nimbran: 121,409,547,033 (2011)]
25 160,422,360,532 JJ (2006) [Note: superseded by Nimbran: 218,123,852,367 (2010)]
26 392,943,720,343 JJ (2006)
27 970,522,492,753 JJ (2006)
MRW := Michael Roby Wetherfield
HCL := Hwang Chien-lih
JJ := Joerg Arndt
```
I am indebted to Michael Roby Wetherfield who supplied a list of arguments X (so that X^2+1 is 761-smooth) beyond the range (10^14) of my exhaustive search. His web site is here. | 1,630 | 4,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-40 | latest | en | 0.724424 |
https://www.doubtnut.com/qna/31089948 | 1,721,133,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00657.warc.gz | 651,354,170 | 33,325 | # Assertion : If ice is floating in water and it melts, then level of water remains unchanged. Reason : When the ice is floating, weight of liquid displaced is equal to the weight of ice.
A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C
If Assertion is true but Reason is false.
D
If Assertion is false but Reason is true.
Video Solution
Text Solution
Verified by Experts
## Both A and R are correct but R is not an explanation for A.
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## Ice floats on water because
Aits density is less than that of water
Bcrystal structure of ice has empty space
CBoth (a) and (b)
DNone of the above
• Question 2 - Select One
## Ice floats on water because
Aits density is less than that of water
Bcrystal structure of ice has empty space
CBoth (a) and (b)
DNone of the above
• Question 3 - Select One
## A piece of ice is floating in a glass vessel filled with water. How will the level of water in the vessel change when the ice melts ?
AIncreases
BDecreases
CRemains the same
DFirst increases then decreases
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 549 | 2,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.902868 |
http://bankersdaily.in/pipes-and-cistern-part-1-learn-series/ | 1,547,924,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583680452.20/warc/CC-MAIN-20190119180834-20190119202834-00038.warc.gz | 23,378,387 | 23,711 | # PIPES AND CISTERN PART 1-LEARN SERIES
Hi Bankersdaily Aspirants,
Learning is a step by step process and you should gradually progress in that to ace the process. As you all know we have started a new Learn Series to help with your preparations for various exams like IBPS PO , IBPS RRB Scale I officer , IBPS RRB Office Assistant , IBPS Clerk , IBPS SO , SBI PO , SBI Clerk , SBI SO and many other exams in this type. Nevertheless this series will also contribute you all the tips and tricks to every miscellaneous sections and also all the important topics which are asked in the exams. If you haven’t looked at the Learn series , Please Click the below link to learn the new concepts.
## INTRODUCTION TO PIPES AND CISTERN:
Today,we are going to look on Pipes and Cisterns.Before we are getting into the topic look into Time and Work because the problems based on Pipes and Cistern is also solved by using LCM Method which we used to solve Time and Work.
## CISTERN:
Cistern is referred to a tank/Reservoir which has some specific capacity to hold liquid.
## PIPES:
There are two types of Pipes,They are
1)Inlet Pipe
2)Outlet Pipe
Inlet Pipe:
Inlet Pipe is used to fill the tank.
Outlet Pipe:
Outlet Pipe is used to empty the tank.
There may be any number of Inlet Pipes and Outlet Pipes that are connected to a cistern.
Before Getting into this topic,let us remember 3 basic terms
Total Capacity=LCM(Given Numbers)(unit-Litres)
Time=Total Capacity/Efficiency(unit-hr,sec,min,day)
Efficiency=Total Capacity/Time(unit Litre/hour)
## #.1.TYPE 1:
### PROBLEMS BASED ON INDIVIDUAL CAPACITY:
There are 3 sub type in this type,they are
1)When both the pipes are Inlet Pipes
2)When both the pipes are Outlet Pipes
3)One Pipe is Inlet Pipe and Outlet Pipe
1)When both the pipes are Inlet Pipes:
1)Pipe A can fill the tank in 10 hrs and Pipe B can fill the tank in 40 hrs.How long will it take to fill the tank if both the pipes are opened?
Explanation
Given that,
Pipe A can fill the tank in 10 hrs and Pipe B can fill the tank in 40 hrs
Total Capacity=LCM(10,40)=40Litres
A’s Efficiency=40/10=4 litre/hour
B’s Efficiency=40/40=1 litre/hour
So,
When both are opened,their total capacity=4+1=5 Litre/hour
Time=40/5=8 hours
So,When Pipe A and B are opened together,they can fill the tank in 8 hours
2)When both the pipes are Outlet Pipes:
1)Pipe A can empty the tank in 3 hours where as Pipe B can empty the tank in 6 hours.When they opened together how,long will it take to empty the tank?
Explanation
Given that,
Pipe A can empty the tank in 3 hours where as Pipe B can empty the tank in 6 hours.
Total Capacity of the tank=LCM(3,6)=6 Litres
A’s Efficiency=6/3=2litres/hour
B’s Efficiency=6/6=1 litre/hour
Total Efficiency of (A+B)=(2+1=3litres/hour)
Time=6/3=2hours
Therefore,When A & B are opened together,they can empty the tank in 2 hours
3)One Pipe is Inlet Pipe and Outlet Pipe:
1)A can fill the tank in 12 hours,B can empty the tank in 15 hours.When both the pipes are opened find how long will it take to empty/fill the tank ?
Explanation
Given that,
A can fill the tank in 12 hours,B can empty the tank in 15 hours
Total Capacity=LCM(12,15)=60 Litres
A’s(Inlet Pipe) Efficiency=60/12=5litre/hour
B’s(Outlet Pipe)Efficiency=60/15=4litre/hour
Note:
If the Efficiency of Inlet Pipe is greater then the Outlet Pipe the tank will be filled,
If the Efficiency of Outlet Pipe is greater then the Inlet Pipe then the tank will be Emptied
Inlet Pipe fill 5 litre in 1 hour at the same time Outlet Pipe empty 4 Litre in 1 hour so,in 1 hour only 1 litre is filled in the tank.
Time taken to fill the tank=60/1=60hours
## #.2.TYPE 2:
### BASED ON PARTIAL WORK:
1)Pipe A can fill the tank in 15 minutes and Pipe B can fill the tank in 20 Minutes.They opened together for 4 Minutes and after that B was turned off.How long will A take to fill the Remaining tank capacity?
Explanation
Given that,
Pipe A=15 minutes(fill)
Pipe B=20 minutes(fill)
Total Capacity=LCM(15,20)=60litres
Efficiency of A=60/15=4litre/min
Efficiency of B=60/20=3litre/min
They were Opened for 4 Minutes,
4+3=7litre/min
7*4=28 Litre.
Remaining 60-28=32 litre
After that B was turned off,then 32 litre is filled by Pipe A
=32/4=8 min | 1,193 | 4,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-04 | longest | en | 0.906808 |
http://www.coursehero.com/file/6722287/multiple-regression/ | 1,397,727,530,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00180-ip-10-147-4-33.ec2.internal.warc.gz | 371,909,724 | 50,392 | 61 Pages
# multiple regression
Course: STAT 3008, Spring 2011
School: CUHK
Word Count: 18161
Rating:
###### Document Preview
MODULE 6 MULTIPLE REGRESSION CONTENTS: MODULE 6 1. Introduction 2. Matrix Notation 3. Forecasting 4. Data Problems 4.1 Multicollinearity 4.2 Measurement Errors 4.3 Outliers and Undue Influence 5. The F Test for Linear Restrictions 5.1 The Redundant Variables Test 5.2 The Linear Restrictions or Wald Test 5.3 The Chow Test 6. Dummy Variables 6.1 Introduction 6.2 Generating Dummy Variables and Time Trends in...
##### Unformatted Document Excerpt
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Britis h His to ryB ritisThis powerpoint was kindly donated towww.worldofteaching.comhttp:/www.worldofteaching.com is home to over athousand powerpoints submitted by teachers. This is acompletely free site and requires no registration. Pleasevisit
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The City-States of GreeceSparta and AthensThe Persian WarsThe Delian LeagueThe Decline of AthensThe City-States ofAncient GreeceSparta and AthensThe Persian WarsThe Delian LeagueThe Decline of AthensNow thats tough! There is a story about a Sp
Boise State - HIST - 100
USCivilRightsMovementBeginnings through the 60sBy J. Aaron CollinsAbolitionistsFrederick Douglas was the editor of anabolitionist newspaper.Onasidenote. Arethey related?HarrietTubmanHelped slaves escape via the UndergroundRailroad.JohnBrown H
Boise State - HIST - 100
ClassicalGreeceClassicalPeriod500339BC "Classical"means: Standardagainstwhichothersarejudgedorevaluated Greatest Enduring Stylisticform(music,art,etc) Stylisticperiod(e.g.afterBaroque) GoldenageofacivilizationClassicalPeriodorTheGoldenAgeofGre
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ClassicalGreekPhilosophySocrates Simple man Stonemason Shrewish wife Loyal service in the war Incredible concentration Wisest man in Athens (oracle) Gad fly (Dialectics/Socratic method) Theunexaminedlifeisnotworthliving.Socrates Convicted of co
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Review of our Presidents fromthe Progressive Era to Cold WarTeddy Roosevelt , William Howard Taft , WoodrowWilson (D)Warren G. Harding , Calvin Coolidge (D),Herbert Hoover , FDR (D)Kamikaze Japanese pilots crashed their bomb-filledplanes into Allie
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The Cold War 1945The1991Two sides of Cold War NATO North Warsaw Pact proAtlantic TreatySoviet countries OrganizationUSSR, and allcountries controlled USA, France, Greatby the USSR.Britain, West COMMUNISMGermany CAPITALISMCold War The Cold
Boise State - HIST - 100
Boise State - HIST - 100
AmericasGreatDisastersMr.RyanGreatDisastersGreatDisasters ManyterriblethingshavehappenedtoAmericaandhercitizens Floods,Hurricanes,Fires,Blizzards,Explosions,Diseases Causesdeath,injuries,lossofhomes,money,jobs,communities,friendships Butmanygoo
Boise State - HIST - 100
Boise State - HIST - 100
Boise State - HIST - 100
Boise State - HIST - 100
EUROPEAN RULERS IN THE AGE OFABSOLUTISMBy CountryA TIME OF EMPIRES1533-1603ENGLANDElizabeth IE N GL A N DJames I The 1 StuartKingstJames sonE N GL A N DCharles ISPAINPhilip II SpainThe GoldenAgeFRANCELouis XIV The Sun KingLOUIS XIVPRUS
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The Fall of Rome and thebeginning of the MiddleAges.Fall of the Roman EmpireRome was the mostpowerful empire theworld had ever seen.Its architecture wasHellenistic and its roadsystem was asimpressive as that ofthe Inca in S. AmericaRoman Empir
Boise State - HIST - 100
Boise State - HIST - 100
Black History Month2008Mr. ClutterVMS LibraryIntroductionCarter G. Woodson was an African American historian, author, journalist andthe Founder of Black History Month. He is considered the first to conduct ascholarly effort to popularize the value
Boise State - HIST - 100
George WashingtonThomas JeffersonTheodoreRooseveltAbrahamLincolnMount Rushmore Facts The carving started in 1927 and finished in1941few injuries but no deaths. They stand at over 5,500 feet Each presidents head is as tall as a 6 storybuilding
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The French Revolution1789Causes of French RevolutionIdeas of liberty and equality fromthe American Revolution (note:Constitution was signed 2 yrsbefore in 1787)Enlightenment ideas of JohnLockeCauses of French Revolution Vastmajority of people w
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Great DepressionBrother can you spare a dime?OBJ #1 - Describe the CAUSES and SPARK of the Great Depression. How didOverproduction affect both farmers and industry? What system collapsed and causedmillions to lose their savings? Explain how buying on
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EarlyGreekScienceandPhilosophyThis Powerpoint is hosted on www.worldofteaching.comPlease visit for 100s more free powerpointsEarly GreeceGreece and Greek ColoniesRome and Roman ColoniesPhoenicia, Carthage and Punic ColoniesThalesofMiletus625BC F
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H alloweenOrigins and TraditionsO rigins Halloween began two thousand years ago inI reland, England, and Northern France with theancient religion of the Celts (Paganism).T heycelebratedtheirNew stYearonNovember1 . This day marked the beginning of
Boise State - HIST - 100
Facts ofHalloweenHalloweenactuallyhasitsoriginsintheCatholicChurch.ItcomesfromacontractedcorruptionofAllHallowsstEve.November1 ,AllHallowsDay(orAllSaintsDay)isaCatholicdayofobservanceinhonorofsaints.In Mexico, theycelebrate El Dia de losMuerto
Boise State - HIST - 100
Boise State - HIST - 100
How did Life in Egypt affectMedicine?Ancient MedicineAncient Medicine3000BC 500ADEgyptians 3000BC-1000BCGreeks 1000BC 500BCRomans 500BC 500 ADKey examine patterns and themesacross the period Focus: EgyptEgypt A Wealthy CountryWealthy countryP
Boise State - HIST - 100
People of the Stone AgeHunters and GatherersCh.1, Lesson 1Mr. Bennetts 6th GradeThe earliest humansprobably lived in Africa.They spread to therest of the world overthe next tens ofthousands of yearsas they hunted andgathered food tosurvive.Ge
Boise State - HIST - 100
Immigrants inAmericaMillions of immigrants moved tothe United States in the late1800s & early 1900s.Immigration Stations Once immigrants arrived in the U.S., theywent through immigration stations, such asEllis Island in New York Harbor.Government
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StandardStandard10.3StudentsanalyzetheeffectsoftheIndustrialRevolutioninEngland,France,Germany,Japan,andtheUnitedStates.IndustrialRevolutionIndustrialRevolution TheIndustrialRevolutionwasthemajorshiftoftechnological,socioeconomicandculturalc
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Industrial RevolutionBy J. CollinsIndustrial RevolutionThe IR is whenpeople stoppedmaking stuff athome and startedmaking stuff infactories.Cottage IndustryFactory systemCotton gin His cotton ginremoved theseeds out ofraw cotton.Steam Engin
Boise State - HIST - 100
IRELAND IN CONFLICT 1909 - 1922Test your knowledge of whos who in the Ireland in Conflicttopic with the following slideshowAs the images of historical personalities from the topic appear,try to work out who they are and what view they held over Irelan
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Islam:History, values and cultureShahbaz YounisPRESENTATION OUTLINEIntroductory RemarksHistorical overviewIslam as a monotheistic religionthe QuranGod or Allahpillars and valuessocial code and reformsrelation with other faithsthe Sunni and Shi
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JerusalemMount Zion, JerusalemAn early historyAncient Canaan 1700 - 1386B.C.EPharaoh Amenhotep ruled overEgypt and CanaanThe Pharoah Ramses III forcedthe Philistines to settle in CanaanIn 1750 B.C.E - The12 tribes ofIsrael settled in Egypt from
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Kennedy AssassinationKennedyCutting through ConspiraciesNovember 22, 1963NovemberJFK was in DallasJFKtrying to getsupport for nextyears election.years Dallas had anDallasunfriendlyreputation towardspoliticians.politicians.Lee Harvey Oswal
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Renaissance Man:Leonardo Da VinciMr. ClutterVMS LibrarySpring 200811/23/111Leonardo the ScientistStudied many topics suchas anatomy, zoology,botany, geology, optics,aerodynamics andhydrodynamics amongothersHe was fascinated by thestudy of p
Boise State - HIST - 100
Lincoln and the EmancipationProclamationProclamationRace Relations in the SouthRace1863-19121863-1912Unit 4What motives lay behind Lincolnsissuing of the EmancipationProclamation?Proclamation?Underlying questions:What was the EmancipationWha
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Louis Pasteur & Germ TheoryBeliefs about disease in19thCentury People knew there was a link between dirtand disease, but could not explain the link. People explained disease as seeds bad seedsin the air known as miasma. 1850s &1860s breakthrough in
Boise State - HIST - 100
MagellanandCoronadoBy: Mrs. MaysFerdinand MagellanMagellan Ferdinand Magellan was a Portugesemaritime (sea) explorer who sailed for thecountry of Spain (just like Columbus andCortes). He led the 1st successful attemptto sail around the entire Ea
Boise State - HIST - 100
Dr. Martin Luther King, Jr.1929-1968Michael Luther King, Jr. wasborn on January 15th toschoolteacher, Alberta Kingand Baptist minister, MichaelLuther King residing at 501Auburn Avenue. His fatherlater changed both their namesto Martin Luther King
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Fully qualified 14years at university, 7of those yrs he studiedmedicine, therealways men.Knowledge: 5Experience: 5Cost of treatment: 1Success rate: 5Knowledge: 4Experience: 4Cost of treatment: 3Success rate: 3He sells andmixesmedicine.Pres
Boise State - HIST - 100
Boise State - HIST - 100
Boise State - HIST - 100
Who were they?Where did they come from?What did they accomplish?Where did they go?The Minoans and MycenaeansMinoan civilization arose on the island of Crete.Legacy (or gift from the past) Their legacy was asmasters of the sea andgreat shipbuilder
Boise State - HIST - 100
Why did the Mormons Manage tosurvive Salt Lake City?The American WestThe Geographical Position of UtahBrigham Young Prior Planning Mormons had faith in theirleader Organisational Skills Setting up supply depotsand workshops No land ownership P
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TheNormanYoke conquest castles war&waste forestlaw rebels&outlaws merrieEnglandHastings,13October1066ThedayEnglandacquiredanewroyaldynasty,anewaristocracy,anewChurch,anewlanguage,anewHaroldkilledDoverburntTheConquerorsfootprintsPlottedby
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PEARLHARBORPEARLHARBORTHEDAYOFINFAMYDecember7,1941USSArizonaCausesCauses The U.S. demanded that Japan withdrawThefrom China and Indochinafrom Japan thought that attacking the U.S. wouldJapanprovide them an easy win, and a territorywith abund
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The Drugs Revolution andthe Development ofPenicillinPenicillinPrevention and CurePreventionPrevention Germ theory VaccinationsSmallpox1906 Tuberculosis1913 DiphtheriaCureKoch discovered thatKochhe could stainbacteriabacteriaDevelopment o
Boise State - HIST - 100 | 3,641 | 13,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2014-15 | longest | en | 0.712174 |
https://ufc246results.com/qa/quick-answer-is-it-better-to-be-in-a-higher-or-lower-percentile.html | 1,621,282,948,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992440.69/warc/CC-MAIN-20210517180757-20210517210757-00158.warc.gz | 608,646,907 | 9,524 | # Quick Answer: Is It Better To Be In A Higher Or Lower Percentile?
## Is 37th percentile good?
“at the percentile”: The 37th percentile is a score that’s higher than 37% of the observed population.
Being at the 37th percentile means you are above that level: you are part of the top 63% (probably just so).
This being said, you will often find “in the percentile” to mean above 37%, so good luck with that..
## What is percentile in simple words?
The most common definition of a percentile is a number where a certain percentage of scores fall below that number. … That means if you scored 156 on the exam, your score was better than 70 percent of test takers. The 25th percentile is also called the first quartile.
## What is a good weight percentile?
There’s a big range of normal on the chart: Anyone who falls between the 5th percentile and the 85th percentile is a healthy weight. If someone is at or above the 85th percentile line on the chart (but less than the 95th percentile), doctors consider that person overweight.
## What SAT score is 99 percentile?
What Are the Percentile Ranges for the SAT?SAT Composite Score RangePercentile Score1550-160099 to 99+1500-155098 to 991450-150096 to 981400-145094 to 9620 more rows•Sep 26, 2020
## What percentile should my baby be in?
A baby on the 50th percentile for weight, for example, is right in the middle of the normal range: 50% of babies their age are lighter, and 50% are heavier. A baby on the 5th percentile weighs less than 95% of other babies of that age. A baby on the 90th percentile weights more than 90% of other babies that age.
## Is 100th percentile possible?
If you use the “percentage below or equal to” definition, then, yes, you can have a 100th percentile. However, if you use the “percentage below” definition, then, no, you cannot have a 100th percentile. … For percentiles, there are actually only 99 equal partitions of the population being ranked: 1 to 99.
## What does 99th percentile mean?
It means the top 1 percent. If something is in the 99th percentile, then it means it is higher than 99% of other things. This is most often used when talking about test results. I scored in the 99th percentile on the standardized test.
## Are baby percentiles accurate?
“Although the charts are commonly used to graphically illustrate the typical growth patterns for boys and girls, it is important to note that they do not accurately reflect the growth of all children,” she says. Also, “there is a large range of normal. . . .
## Do baby percentiles matter?
A healthy child can fall anywhere on the chart. A lower or higher percentile doesn’t mean there is something wrong with your baby. Regardless of whether your child is in the 95th or 15th percentile, what matters is that she or he is growing at a consistent rate over time.
## What is 100th percentile on MCAT?
Anything between a score of 524 and 528 is considered in 100th percentile, and there is no perfect score, as the exam is calculated on a percentage basis and changes from year to year. The Association of American Medical Colleges estimates that about 85,000 people sit for the MCAT every year.
## What does the 98th percentile mean?
To be clear on the math, sometimes you may hear of a child being at the 98th percentile of growth. That means they’re bigger than 98% of children, with just 2% of kids bigger than them (98+2 = 100).
## What is a good percentile rank?
A percentile rank score of 60 or above is considered above average. The National Percentile Rank score (NP) typically follows the Raw Score (RS) as you look across the page of an achievement test report from left to right.
## Is it good to be in the 95th percentile?
A nice property of percentiles is they have a universal interpretation: Being at the 95th percentile means the same thing no matter if you are looking at exam scores or weights of packages sent through the postal service; the 95th percentile always means 95% of the other values lie below yours, and 5% lie above it.
## Is it better to be in the 10th and 90th percentile?
If a candidate scores in the 90th percentile, they have scored higher than 90% of the norm group, putting them in the top 10%. If a candidate scores in the 10th percentile, they have scored higher than 10% of the norm group, putting them in the bottom 10%.
## What is the 90th percentile income?
\$184,292Household income at selected percentiles, 2018PercentileDollar limit50th (median)\$63,17980th\$130,00090th\$184,29295th\$248,7284 more rows•Sep 21, 2019
## What is the 90th percentile of a normal distribution?
When we go to the table, we find that the value 0.90 is not there exactly, however, the values 0.8997 and 0.9015 are there and correspond to Z values of 1.28 and 1.29, respectively (i.e., 89.97% of the area under the standard normal curve is below 1.28)….Computing Percentiles.PercentileZ50th075th0.67590th1.28295th1.6457 more rows•Jul 24, 2016
## What IQ is 95th percentile?
IQ 125 is at the 95th percentile – 95% of people have an IQ equal to or less than 125. This means 5% of the population score higher.
## What IQ is 93rd percentile?
The following is a chart of I.Q. scores from 100 to 202. The rarity column shows how many people in the general population are expected to achieve this percentile or higher….Intelligence interval.I.Q.PercentileRarity (1/x)19699.999 999 901,000,000,00019799.999 999 931,500,000,00019899.999 999 952,000,000,000100 more rows
## How 95th percentile is calculated?
Sort the samples from smallest to largest. Then the number of data samples is multiplied by . 95 and rounded up to the next whole number. The value in that position is the 95th percentile.
## Is percentile better higher or lower?
Percentile “ranks” If you “scored in the 66th percentile”, you scored “as well as or better than” 66% of the group. If your score was the same as “the mean” for that test, you scored in the 50th percentile. (The mean is the middle of a distribution, and the 50th percentile is the same…the middle of the group.)
## What does it mean when your baby is in the 100th percentile?
The first percentile would have the fewest — none. The 100th percentile would include people who own the most turtles. In health care, the term “percentile” is most often used for height and weight. This tells how one individual compares with other individuals in the community — usually, in the United States. | 1,606 | 6,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-21 | longest | en | 0.92581 |
https://mathoverflow.net/questions/232911/countably-generated-sigma-algebra | 1,561,444,711,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999800.5/warc/CC-MAIN-20190625051950-20190625073950-00515.warc.gz | 517,689,629 | 30,860 | # Countably generated $\sigma$-algebra
Let $(\Omega,\Sigma,\mu)$ be a countably generated probability space. Must $(\Omega,\Sigma,\mu)$ be isomorphic modulo null sets to a standard probability space?
I assume not, so here is a more specific question. Let $\Omega$ be an ultraproduct of finite sets and $\Sigma$ a countably generated sub-$\sigma$-algebra of the Loeb $\sigma$-algebra, and let $\mu$ be the Loeb measure. Is $(\Omega,\Sigma,\mu)$ a standard probability space?
I gather from Jin and Keisler. Maharam spectra of Loeb spaces, providing I understand the language, that if your ultraproducts are taken over a countable set in the usual way then the Loeb space is isomorphic modulo null sets to the product $\{0,1\}^\mathbf{R}$.
• About your first question, there's a measure algebra isomorphism between the completion of $(\Omega,\Sigma,\mu)$ and a standard probability space. Do you mean a pointwise isomorphism up to a negligible set ? – Stéphane Laurent Mar 6 '16 at 11:42
For the first question, the answer is yes. There is an isomorphism between measure algebras. Let $B$ be the Boolean algebra of all measurable sets modulo the collection of null sets. Then define a metric $\rho$ on $B$ by letting $\rho(x,y)=\mu((x\wedge y')\vee(y\wedge x'))$.
$\mathbf{Theorem}$:(Caratheodory, see Royden Real Analysis third edition Theorem 15.3.4) Suppose that $(B,\rho)$ is separable and $(B,\mu)$ is a probability space. Then there is an injective measure preserving $\sigma$-complete Boolean algebra homomorphism $\Phi:(B,\rho)\rightarrow(C/I,m)$ where $C$ is the collection of all Borel sets on $[0,1]$ and $I$ is the ideal of all measure zero sets. If $B$ is atomless, then the mapping $\Phi$ can is bijection.
For a proof, if $B$ is generated by a countable subalgebra $A$, then one inductively constructs a homomorphism from $A$ to the Boolean subalgebra of $C$ consisting of all finite unions of open intervals and then one extends this homomorphism from $A$ to $B$. | 539 | 1,981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-26 | longest | en | 0.806697 |
https://crazyproject.wordpress.com/2011/05/09/investigate-the-kernel-of-a-module-homomorphism-to-a-given-tensor-product/ | 1,490,235,613,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186608.9/warc/CC-MAIN-20170322212946-00259-ip-10-233-31-227.ec2.internal.warc.gz | 769,118,511 | 18,336 | ## Investigate the kernel of a module homomorphism to a given tensor product
Let $R$ be an integral domain with field of fractions $Q$ and let $N$ be a left unital $R$-module. Prove that the kernel of the $R$-module homomorphism $\iota : N \rightarrow Q \otimes_R N$ given by $n \mapsto 1 \otimes n$ is $\mathsf{Tor}(N)$.
If $n \in \mathsf{ker}\ \iota$, then $1 \otimes n = 0$. By a previous exercise, there exists a nonzero $r \in R$ such that $r \cdot n = 0$. Thus $n \in \mathsf{Tor}_R(N)$.
Conversely, suppose $n \in \mathsf{Tor}_R(N)$ with $r \in R$ nonzero such that $r \cdot n = 0$. Then $\iota(n) = 1 \otimes n = 0$, again using the previous exercise. | 230 | 662 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-13 | latest | en | 0.590546 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-5-chapters-r-5-cumulative-review-exercises-page-363/7 | 1,524,553,852,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946565.64/warc/CC-MAIN-20180424061343-20180424081343-00179.warc.gz | 428,075,908 | 12,308 | ## Intermediate Algebra (12th Edition)
$x\gt-\dfrac{1}{2}$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $3-2(x+3)\lt4x ,$ use first the Distributive Property. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3-2(x)-2(3)\lt4x \\\\ 3-2x-6\lt4x .\end{array} Using the properties of inequality to combine like terms, the inequality above is equivalent to \begin{array}{l}\require{cancel} -2x-4x\lt-3+6 \\\\ -6x\lt3 .\end{array} Dividing both sides by $-6 ,$ in which the inequality symbol reverses, results to \begin{array}{l}\require{cancel} x\gt\dfrac{3}{-6} \\\\ x\gt-\dfrac{1}{2} .\end{array} | 257 | 794 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2018-17 | latest | en | 0.712232 |
https://www.umsu.de/logic2/08-conditionals.html | 1,718,532,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00897.warc.gz | 902,834,692 | 19,187 | ## 8Conditionals
8.4 Restrictors
### 8.1Material conditionals
We are often interested not just in whether something is in fact the case, but also in whether it is (or would be) the case if something else is (or would be) the case. We might, for example, wonder in what will happen to the climate if we don’t reduce greenhouse gases, or whether World War 2 could have been avoided if certain steps had been taken in the 1930s.
A sentence stating that something is (or would be) the case if something else is (or would be) the case is called a conditional. What exactly, do these statements mean? What is their logic? Philosophers have puzzled over these questions for more than 2000 years, with no agreement in sight.
One attractively simple view is that a conditional ‘if $$A$$ then $$B$$’ is true iff the antecedent $$A$$ is false or the consequent $$B$$ is true. This would make ‘if $$A$$ then $$B$$’ equivalent to ‘not $$A$$ or $$B$$’. Conditionals with these truth-conditions are called material conditionals.
The conditionals $$A \to B$$ of classical logic are material. $$A \to B$$ is equivalent to $$\neg A \lor B$$. The attractively simple view that English conditionals are material conditionals would mean that we can faithfully translate English conditionals into $$\mathfrak {L}_{M}$$-sentences of the form $$A \to B$$. Is this correct?
There are some arguments for a positive answer. Suppose I make the following promise.
• (1) If I don’t have to work tomorrow then I will help you move.
I have made a false promise if the next day I don’t have to work and yet I don’t help you move. Under all other conditions, you could not fault me for breaking my promise. So it seems that (1) is false iff I don’t have to work and I don’t help you move. Generalizing, this suggests that ‘if $$A$$ then $$B$$’ is true iff $$A$$ is false or $$B$$ is true.
Another argument for analysing English conditionals as material conditionals starts with the intuitively plausible assumption that ‘$$A$$ or $$B$$’ entails the corresponding conditional ‘if not-$$A$$ then $$B$$’. (This is sometimes called the or-to-if inference.) Suppose I tell you that Nadia is either in Rome or in Paris. Trusting me, you can infer that if she’s not in Rome then she’s in Paris. Now we can reason as follows.
To begin, suppose that $$A$$ and ‘if $$A$$ then $$B$$’ are both true. Plausibly, we can infer that $$B$$ is true as well: modus ponens is valid for English conditionals. This means that if $$A$$ is true and $$B$$ is false, then ‘if $$A$$ then $$B$$’ is false. Now suppose, alternatively, that $$A$$ is false or $$B$$ is true. Then ‘not-$$A$$ or $$B$$’ is true. By or-to-if, we can infer that ‘if $$A$$ then $$B$$’ is true as well. Thus ‘if $$A$$ then $$B$$’ is true iff $$A$$ is false or $$B$$ is true.
Despite these arguments, most philosophers and linguists don’t think that English conditionals are material conditionals. Consider these facts about logical consequence (in classical propositional logic).
(M1)
$$B \models A \to B$$
(M2)
$$\neg A \models A \to B$$
(M3)
$$\neg (A \to B) \models A$$
(M4)
$$A \to B \models \neg B\to \neg A$$
(M5)
$$A \to B \models (A\land C) \to B$$
If English conditionals were material conditionals then the following inferences, corresponding to (M1)–(M5), would be valid.
(E1)
There won’t be a nuclear war. Therefore: If Russia attacks the US with nuclear weapons then there won’t be a nuclear war.
(E2)
There won’t be a nuclear war. Therefore: If there will be a nuclear war then nobody will die.
(E3)
It is not the case that if it will rain tomorrow then the Moon will fall onto the Earth. Therefore: It will rain tomorrow.
(E4)
If our opponents are cheating, we will never find out. Therefore: If we will find out that our opponents are cheating, then they aren’t cheating.
(E5)
If you add sugar to your coffee, it will taste good. Therefore: If you add sugar and vinegar to your coffee, it will taste good.
These inferences do not sound good. If we wanted to defend the view that English conditionals are material conditionals we would have to explain why they sound bad even though they are valid. We will not explore this option any further.
Exercise 8.1
Can you find a different analysis of English conditionals that, like the material analysis, would make conditionals truth-functional, but that would render all of (E1)–(E5) invalid?
Even those who defend the material analysis of English conditionals admit that it does not work for all English conditionals. Consider (2).
(2)
If water is heated to $$100^\circ$$ C, it evaporates.
This shouldn’t be translated as $$p\to q$$. Intuitively, (2) states that in all (normal) cases where water is heated to $$100^\circ$$ C, it evaporates. It is a quantified, or modal claim.
Another important class of conditionals that can’t be analysed as material conditionals are so-called subjunctive conditionals. Compare the following two statements.
(3)
If Shakespeare didn’t write Hamlet, then someone else did.
(4)
If Shakespeare hadn’t written Hamlet, then someone else would have.
(3) seems true. Someone has written Hamlet; if it wasn’t Shakespeare then it must have been someone else. But (4) is almost certainly false. After all, it is very likely that Shakespeare did write Hamlet. And it is highly unlikely that if he hadn’t written Hamlet – if he got distracted by other projects, say – then someone else would have stepped in to write the exact same piece.
Sentences like (3) are called indicative conditionals. Intuitively, an indicative conditional states that something is in fact the case on the assumption that something else is the case. A subjunctive conditional like (4) states that something would be the case if something else were the case. Typically we know that the “something else” is not in fact the case. We know, for example, that Shakespeare wrote Hamlet and therefore that the antecedent of (4) is false. For this reason, subjunctive conditionals are also called counterfactual conditionals or simply counterfactuals.
It should be clear that subjunctive conditionals are not material conditionals. I said that (4) is almost certainly false. But it almost certainly has a false antecedent. So the corresponding material conditional is almost certainly true.
### 8.2Strict conditionals
One apparent difference between material conditionals $$A \to B$$ and conditionals in natural language is that $$A\to B$$ requires no connection between the antecedent $$A$$ and the consequent $$B$$. Consider (1).
(1)
If we leave after 5, we will miss the train.
Intuitively, someone who utters (1) wants to convey that missing the train is a necessary consequence of leaving after 5 – that it is impossible to leave after 5 and still make it to the train, given certain facts about the distance to the station, the time it takes to get there, etc. This suggests that (1) should be formalized not as $$p \to q$$ but as $$\Box (p \to q)$$ or, equivalently, $$\neg \Diamond (p \land \neg q)$$.
Sentences that are equivalent to $$\Box (A \to B)$$ are called strict conditionals. The label goes back to C.I. Lewis (1918), who also introduced the abbreviation $$A \,\unicode {x297D}\, B$$ for $$\Box (A \to B)$$.
Lewis was not interested in ‘if …then …’ sentences. He introduced $$A \,\unicode {x297D}\, B$$ to formalize ‘$$A$$ implies $$B$$’ or ‘$$A$$ entails $$B$$’. His intended use of $$\,\unicode {x297D}\,$$ roughly matches our use of the double-barred turnstile ‘$$\models$$’. But there are important differences. The turnstile is an operator in our meta-language; Lewis’s $$\,\unicode {x297D}\,$$ is an object-language operator that, like $$\land$$ or $$\to$$, can be placed between any two sentences in a formal language to generate another sentence in the language. $$p \,\unicode {x297D}\, (q \,\unicode {x297D}\, p)$$ is well-formed, whereas $$p \models (q\models p)$$ is gibberish. Moreover, while $$p \models q$$ is simply false – because there are models in which $$p$$ is true and $$q$$ false – Lewis’s $$p \,\unicode {x297D}\, q$$ is true on some interpretation of the sentence letters and false on others. If $$p$$ means that it raining heavily and $$q$$ that it is raining, then $$p \,\unicode {x297D}\, q$$ is true because the hypothesis that it is raining heavily implies that it is raining.
Let’s set aside Lewis’s project of formalizing the concept of implication. Our goal is to find an object-language construction that functions like ‘if …then …’ in English. To see whether ‘$$\ldots \,\unicode {x297D}\,\ldots$$’ can do the job, let’s have a closer look at the logic of strict conditionals.
Since $$A \,\unicode {x297D}\, B$$ is equivalent to $$\Box (A \to B)$$, standard Kripke semantics for the box also provides a semantics for strict conditionals. In Kripke semantics, $$\Box (A \to B)$$ is true at a world $$w$$ iff $$A \to B$$ is true at all worlds $$v$$ accessible from $$w$$. And $$A \to B$$ is true at $$v$$ iff $$A$$ is false at $$v$$ or $$B$$ is true at $$v$$. We therefore have the following truth-conditions for strict conditionals.
Definition 8.1: Kripke semantics for $$\,\unicode {x297D}\,$$
If $$M = \langle W,R,V \rangle$$ is a Kripke model, then
$$M,w \models A \,\unicode {x297D}\, B$$ iff for all $$v$$ such that $$wRv$$, either $$M,v \not \models A$$ or $$M,v \models B$$.
Exercise 8.2
$$A \,\unicode {x297D}\, B$$ is equivalent to $$\Box (A \to B)$$. Can you fill the blank in: ‘$$\Box A$$ is equivalent to —’, using no modal operator other than $$\,\unicode {x297D}\,$$?
As always, the logic of strict conditionals depends on what constraints we put on the accessibility relation. Without any constraints, $$\,\unicode {x297D}\,$$ does not validate modus ponens, in the sense that $$A \,\unicode {x297D}\, B$$ and $$A$$ together do not entail $$B$$. We can see this by translating $$A \,\unicode {x297D}\, B$$ back into $$\Box (A \to B)$$ and setting up a tree. Recall that to test whether some premises entail a conclusion, we start the tree with the premises and the negated conclusion.
With the K-rules, where we don’t make any assumptions about the accessibility relation, node 1 can’t be expanded, so there is nothing more we can do.
Exercise 8.3
Give a countermodel in which $$p \,\unicode {x297D}\, q$$ and $$p$$ are true at some world while $$q$$ is false.
If we assume that the accessibility relation is reflexive, the tree closes:
It is not hard to show that modus ponens for $$\,\unicode {x297D}\,$$ is valid on all and only the reflexive frames. Reflexivity is precisely what we need to render modus ponens valid. Since modus ponens looks plausible for English conditionals (as I’ve argued on p. 216), we’ll probably want the relevant Kripke models to be reflexive.
Exercise 8.4
Using the tree method, and translating $$A \,\unicode {x297D}\, B$$ into $$\Box (A \to B)$$, confirm that following claims hold, for all $$A,B,C$$.
(a)
$$\models _K A \,\unicode {x297D}\, A$$
(b)
$$A \,\unicode {x297D}\, B \models _K \neg B \,\unicode {x297D}\, \neg A$$
(c)
$$A \,\unicode {x297D}\, B \models _K (A \land C) \,\unicode {x297D}\, B$$
(d)
$$A\,\unicode {x297D}\, B, B \,\unicode {x297D}\, C \models _{K} A \,\unicode {x297D}\, C$$
(e)
$$(A \lor B) \,\unicode {x297D}\, C \models _K (A \,\unicode {x297D}\, C) \land (B \,\unicode {x297D}\, C)$$
(f)
$$A \,\unicode {x297D}\, (B \,\unicode {x297D}\, C) \models _T (A \land B) \,\unicode {x297D}\, C$$
(g)
$$A\,\unicode {x297D}\, B \models _{S4} C \,\unicode {x297D}\, (A \,\unicode {x297D}\, B)$$
(h)
$$((A\,\unicode {x297D}\, B) \,\unicode {x297D}\, C) \,\unicode {x297D}\, (A\,\unicode {x297D}\, B) \models _{S5} A\,\unicode {x297D}\, B$$
Which of these schemas do you think should be valid if we assume that $$A \,\unicode {x297D}\, B$$ translates indicative conditionals ‘if $$A$$ then $$B$$’?
We could now look at other conditions on the accessibility relation and decide whether they should be imposed, based on what they would imply for the logic of conditionals. But let’s take a shortcut.
I have suggested that sentence (1) might be understood as saying that it is impossible to leave after 5 and still make it to the train. Impossible in what sense? There are many possible worlds at which we leave after 5 and still make it to the train. There are, for example, worlds at which the train departs two hours later, worlds at which we live right next to the station, and so on. When I say that it is impossible to leave after 5 and still make it to the train, I arguably mean that it is impossible given what we know about the departure time, our location, etc.
Generalizing, a tempting proposal is that the accessibility relation that is relevant for indicative conditionals like (1) is the epistemic accessibility relation that we studied in chapter 5, where a world $$v$$ is accessible from $$w$$ iff it is compatible with what is known at $$w$$. On that hypothesis, the logic of indicative conditionals is determined by the logic of epistemic necessity. We don’t need to figure out the relevant accessibility relation from scratch.
Since knowledge varies from agent to agent, the present idea implies that the truth-value of indicative conditionals should be agent-relative. This seems to be confirmed by the following puzzle, due to Allan Gibbard.
Sly Pete and Mr. Stone are playing poker on a Mississippi riverboat. It is now up to Pete to call or fold. My henchman Zack sees Stone’s hand, which is quite good, and signals its content to Pete. My henchman Jack sees both hands, and sees that Pete’s hand is rather low, so that Stone’s is the winning hand. At this point the room is cleared. A few minutes later, Zack slips me a note which says ‘if Pete called, he won’, and Jack slips me a note which says ‘if Pete called, he lost’.
The puzzle is that Zack’s note and Jack’s note are intuitively contradictory, yet they both seem to be true.
We can resolve the puzzle if we understand the conditionals as strict conditionals with an agent-relative epistemic accessibility relation. Take Zack. Zack knows that Pete knows Stone’s hand. He also knows that Pete would not call unless he has the better hand. So among the worlds compatible with Zack’s knowledge, all worlds at which Pete calls are worlds at which Pete wins. If $$p$$ translates ‘Pete called’ and $$q$$ ‘Pete won’, then $$p \,\unicode {x297D}\, q$$ is true relative to Zack’s information state. Relative to Jack’s information state, however, the same sentence is false. Jack knows that Stone’s hand is better than Pete’s, but he doesn’t know that Pete knows Stone’s hand. Among the worlds compatible with Jack’s knowledge, all worlds at which Pete calls are therefore worlds at which Pete loses. Relative to Jack’s information state, $$p \,\unicode {x297D}\, \neg q$$ is true.
Another advantage of the “epistemically strict” interpretation is that it might explain why indicative conditionals with antecedents that are known to be false seem defective. For example, imagine a scenario in which Jones has gone to work. In that scenario, is (2) true or false?
(2)
If Jones has not gone to work then he is helping his neighbours.
The question is hard to answer – and not because we lack information about the scenario. Once we are told that Jones has gone to work, it is unclear how we are meant to assess whether Jones is helping his neighbours if he has not gone to work. On the epistemically strict interpretation, (2) says that Jones is helping his neighbours at all epistemically accessible worlds at which Jones hasn’t gone to work. Since we know that Jones has gone to work, there are no epistemically accessible worlds at which he hasn’t gone to work. And if there are no $$A$$-worlds then we naturally balk at the question whether all $$A$$-worlds are $$B$$-worlds. (In logic, we resolve to treat ‘all $$A$$s are $$B$$’ as true if there are no $$A$$s. Accordingly, (2) comes out true on the epistemically strict analysis. But we can still explain why it seems defective.)
We have found a promising alternative to the hypothesis that indicative conditionals are material conditionals. According to the present alternative, they are epistemically strict conditionals – strict conditionals with an epistemic accessibility relation.
