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http://everychildcancode.org/lesson-5/ | 1,508,804,571,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00141.warc.gz | 110,710,978 | 7,923 | ## IF, THEN, AND, OR, GOTO
All the codewords in this lesson are for allowing you to set one or more “conditions”, and the computer will decide what to do depending on what conditions you set.
Examples of conditions we come across every day are:
IF it is raining THEN wear a raincoat
IF you are hungry THEN eat something
IF you are tired THEN go to bed
IF and THEN Sometimes in BASIC there are two codewords which must be used together. This happens with the IF and THEN codewords. When you use the IF codeword you are telling the computer to test “something” to find out if it is true. If that “something” is true then the computer carries out one task, but if that “something” is not true then the computer will carry out a different task or it might do nothing.
AND The AND codeword tells the computer to test if two or more different conditions are true. So AND can be used with IF and THEN.
OR The OR codeword tells the computer to test if either of two conditions is true. It can also tell the computer to test if at least one condition is true, out of any number of conditions more than two.
GOTO This codeword tells the computer not to continue running the program at the next line of code. Instead the computer is told to go to a different line number and continue from there.
### The IF and THEN codewords
The IF codeword is a statement/command. The THEN codeword is a statement.
IF is always used with THEN to tell the computer to make a decision and then to act on that decision. To help it make the decision the computer tests something to find out whether or not it is true. If it is true, then the computer does one thing, but if it is not true then the computer does something else (or perhaps nothing at all).
#### How to use the IF and THEN codewords
Here is a simple IF statement:
`80 IF cats = 2 THEN PRINT "I have 2 cats"`
This line of the program tells the computer that if the value of the variable cats is 2 then it should display the message I have 2 cats.
You can also use the IF and THEN codewords to tell the computer to test whether or not the value of a variable is 0. Here is an example of how you can do this:
90 IF dogs THEN PRINT “I have at least one dog.”
This line of the program tells the computer that if the value of dogs is not zero it should print the message I have at least one dog. But if the value of dogs is 0 then the computer does nothing special – it just continues on to the next line of the program.
#### How to compare numbers using IF and THEN
In Lesson 3 we learned a couple of ways to test for true or false. Sometimes it is useful to make an IF . . . THEN statement depend on a true/false test, so that we tell the computer to do something IF the result of the test is true.
Here are some examples. These examples use some symbols that you might already have learned about in your Maths lessons. If you don’t know about them already, now is a good time for you to learn them – they are used to compares two numbers or values.
`l > m means l is greater than m`
`l < m means l is less than m`
`l > m means l is greater than m or it is equal to m`
`l < m means l is less than m or l is equal to m`
`l < > m means l is not equal to m`
The “greater than” comparison:
```50 IF a > b THEN PRINT "a is greater than b" 51 REM In line 50 the test compares the values of the variables a and b. 52 REM We are telling the computer that if a is greater than b then 53 REM it should display the message "a is greater than b".```
The “greater than or equal to” comparison:
```60 IF a > b THEN PRINT "a is greater than or equal to b" 61 REM Line 60 tells the computer that if a is greater than b, 62 REM or if a is equal to b, then it should display the message.```
The “less than” comparison:
```70 IF a < b THEN PRINT "a is less than b" 71 REM Line 70 tells the computer that if a is less than b 72 REM the it should display the message.```
The “less than or equal to” comparison:
```80 IF a < b THEN PRINT "a is less than or equal to b" 81 REM Line 80 tells the computer that if a is less than b 82 REM or if a is equal to b then it should display the message.```
The “not equal to” comparison:
```90 IF a < > b THEN PRINT "a is not equal to b" 91 REM Line 90 tells the computer that if a is not equal to b 92 REM then it should display the message.```
### The AND codeword
The AND codeword is a logical operator.
#### How to use the AND codeword
The AND codeword tells the computer to test if two or more different conditions are true. We use AND together with IF and THEN, to help the computer make slightly more complex decisions than those in the previous examples.
When we are using the AND codeword, if all of the conditions the computer is told about are true then the overall combination of conditions is true. But if one or more of the conditions is not true then the combination of conditions is not true.
Here is an example:
`100 cats = 2 AND dogs THEN PRINT "I have 2 cats and at least one dog"`
This line of the program tells the computer that if the value of the variable cats is 2, and if the value of the variable dogs is not 0, then it should display the message I have 2 cats and at least one dog.
But if the value of the variable cats is not 2 then the computer won’t display that message. And even if the value of cats is 2, then if the value of dogs is not true then the computer still won’t display that message.
### The OR codeword
The OR codeword is a logical operator.
#### How to use the OR codeword
The OR codeword tells the computer to test if any of two or more different conditions are true. We use OR together with IF and THEN.
When we are using the OR codeword, if any of the conditions the computer is told about are true then the overall combination of conditions is true. But if all of the conditions are not true then the combination of conditions is not true.
Here is an example:
`100 IF cats = 2 OR cats = 3 THEN PRINT "I have 2 or 3 cats"`
This line of the program tells the computer that if the value of the variable cats is 2, or if it is 3, then it should display the message I have 2 or 3 cats.
But if the value of the variable cats is not 2 or 3, then the computer won’t display that message. With the OR codeword, it is only necessary for one or more of the conditions to be true for the overall condition to be true.
### The GOTO codeword
The GO TO codeword is a statement/command.
The GO TO codeword tells the computer to jump to a particular line of the program.
#### How to use the GO TO codeword
GO TO is very easy to use. Here is an example:
`60 GO TO 350`
This tells the computer to jump to line 350 of the program and continue running from there.
# Example Program
This program plays a simple game of Rock, Scissors and Paper.
Line 800 to 840 asks the user for their choice.Line 850 to 1030 works out who has won and displays a message. Line 1040 to 1080 asks if you want to play again.
``````800 CLS
810 PRINT "What do you choose?": PRINT : PRINT "Rock (1)": PRINT "Scissors (2)": PRINT "Paper (3)"
820 PRINT : INPUT m
830 IF m>0 AND m<4 THEN GO TO 850
840 PRINT "You must enter a number between 1 and 3": GO TO 820
850 IF c<>m THEN GO TO 870
860 IF m=1 THEN PRINT "We both have Rock we draw": GO TO 1040
861 REM Both players have Rock it's a draw
862 IF m=2 THEN PRINT "We both have Scissors we draw": GO TO 1040
863 REM Both players have Scissors it's a draw
864 PRINT "We both have Paper we draw": GO TO 1040
865 REM Both players have Paper it's a draw
870 IF (c=1) THEN GO TO 930
880 IF (c=2) THEN GO TO 970
890 REM Computer has Paper
895 PRINT "I have Paper ";
896 REM Print the computers choice
900 IF m=1 THEN GO TO 920
910 PRINT "You win ": GO TO 1020
920 PRINT "I Win": GO TO 1030
930 REM Computer has Rock
935 PRINT "I have Rock ";
936 REM Print the computers choice
940 IF m=2 THEN GO TO 960
950 PRINT "You win": GO TO 1030
960 PRINT "I win": GO TO 1010
970 REM Computer has Scissors
975 PRINT "I have Scissors ";
976 REM Print the computers choice
980 IF m=3 THEN GO TO 1000
990 PRINT "You win": GO TO 1010
1000 PRINT "I win": GO TO 1020
1010 PRINT "Rock beats Scissors": GO TO 1040
1020 PRINT "Scissors beats Paper": GO TO 1040
1030 PRINT "Paper beats Rock"
1040 LET c=c+1
1050 IF c=4 THEN LET c=1
1060 INPUT "Do you want to play again? (Y/N)";a\$
1070 IF (a\$="y") OR (A\$="Y") THEN GO TO 800
1080 PRINT : PRINT "That was fun!":PRINT "Thanks for playing"`````` | 2,205 | 8,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-43 | latest | en | 0.93896 |
https://reworkfurnishings.com/fraction-calculator-with-negative-numbers-95 | 1,675,866,707,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00748.warc.gz | 501,720,865 | 6,250 | # Fraction calculator with negative numbers
The denominator number is the one below the fractional dash and the numerator is the one above it. How to add and subtract fractions step by step. It is
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## Online Fraction Calculator. Addition, Subtraction
Using this online calculator with fractions you will be able to add, subtract, multiply or divide fractions or mixed numbers (fractions with
## Fraction Calculator
This calculator converts fractions to decimal numbers and vice versa. Enter a fraction (e.g. 2/3) or a decimal number. Mathepower converts it.