What about subjunctive conditionals? Return to the two Shakespeare conditionals from the previous section. When we evaluate the indicative sentence – ‘If Shakespeare didn’t write Hamlet, then someone else did’ – we hold fixed our knowledge that Hamlet exists; worlds where the play was never written are inaccessible. That’s why the conditional is true. At all accessible worlds at which Shakespeare didn’t write Hamlet, someone else wrote the play. When we evaluate the subjunctive conditional – ‘If Shakespeare hadn’t written Hamlet, then someone else would have’ – we do consider worlds at which Hamlet was never written, even though we know that the actual world is not of that kind. If subjunctive conditionals are strict conditionals, then their accessibility relation does not track our knowledge or information. Unfortunately, as we are going to see in the next section, it is hard to say what else it could track.
This is one problem for the strict analysis of natural-language conditionals. Another problem lies in the logic of strict conditionals. Remember (E1)–(E5) here. If English conditionals are strict conditionals, then (E1)–(E3) are invalid. For example, while $$q$$ entails $$p \to q$$, it does not entail $$p \,\unicode {x297D}\, q$$. But the strict analogs of (M4) and (M5) still hold, no matter what we say about accessibility (see exercise 8.4): \begin {flalign*} \quad & A \,\unicode {x297D}\, B \models \neg B\,\unicode {x297D}\, \neg A; &\\ \quad & A \,\unicode {x297D}\, B \models (A\land C) \,\unicode {x297D}\, B. & \end {flalign*}
So we still predict that the inferences (E4) and (E5) are valid.
(E4)
If our opponents are cheating, we will never find out. Therefore: If we will find out that our opponents are cheating, then they aren’t cheating.
(E5)
If you add sugar to your coffee, it will taste good. Therefore: If you add sugar and vinegar to your coffee, it will taste good.
Exercise 8.5
The badness of (E4) and (E5) suggests that indicative conditionals can’t be analysed as strict conditionals. Can you give a similar argument suggesting that subjunctive conditionals can’t be analysed as strict conditionals?
Exercise 8.6
A plausible norm of pragmatics is that a sentence should only be asserted if it is known to be true. Let’s call a sentence assertable if it is known to be true. Show that if the logic of knowledge is at least S4, then an epistemically strict conditional $$A \,\unicode {x297D}\, B$$ is assertable iff the corresponding material conditional $$A \to B$$ is assertable.
Exercise 8.7
Explain why the ‘or-to-if’ inference from ‘$$p$$ or $$q$$’ to ‘if not $$p$$ then $$q$$’ is invalid on the assumption that the conditional is epistemically strict. How could a friend of this assumption explain why the inference nonetheless looks reasonable, at least in normal situations? (Hint: Remember the previous exercise.)
### 8.3Variably strict conditionals
Let’s have a closer look at subjunctive conditionals. As I am writing these notes, I am sitting in Coombs Building, room 2228, with my desk facing the wall to Al Hájek’s office in room 2229. In light of these facts, (1) seems true.
(1)
If I were to drill a hole through the wall behind my desk, the hole would come out in Al’s office.
There is no logical connection between the antecedent of (1) and the consequent. There are many possible worlds at which I drill a hole through the wall behind my desk and don’t reach Al’s office – for example, worlds at which my desk faces the opposite wall, worlds at which Al’s office is in a different room, and so on. If (1) is a strict conditional then all such worlds must be inaccessible.
Now consider (2).
(2)
If the office spaces had been randomly reassigned yesterday then Al’s office would (still) be next to mine.
(2) seems false, or at least very unlikely. But if (2) is a strict conditional, and worlds at which Al is not in room 2229 or I am not in 2228 are inaccessible – as they seem to be for (1) – then (2) should be true. Among worlds at which I am in 2228 and Al is in 2229, all worlds at which the office spaces have been randomly reassigned yesterday are worlds at which Al’s office is next to mine. When we evaluate (2), it looks like we no longer hold fixed who is in which office. Worlds that were inaccessible for (1) are accessible for (2).
So the accessibility relation, at least for subjunctive conditionals, appears to vary from conditional to conditional. As David Lewis put it, subjunctive conditionals seem to be not strict, but “variably strict”.
Let’s try to get a better grip on how this might work. (What follows is a slightly simplified version of an analysis developed by Robert Stalnaker and David Lewis in the 1960s.)
Intuitively, when we ask what would have been the case if a certain event had occurred, we are looking at worlds that are much like the actual world up to the time of the event. Then these worlds deviate in some minimal way to allow the event to take place. Afterwards the worlds unfold in accordance with the general laws of the actual world.
For example, if we wonder what would have happened if Shakespeare hadn’t written Hamlet, we are interested in worlds that are like the actual world until 1599, at which point some mundane circumstances prevent Shakespeare from writing Hamlet. We are not interested in worlds at which Shakespeare was never born, or in which the laws of nature are radically different from the laws at our world. One might reasonably judge that Shakespeare would have been a famous author even if he hadn’t written Hamlet, although we would hardly be famous in worlds in which he was never born.
Likewise for (1). Here we are considering worlds that are much like the actual world up to now, at which point I decide to drill a hole and find a suitable drill. These changes do not require my office to be in a different room. Worlds where I’m not in room 2228 can be ignored. Figuratively speaking, such worlds are “too remote”: they differ from the actual world in ways that are not required to make the antecedent true.
This suggests that a subjunctive conditional is true iff the consequent is true at the “closest” worlds at which the antecedent is true – where “closeness” is a matter of similarity in certain respects. The closest worlds (to the actual world) at which Shakespeare didn’t write Hamlet are worlds that almost perfectly match the actual world until 1599, then deviate a little so that Shakespeare didn’t write Hamlet, and afterwards still resemble the actual world with respect to the general laws of nature. We will not try to spell out in full generality what the relevant closeness measure should look like.
Let ‘$$v \prec _w u$$’ mean that $$v$$ is closer to $$w$$ than $$u$$, in the sense that $$v$$ differs less than $$u$$ from $$w$$ in whatever respects are relevant to the interpretation of subjunctive conditionals.
We make the following structural assumptions about the world-relative ordering $$\prec$$.
1.
If $$v \prec _w u$$ then $$u \nprec _w v$$. (Asymmetry)
2.
If $$v \prec _w u$$, then for all $$t$$ either $$v \prec _w t$$ or $$t \prec _w u$$. (Quasi-connectedness)
3.
For any non-empty set of worlds $$X$$ and world $$w$$ there is a $$v$$ in $$X$$ such that there is no $$u$$ in $$X$$ with $$u \prec _w v$$.
Asymmetric and quasi-connected relations are known as weak orders. Asymmetry is self-explanatory. Quasi-connectedness is more often called negative transitivity, because it is equivalent to the assumption that if $$t \not < s$$ and $$s\not <r$$ then $$t\not <r$$. It ensures that the “equidistance” relation that holds between $$v$$ and $$u$$ if neither $$v \prec _w u$$ nor $$u \prec _w v$$ is an equivalence relation. With these two assumptions, we can picture each world $$w$$ as associated with nested spheres of worlds; $$v \prec _w u$$ means that $$v$$ is in a more narrow $$w$$-sphere than $$u$$.
Assumption 3 is known as the Limit Assumption. It ensures that for any consistent proposition $$A$$ and world $$w$$, there is a set of closest $$A$$-worlds. Without the Limit Assumption, there could be an infinite chain of ever closer $$A$$-worlds, with no world being maximally close.
Exercise 8.8
Show that asymmetry and quasi-connectedness imply transitivity.
Exercise 8.9
Define $$\preceq _{w}$$ so that $$v \preceq _w u$$ iff $$u \nprec _w v$$ (that is, iff it is not the case that $$u \prec _{w} v$$). Informally, $$v \preceq _w u$$ means that $$v$$ is at least as similar to $$w$$ in the relevant respects as $$u$$. Many authors use $$\preceq$$ rather than $$\prec$$ as their basic notion. Can you express the above three conditions on $$\prec$$ in terms of $$\preceq$$?
We are going introduce a variably strict operator $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$ so that $$A\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ is true at a world $$w$$ iff $$B$$ is true at the closest worlds to $$w$$ at which $$A$$ is true. Models for a language with the $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$ operator must contain closeness orderings $$\prec$$ on the set of worlds.
Definition 8.2
A similarity model consists of
• a non-empty set $$W$$,
• for each $$w$$ in $$W$$ a weak order $$\prec _w$$ that satisfies the Limit Assumption, and
• a function $$V$$ that assigns to each sentence letter a subset of $$W$$.
To formally state the semantics of $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$, we can re-use a concept from section 6.3. Let $$S$$ be an arbitrary set of worlds, and let $$w$$ be some world (that may or may nor be in $$S$$). It will be useful to have an expression that picks out the most similar worlds to $$w$$, among all the worlds in $$S$$. This expression is $$\mathrm {Min}^{\prec _w}(S)$$, which we have defined as follows in section 6.3: $\mathrm {Min}^{\prec _w}(S) =_\text {def} \{ v: v \in S \land \neg \exists u (u \in S \land u \prec _w v) \}.$
Now $$\{ u : M,u\models A \}$$ is the set of worlds (in model $$M$$) at which $$A$$ is true. So $$\mathrm {Min}^{\prec _w}(\{ u : M,u\models A \})$$ is the set of those $$A$$-worlds that are closest to $$w$$. We want $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ to be true at $$w$$ iff $$B$$ is true at the closest $$A$$-worlds to $$w$$.
Definition 8.3: Similarity semantics for $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$
If $$M$$ is a similarity model and $$w$$ a world in $$M$$, then
$$M,w \models A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ iff $$M,v \models B$$ for all $$v$$ in $$\mathrm {Min}^{\prec _w}(\{ u: M,u \models A \})$$.
You may notice that $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ works almost exactly like $$\mathsf {O}(B/A)$$ from section 6.3. There, I said that for any world $$w$$ in any deontic ordering model $$M$$,
$$M,w \models \mathsf {O} (B/A) \text { iff } M,v \models B\text { for all v in \mathrm {Min}^{\prec _w}(\{ u: wRu and M,u\models A \})}$$.
The main difference is that conditional obligation is sensitive to an accessibility relation. If that relation is an equivalence relation then this makes no difference to the logic.
Of course, the order $$\prec$$ in deontic ordering models is supposed to represent degree of conformity to norms, while the order $$\prec$$ in similarity models represents a certain similarity ranking in the evaluation of subjunctive conditionals. A different type of ordering might be in play when we evaluate indicative conditionals, which some have argued should also be interpreted as variably strict. But again, these differences in interpretation don’t affect the logic.
Suppose we add the $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$ operator to the language of standard propositional logic. The set of sentences in this language that are true at all worlds in all similarity models is known as system V. There are tree rules and axiomatic calculi for this system, but they aren’t very user-friendly. We will only explore the system semantically.
To begin, we can check whether modus ponens is valid for $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$. That is, we check whether the truth of $$A$$ and $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ at a world in a similarity model entails the truth of $$B$$.
Assume that $$A$$ and $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ are true at a world $$w$$. By definition 8.3, the latter means that $$B$$ is true at all the closest $$A$$-worlds to $$w$$ (at all worlds in $$\mathrm {Min}^{\prec _w}(\{u: M,u\models A\})$$). The world $$w$$ itself is an $$A$$-world. If we could show that $$w$$ is among the closest $$A$$-worlds to itself then we could infer that $$A$$ is true at $$w$$.
Without further assumptions, however, we can’t show this. If we want to validate modus ponens, we must add a further constraint on our models: that every world is among the closest worlds to itself. More precisely, $\text {for all worlds w and v, v \nprec _{w} w.}$ This assumption is known as Weak Centring. The logic we get if we impose this constraint is system VC.
Exercise 8.10
Should we accept Weak Centring for deontic ordering models?
Exercise 8.11
Explain why $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ entails $$A \to B$$, assuming Weak Centring.
Exercise 8.12
Show that if $$A$$ is true at no worlds, then $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ is true.
None of the problematic inferences (E1)–(E5) are valid if the relevant conditionals are interpreted as variably strict. (E5), for example, would assume that $$p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r$$ entails $$(p \land q) \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r$$. But it does not. We can give a countermodel with two worlds $$w$$ and $$v$$; $$p$$ is true at both worlds, $$q$$ is true only at $$v$$, and $$r$$ only at $$w$$; if $$w$$ is closer to itself than $$v$$, then $$p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r$$ is true at $$w$$ (because the closest $$p$$-worlds to $$w$$ are all $$r$$-worlds), but $$(p \land q) \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r$$ is false at $$w$$ (because the closest $$(p\land q)$$-worlds to $$w$$ aren’t all $$r$$-worlds).
The diagram on the right represents this model. The circles around $$w$$ depict the similarity spheres. $$w$$ is closer to $$w$$ than $$v$$ because it is in the innermost sphere around $$w$$, while $$v$$ is only in the second sphere. (If $$v$$ were also in the innermost sphere then the two worlds would be equally close to $$w$$. That’s allowed.) In general, we can represent the assumption that a world $$v$$ is closer to a world $$w$$ than a world $$u$$ ($$v \prec _w u$$) by putting $$v$$ is in a closer sphere around $$w$$ than $$u$$. I have not drawn any spheres around $$v$$ because it doesn’t matter what these look like.
Exercise 8.13
Draw countermodels showing that (E1)–(E4) are invalid if the conditionals are translated as statements of the form $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$. (Hint: You never need more than two worlds.)
The logic of variably strict conditionals is weaker than the logic of strict conditionals. Some have argued that it is too weak to explain our reasoning with conditionals. It is, for example, not hard to see that the following statements are all false. (The corresponding statements for $$\,\unicode {x297D}\,$$ are true; see exercise 8.4.)
1.
$$p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow q, q \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r \models p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r$$
2.
$$((p \lor q) \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r) \models (p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r) \land (q \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r)$$
3.
$$p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow (q \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r) \models (p \land q) \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow r$$
If English conditionals are variably strict, this means (for example) that we can’t infer ‘if $$p$$ then $$r$$’ from ‘if $$p$$ then $$q$$’ and ‘if $$q$$ then $$r$$’. But isn’t this a valid inference?
Well, perhaps not. Stalnaker gave the following counterexample, using cold-war era subjunctive conditionals.
If J. Edgar Hoover had been born a Russian, he would be a communist.
If Hoover were a communist, he would be a traitor.
Therefore, if Hoover had been born a Russian, he would be a traitor.
Exercise 8.14
Can you find a case where ‘if $$p$$ or $$q$$ then $$r$$’ does not appear to entail ‘if $$p$$ then $$r$$’ and ‘if $$q$$ then $$r$$’? You can use either indicative or subjunctive conditionals. (Hint: Try to find a case in which ‘if $$p$$ or $$q$$ then $$p$$’ sounds acceptable.)
The semantics I have presented for $$\mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow$$ is a middle ground between that of Lewis and Stalnaker. Stalnaker assumes that $$\prec _w$$ is not just quasi-connected, but connected: for any $$w,v,u$$, either $$v \prec _w u$$ or $$v=u$$ or $$u \prec _w v$$. (‘$$v=u$$’ means that $$v$$ and $$u$$ are the same world.) This rules out ties in similarity: no sphere contains more than one world.
Stalnaker’s logic (called C2) is stronger than Lewis’s VC. The following principle of “Conditional Excluded Middle” is C2-valid but not VC-valid: \begin {equation} \tag {CEM}(A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B) \lor (A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow \neg B) \end {equation}
Whether conditionals in natural language satisfy Conditional Excluded Middle is a matter of ongoing debate. On the one hand, it is natural think that ‘it is not the case that if $$p$$ then $$q$$’ entails ‘if $$p$$ then not $$q$$’, which suggests that the principle is valid. On the other hand, suppose I have a number of coins in my pocket, none of which I have tossed. What would have happened if I had tossed one of the coins? Arguably, I might have gotten heads and I might have gotten tails. Either result is possible, but neither would have come about.
Exercise 8.15
Explain why the following statements are true, for all $$A,B,C$$:
(a)
$$A \land B \models _{C2} A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$
(b)
$$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow (B\lor C) \models _{C2} (A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B) \lor (A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow C)$$
Lewis not only rejects connectedness, but also the Limit Assumption. He argued that there might be an infinite chain of ever closer $$A$$-worlds. Definition 8.3 implies that if there are no closest $$A$$-worlds then any sentence of the form $$A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ is true. That does not seem right. Lewis therefore gives a more complicated semantics:
$$M,w \models A \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow B$$ iff either there is no $$v$$ for which $$M,v\models A$$ or there is some world $$v$$ such that $$M,v\models A$$ and for all $$u \prec _w v$$, $$M,w \models A \to B$$.
It turns out that it makes no difference to the logic whether we impose the Limit Assumption and use the old definition or don’t impose the Limit Assumption and use Lewis’s new definition. The same sentences are valid either way.
### 8.4Restrictors
Consider these two statements.
(1)
If it rains we always stay inside.
(2)
If it rains we sometimes stay inside.
On its most natural reading, (1) says that we stay inside at all times at which it rains. We can express this in $$\mathfrak {L}_{M}$$, using the box as a universal quantifier over the relevant times. (So $$\Box A$$ now means ‘always $$A$$’.) The translation would be $$\Box (r \to s)$$.
One might expect that (2) should then be translated as $$\Diamond (r \to s)$$, where the diamond is an existential quantifier over the relevant times (‘sometimes’). But $$\Diamond (r \to s)$$ is equivalent to $$\Diamond (\neg r \lor s)$$. This is true whenever $$\Diamond \neg r$$ is true. (2), however, isn’t true simply because it doesn’t always rain. On its most salient reading, (2) says there are times at which it rains and we stay inside. Its correct translation is $$\Diamond (r \land s)$$.
This is a little surprising, given that (2) seems to contain a conditional. Does the conditional here express a conjunction?
Things get worse if we look at (3).
(3)
If it rains we usually stay inside.
Let’s introduce an operator $$\mathsf {M}$$ for ‘usually’, so that $$\mathsf {M} A$$ is true at a time iff $$A$$ is true at most times. Can you translate (3) with the help of $$\mathsf {M}$$?
You can’t. Neither $$\mathsf {M}(r \to s)$$ nor $$\mathsf {M}(r \land s)$$ capture the intended meaning of (3). $$\mathsf {M}(r \land s)$$ entails that $$r$$ is usually true. But (3) doesn’t entail that it usually rains. $$\mathsf {M}(r \to s)$$ is true as long as $$r$$ is usually false, even if we’re always outside when it is raining. You could try to bring in some of the new kinds of conditional that we’ve encountered in the previous sections. How about $$\mathsf {M}(r \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow s)$$, or $$\mathsf {M}(r \,\unicode {x297D}\, s)$$, or $$r \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow \mathsf {M} s$$, or $$r \,\unicode {x297D}\, \mathsf {M} s$$? None of these are adequate.
The problem is that (3) doesn’t say, of any particular proposition, that it is true at most times. It doesn’t say that among all times, most are such-and-such. Rather, it says that among times at which it rains, most times are times at which we stay inside. The function of the ‘if’-clause in (3) is to restrict the domain of times over which the ‘usually’ operator quantifies.
Now return to (1) and (2). Suppose that here, too, the ‘if’-clause serves to restrict the domain of times, so that ‘always’ and ‘sometimes’ only quantify over times at which it rains. On that hypothesis, (1) says that among times at which it rains, all times are times at which we stay inside, and (2) says that among times at which it rains, some times are times at which we stay inside. This is indeed what (1) and (2) mean, on their most salient interpretation.
As it turns out, ‘among $$r$$-times, all times are $$s$$-times’ is equivalent to ‘all times are not-$$r$$-times or $$s$$-times’. That’s why we can formalize (1) as $$\Box (r \to s)$$. ‘Among $$r$$-times, some times are $$s$$-times’, on the other hand, is equivalent to ‘some times are $$r$$-times and $$s$$-times’. That’s why we can formalize (2) as $$\Diamond (r \land s)$$. It would be wrong to think that the conditional in (1) is material, the conditional in (2) is a conjunction, and the conditional in (3) is something else altogether. A much better explanation is that the ‘if’-clause in (1) does the exact same thing as in (2) and (3). In each case, it restricts the domain of times over which the relevant operators quantify.
We can arguably see the same effect in (4) and (5).
(4)
If the lights are on, Ada must be in her office.
(5)
If the lights are on, Ada might be in her office.
Letting the box express epistemic necessity, we can translate (4) as $$\Box (p \to q)$$. But (5) can’t be translated as $$\Diamond (p \to q)$$, which would be equivalent to $$\Diamond (\neg p \lor q)$$. Nor can we translate (5) as $$p \to \Diamond q$$, which is entailed by $$\Diamond q$$. It is easy to think of scenarios in which (5) is false even though ‘Ada might be in her office’ is true. The correct translation of (5) is plausibly $$\Diamond (p \land q)$$. The sentence is true iff there is an epistemically accessible world at which the lights are on and Ada is in her office.
As before, we can understand what is going if we assume that the ‘if’-clause in (4) and (5) functions as a restrictor. The ‘if’-clause restricts the domain of worlds over which ‘must’ and ‘might’ quantify. (4) says that among epistemically possible worlds at which the lights are on, all worlds are worlds at which Ada is in her office. (5) says that among epistemically possible worlds at which the lights are on, some worlds are worlds at which Ada is in her office.
Exercise 8.16
Translate ‘all dogs are barking’ and ‘some dogs are barking’ into the language of predicate logic. Can you translate ‘most dogs are barking’ if you add a ‘most’ quantifier $$\mathsf {M}$$ so that $$\mathsf {M} x Fx$$ is true iff most things satisfy $$Fx$$?
The hypothesis that ‘if’-clauses are restrictors also sheds light on the problem of conditional obligation.
(6)
Jones ought to help his neighbours.
(7)
If Jones doesn’t help his neighbours, he ought to not tell them that he’s coming.
In chapter 6, we analyzed ‘ought’ as a quantifier over the best of the circumstantially accessible worlds. On this approach, (6) says that among the accessible worlds, all the best ones are worlds at which Jones helps his neighbours. Suppose the ‘if’-clause in (7) serves to restrict the domain of worlds, excluding worlds at which Jones helps his neighbours. We then predict (7) to state that among the accessible worlds at which Jones doesn’t help his neighbours, all the best worlds are worlds at which Jones doesn’t tell his neighbours that he’s coming. This can’t be expressed by combining the monadic $$\mathsf {O}$$ quantifier with truth-functional connectives. Hence we had to introduce a primitive binary operator $$\mathsf {O}(\cdot /\cdot )$$.
The upshot of all this is that we can make sense of a wide range of puzzling phenomena by assuming that ‘if’-clauses are restrictors. Their function is to restrict the domain or worlds or times over which modal operators quantify.
What, then, is the purpose of ‘if’-clauses in “bare” conditionals like (8) and (9), where there are no modal operators to restrict?
(8)
If Shakespeare didn’t write Hamlet, then someone else did.
(9)
If Shakespeare hadn’t written Hamlet, then someone else would have.
Here opinions vary. One possibility, prominently defended by the linguist Angelika Kratzer, is that even bare conditionals contain modal operators. Arguably, ‘would’ in (9) functions as a kind of box. If this box is a simple quantifier over circumstantially accessible worlds, and the ‘if’-clause in (9) restricts its domain, then (9) can be formalized as $$\Box (p \to q)$$. If, on the other hand, ‘would’ in (9) works more light ‘ought’ – if it quantifyies over the closest of the accessible worlds –, and the ‘if’-clause restricts the domain of accessible worlds, then the resulting truth-conditions are those of $$p \mathrel {\mathop \Box }\mathrel {\mkern -2.5mu}\rightarrow q$$. Both the strict and the variably strict analysis of (9) are therefore compatible with the hypothesis that ‘if’-clauses are restrictors.
What about (8)? This sentence really doesn’t appear to contain a relevant modal. Kratzer suggests that it contains an unpronounced epistemic ‘must’: (8) says that if Shakespeare didn’t write Hamlet then someone else must have written Hamlet. Assuming that the ‘if’-clause restricts the domain of this operator, bare indicative conditionals would be equivalent to strict epistemic conditionals.
Exercise 8.17
Suppose bare indicative conditionals like (8) contain a box operator $$\Box$$ whose accessibility relation relates each world to itself and to no other world. (This is a redundant operator insofar as $$\Box A$$ is equivalent to $$A$$.) Assume the ‘if’-clause restricts the domain of that operator. What are the resulting truth-conditions of (8)?
Exercise 8.18
Besides “would counterfactuals” there are also “might counterfactuals” like
(10)
If I had played the lottery, I might have won.
Suppose ‘might’ is the dual of ‘would’, and suppose the ‘if’-clause in (10) restricts the domain of worlds over which ‘might’ quantifies. It follows that ‘if $$A$$ then might $$B$$’ is true iff $$B$$ holds at some of the closest/accessible $$A$$-worlds. (‘Closest’ or ‘accessible’ depending on how we understand the ‘would’/‘might’ operators.) Can you see why this casts doubt on the validity of Conditional Excluded Middle?
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https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-10-exponents-and-radicals-10-2-rational-numbers-as-exponents-10-2-exercise-set-page-641/33 | 1,539,612,209,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509196.33/warc/CC-MAIN-20181015121848-20181015143348-00092.warc.gz | 950,849,895 | 15,332 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$125x^{6}$
Using $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$, the given expression, $(25x^4)^{3/2} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[]{(25x^4)^3} \\\\= \left( \sqrt[]{25x^4} \right)^3 \\\\= \left( \sqrt[]{(5x^2)^2} \right)^3 \\\\= \left( 5x^2 \right)^3 \\\\= 5^3x^{2(3)} \\\\= 125x^{6} .\end{array} | 170 | 398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-43 | longest | en | 0.445216 |
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A quantum channel is a mapping between Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$. In general, we are interested in CPTP maps. The operator spaces can be interpreted as $C^{*}$-algebras and thus we can also view the channel as a mapping between $C^{*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$. Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$ where $C(X)$ is the space of continuous functions on some set $X$ and is also a $C^{*}$-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{*}$-algebras. Note, however, that these are not necessarily the same $C^{*}$-algebras. Since the channels are represented by square matrices, the input and output $C^{*}$-algebras must have the same dimension, $d$. Thus we can consider them both subsets of some $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.
Proof: Consider the quantum channels
$\begin{eqnarray*} r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) & \qquad \textrm{where} \qquad & \sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\ t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) & \qquad \textrm{where} \qquad & \tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} \end{eqnarray*}$
where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ where
\begin{align} \tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\ & = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\ & = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag \end{align}
and the $A_{i}$, $B_{i}$, and $C_{i}$ are Kraus operators.
Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=1$ (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). $\sum_{k} C_{k}^{\dagger}C_{k}=\mathbf{1}.$$For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041). We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state$\rho$in the absence of any channel. Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$. Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$. We have $\begin{align} v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\ & = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\ & = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag \end{align}$ and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either).$\square$Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct. I'm new here, so feel free to change the tags or point out anything that isn't appropriate. 2 Added a note about Kraus operators. A quantum channel is a mapping between Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$. In general, we are interested in CPTP maps. The operator spaces can be interpreted as $C^{*}$-algebras and thus we can also view the channel as a mapping between $C^{*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$. Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$ where $C(X)$ is the space of continuous functions on some set$X$and is also a $C^{*}$-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{*}$-algebras. Note, however, that these are not necessarily the same $C^{*}$-algebras. Since the channels are represented by square matrices, the input and output $C^{*}$-algebras must have the same dimension,$d$. Thus we can consider them both subsets of some$d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from$\mathcal{C}$to itself. Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the$d$-dimensional $C^{*}$-algebra,$\mathcal{C}$, on which it acts, forms a category we call$\mathrm{\mathbf{Chan}}(d)$where$\mathcal{C}$is the sole object and$t$is the sole arrow. Proof: Consider the quantum channels $\begin{eqnarray*} r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) & \qquad \textrm{where} \qquad & \sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\ t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) & \qquad \textrm{where} \qquad & \tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} \end{eqnarray*}$ where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ where $\begin{align} \tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\ & = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\ & = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag \end{align}$ and the $A_{i}$, $B_{i}$, and $C_{i}$ are Kraus operators. Since$A$and$B$are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=1$ (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041). We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state$\rho$in the absence of any channel. Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$. Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$. We have $\begin{align} v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\ & = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\ & = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag \end{align}$ and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either). Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct. I'm new here, so feel free to change the tags or point out anything that isn't appropriate. 1 Quantum channels as categories: question 1. A quantum channel is a mapping between Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$. The operator spaces can be interpreted as $C^{*}$-algebras and thus we can also view the channel as a mapping between $C^{*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$. Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$ where $C(X)$ is the space of continuous functions on some set$X$and is also a $C^{*}$-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{*}$-algebras. Note, however, that these are not necessarily the same $C^{*}$-algebras. Since the channels are represented by square matrices, the input and output $C^{*}$-algebras must have the same dimension,$d$. Thus we can consider them both subsets of some$d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from$\mathcal{C}$to itself. Proposition A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the$d$-dimensional $C^{*}$-algebra,$\mathcal{C}$, on which it acts, forms a category we call$\mathrm{\mathbf{Chan}}(d)$where$\mathcal{C}$is the sole object and$t$is the sole arrow. Proof: Consider the quantum channels $\begin{eqnarray*} r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) & \qquad \textrm{where} \qquad & \sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\ t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) & \qquad \textrm{where} \qquad & \tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger} \end{eqnarray*}$ where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$ where $\begin{align} \tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\ & = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\ & = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag \end{align}$. Since$A$and$B$are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=1$ (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041). We take the identity arrow, $1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$, to be the time evolution of the state$\rho$in the absence of any channel. Since this definition is suitably general we have that $t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$. Consider the three unital quantum channels $r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$, $t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$, and $v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$ where $\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$, $\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$, and $\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$. We have $\begin{align} v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\ & = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\ & = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag \end{align}\$
and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either).
Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.
I'm new here, so feel free to change the tags or point out anything that isn't appropriate. | 4,751 | 13,555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2013-20 | latest | en | 0.808962 |
http://clay6.com/qa/37416/www.clay6.com/qa/37416/in-the-parabola-y-2-4ax-the-length-of-the-chord-passing-through-the-vertex- | 1,596,827,158,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737206.16/warc/CC-MAIN-20200807172851-20200807202851-00263.warc.gz | 23,357,273 | 6,282 | # In the parabola $y^2=4ax$ , the length of the chord passing through the vertex and inclined to the x-axis at an angle $\theta$ is
$\begin{array}{1 1}(A)\;4a \cos \theta /\sin ^2 \theta \\(B)\;4a \sin \theta / \cos ^2 \theta \\(C)\;a \sec^2 \theta\\(D)\;a cosec ^2 \theta \end{array}$ | 110 | 285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-34 | latest | en | 0.617153 |
https://webpala.com/flo-is-a-farmer-and-grows-flowers-on-her-farm-which-is-located-right-next-to-beatrices-property-beatrice-is-a-beekeeper-each-of-beatrices/ | 1,680,030,568,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00401.warc.gz | 665,014,460 | 14,625 | # Flo is a farmer and grows flowers on her farm, which is located right next to Beatrice’s property. Beatrice is a beekeeper. Each of Beatrice’s…
Flo is a farmer and grows flowers on her farm, which is located right next to Beatrice’s property. Beatrice is a beekeeper. Each of Beatrice’s beehives pollinates an acre of Flo’s flowers. Flo’s cost function is Cf (f) = 5(f − 1 3 b) 2 , and Beatrice’s cost function is Cb(b) = 10(b− 1 2 f) 2 , where f is the number of acres of flowers and b is the number of beehives. Each acre of flowers yields \$50 worth of flowers, and each beehive yields \$100 worth of honey.
(a) Calculate f and b at the decentralized equilibrium.
(b) Calculate Flo’s and Beatrice’s profits at the decentralized equilibrium.
(c) Is the decentralized allocation Pareto inefficient? Why?
(d) What conditions must the Pareto efficient allocation of f and b satisfy?
e) Calculate the Pareto efficient allocation of f and b and the corresponding profits at this allocation.
(f) Now suppose that the current allocation is the decentralized equilibrium allocation. Suppose that Flo and Beatrice bargain over f and b as follows: Beatrice offers to pay Flo the amount T if Flo produces a certain quantity f. If Flo rejects, both agents stick to the decentralized equilibrium quantities. What will Beatrice offer be? What will be the resulting allocation? What will be the corresponding profits? | 327 | 1,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.88764 |
https://www.jiskha.com/display.cgi?id=1310260104 | 1,503,545,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886126027.91/warc/CC-MAIN-20170824024147-20170824044147-00481.warc.gz | 921,893,285 | 3,675 | # math
posted by .
At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
this is a cal problem.
• math -
after 4 hours
x = - 30 - 21*4 = - 114
y = 15*4 = 60
dx/dt = -21
dy/dt = 15
D at 4 hr = sqrt(114^2+60^2) = 129
D^2 = x^2 + y^2
2 D dD/dt = 2 x dx/dt + 2 y dy/dt
129 dD/dt = -114(-21) + 60(15)
dD/dt = 25.5 knots
• math -
i think ur answer is wrong
• math -
I got the same answer as Damon.
Damon is right.
• math -
i type it in the computer , its says its wrong. but thanks
## Respond to this Question
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The term “electrical resistance” refers to the reciprocal of the conductance. The specific electrical resistance is the reciprocal of the conductivity.
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Discipline: Mathematics
Type of Paper: Question-Answer
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Paper Format: APA
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Question
Assume a disk rotates at 5,000 RPM with an average rotation of 0.5 round. The average seek time is 5 ms. The data transfer rate is 200MB/sec. There is a controller overhead of 0.1 ms. What is the estimated average latency to read or write a 512B sector?
Expert Answer
```Disk latency to read and write = Seek Time + Avg. Rotation Time + Transfer Time + Controller Overhead
Average Rotational Time = (0.5*60)/(5000) = 6 ms
[On average half rotation is made]
Avg. Seek Time = 5 ms
Transfer Time = 512 / (200 × 106 B/s)
= 2.56 microsec
Controller Overhead = 0.1 ms
Disk latency = Seek Time + Rotation Time + Transfer Time + Controller Overhead
= 5 + 6 + 2.56 * 10-3 + 0.1 milliseconds
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posted by on .
Is there a common ratio for the geometric sequence and what are the missing terms?
16,40,100,?,?
IS there a common ratio for the geometric sequence and what are the missing terms?
81,54,?,?,?
• math - ,
Yes, and 2.5 is the ratio. Following terms are 150, and 225.
For the second question, try 2/3 for the ratio. 36, 24, ...
With only two terms, one really can't say if there is a "common" ratio. There are no other pairs to compare.
• math - ,
if the given sequence is a geometric sequence, then it follows the formula
A,n+1 = r*(A,n)
where
r = ratio between two consecutive terms
A,n = the nth term
A,n+1 = the term after A,n
substituting the first and second term,
40 = r*16
r = 40/16
r = 2.5
to check, let's see if we would obtain 100 if we multiply 40 by the ratio=2.5
40*2.5 = 100
to find the next term after 100, we just multiply by the ratio,, thus the next two terms are:
100*2.5 = 250
250*2.5 = 625
for the second question, if it's a geometric sequence, then it follows the formula above. thus, getting the ratio,
54 = r*81
r = 54/81
r = 2/3
to get the next three terms, we multiply the next terms by 2/3:
54*(2/3) = 36
36*(2/3) = 24
24*(2/3) = 16
hope this helps~ :) | 385 | 1,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-22 | latest | en | 0.913487 |
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Physics Forum Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.
## Help, please - Physics Forum
### Help, please - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.
#1
09-07-2003, 05:48 PM
nebulus Guest Posts: n/a
I was hoping somebody could help me with a problem I've been working on and
possibly point me in the right direction if I've made a mistake. The
From the top of a cliff, a person uses a slingshot to fire a pebble
straight downward, which is the negative direction. The initial speed of
the pebble is 9.0 m/s (a) What is the acceleration (magnitude and
direction) of the pebble during the downward motion? Is the pebble
decelerating? Explain. (b) After .50 s, how far beneath the cliff top is
the pebble.
For part (a) I have the acceleration of the pebble is 9.8 m/s^2 downward
and that it is not decelerating because the vectors do not point in
opposite directions.
For part (b) my math looks as follows:
x = (v[0])(t) + (1/2)(a)(t^2)
x = (9.0 m/s)(.50s) + (1/2)(9.8 m/s^2)((.50s)^2)
x = 4.5m + 1.2m = 5.7m
I have a fair bit of confidence in my answer for part (b), however it is
part (a) that's been troubling me since I'm not really sure what is meant
by "the negative direction." As far as I understand when gravity causes
ojbects to accelerate then it is +9.8m/s^s and when objects are
decelerating due to gravity then it is -9.8m/s^2. Since the object
accelerates all the way to the bottom of the cliff, I don't see why any
values in this problem should be negative. Can someone please explain this
to me?
#2
09-07-2003, 09:43 PM
"nebulus" <[Only registered users see links. ]> wrote in message
news:Xns93EF8219A534Ena@204.127.199.17...
<snip>
The acceleration is the same regardless of the speed or posiiton (which
depend on other factors anyway). The magnitude is 9.8 m/sec^2 and the
direction is [by definition!] "down."
Your convention is that "down" is the negative direction. By this
convention, the velocity is negative AND the acceleration is negative.
not decelerating (which only occurs when they signs of velocity and
acceleration are opposed).
Initial velocity = -9.0 m/sec
Acceleration = -9.8 m/sec^2
Interval = 0.5 sec
Change in velocity = -9.8 m/sec^2 * 0.5 sec = -4.9 m/sec
Final velocity = -9.0 m/sec + -4.9 m/sec = -13.9 m/sec
Average velocity throughout interval = (-9.0 + -13.9)/2 = -22.9/2 = -11.45
m/sec
Distance travelled during interval = -11.45 m/sec * 0.5 sec = -5.725 m
Right.
pretty close...
Whether it is +9.8 or -9.8 m/sec^2 depends on which direction you define as
positive.
"Deceleration" implies a comparison between the directions of the velocity
and acceleration vectors.
this
You said, and I quote: "straight downward, which is the negative direction".
Clear enough???
Tom Davidson
Richmond, VA
#3
09-07-2003, 10:18 PM
tony Guest Posts: n/a
>
This is one of those problems where some prof can be playing word games. If the
negative direction is "down" and the acceleration is "down" I'd wonder if he's
looking for deceleration. Yeah, it's nuts, but if he takes the opposite of
acceleration as deceleration, wll then. . .
Which makes me think the teacher somewhere might have minored in law. The
answer as to displacement is correct, but the word games makes me not sure
l
#4
09-07-2003, 11:44 PM
Double-A Guest Posts: n/a
nebulus <[Only registered users see links. ]> wrote in message news:<Xns93EF8219A534Ena@204.127.199.17>...
justify it. If your teacher marks you down for it, challenge him on
it. I didn't get A's in school by keeping my mouth shut!
Double-A
#5
09-08-2003, 04:33 AM
nebulus Guest Posts: n/a
[Only registered users see links. ]emoove (tony) wrote in
news:[Only registered users see links. ]:
Actually, the problem comes straight from my textbook. I left off the
first sentence which referred to a section of the book that discussed
deceleration but it wasn't quite as clear as the explanation I got from
this newsgroup. The book we are using is Cutnell & Johnson Physics, 5th
Ed. (2001). I purchased a study guide to go with it that has the solutions
and methods of solving maybe 5% of the problems in the book, which means it
really doesn't do me much good for 95% of the material we cover. This
newsgroup has been (and I hope it will continue to be in the future) an
invaluable resource in helping me understand these concepts. I'll search
google... perhaps there is a Cutnell & Johnson Law book floating around
somewhere...
#6
09-08-2003, 04:34 AM
nebulus Guest Posts: n/a
[Only registered users see links. ] (Double-A) wrote in
Always good advice! I'm seriously going for that 4.0 (so far, so good).
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# Linear Programming and the Simplex Method
Answer following problems. The problems solved last week helped me break down all 60 questions I had to answer last week and I would like to do the same thing this week. Here are about 9 problems some multiple parts that are different areas I need help with.
1.
a. Add slack variables or subtract surplus variables.
b. Set up the initial simplex tableau.
2.
a. Add slack variables or subtract surplus variables.
b. Set up the initial simplex tableau.
3.
Use the simplex method to solve the following maximization problem with the given tableau.
4.
Convert into a maximization problem and then solve each problem using both the dual method and the method of section 4.4. You may use an applet.
Dual method:
Section 4.4 method
5.
The following is a final tableau of minimization a problem. State the solution and the minimum value of the objective function.
6. Use the simplex method to solve. (You may need to use artificial variables)
7.
Food Cost A store sells two brands of snacks. A package of Sun Hill costs \$3 and contains 10 oz of peanuts, 4 oz of raisins, and 3 oz of rolled oats. A package of Bear Valley costs \$2 and contains 2 oz of peanuts, 4 oz of raisins, and 8 oz of rolled oats. Suppose you wish to make a mixture that contains at least 20 oz of peanuts, 24 oz of raisins, and 24 oz of rolled oats.
You may use an applet from the internet.
a. Using the method of surplus variables, find how many packages of each you should buy to minimize the cost. What is the minimum cost?
b. Using the method of duals, find how many packages of each you should buy to minimize the cost. What is the minimum cost?
c. Suppose the minimum amount of peanuts is increased to 28. Use shadow costs to calculate the total cost in this case.
d. Explain why it makes sense that the shadow cost for the rolled oats is 0.
8. Give formulas and rules for:
a. Maximization problem
b. Non Standard problem
9. The Aged Wood Winery makes two white wines, Fruity and Crystal, from two kinds of grapes and sugar. One gal of Fruity wine requires 2 bushels of Grape A, 2 bushels of Grape B, 2 lb of sugar, and produces a profit of \$12. One gal of Crystal wine requires 1 bushel of Grape A, 3 bushels of Grape B, 1 lb of sugar, and produces a profit of \$15. The winery has available 110 bushels of grape A, 125 bushels of grape B, and 90 lb of sugar. How much of each wine should be made to maximize profit?
#### Solution Summary
LP problems are solved using the simplex method. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.
\$2.19 | 644 | 2,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-22 | latest | en | 0.886344 |
http://www.avatto.com/engineering/mathematics/mcqs/sets/algebra/35/4.html | 1,430,686,477,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430450458964.67/warc/CC-MAIN-20150501032058-00050-ip-10-235-10-82.ec2.internal.warc.gz | 259,851,943 | 15,955 | +91-9920808017
Home >
# Set Theory and Algebra MCQ >
## Set Theory and Algebra MCQ
Partial Ordering , Lattice and Boolean Algebra
Sets
16:
Match the following
A. Groups I. Associativity
B. Semi groups II. Identity
C. Monoids III. Commutative
D. Abelian Groups IV Left inverse
Codes. A B C D
A. IV I II III B. III I IV II C. II III I IV D. I II III IV
17:
Let (Z, *) be an algebraic structure, where Z is the set of integers and the operation * is defined by n * m = maximum (n, m). Which of the following statements is TRUE for (Z, *) ?
A. (Z, *) is a monoid B. (Z, *) is an abelian group C. (Z, *) is a group D. None of these
18:
Some group (G, 0) is known to be abelian. Then which one of the following is TRUE for G ?
A. g = g-1 for every g ∈ G B. g = g2 for every g ∈ G C. (g o h) 2 = g2o h2 for every g,h ∈ G D. G is of finite order
19:
If the binary operation * is deined on a set of ordered pairs of real numbers as
(a, b) * (c, d) = (ad + bc, bd)
and is associative, then
(1, 2) * (3, 5) * (3, 4) equals
A. (74,40) B. (32,40) C. (23,11) D. (7,11)
20:
If A = (1, 2, 3, 4). Let ~= {(1, 2), (1, 3), (4, 2)}. Then ~ is
A. not anti-symmetric B. transitive C. reflexive D. symmetric | 456 | 1,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2015-18 | longest | en | 0.781178 |
https://fenicsproject.discourse.group/t/i-have-a-square-plate-and-i-want-to-impose-a-gaussian-random-traction/14431 | 1,721,253,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00180.warc.gz | 218,364,030 | 5,143 | # I have a square plate and I want to impose a gaussian random traction
Here sample1 and sample2 are constant number applied throughout the top and the left edges (uniformly distributed load) of the square plate. But now I want to implement a traction which is basically gaussian random noise throughout the edge.
traction_expr = Expression((‘sample1near(x[0],1)', 'sample2near(x[1], 1)’),sigma = 0.5, sample1 = sample1, sample2 = sample2 , degree = 1)
F = inner(stress,grad(v))dx + dot(body_force,v)dx + dot(traction_expr,v)ds + rhoinner(u, v)/(dt**2) * dx - 2rhoinner(u_n_minus_one,v)/(dt2) * dx + rho*inner(u_n_minus_two,v)/(dt2)*dx
problem = NonlinearVariationalProblem(F, u, [bc_left, bc_bottom], J = derivative(F, u ,du))
# Create the Newton solver
solver = NonlinearVariationalSolver(problem)
# Set solver options (optional)
solver.parameters[“newton_solver”][“relaxation_parameter”] = 1.0
solver.parameters[“newton_solver”][“relative_tolerance”] = 1e-10
solver.parameters[“newton_solver”][“maximum_iterations”] = 100
solver.solve() | 299 | 1,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-30 | latest | en | 0.691238 |
https://trustconverter.com/en/volume-conversion/canadian-cups/canadian-cups-to-cubic-centimeters.html | 1,653,624,869,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00669.warc.gz | 647,941,258 | 9,883 | # Canadian cups to Cubic centimeters Conversion
Canadian cup to cubic centimeter conversion allow you make a conversion between canadian cup and cubic centimeter easily. You can find the tool in the following.
to
input
= 227.30450000
= 2.27305 × 102
= 2.27305E2
= 2.27305e2
= 454.60900000
= 4.54609 × 102
= 4.54609E2
= 4.54609e2
= 681.91350000
= 6.81914 × 102
= 6.81914E2
= 6.81914e2
= 909.21800000
= 9.09218 × 102
= 9.09218E2
= 9.09218e2
= 1,136.52250000
= 11.3652 × 102
= 11.3652E2
= 11.3652e2
### Quick Look: canadian cups to cubic centimeters
canadian cup 1 c (CA) 2 c (CA) 3 c (CA) 4 c (CA) 5 c (CA) 6 c (CA) 7 c (CA) 8 c (CA) 9 c (CA) 10 c (CA) 11 c (CA) 12 c (CA) 13 c (CA) 14 c (CA) 15 c (CA) 16 c (CA) 17 c (CA) 18 c (CA) 19 c (CA) 20 c (CA) 21 c (CA) 22 c (CA) 23 c (CA) 24 c (CA) 25 c (CA) 26 c (CA) 27 c (CA) 28 c (CA) 29 c (CA) 30 c (CA) 31 c (CA) 32 c (CA) 33 c (CA) 34 c (CA) 35 c (CA) 36 c (CA) 37 c (CA) 38 c (CA) 39 c (CA) 40 c (CA) 41 c (CA) 42 c (CA) 43 c (CA) 44 c (CA) 45 c (CA) 46 c (CA) 47 c (CA) 48 c (CA) 49 c (CA) 50 c (CA) 51 c (CA) 52 c (CA) 53 c (CA) 54 c (CA) 55 c (CA) 56 c (CA) 57 c (CA) 58 c (CA) 59 c (CA) 60 c (CA) 61 c (CA) 62 c (CA) 63 c (CA) 64 c (CA) 65 c (CA) 66 c (CA) 67 c (CA) 68 c (CA) 69 c (CA) 70 c (CA) 71 c (CA) 72 c (CA) 73 c (CA) 74 c (CA) 75 c (CA) 76 c (CA) 77 c (CA) 78 c (CA) 79 c (CA) 80 c (CA) 81 c (CA) 82 c (CA) 83 c (CA) 84 c (CA) 85 c (CA) 86 c (CA) 87 c (CA) 88 c (CA) 89 c (CA) 90 c (CA) 91 c (CA) 92 c (CA) 93 c (CA) 94 c (CA) 95 c (CA) 96 c (CA) 97 c (CA) 98 c (CA) 99 c (CA) 100 c (CA) cubic centimeter 227.3045 cm3 454.609 cm3 681.9135 cm3 909.218 cm3 1,136.5225 cm3 1,363.827 cm3 1,591.1315 cm3 1,818.436 cm3 2,045.7405 cm3 2,273.045 cm3 2,500.3495 cm3 2,727.654 cm3 2,954.9585 cm3 3,182.263 cm3 3,409.5675 cm3 3,636.872 cm3 3,864.1765 cm3 4,091.481 cm3 4,318.7855 cm3 4,546.09 cm3 4,773.3945 cm3 5,000.699 cm3 5,228.0035 cm3 5,455.308 cm3 5,682.6125 cm3 5,909.917 cm3 6,137.2215 cm3 6,364.526 cm3 6,591.8305 cm3 6,819.135 cm3 7,046.4395 cm3 7,273.744 cm3 7,501.0485 cm3 7,728.353 cm3 7,955.6575 cm3 8,182.962 cm3 8,410.2665 cm3 8,637.571 cm3 8,864.8755 cm3 9,092.18 cm3 9,319.4845 cm3 9,546.789 cm3 9,774.0935 cm3 10,001.398 cm3 10,228.7025 cm3 10,456.007 cm3 10,683.3115 cm3 10,910.616 cm3 11,137.9205 cm3 11,365.225 cm3 11,592.5295 cm3 11,819.834 cm3 12,047.1385 cm3 12,274.443 cm3 12,501.7475 cm3 12,729.052 cm3 12,956.3565 cm3 13,183.661 cm3 13,410.9655 cm3 13,638.27 cm3 13,865.5745 cm3 14,092.879 cm3 14,320.1835 cm3 14,547.488 cm3 14,774.7925 cm3 15,002.097 cm3 15,229.4015 cm3 15,456.706 cm3 15,684.0105 cm3 15,911.315 cm3 16,138.6195 cm3 16,365.924 cm3 16,593.2285 cm3 16,820.533 cm3 17,047.8375 cm3 17,275.142 cm3 17,502.4465 cm3 17,729.751 cm3 17,957.0555 cm3 18,184.36 cm3 18,411.6645 cm3 18,638.969 cm3 18,866.2735 cm3 19,093.578 cm3 19,320.8825 cm3 19,548.187 cm3 19,775.4915 cm3 20,002.796 cm3 20,230.1005 cm3 20,457.405 cm3 20,684.7095 cm3 20,912.014 cm3 21,139.3185 cm3 21,366.623 cm3 21,593.9275 cm3 21,821.232 cm3 22,048.5365 cm3 22,275.841 cm3 22,503.1455 cm3 22,730.45 cm3
Canada now usually employs the metric cup of 250 mL but its conventional cup was somewhat smaller than both American and imperial units.
1 Canadian cup = 8 imperial fluid ounce = 1/20 imperial gallon = 227.3045 millilitres
1 tablespoon = 1/2 imperial fluid ounce
1 teaspoon = 1/6 imperial fluid ounce
Name of unitSymbolDefinitionRelation to SI unitsUnit System
≡ 10 fl oz (imp)
= 284.130625×10−6 m3
Imperial/US
#### conversion table
1= 227.30456= 1363.827
2= 454.6097= 1591.1315
3= 681.91358= 1818.436
4= 909.2189= 2045.7405
5= 1136.522510= 2273.045
The cubic centimetre or cubic centimeter is the unit of volume. Its symbol is cm3. It is the volume of a cube with edges one centimetre in length.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
cubic centimetercm3
≡ 1 cm × 1 cm × 1 cm
≡ 1 cm3
Metric system SI
### conversion table
1= 0.00439938496598186= 0.026396309795891
2= 0.00879876993196357= 0.030795694761872
3= 0.0131981548979458= 0.035195079727854
4= 0.0175975398639279= 0.039594464693836
5= 0.02199692482990910= 0.043993849659818 | 2,071 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.443851 |
https://www.doubtnut.com/qa-hindi/ydi-z-2-3i-tb-z-barz-kaa-maan-hai--234321106 | 1,632,565,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00026.warc.gz | 748,458,710 | 78,552 | Home
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यदि z = 2 - 3i तब z. bar(z) का...
# यदि z = 2 - 3i तब z. bar(z) का मान …………. है ।
उत्तर
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 31-7-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
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5:53 | 426 | 946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-39 | longest | en | 0.17367 |
https://www.geeksforgeeks.org/multivariate-optimization-gradient-and-hessian/?ref=rp | 1,628,091,430,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00435.warc.gz | 826,062,758 | 21,860 | Related Articles
# Multivariate Optimization – Gradient and Hessian
• Last Updated : 17 Jul, 2020
In a multivariate optimization problem, there are multiple variables that act as decision variables in the optimization problem.
z = f(x1, x2, x3…..xn)
So, when you look at these types of problems a general function z could be some non-linear function of decision variables x1, x2, x3 to xn. So, there are n variables that one could manipulate or choose to optimize this function z. Notice that one could explain univariate optimization using pictures in two dimensions that is because in the x-direction we had the decision variable value and in the y-direction, we had the value of the function. However, if it is multivariate optimization then we have to use pictures in three dimensions and if the decision variables are more than 2 then it is difficult to visualize.
Before explaining gradient let us just contrast with the necesaary condition of univariate case. So in case of uni-variate optimization the necessary condition for x to be the minimizer of the function f(x) is:
First-order necessary condition: f'(x) = 0
So, the derivative in a single-dimensional case becomes what we call as a gradient in the multivariate case.
According to the first-order necessary condition in univariate optimization e.g f'(x) = 0 or one can also write it as df/dx. However, since there are many variables in the case of multivariate and we have many partial derivatives and the gradient of the function f is a vector such that in each component one can compute the derivative of the function with respect to the corresponding variable. So, for example, is the first component, is the second component and is the last component.
Note: Gradient of a function at a point is orthogonal to the contours.
Hessian:
Similarly in case of uni-variate optimization the sufficient condition for x to be the minimizer of the function f(x) is:
Second-order sufficiency condition: f”(x) > 0 or d2f/dx2 > 0
And this is replaced by what we call a Hessian matrix in the multivariate case. So, this is a matrix of dimension n*n, and the first component is , the second component is and so on.
Note:
• Hessian is a symmetric matrix.
• Hessian matrix is said to be positive definite at a point if all the eigenvalues of the Hessian matrix are positive.
Attention reader! Don’t stop learning now. Get hold of all the important Machine Learning Concepts with the Machine Learning Foundation Course at a student-friendly price and become industry ready.
My Personal Notes arrow_drop_up | 555 | 2,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-31 | longest | en | 0.919963 |
http://www.astronomy.net/forums/blackholes2/messages/3334.shtml?base=0 | 1,597,354,524,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00153.warc.gz | 102,149,698 | 5,542 | Blackholes2 Forum Message Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem
Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Re: Absolute Nothing Does Not Exist Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Zephram Cochrane/">Zephram Cochrane on February 8, 2000 21:39:29 UTC
: : We do not pretend, but do understand the propagation of electromagnetic waves. Its not a very complicated phenomenon. If you don't that's your problem.
: 0KAY SO YOU KNOW PLEASE ANSWER THIS : WHATS A ELECTROMAGNETIC WAVE 0KAY ? : IF THERE IS A NO MEDIUM (ABSOLUTE NOTHINGNESS IS TERMED USED )TO PROPOGATE THEN HOW DOES IT TRAVEL.
Simple, 1.) ÑXE = -B/t (Translation, a changing magnetic field creates an electric field in vacuum)
2.) ÑXB = m0e0E/t (Translation, a changing electric field creates a magnetic field in vacuum)
3.) Ñ×E = 0 (Translation, we have no electric charge sources of the electric field being in vacuum)
4.) Ñ×B = 0 (Translation, no magnetic monopoles)
Take the curl of (1.) 5.) ÑX(ÑXE) = ÑX(-B/t)
A little manipulation 6.) ÑX(ÑXE) = (-/t)(ÑXB)
Inserting (2) into (6) 7.) ÑX(ÑXE) = (-/t)(m0e0E/t)
A little manipulation 8.) ÑX(ÑXE) =-m0e02E/t2
A little vector calculus 9.) Ñ(Ñ×E) - Ñ2E = -m0e02E/t2
Refer to (3), then (9) becomes Ñ2E = m0e02E/t2
This is a wave equation for an electric wave of speed c = 1/(m0e0)1/2
In other words its because a changing electric field creates a magnetic field and because a changing magnetic field creates an electric field in this manner that the electric field propogates as a wave without a medium. One can use the same kind of method with the same equations 1, 2, 3, 4 to arive at Ñ2B = m0e02B/t2
I have gone far beyond the basics like these things.
• http://hometown.aol.com//zephcochrane/WarpDrives.html | 608 | 2,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-34 | latest | en | 0.735544 |
https://trustconverter.com/en/weight-conversion/megagrams/megagrams-to-yoctograms.html | 1,723,070,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00041.warc.gz | 466,780,939 | 8,850 | # Megagrams to Yoctograms Conversion
Megagram to yoctogram conversion allow you make a conversion between megagram and yoctogram easily. You can find the tool in the following.
### Weight Conversion
to
input
= 1,000,000,000,000,000,019,884,624,838,656.00000000
= 1.0 × 1030
= 1.0E+30
= 1.0e+30
= 2,499,999,999,999,999,908,974,073,741,312.00000000
= 2.5 × 1030
= 2.5E+30
= 2.5e+30
= 4,000,000,000,000,000,079,538,499,354,624.00000000
= 4.0 × 1030
= 4.0E+30
= 4.0e+30
= 5,500,000,000,000,000,250,102,924,967,936.00000000
= 5.5 × 1030
= 5.5E+30
= 5.5e+30
= 6,999,999,999,999,999,294,767,443,738,624.00000000
= 7.0 × 1030
= 7.0E+30
= 7.0e+30
= 8,499,999,999,999,999,465,331,869,351,936.00000000
= 8.5 × 1030
= 8.5E+30
= 8.5e+30
= 9,999,999,999,999,999,635,896,294,965,248.00000000
= 1.0 × 1031
= 1.0E+31
= 1.0e+31
### Quick Look: megagrams to yoctograms
megagram 1 Mg 2 Mg 3 Mg 4 Mg 5 Mg 6 Mg 7 Mg 8 Mg 9 Mg 10 Mg 11 Mg 12 Mg 13 Mg 14 Mg 15 Mg 16 Mg 17 Mg 18 Mg 19 Mg 20 Mg 21 Mg 22 Mg 23 Mg 24 Mg 25 Mg 26 Mg 27 Mg 28 Mg 29 Mg 30 Mg 31 Mg 32 Mg 33 Mg 34 Mg 35 Mg 36 Mg 37 Mg 38 Mg 39 Mg 40 Mg 41 Mg 42 Mg 43 Mg 44 Mg 45 Mg 46 Mg 47 Mg 48 Mg 49 Mg 50 Mg 51 Mg 52 Mg 53 Mg 54 Mg 55 Mg 56 Mg 57 Mg 58 Mg 59 Mg 60 Mg 61 Mg 62 Mg 63 Mg 64 Mg 65 Mg 66 Mg 67 Mg 68 Mg 69 Mg 70 Mg 71 Mg 72 Mg 73 Mg 74 Mg 75 Mg 76 Mg 77 Mg 78 Mg 79 Mg 80 Mg 81 Mg 82 Mg 83 Mg 84 Mg 85 Mg 86 Mg 87 Mg 88 Mg 89 Mg 90 Mg 91 Mg 92 Mg 93 Mg 94 Mg 95 Mg 96 Mg 97 Mg 98 Mg 99 Mg 100 Mg yoctogram 1.0 × 1030 yg 2.0 × 1030 yg 3.0 × 1030 yg 4.0 × 1030 yg 5.0 × 1030 yg 6.0 × 1030 yg 7.0 × 1030 yg 8.0 × 1030 yg 9.0 × 1030 yg 1.0 × 1031 yg 1.1 × 1031 yg 1.2 × 1031 yg 1.3 × 1031 yg 1.4 × 1031 yg 1.5 × 1031 yg 1.6 × 1031 yg 1.7 × 1031 yg 1.8 × 1031 yg 1.9 × 1031 yg 2.0 × 1031 yg 2.1 × 1031 yg 2.2 × 1031 yg 2.3 × 1031 yg 2.4 × 1031 yg 2.5 × 1031 yg 2.6 × 1031 yg 2.7 × 1031 yg 2.8 × 1031 yg 2.9 × 1031 yg 3.0 × 1031 yg 3.1 × 1031 yg 3.2 × 1031 yg 3.3 × 1031 yg 3.4 × 1031 yg 3.5 × 1031 yg 3.6 × 1031 yg 3.7 × 1031 yg 3.8 × 1031 yg 3.9 × 1031 yg 4.0 × 1031 yg 4.1 × 1031 yg 4.2 × 1031 yg 4.3 × 1031 yg 4.4 × 1031 yg 4.5 × 1031 yg 4.6 × 1031 yg 4.7 × 1031 yg 4.8 × 1031 yg 4.9 × 1031 yg 5.0 × 1031 yg 5.1 × 1031 yg 5.2 × 1031 yg 5.3 × 1031 yg 5.4 × 1031 yg 5.5 × 1031 yg 5.6 × 1031 yg 5.7 × 1031 yg 5.8 × 1031 yg 5.9 × 1031 yg 6.0 × 1031 yg 6.1 × 1031 yg 6.2 × 1031 yg 6.3 × 1031 yg 6.4 × 1031 yg 6.5 × 1031 yg 6.6 × 1031 yg 6.7 × 1031 yg 6.8 × 1031 yg 6.9 × 1031 yg 7.0 × 1031 yg 7.1 × 1031 yg 7.2 × 1031 yg 7.3 × 1031 yg 7.4 × 1031 yg 7.5 × 1031 yg 7.6 × 1031 yg 7.7 × 1031 yg 7.8 × 1031 yg 7.9 × 1031 yg 8.0 × 1031 yg 8.1 × 1031 yg 8.2 × 1031 yg 8.3 × 1031 yg 8.4 × 1031 yg 8.5 × 1031 yg 8.6 × 1031 yg 8.7 × 1031 yg 8.8 × 1031 yg 8.9 × 1031 yg 9.0 × 1031 yg 9.1 × 1031 yg 9.2 × 1031 yg 9.3 × 1031 yg 9.4 × 1031 yg 9.5 × 1031 yg 9.6 × 1031 yg 9.7 × 1031 yg 9.8 × 1031 yg 9.9 × 1031 yg 1.0 × 1032 yg
Megagram or megagramme is equal to 1 000 000 gram (unit of mass), comes from a combination of the metric prefix mega (M) with the gram (g). Plural name is megagrams.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
megagramMg
≡ 1 000 000 g
≡ 1 000 kg
Metric system SI
#### conversion table
megagramsyoctogramsmegagramsyoctograms
1≡ 1.0E+3011≡ 1.1E+31
2.5≡ 2.5E+3012.5≡ 1.25E+31
4≡ 4.0E+3014≡ 1.4E+31
5.5≡ 5.5E+3015.5≡ 1.55E+31
7≡ 7.0E+3017≡ 1.7E+31
8.5≡ 8.5E+3018.5≡ 1.85E+31
10≡ 1.0E+3120≡ 2.0E+31
yoctogram or yoctogramme is equal to 10-24 gram (unit of mass), comes from a combination of the metric prefix yocto (y) with the gram (g). Plural name is yoctograms.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
yoctogramyg
≡ 10-24 g
≡ 10-27 kg
Metric system SI
### conversion table
yoctogramsmegagramsyoctogramsmegagrams
1≡ 1.0E-3011≡ 1.1E-29
2.5≡ 2.5E-3012.5≡ 1.25E-29
4≡ 4.0E-3014≡ 1.4E-29
5.5≡ 5.5E-3015.5≡ 1.55E-29
7≡ 7.0E-3017≡ 1.7E-29
8.5≡ 8.5E-3018.5≡ 1.85E-29
10≡ 1.0E-2920≡ 2.0E-29
### Conversion table
megagramsyoctograms
1≡ 1.0 × 1030
1.0 × 10-30≡ 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …) | 2,244 | 4,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.575156 |
http://www.jiskha.com/display.cgi?id=1168549970 | 1,454,785,074,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701147492.21/warc/CC-MAIN-20160205193907-00237-ip-10-236-182-209.ec2.internal.warc.gz | 479,361,202 | 3,915 | Saturday
February 6, 2016
# Homework Help: Chemistry
Posted by Bryan on Thursday, January 11, 2007 at 4:12pm.
How you do this? can you do the first one, i'll try the second one if you can explain
Perform the following operations and give the answear in standard exponential form with the correct number of significant figures
1) 37.2mL + 18.0mL + 380mL=
2) 0.57cm x 0.86cm x 17.1cm=
add the first. line up the decimals. Now notice that in the last, the zero in the units position is not a significant digit. | 151 | 513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-07 | longest | en | 0.906702 |
http://www.wiki-teacher.com/resourceView.php?id=2488 | 1,519,522,921,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00305.warc.gz | 539,093,632 | 8,675 | # Wiki-Teacher
## The Doorbell Rang Student Book
This student book is based on "The Doorbell Rang" by Pat Hutchins. Students can create their own addition book following the pattern of the original story.
User:
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#### Content Area
• NVACS Mathematics
#### Resource Type(s)
• Miscellaneous
• Kindergarten
#### Course(s)
• No Courses Identified
#### Standard(s)/Course Objective(s)
• K.OA.A.1 - Represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations. (Drawings need not show details, but should show the mathematics in the problem. [This applies wherever drawings are mentioned in the Standards.])
• 1.OA.C.6 - Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
#### Textbook Unit(s)
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http://www.dictionary.com/browse/ordinal--numbers | 1,532,022,427,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00192.warc.gz | 465,273,446 | 19,618 | definitions
• synonyms
# ordinal number
See more synonyms on Thesaurus.com
noun
1. Also called ordinal numeral. any of the numbers that express degree, quality, or position in a series, as first, second, and third (distinguished from cardinal number).
2. Mathematics. a symbol denoting both the cardinal number and the ordering of a given set, being identical for two ordered sets having elements that can be placed into one-to-one correspondence, the correspondence preserving the order of the elements.
## Origin of .css-1fxfie5{font-size:22px;}@media (max-width:768px){.css-1fxfie5{font-size:18px;margin:0 10px 10px 0;word-break:break-all;word-wrap:break-word;-webkit-hyphens:auto;-moz-hyphens:auto;-ms-hyphens:auto;hyphens:auto;line-height:22px;}}ordinal number
First recorded in 1600–10
Dictionary.com Unabridged Based on the Random House Unabridged Dictionary, © Random House, Inc. 2018
British Dictionary definitions for ordinal numbers
# ordinal number
noun
1. a number denoting relative position in a sequence, such as first, second, thirdSometimes shortened to: ordinal
2. logic maths a measure of not only the size of a set but also the order of its elements
Collins English Dictionary - Complete & Unabridged 2012 Digital Edition © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
ordinal numbers in Science
# ordinal number
[ôrdn-əl]
1. A number, such as 3rd, 11th, or 412th, used in counting to indicate position in a series but not quantity. Compare cardinal number. | 421 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.748335 |
http://toughstem.com/problem-answer-solution/1669/part-two-electrons-proton-shown-figure-find-magnitude-net-electrical-force-exert | 1,585,539,526,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496523.5/warc/CC-MAIN-20200330023050-20200330053050-00106.warc.gz | 143,112,680 | 19,245 | ToughSTEM
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Part A
If two electrons are each 2.30x10^-10m from a proton, as shown in the figure, find the magnitude and of the net electrical force they will exert on the proton.
Part B
Find the direction of the net electrical force electrons will exert on the proton.
Enter your answer as an angle measured relative to the line connecting the proton and the right electron with counterclockwise being positive.
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The force between one electron and proton
= K * Q^2/R
Q = charge of electron or proton (both have the same charge )
R = distance between them
K = 9 * 10^9
so
= 9 * 10^9 * 1.6 * 10^-19 * 1.6 * 10^-19 /(2.3 * 10^-10)^2
= 4.36 * 10^-9 N
same force with other electron and proton
so
Net force = 1.685 * 4.36 * 10^-9 N
= 7.347 * 10^-9 N
And the direction is 65/2 = 32.5 degrees counterclockwise
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# 1.5 Volt Light Bulb Is How Many Watts? Update New
Let’s discuss the question: 1.5 volt light bulb is how many watts. We summarize all relevant answers in section Q&A of website Activegaliano.org in category: Blog Marketing. See more related questions in the comments below.
## What is 1.5 W LED equivalent to?
The 1.5W light output is equivalent to the light of a traditional 15 Watt halogen light bulb and is extremely energy saving with low power consumption.
## Can a 1.5 V battery power a light bulb?
Light bulb with threaded, screw-style, base burns brightly with 1.5-volt batteries.
### joule thief circuit | how to make 1.5 volt led circuit | glow leds on 1.5 volt battery
joule thief circuit | how to make 1.5 volt led circuit | glow leds on 1.5 volt battery
joule thief circuit | how to make 1.5 volt led circuit | glow leds on 1.5 volt battery
## How many watts are in a 1.5 volt battery?
The power of a 1.5 V battery varies with the number of hours its in service. According to the chart above, the power discharge for a 1.5 V “D”battery at approximately 210 hours is 0.1 Watts (W).
## How bright is a 15W LED light?
LED light bulbs for general use around the home will typically have a wattage between 5W-15W, and will emit between 300-500 lumens.
## How bright is a 1 watt LED bulb?
How bright is a 1 watt LED bulb? 1 watt LED bulb is 90 Lumens. 1 watt LED bulb is the same as 9 watts in an incandescent light bulb.
## Will 1.5 V light an LED?
Their voltage drop (~2 V) is higher than the supply voltage (1.5 V). If the supply voltage is lower than the voltage drop, any diode (including LED) doesn’t conduct at all.
## How many watts does a light bulb have?
The most common types of light bulb used in homes are 25, 40, 60, 75 and 100 watts. For most rooms, a 60 watt bulb is the standard. A 25 watt bulb gives off low levels of light, where as a 100 watt bulb is very bright.
## How many watts does a light bulb use?
Light Bulbs
A LED light bulb uses just seven to ten watts while a fluorescent light bulb consumes 16-20 watts, an incandescent light bulb will use 60 watts typically and cost about 0.6 cents an hour to run, according to the energy use chart. Here’s how to pick LED lights if you’ve been on the fence about them.
## How many volts does a light bulb need?
The most common voltage for electric light bulbs is 120 volts (120V). This is the default voltage for most lighting fixtures. However, some lighting fixtures are low voltage. Low voltage lighting is more energy-efficient, but it requires low voltage bulbs to work.
## Can I use 120V instead of 130V?
For example 120V and 130V bulbs are interchangeable and the pros in using a 130 Volt bulb are: cooler burning bulb, less energy use, longer life and better handling of voltage surges. The cons are lower lumens output (not as bright) and lower color temperature.
## What is the voltage of LED bulb?
Typically, the forward voltage of an LED is between 1.8 and 3.3 volts. It varies by the color of the LED. A red LED typically drops around 1.7 to 2.0 volts, but since both voltage drop and light frequency increase with band gap, a blue LED may drop around 3 to 3.3 volts.
### 20,000 Watt Light Bulb Test
20,000 Watt Light Bulb Test
20,000 Watt Light Bulb Test
## Is a 1.5 volt battery the same as AAA battery?
A 1.5-volt battery is the same size as AA batteries, but is rated for lower voltage to suit a device with a lighter power requirement such as a flashlight. Just be sure not to mix the two up—because they’re exactly the same battery. Mixing them together will result in low-voltage operation of electronic devices.
## How do I convert volts to watts?
The formula to convert voltage to watts is watts = amps x volts.
## How many volts is 9 watts?
Equivalent Volts and Watts Measurements
Voltage Power Current
9 Volts 9 Watts 1 Amps
9 Volts 18 Watts 2 Amps
9 Volts 27 Watts 3 Amps
9 Volts 36 Watts 4 Amps
## What is 18 watt LED equivalent to?
LED tube lights use between 40 and 50% less energy than fluorescent tube lights, which is of course comparable to CFLs that utilise the same technology.
LED equivalents to fluorescent tube lights.
Fluorescent Tube Light Wattage LED Equivalent Wattage
70 Watt 24 Watt
58 Watt 22 Watt
35 Watt 18 Watt
20 Watt 9 Watt
9 thg 3, 2018
## What is 10W LED equivalent to?
Only 10W in power, incandescent equivalent to 80W. These Cool White (6000K) LED bulbs are ideal for use in table and floor lamps, living rooms, kitchens, bedrooms and hallways or ceiling fixtures. Radiates a bright light of up to 1000 lumens with good heat dissipation and no flicker.
## How bright is a 7W bulb?
Lumen & Wattage Comparison
Lumens (Brightness) Incandescent Watts LED Watts (Viribright)
400 – 500 40W 6 – 7W
650 – 850 60W 7 – 10W
1000 – 1400 75W 12 – 13W
1450-1700+ 100W 14 – 20W
## Is a 5 watt light bulb bright?
5W LED VS 5W CFL bulb brightness comparision in terms of brightness. 5W LED bulb gives 550 lumens brightness whereas 5W CFL bulb gives only 400 lumens brightness. For tye same power consumption, LED gives more brightness than CFL bulb.
## How bright is 3 watts?
How bright is 3 watt LED light? This 3-watt outdoor LED spotlight emits up to 150 lumens of cool or warm white illumination in a 30° beam pattern.
## Can a 1.5 volt battery power an LED?
In this instructable you will able to make a simple led hack which can power a led with single a AA 1.5v battery. A LED need at least 2.4 v or more voltage. This circuit will boost the voltage up to drive a single led. But it will consume more current than a led need due to conversion of the voltage.
### Volts, Amps, and Watts Explained
Volts, Amps, and Watts Explained
Volts, Amps, and Watts Explained
## What are 1.5 volt batteries used for?
Batteries like this are great for powering devices such as clocks, remotes, smoke alarms, and flashlights. Other varieties of the 1.5v battery power devices like pen flashlights, active styluses, and laser pointers, cameras, calculators, pencil sharpeners, and so much more.
## What is the lowest voltage LED bulb?
Therefore, the lowest voltage limit of white LED lights is 24V. 2. Red LEDs: According to the minimum driving voltage: 6*1.8V=10.8V, based on the common power supply specification, 12V is fine.
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alrededor de 1 año ago | 0 | 1,478 | 5,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.794147 |
https://www.airmilescalculator.com/distance/avp-to-tbn/ | 1,606,857,229,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00276.warc.gz | 513,673,553 | 44,593 | # Distance between Wilkes-Barre, PA (AVP) and Fort Leonard Wood, MO (TBN)
Flight distance from Wilkes-Barre to Fort Leonard Wood (Wilkes-Barre/Scranton International Airport – Waynesville-St. Robert Regional Airport) is 910 miles / 1464 kilometers / 791 nautical miles. Estimated flight time is 2 hours 13 minutes.
Driving distance from Wilkes-Barre (AVP) to Fort Leonard Wood (TBN) is 1017 miles / 1637 kilometers and travel time by car is about 17 hours 46 minutes.
## Map of flight path and driving directions from Wilkes-Barre to Fort Leonard Wood.
Shortest flight path between Wilkes-Barre/Scranton International Airport (AVP) and Waynesville-St. Robert Regional Airport (TBN).
## How far is Fort Leonard Wood from Wilkes-Barre?
There are several ways to calculate distances between Wilkes-Barre and Fort Leonard Wood. Here are two common methods:
Vincenty's formula (applied above)
• 909.774 miles
• 1464.139 kilometers
• 790.572 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 907.795 miles
• 1460.954 kilometers
• 788.852 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Wilkes-Barre/Scranton International Airport
City: Wilkes-Barre, PA
Country: United States
IATA Code: AVP
ICAO Code: KAVP
Coordinates: 41°20′18″N, 75°43′24″W
B Waynesville-St. Robert Regional Airport
City: Fort Leonard Wood, MO
Country: United States
IATA Code: TBN
ICAO Code: KTBN
Coordinates: 37°44′29″N, 92°8′26″W
## Time difference and current local times
The time difference between Wilkes-Barre and Fort Leonard Wood is 1 hour. Fort Leonard Wood is 1 hour behind Wilkes-Barre.
EST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 144 kg (318 pounds).
## Frequent Flyer Miles Calculator
Wilkes-Barre (AVP) → Fort Leonard Wood (TBN).
Distance:
910
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
910
Round trip? | 566 | 2,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-50 | latest | en | 0.791044 |
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Shirley is drawing triangles that have the same area. The base of each triangle varies inversely with the height. What are the possible base and height of a second triangle if the first triangle's base is 10 and its height is 6
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1. any pair of numbers whose product is 60
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# OG 12.problem 130
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OG 12.problem 130 [#permalink]
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05 Dec 2009, 15:20
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–5 –4 –3 –2 –1 0 1 2 3 (number line shaded portion and X can take any value from -5 until 3)
130. Which of the following inequalities is an algebraic
expression for the shaded part of the number line
above?
(A) |x| ≤ 3
(B) |x| ≤ 5
(C) |x − 2| ≤ 3
(D) |x − 1| ≤ 4
(E) |x + 1| ≤ 4
The answer is E, why not B?
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Re: OG 12.problem 130 [#permalink]
### Show Tags
05 Dec 2009, 15:34
ISBtarget wrote:
–5 –4 –3 –2 –1 0 1 2 3 (number line shaded portion and X can take any value from -5 until 3)
130. Which of the following inequalities is an algebraic
expression for the shaded part of the number line
above?
(A) |x| ≤ 3
(B) |x| ≤ 5
(C) |x − 2| ≤ 3
(D) |x − 1| ≤ 4
(E) |x + 1| ≤ 4
The answer is E, why not B?
b is saying -5≤x≤5
e is saying -5≤x≤3
x+1≤4 = x≤3
-x -1 ≤ 4
-x ≤ 5
x>=-5
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Re: OG 12.problem 130 [#permalink]
### Show Tags
06 Dec 2009, 07:29
take both conditions
if x>0
ie. if x+1>0
or x>-1
then |x+1|<=4
x+1<= 4
x<= 3
If x<0
i.e if x+1<0
or x<-1
then -|x+1|<=4
-x-1<= 4
-x<= 5
x>= -5
combining the limits we get
-5<x<3 therefore E
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Re: OG 12.problem 130 [#permalink]
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22 Feb 2010, 01:16
ISB target ....try and put x=4 in eqn 2. It still holds good but doesnt satisfies your number options for the line. Eqn 5 does that for x=4 and is justified only for the numbers given in the quesn.
Hope its clear now.
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Re: OG 12.problem 130 [#permalink]
### Show Tags
22 Feb 2010, 07:49
ISBtarget wrote:
–5 –4 –3 –2 –1 0 1 2 3 (number line shaded portion and X can take any value from -5 until 3)
130. Which of the following inequalities is an algebraic
expression for the shaded part of the number line
above?
(A) |x| ≤ 3
(B) |x| ≤ 5
(C) |x − 2| ≤ 3
(D) |x − 1| ≤ 4
(E) |x + 1| ≤ 4
The answer is E, why not B?
Length = 3-(-5) = 8
center = (-5)+8/2 = -1
Now only D & E has 8/2 = 4 on right side.
Also at x=-5 only E is satisfied and not D so answer is E.
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Re: OG 12.problem 130 [#permalink]
### Show Tags
22 Feb 2010, 08:39
|x+a| <= b
you can take the above case and simplify as below
if the origin is at (-a,0) then x will have as many values to the right of -a as to the left of -a. Bacially the values of x are symetric at the value (-a,0)
for the expression given in the question, the values of x are -5<=x<=3 .. the values are symetric at x=-1
so the expression would be a = 1 and b =4 ... |x+1|<=4
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Re: OG 12.problem 130 [#permalink]
### Show Tags
26 Feb 2010, 14:40
|x + 1| ≤ 4
remove the absolute value by
-4 ≤ x+1 ≤ 4
add -1 to all side
-1 -4 ≤ x+1 -1 ≤ 4 -1
-5 ≤ x ≤ 3
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Re: OG 12.problem 130 [#permalink]
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01 Mar 2010, 01:28
E is correct.
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Re: OG 12.problem 130 [#permalink]
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21 Nov 2011, 10:06
1
KUDOS
All the answers above are back solving from answer E.
But if we were to solve directly from question to get the answer,
We need to find the center point in the shaded part of the number line.
Here it is easy to identify that -1 is the center point and the region is 4 units from it on either side.
So x is less than 4 units on either side of -1
this gives |x+1|<=4.
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Re: OG 12.problem 130 [#permalink]
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23 Nov 2011, 06:07
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 08 Sep 2011 Posts: 75 Concentration: Finance, Strategy Followers: 3 Kudos [?]: 2 [0], given: 5 Re: OG 12.problem 130 [#permalink] ### Show Tags 23 Nov 2011, 09:13 E. It is not B because B implies that it could be 4 or 5 as well when we know that it 3 is the highest positive number Re: OG 12.problem 130 [#permalink] 23 Nov 2011, 09:13 Similar topics Replies Last post Similar Topics: 7 If the time is currently 1:30 pm, what time will it be in exactly 647 8 21 Jul 2015, 03:19 1 The sum of 5 different positive 2-digit integers is 130. What is the 4 22 Jan 2015, 08:01 If the time is currently 1:30pm, what time will it be in 1 23 Apr 2012, 06:55 Bottle R contains 250 capsules and costs$6.25. Bottle T contains 130 5 07 Jan 2008, 03:37
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# OG 12.problem 130
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What is Principal Component Analysis?Objectives Assumptions When to use PCA?How does PCA algorithm work?Steps of dimentionality reductionApplications of PCA
## Understanding Learning Rate in Machine Learning
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## Design thinking is Apple’s Success Mantra
Design Thinking has become extremely popular in recent years. All the top brands such as Google, Apple, Airbnb, and Adidas are using... | 1,557 | 7,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-34 | latest | en | 0.795497 |
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# Student Learning Groups: Homogeneous or Heterogeneous?
Updated 01/2014
OK kids, we are going to be learning in groups today! Each group needs a math checker, a presenter, a writer/editor, and an illustrator. You decide who does what. You will be reviewing the best ways to solve polynomial problems.
Please pull out the instructions and the rubric for this assignment. As a group, your task is to create a one page, step-by-step process that some one could follow to arrive at a solution...
...You have 15 minutes to complete this task according to the rubric that I have handed out.
The teacher then spends the next 15 minutes roving about the classroom, reviewing the progress of each group, and asking probing questions to help the individual groups clarify their thinking.
Grouping sounds so easy. What we don't see in the above example is how the teacher has organized students in the groups in order to achieve the best results. Some educators firmly believe that a teacher must mix the groups so that students of all levels are represented in each group (heterogeneous grouping of students), while others believe that a teacher must organize the students by ability levels (homogeneous grouping of students). Robert Marzano, Debra Pickering, and Jane Pollock, in their familiar work, Classroom Instruction That Works, explain that there are advantages to both methods depending on what the teacher wants to do.
### Identifying Purposes
If the purpose of the group learning activity is to help struggling students, then the research shows that heterogeneous groups may help most. On the other hand, if the purpose of the group learning activity is to encourage medium ability groups to learn at high levels then homogeneous grouping would be better.
I learned this as a teacher when one of my gifted and talented students told me in confidence that she really hated being in heterogeneous groups (she said it differently of course) all the time because by default, the other members of the group expected her to be the leader, organize things and do all of the work.
This was my "tipping point" because it made me realize that I wasn't grouping students for increased learning. I was using grouping mainly as a discipline management tool and that in actuality my attempt to increase student engagement had completely backfired. By always making sure that the "smart" students and the struggling students were equally divided in the groups, I was actually limiting the student participation to the defacto leaders of the groups.
### Deciding Which is Best
Because of this epiphany, I remember vowing that I would further differentiate my teaching by also seeking ways to give the upper-level students challenging and engaging learning activities. I promised to stop using the "good kids" in the hopes that some of their "goodness" would rub off on the other students. An interesting thing happened when I ability grouped the students. New leadership structures formed, and students who had never actively participated in groups before, all of the sudden demonstrated skills and creativity that I never knew they had.
Students are smart and they can easily figure out what we are really doing. Students, in our classrooms, know when they are being grouped to mainly tutor and remediate less capable students and... most of the time they resent it. We can also "tick them off" when we form groups solely for discipline purposes by placing the calm, obedient students in each group to separate and calm down the unruly ones. My daughter Mercedes, who falls in both categories above, stated that when teachers do this to her, she doesn't learn and it is not fun for her or the other students. Perhaps more often than not, students are savvy enough to play along when they recognize that the grouping is nothing more than a routine way to spend the time and has no real learning purpose at all.
If given a choice, students prefer to learning in groups of their peers and friends (homogeneous groups), but they also appreciate getting to know and learn from other members of the classroom. This requires that we trust students to make good decisions and we hold them accountable for following the norms of learning in groups.
Effective learning in groups must have at least the following elements (Marzano, et. al, pages 85-86):
• They must include every member of the group
• Each person has a valid job to perform with a known standard of completion
• Each member is invested in completing the task or learning goal
• Each member is accountable individually and collectively
Remember that the desks are not attached to the floor and we can mix things up in heterogeneous and homogeneous groups in interesting and creative ways: eye color, left or right handedness, preferred pizza toppings, number of siblings, music preferences, gender, nationality, hair length, shoe laces, genetic traits, learning styles, etc.
How do your students find success in group-learning?
•
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Last comment 6 days 15 hours ago in Student Engagement | 1,143 | 5,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2015-18 | latest | en | 0.950733 |
http://ciphersbyritter.com/ARTS/FEISNONL.HTM | 1,516,124,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00688.warc.gz | 73,983,763 | 10,688 | # Measured Boolean Function Nonlinearity in Feistel Cipher Constructions
## Terry Ritter
Nonlinearity is the number of bits which must change in the truth table of a Boolean function to reach the closest affine function, and thus is a measure of one kind of cipher strength. By measuring the nonlinearity of Feistel constructions with various numbers of rounds and tables, we hope to gain insight into how Feistel mixing compares to other alternatives.
```From: ritter@io.com (Terry Ritter)
Newsgroups: sci.crypt
Subject: Measured Nonlinearity in Feistel Constructions (LONG!)
Date: Thu, 26 Feb 1998 06:39:28 GMT
Lines: 846
Message-ID: <34f50e19.599891@news.io.com>
MEASURED NONLINEARITY IN FEISTEL CONSTRUCTIONS
Terry Ritter
Ritter Software Engineering
http://www.io.com/~ritter/
1998-02-25
Abstract
Nonlinearity is the number of bits which must change in the truth
table of a Boolean function to reach the closest affine function,
and thus is a measure of one kind of cipher strength. By measuring
the nonlinearity of Feistel constructions with various numbers of
rounds and tables, we hope to gain insight into how Feistel mixing
compares to other alternatives.