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## Fraction Calculator 【ONLINE】 all operations
Add, subtract, simplify, divide and multiply fractions step by step - Most used actions - Related - Line number - Graph - Examples - Inputs from
## Fraction Calculator
This is a fraction calculator with the steps shown in the solution. If you have negative fractions, insert a minus sign before the numerator.
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If you're struggling with your homework, our Homework Help Solutions can help you get back on track.
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Fraction calculator: Add, subtract, divide, multiply mixed fractions, proper fractions, improper fractions, whole numbers, and numbers
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• Clarify math equations | 543 | 2,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-06 | latest | en | 0.879273 |
https://davidwees.com/content/2024/03/ | 1,726,383,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00527.warc.gz | 169,053,510 | 25,863 | Education ∪ Math ∪ Technology
# Month: March 2024(page 1 of 1)
Here is a transcript from part of a classroom discussion.
Student: “So, I think.. well we believe, we noticed that expression number two combines with vision number B because it’s parenthesis x plus two, right?So all the visions and all the squares are repeated three times because it’s being multiplied by three. And so the x I believe, well we believe, would be the rectangle and the squares would be the two.”
What do you think is being discussed in this example? It’s impossible to understand this conversation without a visual representation. So, let’s look at the task the students are talking about.
Now, consider the same questions. What mathematics is being discussed?
Let’s look at the task again but with some annotation meant to draw students’ attention to the salient details of the explanation.
Now imagine the conversation. What is being discussed? What strategy is being shared?
Here is a generalizable principle: if one doesn’t have a visual way for students to track a conversation, one should assume they cannot.
An instructional routine is a consistent way of interacting with learners. The primary benefits of a routine emerge when the teacher and the learners know the routine well enough that the steps fade into the background, and everyone can focus more on the ideas being shared.
All successful teachers use instructional routines, but not all routines are equally effective. Below are some routines that educators have tested, and all have one important trait in common—they treat learners as sense-makers.
## Choral Counting
In this routine, students look for structure in numbers that emerge as they count those numbers as a whole group. First, given a starting number and an amount to count by, the class counts together as an entire group, then the teacher pauses the class to give students opportunities to look for structure. Students then share some of the patterns and relationships they noticed.
More detail on this routine is available here: https://tedd.org/choral-counting/
## Connecting Representations
This routine, developed by Amy Lucenta and Grace Kelemanik, asks students to use mathematical structure to connect two visualizations of the same mathematical idea represented differently.
After this activity is launched, students are presented with two different kinds of representations. They then make matches between one type of representation and the other. Once they have matched made with their partner, the teacher orchestrates a classroom discussion where students describe how they made their matches. Typically, a representation goes unmatched, so students create the missing representation next. Finally, students reflect on what helped them make their matches — what might be helpful to pay attention to next time.
## Contemplate then Calculate
This routine, also developed by Amy Lucenta and Grace Kelemanik, asks students to use mathematical structure to create shortcuts for calculations.
After this activity is launched, students are given a glance at a mathematical image. They are asked to share what they notice while their teacher records these noticings for everyone. Students are then given the mathematical image again, this time the remainder of the class, along with a question to address related to the image.
## Counting Collections
Angela Chan, Megan L Franke, and Elham Kazemi describe this routine in detail in their book Choral Counting & Counting Collections. In it, students count objects and then represent how they count them.
The magic of this routine is in the time spent by educators watching students count and keeping track of students’ ability to use one-to-one correspondence, mathematical structure, place value, skip counting, etc… as they count.
## Number Talks
The goal of a number talk is usually to expose students to multiple strategies for solving the same mathematical problem. Students are given a mathematical problem to solve, usually a problem involving arithmetic or counting, and asked to come up with a solution in their heads. Once sufficient numbers of students have indicated that they have a solution, the teacher leads a discussion where several solutions are shared.
This routine is terrific for celebrating learning that has occurred but doesn’t necessarily press students into using new strategies. For that, one wants to use a problem string instead (see below).
There’s lots of great information on number talks on this page: https://brownbagteacher.com/number-talks-how-and-why/
## Problem Strings
A problem string is a deliberately selected sequence of problems given to students, one at a time, to help students develop a new mathematical strategy. They are similar in how they are run to number talks but are not necessarily restricted to arithmetic or counting problems. Problem strings lend themselves well to all areas of mathematics.
Pam Harris has some resources for problem strings on her website here. She’s also written a terrific book on problem strings in high school. There are resources here on implementing a specific kind of problem string called a number string here: https://tedd.org/number-strings/
Three Reads is a routine used to help students make sense of contextual problems and learn how to deconstruct mathematical problems they have read. First, the routine is launched with an explanation of what students will work on, why they are working on it, and how they will work on it. Next, students read the same mathematical problem three times, each time for different types of information. Finally, they share what they understood from their reading with each other. Over time, students get better at mathematical reading.
Amy and Grace wrote a chapter on this routine in their book Routines for Reasoning, and I highly recommend reading their book to learn more about it.
## Which One Doesn’t Belong?
The goal of this routine is to spark mathematical creativity, give students opportunities to construct mathematical arguments and show them that there are many mathematical questions one can ask that do not have a single correct answer. The most important part of this routine is the argumentation students develop as they justify their choice.
Students are given four different mathematical objects in a collection and asked to explain which one doesn’t belong. Usually, students do this first on their own, share their ideas with a partner, and then their teacher leads a mathematical discussion based on their ideas.
For more information on this routine, visit this website or read Christopher Danielson’s wonderful book on the routine.
## Examples – Non-Examples
In this routine, students are presented with examples and non-examples. They analyze each pair and come to a better understanding of the concept presented by the examples.
Using examples and non-examples is perfect when you have a mathematical concept for which you want students to have a definition. This is especially helpful as students often struggle to understand definitions given to them without sufficient explanation.
A guide for getting started with this routine is available here.
## Sharing Skepticism
The Sharing Skepticism routine is intended to support students in constructing arguments, critiquing each other’s arguments, and reflecting on what makes an argument good. Over time, as students debate which arguments are more convincing, they will develop the habits of mind necessary to construct mathematical arguments. The overall structure of the routine is for students to convince themselves, convince a friend, and then convince a skeptic.
First, the routine is launched so that students know why they are sharing skepticism today, what they will learn, and how the routine proceeds. Next, students solve a problem independently and then share their solution with a partner. Two or three solutions are presented to the whole class; then, students work with a partner to select their favourite argument and try to improve it somehow. Some ideas for improvement are shared with the class, and then students reflect on their experience and consider what makes a good argument.
Any task that every student can devise a strategy for solving, which requires some level of mathematical thinking and has multiple strategies for solving, can be used as a task for this routine.
## Paired Examples
This routine aims to help students make connections between different mathematical concepts, allowing them to explicitly build a procedure or algorithm from something they already know.
First, the routine is launched, and then students are presented with the first step in each pair of mathematical procedures. Students look for and name connections between each step. Each step is unveiled in the same way, with students sharing the connections they notice in a structured way with the whole class. Once the entire procedure is unveiled, students are presented with another example of the new procedure for them to try, first independently and then working with a partner. A couple of these new procedures are selected to be shared with the whole class. Finally, students reflect on what they learned from the activity.
This routine is ideal anytime there is a mathematical procedure one wants students to learn when there is an early similar mathematical procedure students have already learned. The earlier example does not need to be identical to the new procedure. Students noticing the differences between the two can be as powerful as noticing the similarities.
Further resources related to this routine are available here.
## Language Routines
There are times when the focus of a mathematics classroom is on building language, both the academic language of the discipline of mathematics and the everyday language needed to converse about mathematical ideas with each other.
The eight language routines developed by the Standard Center for Assessment, Learning, and Equity are ideal for these moments.