Introduction
The ideal block cipher is a keyed simple substitution table of
sufficient size. Unfortunately, with 128-bit blocks, there would
be 2**128 entries in that table, which is completely out of the
question. So the modern block cipher is a *construction* intended
to *simulate* a keyed substitution of the desired size. At issue
is the effectiveness of the construction technology. One way to
investigate this is by using Boolean function theory, since a
substitution table -- or cipher -- can be considered a set of
independent Boolean functions, one for each output bit.
A Boolean function produces a single-bit result for each possible
combination of values from perhaps many Boolean variables. The
*nonlinearity* of a Boolean function is the Hamming distance to
the closest affine function [e.g., PIE88, PIE89, PIE89B]. (Also
see:
http://www.io.com/~ritter/ARTS/MEASNONL.HTM
). That is, nonlinearity is the number of bits which must change
in the truth table of a Boolean function to reach the closest affine
function. If "linearity" is considered a significant cryptographic
weakness, nonlinearity is an explicit measure of the *lack* of that
weakness. So nonlinearity measures one form of cipher "strength."
For cryptographic purposes, it is desired to take the nonlinearity
of a substitution table to be the *minimum* of the nonlinearity
values for each output bit in that table. Nonlinearity is measured
by forming the (one-bit-wide) truth table for a particular output
bit, then performing a Fast Walsh-Hadamard Transform (FWT) on that
array. Each result value is essentially a correlation count to a
particular affine function, and the minimum distance (the maximum
correlation) is found by scanning the transform results.
In measuring nonlinearity, it is generally necessary to record the
function result for each possible combination of input variables.
If we have an "8-bit" table, we must record and then transform
256 elements, and if we have a "16-bit" table, we must record and
transform 64K elements. Thus, measuring large functions rapidly
becomes impossible. So, although we cannot hope to measure the
nonlinearity of a real 64-bit or 128-bit block cipher, we *can*
measure nonlinearity in substitution tables and small block
constructions. Here we deal with 5-bit tables and the resulting
10-bit blocks.
Ideal Nonlinearity Distributions
By properly shuffling tables using a large-state cryptographic RNG
(in particular:
http://www.io.com/~ritter/KEYSHUF.HTM
) we can sample or construct a presumably uniform distribution of
different table arrangements. We can measure the nonlinearity of
each table, and accumulate the results. For 5-bit tables, we get
a nonlinearity distribution something like that shown in Figure 1.
Nonlinearity Distribution in 5-Bit Tables Fig. 1
0.7 | *
0.6 | *
0.5 | *
0.4 | *
0.3 | *
0.2 | * *
0.1 | * * *
0.0 | * * *
Prob +--+--+--+--+--+--+--
0 2 4 6 8 10 Nonlinearity
Our block cipher construction problem essentially involves using
small random tables (here 5-bit tables with 32 entries and a mean
nonlinearity of 7.8) to somehow emulate a table which is twice as
wide (here a 10-bit table with 1024 entries and a mean nonlinearity
of 447.7).
For 10-bit tables, we get a nonlinearity distribution
something like that in Figure 2.
Nonlinearity Distribution in 10-Bit Tables Fig. 2
0.2 |
0.175 | * *
0.15 | * * *
0.125 | * * * *
0.1 | * * * * *
0.075 | * * * * * *
0.05 | * * * * * * * *
0.025 | * * * * * * * * * *
0.00 | * * * * * * * * * * *
Prob +--+--+--+--+--+--+--+--+--+--+--+--+--
436 440 444 448 452 456 Nonlinearity
Feistel Mixing
The classic Feistel construction of Figure 3 is a way of mixing
two small blocks into a large block. Here, + is the exclusive-OR
function. As used in the U.S. Data Encryption Standard (DES),
L and R are each 32 bits; here we have only 5 bits in each.
Feistel Construction Fig. 3
L R
| |
|--> F --> + round 1
| |
+ <-- F <--| round 2
| |
v v
L' R'
In DES, ideal properties for F are much-discussed topics. The
Feistel construction itself supports the ability to encipher and
decipher using a random function F, not necessarily a permutation.
But the 8 different "S-boxes" used in DES *are* permutations, if
we see each as a set of 4 permutations per box, so using random
permutations in Feistel mixing is not at all unreasonable. And by
using permutation tables as components we have a single compatible
measure both for the component tables and the simulated larger table,
and thus can draw conclusions about the construction. Invertible
tables also support a direct comparison to earlier work on
nonlinearity with respect to Mixing and VSBC constructions.
Unlike the earlier work, the decision to use random invertible
substitutions for function F still leaves us with a couple of
remaining parameters: The number of rounds, and the number of
tables. Here we investigate the mixing obtained from the Feistel
construction with two variations: The use of a single table for
various numbers of rounds, and the use of a new random table on
every other round, for various numbers of rounds.
Feistel Mixing and DES
Unfortunately, the Feistel construction is *not* the only form
of mixing found in DES. Indeed, the practicality of the Feistel
construction depends upon having (that is, *constructing*) a wide
function F. And while F *might* be realized as a recursive
sequence of smaller Feistel constructions, any such structure
would be impractically slow. So the very existence of practical
Feistel ciphers depends upon the construction of a random function
F which itself mixes and diffuses information across its width.
In each DES "S-box," the outside input bits (b0 and b5) -- which
are set by expansion "E" -- just select which of the 4 boxes to use.
This is a data-dependent dynamic table selection which is, in itself,
a form of mixing. Of course, the simple combination of expansion "E"
and the array of "S-boxes" by themselves probably would not produce
good mixing: A single bit-change at one end of the input block
could propagate no faster than one table per round, *if* all data
were kept aligned. But this is not the case, since "Permutation P"
transposes the data bits. So, other than the dynamic table selection
(the two extra inputs for each S-box), function F is itself a classic
substitution-permutation design.
This means that the overall mixing in DES is a complex combination
of the Feistel construction *plus* the expansion, table-selection,
table lookup and permutation in each round. This complexity makes
it difficult to independently extract the importance and consequences
of each feature, as we surely must if we are to deeply understand
the design. This confluence of multiple features also makes it
impossible to scale DES to a tiny testable size. Thus, we seem
to be limited to testing a tiny Feistel construction -- which *is*
scalable -- instead of a tiny DES.
Feistel Mixing with Random Tables
A common assumption about Feistel mixing is that some fixed small
set of optimized tables will be used repeatedly. This concept
presumably comes from DES. But in the special context of DES:
1. DES S-boxes each contain 4 separate *permutations*,
and are *not* random constructions; and
2. The DES permutations are only 4 bits wide, and 4-bit-wide
permutations are just about the weakest possible tables
that have any nonlinear strength at all.
Accordingly, it should come as no surprise that replacing DES
S-boxes with random tables is not a good idea. And random 4-bit
permutations are often weak, so this is *also* not a good idea.
In marked contrast, random 8-permutations are almost *never* that
weak (the chance of finding even one linear output is 1 in 2**239).
It thus appears that the assumption that random tables are
necessarily bad is based on the special context of DES, and is not
valid in general.
Comparisons to Mixing and VSBC Constructions
Earlier nonlinearity measurement experiments on tiny Mixing (
http://www.io.com/~ritter/ARTS/MIXNONLI.HTM
) and VSBC (
http://www.io.com/~ritter/ARTS/VSBCNONL.HTM
) constructions also used random tables, thus potentially making
it possible to see how Feistel mixing compares.
Unfortunately, the comparison is not exact, because both the Mixing
and VSBC constructions do in some sense mix multiple tables in a
single mixing level. With Feistel mixing, we would normally think
of multiple "rounds" of mixing, rather than having several tables
in each round. DES does of course manage this, but does not give
us a scalable rule which we could use to build a tiny version with
multiple tables. And even in the best possible situation we would
be looking at the experimental measurement of a 16-bit simulated
table, for which it would be very difficult to collect an accurate
ideal distribution.
So, given the limitations of reality, the first thing we might want
to look at is the nonlinearity measure from Feistel mixing with a
single table, under a varying the number of rounds. Then we might
look at using multiple tables, again for a varying number of rounds.
The Experiments
We construct 10,000 random 5-bit tables, which are then used in
10-bit Feistel constructions. Each of these constructions is
measured for nonlinearity, and the result accumulated.
While Figure 2 gives a general idea of the desired distribution, we
eventually want to use a chi-square comparison, for which we need
an ideal reference distribution. The ideal model in Figure 4 is
thus constructed from three different 15-hour trials of 1,000,000
random 10-bit tables each, with the counts reduced by a factor of
100 and rounded to an integer.
The Ideal 10-Bit Nonlinearity Distribution Fig. 4
NL Trial 1 Trial 2 Trial 3 Ideal
462 0 1 0 0
460 55 75 55 1
458 2145 2269 2234 22
456 20537 20938 20882 208
454 76259 76344 75980 763
452 148847 148135 148510 1485
450 188329 187802 187585 1878
448 177775 178208 178616 1782
446 140283 140219 140407 1403
444 97415 97317 97465 974
442 61999 62282 62213 622
440 37829 37685 37382 376
438 21675 21789 21753 218
436 12417 12320 12408 124
434 6826 6818 6912 68
432 3740 3693 3584 37
430 1912 2050 1884 19
428 979 1043 1073 10
426 490 537 526 5
424 249 241 265 2
422 119 133 143 1
420 64 55 69 0
418 26 25 30 0
416 12 14 12 0
414 9 5 6 0
412 3 2 4 0
410 0 0 1 0
In each experimental trial, we first initialize the substitutions
from a random key. This produces a particular 10-bit cipher -- a
simulated 10-bit simple substitution. We then traverse the complete
transformation -- all 1024 possible data values -- and collect all
1024 10-bit ciphertext results. We then traverse all 10 ciphertext
bit-columns, one-by-one, and perform a nonlinearity computation on
each. The minimum nonlinearity value found across all columns is
accumulated into the growing distribution.
We see the results for 2 rounds of Feistel mixing in Figure 5.
Feistel Nonlinearity Distribution
One Table, Two Rounds Fig. 5
NL Expect Found
64 0 2
128 0 78
192 0 1494
256 0 7387
320 0 1039
Clearly, 2 Feistel mixing rounds do not approximate the nonlinearity
distribution from random substitutions, so we move on to 4 rounds
as in Figure 6.
Feistel Nonlinearity Distribution
One Table, Four Rounds Fig. 6
NL Expect Found
224 0 1
312 0 56
320 0 1
344 0 1
352 0 35
364 0 1
368 0 8
384 0 696
388 0 3
392 0 187
396 0 2
400 0 11
404 0 15
406 0 3
408 0 13
410 0 2
412 0 38
414 0 2
416 0 1709
418 0 4
420 0 66
422 1 11
424 2 126
426 5 13
428 10 256
430 19 20
432 37 780
434 68 51
436 124 660
438 218 109
440 376 3566
442 622 70
444 974 589
446 1403 91
448 1782 639
450 1878 43
452 1485 107
454 763 12
456 208 2
458 22 1
460 1 0
462 0 0
Again, the is pretty brutal: So even four Feistel mixing rounds do
not approximate random substitutions. And neither do six (Figure 7),
eight (Figure 8), ten (Figure 9) or twelve (Figure 10):
Feistel Nonlinearity Distribution
One Table, Six Rounds Fig. 7
NL Expect Found
296 0 1
316 0 1
324 0 1
332 0 1
336 0 2
352 0 1
372 0 1
384 0 2
388 0 3
392 0 1
394 0 1
396 0 2
398 0 1
400 0 3
402 0 1
404 0 10
406 0 0
408 0 23
410 0 2
412 0 22
414 0 6
416 0 22
418 0 14
420 0 46
422 1 25
424 2 85
426 5 21
428 10 114
430 19 70
432 37 192
434 68 138
436 124 386
438 218 269
440 376 641
442 622 662
444 974 1197
446 1403 1238
448 1782 1612
450 1878 1394
452 1485 1151
454 763 492
456 208 133
458 22 13
460 1 0
462 0 0
Feistel Nonlinearity Distribution
One Table, Eight Rounds Fig. 8
NL Expect Found
346 0 1
370 0 1
372 0 1
402 0 2
404 0 3
406 0 3
408 0 6
410 0 1
412 0 6
414 0 10
416 0 17
418 0 18
420 0 30
422 1 36
424 2 38
426 5 53
428 10 74
430 19 95
432 37 137
434 68 194
436 124 261
438 218 406
440 376 519
442 622 808
444 974 1080
446 1403 1365
448 1782 1569
450 1878 1527
452 1485 1063
454 763 519
456 208 136
458 22 21
460 1 0
462 0 0
Feistel Nonlinearity Distribution
One Table, Ten Rounds Fig. 9
NL Expect Found
362 0 1
396 0 1
398 0 2
400 0 1
402 0 1
404 0 0
406 0 4
408 0 3
410 0 10
412 0 10
414 0 11
416 0 28
418 0 16
420 0 25
422 1 39
424 2 63
426 5 63
428 10 78
430 19 116
432 37 155
434 68 247
436 124 314
438 218 380
440 376 584
442 622 811
444 974 1034
446 1403 1353
448 1782 1556
450 1878 1428
452 1485 1046
454 763 480
456 208 126
458 22 14
460 1 0
462 0 0
Feistel Nonlinearity Distribution
One Table, Twelve Rounds Fig. 10
NL Expect Found
398 0 1
400 0 3
402 0 2
404 0 2
406 0 6
408 0 4
410 0 8
412 0 12
414 0 13
416 0 12
418 0 30
420 0 23
422 1 34
424 2 64
426 5 62
428 10 96
430 19 113
432 37 179
434 68 234
436 124 327
438 218 462
440 376 597
442 622 797
444 974 1054
446 1403 1347
448 1782 1467
450 1878 1450
452 1485 1011
454 763 450
456 208 128
458 22 12
460 1 0
462 0 0
The twelve rounds with a fixed table of Figure 10 is about as good
as it gets. The distribution from a single table never gets close
enough to the ideal distribution to make a chi-square computation
reasonable.
Feistel Mixing with Random Multiple Tables
On the other hand, things improve if we have a new table every couple
of rounds (Figure 11). (There would seem to be little advantage in
having a unique table for each round, since only half of the block
would be affected by that table.)
Feistel Nonlinearity Distribution
Two Tables, Four Rounds Fig. 11
NL Expect Found
320 0 3
352 0 39
360 0 2
364 0 1
368 0 6
382 0 1
384 0 156
386 0 1
388 0 3
390 0 0
392 0 225
394 0 1
396 0 4
398 0 1
400 0 6
402 0 1
404 0 13
406 0 3
408 0 22
410 0 4
412 0 42
414 0 2
416 0 1951
418 0 3
420 0 81
422 1 5
424 2 143
426 5 13
428 10 276
430 19 29
432 37 853
434 68 47
436 124 788
438 218 90
440 376 3175
442 622 98
444 974 805
446 1403 130
448 1782 833
450 1878 33
452 1485 95
454 763 10
456 208 5
458 22 1
460 1 0
462 0 0
While hardly ideal, we can see that the results in Figure 11
are somewhat better than the same number of rounds with a single
table (Figure 6). But things perk up rather nicely with three
tables and six rounds as shown in Figure 12.
Feistel Nonlinearity Distribution
Three Tables, Six Rounds Fig. 12
NL Expect Found Chi-Sq DF
420 0 0 }
422 1 3 }
424 2 3 }
426 5 3 }
428 10 10 } 0.056 0
430 19 13 1.950 1
432 37 40 2.194 2
434 68 59 3.385 3
436 124 123 3.393 4
438 218 200 4.879 5
440 376 381 4.946 6
442 622 633 5.140 7
444 974 990 5.403 8
446 1403 1390 5.523 9
448 1782 1780 5.526 10
450 1878 1952 8.441 11
452 1485 1441 9.745 12
454 763 754 9.851 13
456 208 208 9.851 14
458 22 17 }
460 1 0 }
462 0 0 } 11.417 15
With 15 "degrees of freedom," the chi-square value of 11.417 in
Figure 11 is very believable. The critical chi-square values for
DF = 15 are given in Figure 13.
Chi-Square Critical Values, for DF = 15 Fig. 13
1% 5% 25% 50% 75% 95% 99%
5.2293 7.2609 11.0365 14.3389 18.2451 24.9958 30.5779
And another couple of rounds with yet another table continues the
success story in Figure 14.
Feistel Nonlinearity Distribution
Four Tables, Eight Rounds Fig. 14
NL Expect Found Chi-Sq DF
410 0 1 }
412 0 0 }
414 0 0 }
416 0 0 }
418 0 1 }
420 0 2 }
422 1 2 }
424 2 1 }
426 5 4 }
428 10 10 } 0.500 0
430 19 19 0.500 1
432 37 30 1.824 2
434 68 65 1.957 3
436 124 119 2.158 4
438 218 222 2.232 5
440 376 375 2.234 6
442 622 634 2.466 7
444 974 982 2.532 8
446 1403 1365 3.561 9
448 1782 1750 4.135 10
450 1878 1928 5.467 11
452 1485 1504 5.710 12
454 763 750 5.931 13
456 208 212 6.008 14
458 22 23 }
460 1 1 }
462 0 0 } 6.052 15
While the chi-square value of 6.052 may seem small, this means the
distribution is close to the ideal. When comparing distributions,
we are rarely worried about a chi-square value that seems "too
good," for the alternative is far *less* believable: Either we
have had the good luck of sampling a close distribution so it looks
better than it should, or we have had the absolutely unbelievable
luck of sampling a fundamentally different distribution so it looks
not just good, but actually *too* good. Thus, these tests are
basically "one-tail" statistical tests.
Conclusions
Feistel mixing seeks to produce a block cipher which is double
the size of a smaller "block cipher" component; in this respect,
it is similar to the Mixing constructions measured earlier. When
using a single fixed table, it appears that Feistel mixing is
never good enough, no matter how many rounds are applied. But
even 6 rounds of Feistel mixing with 3 *different* tables
apparently does approximate the nonlinearity of a larger
substitution table.
These experiments show an ultimately successful result using random
invertible tables. Thus, the experiments give no indication that
Feistel mixing requires the ideal tables that so much of the
literature seems directed at producing. In contrast to the unending
search for the qualities of an ideal table, it appears that we can
produce an effective Feistel cipher by choosing tables "at random"
(that is, we can construct tables based on the cipher key). Note
that Differential and Linear Cryptanalysis style attacks generally
depend upon knowledge of the contents of the tables, knowledge
which is unavailable when keyed tables are used.
When compared to the Mixing and VSBC constructions investigated
previously, the problem with Feistel mixing is that, by itself, it
only doubles the size of the nonlinear component, and we assume
that repeated doubling requires exponential effort: That is, if
6 Feistel rounds are required to take random 8-bit tables to a
16-bit emulated table, 6 rounds of the emulation (36 total
rounds) should be required to get a 32-bit table, and 216 total
rounds should be required for a 64-bit cipher. (It would be
interesting to check these assumptions on modern fast equipment.)
Now, we know that DES (an emulated 64-bit table) does *not* require
exponential effort. But Feistel mixing is *not* the only form of
mixing in DES. We avoid addressing the other structures, because
they do not scale well, and because we wish to examine each
contribution as independently as possible.
Here we have measured the Feistel mixing alone, which we assume
requires effort exponential with the block size. In contrast,
Mixing constructions require only effort proportional to n log n,
and VSBC constructions require only effort linear with block size.
Thus, these alternative constructions offer the possibility of
far more efficient cryptographic mixing.
References and Bibliography
[AY82] Ayoub, F. 1982. Probabilistic completeness of
substitution-permutation encryption networks. IEE Proceedings,
Part E. 129(5): 195-199.
[DAE94] Daemen, J., R. Govaerts and J. Vandewalle. 1994.
Correlation Matrices. Fast Software Encryption. 275-285.
[FOR88] Forre, R. 1988. The Strict Avalanche Criterion:
Spectral Properties of Boolean Functions and an Extended
Definition. Advances in Cryptology -- CRYPTO '88. 450-468.
[GOR82] Gordon, J. and H. Retkin. 1982. Are Big S-Boxes Best?
Cryptography. Proceedings of the Workshop on Cryptography,
Burg Feuerstein, Germany, March 29-April 2, 1982. 257-262.
[HEY94] Heys, H. and S. Tavares. 1994. On the security of the
CAST encryption algorithm. Canadian Conference on Electrical and
Computer Engineering. Halifax, Nova Scotia, Canada, Sept. 1994.
332-335.
[HEY95] Heys, H. and S. Tavares. 1995. Known plaintext
cryptanalysis of tree-structured block ciphers. Electronics
Letters. 31(10): 784-785.
[MEI89] Meier, W. and O. Staffelbach. 1989. Nonlinearity Criteria
for Cryptographic Functions. Advances in Cryptology --
Eurocrypt '89. 549-562.
[MIR97] Mirza, F. 1997. Linear and S-Box Pairs Cryptanalysis
of the Data Encryption Standard.
[OC91] O'Connor, L. 1991. Enumerating nondegenerate permutations.
Advances in Cryptology -- Eurocrypt '91. 368-377.
[OC93] O'Connor, L. 1993. On the Distribution Characteristics in
Bijective Mappings. Advances in Cryptology -- EUROCRYPT '93.
360-370.
[PIE88] Pieprzyk, J. and G. Finkelstein. 1988. Towards effective
nonlinear cryptosystem design. IEE Proceedings, Part E.
135(6): 325-335.
[PIE89] Pieprzyk, J. and G. Finkelstein. 1989. Permutations
that Maximize Non-Linearity and Their Cryptographic Significance.
Computer Security in the Age of Information. 63-74.
[PIE89B] Pieprzyk, J. 1989. Non-linearity of Exponent Permutations.
Advances in Cryptology -- EUROCRYPT '89. 80-92.
[PIE93] Pieprzyk, J., C. Charnes and J. Seberry. 1993. Linear
Approximation Versus Nonlinearity. Proceedings of the Workshop on
Selected Areas in Cryptography (SAC '94). 82-89.
[PRE90] Preneel, B., W. Van Leekwijck, L. Van Linden, R. Govaerts
and J. Vandewalle. 1990. Propagation Characteristics of Boolean
Functions. Advances in Cryptology -- Eurocrypt '90. 161-173.
[RUE86] Rueppel, R. 1986. Analysis and Design of Stream Ciphers.
Springer-Verlag.
[XIO88] Xiao, G-Z. and J. Massey. 1988. A Spectral Characterization
of Correlation-Immune Combining Functions. IEEE Transactions on
Information Theory. 34(3): 569-571.
[YOU95] Youssef, A. and S. Tavares. 1995. Resistance of Balanced
S-boxes to Linear and Differential Cryptanalysis. Information
Processing Letters. 56: 249-252.
[YOU95B] Youssef, A. and S. Tavares. 1995. Number of Nonlinear
Regular S-boxes. Electronics Letters. 31(19): 1643-1644.
[ZHA95] Zhang, X. and Y. Zheng. 1995. GAC -- the Criterion for
Global Avalanche Characteristics of Cryptographic Functions.
Journal for Universal Computer Science. 1(5): 316-333.
---
Terry Ritter ritter@io.com http://www.io.com/~ritter/
The Crypto Glossary: http://www.io.com/~ritter/GLOSSARY.HTM
```
Also see: Measured Nonlinearity in Variable Size Block Ciphers (1998) (29K),
Measuring Nonlinearity by Walsh Transform (1998) (20K), and
Measured Nonlinearity in Mixing Constructions (1997) (25K).
Terry Ritter, his current address, and his top page.
Last updated: 1998-02-26 | 8,448 | 29,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-05 | latest | en | 0.840109 |
https://help.synerise.com/use-cases/rfm-analysis/ | 1,675,608,697,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00340.warc.gz | 319,744,225 | 14,856 | # RFM analysis
Published December 20, 2021
Modules
Difficulty
Selected Clients
Industry
RFM segmentation is a method used to identify clusters of customers for special treatment. It is commonly used in database marketing and direct marketing, and has received particular attention in many different industries. In order to differentiate customers from data sets, the RFM method uses three different attributes:
• Recency of the last purchase (R) - refers to the interval between the time of the latest customer purchase and the current date. The shorter the interval between current date and last purchase, the higher R score is.
• Frequency of the purchases (F) - refers to the number of transactions in a particular period. The higher number of transactions, the higher the F score is.
• Monetary value of the purchases (M) - refers to monetary value of products purchased by the customer. The more the customer spends, the higher the M score is.
This lets you isolate groups, identify new or most active customers, and target personalized messages based on transaction data.
## Prerequisites
• Come up with the categories you want to organize your customers, for example: Top customers, Recent customers, Churn risk, Heavy buyers, Lost heavy buyers, Lost customers.
## Process
1. Create a number of basic analyses, so you can later use them in more complex calculations.
2. Create a set of metrics for Recency, Frequency, and Monetary scores to return the value of 0.20, 0.40, 0.60, 0.80 quantiles. This way you receive four thresholds for all scores.
3. Create a dashboard to conveniently display the results of the metrics.
4. Prepare segments for Recency, Frequency, and Monetary scores based on the values returned by the metrics you created before. The sub-segmentation are named in the following way:
• For Recency score: 500, 400, 300, 200, 100
• For Frequency score: 50, 40, 30, 20, 10
• For Monetary score: 5, 4, 3, 2, 1
5. Create an expression that adds all three segmentations, the example result is 555, which is the highest score a customer can get.
6. Create a RFM segmentation that contains the sub-segmentations which correspond to the customer categories (for example, Top customers, Heavy buyers, Lost customers, and so on). Decide the range of scores for each category.
Note: You can find furhter explanation in the corresponding sections in the article.
## Create a set of basic analyses
In this part of the process, you must create a number of analyses which will be reused in the further part of the process.
### Number of transactions
Create the aggregate that calculates the number of transactions for a single customer in the last 90 days. It will be reused in the further analyses.
1. Go to Analytics > Aggregates > New aggregate.
2. Enter a meaningful name of the aggregate.
3. Set the Aggregate option to Count.
4. Click the Choose event dropdown list.
5. From the dropdown list, select the event that signifies a purchase.
Tip: Events may have different labels between workspaces, but you can always find them by their action name (in this step, it’s transaction.charge).
6. Using the date picker in the lower-right corner, set the time range to Relative time range > Custom > last 90 days.
7. Save the aggregate.
### Time of the first purchase
Create the aggregate which returns the time when a customer made the first purchase in the last 90 days. It will be reused in the further analyses.
1. Go to Analytics > Aggregates > New aggregate.
2. Enter a meaningful name of the aggregate.
3. Set the Aggregate option to First.
4. Click the Choose event dropdown list.
5. From the dropdown list, select the event that signifies a purchase.
Tip: Events may have different labels between workspaces, but you can always find them by their action name (in this step, it’s transaction.charge).
6. Click + where button.
7. On the dropdown list, click > Specials.
8. Choose TIMESTAMP.
9. Using the date picker in the lower-right corner, set the time range to Relative time range > Custom > last 90 days.
10. Save the aggregate.
### Time of the last purchase
Create the aggregate which returns the time when a customer made the latest purchase in the last 90 days. It will be reused in the further analyses.
1. Go to Analytics > Aggregates > New aggregate.
2. Enter a meaningful name of the aggregate.
3. Set the Aggregate option to Last.
4. Click the Choose event dropdown list.
5. From the dropdown list, select the event that signifies a purchase.
Tip: Events may have different labels between workspaces , but you can always find them by their action name (in this step, it’s transaction.charge).
6. Click + where button.
7. On the dropdown list, click > Specials.
8. Choose TIMESTAMP.
9. Using the date picker in the lower-right corner, set the time range to Relative time range > Custom > last 90 days.
10. Save the aggregate.
### Value of purchase
Create the aggregate that returns the total amount of money a single customer spent in the last 90 days.
1. Go to Analytics > Aggregates > New aggregate.
2. Enter a meaningful name of the aggregate.
3. Set the Aggregate option to Sum.
4. Click the Choose event dropdown list.
5. From the dropdown list, select the event that signifies a purchase.
Tip: Events may have different labels between workspaces, but you can always find them by their action name (in this step, it’s transaction.charge).
6. Click + where button.
7. On the dropdown list, select the parameter that signifies the total amount of a transaction, for example, `\$totalAmount`
8. Using the date picker in the lower-right corner, set the time range to Relative time range > Custom > last 90 days.
9. Save the aggregate.
### Number of days since last transaction
Create an attribute expression that calculates how many days passed since the last transaction. This expression reuses the aggregate that returns the timestamp of the last transaction in the last 90 days.
1. Go to Analytics > Expressions > New expression.
2. Enter a meaningful name of the expression.
3. Leave the expression type at default (Attribute).
4. Build the following expression formula:
In the formula of the expression, you must deduct the date of the last transaction from the current date and divide the result by 86400000 to receive the number of days which passed since the last purchase.
5. Save the expression.
### Time from the first transaction
Create an attribute expression that calculates how much time passed since the last transaction. This expression reuses the aggregate that returns the timestamp of the first transaction in the last 90 days and it will be reused in the expression that returns the number of weeks that passed from the first transaction.
1. Go to Analytics > Expressions > New expression.
2. Enter a meaningful name of the expression.
3. Leave the expression type at default (Attribute).
4. Build the following expression formula:
In the formula of the expression, you must deduct the date of the first transaction from the current date. The result of the expression is given in milliseconds.
5. Save the expression.
### Weeks from the first transaction
Create an expression that calculates the number of weeks since the first transaction of a customer in the last 90 days. This expression reuses the expression that calculates the time from the first transaction.
1. Go to Analytics > Expressions > New expression.
2. Enter a meaningful name of the expression.
3. Leave the expression type at default (Attribute).
4. Build the following expression formula:
1. Click the Select node.
2. From the dropdown list, select Customer.
3. Click the unnamed node that has been added to the canvas.
4. Scroll down the page and click Choose attribute.
5. Search and select the expression that calculates the time from the first transaction.
6. Next to the expression on the canvas, click the plus button icon.
7. Select Constant.
8. Click 0 that has been added to the canvas.
9. Change the number to `604800000`.
10. Click the mathematical operator between the nodes and change it to a division sign.
5. Save the expression.
### Average number of transactions per week
Create an expression that calculates the average number of transactions a customer made per week in the last 90 days. This expression reuses the aggregate that calculates the number of transactions in the last 90 days and the expression that calculates the number of weeks since the first transaction in the last 90 days.
1. Go to Analytics > Expressions > New expression.
2. Enter a meaningful name of the expression.
3. Leave the expression type at default (Attribute).
4. Build the following expression formula:
1. Click the Select node.
2. From the dropdown list, select Customer.
3. Click the unnamed node that has been added to the canvas.
4. Scroll down the page and click Choose attribute.
5. Search and select the aggregate that calculates the number of transactions.
6. Next to the expression on the canvas, click the plus button icon.
7. Select Customer.
8. Click the unnamed node that has been added to the canvas.
9. Scroll down the page and click Choose attribute.
10. Search and select the expression that calculates weeks from first transactions.
11. Click the mathematical operator between the nodes and change it to a division sign.
5. Save the expression.
### Segment of loyal customers
Create a group of loyal customers. The business asumptions about loyal customers may slightly differ in each organization, they can include the defined number of transactions, the loyalty card, frequent usage of moble application, and so on. Construct the condition of a segmentation in accordance with your requirements for loyal customers.
1. Go to Analytics > Segmentations > New segmentation.
2. Enter a meaningful name of the segmentation.
3. In the condition of the segmentation include customers who according to your busisness assumptions.
4. Save the segmentation.
## Create metrics
In this part of the process, prepare metrics which will calculate the qunatiles: `0.80`, `0.60`, `0.40`, `0.20` for the values of the expressions and an aggregate:
Create four metrics for each score (which means that in total you will create 12 metrics). This way, you can distinguish tresholds for Recency, Frequency, and Monetary scores.
### Recency score
1. Go to Analytics > Metrics > New metric.
2. Enter a meaningful metric name.
3. Leave the metric kind at default (Simple).
4. Change the Metric type to Customer.
5. Change the Aggregator option to Quantile.
6. Next to the Quantile option, in the text field, enter `0.80`.
7. Click Choose value.
8. Search and select the expression that returns the number of days that passed since the last transaction.
9. Click Enable filter.
10. Click Choose filter.
11. Search and select the segmentation of loyal customers.
12. As the logical operator, select Is true.
13. Confirm the filter settings by clicking Apply.
14. Using the date picker in the lower-right corner, set the time range to Relative time range > Lifetime.
15. Save the metric.
16. Create three metrics with the same settings for the following quantile values: `0.60`, `0.40`, and `0.20`.
### Frequency score
1. Go to Analytics > Metrics > New metric.
2. Enter a meaningful metric name.
3. Leave the metric kind at default (Simple).
4. Change the Metric type to Customer.
5. Change the Aggregator option to Quantile.
6. Next to the Quantile option, in the text field, enter `0.20`.
7. Click Choose value.
8. Search and select the expression that returns the average number of transactions in a week.
9. Click Enable filter.
10. Click Choose filter.
11. Search and select the segmentation of loyal customers.
12. As the logical operator, select Is true.
13. Confirm the filter settings by clicking Apply.
14. Using the date picker in the lower-right corner, set the time range to Relative time range > Lifetime.
15. Save the metric.
16. Create three metrics with the same settings for the following quantile values: `0.40`, `0.60`, and `0.80`.
### Monetary score
1. Go to Analytics > Metrics > New metric.
2. Enter a meaningful metric name.
3. Leave the metric kind at default (Simple).
4. Change the Metric type to Customer.
5. Change the Aggregator option to Quantile.
6. Next to the Quantile option, in the text field, enter `0.20`.
7. Click Choose value.
8. Search and select the aggregate that returns the value of purchases in last 90 days.
9. Click Enable filter.
10. Click Choose filter.
11. Search and select the segmentation of loyal customers.
12. As the logical operator, select Is true.
13. Confirm the filter settings by clicking Apply.
14. Using the date picker in the lower-right corner, set the time range to Relative time range > Lifetime.
15. Save the metric.
16. Create three metrics with the same settings for the following quantile values: `0.40`, `0.60`, and `0.80`.
## Create a dashboard
In this part of the process, to conveniently preview the results of all metrics you created in the previous part of the process, create a dashboard.
1. Go to Analytics > Dashboard > New dashboard.
2. Enter a meaningful name of the dashboard.
3. Add the 12 metrics you created in the previous part of the process.
4. Save the dashboard.
## Create segmentations
Based on the values returned by the metrics you created before, create three segmentations for Recency, Frequency and Monetary scores. Each of the segmentation contains 5 sub-segmentations. Each sub-segmentation reuses the result of a quantile calculated by the metrics.
The sub-segmentation are named in the following way:
• For Recency score: 500, 400, 300, 200, 100
Sub-segmentation 500 400 300 200 100
Conditions The value lower than
the 0.20 quantile
The value lower than
the 0.40 quantile but higher than 0.20 quantile
The value lower than
the 0.60 quantile but higher than 0.40 quantile
The value lower than
the 0.80 quantile but higher than 0.60 quantile
The value higer than 0.80
quantile
• For Frequency score: 50, 40, 30, 20, 10
Sub-segmentation 50 40 30 20 10
Conditions The value higher than
0.80 quantile
The value lower than
0.80 quantile but higher than 0.60 quantile
The value lower than
0.60 quantile but higher than 0.4 quantile
The value lower than
0.40 quantile but higher than 0.20 quantile
The value lower than 0.20
quantile
• For Monetary score: 5, 4, 3, 2, 1
Sub-segmentation 5 4 3 2 1
Conditions The value higher than
0.80 quantile
The value lower than
0.80 quantile but higher than 0.60 quantile
The value lower than
0.60 quantile but higher than 0.4 quantile
The value lower than
0.40 quantile but higher than 0.20 quantile
The value lower than 0.20
quantile
Customers belong to a specific sub-segmentation under defined circumstances and based on the sub-segmentation classification the overall score will be created, for example, a customer can belong to the following sub-segmentations: 500 (R), 20 (F), 3 (M). In the next part of the process these values will be added in an expression to produce the final RFM score, which in this example will be 523.
### Recency
1. Go to Analytics > Segmentations > New segmentation.
2. Enter a meaningful name of the segmentation.
3. As the name of the sub-segmentation, enter `500`.
4. Click Choose filter.
5. Search and select the expression that returns the number of days since the last transaction.
6. As the logical operator select Less than.
7. In the text field, next to the logical operator, enter the number returned by the metrics that calculate the `0.20` quantile. In this use case, the quantile result amounts to `3.89`, so the customers who made a transaction less than 3 days after the date of the last transaction in the last 90 days, belong to 500. It means that the R score of the customer is 500, which is the highest.
8. Click Choose filter.
9. Select the segment of loyal customers.
10. Join the two conditions by selecting the AND operator.
11. Add the rest of sub-segmentations by clicking the icon.
12. In the conditions of the sub-segmentations:
• For the 400 sub-segmentation: more than the value of the `0.20` quantile AND less than `0.40` quantile AND segment of loyal customers
• For the 300 sub-segmentation: more than the value of the `0.40` quantile AND less than `0.60` quantile AND segment of loyal customers
• For the 200 sub-segmentation: more than the value of the `0.60` quantile AND less than `0.80` quantile AND segment of loyal customers
• For the 100 sub-segmentation: less than the value of the `0.80` quantile AND segment of loyal customers
### Frequency
1. Go to Analytics > Segmentations > New segmentation.
2. Enter a meaningful name of the segmentation.
3. As the name of the sub-segmentation, enter `50`.
4. Click Choose filter.
5. Search and select the expression that returns the average number of transactions per week.
6. As the logical operator select More than or equal.
7. In the text field, next to the logical operator, enter the number returned by the metrics that calculate the `0.80` quantile. In this use case, the quantile result amounts to `0.98`, so the customers who make more than 0.98 transaction a week in the last 90 days, belong to 50. It means that the F score of the customer is 50, which is the highest.
8. Click Choose filter.
9. Select the segment of loyal customers.
10. Join the two conditions by selecting the AND operator.
11. Add the rest of sub-segmentations by clicking the icon.
12. In the conditions of the sub-segmentations:
• For the 40 sub-segmentation: more than the value of the `0.60` quantile AND less than `0.80` quantile AND segment of loyal customers
• For the 30 sub-segmentation: more than the value of the `0.40` quantile AND less than `0.60` quantile AND segment of loyal customers
• For the 20 sub-segmentation: more than the value of the `0.20` quantile AND less than `0.40` quantile AND segment of loyal customers
• For the 10 sub-segmentation: less than the value of the `0.20` quantile AND segment of loyal customers AND transaction event in last 90 days
### Monetary
1. Go to Analytics > Segmentations > New segmentation.
2. Enter a meaningful name of the segmentation.
3. As the name of the sub-segmentation, enter `5`.
4. Click Choose filter.
5. Search and select the aggregate that returns the value of transactions in last 90 days.
6. As the logical operator select More than or equal.
7. In the text field, next to the logical operator, enter the number returned by the metrics that calculate the `0.80` quantile. In this use case, the quantile result amounts to `967,30 PLN`, so the customers who spent this amount of money or more in the last 90 days, belong to 5. It means that the M score of the customer is 5, which is the highest.
8. Click Choose filter.
9. Select the segment of loyal customers.
10. Join the two conditions by selecting the AND operator.
11. Add the rest of sub-segmentations by clicking the icon.
12. In the conditions of the sub-segmentations:
• For the 4 sub-segmentation: more than the value of the `0.60` quantile AND less than `0.80` quantile AND segment of loyal customers
• For the 3 sub-segmentation: more than the value of the `0.40` quantile AND less than `0.60` quantile AND segment of loyal customers
• For the 2 sub-segmentation: more than the value of the `0.20` quantile AND less than `0.40` quantile AND segment of loyal customers
• For the 1 sub-segmentation: less than the value of the `0.20` quantile AND segment of loyal customers AND transaction event in the last 90 days.
## Create an expression
In this part of the process, create an expression that adds the sub-segmentations a customer belongs to, in order to produce the final RFM score.
1. Go to Analytics > Expressions > New expression.
2. Enter a meaningful name of the expression.
3. Leave the expression type at default (Attribute).
4. Build the formula of the expression that adds the three segmentations you created in the previous part of the procedure.
1. Click the Select node.
2. From the dropdown list, select Customer.
3. Click the unnamed node that appeared on the canvas.
4. Scroll down the page and click Choose attribute.
5. On the dropdown list select the Recency segmentation.
6. Next to the segmentation added to the canvas, click the plus button.
7. Repeat steps from 1 to 6 for the Frequency and Monetary segmentations.
8. Click the mathematical operator between the nodes and change it to a plus sign.
5. Save the expression.
Result: The table below contains all possible results:
Possible results
111, 112, 113, 114, 115, 121, 122, 123, 124, 125,131, 132, 133, 134, 135, 141, 142, 143, 144, 145, 151, 152, 153, 154, 155, 211, 212, 213, 214, 215, 221, 222, 223, 224, 225, 231, 232, 233, 234, 235, 241, 242, 243, 244, 245, 251, 252, 253, 254, 255, 311, 312, 313, 314, 315, 321, 322, 323, 324, 325, 331, 332, 333, 334, 335, 341, 342, 343, 344, 345, 351, 352, 353, 354, 355, 411, 412, 413, 414, 415, 421, 422, 423, 424, 425, 431, 432, 433, 434, 435, 441, 442, 443, 444, 445, 451, 452, 453, 454, 455, 511, 512, 513, 514, 515, 521, 522, 523, 524, 525, 531, 532, 533, 534, 535, 541, 542, 543, 544, 545, 551, 552, 553, 554, 555
You need to come up with categories of the customers and assign the results to the specific category. Example categories can be: Top customers, Recent customers, Churn risk, Heavy buyers, Lost heavy buyers, Lost customers.
## Create a RFM segmentation
In the final part of the process, create a segmentation that contains sub-segments. Each sub-segment represents a category of customers. As the condition for each sub-segmentation, use the expression created in the previous part of the process and define the scores that fall into a particular category.
1. Go to Analytics > Segmentations > New segmentation.
2. Enter a meaningful title of the segmentation.
3. Click Choose filter.
4. On the dropdown list, search and select the expression created in the previous part of the process.
5. As the logical operator, select In.
6. Enter the score that fall into a category.
7. Click the icon.
8. Repeat steps from 3 to 7 for the rest of the categories.
9. Save the segmentation.
## Check the use case set up on the Synerise demo workspace
Check the analytics from our use case in Synerise demo workspace:
• Aggregate which counts transactions,
• Aggregate counts first transactions,
• Aggregate counts last transactions,
• Aggregate presented monetary value,
• Expression counts days from last transaction,
• Expression that calculates how much time passed since the last transaction,
• Expression that calculates the number of weeks since the first transaction of a customer,
• Expression that calculates the average number of transactions a customer made per week.
See the metrics which will calculate the qunatiles for the values of the expressions and an aggregate"
Check the dashboard presenting RFM summary.
Check three segments, created in this case:
Check the RFM general score.
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Number of inequalities to solve: 23456789
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solving equations by square root property
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Author Message
Cinvil Dlajon Briodx
Registered: 16.10.2003
From: NL
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Registered: 21.03.2005
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Registered: 24.10.2003
From: Where the trout streams flow and the air is nice
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Vild
Registered: 03.07.2001
From: Sacramento, CA
Posted: Monday 01st of Jan 09:24 I am a regular user of Algebrator. It not only helps me complete my homework faster, the detailed explanations offered makes understanding the concepts easier. I recommend using it to help improve problem solving skills. | 1,010 | 4,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-13 | latest | en | 0.890003 |
https://piping-designer.com/index.php/properties/546-surface-tension | 1,686,397,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00498.warc.gz | 515,229,163 | 8,298 | # Surface Tension
on . Posted in Classical Mechanics
Surface tension, abbreviated as $$\sigma$$ (Greek symbol sigma), is the force that acts on the surface of a liquid, tending to minimize its surface area. It is caused by the cohesive forces between the liquid molecules, which are stronger at the surface than in the bulk of the liquid.
The surface tension of a liquid is a measure of its ability to resist deformation and is dependent on the type of liquid, temperature, and pressure. Surface tension is responsible for many interesting phenomena, such as the formation of droplets, the shape of soap bubbles, and the ability of insects to walk on water. The surface tension formula is widely used in physics, chemistry, and engineering to describe the behavior of liquids at interfaces, and to design and optimize various applications involving fluid interfaces, such as emulsions, foams, and coatings.
## Surface tension formula
$$\large{ \sigma = \frac{F}{l} }$$
Symbol English Metric
$$\large{ \sigma }$$ (Greek symbol sigma) = surface tension $$\large{\frac{lbf}{ft}}$$ $$\large{\frac{N}{m}}$$
$$\large{ F }$$ = force acting perpendicular to the surface $$\large{ lbf }$$ $$\large{N}$$
$$\large{ l }$$ = length of the surface over which the force acts $$\large{ ft }$$ $$\large{ m }$$ | 313 | 1,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-23 | longest | en | 0.904103 |
https://www.raiseupwa.com/users-questions/what-is-a-product-of-1/ | 1,718,461,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00695.warc.gz | 859,450,871 | 32,153 | Categories :
## What is a product of 1?
THE PRODUCT OF 1 AND ANY NUMBER IS THE NUMBER ITSELF. FOR EXAMPLE, 1×2=2.
## What is the product between 12 and 3?
Answer: The difference between 12 and 3 is 12 subtract 3 which is 9. The quotient of 12 and 3 is 12 divided by 3 which is 4. Now work out the product of 4 and 9 by multiplying them together to give 36.
How do you calculate product?
The product of two numbers is the result you get when you multiply them together. So 12 is the product of 3 and 4, 20 is the product of 4 and 5 and so on.
Why is product of No 1?
Allowing a “product” with zero factors reduces the number of cases to be considered in many mathematical formulas. For these reasons, the “empty product is one” convention is common practice in mathematics and computer programming.
### What is the product of anything times zero?
Multiplication by Zero Multiplying by 0 makes the product equal zero. The product of any real number and 0 is 0 .
### How do you write a product?
8 Easy Rules to Write Product Descriptions That Sell
1. Know Who Your Target Audience is.
2. Focus on the Product Benefits.
3. Tell the Full Story.
4. Use Natural Language and Tone.
5. Use Power Words That Sell.
6. Make it Easy to Scan.
7. Optimize for Search Engines.
8. Use Good Product Images.
What is the product of 59 736 and 600?
The product of 59.736 and 600 is – Gauthmath.
What number gives the same amount when it is added to 3 as when it is multiplied by 3?
Step-by-step explanation: 3/2 is the number that gives the same amount when it is added to 3 as when it is multiplied by 3.
#### What is the formula for total product cost?
Total product costs can be determined by adding together the total direct materials and labor costs as well as the total manufacturing overhead costs.
#### What is the formula of total product?
It refers to the total amount of output that a firm produces within a given period, utilising given inputs. It is output per unit of inputs of variable factors. Average Product (AP)= Total Product (TP)/ Labour (L). It denotes the addition of variable factor to total product.
Why is empty sum zero?
with no terms evaluates to 0. Allowing a “sum” with only 1 or 0 terms reduces the number of cases to be considered in many mathematical formulas. For these reasons, the “empty sum is zero” extension is standard practice in mathematics and computer programming (assuming the domain has a zero element).
Which is the final product 1 / 3 or 3 / 5?
Multiplying 1/3 and 3/5 together, three’s will cancel and you’ll be left with 1/5. 1/5 is the final product. The Rock admits this was the best decision he ever made. The big companies don’t want you to know his secrets.
## What do you need to know about 3 in one products?
With products that call out their specific function and area of use, 3-IN-ONE ® makes it easy to select the right solution for the project at hand. See where 3-IN-ONE ® is available near you.
## What is the value of the product in n terms?
The neat thing as has already been pointed out is for the general expression in n terms. The product will be n!/ (n+1)! or 1/ (n+1) for 4 terms, (n=4) this is 1/5, for 9236 terms, it will be 1/9237. It’s quite a cute problem when generalised.
How to calculate the value of a product?
Multiplying 1/2 and 2/3 together, two’s will cancel and you’ll be left with a product of 1/3. Multiplying 3/4 and 4/5 together, four’s will cancel and you’ll be left with 3/5. Now, I’ll multiply both of those products together. Multiplying 1/3 and 3/5 together, three’s will cancel and you’ll be left with 1/5. 1/5 is the final product. | 932 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.9369 |
https://www.jiskha.com/display.cgi?id=1457477397 | 1,529,885,664,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00089.warc.gz | 836,838,535 | 3,796 | # math
posted by jacobe
Graysen has a green number cube and a white number cube. the faces of the cubes are numbered 1 through 6. Graysen will roll each cube one time, what is the probability that the green cube will land with even number face up and the white cube will land with a number greater than 2 face up
1. John
And so we will multiply the two probabilities
green even 3/6 or 1/2
white greater than 2 4/6 or 2/3
probability is 1/2 times 2/3 = 2/6 or 1/3 or as a decimal .333
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More Similar Questions | 765 | 2,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-26 | latest | en | 0.916062 |
https://www.jiskha.com/questions/568/27-letters-in-1988-18-letters-in-1989-what-is-of-change-need-to-know-how-you | 1,556,232,598,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578742415.81/warc/CC-MAIN-20190425213812-20190425235812-00328.warc.gz | 729,191,012 | 5,292 | math
27 letters in 1988, 18 letters in 1989 what is % of change?
*need to know how you arrive at the answer
Didide the change (-9) by the original value (27), and convert that fraction to a percentage.
1. 👍 0
2. 👎 0
3. 👁 27
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thesolarpanelcalculator.com | 1,480,948,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541697.15/warc/CC-MAIN-20161202170901-00279-ip-10-31-129-80.ec2.internal.warc.gz | 269,973,683 | 5,782 | # Calculating The Cost Of Going With Solar Energy
There are so many things to know about using solar energy. Technology is changing and improving so quickly that many people have only a vague notion of how solar energy works and just how efficient it really is. In this article we will describe some of the ideal conditions for using solar energy to power your home, vacation cabin, recreation vehicle and more. We will also discuss the components of a good solar energy system and talk about how to use a solar panel calculator. Read on to learn more.
What are ideal conditions, and what can solar do?
Many people harbor glaring misconceptions about solar energy. On one end of the spectrum are people who believe that solar is wildly inefficient and cannot be used to power a whole house or other large application. On the other end are people who want to use inadequate solar energy systems to recharge spent batteries and perform other very challenging tasks quickly. Both of these points of view are incorrect. The best use of a properly planned and installed solar energy system is to provide for ongoing, sensible use.
Naturally, with even the best system, the most electricity will be produced on days when the panels receive full sun for many hours. Remember that the ratings for solar panels are calculated considering direct, bright sunshine. If your circumstances differ from this ideal (e.g. partial shade, short days, fog, overcast, etc.) your results will vary.
Energy storage is possible to an extent; however, you cannot expect massive bursts of energy from your system. Making use of a solar panel calculator to determine exactly how many panels you will need for your intended use can help you plan your system properly to get efficient, satisfactory use. Below are some of the other considerations you must take into account when planning your system.
What you need to know about voltage rating
When you look at solar energy products, you will see that the majority of them are designed to accommodate 12 volts direct current (VDC). You may be surprised to know that there are also smaller 24 volt panels available. If this size is needed, solar panels can be wired in a series.
Alternately, even greater sizes, such as 24, 36 and 48 direct current (DC) volts can be had. Solar controllers prevent overcharge and battery discharge as well as improving charge quality.
If you are using panels that are rated over 5 or 6 watts output, you should use a charge controller. This device acts as and ON/OFF switch. It lets power pass through when it is needed by the battery and shuts it off after the battery has become fully charged. Be aware that controllers are rated using AMPS. Solar panels are rated using watts. Generally speaking a 6 AMP controller will work with panels up to 70 watts in power.
Determining the size of solar panels you will need
When you are trying to figure out what size solar panels to purchase, you must understand that you simply have to do the math. You must determine how much power you need and compare this with the amount of power various sizes of solar panels are capable of producing. You will need to know the number of AMP hours or watts your system is able to produce in a given time-frame. You could measure by hours, or you could measure by days. Using a twenty four hour day as a baseline is a practical approach. Figure out how much electricity you will need in a period of twenty four hours. Determine how much direct sunlight your solar energy system will receive during that same twenty four hour period. When you have come up with some numbers, you are ready to use a good system to determine the exact size of panels you will need.
This article has covered some basic information about selecting and using a solar energy system. When you are wondering about the efficiency and cost of switching to solar, once you have determined your energy needs, the smartest thing you can do is seek out a reliable online calculator to help you determine how much your ideal system will cost and what you can expect in terms of performance. Some calculators also provide you comparisons between your existing system and solar energy in terms of environmental impact. Once you have gotten some solid data on the cost, efficiency and impact of making this smart switch, you should be good to go and well prepared to get started using energy from the sun. | 872 | 4,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-50 | longest | en | 0.944116 |
https://math.stackexchange.com/questions/3261733/properties-of-functions-in-w1-p-omega-and-their-weak-derivatives | 1,563,417,432,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.64/warc/CC-MAIN-20190718022001-20190718044001-00189.warc.gz | 460,559,981 | 36,169 | # Properties of Functions in $W^{1,p}(\Omega)$ and Their Weak Derivatives
Let $$\Omega \subset \mathbb{R}^{N}$$ and $$W^{1,p}(\Omega)$$ be Sobolev Space. Then we let $$u\in W^{1,p}(\Omega)$$ and define
(i) $$u^{+} := \max\{0,u\}$$
(ii) $$u^{-} := \max\{0,-u\}$$
Then, we claim that $$u^{+}, u^{-}\in W^{1,p}(\Omega)$$ and we have $$$$\tag{1} \nabla(u^{+}) = \begin{cases} \nabla u &\text{ if }u>0 \\ 0 &\text { if }u\leq0 \end{cases}$$$$ and $$$$\tag{2} \nabla(u^{-}) = \begin{cases} \nabla u &\text{ if }u<0 \\ 0 &\text { if }u\geq0 \end{cases}$$$$ almost everywhere in $$\Omega$$.
How to show that $$u^{+},u^{-}\in W^{1,p}(\Omega)$$? I know that $$||u^{+}||_{p} \leq ||u||_{p}$$ and $$||u^{-}||_{p} \leq ||u||_{p}$$ but I do not know how to show that $$||\nabla u^{+}||_{p}$$ and $$||\nabla u^{-}||_{p}$$ are bounded. Also how to show that (1) and (2) hold true? Any hint is much appreciated
Thank you very much!
It is easy to see that $$u^+$$ or $$u^-$$ have as much weak derivatives as $$u$$, just set
$$\Omega^+:=\{x\in\Omega: u(x)\ge 0\},\quad\Omega^-:=\{x\in\Omega: u(x)\le 0\}$$
Hence
$$\int_{\Omega} u^+(x)\partial_j\varphi(x)\, dx=\int_{\Omega} u(x)\partial_j\varphi(x)\chi_{\Omega^+}(x)\, dx=-\int_{\Omega}\partial_j u(x)\chi_{\Omega^+}(x)\varphi(x)\, dx$$
where $$\partial_j u$$ is the $$j$$-weak partial derivative of $$u$$, and $$\varphi$$ is any test function (it holds because we can choose any test function with compact support in $$\Omega^+$$, hence its easy to see that it holds for any test function).
Thus $$\partial_j u\chi_{\Omega^+}$$ is the $$j$$-weak partial derivative of $$u^+$$, hence it follows that $$\nabla u^+=\nabla u\chi_{\Omega^+}$$.
Thus clearly $$\|\nabla u^+\|_p\le\|\nabla u\|_p$$, just note that
$$|\nabla u^+(x)|^p=|\nabla u(x)|^p|\chi_{\Omega^+}(x)|^p\le |\nabla u(x)|^p$$
for all $$x\in\Omega$$ and all $$p\ge 0$$.
The same can be shown for the other case.
• I dont think $\partial_{j}(u(x))\chi_{\Omega^{+}}(x)= \partial_{j} (u(x)\chi_{\Omega^{+}}(x))$ – Evan William Chandra Jun 14 at 3:41
• @EvanWilliamChandra me neither, I didn't said that. Note that $$\int_\Omega f\chi_A=\int_{\Omega\cap A} f$$ – Masacroso Jun 14 at 3:42
• Now I get it! Thank you very much! – Evan William Chandra Jun 14 at 3:44
• @EvanWilliamChandra I edited to add some more explanation to see clearly why the identity with the integrals holds. – Masacroso Jun 14 at 3:51 | 947 | 2,407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 32, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-30 | latest | en | 0.653753 |
https://www.coursehero.com/file/6784153/Smore-Stoichiometry-Pre-ap/ | 1,513,541,707,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597485.94/warc/CC-MAIN-20171217191117-20171217213117-00273.warc.gz | 711,559,728 | 64,463 | S'more Stoichiometry Pre-ap
# S'more Stoichiometry Pre-ap - Name Date Lab Station...
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Name ___________________________ Date _________________ Lab Station ______ Sign-Off ______ Group Members _______________________________________________________________________ S’more Stoichiometry Introduction : The building of s’mores is using the same process of stoichiometry as in any chemical reaction. The amount of ingredients determines how many s’mores can be made. The amount of the limiting reactant determines how many s’mores or the amount of product that you can make. It’s like 10 hot dogs but only eight hot dog buns! What is up with that? The limiting reactant (reagent) is the reactant that ___________ the amounts of the other reactants that can combine – and the amount of product that can form – in a chemical reaction (the reactant that runs out __________). The reaction will stop once the limiting reactant is used up. The _____________ reactant in a chemical reaction is the substance that is _______ used up completely in a chemical reaction (the reactant that you have leftovers of). Example: For the reaction: ___ K + ___ O 2 ___ K 2 O; if 0.50 g of K are allowed to react with 0.10 g of O 2 , what is the limiting reagent, and what is the reactant in excess? The following are the symbols you will use for your ingredients: Gc = Graham Crackers Cs = Chocolate squares Mm = Marshmallows Sm = S’more The following ingredients are needed to make one s’more: 2 Graham Crackers 3 Squares of chocolate 1 Marshmallow The balanced equation to make 1 smore is as follows: 2 Gc + 3 Cs + 1 Mm 1 Gc 2 Cs 3 Mm Data : Bag # __________ Number of Graham Crackers = ___________ # of S’mores = ___________ Number of Marshmallows = ____________ # of S’mores = ___________ Number of chocolate pieces = ___________
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Ask a homework question - tutors are online | 547 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-51 | latest | en | 0.878085 |
https://www.nottingham.ac.uk/xpert/scoreresults.php?keywords=French%20year%201%20semester%252&start=11220&end=11240 | 1,571,334,715,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00121.warc.gz | 1,032,217,229 | 24,880 | So far you have come across exercises and brain stretcher puzzles. The exercises are designed to help you practise certain skills and the puzzles are to get you to think a bit more generally. However now here is a different type of activity: an investigation.
Investigations are usually much more open-ended than exercises. With an investigation you often need to stop and think about what information you will need to tackle the task, and how you should interpret the results of your calcul
Author(s): The Open University
While reading a newspaper article, I noticed some examples of how prices changed in the 10 years from 1984 to 1994. The table below shows the typical prices that you would have expected to pay in 1984 and in 1994 for a pint of milk and a Ford Fiesta motor car.
Price in 1984 Price in 1994
Pint o
Author(s): The Open University
Many people see calculators only as a way of producing answers—indeed some people see them almost as a means of cheating, of short-cutting procedures that can and should be carried out in one's head or on paper. However, the calculator can also be a means of learning mathematics more effectively, something you will come to appreciate more. Many previous mathematics students have found that their graphics calculator, used with understanding and intelligence, has become a most effective aid t
Author(s): The Open University
Now try the quiz and see if there are any areas you need to work on.
Author(s): The Open University
By the end of this unit you should be able to:
• evaluate the squares, cubes and other powers of positive and negative numbers with or without your calculator;
• estimate square roots and calculate them using your calculator;
• describe the power notation for expressing numbers;
• use your calculator to find powers of numbers;
• multiply and divide powers of the same number;
• understand and apply negative powers, t
Author(s): The Open University
1 Look at the diagram below and answer the following questions:
• (a) Write down the coordinates of the points P, Q, R, S and T.
• (b) On this diagram,
Author(s): The Open University
Working in mathematics education involves a sense of both past and future, and how the two combine to influence the present. It may seem that, because the past has already happened, it cannot be altered; however, you can alter how you perceive the past, and what lessons you take from it. Each of us has a personal past in mathematics education—the particular events of our personal lives, who taught us, where, what and how they taught us, and what we took from the experiences. Each of us also
Author(s): The Open University
This unit focuses on your initial encounters with research. It invites you to think about how perceptions of mathematics have influenced you in your prior learning, your teaching and the attitudes of learners.
Author(s): The Open University
In this unit you have been introduced to the difference between mathematical content and processes. You have worked on the do–talk–record (DTR) framework for learning mathematics.
Author(s): The Open University
Assuming that both the content of mathematics and the processes need to be included in programmes and curricula, the problem becomes one of how a suitable curriculum can be structured. One possibility is to construct a very specific curriculum with clearly defined objectives for both content and processes separately, and possibly with suggested learning activities. However, content and process are two complementary ways of viewing the subject.
An alternative is to see the curriculum in
Author(s): The Open University
After studying this unit you should:
• be able to perform basic algebraic manipulation with complex numbers;
• understand the geometric interpretation of complex numbers;
• know methods of finding the nth roots of complex numbers and the solutions of simple polynomial equations.
Author(s): The Open University
Mailing or discussion lists are email-based discussion groups. When you send an email to a mailing list address, it is sent automatically to all the other members of the list.
The majority of academic-related mailing lists in the UK are maintained by Jiscmail. You will find details of joining these mailing lists on the Jiscmail website. Mailing lists are useful for getting in touch with like-minded colleagues. They are also handy for keeping up to date with current thinking and research
Author(s): The Open University
Online bookshops and some of the major search engines offer ‘Alerts’ services. These work by allowing you to set up a profile once you have registered on their site, and when there are items meeting your criteria you receive an email. The good thing about alerts is that you don’t have to do anything once you have set up your profile. The downside, particularly with alerts services from the search engines, is that given the extent to which internet traffic is on the increase whether new
Author(s): The Open University
Referencing is not only useful as a way of sharing information, but also as a means of ensuring that due credit is given to other people’s work. In the electronic information age, it is easy to copy and paste from journal articles and web pages into your own work. But if you do use someone else’s work, you should acknowledge the source by giving a correct reference.
Taking someone's work and not indicating where you took it from is termed plagiarism and is regarded as an infringemen
Author(s): The Open University
If you are considering taking your studies further you might like to consider using bibliographic software. Bibliographic software can be used to sort references, annotate them, manage quotations or create reading lists.
There are several software packages on the market. Some are listed below.
• BibTex
• EndNote
• Procite
• Reference Manager
• RefWorks
If you are not sure
Author(s): The Open University
We mentioned above that we need to reference sources to ensure we abide by copyright legislation. But there is another reason we need to give accurate references to items we use – so we can share it.
Consider this scenario. A friend says they’ve just read an interesting article where Joshua Schachter, founder of del.icio.us has spoken about why it isn’t a faceted search system, and you should read it. How would you go about finding it? Would you start looking in a news database, a
Author(s): The Open University
If you find you have a long unmanageable list of favourites/bookmarks you might like to try social bookmarks as an alternative.
## Activity – what you need to know about social bookmarks
Author(s): The Open University
1.5.4The 5 Ds
If you don’t use a system at all, then you could suffer from the effects of information overload:
• losing important information
• wasting time on trying to find things
• ending up with piles of physical and virtual stuff everywhere
One technique you might like to apply to your files (be they paper or electronic) is the 5Ds. Try applying these and see if you can reduce your information overload.
Author(s): The Open University
1.5.3 Desktop search tools
Finding your paperwork or electronic files can be a problem. You may find that even if you do have some sort of filing system, your structure soon gets quite large with files in multiple locations, which can be hard to navigate. You may find yourself making arbitrary decisions about which folder to place a document in. It may make sense now but in the future, when you look where you think it should be, it’s not there.
At times like this you may resort to the search command from the Wi
Author(s): The Open University
1.5.1 Why is it important to be organised?
• 87% of items that are filed into a filing cabinet are never looked at again. STANFORD UNIVERSITY
• The world is producing nearly two exabytes of new and unique information every year – an exabyte is a new term that had to be coined for a billion gigabytes. All the words ever spoken by human beings comes to five exabytes. UNIVERSITY OF CALIFORNIA (BERKELEY)
• More new information has been produc
Author(s): The Open University | 1,705 | 8,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-43 | latest | en | 0.956647 |
https://ideal.accelerate-ed.com/modulecourseguide/index/e35d6cf3-0bd6-4d4e-a103-62bf8bc87d00 | 1,607,158,618,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00378.warc.gz | 319,745,442 | 7,426 | # Module 4 - Math 4A - Course Guide
## Module Overview:
In this module students will learn about how to multiply using clustering and how to use regrouping in multiplication. They will learn about how to multiply with arrays and how to multiply large numbers many different ways. Students will learn about long division, dividing using grouping, using an area model to solve division problems, and how to apply division strategies. Students will become familiar with the parts of fractions and writing fractions.
## Module Materials:
Lesson # Lesson Title Material(s)
1 Clustering Scissors
2 Regrouping Multiplication Math Notebook
3 Multiplying with Arrays Graph paper
4 Applications of Multiplication Math Notebook
5 Long Division Math Notebook
6 Dividing Using Grouping Math Notebook
7 Area Model Graph paper Markers
8 Application of Division Strategies Math Notebook
9 Fraction Parts Scissors
10 Fraction Shape Crayons or Markers
11 Writing Fractions Math Notebook
Module Objectives:
Lesson # Lesson Title Objective(s)
1 Clustering
1. Multiply numbers using clustering
2 Regrouping Multiplication
1. Multiply numbers using re-grouping.
3 Multiplying with Arrays
1. Multiply numbers using the arrays model.
4 Applications of Multiplication
1. Apply any of the multiplication strategies to solve real-life problems.
5 Long Division
1. Divide numbers using a traditional algorithm.
6 Dividing Using Grouping
1. Divide numbers using groups for the divisor.
7 Area Model
1. Divide numbers using area models.
8 Application of Division Strategies
1. Apply any of the division strategies to solve real-life problems.
9 Fraction Parts
1. Define fraction; numerator; denominator; fraction bar.
10 Fraction Shape
1. Identify the number of shaded parts and the number of equal parts in a shape (circle; rectangle).
11 Writing Fractions
1. Write a fraction using mathematical notation and using words.
Module Key Words:
Key Words
multiplication
clustering
commutative
re-grouping
array
math
large numbers
division
quotient
dividend
divisor
division bracket
grouping
area model
fraction
equivalent
numerator
denominator
fraction bar
Fraction Line
Modeling
Module Assignments:
Lesson # Lesson Title Page # Assignment Title
1 Clustering 7 Clustering with Tiles Activity
2 Regrouping Multiplication 4 Re-grouping Activity
3 Multiplying with Arrays 8 Array Practice
4 Applications of Multiplication 2 Applications Practice
7 Area Model 7 Modeling Practice
10 Fraction Shape 3 Modeling Fractions Practice
## Learning Coach Notes:
Lesson # Lesson Title Notes
1 Clustering In this lesson have your student complete the practice activity first before looking at the answer key. Be sure to monitor this. When they check their answers using the key have them circle what they got wrong and talk about why they got it wrong and make corrections.
2 Regrouping Multiplication In this lesson have your student complete the practice activity first before looking at the answer key. Be sure to monitor this. When they check their answers using the key have them circle what they got wrong and talk about why they got it wrong and make corrections. In their Math notebook have your student describe regrouping and the write the steps to multiplying four digit numbers.
3 Multiplying with Arrays This lesson will have your student using graph paper there is a downloadable version to print if you do not have graph paper on hand. In this lesson have your student complete the practice activity first before looking at the answer key. Be sure to monitor this. When they check their answers using the key have them circle what they got wrong and talk about why they got it wrong and make corrections.
4 Applications of Multiplication In this lesson have your student complete the practice activity first before looking at the answer key. Be sure to monitor this. When they check their answers using the key have them circle what they got wrong and talk about why they got it wrong and make corrections. In their math notebook have your student write down the strategies they used to solve the problems in this lesson.
5 Long Division In their Math notebook have your student write down the 4 steps to solving a long division problem and have them diagram the parts of a division problem.
6 Dividing Using Grouping In their Math notebook have them describe how to use grouping strategies to solve division problems.
7 Area Model This lesson will have your student using graph paper there is a downloadable version to print if you do not have graph paper on hand. In this lesson have your student complete the practice activity first before looking at the answer key. Be sure to monitor this. When they check their answers using the key have them circle what they got wrong and talk about why they got it wrong and make corrections.
8 Application of Division Strategies In this lesson your student will be given practice problems to solve using any of the division strategies they have learned. Sit with them while they solve the problem. Do not allow them to click on the answer until after they have solved the problem and discussed what strategy they used to solve the problem with you. When they have finished solving the problem and discussed their strategy allow them to click on the answer to compare. In their Math notebook have your student write down the division strategies they can use to solve division problems.
9 Fraction Parts One of the activities in this lesson ask the student to practice writing fractions. As your student writes the fractions have them identify for you the numerator and the denominator in each fraction they write.
10 Fraction Shape Give your student 5 fractions and ask them to create a model for each fraction in their Math notebook.
11 Writing Fractions Write 5 fractions in your students math notebook and have them read them and label the parts. Then say a fraction and ask your student to write the fractions you tell them.
## Module Guiding Questions:
When a student starts a lesson ask them questions to check for prior knowledge and understanding and to review concepts being taught. At the end of the lesson ask the questions again to see if their answer changes.
Lesson Title Question
Clustering
1. What is clustering?
Regrouping Multiplication
1. What is re-grouping in multiplication?
2. What is the commutative property?
Multiplying with Arrays
1. What is an array?
Applications of Multiplication
1. What are the different ways you can multiply large numbers?
Long Division
1. How do you solve long division problems?
Dividing Using Grouping
1. How do you divide using grouping?
Area Model
1. What is an area model?
2. What does it mean that division is known as the inverse of multiplication?
Application of Division Strategies
1. What are some strategies you can use to solve division problems?
Fraction Parts
1. What are the parts of a fraction?
Fraction Shape
1. How does modeling help with understanding fractions?
Writing Fractions
1. How do you write fractions?
## Module Video Questions:
When a student watches a video take time to ask them questions about what they watched. Suggested questions for the videos in this module are listed here. Suggestion: Have the student watch the entire video first all the way through. Then have them watch the video a second time, as they watch it pause the video and ask the questions.
Lesson Title Video Question
Clustering Maths Mansion
1. What are the problems the kids have to solve to get out of the Maths Mansion?
2. Can you solve them before the kids do?
Regrouping Multiplication Four Digit Multiplication
1. What are the steps involved in multiplying four digit numbers?
Multiplying with Arrays 1001 Naughts
1. How are arrays used to multiply in this video?
Multiplying with Arrays Completing Multiplication Sentences Using Arrays
1. How does an array represent multiplication sentences?
Long Division Division and Multiplication - Beads
1. How can beads teach us the properties of division?
Long Division Division Procedures
1. What are the steps to long division?
Dividing Using Grouping Dividing Using Grouping
1. What is a grouping strategy?
Area Model Understanding the Relationship Between the Remainder and the Divisor
1. How can you use base ten blocks to understand the relationship between the remainder and the divisor?
Application of Division Strategies Dividend, Divisor and Quotient
1. What is the dividend?
2. What is the divisor?
3. What is the quotient?
Application of Division Strategies Division Word Problems
1. What are the steps to solving a word problem?
2. What are some strategies you can use to solve division word problems?
Fraction Parts Fractions for Food
1. What food was broken into fractions?
Fraction Parts Parts of a Whole
1. What is a fraction?
2. What is the numerator of a fraction?
3. What is the denominator of a fraction?
Fraction Shape Modeling Fractions
1. Describe how you model fractions.
Writing Fractions Read and Write Fractions
1. What is a fraction?
2. What is the denominator?
3. What is the numerator?
4. How do you read a fraction?
5. How do you write a fraction?
## Module Suggested Read Aloud Books:
Take time to read to your student or have them read aloud to you. Read a different book each day. While reading the book point out concepts being taught. You may purchase these books or find them at your local library. Suggested things to discuss while reading the book:
• What is the main idea?
• What are three things new you learned?
• How does this book relate to what you are learning about?
# Book Author Lexile Level
1 Amanda Bean's Amazing Dream Cindy Neuschwander AD520L
2 How Do You Count a Dozen Ducklings? Seon Chae
3 Bean Thirteen Matthew McElligott 470L
4 The Lion's Share Matthew McElligott 550L
5 Full House: An Invitation to Fractions Dayle Ann Dodds 530L
6 A Fraction's Goal ― Parts of a Whole Brian P. Cleary 620L
## Module Outing:
Take some time to apply what your student is learning to the real world. Suggested outings are below.
# Outing | 2,085 | 10,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-50 | longest | en | 0.695804 |
http://www.extremeoptimization.com/QuickStart/FSharp/NonlinearCurveFitting.aspx | 1,653,431,728,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00738.warc.gz | 75,690,927 | 7,694 | Data Analysis Mathematics Linear Algebra Statistics
New Version 8.1!
Supports .NET 6.0. Try it for free with our fully functional 60-day trial version.
QuickStart Samples
# Nonlinear Curve Fitting QuickStart Sample (F#)
Illustrates nonlinear least squares curve fitting of predefined and user-defined curves using the NonlinearCurveFitter class in F#.
```// Illustrates nonlinear least squares curve fitting using the
// NonlinearCurveFitter class in the
// Extreme.Mathematics.Curves namespace of the Extreme
// Optimization Numerical Libraries for .NET.
#light
open System
open Extreme.Mathematics
// The curve fitting classes reside in the
// Extreme.Mathematics.Curves namespace.
open Extreme.Mathematics.Curves
// The predefined non-linear curves reside in the
// Extreme.Mathematics.Curves namespace.
open Extreme.Mathematics.Curves.Nonlinear
// Vectors reside in the Extreme.Mathemaics.LinearAlgebra
// namespace
open Extreme.Mathematics.LinearAlgebra
// The non-linear least squares optimizer resides in the
// Extreme.Mathematics.Optimization namespace.
open Extreme.Mathematics.Optimization
// Nonlinear least squares fits are calculated using the
// NonlinearCurveFitter class:
let fitter = NonlinearCurveFitter()
// In the first example, we fit a dose response curve
// to a data set that includes error information.
// The data points must be supplied as Vector objects:
let dose =
Vector.Create(1.46247, 2.3352, 4.0,
7.0, 12.0, 18.0, 23.0, 30.0, 40.0, 60.0, 90.0, 160.0, 290.0, 490.0, 860.0)
let response =
Vector.Create(95.49073, 95.14551, 94.86448,
92.66762, 85.36377, 74.72183, 62.76747, 51.04137, 38.20257,
28.01712, 19.40086, 13.18117, 9.87161, 7.64622, 7.21826)
let error =
Vector.Create(4.74322, 4.74322, 4.74322,
4.63338, 4.26819, 3.73609, 3.13837, 3.55207, 3.91013,
2.40086, 2.6, 3.65906, 2.49358, 2.38231, 2.36091)
// You must supply the curve whose parameters will be
// fit to the data. The curve must inherit from NonlinearCurve.
// The FourParameterLogistic curve is one of several
// predefined nonlinear curves:
let doseResponseCurve = FourParameterLogisticCurve()
// Now we set the curve fitter's Curve property:
fitter.Curve <- doseResponseCurve
// The GetInitialFitParameters method sets the curve parameters
// to initial values appropriate for the data:
fitter.InitialGuess <- doseResponseCurve.GetInitialFitParameters(dose, response)
// and the data values:
fitter.XValues <- dose
fitter.YValues <- response
// The GetWeightVectorFromErrors method of the WeightFunctions
// class lets us convert the error values to weights:
fitter.WeightVector <- WeightFunctions.GetWeightVectorFromErrors(error)
// The Fit method performs the actual calculation.
let result = fitter.Fit()
// The standard deviations associated with each parameter
// are available through the GetStandardDeviations method.
let s = fitter.GetStandardDeviations()
// We can now print the results:
printfn "Dose response curve"
printfn "Initial value: %10.6f +/- %.4f" doseResponseCurve.InitialValue s.[0]
printfn "Final value: %10.6f +/- %.4f" doseResponseCurve.FinalValue s.[1]
printfn "Center: %10.6f +/- %.4f" doseResponseCurve.Center s.[2]
printfn "Hill slope: %10.6f +/- %.4f" doseResponseCurve.HillSlope s.[3]
// We can also show some statistics about the calculation:
printfn "Residual sum of squares: %A" (fitter.Residuals.Norm())
// The Optimizer property returns the MultidimensionalOptimization object
// used to perform the calculation:
printfn "# iterations: %d" fitter.Optimizer.IterationsNeeded
printfn "# function evaluations: %d" fitter.Optimizer.EvaluationsNeeded
printfn ""
//
// Defining your own nonlinear curve
//
// In this example, we use one of the datasets (MGH10)
// from the National Institute for Statistics and Technology
// (NIST) Statistical Reference Datasets.
// See http://www.itl.nist.gov/div898/strd for details
let fitter2 = NonlinearCurveFitter()
// Here, we need to define our own curve.
// The MyCurve class is defined below.
// This is our nonlinear curve implementation. For details, see
// http://www.itl.nist.gov/div898/strd/nls/data/mgh10.shtml
// You must inherit from NonlinearCurve:
type MyCurve() as this =
// Call the base constructor with the number of parameters.
inherit NonlinearCurve(3)
do
// It is convenient to set common starting values
// for the curve parameters in the constructor:
this.Parameters.[0] <- 0.2
this.Parameters.[1] <- 40000.0
this.Parameters.[2] <- 2500.0
override this.ValueAt x =
this.Parameters.[0] * exp(this.Parameters.[1] / (x + this.Parameters.[2]))
override this.SlopeAt x =
this.Parameters.[0] * this.Parameters.[1] * exp(this.Parameters.[1] / (x + this.Parameters.[2]))
/ (pown (x + this.Parameters.[2]) 2)
// The FillPartialDerivatives evaluates the partial derivatives
// with respect to the curve parameters, and returns
// the result in a vector. If you don't supply this method,
// a numerical approximation is used.
override this.FillPartialDerivatives (x, f) =
let exp = Math.Exp(this.Parameters.[1] / (x + this.Parameters.[2]))
f.[0] <- exp
f.[1] <- this.Parameters.[0] * exp / (x + this.Parameters.[2])
f.[2] <- -this.Parameters.[0] * this.Parameters.[1] * exp / (pown (x + this.Parameters.[2]) 2)
fitter2.Curve <- MyCurve()
// The data is provided as Vector objects.
// X values go into the XValues property...
fitter2.XValues <- Vector.Create(
[|
5.000000E+01; 5.500000E+01; 6.000000E+01; 6.500000E+01;
7.000000E+01; 7.500000E+01; 8.000000E+01; 8.500000E+01;
9.000000E+01; 9.500000E+01; 1.000000E+02; 1.050000E+02;
1.100000E+02; 1.150000E+02; 1.200000E+02; 1.250000E+02
|])
// ...and Y values go into the YValues property:
fitter2.YValues <- Vector.Create(
[|
3.478000E+04; 2.861000E+04; 2.365000E+04; 1.963000E+04;
1.637000E+04; 1.372000E+04; 1.154000E+04; 9.744000E+03;
8.261000E+03; 7.030000E+03; 6.005000E+03; 5.147000E+03;
4.427000E+03; 3.820000E+03; 3.307000E+03; 2.872000E+03
|])
fitter2.InitialGuess <- Vector.Create(fitter2.Curve.Parameters.ToArray())
fitter2.WeightVector <- null
// The Fit method performs the actual calculation:
let result2 = fitter2.Fit()
// A Vector containing the parameters of the best fit
// can be obtained through the
// BestFitParameters property.
let solution2 = fitter2.BestFitParameters
let s2 = fitter2.GetStandardDeviations()
printfn "NIST Reference Data Set"
printfn "Solution:"
printfn "b1: %20f %20f" solution2.[0] s2.[0]
printfn "b2: %20f %20f" solution2.[1] s2.[1]
printfn "b3: %20f %20f" solution2.[2] s2.[2]
printfn "Certified values:"
printfn "b1: %20f %20f" 5.6096364710E-03 1.5687892471E-04
printfn "b2: %20f %20f" 6.1813463463E+03 2.3309021107E+01
printfn "b3: %20f %20f" 3.4522363462E+02 7.8486103508E-01
// Now let's redo the same operation, but with observations weighted
// by 1/Y^2. To do this, we set the WeightFunction property.
// The WeightFunctions class defines a set of ready-to-use weight functions.
fitter2.WeightFunction <- WeightFunctions.OneOverX
// Refit the curve:
let result3 = fitter2.Fit()
let solution3 = fitter2.BestFitParameters
let s3 = fitter2.GetStandardDeviations()
// The solution is slightly different:
printfn "Solution (weighted observations):"
printfn "b1: %20f %20f" solution3.[0] s3.[0]
printfn "b2: %20f %20f" solution3.[1] s3.[1]
printfn "b3: %20f %20f" solution3.[2] s3.[2]
printf "Press Enter key to exit..." | 2,263 | 7,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-21 | latest | en | 0.55465 |
http://freetofindtruth.blogspot.ca/2017/07/93-112-147-days-of-week-and-classical.html | 1,527,072,247,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00444.warc.gz | 110,755,143 | 23,539 | ## Friday, July 28, 2017
### 68 93 112 147 | The days of the week and the classical planets +What do you see about the dates and planets I don't?
This knowledge traces back to the time the sun was discovered to be what earth orbited around, the Greeks, known as the Hellenistic Period.
In this time they discovered the distance of the sun, 93 million miles away as we know it.
Saturn = 93 (The only planet summing to 93; the 6th and most distance to the ancients)
The 7 Luminaries are the 7 planets to the Greeks, the Sun, Moon, Mercury, Venus, Mars, Jupiter, Saturn.
Sunday - Sun
Monday - Moon
Tuesday - Mars
Wednesday - Mercury
Thursday - Jupiter
Friday - Venus
Saturday - Saturn
They are known as the Classical Planets.
Freemason = 48/96/147
Outer Space = 147
Space Ship = 147
Conspiracy = 147
God = 26
YHVH = 26 (In Hebrew)
Sunday = Sun = Apollo
Monday = Moon = Artemis
Tuesday = Mars = Ares
Wednesday = Mercury = Hermes
Thursday = Jupiter = Zeus
Friday = Venus = Aphrodite
Saturday = Saturn = Cronus
Here are a list of the days of the week and corresponding classical planet match.
Sunday = 3 / 6
Sun = = 9 / 9
Monday = 9 / 9
Moon = 3 / 6
Tuesday = 5 / 4
Mars = 6 / 3
Wednesday = 1 / 8
Mercury = 4 / 5
Thursday = 8 / 1
Jupiter = 9 / 9
Friday 9 / 9
Venus = 9 / 9
Saturday = 1 / 8
Saturn = 3 / 6
I don't see the relevance between the numbers, planets and days of the week that I would expect to see. That's probably because I don't know enough.
Mathematics = 112; Circle = 112; 1+1 = 2
Last, notice the overlap between Helios and Planet.
Mathematics = 68
Sacred Geometry = 68
Ninety-Four = 147
Classical Planet = 147
NASA was established on a date with 94 numerology | 519 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-22 | latest | en | 0.910452 |
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# sport-utility vehicles have become extremely popular because
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sport-utility vehicles have become extremely popular because [#permalink]
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07 Jan 2004, 07:52
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sport-utility vehicles have become extremely popular because of the robust and energetic image they project. although these vehicles look sturdy, they are subject to only a small fraction of the safety standards the government imposes on ordinarly passenger cars. consequently, a high-impact collision involving both a passenger car and a sport-utility vehicle is much more likely to injure the latter and not the former.
each of the following serves to strengthen the conclusion above except
1. those who design vehicles are inclined to make them safe only if government rules dictate that they must
2. sport-utility vehicles have a higher center of gravity, which makes them more susceptible to turning over in a collision
3. the government rigorously enforces its standards for maximum roof strength and impact resistance in all passenger cars
4. sport-utility vehicles are less aerodynamic than passenger cars, and this extra bulk hinders their ability to accelerate
5. people who drive sport-utility vehicles are often instilled with a false sense of security and therefore neglect to wear their seatbelts
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### Show Tags
07 Jan 2004, 20:40
5)
The premise is that SUV are subject to only a small fraction of the safety standards the government imposes on ordinarly passenger cars.
The conclusion is that therefore, in a high-impact collision, SUV passengers are more likely to be injured.
5 says that the higher probability of SUV passengers be harmed may be due to some cause ( false sense of security ) other than SUVs' "non-compliance" to government standards. This may weaken the argument, or at least, not strenghten it by giving an alternative explanation.
4 does somewhat strenghten the argument. If SUVs are bulkier, then their manoeuverability can be reduced. Also, if the acceleration is hindered, then the ability to avoid such a collision (by quickly bifurcating elsewhere before the impact), may be hindered as well. Agree, that this is more inference but IMO, it could strenghten the argument more than 5 does.
1 is no good because it clearly strenghtens the argument by saying that because SUV car makers are not under the same safety standards, then they will no be inclined to make their cars safe ( they don't have to )
_________________
Best Regards,
Paul
Last edited by Paul on 07 Jan 2004, 22:22, edited 1 time in total.
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### Show Tags
07 Jan 2004, 21:14
Good explanation Paul.
Official ans. ?
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### Show Tags
08 Jan 2004, 08:24
Paul wrote:
5)
The premise is that SUV are subject to only a small fraction of the safety standards the government imposes on ordinarly passenger cars.
The conclusion is that therefore, in a high-impact collision, SUV passengers are more likely to be injured.
5 says that the higher probability of SUV passengers be harmed may be due to some cause ( false sense of security ) other than SUVs' "non-compliance" to government standards. This may weaken the argument, or at least, not strenghten it by giving an alternative explanation.
4 does somewhat strenghten the argument. If SUVs are bulkier, then their manoeuverability can be reduced. Also, if the acceleration is hindered, then the ability to avoid such a collision (by quickly bifurcating elsewhere before the impact), may be hindered as well. Agree, that this is more inference but IMO, it could strenghten the argument more than 5 does.