• MLR 1: Stronger and Clearer Each Time
• MLR 2: Collect and Display
• MLR 3: Critique, Correct, and Clarify
• MLR 4: Information Gap
• MLR 5: Co-Craft Questions and Problems
• MLR 7: Compare and Connect
• MLR 8: Discussion Supports
Each routine is described in a little detail in the teacher guide for the Illustrative Mathematics curriculum. The IM curriculum also contains many examples of when and why one might want to use the math language routines.
## Group Learning Routines
For those teachers who teach multiple subjects, the subject-agnostic group learning routines developed by All Ed may be ideal. Each routine is described in detail, and different routines are provided for different use cases.
One of my favourite group learning routines is Idea Carousel, and I frequently use it in professional development sessions. In this routine, learners have a big idea they are unpacking, and the components of this big idea have been separated into separate posters. Learners are given one marker for each group and rotate through the posters around the room, swapping the marker to another recorder on each rotation. As learners circulate, they add detail and annotate the previous ideas added by other groups. At the end of the carousel, the teacher leads a whole group conversation based on the posters.
## Routines for Teacher Learning
While all of the routines I’ve described before lead to educator learning, there are specific routines designed for people who facilitate teacher learning to use.
I asked ChatGPT to write a TED talk about why we need to change how we teach math. Here’s what it came up with.
Title: The Mathematics Revolution: Unlocking the Power of Learning
[Introduction]
Speaker: Good morning, ladies and gentlemen! Today, I stand before you to ignite a spark of curiosity, to challenge the conventional wisdom, and to advocate for a fundamental change in how we teach one of the most critical subjects in education: mathematics.
[Opening Hook]
Speaker: Imagine a world where mathematics is no longer a daunting, mysterious puzzle, but rather an open gateway to creativity, critical thinking, and problem-solving. Envision classrooms buzzing with excitement, where students are not merely memorizing formulas but actively engaging in the beauty and applicability of mathematics. This is the vision I propose we embrace today.
[Establishing the Problem]
Speaker: We live in an era of unprecedented innovation and technological advancements, yet our traditional approach to teaching mathematics has remained largely unchanged. This outdated method focuses on rote memorization, stifles creativity, and fails to inspire a genuine passion for the subject. It’s time for us to acknowledge that this approach is limiting the potential of our students and hindering progress in a rapidly evolving world.
[Point 1: Shifting Focus from Answers to Understanding]
Speaker: Our current education system fixates on finding the right answers, often neglecting the importance of deep comprehension. Mathematics is not merely a series of procedures to reach a solution; it is a language that empowers us to decode the complexities of the universe. We must shift our focus from answers to understanding, encouraging students to explore the “why” behind mathematical concepts and nurturing their ability to think critically.
[Point 2: Embracing Real-World Applications]
Speaker: Mathematics is not an isolated subject confined within the pages of a textbook. It permeates every aspect of our daily lives, from calculating the optimal route to work to analyzing statistical trends in healthcare. By integrating real-world applications into the curriculum, we can show students the tangible relevance of mathematics and inspire them to apply their skills in meaningful ways. Let’s bridge the gap between theory and practice, allowing students to see the direct impact of mathematics in fields like technology, finance, and even art.
[Point 3: Fostering Collaboration and Creativity]
Speaker: Mathematics is often wrongly associated with solitary and rigid thinking. However, it thrives when we encourage collaboration and embrace creativity. By introducing group projects, open-ended problem-solving, and mathematical investigations, we can cultivate an environment that nurtures teamwork, communication, and innovation. Let’s unleash the power of collective thinking and teach our students that mathematics is a dynamic, collaborative endeavor.
[Point 4: Redefining Assessment]
Speaker: The conventional method of assessing mathematical proficiency through standardized tests falls short in capturing the true essence of mathematical competence. We need to redefine assessment to encompass a broader range of skills, including problem-solving, critical thinking, and logical reasoning. Embracing alternative assessment methods, such as portfolios, projects, and presentations, will provide a more accurate representation of a student’s mathematical abilities and foster a holistic understanding of the subject.
[Conclusion]
Speaker: Ladies and gentlemen, the time for change is now. We have the opportunity to revolutionize mathematics education and unlock the immense potential within our students. Let’s move away from a system that stifles creativity, instills fear, and perpetuates the myth that math is only for a select few. Instead, let us embark on a journey of discovery, exploration, and collaboration, where mathematics becomes a tool for empowerment and innovation.
Speaker: Together, let’s create a future where the beauty and practicality of mathematics are celebrated, where every student is equipped with the skills and confidence to tackle the challenges of an ever-changing world. Thank you!
[End of TED Talk]
Here’s a video of the TED talk that AI also created using the same script above.
Neither the script nor the video are amazing, but one doesn’t need much imagination to see that some of our ways of working are about to change pretty dramatically.
I’ve seen a lot of interest in applying artificial intelligence to education, but the efforts appear to be starting in the wrong places.
What is being built
We have AI being applied to write lessons for educators.
These early efforts create lessons that are not likely helpful for their target audience — educators who need lessons targeting specific concepts for which the educators do not already have lessons. The most likely educator in this situation is an early career educator or an educator teaching a new grade level or course. In the long run, neither of these groups is well-served by poorly written lessons by a “novice” AI curriculum designer.
This sample prompt might give an educator an idea of how to generate this task, but it isn’t something that anyone should put directly in front of students.
We have AI being used to tutor students.
Dan Meyer wrote a great post outlining one of these efforts. The critical issue is that the AI tutors are fumbling around in the dark, trying to support students without understanding their needs. Where are the supportive visuals? Where are the simplifying examples? Why does every tutoring program rely almost entirely on questions?
We have AI being used to grade students.
This is a time-consuming task for educators, to be sure, but auto-grading forgets the key reason educators do the task in the first place—to learn more about their students! A summary report reveals little about student learning. By removing educators from the job of looking through their student work, we make educators blind when they plan future lessons for those students.
This is my first attempt at using a custom AI to create an assessment on the Common Core standards K.CC.A.2, K.CC.C.7, and K.MD.B.3. This isn’t even in the ballpark of useful.
What should be built
AI categorization of student thinking
I read a fascinating tweet the other day about an AI being used to translate between two languages. It made me wonder—can AI be used to translate between the language students use when initially approaching a concept and the language educators as people who have already mastered the concept?
Students use their partial understanding to make sense of concepts, which results in them using novice language to describe their ideas. Frequently, the best person to explain a concept is someone who just learned it rather than someone who knows it so well that they forget what it felt like when learning it. This is part of the reason group work works.
This novice language can be hard for educators to understand, especially when they have many students in their classes with different ways of understanding an idea. Decision-making about what to do to support students at the moment is hampered by this challenge in translation. What if AI could be leveraged to translate what students say into how it connects to what educators know about the topic? This would help educators build what Deborah Ball calls Mathematical Pedagogical Knowledge for Teaching, but more immediately, it would help them make better decisions about supporting their students.
Another main reason to start with a project like this is that AI tutoring and other educational uses of AI almost all depend on this capacity.
Insights about student learning from student work
What are these students thinking? What strategies are they using?
(source: NCTM Blog)
This is a task every mathematics educator needs to know how to do. Take a sample of student work, figure out what students are doing, and then make a plan of action around this thinking.
It is also a task on which AI could shine. Most AI programs are well suited to categorization problems, which is exactly what this is. Large samples of similar student work on similar problems are impossible for educators to analyze, except over a lifetime of working with many students, but AI would trivialize this task.
Note that this task is fundamentally different than auto-grading work. Grading is about evaluating work; here, we want to understand the work and how it connects to our goals for teaching.
The Future
It is obvious that AI has great potential in education. Who is working to realize that potential?
One of the most challenging topics to teach in high school mathematics is “Completing the Square.” This is because educators do not always fully understand the topic and because it includes several algebraic steps that are incredibly challenging for students.
An eye-opening experience was when I first generated visuals for each step in completing the square using an area model.
The name “Completing the Square” is not arbitrary! Visually, one can see that we are literally taking an incomplete square (at least in cases like the one above) and making it complete. This visualization makes the algorithm’s goal obvious and helps students see what they are trying to accomplish.
However, the steps above are not sufficient for students. The area model above is much easier to understand if you already know the mathematics it represents. Since children don’t, we need to introduce the area model with more straightforward examples before using it for more complex ones. Here’s a worksheet that aims to do this.
Here’s the link to the full worksheet.
Once we have used the area model to establish the purpose of the steps when completing the square, we gradually remove the visuals. This is because we don’t want students to need to draw out the visuals each time to help them solve the equation. The goal is for students to understand conceptually what completing the square is and what steps are needed to complete it. The visuals are an aid for this goal.
There are other cases to consider (for example, expressions like x² + 6x + 10 and 2x² + 6x + 10). Once students have a handle on the simpler cases, these examples will be easier for them to manipulate algebraically.
The key idea here is that some ideas are much more obvious when represented visually than when we focus purely on a symbolic approach. | 4,185 | 22,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-38 | latest | en | 0.950462 |
https://mathoverflow.net/questions/118503/d-module-that-is-coherent-as-o-module | 1,660,337,810,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00264.warc.gz | 376,243,443 | 26,233 | # D-module that is coherent as O-module
Suppose that $X$ is an algebraic variety over $\mathbb C$, not necessarily smooth. Is it still true that each $\mathcal D_X$-module ($\mathcal D_X$ is of course the sheaf of differential operators) that is coherent as an $\mathcal O_X$-vodule must be locally free as an $\mathcal O_X$-module?
Serge
[Edited to correct errors pointed out by David Ben-Zvi and to answer a query by serge_I. These corrections reduce this answer'' to the status of a comment.]
(1) (Over $\mathbb C$) Under strong assumptions on the singularities of $X$ (namely, that $X$ should be cuspidal, Ben-Zvi and Nevins, arXiv 0212094v3), if $\mathcal E$ is coherent on $X$, then giving it a structure as a $\mathcal D_X$-module means giving an isomorphism $\phi:p_1^*\mathcal E\to p_2^*\mathcal E$, where $\mathcal X_1$ is the completion of $X\times X$ along the diagonal and $p_1,p_2:\mathcal X_1\to X$ are the projections. (There is also the cocycle condition, $p_{31}^*\phi=p_{32}^*\phi\circ p_{21}^*\phi,$ where $p_{ij}:\mathcal X_2\to\mathcal X_1$ are the projections from the completion $\mathcal X_2$ of $X\times X\times X$ along the diagonal, but we don't need this here.)
Since $X$ is noetherian, there is a unique flattening stratification $X=\coprod X_i$ associated to $\mathcal E$: each $X_i$ is locally closed in $X$ and for any $f:Y\to X$, $f^*\mathcal E$ is locally free if and only if $f$ factors through some $X_i$. The existence of $\phi$ shows that $p_1^*\mathcal E$ and $p_2^*\mathcal E$ have the same flattening stratification, while $\{p_1^{-1}(X_i)\}_i$ is the flattening stratification for $p_1^*\mathcal E$ and $\{p_2^{-1}(X_i)\}_i$ is the flattening stratification for $p_2^*\mathcal E$. Since $X$ is irreducible, there is a unique stratum $X_0$ of maximal dimension, so that $p_1^{-1}(X_0)=p_2^{-1}(X_0)$. Now think in terms of a tubular neighborhood of the diagonal in $X\times X$ to see that this forces $X_0=X$, so that $\mathcal E$ is locally free provided that $X$ is cuspidal.
(2) (In char. $p$, or over any base) If $X$ is smooth and $\mathcal D_X$ is taken to be the full ring of differential operators rather than the subring generated by those operators of order at most $1$, then the same argument applies.
• This is a very nice argument, but is it clear it applies to D-modules? your definition is that of a stratification (or comodule over jets), which I don't think are the same as modules over the ring of differential operators in general (yours is the much better behaved one in general). I only know they are the same for varieties with only cusp singularities, for which all the various notions of D-module coincide. Jan 10, 2013 at 18:21
• I am sorry, could you, please, explain in more detail how you conclude that for every $i$ there is a $j$ such that $p_1^{-1}(X_i)=p_2^{-1}(X_j)$? Jan 10, 2013 at 18:43
• Note that in characteristic p the statement is false even for X smooth if by $D_X$ you mean the ring of "crystalline diffops" (generated by functions and vector fields), rather than the full divided power ring (whose modules are the same as stratifications) -- eg Frobenius pullback of any coherent sheaf is a D-module. Jan 10, 2013 at 19:04
• Also projectivity should not be necessary, the argument you give (and construction of flattening stratification) is local.. maybe Noetherian?? Jan 10, 2013 at 19:04
No, you can generally only get locally freeness on an open dense set. This is Lemma VII.9.3 in Algebraic D-modules by Armand Borel et al.
EDIT: just realised that Lemma talks about where $\mathcal{D}_X$-coherent modules are $\mathcal{O}_X$-locally free, not whether $\mathcal{O}_X$-coherent ones are. A $\mathcal{D}_X$-coherent module needs a good filtration by $\mathcal{O}_X$-coherent modules to be $\mathcal{O}_X$-coherent, so you may have to check that. Section II of Borel et al. should cover this, have a look there.
• Well, to tell the truth, I did not manage to find anything to the point in Chapter 2 of "Algebraic D-modules" :( Thanks anyway for your time. Jan 10, 2013 at 14:35
• You could also check J.E. Björk, Analytic D-modules and their applications, he spells out the coherence stuff in a little more detail. Might be hard to find a copy, though. Jan 10, 2013 at 16:37 | 1,268 | 4,271 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | longest | en | 0.84296 |
https://davida.davivienda.com/viewer/printable-3-times-table-worksheet.html | 1,716,981,750,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00632.warc.gz | 162,603,474 | 8,756 | # Printable 3 Times Table Worksheet
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Web download and print these 3 times table charts for free. 4.6 (37 reviews) maths times tables 3 times tables. Homeschoolers can also use these multiplication table sheets to practice at. Printable worksheet on 3 times table can be used from everywhere. This printable gives students a chance to practice multiplication problems where at least one of the factors is.
Ad bring learning to life with thousands of worksheets, games, and more from education.com. Web in this free printable times tables worksheet students have to learn and practice 3 times table multiplication. Web download and print these 3 times table charts for free. Web worksheet on 3 times table.
## Ad Bring Learning To Life With Thousands Of Worksheets, Games, And More From Education.com.
Web 3 times tables worksheets. Free | worksheets | math drills | multiplication facts | printable These free worksheets emphasize basic multiplication. If your children are starting to learn.
## Help Them Begin By Counting In 3S.
Web students multiply 3 times numbers between 1 and 12. Web 3 times tables. And it is imperative to ensure that they. Learning through a multiplication table enables faster calculation in.
## Web In This Free Printable Times Tables Worksheet Students Have To Learn And Practice 3 Times Table Multiplication.
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## A Pretty 3 Times Table Chart In A4 Format (Pdf) That Will Help You Learn Your 3 Times Table.
Web 3 times table worksheets. It increases their confidence and with the practice of more and more problems based on 3. Web 3 times table worksheets will help kids to improve their multiplication knowledge. 3 x 10 = 30 3 x 10 =. | 1,097 | 4,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.868 |
http://www.chegg.com/homework-help/questions-and-answers/exploring-newly-discovered-planet-radius-planet-730-10-7m--suspend-lead-weight-lower-end-l-q3519457 | 1,472,572,241,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982984973.91/warc/CC-MAIN-20160823200944-00122-ip-10-153-172-175.ec2.internal.warc.gz | 371,102,976 | 13,966 | You are exploring a newly discovered planet. The radius of the planet is 7.30�10^7m . You suspend a lead weight from the lower end of a light string that is 4.00m long and has mass 0.0280kg . You measure that it takes 0.0580s for a transverse pulse to travel from the lower end to the upper end of the string. On earth, for the same string and lead weight, it takes 0.0380s for a transverse pulse to travel the length of the string. The weight of the string is small enough that its effect on the tension in the string can be neglected. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass? | 157 | 631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-36 | latest | en | 0.934831 |
http://webcatplus.nii.ac.jp/webcatplus/details/book/isbn/9780470740095.html?txt_isbn=9780470740095 | 1,660,878,258,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573540.20/warc/CC-MAIN-20220819005802-20220819035802-00697.warc.gz | 50,199,737 | 6,365 | ## Design of Rotating Electrical Machines
By (author) Pyrhonen, Juha; By (author) Jokinen, Tapani; By (author) Hrabovcova, Valeria
In one complete volume, this essential reference presents an in-depth overview of the theoretical principles and techniques of electrical machine design. This book enables you to design rotating electrical machines with its detailed step-by-step approach to machine design and thorough treatment of all existing and emerging technologies in this field. Senior electrical engineering students and postgraduates, as well as machine designers, will find this book invaluable. In depth, it presents the following:* Machine type definitions; different synchronous, asynchronous, DC, and doubly salient reluctance machines.* An analysis of types of construction; external pole, internal pole, and radial flux machines.* The properties of rotating electrical machines, including the insulation and heat removal options. Responding to the need for an up-to-date reference on electrical machine design, this book includes exercises with methods for tackling, and solutions to, real design problems. A supplementary website hosts two machine design examples created with MATHCAD: rotor surface magnet permanent magnet machine and squirrel cage induction machine calculations. Classroom tested material and numerous graphs are features that further make this book an excellent manual and reference to the topic.