1 is no good because it clearly strenghtens the argument by saying that because SUV car makers are not under the same safety standards, then they will no be inclined to make their cars safe ( they don't have to )
i have broken my head, while thinking about this question. i chose (b), thinking that the word "susceptable" means "stable" (not native speaker, shame on me.. ). the answer is (d). explanation arrives:
before we discuss the answer to this strengthen EXCEPT question, it's important to make a point: four of the answer choices strengthen the argument, and one does not. that doesn't necessarily mean that this fifth choice weakens the argument. it just means that it does not strengthen it (it could be out of the scope). the argument says that people who drive SUVs are more likely to be injured in a car wreck, and (b), (c) and (e) strengthen the idea by asserting either that cars are safer or that SUVs are more dangerous. (d) is irrelevant (and therefore the best answer) because we don't know if a lack of acceleration makes a car any less safe to drive. (a) strengthens because the argument says that SUVs don't have as many safety standards: thus, SUV designers will cut corners if they can.
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Re: sport-utility vehicles have become extremely popular because [#permalink]
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11 Apr 2016, 04:05
I don't have explanation for it, but i think it's 5.
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Re: sport-utility vehicles have become extremely popular because [#permalink] 11 Apr 2016, 04:05
Display posts from previous: Sort by | 1,554 | 6,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-43 | latest | en | 0.934952 |
http://www.docstoc.com/docs/45582318/Transaction-Scale-Profit | 1,438,052,398,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981460.12/warc/CC-MAIN-20150728002301-00156-ip-10-236-191-2.ec2.internal.warc.gz | 406,978,814 | 42,090 | # Transaction Scale Profit by wzw10454
VIEWS: 14 PAGES: 26
• pg 1
``` The Organization of the Firm
The Multi-Plant Problem
Cost-Volume-Profit Analysis
Managerial Decisions
– Securing Inputs and Transaction Costs
– Managerial Compensation
– Worker-Management Relations
Cost Elasticity
Cost Elasticity = Percentage change in total
cost associated with a 1% change in output.
% Change in TC / % Change in Q
Note the dependent and independent variable.
Interpretation
Ec < 1 Increasing Returns to Scale
Ec = 1 Constant Returns to Scale
Ec > 1 Decreasing Returns to Scale
Long-Run Average Cost
Capacity – Output level at which short-run
average costs are minimized.
If a firm moves beyond ‘capacity’ the firm may
want to consider building a larger plant.
BE ABLE TO ILLUSTRATE THIS STORY IN THE
SHORT-RUN
Minimum Efficient Scale – Output level at which
long-run average costs are minimized.
BE ABLE TO ILLUSTRATE LONG-RUN AVERAGE
COST AND IDENTIFY THE LEVEL OF OUTPUT
CORRESPONDING TO MES.
Firm Size and Plant Size
Multi-plant Economies of Scale – Cost
advantages from operating multiple
facilities in the same line of business
or industry.
Multi-plant Diseconomies of Scale –
Cost disadvantages from operating
multiple facilities in the same line of
The Economics of Multi-Plant Operations
Elements needed for problem
Equation for Demand Curve
Short-run Total Cost Function
Steps in Solving the Problem
Solve for profit maximizing output, price, and profit.
Solve for average cost minimizing output.
Solve for MC when firm produces at capacity.
Set MR equal to MC at capacity to determine optimal
multi-plant operation.
Determine the optimal number of plants.
Determine price and profit when firm employs the
optimal number of plants.
Cost-Volume-Profit
Analysis
• Cost-Volume-Profit Analysis – Analytical technique
used to study the relations among cost, revenues, and
profits.
• Breakeven Quantity – A zero profit activity level
• TR = TC
• P*Q = TFC + AVC*Q
• TFC = [P – AVC] *Q
• Q = TFC / [P-AVC]
• Profit Contribution = P – AVC
CVP Analysis Example
• Price = \$80 AVC = \$60
• TFC = \$20K
• Desired Profit = \$40K
• Q = [Fixed Cost + Profit
Requirement] / Profit
Contribution
• Q = [\$20,000 + \$40,000] / \$20
• Q = 3,000 units
• Given price, cost conditions,
and desired profit, firm will need
to produce 3,000 units.
Degree of Operating
Leverage
• Degree of Operating Leverage –
Percentage change in profit from a 1%
change in output.
• DOL = % change in profit / % change in Q
• DOL = Elasticity of Profit
• DOL at a given level of output =
[Q(profit contribution)] / {[Q(profit
contribution)] – Total Fixed Cost}
• OR [P-AVC] / [P-ATC]
Limitations of CVP
Analysis
• Assumes selling price is
constant. Each time the selling
price changes the analysis must
be completed again.
• Assumes that average variable
cost is also constant. If this is
not true, then the analysis is not
particularly useful.
Methods of Acquiring Inputs:
Spot Exchange
Spot Exchange – an informal
relationship between a buyer and seller
in which neither party is obligated to
adhere to specific terms for exchange.
This is often used when inputs are
standardized so effort in finding the
‘best’ input is not needed.
Methods of Acquiring Inputs:
Acquiring Inputs Via Contract
Contract – a formal relationship
between a buyer and seller that
obligates the buyer and seller to
exchange at terms specified in a legal
document.
Contracts can reduce uncertainty, but
increase the transaction costs incurred
by the firm.
Methods of Acquiring Inputs:
Internal Production
Vertical Integration – a situation where
a firm produces the inputs required to
make its final product.
Vertical integration (alternative
definition)- various stages of
production of a single product are
conducted by a single firm.
Motivation: Reduces Transaction
Costs
Transaction Costs
Transaction costs - the expenses of
trading with others above and beyond
the price. i.e. the cost of writing and
enforcing contracts.
Transaction costs determine whether
markets are internalized or allowed to
remain external to the firm.
More on Transaction Costs: The Work of
Oliver Williamson
Four basic concepts that underlie
transaction costs analysis.
1. Markets and firms are alternative means for completing related sets
of transactions.
2. The relative cost of using markets or a firm’s own resources should
determine the choice.
3. The transaction cost of writing and executing contracts across a
market is a function of
1. the characteristics of the involved human actors
2. the objective properties of the market
4. In sum, both human and environmental factors impact the
transaction costs across firms and markets.
More from Williamson
Purpose of this analysis is to identify
the set of environmental and human
factors that explain both internal firm
and industrial organization.
Key environmental factors:
Uncertainty and number of firms
Key human factors:
Bounded rationality and opportunism
Bounded Rationality and
Opportunism
Bounded rationality - the limited human capacity to
anticipate and solve complex problems.
Opportunistic behavior - Taking advantage of another
when allowed by circumstances.
High transaction costs:
Specialized products: The creation of specialized
products, where only a single buyer and/or seller
exists, can lead to opportunistic behavior. This provides
an incentive for vertical integration.
Changing market conditions: Bounded rationality and
uncertain market conditions make the writing and
enforcement of contracts involving future conditions
undesirable for both parties. Such high transaction costs
increases the likelihood of vertical integration.
Market vs. Internal Production
Labor theory: Wages = Marginal Revenue
Product
Marginal Revenue Product = Marginal
Revenue of Output (MR) * Marginal Product of
Labor (MP)
However, for this to be true for each worker a
firm would need to measure MP.
What if a firm cannot measure MP? Then a
worker can reduce effort an still maintain the
same wage.
When monitoring costs are high, a firm has an
incentive to sub-contract work.
Why? For independent workers the wage
(profit) is closer linked to productivity.
More Benefits from Vertical Integration
In addition to transaction costs, vertical
integration is also motivated by two
Vertical integration provides assurance
of supplying inputs/outputs in a market
that may be unstable.
Threatens potential entrants by raising
entry barriers (aluminum example)
The Principal-Agent Problem
A principal is the person who wants an action taken.
In the work environment, this is the owner of the firm.
The agent is the person who takes the action. In the
work environment, this is the worker.
If motivations differ between the principal and
agent, and information is not perfect, a principal-agent
problem exists.
A specific example is the issue of moral hazard. Moral
hazard occurs when the agent can take actions that the
principal cannot directly observe that will reduce the
welfare of the principal. For example, consider
shirking.
How can the firm limit shirking?
Difficulty of Vertical Integration
Shirking of Workers
Shirking - the behavior of a worker who is putting
forth less than the agreed to effort.
Efficiency Wages – Paying the worker a wage above the
market wage.
Why is this necessary? Because workers can vary
productivity, a firm may need to pay higher wages to
ensure higher levels of output.
Why would firms pay efficiency wages? In other words,
why do higher wages elicit higher productivity.
a. The Gift exchange hypothesis
b. Worker turnover
c. Worker quality
Shirking Defense
How do firms prevent the manager from shirking?
Make the manager a residual claimant.
• Residual claimant - persons who share in the profits of the
firm.
How do firms prevent workers from shirking?
• Profit sharing – mechanism used to enhance workers’ efforts
that involve tying compensation to the underlying profitability
of the firm
STOCK OPTIONS, etc..
• Revenue sharing – mechanism used to enhance workers’
efforts that involve tying compensation to the underlying
revenues of the firm
SALES COMMISSIONS, TIPS, etc...
NO INCENTIVE TO LOWER COSTS
Teams and Productivity
Teamwork is employed when a team of
individuals can produce more than the
sum of individuals working alone.
Observing individual productivity is
difficult, so shirking can occur: The Free
Rider Problem
Profit Sharing: If team members share in
the profits of the firm, then they have an
incentive to monitor other team
members. If the incentive to monitor
exceeds the free-rider effect, profit
More Defense: Piece Rates
Piece-Rate Compensation –
Employee is paid according to
productivity.
Such a compensation plan will
increase productivity.
Will only work if productivity can
be measured.
Problems
• Teamwork will diminish.
• Quantity is easy to measure, quality is
not. Thus quality can suffer with this
compensation plan.
Subjective Evaluations
Why are subjective evaluations
employed? To encourage
innovation, dependability,
cooperation, etc...
Subjective evaluations can lead
to rent-seeking by workers, or
actions taken to re-distribute
resources from others.
Subjective evaluations can also
be quite inaccurate.
Inaccurate evaluations can
The Role of Management
What is the primary role of the manager?
To prevent shirking, which limits the
production of the firm.
In essence, employees employ the manager
to raise the return to the firm.
Implications: If the manager is poor,
employees will leave. If the returns of the
firm do not accrue to the employees, the
employees will leave.
Remember, the labor market is like any
other market. Exchange takes place by both
parties because benefits exceed the costs.
The Objectives of Management
Managers seek to maximize utility (A.A. Berle and Gardner
Means)
Focus of these authors is on the separation of ownership
and management, which arose due to the rise of the
corporation.
How would this impact market behavior? Studies have
shown that managerial control is less profitable than owner
control. Manager’s are more risk adverse, due to an
inability to diversify.
A related view.... Managers seek to satisfice (Richard Cyret,
James March and Herbert Simon)
In this class we assume that firms seek to maximize profits.
This is a simplification.
WHY DO WE NEED TO ANSWER THIS QUESTION? We need
to know the motivation of the people we study.
```
To top | 2,663 | 10,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2015-32 | longest | en | 0.803971 |
https://www.kodytools.com/units/area/from/kanal/to/pari | 1,723,022,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00372.warc.gz | 650,660,543 | 18,819 | # Kanal to Pari Converter
Convert → Pari to Kanal
1 Kanal = 0.05 Pari
## One Kanal is Equal to How Many Pari?
The answer is one Kanal is equal to 0.05 Pari and that means we can also write it as 1 Kanal = 0.05 Pari. Feel free to use our online unit conversion calculator to convert the unit from Kanal to Pari. Just simply enter value 1 in Kanal and see the result in Pari.
Manually converting Kanal to Pari can be time-consuming,especially when you don’t have enough knowledge about Area units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Kanal to Pari converter tool to get the job done as soon as possible.
We have so many online tools available to convert Kanal to Pari, but not every online tool gives an accurate result and that is why we have created this online Kanal to Pari converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Kanal to Pari (kanal to pari)
By using our Kanal to Pari conversion tool, you know that one Kanal is equivalent to 0.05 Pari. Hence, to convert Kanal to Pari, we just need to multiply the number by 0.05. We are going to use very simple Kanal to Pari conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Kanal} = 1 \times 0.05 = \text{0.05 Pari}$$
## What Unit of Measure is Kanal?
Kanal is a unit of area measurement and mostly used in various parts of India and Pakistan for land measurement. One kanal is equal to 605 square yards.
## What is the Symbol of Kanal?
The symbol of Kanal is kanal. This means you can also write one Kanal as 1 kanal.
## What Unit of Measure is Pari?
Pari is a unit of measurement for area. Pari is one of the traditional units for land measurement in Indian state of Manipur. One pari is equal to 108900 square feet.
## What is the Symbol of Pari?
The symbol of Pari is pari. This means you can also write one Pari as 1 pari.
## How to Use Kanal to Pari Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Kanal and in the first input field, enter a value.
• From the second dropdown, select Pari.
• Instantly, the tool will convert the value from Kanal to Pari and display the result in the second input field.
Kanal
1
Pari
0.05
# Kanal to Pari Conversion Table
Kanal [kanal]Pari [pari]Description
1 Kanal0.05 Pari1 Kanal = 0.05 Pari
2 Kanal0.1 Pari2 Kanal = 0.1 Pari
3 Kanal0.15 Pari3 Kanal = 0.15 Pari
4 Kanal0.2 Pari4 Kanal = 0.2 Pari
5 Kanal0.25 Pari5 Kanal = 0.25 Pari
6 Kanal0.3 Pari6 Kanal = 0.3 Pari
7 Kanal0.35 Pari7 Kanal = 0.35 Pari
8 Kanal0.4 Pari8 Kanal = 0.4 Pari
9 Kanal0.45 Pari9 Kanal = 0.45 Pari
10 Kanal0.5 Pari10 Kanal = 0.5 Pari
100 Kanal5 Pari100 Kanal = 5 Pari
1000 Kanal50 Pari1000 Kanal = 50 Pari
# Kanal to Other Units Conversion Table
ConversionDescription
1 Kanal = 75.63 Ankanam1 Kanal in Ankanam is equal to 75.63
1 Kanal = 15.91 Aana1 Kanal in Aana is equal to 15.91
1 Kanal = 0.12499988949457 Acre1 Kanal in Acre is equal to 0.12499988949457
1 Kanal = 0.14795915553666 Arpent1 Kanal in Arpent is equal to 0.14795915553666
1 Kanal = 5.06 Are1 Kanal in Are is equal to 5.06
1 Kanal = 5.058570528e+30 Barn1 Kanal in Barn is equal to 5.058570528e+30
1 Kanal = 0.378125 Bigha [Assam]1 Kanal in Bigha [Assam] is equal to 0.378125
1 Kanal = 0.378125 Bigha [West Bengal]1 Kanal in Bigha [West Bengal] is equal to 0.378125
1 Kanal = 0.20166666666667 Bigha [Uttar Pradesh]1 Kanal in Bigha [Uttar Pradesh] is equal to 0.20166666666667
1 Kanal = 0.2 Bigha [Rajasthan]1 Kanal in Bigha [Rajasthan] is equal to 0.2
1 Kanal = 0.20003673769287 Bigha [Bihar]1 Kanal in Bigha [Bihar] is equal to 0.20003673769287
1 Kanal = 0.3125 Bigha [Gujrat]1 Kanal in Bigha [Gujrat] is equal to 0.3125
1 Kanal = 0.625 Bigha [Himachal Pradesh]1 Kanal in Bigha [Himachal Pradesh] is equal to 0.625
1 Kanal = 0.074691358024691 Bigha [Nepal]1 Kanal in Bigha [Nepal] is equal to 0.074691358024691
1 Kanal = 4.03 Biswa [Uttar Pradesh]1 Kanal in Biswa [Uttar Pradesh] is equal to 4.03
1 Kanal = 0.00843095088 Bovate1 Kanal in Bovate is equal to 0.00843095088
1 Kanal = 0.05058570528 Bunder1 Kanal in Bunder is equal to 0.05058570528
1 Kanal = 0.001124126784 Caballeria1 Kanal in Caballeria is equal to 0.001124126784
1 Kanal = 0.0037694266229508 Caballeria [Cuba]1 Kanal in Caballeria [Cuba] is equal to 0.0037694266229508
1 Kanal = 0.001264642632 Caballeria [Spain]1 Kanal in Caballeria [Spain] is equal to 0.001264642632
1 Kanal = 0.039213725023256 Carreau1 Kanal in Carreau is equal to 0.039213725023256
1 Kanal = 0.0010408581333333 Carucate1 Kanal in Carucate is equal to 0.0010408581333333
1 Kanal = 0.093677232 Cawnie1 Kanal in Cawnie is equal to 0.093677232
1 Kanal = 12.5 Cent1 Kanal in Cent is equal to 12.5
1 Kanal = 505.86 Centiare1 Kanal in Centiare is equal to 505.86
1 Kanal = 6932.78 Circular Foot1 Kanal in Circular Foot is equal to 6932.78
1 Kanal = 998320.76 Circular Inch1 Kanal in Circular Inch is equal to 998320.76
1 Kanal = 0.5058570528 Cong1 Kanal in Cong is equal to 0.5058570528
1 Kanal = 0.18749334796145 Cover1 Kanal in Cover is equal to 0.18749334796145
1 Kanal = 0.12871680732824 Cuerda1 Kanal in Cuerda is equal to 0.12871680732824
1 Kanal = 121 Chatak1 Kanal in Chatak is equal to 121
1 Kanal = 12.5 Decimal1 Kanal in Decimal is equal to 12.5
1 Kanal = 0.50585738651412 Dekare1 Kanal in Dekare is equal to 0.50585738651412
1 Kanal = 12.5 Dismil1 Kanal in Dismil is equal to 12.5
1 Kanal = 1512.5 Dhur [Tripura]1 Kanal in Dhur [Tripura] is equal to 1512.5
1 Kanal = 29.88 Dhur [Nepal]1 Kanal in Dhur [Nepal] is equal to 29.88
1 Kanal = 0.5058570528 Dunam1 Kanal in Dunam is equal to 0.5058570528
1 Kanal = 0.019694010416667 Drone1 Kanal in Drone is equal to 0.019694010416667
1 Kanal = 0.078671392348367 Fanega1 Kanal in Fanega is equal to 0.078671392348367
1 Kanal = 0.49985874782609 Farthingdale1 Kanal in Farthingdale is equal to 0.49985874782609
1 Kanal = 0.12135911601415 Feddan1 Kanal in Feddan is equal to 0.12135911601415
1 Kanal = 6.3 Ganda1 Kanal in Ganda is equal to 6.3
1 Kanal = 605 Gaj1 Kanal in Gaj is equal to 605
1 Kanal = 605 Gajam1 Kanal in Gajam is equal to 605
1 Kanal = 5 Guntha1 Kanal in Guntha is equal to 5
1 Kanal = 0.125 Ghumaon1 Kanal in Ghumaon is equal to 0.125
1 Kanal = 2.27 Ground1 Kanal in Ground is equal to 2.27
1 Kanal = 0.0000056457260357143 Hacienda1 Kanal in Hacienda is equal to 0.0000056457260357143
1 Kanal = 0.05058570528 Hectare1 Kanal in Hectare is equal to 0.05058570528
1 Kanal = 0.0010408581333333 Hide1 Kanal in Hide is equal to 0.0010408581333333
1 Kanal = 0.35592070837332 Hout1 Kanal in Hout is equal to 0.35592070837332
1 Kanal = 0.000010408581333333 Hundred1 Kanal in Hundred is equal to 0.000010408581333333
1 Kanal = 0.25022977941176 Jerib1 Kanal in Jerib is equal to 0.25022977941176
1 Kanal = 0.087898705960035 Jutro1 Kanal in Jutro is equal to 0.087898705960035
1 Kanal = 7.56 Katha [Bangladesh]1 Kanal in Katha [Bangladesh] is equal to 7.56
1 Kanal = 0.31510416666667 Kani1 Kanal in Kani is equal to 0.31510416666667
1 Kanal = 25.21 Kara1 Kanal in Kara is equal to 25.21
1 Kanal = 3.28 Kappland1 Kanal in Kappland is equal to 3.28
1 Kanal = 0.125 Killa1 Kanal in Killa is equal to 0.125
1 Kanal = 75.63 Kranta1 Kanal in Kranta is equal to 75.63
1 Kanal = 37.81 Kuli1 Kanal in Kuli is equal to 37.81
1 Kanal = 1.25 Kuncham1 Kanal in Kuncham is equal to 1.25
1 Kanal = 37.81 Lecha1 Kanal in Lecha is equal to 37.81
1 Kanal = 0.00070567185379918 Labor1 Kanal in Labor is equal to 0.00070567185379918
1 Kanal = 0.000028226874151967 Legua1 Kanal in Legua is equal to 0.000028226874151967
1 Kanal = 0.05058570528 Manzana [Argentina]1 Kanal in Manzana [Argentina] is equal to 0.05058570528
1 Kanal = 0.07237944598338 Manzana [Costa Rica]1 Kanal in Manzana [Costa Rica] is equal to 0.07237944598338
1 Kanal = 20 Marla1 Kanal in Marla is equal to 20
1 Kanal = 0.20234282112 Morgen [Germany]1 Kanal in Morgen [Germany] is equal to 0.20234282112
1 Kanal = 0.059047163861328 Morgen [South Africa]1 Kanal in Morgen [South Africa] is equal to 0.059047163861328
1 Kanal = 0.75878557540607 Mu1 Kanal in Mu is equal to 0.75878557540607
1 Kanal = 0.0049999955797828 Murabba1 Kanal in Murabba is equal to 0.0049999955797828
1 Kanal = 40.33 Mutthi1 Kanal in Mutthi is equal to 40.33
1 Kanal = 1.26 Ngarn1 Kanal in Ngarn is equal to 1.26
1 Kanal = 2.52 Nali1 Kanal in Nali is equal to 2.52
1 Kanal = 0.00843095088 Oxgang1 Kanal in Oxgang is equal to 0.00843095088
1 Kanal = 63.64 Paisa1 Kanal in Paisa is equal to 63.64
1 Kanal = 14.8 Perche1 Kanal in Perche is equal to 14.8
1 Kanal = 2 Parappu1 Kanal in Parappu is equal to 2
1 Kanal = 153.01 Pyong1 Kanal in Pyong is equal to 153.01
1 Kanal = 0.316160658 Rai1 Kanal in Rai is equal to 0.316160658
1 Kanal = 0.5 Rood1 Kanal in Rood is equal to 0.5
1 Kanal = 0.99433893352812 Ropani1 Kanal in Ropani is equal to 0.99433893352812
1 Kanal = 12.5 Satak1 Kanal in Satak is equal to 12.5
1 Kanal = 0.0001953125 Section1 Kanal in Section is equal to 0.0001953125
1 Kanal = 0.0000281031696 Sitio1 Kanal in Sitio is equal to 0.0000281031696
1 Kanal = 54.45 Square1 Kanal in Square is equal to 54.45
1 Kanal = 5.058570528e+22 Square Angstrom1 Kanal in Square Angstrom is equal to 5.058570528e+22
1 Kanal = 2.2603567234208e-20 Square Astronomical Units1 Kanal in Square Astronomical Units is equal to 2.2603567234208e-20
1 Kanal = 5.058570528e+38 Square Attometer1 Kanal in Square Attometer is equal to 5.058570528e+38
1 Kanal = 5.058570528e+26 Square Bicron1 Kanal in Square Bicron is equal to 5.058570528e+26
1 Kanal = 5058570.53 Square Centimeter1 Kanal in Square Centimeter is equal to 5058570.53
1 Kanal = 1.25 Square Chain1 Kanal in Square Chain is equal to 1.25
1 Kanal = 2420 Square Cubit1 Kanal in Square Cubit is equal to 2420
1 Kanal = 50585.71 Square Decimeter1 Kanal in Square Decimeter is equal to 50585.71
1 Kanal = 5.06 Square Dekameter1 Kanal in Square Dekameter is equal to 5.06
1 Kanal = 1393920 Square Digit1 Kanal in Square Digit is equal to 1393920
1 Kanal = 5.058570528e-34 Square Exameter1 Kanal in Square Exameter is equal to 5.058570528e-34
1 Kanal = 151.25 Square Fathom1 Kanal in Square Fathom is equal to 151.25
1 Kanal = 5.058570528e+32 Square Femtometer1 Kanal in Square Femtometer is equal to 5.058570528e+32
1 Kanal = 5.058570528e+32 Square Fermi1 Kanal in Square Fermi is equal to 5.058570528e+32
1 Kanal = 5445 Square Feet1 Kanal in Square Feet is equal to 5445
1 Kanal = 0.012499988949457 Square Furlong1 Kanal in Square Furlong is equal to 0.012499988949457
1 Kanal = 5.058570528e-16 Square Gigameter1 Kanal in Square Gigameter is equal to 5.058570528e-16
1 Kanal = 0.05058570528 Square Hectometer1 Kanal in Square Hectometer is equal to 0.05058570528
1 Kanal = 784080 Square Inch1 Kanal in Square Inch is equal to 784080
1 Kanal = 0.000021701302300224 Square League1 Kanal in Square League is equal to 0.000021701302300224
1 Kanal = 5.6516873700825e-30 Square Light Year1 Kanal in Square Light Year is equal to 5.6516873700825e-30
1 Kanal = 0.0005058570528 Square Kilometer1 Kanal in Square Kilometer is equal to 0.0005058570528
1 Kanal = 5.058570528e-10 Square Megameter1 Kanal in Square Megameter is equal to 5.058570528e-10
1 Kanal = 505.86 Square Meter1 Kanal in Square Meter is equal to 505.86
1 Kanal = 784079308318430000 Square Microinch1 Kanal in Square Microinch is equal to 784079308318430000
1 Kanal = 505857052800000 Square Micrometer1 Kanal in Square Micrometer is equal to 505857052800000
1 Kanal = 5.058570528e+26 Square Micromicron1 Kanal in Square Micromicron is equal to 5.058570528e+26
1 Kanal = 505857052800000 Square Micron1 Kanal in Square Micron is equal to 505857052800000
1 Kanal = 784080000000 Square Mil1 Kanal in Square Mil is equal to 784080000000
1 Kanal = 0.0001953125 Square Mile1 Kanal in Square Mile is equal to 0.0001953125
1 Kanal = 505857052.8 Square Millimeter1 Kanal in Square Millimeter is equal to 505857052.8
1 Kanal = 505857052800000000000 Square Nanometer1 Kanal in Square Nanometer is equal to 505857052800000000000
1 Kanal = 0.000016387146462408 Square Nautical League1 Kanal in Square Nautical League is equal to 0.000016387146462408
1 Kanal = 0.00014748418805891 Square Nautical Mile1 Kanal in Square Nautical Mile is equal to 0.00014748418805891
1 Kanal = 4794.85 Square Paris Foot1 Kanal in Square Paris Foot is equal to 4794.85
1 Kanal = 5.3128383492245e-31 Square Parsec1 Kanal in Square Parsec is equal to 5.3128383492245e-31
1 Kanal = 20 Perch1 Kanal in Perch is equal to 20
1 Kanal = 9.9 Square Perche1 Kanal in Square Perche is equal to 9.9
1 Kanal = 5.058570528e-28 Square Petameter1 Kanal in Square Petameter is equal to 5.058570528e-28
1 Kanal = 5.058570528e+26 Square Picometer1 Kanal in Square Picometer is equal to 5.058570528e+26
1 Kanal = 20 Square Pole1 Kanal in Square Pole is equal to 20
1 Kanal = 20 Square Rod1 Kanal in Square Rod is equal to 20
1 Kanal = 5.058570528e-22 Square Terameter1 Kanal in Square Terameter is equal to 5.058570528e-22
1 Kanal = 784080000000 Square Thou1 Kanal in Square Thou is equal to 784080000000
1 Kanal = 605 Square Yard1 Kanal in Square Yard is equal to 605
1 Kanal = 5.058570528e+50 Square Yoctometer1 Kanal in Square Yoctometer is equal to 5.058570528e+50
1 Kanal = 5.058570528e-46 Square Yottameter1 Kanal in Square Yottameter is equal to 5.058570528e-46
1 Kanal = 0.18673202392027 Stang1 Kanal in Stang is equal to 0.18673202392027
1 Kanal = 0.5058570528 Stremma1 Kanal in Stremma is equal to 0.5058570528
1 Kanal = 180 Sarsai1 Kanal in Sarsai is equal to 180
1 Kanal = 0.80448004580153 Tarea1 Kanal in Tarea is equal to 0.80448004580153
1 Kanal = 306.04 Tatami1 Kanal in Tatami is equal to 306.04
1 Kanal = 0.091707224945613 Tonde Land1 Kanal in Tonde Land is equal to 0.091707224945613
1 Kanal = 153.02 Tsubo1 Kanal in Tsubo is equal to 153.02
1 Kanal = 0.0000054253424259796 Township1 Kanal in Township is equal to 0.0000054253424259796
1 Kanal = 0.10247489117576 Tunnland1 Kanal in Tunnland is equal to 0.10247489117576
1 Kanal = 605 Vaar1 Kanal in Vaar is equal to 605
1 Kanal = 0.00421547544 Virgate1 Kanal in Virgate is equal to 0.00421547544
1 Kanal = 0.063020833333333 Veli1 Kanal in Veli is equal to 0.063020833333333
1 Kanal = 0.05 Pari1 Kanal in Pari is equal to 0.05
1 Kanal = 0.2 Sangam1 Kanal in Sangam is equal to 0.2
1 Kanal = 7.56 Kottah [Bangladesh]1 Kanal in Kottah [Bangladesh] is equal to 7.56
1 Kanal = 5 Gunta1 Kanal in Gunta is equal to 5
1 Kanal = 12.5 Point1 Kanal in Point is equal to 12.5
1 Kanal = 0.1 Lourak1 Kanal in Lourak is equal to 0.1
1 Kanal = 0.4 Loukhai1 Kanal in Loukhai is equal to 0.4
1 Kanal = 0.8 Loushal1 Kanal in Loushal is equal to 0.8
1 Kanal = 1.6 Tong1 Kanal in Tong is equal to 1.6
1 Kanal = 37.81 Kuzhi1 Kanal in Kuzhi is equal to 37.81
1 Kanal = 54.45 Chadara1 Kanal in Chadara is equal to 54.45
1 Kanal = 605 Veesam1 Kanal in Veesam is equal to 605
1 Kanal = 2 Lacham1 Kanal in Lacham is equal to 2
1 Kanal = 1.49 Katha [Nepal]1 Kanal in Katha [Nepal] is equal to 1.49
1 Kanal = 1.89 Katha [Assam]1 Kanal in Katha [Assam] is equal to 1.89
1 Kanal = 4 Katha [Bihar]1 Kanal in Katha [Bihar] is equal to 4
1 Kanal = 80.01 Dhur [Bihar]1 Kanal in Dhur [Bihar] is equal to 80.01
1 Kanal = 1600.29 Dhurki1 Kanal in Dhurki is equal to 1600.29 | 5,790 | 15,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.922624 |
https://studylib.net/doc/5722217/chapter-8-slutsky-equation | 1,656,682,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00137.warc.gz | 611,506,102 | 12,373 | # Chapter 8 Slutsky Equation
```Course: Microeconomics
Text: Varian’s Intermediate
Microeconomics
In Chapter 6, we talk about how demand
changes when price and income change
individually.
In this chapter, we want to further analyze how
the change in price changes the demand.
In particular, we decompose the change in
quantity demanded due to price change into
substitution effect and income effect.
2
What happens when a commodity’s price
decreases?
Substitution effect: the commodity is
relatively cheaper, so consumers use more of
it, instead of other commodities, which are
now relatively more expensive.
Income effect: the consumer’s budget of \$m
can purchase more than before, as if the
consumer’s income rose, with consequent
income effects on quantities demanded.
3
x2
m
p2
Consumer’s budget is \$m.
Original choice
x1
4
x2
m
p2
Consumer’s budget is \$y.
Lower price for commodity 1
pivots the constraint outwards.
x1
5
x2
m
p2
m'
p2
Consumer’s budget is \$m.
Lower price for commodity 1
pivots the constraint outwards.
Now only \$m’ are needed to buy the
original bundle at the new prices,
as if the consumer’s income has
increased by \$m -- \$m’.
x1
6
Slutsky asserted that if, at the new prices,
If less income is needed to buy the original
bundle then “real income” is increased
If more income is needed to buy the original
bundle then “real income” is decreased
7
Changes to quantities demanded due to the
change in relative prices, keeping income just
enough to buy the original bundle, are the
(pure) substitution effect of the price change.
Changes to quantities demanded due to the
change in ‘real income’ are the income effect of
the price change.
8
Slutsky discovered that changes to demand
from a price change are always the sum of a
pure substitution effect and an income effect.
xi x x
s
i
n
i
9
x2
x2’
x1’
x1
10
x2
x2’
x1’
x1
11
x2
x2’
x1’
x1
12
x2
x2’
x2’’
x1’
x1’’
x1
13
x2
x2’
x2’’
x1’
x1’’
x1
14
x2
Lower p1 makes good 1 relatively
cheaper and causes a substitution
from good 2 to good 1.
(x1’,x2’) (x1’’,x2’’) is the
pure substitution effect.
x2’
x2’’
x1’
x1’’
x1
15
Substitution effect is always negatively related to
the price change.
Note that the portion of the yellow compensated
budget line below x’1 is inside the budget set of
the original budget, thus these bundles should be
less preferred than the original bundle.
As a result, the consumer must choose a point at
or more than x’1 with the compensated budget,
and as a result, the substitution effect is positive for
a price decrease.
16
x2
(x1’’’,x2’’’)
x2’
x2’’
x1’
x1’’
x1
17
x2
The income effect is
(x1’’,x2’’) (x1’’’,x2’’’).
(x1’’’,x2’’’)
x2’
x2’’
x1’
x1’’
x1
18
The change to demand due to
lower p1 is the sum of the
income and substitution effects,
(x1’,x2’) (x1’’’,x2’’’).
x2
(x1’’’,x2’’’)
x2’
x2’’
x1’
x1’’
x1
19
Most goods are normal (i.e. demand increases
with income).
The substitution and income effects reinforce
each other when a normal good’s own price
changes.
20
x2
Good 1 is normal because
higher income increases
demand, so the income
and substitution
(x1’’’,x2’’’)
effects reinforce
each other.
x2’
x2’’
x1’
x1’’
x1
21
Since both the substitution and income effects
increase demand when own-price falls, a normal
good’s ordinary demand curve slopes down.
The Law of (Downward-Sloping) Demand
therefore always applies to normal goods.
22
Some goods are inferior (i.e. demand is reduced
when income is higher.)
The substitution and income effects oppose
each other when an inferior good’s own price
changes.
23
x2
x2’
x1’
x1
24
x2
x2’
x1’
x1
25
x2
x2’
x1’
x1
26
x2
x2’
x2’’
x1’
x1’’
x1
27
x2
The pure substitution effect is as for
a normal good. But, ….
x2’
x2’’
x1’
x1’’
x1
28
The pure substitution effect is as for a
normal good. But, the income effect is
in the opposite direction.
x2
(x1’’’,x2’’’)
x2’
x2’’
x1’
x1’’
x1
29
x2
The overall changes to demand are
the sums of the substitution and
income effects.
(x1’’’,x2’’’)
x2’
x2’’
x1’
x1’’
x1
30
In rare cases of extreme income-inferiority, the
income effect may be larger than the
substitution effect, causing quantity demanded
to fall as own-price rises.
Such goods are Giffen goods.
31
x2
A decrease in p1 causes
quantity demanded of
good 1 to fall.
x2’
x1’
x1
32
x2
A decrease in p1 causes
quantity demanded of
good 1 to fall.
x2’’’
x2’
x1’’’
x1’
x1
33
x2
A decrease in p1 causes
quantity demanded of
good 1 to fall.
x2’’’
x2’
x2’’
x1’’’
x1’
x1’’
Substitution effect
Income effect
x1
34
Giffen good can only result when the income effect
of an inferior good is so strong that it dominates the
substitution effect.
This may be possible for poor households where
the low-quality necessity has taken up a large
portion of expenditure.
This case is very rare, even if exists, so we have
confidence that the Law of Demand almost always
holds.
35
If we denote m’ as the income required to
obtain the original bundle at the new prices, so
that
m’=p’1 x1 + p2 x2 and m=p1 x1 + p2 x2 .
Thus the change in real income is
m’– m = (p’1 – p1 ) x1
Or
m p1 x1
36
The substitution effect is
x x1 ( p'1 , m' ) x1 ( p1 , m)
s
1
The Income effect is
x1n x1 ( p1 ' , m) x1 ( p1 ' , m' )
Total Effect
x1 x1 ( p1 ' , m) x1 ( p1, m) x x
s
1
n
1
37
In terms of derivative (or rate of change):
x1
x1s x1 m
p1
p1 m p1
x1
x1s x1
( x1 )
p1
p1 m
x1
x1s x1
( x1 )
p1
p1 m
Which is known as the Slutsky Equation.
(This is just a rough presentation. The tools need
for formal derivations is not covered in this class.)
38
39
40
41
Slutsky’s method of decomposition is not the
only reasonable way.
Hicks proposed another way of holding “real
income” constant.
back the original bundle, Hicks method
compensates the consumer to buy back a
bundle that gives him the same utility as before.
42
43
Hicks Substitution Effect is also negative,
because of the convex preference. (It can also be
shown by revealed preference.)
The nominal income required to maintain the
utility constant is less than the one required to
buy back the same bundle. It implies a larger
income effect for a price decrease, but a smaller
income effect for a price increase.
44
If government wants to impose tax to support
public expenditure, or to ‘punish’ consumption of
a good, say, due to pollution, various means can be
used.
Here, given the revenue are the same in
equilibrium, how do the effects of income tax and
quantity tax on good 1 differ?
(Note: Income and tax rates are regarded as given
for consumers. The tax rate is adjusted so that at
equilibrium the tax revenue is the same.)
45
46
Income tax corresponds to an inward shift of
budget line.
Quantity tax corresponds to an inward rotation
of the budget line.
When the revenues are held the same for
comparison, the budget line for the income tax
must pass through the optimal point for
quantity tax.
(Note: The tax revenue is tx1*, where x1* is the
optimal quantity under quantity tax.)
47
With the same tax revenue, the utility level
attained is higher with income tax than with
the quantity tax.
But quantity tax has a stronger effect in
reducing the consumption of good 1 than
income tax.
48
Consider a similar case. Now a tax is imposed
to reduce consumption of certain good, but at
the same time, an equivalent amount is rebated
(given back) to the consumer.
Again, consumer has to take the rebate and tax
rate constant for his decision.
49
50
Note that the new consumption bundle must
be on the original budget line, because in
equilibrium, the tax amount and rebate are the
same.
The consumer has become worse off after this
quantity tax and rebate program.
51
In this chapter, a decomposition of price effect on
quality demand is introduced.
Substitution effect: effect of change of price
holding ‘real income’ constant.
Income effect: effect of change in real income.
For normal goods, both effects are negative w.r.t. a
price rise.
For inferior goods, sub. effect is negative, but income
effect is positive w.r.t. a price rise.
Giffen goods can only be inferior goods with very
strong income effect.
52
``` | 2,652 | 8,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-27 | latest | en | 0.899618 |
https://www.teacherspayteachers.com/Product/4th-Grade-Math-Explain-why-a-fraction-is-equivalent-to-a-fraction-4NFA1-1297173 | 1,513,179,153,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00283.warc.gz | 808,167,729 | 24,553 | Total:
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# 4th Grade Math - Explain why a fraction is equivalent to a fraction 4.NF.A.1
Common Core Standards
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Fourth Grade Common Core Math - Explain why a fraction is equivalent to a fraction 4.NF.1
Practice provides two ways for students to practice and show mastery of their ability to explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models.
It includes 40 distinct problems in two sets (129 slides in all!), with and without following answer slides. The PowerPoint file can be used on computers, or Promethean Activboard and Smart boards.
Take a look at the preview file and buy today for your students benefit!
Standard Extend understanding of fraction equivalence and ordering. CCSS.MATH.CONTENT.4.NF.A.1
Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
4 NF.A.1 4 NF.1
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55 minutes
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\$4.50 | 334 | 1,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-51 | latest | en | 0.832148 |
https://www.cpalms.org/PreviewResourceUrl/Preview/7019 | 1,656,990,530,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00266.warc.gz | 758,304,932 | 28,133 | # Who Would Have Figured? (Probability)
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### General Information
Subject(s): Mathematics
Intended Audience: Educators
Instructional Time: 1 Hour(s)
Keywords: probability, experimental, coin tossing
Instructional Component Type(s):
Instructional Design Framework(s): Confirmation Inquiry (Level 1)
Resource Collection: General Collection
## Aligned Standards
This vetted resource aligns to concepts or skills in these benchmarks.
## 1 Lesson Plan
M & M Candy: I Want Green "Students compare mathematical expectations and experimental probability; then explain any difference in the two numbers. Students use colored candy pieces (such as M & M's) for their data collection, comparisons, and explanations." from Beacon Learning Center.
## Related Resources
Other vetted resources related to this resource.
## Lesson Plans
Marble Mania:
In this lesson, "by flipping coins and pulling marbles out of a bag, students begin to develop a basic understanding of probabilities, how they are determined, and how the outcome of an experiment can be affected by the number of times it is conducted." (from Science NetLinks)
Type: Lesson Plan
M & M Candy: I Want Green:
"Students compare mathematical expectations and experimental probability; then explain any difference in the two numbers. Students use colored candy pieces (such as M & M's) for their data collection, comparisons, and explanations." from Beacon Learning Center.
Type: Lesson Plan | 309 | 1,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-27 | longest | en | 0.867941 |
http://www.maths.manchester.ac.uk/undergraduate/ugstudies/units/2008-09/level3/MATH31061/ | 1,369,443,054,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705305291/warc/CC-MAIN-20130516115505-00029-ip-10-60-113-184.ec2.internal.warc.gz | 511,395,456 | 4,605 | You are here: Mathematics > undergraduate > undergraduate studies > course units > level 3 units > MATH31061
School of Mathematics
MATH31061/MATH41016 - 2008/2009
General Information
• Title: Differentiable Manifolds
• Unit code: MATH31061/MATH41061
• Credits: 10 (MATH31061), 15 (MATH41061)
• Prerequisites: MATH20222 Introduction to Geometry, MATH20132 Calculus of Several Variables
• Co-requisite units: MATH31051 Introduction to Topology (optional). Although taking MATH31051 in parallel may be beneficial; it is not required.
• School responsible: Mathematics
• Member of staff responsible:
Page Contents
Other Resources
Specification
Aims
The unit aims to introduce the basic ideas of differentiable manifolds.
Brief Description of the unit
Differentiable manifolds are among the most fundamental notions of modern mathematics. Roughly, they are geometrical objects that can be endowed with coordinates; using these coordinates one can apply differential and integral calculus, but the results are coordinate-independent.
Examples of manifolds start with open domains in Euclidean space Rn , and include "multidimensional surfaces" such as the n-sphere Sn and n-torus Tn , the projective spaces RPn and CPn , and their generalizations, matrix groups such as the rotation group SO(n), etc. Differentiable manifolds naturally appear in various applications, e.g., as configuration spaces in mechanics. They are arguably the most general objects on which calculus can be developed. On the other hand, differentiable manifolds provide for calculus a powerful invariant geometric language, which is used in almost all areas of mathematics and its applications.
In this course we give an introduction to the theory of manifolds, including their definition and examples; vector fields and differential forms; integration on manifolds and de Rham cohomology.
Learning Outcomes
On completion of this unit successful students will be able to:
• deal with various examples of differentiable manifolds and smooth maps;
• have familiarity with tangent vectors, tensors and differential forms;
• work practically with vector fields and differential forms;
• appreciate the basic ideas of de Rham cohomology and its examples;
• apply the ideas of differentiable manifolds to other areas.
Future topics requiring this course unit
Differentiable manifolds are used in almost all areas of mathematics and its applications, including physics and engineering. The following course units are specifically based on MATH31061/MATH41061 Differentiable Manifolds:
MATH41122 Differential Geometry; MATH41101 Geometric Cobordism Theory
Syllabus
1. Manifolds and smooth maps. Coordinates on familiar spaces. Charts and atlases. Definitions of manifolds and smooth maps. Products. Specifying manifolds by equations. More examples of manifolds.
2. Tangent vectors. Velocity of a curve. Tangent vectors. Tangent bundle. Differential of a map.
3. Topology of a manifold. Topology induced by manifold structure. Identification of tangent vectors with derivations. Bump functions and partitions of unity. Embedding manifolds in RN.
4. Tensor algebra. Dual space, covectors and tensors. Einstein notation. Behaviour under maps. Tensors at a point. Example: differential of a function as covector.
5. Vector fields. Tensor and vector fields. Examples. Vector fields as derivations. Flow of a vector field. Commutator.
6. Differential forms. Antisymmetric tensors. Exterior multiplication. Forms at a point. Bases and dimensions. Exterior differential: definition and properties.
7. Integration. Orientation. Integral over a compact oriented manifold. Independence of atlas and partition of unity. Integration over singular manifolds and chains. Stokes theorem.
8. De Rham cohomology. Definition of cohomology and examples of nonzero classes. Poincaré Lemma. Examples of calculation.
Textbooks
No particular textbook is followed. Students are advised to keep their own lecture notes. There are many good sources available treating various aspects of differentiable manifolds on various levels and from different viewpoints. Below is a list of texts that may be useful. More can be found by searching library shelves.
• R. Abraham, J. E. Marsden, T. Ratiu. Manifolds, tensor analysis, and applications.
• B.A. Dubrovin, A.T. Fomenko, S.P. Novikov. Modern geometry, methods and applications.
• A. S. Mishchenko, A. T. Fomenko. A course of differential geometry and topology
• S. Morita. Geometry of differential forms.
• Michael Spivak. Calculus on manifolds.
• Frank W. Warner. Foundations of differentiable manifolds and Lie groups
Teaching and learning methods
Two lectures per week plus one weekly examples class.
Assessment
Coursework; Weighting within unit 20%
2 hours end of semester examination; Weighting within unit 80%
Arrangements
Online course materials are available for this unit. | 1,070 | 4,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2013-20 | latest | en | 0.879228 |
https://brane-space.blogspot.com/2021/08/examining-fields-and-vector-spaces.html | 1,653,743,468,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016853.88/warc/CC-MAIN-20220528123744-20220528153744-00717.warc.gz | 197,970,668 | 15,525 | ## Thursday, August 12, 2021
### Examining Fields and Vector Spaces
The fundamental starting point for linear algebra is the concept of the vector space, and this in turn requires familiarity with fields.
By a field (F, +, ·) we mean a commutative ring with unity which satisfies the additional axiom: (A1): For every non-zero element a
F there exists an element a-1 F such that: a· a-1 = 1. The element a-1 = 1/a is called the reciprocal or multiplicative inverse of a. Thus, the key thing about a field is that it comprises a set of elements which can be added or multiplied in such a way that addition and multiplications satisfy the ordinary rules of arithmetic, and in such a way one can divide by non-zero elements.
Of course, if one has a field, one can also have a sub-field. Thus, say P and L are fields and L is contained in P (e.g. L
P ), then L is a sub-field of P.
A vector space V over the field K, say, is a set of objects that can be added and multiplied by elements of K in such a way that the sum of any 2 elements of V is again, an element of V.
One can also have some set W V , i.e. W is a subset of V, which satisfies specific conditions: i) the sum of any 2 elements of W is also an element of W, ii) the multiple m(w1) of an element of W is also an element of W, the element 0 of V is also an element of W, then we call W a subspace of V.
As an illustration, let V = R n and let W be the set of vectors in V whose last coordinate = 0. Then, W is a subset of V, which we could identify with R n-1 .
As another illustration, let V be an arbitrary vector space and let v 1, v2v3........v n be elements of V. Also let x 1, x 2, x 3...... x n be numbers. Then it is possible to form an expression of the type:
x 1 v 1 + x 2 v 2 + x 3 v 3 +.............x n v n
which is called a linear combination of v 1, v2v3........v n. The set of all linear combinations of v 1, v2v3........v n is a subspace of V,
Yet another example: let A be a vector in R 3. Let W be the set of all elements B in R 3 such that B · A = 0, i.e. such that B is perpendicular to A. Then W is a subspace of R 3.
An additional important consideration is whether elements of a vector space are linearly dependent or linearly independent. We say the elements v 1, v2v3........v n are linearly dependent over a field F if there exist elements in F not all equal to zero such that:
a 1 v 1 + a 2 v 2 + ..............a n v n = 0
If, on the other hand, there do not exist such numbers a1, a2 etc. we say that the elements v 1, v2v3........v n are linearly independent.
Now, if elements v 1, v2v3........v n of the vector space V generate V and also are linearly independent, then (v 1, v2v3........v n) is called a basis of V. One can also say that those elements v 1, v2v3........v n form a basis of V.
Example: Let W f be a vector space of functions generated by the two functions: exp(t) and exp(2t), then {exp(t), exp(2t)} is a basis of W f
As a further illustration, let V be a vector space and let (v 1, v2v3........v n) be a basis of V. The elements of V can be represented by n-tuples relative to this basis, e.g. if an element v of V is written as a linear combination:
v = x 1 v 1 + x 2 v 2 + x 3 v 3 +.............x n v n
then we call (x 1, x 2, .......x n) the coordinates of v with respect to our basis.
Let V be the vector space of functions generated by the two functions: exp(t) and exp(2t), then what are the coordinates for f(V) = 3 exp(t) + 5 exp(2t)?
Ans. The coordinates are (3, 5) with respect to the basis {exp(t), exp(2t)} .
Example Problem :
Show that the vectors (1, 1) and (-3, 2) are linearly independent.
Solution: Let a, b be two numbers associated with some vector space - call it W- such that:
a(1,1) + b(-3,2) = 0
Writing out the components as linear combinations:
a - 3b = 0 and a + 2b = 0
Then solve simultaneously:
a - 3b = 0
a + 2b = 0
----------
0 -5b = 0
or b = 0, so a = 0
Both a and b are equal to zero so the vectors are linearly independent.
Suggested Problem:
Show that the vectors (1, 1) and (-1, 2) form a basis of R 2. | 1,206 | 4,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-21 | longest | en | 0.922793 |
https://www.geeksforgeeks.org/maximum-profit-by-buying-and-selling-a-share-at-most-twice/?ref=rp | 1,642,986,699,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00007.warc.gz | 801,030,119 | 35,443 | # Maximum profit by buying and selling a share at most twice
• Difficulty Level : Hard
• Last Updated : 20 Jan, 2022
In daily share trading, a buyer buys shares in the morning and sells them on the same day. If the trader is allowed to make at most 2 transactions in a day, whereas the second transaction can only start after the first one is complete (Buy->sell->Buy->sell). Given stock prices throughout the day, find out the maximum profit that a share trader could have made.
Examples:
```Input: price[] = {10, 22, 5, 75, 65, 80}
Output: 87
Trader earns 87 as sum of 12, 75
Buy at 10, sell at 22,
Buy at 5 and sell at 80
Input: price[] = {2, 30, 15, 10, 8, 25, 80}
Output: 100
Trader earns 100 as sum of 28 and 72
Buy at price 2, sell at 30, buy at 8 and sell at 80
Input: price[] = {100, 30, 15, 10, 8, 25, 80};
Output: 72
Buy at price 8 and sell at 80.
Input: price[] = {90, 80, 70, 60, 50}
Output: 0
Not possible to earn.```
A Simple Solution is to consider every index ‘i’ and do the following
Max profit with at most two transactions =
MAX {max profit with one transaction and subarray price[0..i] +
max profit with one transaction and subarray price[i+1..n-1] }
i varies from 0 to n-1.
Maximum possible using one transaction can be calculated using the following O(n) algorithm
The maximum difference between two elements such that the larger element appears after the smaller number
The time complexity of the above simple solution is O(n2).
We can do this O(n) using the following Efficient Solution. The idea is to store the maximum possible profit of every subarray and solve the problem in the following two phases.
1) Create a table profit[0..n-1] and initialize all values in it 0.
2) Traverse price[] from right to left and update profit[i] such that profit[i] stores maximum profit achievable from one transaction in subarray price[i..n-1]
3) Traverse price[] from left to right and update profit[i] such that profit[i] stores maximum profit such that profit[i] contains maximum achievable profit from two transactions in subarray price[0..i].
4) Return profit[n-1]
To do step 2, we need to keep track of the maximum price from right to left side, and to do step 3, we need to keep track of the minimum price from left to right. Why we traverse in reverse directions? The idea is to save space, in the third step, we use the same array for both purposes, maximum with 1 transaction and maximum with 2 transactions. After iteration i, the array profit[0..i] contains the maximum profit with 2 transactions, and profit[i+1..n-1] contains profit with two transactions.
Below are the implementations of the above idea.
## C++
`// C++ program to find maximum``// possible profit with at most``// two transactions``#include ``using` `namespace` `std;` `// Returns maximum profit with``// two transactions on a given``// list of stock prices, price[0..n-1]``int` `maxProfit(``int` `price[], ``int` `n)``{`` ``// Create profit array and`` ``// initialize it as 0`` ``int``* profit = ``new` `int``[n];`` ``for` `(``int` `i = 0; i < n; i++)`` ``profit[i] = 0;` ` ``/* Get the maximum profit with`` ``only one transaction`` ``allowed. After this loop,`` ``profit[i] contains maximum`` ``profit from price[i..n-1]`` ``using at most one trans. */`` ``int` `max_price = price[n - 1];`` ``for` `(``int` `i = n - 2; i >= 0; i--) {`` ``// max_price has maximum`` ``// of price[i..n-1]`` ``if` `(price[i] > max_price)`` ``max_price = price[i];` ` ``// we can get profit[i] by taking maximum of:`` ``// a) previous maximum, i.e., profit[i+1]`` ``// b) profit by buying at price[i] and selling at`` ``// max_price`` ``profit[i]`` ``= max(profit[i + 1], max_price - price[i]);`` ``}` ` ``/* Get the maximum profit with two transactions allowed`` ``After this loop, profit[n-1] contains the result */`` ``int` `min_price = price[0];`` ``for` `(``int` `i = 1; i < n; i++) {`` ``// min_price is minimum price in price[0..i]`` ``if` `(price[i] < min_price)`` ``min_price = price[i];` ` ``// Maximum profit is maximum of:`` ``// a) previous maximum, i.e., profit[i-1]`` ``// b) (Buy, Sell) at (min_price, price[i]) and add`` ``// profit of other trans. stored in profit[i]`` ``profit[i] = max(profit[i - 1],`` ``profit[i] + (price[i] - min_price));`` ``}`` ``int` `result = profit[n - 1];` ` ``delete``[] profit; ``// To avoid memory leak` ` ``return` `result;``}` `// Driver code``int` `main()``{`` ``int` `price[] = { 2, 30, 15, 10, 8, 25, 80 };`` ``int` `n = ``sizeof``(price) / ``sizeof``(price[0]);`` ``cout << ``"Maximum Profit = "` `<< maxProfit(price, n);`` ``return` `0;``}`
## Java
`class` `Profit {`` ``// Returns maximum profit`` ``// with two transactions on a`` ``// given list of stock prices,`` ``// price[0..n-1]`` ``static` `int` `maxProfit(``int` `price[], ``int` `n)`` ``{`` ``// Create profit array`` ``// and initialize it as 0`` ``int` `profit[] = ``new` `int``[n];`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``profit[i] = ``0``;` ` ``/* Get the maximum profit`` ``with only one transaction`` ``allowed. After this loop,`` ``profit[i] contains`` ``maximum profit from`` ``price[i..n-1] using at most`` ``one trans. */`` ``int` `max_price = price[n - ``1``];`` ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) {`` ``// max_price has maximum`` ``// of price[i..n-1]`` ``if` `(price[i] > max_price)`` ``max_price = price[i];` ` ``// we can get profit[i]`` ``// by taking maximum of:`` ``// a) previous maximum,`` ``// i.e., profit[i+1]`` ``// b) profit by buying`` ``// at price[i] and selling`` ``// at`` ``// max_price`` ``profit[i] = Math.max(profit[i + ``1``],`` ``max_price - price[i]);`` ``}` ` ``/* Get the maximum profit`` ``with two transactions allowed`` ``After this loop, profit[n-1]`` ``contains the result`` ``*/`` ``int` `min_price = price[``0``];`` ``for` `(``int` `i = ``1``; i < n; i++) {`` ``// min_price is minimum`` ``// price in price[0..i]`` ``if` `(price[i] < min_price)`` ``min_price = price[i];` ` ``// Maximum profit is maximum of:`` ``// a) previous maximum, i.e., profit[i-1]`` ``// b) (Buy, Sell) at (min_price, price[i]) and`` ``// add`` ``// profit of other trans.`` ``// stored in profit[i]`` ``profit[i] = Math.max(`` ``profit[i - ``1``],`` ``profit[i] + (price[i] - min_price));`` ``}`` ``int` `result = profit[n - ``1``];`` ``return` `result;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String args[])`` ``{`` ``int` `price[] = { ``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80` `};`` ``int` `n = price.length;`` ``System.out.println(``"Maximum Profit = "`` ``+ maxProfit(price, n));`` ``}` `} ``/* This code is contributed by Rajat Mishra */`
## Python3
`# Returns maximum profit with``# two transactions on a given``# list of stock prices price[0..n-1]` `def` `maxProfit(price, n):` ` ``# Create profit array and initialize it as 0`` ``profit ``=` `[``0``]``*``n` ` ``# Get the maximum profit`` ``# with only one transaction`` ``# allowed. After this loop,`` ``# profit[i] contains maximum`` ``# profit from price[i..n-1]`` ``# using at most one trans.`` ``max_price ``=` `price[n``-``1``]` ` ``for` `i ``in` `range``(n``-``2``, ``0``, ``-``1``):` ` ``if` `price[i] > max_price:`` ``max_price ``=` `price[i]` ` ``# we can get profit[i] by`` ``# taking maximum of:`` ``# a) previous maximum,`` ``# i.e., profit[i+1]`` ``# b) profit by buying at`` ``# price[i] and selling at`` ``# max_price`` ``profit[i] ``=` `max``(profit[i``+``1``], max_price ``-` `price[i])` ` ``# Get the maximum profit`` ``# with two transactions allowed`` ``# After this loop, profit[n-1]`` ``# contains the result`` ``min_price ``=` `price[``0``]` ` ``for` `i ``in` `range``(``1``, n):` ` ``if` `price[i] < min_price:`` ``min_price ``=` `price[i]` ` ``# Maximum profit is maximum of:`` ``# a) previous maximum,`` ``# i.e., profit[i-1]`` ``# b) (Buy, Sell) at`` ``# (min_price, A[i]) and add`` ``# profit of other trans.`` ``# stored in profit[i]`` ``profit[i] ``=` `max``(profit[i``-``1``], profit[i]``+``(price[i]``-``min_price))` ` ``result ``=` `profit[n``-``1``]` ` ``return` `result` `# Driver function``price ``=` `[``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80``]``print` `(``"Maximum profit is"``, maxProfit(price, ``len``(price)))` `# This code is contributed by __Devesh Agrawal__`
## C#
`// C# program to find maximum possible profit``// with at most two transactions``using` `System;` `class` `GFG {` ` ``// Returns maximum profit with two`` ``// transactions on a given list of`` ``// stock prices, price[0..n-1]`` ``static` `int` `maxProfit(``int``[] price, ``int` `n)`` ``{` ` ``// Create profit array and initialize`` ``// it as 0`` ``int``[] profit = ``new` `int``[n];`` ``for` `(``int` `i = 0; i < n; i++)`` ``profit[i] = 0;` ` ``/* Get the maximum profit with only`` ``one transaction allowed. After this`` ``loop, profit[i] contains maximum`` ``profit from price[i..n-1] using at`` ``most one trans. */`` ``int` `max_price = price[n - 1];` ` ``for` `(``int` `i = n - 2; i >= 0; i--) {` ` ``// max_price has maximum of`` ``// price[i..n-1]`` ``if` `(price[i] > max_price)`` ``max_price = price[i];` ` ``// we can get profit[i] by taking`` ``// maximum of:`` ``// a) previous maximum, i.e.,`` ``// profit[i+1]`` ``// b) profit by buying at price[i]`` ``// and selling at max_price`` ``profit[i] = Math.Max(profit[i + 1],`` ``max_price - price[i]);`` ``}` ` ``/* Get the maximum profit with two`` ``transactions allowed After this loop,`` ``profit[n-1] contains the result */`` ``int` `min_price = price[0];` ` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// min_price is minimum price in`` ``// price[0..i]`` ``if` `(price[i] < min_price)`` ``min_price = price[i];` ` ``// Maximum profit is maximum of:`` ``// a) previous maximum, i.e.,`` ``// profit[i-1]`` ``// b) (Buy, Sell) at (min_price,`` ``// price[i]) and add profit of`` ``// other trans. stored in`` ``// profit[i]`` ``profit[i] = Math.Max(`` ``profit[i - 1],`` ``profit[i] + (price[i] - min_price));`` ``}`` ``int` `result = profit[n - 1];` ` ``return` `result;`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int``[] price = { 2, 30, 15, 10, 8, 25, 80 };`` ``int` `n = price.Length;` ` ``Console.Write(``"Maximum Profit = "`` ``+ maxProfit(price, n));`` ``}``}` `// This code is contributed by nitin mittal.`
## PHP
`= 0; ``\$i``--)`` ``{`` ``// max_price has maximum`` ``// of price[i..n-1]`` ``if` `(``\$price``[``\$i``] > ``\$max_price``)`` ``\$max_price` `= ``\$price``[``\$i``];` ` ``// we can get profit[i] by`` ``// taking maximum of:`` ``// a) previous maximum,`` ``// i.e., profit[i+1]`` ``// b) profit by buying at`` ``// price[i] and selling at`` ``// max_price`` ``if``(``\$profit``[``\$i` `+ 1] >`` ``\$max_price``-``\$price``[``\$i``])`` ``\$profit``[``\$i``] = ``\$profit``[``\$i` `+ 1];`` ``else`` ``\$profit``[``\$i``] = ``\$max_price` `-`` ``\$price``[``\$i``];`` ``}` ` ``// Get the maximum profit with`` ``// two transactions allowed.`` ``// After this loop, profit[n-1]`` ``// contains the result`` ``\$min_price` `= ``\$price``[0];`` ``for` `(``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++)`` ``{`` ``// min_price is minimum`` ``// price in price[0..i]`` ``if` `(``\$price``[``\$i``] < ``\$min_price``)`` ``\$min_price` `= ``\$price``[``\$i``];` ` ``// Maximum profit is maximum of:`` ``// a) previous maximum,`` ``// i.e., profit[i-1]`` ``// b) (Buy, Sell) at (min_price,`` ``// price[i]) and add`` ``// profit of other trans.`` ``// stored in profit[i]`` ``\$profit``[``\$i``] = max(``\$profit``[``\$i` `- 1],`` ``\$profit``[``\$i``] +`` ``(``\$price``[``\$i``] - ``\$min_price``));`` ``}`` ``\$result` `= ``\$profit``[``\$n` `- 1];`` ``return` `\$result``;``}` `// Driver Code``\$price` `= ``array``(2, 30, 15, 10,`` ``8, 25, 80);``\$n` `= sizeof(``\$price``);``echo` `"Maximum Profit = "``.`` ``maxProfit(``\$price``, ``\$n``);`` ` `// This code is contributed``// by Arnab Kundu``?>`
## Javascript
``
Output
`Maximum Profit = 100`
The time complexity of the above solution is O(n).
There is one more approach for calculating this problem using the Valley-Peak approach i.e. take a variable profit and initialize it with zero and then traverse through the array of price[] from (i+1)th position whenever the initial position value is greater than the previous value add it to variable profit.
But this approach does not work in this problem, that you can only buy and sell a stock two times. For example, if price [] = {2, 4, 2, 4, 2, 4} then this particular approach will give result 6 while in this given problem you can only do two transactions so answer should be 4. Hence this approach only works when we have a chance to do infinite transaction.
## C++
`#include ``using` `namespace` `std;`` ` `int` `main()``{`` ``int` `price[] = { 2, 30, 15, 10, 8, 25, 80 };`` ``int` `n = 7;`` ` ` ``// adding array`` ``int` `profit = 0;`` ` ` ``// Initializing variable`` ``// valley-peak approach`` ``/*`` ``80`` ``/`` ``30 /`` ``/ \ 25`` ``/ 15 /`` ``/ \ /`` ``2 10 /`` ``\ /`` ``8`` ``*/`` ``for` `(``int` `i = 1; i < n; i++)`` ``{`` ` ` ``// traversing through array from (i+1)th`` ``// position`` ``int` `sub = price[i] - price[i - 1];`` ``if` `(sub > 0)`` ``profit += sub;`` ``}`` ` ` ``cout << ``"Maximum Profit="` `<< profit;`` ``return` `0;``}`` ` `// This code is contributed by RohitOberoi.`
## Java
`import` `java.util.*;` `class` `GFG {` ` ``public` `static` `void` `main(String[] args) {`` ``int` `price[] = { ``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80` `};`` ``int` `n = ``7``;` ` ``// adding array`` ``int` `profit = ``0``;` ` ``// Initializing variable`` ``// valley-peak approach`` ``/*`` ` ` ``80`` ``/`` ``30 /`` ``/ \ 25`` ``/ 15 /`` ``/ \ /`` ``2 10 /`` ``\ /`` ``8`` ``*/`` ``for` `(``int` `i = ``1``; i < n; i++) {` ` ``// traversing through array from (i+1)th`` ``// position`` ``int` `sub = price[i] - price[i - ``1``];`` ``if` `(sub > ``0``)`` ``profit += sub;`` ``}` ` ``System.out.print(``"Maximum Profit="` `+ profit);`` ``}``}` `// This code is contributed by umadevi9616`
## Python3
`if` `__name__ ``=``=` `'__main__'``:`` ``price ``=` `[ ``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80``] ;`` ``n ``=` `7``;` ` ``# adding array`` ``profit ``=` `0``;` ` ``# Initializing variable`` ``# valley-peak approach`` ``'''`` ``80`` ``/`` ``30 /`` ``/ \ 25`` ``/ 15 /`` ``/ \ /`` ``2 10 /`` ``\ /`` ``8`` ``'''`` ``for` `i ``in` `range``(``1``,n):` ` ``# traversing through array from (i+1)th`` ``# position`` ``sub ``=` `price[i] ``-` `price[i ``-` `1``];`` ``if` `(sub > ``0``):`` ``profit ``+``=` `sub;`` ` ` ``print``(``"Maximum Profit="` `,profit);` `# This code is contributed by gauravrajput1`
## C#
`using` `System;``public` `class` `GFG {` ` ``public` `static` `void` `Main(String[] args) {`` ``int` `[]price = { 2, 30, 15, 10, 8, 25, 80 };`` ``int` `n = 7;` ` ``// adding array`` ``int` `profit = 0;` ` ``// Initializing variable`` ``// valley-peak approach`` ``/*`` ` ` ``80`` ``/`` ``30 /`` ``/ \ 25`` ``/ 15 /`` ``/ \ /`` ``2 10 /`` ``\ /`` ``8`` ``*/`` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// traversing through array from (i+1)th`` ``// position`` ``int` `sub = price[i] - price[i - 1];`` ``if` `(sub > 0)`` ``profit += sub;`` ``}` ` ``Console.Write(``"Maximum Profit="` `+ profit);`` ``}``}` `// This code is contributed by gauravrajput1`
## Javascript
``
Output
`Maximum Profit=100`
The time and space complexity is O(n) and O(1) respectively.
Another approach:
Initialize four variables for taking care of the first buy, first sell, second buy, second sell. Set first buy and second buy as INT_MIN and first and second sell as 0. This is to ensure to get profit from transactions. Iterate through the array and return the second sell as it will store maximum profit.
## C++
`#include ``#include``using` `namespace` `std;` `int` `maxtwobuysell(``int` `arr[],``int` `size) {`` ``int` `first_buy = INT_MIN;`` ``int` `first_sell = 0;`` ``int` `second_buy = INT_MIN;`` ``int` `second_sell = 0;`` ` ` ``for``(``int` `i=0;i
## Java
`import` `java.util.*;``class` `GFG{` `static` `int` `maxtwobuysell(``int` `arr[],``int` `size) {`` ``int` `first_buy = Integer.MIN_VALUE;`` ``int` `first_sell = ``0``;`` ``int` `second_buy = Integer.MIN_VALUE;`` ``int` `second_sell = ``0``;`` ` ` ``for``(``int` `i = ``0``; i < size; i++) {`` ` ` ``first_buy = Math.max(first_buy,-arr[i]);`` ``first_sell = Math.max(first_sell,first_buy+arr[i]);`` ``second_buy = Math.max(second_buy,first_sell-arr[i]);`` ``second_sell = Math.max(second_sell,second_buy+arr[i]);`` ` ` ``}`` ``return` `second_sell;``}` `public` `static` `void` `main(String[] args)``{`` ``int` `arr[] = {``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80``};`` ``int` `size = arr.length;`` ``System.out.print(maxtwobuysell(arr,size));``}``}` `// This code is contributed by gauravrajput1`
## Python3
`import` `sys` `def` `maxtwobuysell(arr, size):`` ``first_buy ``=` `-``sys.maxsize;`` ``first_sell ``=` `0``;`` ``second_buy ``=` `-``sys.maxsize;`` ``second_sell ``=` `0``;` ` ``for` `i ``in` `range``(size):` ` ``first_buy ``=` `max``(first_buy, ``-``arr[i]);`` ``first_sell ``=` `max``(first_sell, first_buy ``+` `arr[i]);`` ``second_buy ``=` `max``(second_buy, first_sell ``-` `arr[i]);`` ``second_sell ``=` `max``(second_sell, second_buy ``+` `arr[i]);` ` ` ` ``return` `second_sell;` `if` `__name__ ``=``=` `'__main__'``:`` ``arr ``=` `[ ``2``, ``30``, ``15``, ``10``, ``8``, ``25``, ``80` `];`` ``size ``=` `len``(arr);`` ``print``(maxtwobuysell(arr, size));` `# This code is contributed by gauravrajput1`
## C#
`using` `System;` `public` `class` `GFG{` `static` `int` `maxtwobuysell(``int` `[]arr,``int` `size) {`` ``int` `first_buy = ``int``.MinValue;`` ``int` `first_sell = 0;`` ``int` `second_buy = ``int``.MinValue;`` ``int` `second_sell = 0;`` ` ` ``for``(``int` `i = 0; i < size; i++) {`` ` ` ``first_buy = Math.Max(first_buy,-arr[i]);`` ``first_sell = Math.Max(first_sell,first_buy+arr[i]);`` ``second_buy = Math.Max(second_buy,first_sell-arr[i]);`` ``second_sell = Math.Max(second_sell,second_buy+arr[i]);`` ` ` ``}`` ``return` `second_sell;``}` `public` `static` `void` `Main(String[] args)``{`` ``int` `[]arr = {2, 30, 15, 10, 8, 25, 80};`` ``int` `size = arr.Length;`` ``Console.Write(maxtwobuysell(arr,size));``}``}` `// This code is contributed by gauravrajput1`
## Javascript
``
Output
`100`
Time Complexity: O(N)
Auxiliary Space: O(1)
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# Teen Numbers - PowerPoint PPT Presentation
Teen Numbers. What are teen numbers?. 11. 12. 13. 14. 15. 16. 17. 18. 19. Teen Numbers numbers made of tens and units. 11. 15. 19. tens = =10 units = =1. Teen numbers are 2-digit numbers. Use nursery rhymes to recognize that teen numbers
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### Teen Numbers
11
12
13
14
15
16
17
18
19
Teen Numbersnumbers made of tens and units
11
15
19
tens = =10
units = =1
• Use nursery rhymes to recognize that teen numbers
• Are made of tens and units
• Are ten and some more.
• Use Jack Be Nimble to recognize that teen numbers
• Are ten and some more.
Jack be Nimble
Jack be quick
Jack jump over the candlestick.
figure out how many candles he has to jump over.
• Count the number of candles.
• Match the number of candles to the correct numeral.
• Choose the correct numeral from the circles.
1 candle
10
13
11
Try Again
go back
next
4 candles
16
12
14
Try Again
go back
next
5 candles
15
17
19
Try Again
go back
next | 450 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-47 | latest | en | 0.814983 |
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## Transcription
1 CS 494/594 Computer and Network Security Dr. Jinyuan (Stella) Sun Dept. of Electrical Engineering and Computer Science University of Tennessee Fall
2 Introduction to Cryptography What is cryptography? Types of cryptography Attacks on cryptosystem Early ciphers SKC, PKC, Hash Security notions 2
3 What Is Cryptography? Cryptography is the art of secret writing Traditional use: encryption Cryptographic systems: algorithm+secret Cryptology: cryptography+cryptanalysis 3
4 Types of Cryptography Operations used to transform plaintext to ciphertext: - Substitution: each element in plaintext is mapped into another element - Transposition: elements in plaintext are rearranged Number of keys used: - Single-key: secret-key cryptography - Two-key: public-key cryptography - Zero-key: hash functions The way plaintext is processed: - Block cipher: input & output in blocks - Stream cipher: input & output in bits 4
5 Conventional Cryptography: Symmetric Encryption Plaintext: original message Encryption algorithm: substitutions/transpositions Secret key: independent of plaintext and algorithm Ciphertext: depends on plaintext and key, appears random Decryption algorithm: reverse of encryption 5
6 Attacking Cryptosystems Cryptanalysis: attempts to deduce the plaintext and/or the key being used, with knowledge of the nature of the algorithm + general characteristics of plaintext + some sample plaintext-ciphertext pairs Brute-force attack: attacker tries every possible key on a piece of ciphertext until an intelligible translation into plaintext is obtained. On average, half of all possible keys must be tried to achieve success. 6
7 Types of Cryptanalytic Attacks 7
8 Brute-force Attacks Average Time Required for Exhaustive Key Search 8
9 Early Ciphers: Substitution Substitution: letters of plaintext are replaced by other letters or by numbers or symbols, involves replacing plaintext bit patterns with ciphertext bit patterns Caesar cipher Captain Midnight Secret Decoder Rings Monoalphabetic cipher Hill cipher Polyalphabetic cipher (Vigenere) One-time pad 9
10 Caesar Cipher Replacing each letter of the alphabet with the letter three places further down the alphabet, e.g., Captain Midnight Secret Decoder Ring (slightly enhanced): If a numerical value is assigned to each letter, for each plaintext letter p, the ciphertext letter C=E(k, p)=(p+k) mod 26 where k takes on the value in [1,25]: Subject to brute-force attack: simply try all 25 possible k 10
11 Monoalphabetic Cipher The "cipher" line can be any permutation of the 26 alphabetic characters, 26! or around 4x10 26 possible keys: Is it secure enough to resist cryptanalysis? Consider: By exploiting the regularities of the language and counting letter frequencies: The following can be recovered: 11
12 Hill Cipher Takes m successive plaintext letters and substitutes for them m ciphertext letters The substitution is determined by m linear equations in which each character is assigned a numerical value (a = 0, b = 1... z = 25) For example: m = 3 for plaintext paymoremoney and encryption key we have ciphertext: LNSHDLEWMTRW 12
13 Polyalphabetic Cipher Use different monoalphabetic substitutions as one proceeds through the plaintext message Vigenere cipher: best known polyalphabetic cipher. The set of related monoalphabetic substitution rules consists of the 26 Caesar ciphers, with shifts of 0 through 25. For keyword deceptive and plaintext we are discovered save yourself, the encryption works as: Vulnerable to cryptanalysis: the key and plaintext share the same language regularities and frequency distribution of letters. Solutions? 13
14 One-time Pad Use a random key that is as long as the message so that the key need not be repeated The key is used to encrypt and decrypt a single message, and then is discarded Perfectly secure: unbreakable because it produces random output (from the random key) that bears no statistical relationship to the plaintext Drawbacks: large quantities of random keys needed, key distribution and protection (both sender and receiver) 14
15 Early Ciphers: Transposition Different from substituting a ciphertext symbol for a plaintext symbol Transposition cipher: performs some sort of permutation/rearrangement on plaintext letters Cryptanalysis is straightforward: a transposition cipher has the same letter frequency as plaintext Can be made more secure by performing more than one stage of transposition (result is not easily reconstructed) 15
16 Rail Fence Cipher Simplest transposition cipher Plaintext is written down as a sequence of diagonals and then read off as a sequence of rows Plaintext: Ciphertext: Trivial to attack 16
17 Double Transposition A more complex scheme: permute the order of the columns with a key Double transposition: more than one permutation, number the above plaintext letters 1-28 and after the first permutation we have After the 2 nd permutation? 17
18 Secret Key Cryptography (SKC) Aka: conventional cryptography, symmetric cryptography Use a single key Ciphertext about the same length as plaintext Examples: Captain Midnight code, monoalphabetic cipher, DES, AES, RC4 plaintext encryption shared secret key ciphertext ciphertext decryption plaintext 18
19 SKC Applications Transmitting over an insecure channel Secure storage on insecure media Authentication: Integrity check: message authentication code (MAC), aka, message integrity check (MIC) 19
20 Public Key Cryptography (PKC) Aka: asymmetric cryptography, invented in 1970s Use two keys: a public key known to everyone, a private key kept secret to the owner Encryption/decryption: encryption can be done by everyone using the recipient s public key, decryption can be done only by the recipient with his/her private key encryption plaintext ciphertext pub B ( puba, pri A) ( pubb, prib) pri B ciphertext decryption plaintext 20
21 PKC Applications Everything that SKC does can be done by PKC Transmitting over an insecure channel Secure storage over insecure media Authentication Key exchange: establish a shared session key with PKC 21
22 PKC Applications (Cont d) Digital signature: non-repudiation plaintext sign Signed message pri A ( puba, pri A) ( pubb, prib) pub A Signed message verify True or false 22
23 Hash Functions Aka: message digests, one-way transformations Take a message m of arbitrary length (transformed into a string of bits) and computes from it a fixed-length (short) number h(m) Properties: - easy-to-compute: for any message m, it is relatively easy to compute h(m) - non-reversible: given h(m), there is no way to find an m that hashes to h(m) except trying all possibilities of m - computationally infeasible to find m and m such that h(m)=h(m ) and m!=m 23
24 Applications of Hash Functions Password hashing Message integrity: keyed hash File fingerprint Downline load security Digital signature efficiency 24
25 Security Notions Unconditionally secure: ciphertext generated by the scheme does not contain enough information to determine uniquely the corresponding plaintext, no matter how much ciphertext is available. Perfectly secure, unlimited power of adversary Provably secure: under the assumption of well-known hard mathematical problem, e.g., factoring large numbers, discrete logarithm problem Computationally secure: if cost of breaking the cipher exceeds the value of the encrypted information, or time required to break the cipher exceeds the useful lifetime of the information, practical security 25
26 Cryptography vs. Steganography Cryptography conceals the context of message Steganography conceals the existence of message, useful when the fact of secret communication should be concealed - an arrangement of words/letters of the overall message spells out the hidden message - character marking: selected letters overwritten in pencil, not visible unless the paper is held at an angle to bright light - invisible ink: substances used for writing but leave no visible trace until heat or some chemical is applied to the paper - pin punctures: small pin punctures on selected letters are not visible unless the paper is held up in front of a light Drawbacks of steganography: high overhead to hide a relatively few bits of information, becomes worthless once the system is discovered (can make insertion depend on key) 26
27 Reading Assignments [Kaufman] Chapter 2 27
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Security and Cryptography just by images Pascal Lafourcade 2009 pascal.lafourcade@imag.fr 1 / 52 Motivations Applications 2 / 52 Motivations Secrecy or Confidentiality Alice communicates with the White | 7,953 | 35,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-51 | latest | en | 0.754205 |
https://hopsblog-hop.blogspot.com/2013/06/even-mit-students-can-make-misteaks.html | 1,723,312,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00351.warc.gz | 234,169,024 | 12,738 | ## Sunday, June 30, 2013
### Even MIT students can make misteaks.
Finding Delta V between various places in the solar system has been one of my hobbies. I like to learn as well as share what I know about this topic. So I occasionally Google search strings that include the term "Delta V".
Googling: Delta V Mars. The first hit is a Wikipedia article, the second hit is How Much Delta V do you need to get to Mars - Yahoo! Answers.
A fellow who calls himself ronwizfr confidently asserts that the delta V from earth orbit to Mars transfer is 6.6 km/s. Ronwizfr cites an MIT student project for a class called Solving Complex Problems.
The MIT page cited is a well done presentation until the students get to constants:
Then the students made an arithmetic error. The distance from the earth to the sun is 1.49X1011 meters, not 1.49X1010 meters. Their radius for Mars' orbit is also off by a factor of 10.
The MIT students correctly plug in these wrong quantities to get speeds that are off by sqrt(10):
The velocity of the earth is not 94,384 m/s but closer to 29,846 m/s. Likewise velocity of Mars isn't 74,467 m/s but closer to 23,548 m/s. Their quantities would have been correct if the earth and Mars were .1 A.U. and .152 A.U. from the sun.
Even the very accomplished can make errors. So we should examine every assertion, no matter the source. My dad used to tell me we all put our pants on one leg at a time. His way of saying we're all human and capable of making mistakes.
Tage said...
Hi Mr. David,
It seems you didn't do any arithmetic errors in this blog post either, so I'll just mention that when you wrote "if the earth and moon were .1 A.U. and .152 A.U. from the sun", you probably meant "Mars" and not "moon". Excellent post either way, though. :)
The last days I have spent several hours going through the calculations that you did in your Hohmann Excel spreadsheet, planning to use them for my astrophysics project on Mars colonization. When going through your spreadsheet I assumed (by the quality of the spreadsheet) that you were working for NASA or something, but you are refering to delta V calculations as a hobby? Are you completely self-taught? No academic background in natural sciences? I think you've done a very impressive piece of work either way.
Best regards,
Tage Augustson
MSc of Mechanical Engineering
Hop David said...
I'm strictly amateur. My credentials: I graduated in the top 60% of my high school class.
I am passionate about space exploration. Maybe we are on the cusp of opening a new frontier. If so, that'd be a major turning point in human history. | 617 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.959742 |
https://www.convertunits.com/from/kilometer/to/line | 1,627,713,700,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00453.warc.gz | 733,024,967 | 13,042 | ## ››Convert kilometre to line
kilometer line
Did you mean to convert kilometer to line line [small]
How many kilometer in 1 line? The answer is 2.1167E-6.
We assume you are converting between kilometre and line.
You can view more details on each measurement unit:
kilometer or line
The SI base unit for length is the metre.
1 metre is equal to 0.001 kilometer, or 472.43350498417 line.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilometres and lines.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kilometer to line
1 kilometer to line = 472433.50498 line
2 kilometer to line = 944867.00997 line
3 kilometer to line = 1417300.51495 line
4 kilometer to line = 1889734.01994 line
5 kilometer to line = 2362167.52492 line
6 kilometer to line = 2834601.02991 line
7 kilometer to line = 3307034.53489 line
8 kilometer to line = 3779468.03987 line
9 kilometer to line = 4251901.54486 line
10 kilometer to line = 4724335.04984 line
## ››Want other units?
You can do the reverse unit conversion from line to kilometer, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kilometer
A kilometre (American spelling: kilometer, symbol: km) is a unit of length equal to 1000 metres (from the Greek words khilia = thousand and metro = count/measure). It is approximately equal to 0.621 miles, 1094 yards or 3281 feet.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 528 | 1,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-31 | latest | en | 0.835877 |
https://www.mathworksheets4kids.com/direct-inverse-variation-word-problems.php | 1,723,485,050,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641045630.75/warc/CC-MAIN-20240812155418-20240812185418-00821.warc.gz | 676,303,145 | 7,730 | Variation Word Problems Worksheets | Direct, Inverse, Joint, Combined Variation
Learn how to apply the concept of variation in real-life situations with these 15 pdf worksheets exclusively focusing on word problems, involving direct variation, inverse variation, joint variation and combined variation. A knowledge in solving direct and inverse variation is a prerequisite to solve these word problems exclusively designed for high school students. Try some of these worksheets for free!
Direct Variation Word Problems
The key to solve these word problems is to comprehend the problem, figure out the relationship between two entities and formulate an equation in the form y = kx. Find the constant of variation, substitute the value and solve.
Inverse Variation Word Problems
In this set of inverse variation worksheet pdfs, read the word problem and formulate an equation in the form y = k / x. Find the constant of variation, plug in the values and solve the word problems.
Direct and Inverse Variation: Mixed Word Problems
This collection of printable worksheets is packed with exercises involving a mix of direct and inverse variation word problems. The learner should identify the type of variation and then solves accordingly.
Joint and Combined Variation
The self-explanatory word problems here specifically deal with joint and combined variations.
Mixed Word Problems: Direct, Inverse, Joint and Combined
Master the four types of variation with this potpourri of 15 word problems, perfect for high schoolers to recapitulate the concepts learnt.
Related Worksheets | 291 | 1,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.904287 |
https://the-equivalent.com/what-is-4-tablespoons-equivalent-to/ | 1,679,654,043,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00103.warc.gz | 638,958,205 | 13,881 | # What is 4 tablespoons equivalent to
## What is 4 tablespoons equivalent to in a cup?
So when you need 4 tablespoons, you can use ¼ cup.
## Does 4 Tbs equal 1 4 cup?
How Many Tablespoons are in a 1/4 cup? ¼ cup can be converted to 4 tablespoons.
## Does 4 tablespoons equal 1 8 cup?
Volume Equivalents (liquid)*2 tablespoons1/8 cup1 fluid ounce4 tablespoons1/4 cup2 fluid ounces5 1/3 tablespoons1/3 cup2.7 fluid ounces8 tablespoons1/2 cup4 fluid ounces6 more rows
## What is another measurement for 4 tablespoons?
¼ U.S. cup¼ U.S. cup = 4 tablespoons.