「Nielsen BookData」より
[目次]
• About the Authors. Preface. Abbreviations and Symbols. 1 Principal Laws and Methods in Electrical Machine Design. 1.1 Electromagnetic Principles. 1.2 Numerical Solution. 1.3 The Most Common Principles Applied to Analytic Calculation. 1.4 Application of the Principle of Virtual Work in the Determination of Force and Torque. 1.5 Maxwell's Stress Tensor
• Radial and Tangential Stress. 1.6 Self-Inductance and Mutual Inductance. 1.7 Per Unit Values. 1.8 Phasor Diagrams. Bibliography. 2 Windings of Electrical Machines. 2.1 Basic Principles. 2.2 Phase Windings. 2.3 Three-Phase Integral Slot Stator Winding. 2.4 Voltage Phasor Diagram and Winding Factor. 2.5 Winding Analysis. 2.6 Short Pitching. 2.7 Current Linkage of a Slot Winding. 2.8 Poly-Phase Fractional Slot Windings. 2.9 Phase Systems and Zones of Windings. 2.10 Symmetry Conditions. 2.11 Base Windings. 2.12 Fractional Slot Windings. 2.13 Single- and Two-Phase Windings. 2.14 Windings Permitting a Varying Number of Poles. 2.15 Commutator Windings. 2.16 Compensating Windings and Commutating Poles. 2.17 Rotor Windings of Asynchronous Machines. 2.18 Damper Windings. Bibliography. 3 Design of Magnetic Circuits. 3.1 Air Gap and its Magnetic Voltage. 3.2 Equivalent Core Length. 3.3 Magnetic Voltage of a Tooth and a Salient Pole. 3.4 Magnetic Voltage of Stator and Rotor Yokes. 3.5 No-Load Curve, Equivalent Air Gap and Magnetizing Current of the Machine. 3.6 Magnetic Materials of a Rotating Machine. 3.7 Permanent Magnets in Rotating Machines. 3.8 Assembly of Iron Stacks. 3.9 Magnetizing Inductance. Bibliography. 4 Flux Leakage. 4.1 Division of Leakage Flux Components. 4.2 Calculation of Flux Leakage. Bibliography. 5 Resistances. 5.1 DC Resistance. 5.2 Influence of Skin Effect on Resistance. 6 Main Dimensions of a Rotating Machine. 6.1 Mechanical Loadability. 6.2 Electrical Loadability. 6.3 Magnetic Loadability. 6.4 Air Gap. Bibliography. 7 Design Process and Properties of Rotating Electrical Machines. 7.1 Asynchronous Motor. 7.2 Synchronous Machine. 7.3 DC Machines. 7.4 Doubly Salient Reluctance Machine. Bibliography. 8 Insulation of Electrical Machines. 8.1 Insulation of Rotating Electrical Machines. 8.2 Impregnation Varnishes and Resins. 8.3 Dimensioning of an Insulation. 8.4 Electrical Reactions Ageing Insulation. 8.5 Practical Insulation Constructions. 8.6 Condition Monitoring of Insulation. 8.7 Insulation in Frequency Converter Drives. Bibliography. 9 Heat Transfer. 9.1 Losses. 9.2 Heat Removal. 9.3 Thermal Equivalent Circuit. Bibliography. Appendix A. Appendix B. Index.
「Nielsen BookData」より
書名 Design of Rotating Electrical Machines Hrabovcova, Valeria Jokinen, Tapani Pyrhonen, Juha Wiley-Blackwell (an imprint of John Wiley & Sons Ltd) 2008.12.17 538p H250 x W175 9780470740095 英語 イギリス
この本を： | 1,098 | 4,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.834013 |
https://interviewquestions.ap6am.com/tag/hyperion-essbase-5/ | 1,603,807,847,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00130.warc.gz | 355,196,010 | 9,291 | ## Hyperion Essbase 5 Interview Questions & Answers
1. Question 1. What Are The Two Storage Options Available In Essbase And What Are The Differences?
ASO (Aggregate Storage Option) Used for
• If we have large number of dimensions (generally more than 10).
• Most of the dimensions are sparse.
• We cannot write back in ASO. I heard recently that ASO will also have the write back capability.
BSO (Block Storage Option)
• Dimensions are less and dense (recommended values for no of dim are between 4 and 8) .
• We can write back hence much suited for planning applications.Financial applications are generally dense structures and normal analytical cubes are sparse. Here we also need to understand what these dense and sparse members are. When the intersections or cells of two dimensions mostly contain a value it is dense.
• Say for example we have two dimensions period(Month or quarter) and region and we are calculation the sales amount. Generally maximum regions(countries, cities) will have some sales value for a particular month or quarter. Hence we can say that our period dimensions would be dense. But now instead of period, consider another dimension product. Now there are some products which are sold in only some particular regions hence for them there will be no values in other regions or will have very low percentage of data values hence the structure will become sparse.
2. Question 2. Can We Build Dimensions Directly From Data Sources Without Using Rule Files.
No. we cannot build dimensions directly from data sources without using rule files.
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4. Question 3. When Do We Generally Use Build Rules Files?
To automate the process of creating dimensions with thousands of members.
5. Question 4. What Are The Three Primary Build Methods For Building Dimensions?
1. Generation references.
2. level references.
3. ParentChild references.
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7. Question 5. In What Case We Can Define Generation 1 To A Field In Generation Build Method.
We cannot define that as Generation 1 is not valid.
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9. Question 6. Suppose We Have Assigned Generation 2 And Generation 4 As Of Now And Think Of Adding Generation 3 Later Some Time. Can We Build The Dimension.
No. If gen 2 and gen 4 exists, we must assign gen 3.
10. Question 7. Can We Create More Than 2 Dimensions Using Only One Build Rule File.
Yes, we can do that but it is recommended to use separate rule file for each dimension.
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Hyperion Financial Management Interview Questions
12. Question 8. What Is Uda( User Defined Attributes). How Are They Different Than Aliases.
UDA represents the class of the members. Aliases are just another names of the members. both are different and has different usage.
13. Question 9. Can We Query A Member For Its Uda In A Calculation Script.
Yes. You can query a member for its UDA in a calculation script.
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15. Question 10. How Does Uda’s Impact Database Size?
Absolutely no impact as UDA’s does not require additional storage space.
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17. Question 11. What Is The Difference Between Uda’s And Attribute Dimensions?
Attribute dimensions provides more flexibility than UDA’s. Attribute calculations dimensions which include five members with the default names sum, count, min, max and avg are automatically created for the attribute dimensions and are calculated dynamically.
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19. Question 12. How Does Attribute Dimensions And Uda’s Impact Batch Calculation Performance?
UDA’s No Impact as they do not perform any inherent calculations.Attribute dim No Impact as they perform only dynamic calculations.
20. Computer Technical Support Interview Questions
21. Question 13. How Can We Display Uda’s In Reports? How Do They Impact Report Report Performance.
UDA’s values are never displayed in the reports and hence do not impact report performance.
22. Question 14. How Does Attribute Dim Impact Report Performance?
They highly impact the report performance as the attributes are calculated dynamically when referenced in the report. For very large number of att dim displayed in the report, the performance could drastically reduce.
23. Question 15. While Loading The Data, You Have Applied Both The Selection Criteria As Well As Rejection Criteria To A Same Record.what Will Be The Outcome.
The record will be rejected.
24. Oracle DBA Troubleshooting Interview Questions
25. Question 16. How Is Data Stored In The Essbase Database?
Essbase is an file based database where the data is stored in PAG files of 2 GB each and grows sequentially.
26. Question 17. Can We Have Multiple Databases In One Single Application?
Yes. But only one database per application is recommended.
27. MongoDB Interview Questions
28. Question 18. Can We Have One Aso Database And One Bso Database In One Single Application. If Yes, How And If No, Why.
No. Because we define ASO or BSO option while creating the application and not database. Hence if the application is ASO, the databases it contains will be that type only.
29. Qlik View Interview Questions
30. Question 19. What Are The File Extensions For An Outline, Rule File And A Calc Script.
.OTL, .RUL and .CSC.
31. Question 20. What Is The Role Of Provider Services.
To communicate between Essbase and Microsoft office tools.
32. Oracle Hyperion Planning Interview Questions
33. Question 21. What Is An Alternative To Create Meta Outline, Rule File And Load Data.
Integration services and in version 11, we have Essbase studio.
34. Question 22. Can We Start And Stop An Application Individually. How Can This Be Used To Increase Performance.
Yes. We can manage our server resources by starting only the applications which receive heavy user traffic. When an application is started, all associated databases are brought to the memory.
35. Question 23. We Have Created An Application As Unicode Mode. Can We Change It Later To Nonunicode Mode.
No.We cannot change to nonunicode mode.
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37. Question 24. How Can I Migrate An Application From My Test Environment To The Production Environment?
Hyperion Administrative services console provides a migration utility to do this but only the application, database objects are migrated and no data is transferred.
38. Hyperion Financial Management Interview Questions
39. Question 25. A Customer Wants To Run Two Instances Of An Essbase Server On A Same Machine To Have Both Test Env And Development Env On The Same Server. Can He Do That?
Yes. We can have multiple instances of an Essbase server on a single machine and there will be different sets of windows services for all these instances.
40. Question 26. Suppose I Have A Dimension A With Members B And C And I Do Not Want B And C To Roll Up To A. How Can I Do This.
Using (~) exclude from consolidation operator.
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42. Question 27. What Does Never Consolidate Operator (^) Do?
It prevents members from being consolidate across any dimension.
43. Oracle Data Integrator (ODI) Interview Questions
44. Question 28. What Is Hybrid Analysis?
Lower level members and associated data remains in relational database where as upper level members and associated data resides in Essbase database.
45. Question 29. Why Topdown Calculation Less Efficient Than A Bottomup Calculation?being Less Efficient, Why Do We Use Them.
In the process it calculates more blocks than is necessary. Sometimes it is necessary to perform topdown calculation to get the correct calculation results.
46. Question 30. On What Basis You Will Decide To Invoke A Serial Or Parellel Calculation Method.
If we have a single processor, we will use serial calculation but if we have multiple processors we can break the task into threads and make them run on different processors.
47. Question 31. What Is Block Locking System?
Analytic services (or Essbase Services) locks the block and all other blocks which contain the Childs of that block while calculating this block is block locking system.
48. Question 32. What Are The Types Of Partitioning Options Available In Essbase?
1. Replicated partition.
2. Transparent partition
49. Question 33. Dynamic Calc Decreases The Retrieval Time And Increases Batch Database Calculation Time. How True Is The Statement?
The statement should be just opposite. As dynamic calc members are calculated when requested, the retrieval time should increase.
50. IBM BPM Interview Questions
51. Question 34. Can We Have Multiple Meta Outlines Based On One Olap Model In Integration Services?
Yes.we have multiple meta outlines based on one OLAP model in Integration services.
52. Question 35. What Are Lro’s( Linked Reporting Objects)?
They are specific objects like files, cell notes or URL’s associated with specific data cells of Essbase database. You can link multiple objects to a single data cell.These linked objects are stored in the server. These LRO’s can be exported or imported with the database for backup and migration activities.
53. Question 36. What Is Custom Based Macro?
Essbase macros that we write with Essbase calculator functions and special macro functions. Custom-defined macros use an internal Essbase macro language that enables you to combine calculation function and operate on multiple input parameters.
54. Oracle DBA Troubleshooting Interview Questions
55. Question 37. What Do You Mean By Dirty Block?
A data block containing cells that have been changed since the last calculation. Upper level blocks are marked as dirty if their child blocks are dirty (that is, have been updated)
56. Question 38. How Does Uda’s Impact Database Size?
Absolutely no impact as UDA’s do not require additional storage space.
57. Question 39. What Does Never Consolidate Operator(^) Do?
It prevents members from being consolidate across any dimension.
58. Question 40. Why Are Filters Used?
If you want to grant access to all dimensions then you wouldn’t use a filter. Just grant read access for the database to the user and/or group. The only reason you would use a filter is if you wanted to restrict access.
59. MongoDB Interview Questions
60. Question 41. Can We Have Multiple Metaoutlines Based On One Olap Model In Integration Services? | 2,192 | 10,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | longest | en | 0.858167 |
https://www.jiskha.com/similar?question=Hybrid+English+course+is+good+but+it+has+also+bad+sides.+If+a+student+is+looking+for+flexible+time+and+busy+of+work+hybrid+class+is+your+choice.+I+enjoyed+this+class+and+my+writing+has+improved.+I+found+out&page=52 | 1,563,909,651,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529481.73/warc/CC-MAIN-20190723172209-20190723194209-00328.warc.gz | 739,346,217 | 19,009 | Hybrid English course is good but it has also bad sides. If a student is looking for flexible time and busy of work hybrid class is your choice. I enjoyed this class and my writing has improved. I found out
28,601 questions, page 52
1. elem math
can some help me with my daughters homework. what figure is part of a line and has two endpoints? does a door have angles like a right , smaller or lager then a right angle? Seth made a picture frame with 4 equal sides and a right angle at each corner,
asked by Mrs.Green on February 19, 2008
2. PHYSICS ( PERIODIC MOTION)
In a lab, a student measures the unstretched length of a spring as 12.9 cm. When a 113.6-g mass is hung from the spring, its length is 26.3 cm. The mass-spring system is set into oscillatory motion, and the student observes that the amplitude of the
asked by MUSTAFA GHANIM on March 9, 2016
3. programming
trying to debug pseudocode 2. Th e Springwater Township School District has two high schools—Jeff erson and Audubon. Each school maintains a student fi le with fi elds containing student ID, last name, fi rst name, and address. Each fi le is in student
asked by Andre on September 4, 2013
4. math
sin2x-cotx = -cotxcos2x Using the various trigonometric identities(i.e. double angle formulas, power reducing formulas, half angle formulas, quotient identities, etc.) verify the identity. I first added cotx to both sides to get sin2x = -cotxcos2x+cotx
asked by sam on April 4, 2008
5. English
1. Talking and singing to the vegetables is good for the vegetables. 2. Talking and singing to the vegetables is good for the vegetables and produce. 3. Talking and singing to the vegetables is good for the produce. 4. Talking and singing to the vegetables
asked by rfvv on November 5, 2018
6. Poem
is this a good poem title, and is it good. Broken Heart Pieces torn apart My heart is empty Nothing of compassion, or joy Just the pieces that were many Joined together to form As one as all Are now just two halfs They are very small Love, and careness Was
asked by Spongebob on January 14, 2009
7. tv/media analysis
I need a good topic for an advertisement parody. heres the assignment: Create your own parody of a national print advertisement, one that will not only get the audience's attention but will subvert the original purpose of the ad. Most good ad parodies are
asked by brandon duncan on January 19, 2011
8. Algebra