## What is 1/2 a cup in tablespoons?
Cup and tablespoon conversionsCupsTablespoonsFluid Ounces1/2 cup8 tablespoons4 fluid ounces1/3 cup5 tablespoons plus 1 teaspoon2 ⅓ fluid ounces1/4 cup4 tablespoons2 fluid ounces1/6 cup2 tablespoons plus 2 teaspoons1 ⅓ fluid ounce6 more rows
## Does 2 tablespoons equal 1 4 cup?
There are 16 tablespoons in one cup, so the conversion formula is Tablespoons = Cups x 16….How to Convert Tbsp to Cups.TablespoonsCups2 tablespoons⅛ cup4 tablespoons¼ cups5 tablespoons + 1 teaspoon⅓ cup8 tablespoons½ cup3 more rows•Apr 22, 2022
## How do you measure 1/8 of a cup?
Even if you have all the right measuring tools, a lot of people can have difficulty using the right cooking measurements….Some handy kitchen knowledge.1 tablespoon (tbsp) =3 teaspoons (tsp)1/8 cup =2 tablespoons1/6 cup =2 tablespoons + 2 teaspoons1/4 cup =4 tablespoons1/3 cup =5 tablespoons + 1 teaspoon16 more rows
## What is half of 1/4 cup in tablespoons?
1. Half of ¼ cup is equivalent to 2 tbsp.
## How many dry tablespoons are in a cup?
161 US Cup = 16 US tablespoons (tbsp)
## How do you measure a tablespoon?
The most accurate way to measure a tablespoon is to use a measuring spoon. If you don’t have a measuring spoon, you can use measurement equivalents. For example, 3 level teaspoons makes 1 tablespoon. 1/16 of a cup is also equivalent to 1 tablespoon, and 15 ml of any liquid is equal to 1 tablespoon.
## How big is a tbsp?
A tablespoon is a unit of measure equivalent to 1/16 cup, three teaspoons, or 1/2 fluid ounce in the United States. It is approximately or (in some countries) precisely equal to 15 ml.
## What is 3 tablespoons equivalent to in cups?
Tablespoon to Cup Conversion TableTablespoonsCups3 tbsp0.1875 c4 tbsp0.25 c5 tbsp0.3125 c6 tbsp0.375 c36 more rows
## What is a 1/4 cup of butter?
Butter Measurements Conversion ChartCupsSticksTablespoons1/4 cup1/2 stick4 tablespoons1/2 cup1 stick8 tablespoons3/4 cup1 1/2 sticks12 tablespoons1 cup2 sticks16 tablespoons4 more rows•May 19, 2022
## How many sticks of butter is a 1/4 cup?
Our butter sticks are easy to measure! One full stick of butter equals 1/2 cup, or 8 tablespoons. Our half sticks equal 1/4 cup of butter, or 4 tablespoons. They can be used interchangeably in recipes.
## What is half of 1/4 cup in cooking?
2 TablespoonsSince 1/4 cup is 4 Tablespoons, we can easily cut that in half to 2 Tablespoons. So half of 1/4 cup is 2 Tablespoons.
## What’s half of 1/4 cup of butter?
2 tbsp 2Half of ¼ cup is equivalent to 2 tbsp. 2. Half of ⅓ cup is equivalent to 2 tbsp + 2 tsp. 3.
## What is a tablespoon?
A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating. By extension, the term is used as a measure of volume in cooking.
## How many cups is 1 tablespoon?
The conversion factor from tablespoons to cups is 0.062500000000211, which means that 1 tablespoon is equal to 0.062500000000211 cups:
## How many tablespoons are in a cup?
An alternative is also that one cup is approximately four times four tablespoons.
## How many ounces are in a cup?
In the United States, the customary cup is half of a liquid pint or 8 U.S. customary fluid ounces. One customary cup is equal to 236.5882365 millilitres.
## What is a tablespoon?
A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating. By extension, the term is used as a measure of volume in cooking.
## How many ounces are in a cup?
In the United States, the customary cup is half of a liquid pint or 8 U.S. customary fluid ounces. One customary cup is equal to 236.5882365 millilitres.
## Convert 4 Tablespoons to Teaspoons
To calculate 4 Tablespoons to the corresponding value in Teaspoons, multiply the quantity in Tablespoons by 3.0000000000122 (conversion factor). In this case we should multiply 4 Tablespoons by 3.0000000000122 to get the equivalent result in Teaspoons:
## How to convert from Tablespoons to Teaspoons
The conversion factor from Tablespoons to Teaspoons is 3.0000000000122. To find out how many Tablespoons in Teaspoons, multiply by the conversion factor or use the Volume converter above. Four Tablespoons is equivalent to twelve Teaspoons.
## Definition of Tablespoon
In the United States a tablespoon (abbreviation tbsp) is approximately 14.8 ml (0.50 US fl oz). A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating.
## Definition of Teaspoon
A teaspoon (occasionally “teaspoonful”) is a unit of volume, especially widely used in cooking recipes and pharmaceutic prescriptions. It is abbreviated as tsp. or, less often, as t., ts., or tspn.
## Tablespoon. Conversion Chart
This converter features contemporary units of capacity. There is also a special converter for historical units of volume you might want to visit for ancient, medieval and other old units that are no longer used.
## Conversion settings explained
First of all, you don’t have to change any settings to use the converter. It’s absolutely optional.
## Number of significat figures
Do you want rounded off figures or scientifically precise ones? For everyday conversions we recommend choosing 3 or 4 significant digits. If you want maximum precision, set the number to 9
## Digit groups separator
Choose how you want to have your digit groups separated in long numbers:
## Apothecaries
The traditional English apothecaries’ system defines weights, not volumes. However, there was apothecaries’ system for volumes too, though less commonly used. Before introduction of the imperial units, all apothecaries’ measures were based on the wine gallon which later became the base of U.S. liquid gallon.
Imperial and U.S. definitions of these units slightly differ. We use Imperial definitions here.
Historically the terms beer and ale referred to distinct brews. From the mid 15th century until 1803 in Britain “ale” casks and “beer” casks differed in the number of gallons they contained. We provide conversions for the latest standard (1824) of units that are based on imperial gallons.
## What is a tablespoon?
A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating. By extension, the term is used as a measure of volume in cooking.
## How many teaspoons are in a tablespoon?
The conversion factor from tablespoons to teaspoons is 3.0000000000122, which means that 1 tablespoon is equal to 3.0000000000122 teaspoons:
## How many ml is a teaspoon?
or, less often, as t., ts., or tspn. In the United States one teaspoon as a unit of culinary measure is 1⁄3 tablespoon, that is, 4.92892159375 ml; it is exactly 1 1⁄3 US fluid drams, 1⁄6 US fl oz, 1⁄48 US cup, and 1⁄768 US liquid gallon and 77⁄256 or 0.30078125 cubic inches. For nutritional labeling on food packages in the US, the teaspoon is defined as precisely 5 ml. | 2,052 | 8,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-14 | latest | en | 0.846013 |
https://ios.lisisoft.com/s/function-evaluated.html | 1,701,931,365,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00169.warc.gz | 372,467,809 | 58,690 | # Function Evaluated iOS Apps
## Best Apps Error Function Beta Function
1
Function Well
By Darren Bain ( Free )
Located in the iconic Woolstore Building in the heart of Newstead, our gym encompass a holistic view on health and fitness through our three separate zones. Download the App to access your profile, program, results ...
2
Function
By Cardiff and Vale University Local Health Board ( Free )
WHAT IS FUNCTIONAL ANALYSIS? A child s behaviour is like an Iceberg. We just see the behaviour. We don t see what is beneath the surface of the behaviour. This stops us understanding the reason for ...
3
Function Cards
By Lisa Englard ( \$0.99 )
Conceptual understanding of functions is a requisite foundation for success in Algebra and beyond. Students need to understand that a function is a rule that assigns to each input exactly one output. The Function Cards app ...
4
Function Integrator
By Polytope Media ( \$0.99 )
Calculate definite integrals on your iOS device Key in the formula you want to integrate, and the bounds, and Function Integrator will calculate the integral using a fast adaptive numerical technique. The `Options` tab allows ...
5
Function Express
By Bermotech ( Free )
Function Express is a handy mobile app, created by a group of teens aged 13-15 during a week at Bermotech app development camp. Function Express is not only fun, but educational too. The mobile app offers ...
6
Function-iP
By Dirk Hoffmann ( \$2.99 )
The calculation and visualisation app for linear functions and integrals. Function-iP is the choice for lab workers,scientists,students teachers.... Enter the a,x,b parameters and the left and right borders for your integral. The graph is drawn and real time ...
7
Function Generator
By Thomas Gruber ( Free )
This function generator plays various audio test tones (sine, rectangular, triangular, saw tooth) from 20Hz to 20kHz. The frequency slider is used to set the required output frequency. The tool can also play a sweep from ...
8
function.repair
By Demergo Studios ( \$1.99 )
Function.repair is a fresh new 2D platform shooter with a unique, no-gravity movement style. You assume the role of Fixbot, one of the many fixbots, who awakens to find himself in the middle of a massive, ...
9
Temple Function
By SHREE CEMENT LIMITED ( Free )
Shree Cement is the fastest growing Cement Manufacturer in India. This app would help you get the latest updates on the events and celebrations at Shree. Shree Parivar is pleased to welcome you all for ...
10
Form Plus Function
By Lokolapps, LLC ( Free )
Welcome to the new era of customer service Have a question or concern? Simple, just tap on the concierge button in the app and ask the question. We will respond shortly to you with an ...
11
Function Studios Inc.
By MINDBODY, Incorporated ( Free )
Download the Function Studios Inc. app today to plan and schedule your classes From this mobile app you can view class schedules, sign-up for classes, view ongoing promotions, as well as view the studio s ...
12
Error Function
By Business Compass LLC ( \$0.99 )
1. Computes the value of the error function (a.k.a the Gauss error function) evaluated at a value of x. The value returned will be the error function evaluated along the integral from zero to x. 2. ...
13
Gamma Function
By Business Compass LLC ( \$4.99 )
Key Features: 1. Gamma Function 2. Incomplete Gamma Function 3. Regularized Incomplete Gamma Function 4. Print result Requirement: This application requires active data connection to work properly.
14
Wedding and Function
By Realm Computers (Pty) Ltd. ( Free )
South Africa`s largest wedding publication with thousands of wedding service providers in each region of the country.
15
Beta Function
By Business Compass LLC ( \$2.99 )
Key Features: 1. Beta Function Calculator 2. Incomplete Beta Function Calculator 3. Regularized Incomplete Beta Function Calculator 4. Print result Requirement: Active data connection is required for the application to function properly.
### Gamma Function Error Function Error Function Evaluated
1
Error Function
By Business Compass LLC ( \$0.99 )
1. Computes the value of the error function (a.k.a the Gauss error function) evaluated at a value of x. The value returned will be the error function evaluated along the integral from zero to x. 2. ...
2
K-12 Resource Collection
By Educational Resource Acquisition Consortium ( Free )
ERAC's Evaluated & Approved K-12 Resource Collection is an online, searchable compilation of resources evaluated by BC educators for use in classrooms. Educators and administrators can quickly find products to meet their resource needs.
3
TCY-CMS
By TCY LEARNING SOLUTIONS PRIVATE LIMITED ( Free )
Salient feature of this app is the uploading of evaluated test sheets of the students and then submitting the scores for individual student as per the designed rubrics. This helps the students to see their ...
4
Japanese Hot Springs
By info arata ( Free )
It is an application that can search hot springs all over Japan. You can search hot springs from your current location and display the route. You can display multiple hot springs near the current location. Explanation about hot ...
5
SmartBridge - Sensor Technology 4.0
By Pepperl-Fuchs GmbH ( Free )
SmartBridge eases the commissioning of sensors with its clear and easy access to parameters, identification data, and measured values. It is also important that the sensor function can be evaluated for maintenance tasks without the ...
6
Mobile Statistics Professor
By Business Compass LLC ( \$29.99 )
Key Features: 1. Analysis of Variance (ANOVA) 2. Beta Function 3. Advanced Scientific Calculator 4. Covariance & Correlation 5. Effective Size 6. Error Function 7. Gamma Function 8. Probability a. CDF & PDF - Chi-Square, Fisher F-Distribution, Normal Distribution, Standard Normal Distribution, Student`s t-Distribution c. ...
7
Become Indispensable
By Aimee Robertson ( Free )
We need a place or way to plan some of our own things. Especially for the good habit of regular exercise, Become Indispensable App is very simple and practical. The function is introduced as follows: [Function 1] ...
8
CB Strategies Tactile
By Carl Zeiss AG ( 89.990 )
The app contains the digital version of the Cookbook Measuring Strategies for tactile Coordinate Metrology and includes some of the most common measuring tasks (as evaluated in a study by the Global Application Knowledge ...
9
DCircuit Lab
By Fabrizio Boco ( 5.990 )
DCircuit Lab is a tool for simulating combinatorial and sequential digital circuits. Circuits can be simulated in two ways: 1) one single step: all the outputs values are evaluated at the same time 2) step by step: outputs ...
10
All Pretty Good
By Duc Thang Nguyen ( Free )
We have to face and deal with many things every day. How to make yourself comfortable. All Pretty Good App can help you. The function is introduced as follows: Function 1: Record and plan them according to ...
11
EasyGroup Plus (Full): Send SMS/E-Mail to Group/Contacts, Group Contacts
By Wenhsin Cheng ( Free )
Send SMS/E-Mail to Group/Contacts, Group Contacts Do you know a HIDDEN function of the Contacts(AddressBook)? Do you know you can GROUP your contacts(AddressBook)? If you desire to have a try, just download it - Sync with Contacts(AddressBook) ...
12
Quick PASI3
By Aoi Ogihara ( \$1.99 )
The PASI score is very useful for evaluating the severity of psoriasis. However, it is difficult to do a sum in our head at outpatient care. QuickPASI tells you the score in a few steps. The ...
13
Checkup Study
By Digisight ( Free )
Checkup Study app is a vision testing application being evaluated in a clinical trial. The app enables patients to test their visual function at home on a smart device (supported iPod, iPhone, or iPad) and ...
14
Simple Colorful Calculator
By thomas Perks ( 0.99 )
Make maths fun with this beautiful and colorful, basic calculator Features: - Addition function - Subtraction function - Division function - Multiplication function - Cancel/reset all function Enjoy
15
Clockface Test
By Kevin Lease ( Free )
Clockface Test is a medical app in which the ability to create a normal clock face is evaluated. This cognitive test can be used to evaluate for dementia such as Alzheimer`s disease, delirium, and brain ... | 1,841 | 8,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.847083 |
https://de.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/1746203 | 1,591,435,602,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00166.warc.gz | 295,439,206 | 15,585 | Cody
# Problem 8. Add two numbers
Solution 1746203
Submitted on 8 Mar 2019 by Ahsanul Islam
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) | 167 | 517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-24 | latest | en | 0.663361 |
https://www.maximintegrated.com/cn/design/technical-documents/app-notes/1/187.html | 1,642,565,099,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00205.warc.gz | 841,873,887 | 39,994 | # 1-Wire搜索算法
## 绪论
Maxim的每片1-Wire器件都有唯一的64位注册码，它存储在只读存储器(ROM)中。在1-Wire网络中，注册码用于1-Wire主机对从机器件进行逐一寻址。如果1-Wire网络中从机 器件的ROM码是未知的，可以通过搜索算法来找到此码。本文不仅详细地解释了搜索算法，而且 还提供了实现快速整合的例程。该算法适用于任何具有1-Wire接口特性的现有产品及未来产品。
## 搜索算法
Bit (true) Bit (complement) Information Known 0 0 There are both 0s and 1s in the current bit position of the participating ROM numbers. This is a discrepancy. 0 1 There are only 0s in the bit of the participating ROM numbers. 1 0 There are only 1s in the bit of the participating ROM numbers. 1 1 No devices participating in search.
Master Slave 1-Wire reset stimulus Produce presence pulse Write search command (normal or alarm) Each slave readies for search. Read 'AND' of bit 1 Each slave sends bit 1 of its ROM number. Read 'AND' of complement bit 1 Each slave sends complement bit 1 of its ROM number. Write bit 1 direction (according to algorithm) Each slave receives the bit written by master, if bit read is not the same as bit 1 of its ROM number then go into a wait state. Read 'AND' of bit 64 Each slave sends bit 64 of its ROM number. Read 'AND' of complement bit 64 Each slave sends complement bit 64 of its ROM number. Write bit 64 direction (according to algorithm) Each slave receives the bit written by master, if bit read is not the same as bit 64 of its ROM number then go into a wait state.
Search Bit Position vs Last Discrepancy Path Taken = take the '1' path < take the same path as last time (from last ROM number found) > take the '0' path
DS2480B系列串口到1-Wire线路的驱动程序在硬件中实现了部分与本文档中相同的搜索算法；详 细资料请参阅DS2480B数据资料和应用笔记192，DS2480B串行接口1-Wire线驱动器的使用的详细介绍；从DS2490 USB口到1-Wire桥接器硬件电路 中实现了整个搜索过程。
## First
‘FIRST’操作是搜索1-Wire总线上的第一个从机器件。该操作是通过将LastDiscrepancy、 LastFamilyDiscrepancy和LastDeviceFlag置零，然后进行搜索完成的。最后ROM码从ROM_NO寄存器中读出。若1-Wire总线上没有器件，复位序列就检测不到应答脉冲，搜索过程中止。
## Next
‘NEXT’操作是搜索1-Wire总线上的下一个从机器件；一般情况下，此搜索操作是在‘FIRST’操作之后或上一次‘NEXT’操作之后进行；保持上次搜索后这些值的状态不变、执行又一次搜 索即可实现‘NEXT’操作。之后从ROM_NO寄存器中来读出新一个ROM码。若前一次搜索到 的是1-Wire上的最后一个器件，则返回一个无效标记FALSE，并且把状态设置成下一次调用搜 索算法时将是‘FIRST’操作的状态。
## 高级变量搜索
### Verify
‘VERIFY’操作用来检验已知ROM码的器件是否连接在1-Wire总线上，通过提供ROM码并对 该码进行目标搜索就可确定此器件是否在线。首先，将ROM_NO寄存器值设置为已知的ROM码 值，然后将LastDiscrepancy和LastDeviceFlag标志位分别设置为64 (40h)和0； 进行搜索操作， 然后读ROM_NO的输出结果；如果搜索成功并且ROM_NO中存储的仍是要搜索器件的ROM码 值，那么此器件就在1-Wire总线上。
### Target Setup
‘TARGET SETUP’操作就是用预置搜索状态的方式首先查找一个特殊的家族类型，每个1-Wire器件都有一个字节的家族码内嵌在ROM码中(参见图1)，主机可以通过家族码来识别器件所具 有的特性和功能。若1-Wire总线上有多片器件时，通常是将搜索目标首先定位在需注意的器件类 型上。为了将一个特殊的家族作为搜索目标，需要将所希望的家族码字节放到ROM_NO寄存器 的第一个字节中，并且将ROM_NO寄存器的复位状态置零，然后将LastDiscrepancy设置为64 (40h)；把LastDeviceFlag和LastFamilyDiscrepancy设置为0。在执行下一次搜索算法时就能找出 所期望的产品类型的第一个器件；并将此值存入ROM_NO寄存器。需要注意的是如果1-Wire总 线上没有挂接所期望的产品类型的器件，就会找出另一类型的器件，所以每次搜索完成后，都要 对ROM_NO寄存器中存储的结果进行校验。
### Family Skip Setup
‘FAMILY SKIP SETUP’操作用来设置搜索状态以便跳过搜索到的指定家族中的所有器件，此操 作只有在一个搜索过程结束后才能使用。通过把LastFamilyDiscrepancy复制到LastDiscrepancy， 并清除LastDeviceFlag即可实现该操作；在下一搜索过程就会找到指定家族中的下一个器件。如 果当前家族码分组是搜索过程中的最后一组，那么搜索过程结束并将LastDeviceFlag置位。
LastDiscrepancy LastFamily- Discrepancy LastDeviceFlag ROM_NO FIRST 0 0 0 result NEXT leave unchanged leave unchanged leave unchanged result VERIFY 64 0 0 set with ROM to verify, check if same after search TARGET SETUP 64 0 0 set first byte to family code, set rest to zeros FAMILY SKIP SETUP copy from LastFamilyDiscrepancy 0 0 leave unchanged
## 附录
```// TMEX API TEST BUILD DECLARATIONS
#define TMEXUTIL
#include "ibtmexcw.h"
long session_handle;
// END TMEX API TEST BUILD DECLARATIONS
// definitions
#define FALSE 0
#define TRUE 1
// method declarations
int OWFirst();
int OWNext();
int OWVerify();
void OWTargetSetup(unsigned char family_code);
void OWFamilySkipSetup();
int OWReset();
void OWWriteByte(unsigned char byte_value);
void OWWriteBit(unsigned char bit_value);
int OWSearch();
unsigned char docrc8(unsigned char value);
// global search state
unsigned char ROM_NO[8];
int LastDiscrepancy;
int LastFamilyDiscrepancy;
int LastDeviceFlag;
unsigned char crc8;
//--------------------------------------------------------------------------
// Find the 'first' devices on the 1-Wire bus
// Return TRUE : device found, ROM number in ROM_NO buffer
// FALSE : no device present
//
int OWFirst()
{
// reset the search state
LastDiscrepancy = 0;
LastDeviceFlag = FALSE;
LastFamilyDiscrepancy = 0;
return OWSearch();
}
//--------------------------------------------------------------------------
// Find the 'next' devices on the 1-Wire bus
// Return TRUE : device found, ROM number in ROM_NO buffer
//
int OWNext()
{
// leave the search state alone
return OWSearch();
}
//--------------------------------------------------------------------------
// Perform the 1-Wire Search Algorithm on the 1-Wire bus using the existing
// search state.
// Return TRUE : device found, ROM number in ROM_NO buffer
//
int OWSearch()
{
int id_bit_number;
int last_zero, rom_byte_number, search_result;
int id_bit, cmp_id_bit;
// initialize for search
id_bit_number = 1;
last_zero = 0;
rom_byte_number = 0;
search_result = 0;
crc8 = 0;
// if the last call was not the last one
if (!LastDeviceFlag)
{
// 1-Wire reset
if (!OWReset())
{
// reset the search
LastDiscrepancy = 0;
LastDeviceFlag = FALSE;
LastFamilyDiscrepancy = 0;
return FALSE;
}
// issue the search command
OWWriteByte(0xF0);
// loop to do the search
do
{
// read a bit and its complement
// check for no devices on 1-wire
if ((id_bit == 1) && (cmp_id_bit == 1))
break;
else
{
// all devices coupled have 0 or 1
if (id_bit != cmp_id_bit)
search_direction = id_bit; // bit write value for search
else
{
// if this discrepancy if before the Last Discrepancy
// on a previous next then pick the same as last time
if (id_bit_number < LastDiscrepancy)
search_direction = ((ROM_NO[rom_byte_number] & rom_byte_mask) > 0);
else
// if equal to last pick 1, if not then pick 0
search_direction = (id_bit_number == LastDiscrepancy);
// if 0 was picked then record its position in LastZero
if (search_direction == 0)
{
last_zero = id_bit_number;
// check for Last discrepancy in family
if (last_zero < 9)
LastFamilyDiscrepancy = last_zero;
}
}
// set or clear the bit in the ROM byte rom_byte_number
if (search_direction == 1)
else
// serial number search direction write bit
OWWriteBit(search_direction);
// increment the byte counter id_bit_number
id_bit_number++;
// if the mask is 0 then go to new SerialNum byte rom_byte_number and reset mask
{
docrc8(ROM_NO[rom_byte_number]); // accumulate the CRC
rom_byte_number++;
}
}
}
while(rom_byte_number < 8); // loop until through all ROM bytes 0-7
// if the search was successful then
if (!((id_bit_number < 65) || (crc8 != 0)))
{
// search successful so set LastDiscrepancy,LastDeviceFlag,search_result
LastDiscrepancy = last_zero;
// check for last device
if (LastDiscrepancy == 0)
LastDeviceFlag = TRUE;
search_result = TRUE;
}
}
// if no device found then reset counters so next 'search' will be like a first
if (!search_result || !ROM_NO[0])
{
LastDiscrepancy = 0;
LastDeviceFlag = FALSE;
LastFamilyDiscrepancy = 0;
search_result = FALSE;
}
return search_result;
}
//--------------------------------------------------------------------------
// Verify the device with the ROM number in ROM_NO buffer is present.
// Return TRUE : device verified present
// FALSE : device not present
//
int OWVerify()
{
unsigned char rom_backup[8];
int i,rslt,ld_backup,ldf_backup,lfd_backup;
// keep a backup copy of the current state
for (i = 0; i < 8; i++)
rom_backup[i] = ROM_NO[i];
ld_backup = LastDiscrepancy;
ldf_backup = LastDeviceFlag;
lfd_backup = LastFamilyDiscrepancy;
// set search to find the same device
LastDiscrepancy = 64;
LastDeviceFlag = FALSE;
if (OWSearch())
{
// check if same device found
rslt = TRUE;
for (i = 0; i < 8; i++)
{
if (rom_backup[i] != ROM_NO[i])
{
rslt = FALSE;
break;
}
}
}
else
rslt = FALSE;
// restore the search state
for (i = 0; i < 8; i++)
ROM_NO[i] = rom_backup[i];
LastDiscrepancy = ld_backup;
LastDeviceFlag = ldf_backup;
LastFamilyDiscrepancy = lfd_backup;
// return the result of the verify
return rslt;
}
//--------------------------------------------------------------------------
// Setup the search to find the device type 'family_code' on the next call
// to OWNext() if it is present.
//
void OWTargetSetup(unsigned char family_code)
{
int i;
// set the search state to find SearchFamily type devices
ROM_NO[0] = family_code;
for (i = 1; i < 8; i++)
ROM_NO[i] = 0;
LastDiscrepancy = 64;
LastFamilyDiscrepancy = 0;
LastDeviceFlag = FALSE;
}
//--------------------------------------------------------------------------
// Setup the search to skip the current device type on the next call
// to OWNext().
//
void OWFamilySkipSetup()
{
// set the Last discrepancy to last family discrepancy
LastDiscrepancy = LastFamilyDiscrepancy;
LastFamilyDiscrepancy = 0;
// check for end of list
if (LastDiscrepancy == 0)
LastDeviceFlag = TRUE;
}
//--------------------------------------------------------------------------
// 1-Wire Functions to be implemented for a particular platform
//--------------------------------------------------------------------------
//--------------------------------------------------------------------------
// Reset the 1-Wire bus and return the presence of any device
// Return TRUE : device present
// FALSE : no device present
//
int OWReset()
{
// platform specific
// TMEX API TEST BUILD
return (TMTouchReset(session_handle) == 1);
}
//--------------------------------------------------------------------------
// Send 8 bits of data to the 1-Wire bus
//
void OWWriteByte(unsigned char byte_value)
{
// platform specific
// TMEX API TEST BUILD
TMTouchByte(session_handle,byte_value);
}
//--------------------------------------------------------------------------
// Send 1 bit of data to teh 1-Wire bus
//
void OWWriteBit(unsigned char bit_value)
{
// platform specific
// TMEX API TEST BUILD
TMTouchBit(session_handle,(short)bit_value);
}
//--------------------------------------------------------------------------
// Read 1 bit of data from the 1-Wire bus
// Return 1 : bit read is 1
// 0 : bit read is 0
//
{
// platform specific
// TMEX API TEST BUILD
return (unsigned char)TMTouchBit(session_handle,0x01);
}
// TEST BUILD
static unsigned char dscrc_table[] = {
0, 94,188,226, 97, 63,221,131,194,156,126, 32,163,253, 31, 65,
157,195, 33,127,252,162, 64, 30, 95, 1,227,189, 62, 96,130,220,
35,125,159,193, 66, 28,254,160,225,191, 93, 3,128,222, 60, 98,
190,224, 2, 92,223,129, 99, 61,124, 34,192,158, 29, 67,161,255,
70, 24,250,164, 39,121,155,197,132,218, 56,102,229,187, 89, 7,
219,133,103, 57,186,228, 6, 88, 25, 71,165,251,120, 38,196,154,
101, 59,217,135, 4, 90,184,230,167,249, 27, 69,198,152,122, 36,
248,166, 68, 26,153,199, 37,123, 58,100,134,216, 91, 5,231,185,
140,210, 48,110,237,179, 81, 15, 78, 16,242,172, 47,113,147,205,
17, 79,173,243,112, 46,204,146,211,141,111, 49,178,236, 14, 80,
175,241, 19, 77,206,144,114, 44,109, 51,209,143, 12, 82,176,238,
50,108,142,208, 83, 13,239,177,240,174, 76, 18,145,207, 45,115,
202,148,118, 40,171,245, 23, 73, 8, 86,180,234,105, 55,213,139,
87, 9,235,181, 54,104,138,212,149,203, 41,119,244,170, 72, 22,
233,183, 85, 11,136,214, 52,106, 43,117,151,201, 74, 20,246,168,
116, 42,200,150, 21, 75,169,247,182,232, 10, 84,215,137,107, 53};
//--------------------------------------------------------------------------
// Calculate the CRC8 of the byte value provided with the current
// global 'crc8' value.
// Returns current global crc8 value
//
unsigned char docrc8(unsigned char value)
{
// See Application Note 27
// TEST BUILD
crc8 = dscrc_table[crc8 ^ value];
return crc8;
}
//--------------------------------------------------------------------------
// TEST BUILD MAIN
//
int main(short argc, char **argv)
{
short PortType=5,PortNum=1;
int rslt,i,cnt;
// TMEX API SETUP
// get a session
session_handle = TMExtendedStartSession(PortNum,PortType,NULL);
if (session_handle <= 0)
{
printf("No session, %d\n",session_handle);
exit(0);
}
// setup the port
rslt = TMSetup(session_handle);
if (rslt != 1)
{
printf("Fail setup, %d\n",rslt);
exit(0);
}
// END TMEX API SETUP
// find ALL devices
printf("\nFIND ALL\n");
cnt = 0;
rslt = OWFirst();
while (rslt)
{
// print device found
for (i = 7; i >= 0; i--)
printf("%02X", ROM_NO[i]);
printf(" %d\n",++cnt);
rslt = OWNext();
}
// find only 0x1A
printf("\nFIND ONLY 0x1A\n");
cnt = 0;
OWTargetSetup(0x1A);
while (OWNext())
{
// check for incorrect type
if (ROM_NO[0] != 0x1A)
break;
// print device found
for (i = 7; i >= 0; i--)
printf("%02X", ROM_NO[i]);
printf(" %d\n",++cnt);
}
// find all but 0x04, 0x1A, 0x23, and 0x01
printf("\nFIND ALL EXCEPT 0x10, 0x04, 0x0A, 0x1A, 0x23, 0x01\n");
cnt = 0;
rslt = OWFirst();
while (rslt)
{
// check for incorrect type
if ((ROM_NO[0] == 0x04) || (ROM_NO[0] == 0x1A) ||
(ROM_NO[0] == 0x01) || (ROM_NO[0] == 0x23) ||
(ROM_NO[0] == 0x0A) || (ROM_NO[0] == 0x10))
OWFamilySkipSetup();
else
{
// print device found
for (i = 7; i >= 0; i--)
printf("%02X", ROM_NO[i]);
printf(" %d\n",++cnt);
}
rslt = OWNext();
}
// TMEX API CLEANUP
// release the session
TMEndSession(session_handle);
// END TMEX API CLEANUP
}
```
01/30/02版1.0—初始版本。
05/16/03版1.1—修改内容：Search ROM命令修改为F0 (十六进制)。 | 4,451 | 13,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.469211 |
http://www.enotes.com/homework-help/which-pairs-consists-equivalent-rational-numbers-9-89583 | 1,477,442,588,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720471.17/warc/CC-MAIN-20161020183840-00033-ip-10-171-6-4.ec2.internal.warc.gz | 427,301,205 | 9,161 | # Which of the pairs consists of equivalent rational numbers?-9 /15 and -3 /5
neela | High School Teacher | (Level 3) Valedictorian
Posted on
We get an eqivalent fraction (rational number )if we multiply both numerator and dinominator of the fraction bythe same number
The given numbers are -9/15 , -3/5, only one pair, and they are equivakent forhe following reason:
(-3*3)/(5*3) = -9/15, both the numerator and dinominators of the fraction -3/5 are multipled by 3 and we got the -9/15.
Therefore, -3/5 and -9/15 are equivalent fraction (rational numbers).
Also if both numerator and dinominator of a rational number are divided by the same number, then also we get arational number.
(-9/3)/(15/3) =-3/5. Therefore, -9/15 and -3 /5 are fractions (rational numbers). | 222 | 777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-44 | latest | en | 0.84123 |
https://gmatclub.com/forum/eight-friends-buy-chocolates-such-that-everyone-buys-at-least-one-choc-288297.html | 1,550,287,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479838.37/warc/CC-MAIN-20190216024809-20190216050809-00192.warc.gz | 582,172,936 | 152,477 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Eight friends buy chocolates such that everyone buys at least one choc
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Joined: 25 Dec 2018
Posts: 146
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09 Feb 2019, 09:07
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Eight friends buy chocolates such that everyone buys at least one chocolate. If at the most of 3 friends purchased equal number of chocolates and all others purchased a different number of chocolates, what is the minimum number of chocolates they purchased?
A. 15
B. 20
C. 23
D. 25
E. 28
Math Expert
Joined: 02 Aug 2009
Posts: 7334
### Show Tags
09 Feb 2019, 09:23
1
akurathi12 wrote:
Eight friends buy chocolates such that everyone buys at least one chocolate. If at the most of 3 friends purchased equal number of chocolates and all others purchased a different number of chocolates, what is the minimum number of chocolates they purchased?
A. 15
B. 20
C. 23
D. 25
E. 28
We are looking for the MINIMUM total...
Since three can buy sane and rest 5 have different, let us take the least values..
So three of them have 1 each and remaining five have 2,3,4,5,6 respectively..
Total = 3*1+2+3+4+5+6=23
C
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### Show Tags
09 Feb 2019, 09:27
akurathi12 wrote:
Eight friends buy chocolates such that everyone buys at least one chocolate. If at the most of 3 friends purchased equal number of chocolates and all others purchased a different number of chocolates, what is the minimum number of chocolates they purchased?
A. 15
B. 20
C. 23
D. 25
E. 28
KeyWord =
3 friends purchased equal number of chocolates
rest purchased a different number of chocolates,
everyone buys at least one chocolate
a + b +c+d +e + f+ g+h
3h + 2+3+4+5+6
3*1 +20
23
C
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### Show Tags
10 Feb 2019, 04:26
akurathi12 wrote:
Eight friends buy chocolates such that everyone buys at least one chocolate. If at the most of 3 friends purchased equal number of chocolates and all others purchased a different number of chocolates, what is the minimum number of chocolates they purchased?
A. 15
B. 20
C. 23
D. 25
E. 28
assume three friends bought 1 chocoloate each = 3
and other bought = 2,3,4,5,6 ; sum = 20
total 23
IMO C
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Re: Eight friends buy chocolates such that everyone buys at least one choc [#permalink] 10 Feb 2019, 04:26
Display posts from previous: Sort by | 1,280 | 4,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-09 | longest | en | 0.88371 |
https://www.experts-exchange.com/questions/28238144/Excel-Array-formula-Index-Match-based-on-Max-value.html | 1,537,304,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155702.33/warc/CC-MAIN-20180918205149-20180918225149-00530.warc.gz | 732,364,658 | 18,560 | # Excel Array formula Index/Match based on Max value
In attached, top table should match the values of bottom table.
I know I'm mostly right, but can't crack it.
QueryTable.xlsm
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Commented:
I assume you can get the correct dates in columns O and P using your existing O18 formula copied across to P18 and down both columns. Once you have those try this formula in N18
=INDEX(otr!\$A\$3:\$I\$42,MATCH(1,(\$D\$3:\$D\$42=\$O18)*(\$A\$3:\$A\$42=\$M18),0),MATCH(otr!N\$17,otr!\$A\$1:\$I\$1,0))
confirmed with CTRL+SHIFT+ENTER and copied down column
regards, barry
0
Commented:
.......no actually I see you won't get the right results for column P, so you need this version for P18 copied down
=INDEX(otr!\$A\$3:\$I\$42,MATCH(MAX(IF(\$A\$3:\$A\$42=\$M18,\$E\$3:\$E\$42)),\$E\$3:\$E\$42,0),MATCH(otr!P\$17,otr!\$A\$1:\$I\$1,0))
Note that if you take the max date for "Issue" and the max date for "Maturity" theoretically they might not belong to the same row. My suggested N18 formula just matches the max Issue date - if you want you can match both with this version
=INDEX(otr!\$A\$3:\$I\$42,MATCH(1,(\$D\$3:\$D\$42=\$O18)*(\$E\$3:\$E\$42=\$P18)*(\$A\$3:\$A\$42=\$M18),0),MATCH(otr!N\$17,otr!\$A\$1:\$I\$1,0))
In this sample it gives you the same results as the previously suggested N18 formula, but depending on your data it may not do so (and may even return #N/A if there is no matching row).
regards, barry
0
Commented:
The attached has two versions. The first (yellow) uses a helper column ("Row") to find the relevant row...
``````=OFFSET(\$A\$1,\$Q18-1,MATCH(N\$17,\$A\$1:\$I\$1,0)-1)
``````
...and the helper...
``````{=MIN(IF(\$A\$3:\$A\$42&":"&\$D\$3:\$D\$42=\$M18&":"&MAX(IF(\$A\$3:\$A\$42=\$M18,\$D\$3:\$D\$42)),ROW(\$A\$3:\$A\$42),9^10))}
``````
The second version (red) doesn't use a helper...
``````=OFFSET(\$A\$1,MIN(IF(\$A\$3:\$A\$42&":"&\$D\$3:\$D\$42=\$M38&":"&MAX(IF(\$A\$3:\$A\$42=\$M38,\$D\$3:\$D\$42)),ROW(\$A\$3:\$A\$42),9^10))-1,MATCH(N\$17,\$A\$1:\$I\$1,0)-1)
``````
Regards,
Brian.QueryTable-V2.xlsm
0
Author Commented:
@barry, you were right on the first count, only based off max of Issue, but you're not using max formula? also, when I apply it to other fields, getting circular reference errors.
@redmond, yours works consistently, but do you think it's possible to just use index/match/max combo, like what I was trying to attempt?
0
Commented:
I'm sorry that mine wasn't what you wanted, but, trust me, it would be a complete waste of your time and mine to work on it while Barry's involved!
Regards,
brian.
0
Commented:
=INDEX(otr!\$A\$3:\$I\$42,MATCH(1,(\$D\$3:\$D\$42=\$O18)*(\$A\$3:\$A\$42=\$M18),0),MATCH(otr!N\$17,otr!\$A\$1:\$I\$1,0))
It doesn't use MAX but it uses the result from O18 so it is using MAX indirectly because that is used in O18.
Where do you get a circular reference?
Perhaps you can replace O18 with the MAX(IF part, i.e.
=INDEX(otr!\$A\$3:\$I\$42,MATCH(1,(\$D\$3:\$D\$42=MAX(IF(\$A\$3:\$A\$42=\$M18,\$D\$3:\$D\$42)))*(\$A\$3:\$A\$42=\$M18),0),MATCH(otr!N\$17,otr!\$A\$1:\$I\$1,0))
regards, barry
0
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Mechanical EngineerCommented:
Which version of Excel are you using?
If the answer is Excel 2010 or later, then you can use the AGGREGATE function. It will let you create a formula that has the benefits of Boolean expressions without needing to be array-entered. Here is a formula for P18 that you may copy both across and down:
=AGGREGATE(14,6,INDEX(otr!\$A\$3:\$I\$42,,MATCH(P\$17,otr!\$A\$1:\$I\$1,0))/(otr!\$A\$3:\$A\$42=\$M18),1)
The first parameter (14) means to find the largest values. The last parameter (1) means to find the first largest value. The combination is equivalent to MAX. The second parameter (6) means to ignore error values (the Boolean expression returns an error value if no match in column A for M18).
0
Author Commented:
@ barry, yes!, last formula works perfect!
The circular reference was because formula should look in the main data table, and copied across, so when I put it in O18, it referenced itself.
While I don't fully understand the Match nesting, I can 'see' that it is acting on the max of Issue date and lookup value.
@ red, yours also works perfect, just less intuitive for me to understand.
@ byundt, thanks, using 2007, but good to know!
0
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Microsoft Excel
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https://finance.yahoo.com/news/victory-capital-holdings-inc-nasdaq-152702989.html?.tsrc=rss | 1,563,396,372,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525402.30/warc/CC-MAIN-20190717201828-20190717223828-00471.warc.gz | 406,075,464 | 103,531 | U.S. Markets closed
# Do You Like Victory Capital Holdings, Inc. (NASDAQ:VCTR) At This P/E Ratio?
This article is for investors who would like to improve their understanding of price to earnings ratios (P/E ratios). We'll look at Victory Capital Holdings, Inc.'s (NASDAQ:VCTR) P/E ratio and reflect on what it tells us about the company's share price. Victory Capital Holdings has a P/E ratio of 15.86, based on the last twelve months. That means that at current prices, buyers pay \$15.86 for every \$1 in trailing yearly profits.
### How Do You Calculate A P/E Ratio?
The formula for P/E is:
Price to Earnings Ratio = Share Price ÷ Earnings per Share (EPS)
Or for Victory Capital Holdings:
P/E of 15.86 = \$15.24 ÷ \$0.96 (Based on the year to December 2018.)
### Is A High Price-to-Earnings Ratio Good?
The higher the P/E ratio, the higher the price tag of a business, relative to its trailing earnings. That is not a good or a bad thing per se, but a high P/E does imply buyers are optimistic about the future.
### How Growth Rates Impact P/E Ratios
Probably the most important factor in determining what P/E a company trades on is the earnings growth. If earnings are growing quickly, then the 'E' in the equation will increase faster than it would otherwise. And in that case, the P/E ratio itself will drop rather quickly. So while a stock may look expensive based on past earnings, it could be cheap based on future earnings.
Victory Capital Holdings's earnings made like a rocket, taking off 104% last year. The sweetener is that the annual five year growth rate of 64% is also impressive. So I'd be surprised if the P/E ratio was not above average.
### How Does Victory Capital Holdings's P/E Ratio Compare To Its Peers?
One good way to get a quick read on what market participants expect of a company is to look at its P/E ratio. If you look at the image below, you can see Victory Capital Holdings has a lower P/E than the average (30.5) in the capital markets industry classification.
Its relatively low P/E ratio indicates that Victory Capital Holdings shareholders think it will struggle to do as well as other companies in its industry classification. Since the market seems unimpressed with Victory Capital Holdings, it's quite possible it could surprise on the upside. If you consider the stock interesting, further research is recommended. For example, I often monitor director buying and selling.
### Remember: P/E Ratios Don't Consider The Balance Sheet
Don't forget that the P/E ratio considers market capitalization. Thus, the metric does not reflect cash or debt held by the company. Hypothetically, a company could reduce its future P/E ratio by spending its cash (or taking on debt) to achieve higher earnings.
Such spending might be good or bad, overall, but the key point here is that you need to look at debt to understand the P/E ratio in context.
### Victory Capital Holdings's Balance Sheet
Victory Capital Holdings's net debt is 21% of its market cap. This could bring some additional risk, and reduce the number of investment options for management; worth remembering if you compare its P/E to businesses without debt.
### The Verdict On Victory Capital Holdings's P/E Ratio
Victory Capital Holdings has a P/E of 15.9. That's below the average in the US market, which is 18. The company does have a little debt, and EPS growth was good last year. If it continues to grow, then the current low P/E may prove to be unjustified. Since analysts are predicting growth will continue, one might expect to see a higher P/E so it may be worth looking closer.
When the market is wrong about a stock, it gives savvy investors an opportunity. If it is underestimating a company, investors can make money by buying and holding the shares until the market corrects itself. So this free visualization of the analyst consensus on future earnings could help you make the right decision about whether to buy, sell, or hold.
Of course you might be able to find a better stock than Victory Capital Holdings. So you may wish to see this free collection of other companies that have grown earnings strongly.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 1,017 | 4,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-30 | latest | en | 0.942571 |
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