x = adult tkts y = student tkts 3x = value of adult tkts 1.5y = value of student tkts x + y = 105 3x + 1.5y = 250
asked by Robert on January 14, 2011
9. Algebra
x = adult tkts y = student tkts 3x = value of adult tkts 1.5y = value of student tkts x + y = 105 3x + 1.5y = 250
asked by Robert on January 14, 2011
10. chemistry
A student is performing an experiment to find the density of water. She obtains a 100ml beaker, which weighs 53.257 grams, and she fills it with 60 mL of water. The temperature of the water is 25.5 degrees celsius (room temperature). Again the student
asked by Samantha on February 16, 2015
11. math
In a class of 50 student, 25 offer mathematics, 22 offer physcis, 30 offer chemistry and all the student take at least one of these subjects , 10 offer physics and mathematics, 8 offer chemistry and mathematics , 16 offer physics and chemistry. (a) draw a
asked by kunfayakun on December 24, 2017
12. Poetry
i need help translating this poem into modern english. it is by John Donne I wonder, by my troth, what thou and I Did, till we loved?were we not weaned till then, But sucked on country pleasures, childishly? Or snorted we in the seven sleepers' den? Twas
asked by jake on January 14, 2010
13. Algebra
On a standardized test, the distribution of scores is normal, the mean of the scores is 75, and the standard deviation is 5.8. If a student scored 83, the student's score ranks 1. below the 75th percentile 2. above the 97th percentile 3. between the 75th
asked by Priscilla on June 8, 2010
14. english lit
The feeling that things weren't as good as they seemed is described best by which image in "Borders"? A. "...as if she were trying to see through a bad storm or riding high on black ice." *** B. "...if she wasn't spreading jelly on the truth..." C. "coyote
asked by TeaCup on December 14, 2015
15. word problems in algebra
A square has sides of length 3x - 2 cm. Express the area of the square as a polynomial. would i write this as 3x-2cm^4 because there are foru sides sorry post went before i was done polynominial confuse me after thinking about what polynomial is suppose to
asked by TOMMY on April 22, 2007
16. COLLEGE
an you please tell me if this passage is correct? I am looking for 20 grammatical errors, comma(six); capitalization(six); question marks,apostrophes, and quotation marks(five); colons,semi-colon and common misused words(three) Why are great numbers of
asked by Queen on November 21, 2010
17. AP CHEMISTRY HELP
a) A student fails to wash the weighing paper when transferring the KHP sample into the beaker. What effect does this error have on the calculated molarity of the NAOH solution? Mathematically justify your answer. b) A student failed to notice an air
asked by Amber on December 20, 2012
18. Early Child Ed.
Would you please check these questions and my answers? Please let me know the ones you seem to think are wrong. 1. You want children to walk in the hallway. If they begin to run, the best thing to say to redirect them is: A. "Stop Running." B. "Walk in the
asked by Tamera on August 1, 2007
19. Algebra
A gardener has 60 feet of fencing with which to enclose a garden adjacent to a long existing wall. The gardener will use the wall for one side and the available fencing for the remaining three sides. (a) If the sides perpendicular to the wall have length x
asked by Rob on November 21, 2010
20. math
Can someone give me an example of a statement in English representing an algebraic relationship.I have to create one but i don't know what this refers to. Well it is just a written algebraic statement sort of like: a squared plus b squared is equal to c
asked by jasmine20 on December 30, 2006
21. French
I'm trying to figure out how you would say "he is being" in French. I know it has something to do with the present participle "étant," but we haven't really gotten very far into this yet in class. Would you say "il est étant" or "il étant?" Ok I found
asked by Alice on March 4, 2007
22. english
this is a portion of my english homework that i cant do. i was assigned over 100 questions and these are the ones i need the most help with and i need to turn it in in like an hour so any help would be great :-) 1. Which is not a sentence fragment? A.
asked by alan on November 30, 2007
23. Math
The mean of Susan's math and science scores is 74 points. The mean of her math and English scores is 83 points. How many more points did Susan score in English than in science?
asked by Chris on December 17, 2014
24. TRIANGLE MATH HELP!!!
Are the figures below similar? Why or why not? Determine whether the triangles below are similar. (Triangle M S I is shown on the left. Side M S is labeled 4. Side S I is labeled 6. Side M I is labeled 8. Angle M is 47 degrees. Angle S is 104 degrees.
asked by Agal on November 16, 2016
25. English
I forgot to include the following sentences. Thank you a lot! 1.Avoid repeating concepts over and over again and making personal comments which you cannot substantiate .. (?). Try to stick to the question. 2.Antony doesn’t care about the good of Rome (is
asked by Franco on February 22, 2010
26. English
Writeacher, I included a few sentences for you to check. I wonder if you coud help me organize my ideas more clearly. 1) Winston is tortoured by the Thoughtpolice and threatened that he will be attacked by rats, the thing which he hates most. 2) They make
asked by Henry on March 5, 2012
27. Probabitilty
For hair color, Brown (B) is a dominant trait and Blond (b) is a recessive trait. If one parent has pure blond hair and one parent has hybrid brown hair, complete the Punnett square and questions that follow. a) What is the probability that their first
asked by Jessica on April 30, 2014
28. statistics
For hair color, Brown (B) is a dominant trait and Blond (b) is a recessive trait. If one parent has pure blond hair and one parent has hybrid brown hair, complete the Punnett square and questions that follow. a) What is the probability that their first
asked by Anonymous on April 30, 2014
29. Probability
For hair color, Brown (B) is a dominant trait and Blond (b) is a recessive trait. If one parent has pure blond hair and one parent has hybrid brown hair, complete the Punnett square and questions that follow. a) What is the probability that their first
asked by Dawn on April 30, 2014
30. AOU
How should the decision making process be followed in order to make good decisions to respond to the situation faced by the BBC in 2001? #1 - Remember that we don't do students' homework for them. #2 - Your question is not clear, which may be why you're
asked by mohd on April 26, 2007
31. History
Plato and Aristotle both argued that political life should reflect humans' aspirations to live in a genuinely "good" society. However, their visions and ideas were very distinct. How were these "good societies" similar, and how were they different? Whose
asked by Cat on September 26, 2012
32. lovely bones
i was wondering if someone could either suggest to me some good websites or actually post the info about the gist of the book "lovely bones" esp. the main character...i have no idea what this book is about or anything, but there is an attempt at a
asked by B on September 3, 2007
33. Math Check
Solve the inequality. So i have done this problem at least three times and I keep getting the same answer. But in the back of the book the correct answer for this problem is x
asked by Heather on August 25, 2010
34. English
I have pictures of national flags of six countries. Which national flag is French natonal flag? Which country's national flag is this picture? Which national flag is related to French? Which national flag is French's. I want you to find Chinese national
asked by Tom on October 28, 2009
35. English - Macbeth
I have a little trouble understanding my essay question. This question is: Is Macbeth a tragic hero according to the classical definition of the term or is he merely a monster? Does Shakespeare succeed in creating sympathy for Macbeth? To repsond to this
asked by TP on August 17, 2008
36. algebra
As isosceles triangle has two sides of equal length. If the third side of an isosceles triangle is five inches longer than one of the equal sides, and the perimeter is 26 inches, find the length of each side.
asked by gigi on July 10, 2015
37. English, about my essay
Hi, thanks for your kind reply. I am following MLA format, except some parts need to be underlined ("Welcome" IRRI) -> IRRI has to be underlined, but this site doesn't have underlines, i think. My teacher told me that what I did was write, so that's ok.
asked by Jennifer on February 11, 2009
38. People
who is a famous person who didn't go to college or didn't graduate high school and is in bad shape because of it. (aka is poor or cant get a job or turned into a criminal etc)
asked by Samantha on December 12, 2012
39. english
poof read for me I am very sorry about her mother who has killed by drunk driver Zeke and I felt bad about Kate because she lost her mom. I wish her mom was alive and never get upset.
asked by anu on December 17, 2009
40. geometry
In the accompying diagram of circleo, diameter ad,chord AE,ans secantsCAB and CDE are drawn ; m angle BAD=40;and m arch AE=5((m arch ED) what is the measure of ACE?
asked by Vanna on March 10, 2012
41. english
12. In which one of the following sentences is the word too used like the word also? A. John had too much homework over the weekend. B. Would you care if I went too? C. The dessert was too rich for his taste. D. Harriman told Jake it was too bad about his
asked by Debbie on December 26, 2014
42. Health Care
Give an example of ineffective communication that resulted in bad customer service in the medical office and discuss strategies for overcoming perceived service issues.
asked by Anonymous on February 18, 2008
43. easy geometry
1)If diagonals of a rhombus are 10 cm and 24 cm. find the area and perimeter of the rhombus. 2)A regular hexagon with a perimeter of 24 units is inscribed in a circle. Find the radius of the circle. 3)Find the altitude,perimeter and area of an isosceles
asked by cyrus on December 13, 2014
44. english
15. Identify the sentence in which the underlined pronoun is used incorrectly. There are only six days left in the school year. It will be fun to visit their house in the summer. Their just going to have to wait to see the movie. Greg refused to go there
asked by john on April 10, 2015
45. English
Identify the sentence in which the underlined pronoun is used incorrectly. A. There are only six days left in the school year. B. It will be fun to visit their house in the summer. C.Their just going to have to wait to see the movie. D. Greg refused to go
asked by Connor on May 20, 2015
46. english
15. Identify the sentence in which the underlined pronoun is used incorrectly. There are only six days left in the school year. It will be fun to visit their house in the summer. Their just going to have to wait to see the movie. Greg refused to go there
asked by brian on April 10, 2015
47. spanish
what does the question and answers mean? "¿Cómo iba Ud. a la escuela primaria?" "Generalmente _____." a. caminaba b. lloraba c. me portaba mal d. molestaba a--walk? b--cry? c--behave bad? d--annoy/bother?
asked by y912f on May 5, 2009
48. History
-this was the second question -when FDR decided not to take part in helping the global depression, I was wondering why when the US withdrew from exchange-rate stabilization, how was this a bad thing for other countries? why couldn't they continue what they
asked by Amy on October 27, 2010
49. British Literature
"This supernatural soliciting. Cannot be ill, cannot be good. If ill, Why hath it given me earnest of success Commencing in a truth? I am Thane of Cawdor. If good, why do I yield to that suggestion Whose horrid image doth unfix my hair And make my seated
asked by Scooby10 on February 11, 2012
50. AP English
I have to do this summer pre-course work for AP English. One part is reading a book on Greek and Roman mythology and picking three people out of newspapers or magazines and writing a paragraph for each about how they are like a character in the book. Any
asked by Emily on August 11, 2007
51. english
Who is the intended audience for the novel The Scarlet Letter? That is a good question. He wrote of historical times, of sin, guilt, and repentance. He wrote of social mores, social pressure, and the pain of reconciling this with personal morals. I think
asked by anonymous on October 26, 2006
52. English
I forgot to include the following sentences. I included my doubts in brackets. Thank you. 1) Do you think we could possibly talk through Skype? 2)Let me know when you are available for talking (?) through Skypr. 3) Samantha tried to use a translation
asked by Mike1 on April 11, 2011
53. english
- Computers - English - Foreign Languages - Health - Home Economics - Math - Music - Physical Education - Science - Social Studies GRADE LEVELS - Preschool - Kindergarten - Elementary School - 1st Grade - 2nd Grade - 3rd Grade - 4th Grade - 5th Grade - 6th
asked by peter on January 8, 2010
54. physics
A rope of negligible mass passes over a puley of a negligible mass attached to the ceiling. One end of the rope is held by stdent A of mass 70 kg who is at rest on the floor. The opposite end of the rope is held by student B of mass 60 kg who is suspended
asked by physics on September 11, 2009
55. Phsical Science
Other than familiarity, what is the advantage of the English System of measurement?
asked by Rashida on August 23, 2014
56. Social Studies
how did the representatives became independent from the english parliament? And i tried wikipedia
asked by Karann Help!!! on November 6, 2011
57. English
I just learning more accuracy English important with communications.
asked by Chrysta E Banks on August 27, 2015
58. English
Help!!!! I need a sample of English formal letter. immediate plzzzzz
asked by Ali on January 23, 2016
59. French
Could you help me find the story of Dantés in english, or french. Thanks
asked by Sab on May 15, 2008
60. law
english law. what are the powers of the court of appeal.
asked by elizabeth on April 9, 2011
61. spanish
What does Porque siempre me duele la cabeza mean in English?
asked by Cherylynn on January 11, 2010
62. social
what are two actions outlawed by the english bill of rights?
asked by abby on December 2, 2009
63. 7th grade
Formulas for changing metric into English measurements
asked by Anonymous on March 27, 2009
64. english
are there any objects that represent perseverence? and why? i need this for an english projcet
asked by stephanie on November 13, 2010
65. question
Tomorrow is my English exam. do you think I can get high marks?
asked by verver on May 5, 2015
66. spanish
what does corregir los campos senalados mean in english
asked by judy on October 26, 2009
67. English
Hello!! I have to write paragraph. What I do, english is my second language. Help thank yuo.
asked by Rosita de la Cruz on February 26, 2016
68. French
How do you give orders to someone in French using their name? Is this done in the same construction as in English?
asked by Anonymous on September 26, 2014
69. spainsh
what does me gusta muy tu. tu hacer me algre mean in english
asked by jon on February 23, 2010
70. English
Can somebody please suggest any essay topics that will help me for my English provincial?
asked by Canada 05 on January 21, 2016
71. Social Studies
what are two actions outlawed by the English Bill of rights.
asked by Anonymous on December 8, 2015
72. ENGLISH
How English 002 help me in achieving the T.I.P Graduate Attributes?
asked by :)) on March 17, 2013
73. spanish
how would we say this spanish word in english? dona Josefina?
asked by hunter on August 18, 2010
74. languages
Is there a web site that can help students out, if they can not speak English?
asked by sally on October 23, 2008
75. english
what opportunities does english makes to peaple who study it
asked by thamsanqa on April 13, 2019
76. Spanish
What does "me gusta la clase de cinecias naturales" mean in english?
asked by No name on October 12, 2018
77. chemistry
Please explain, in simple English, why is milk a colloid?
asked by Maddie on January 8, 2014
78. comp 150
English, should be the official language of the united states
asked by Anonymous on October 18, 2010
79. social studies
what are the steps toward democaracy of the early English settlers?
asked by Chelsea on November 5, 2008
80. Siddhartha Higher Secondary School
English what will you do if your parents were unable to send you school
asked by swachitta on October 29, 2015
81. French
What is the English translation of: Que veut dire VTT?
asked by Erika on May 25, 2013
82. iran national anthem
I need the English lyrics of Iran's National Anthem...please help....
asked by klar maesen on February 11, 2008
83. parts of speech
What parts of speeech in English express huour
asked by Ani on January 31, 2011
84. french
what does Elle travaille dans un hotel mean in english?
asked by sufain on December 5, 2010
85. French 1
what does "Je rentre de l'ecole et je vais au restaurant." mean in English?
asked by Helpneeded.love on February 25, 2010
86. To Writeacher
Writeacher I have a post on my English assignment, and I have it addressed to you, may you please take a look at it, cause I really need help, thanks.
asked by Sara on October 5, 2010
87. French
In "le maïs", what is the two dots on top of the i called in english?
asked by Angie on June 8, 2009
88. Geography
which of the following statements best characteristics the english colony of roanoke? A B C D
asked by Student on October 11, 2017
89. french
what does virage mean? i cannot find it in my english to french dictionary....
asked by mellisa on February 1, 2012
90. Spanish
What does siete niños translate to in English? and what does primaria mean?
asked by Miley on March 4, 2008
91. spanish
What does Que llevan Los jugadores de futbol? Mean in english
asked by kk on May 14, 2010
92. english- please help me edit
Hi I was wondering if anyone can help me edit/proofread my english essay =)
asked by Dina on May 1, 2008
93. 5th grade
need the biography ofEmily Hutchinson for my English class.
asked by ricky on September 10, 2008
94. French
How do you translate "I don't understand English?" in French?
asked by Anonymous on May 29, 2009
95. English-For Jiskha Tutors
Can some of you please take a look at my previous English posts. They are all on this page, and I need all of them to be checked. Please and thank you:-)
asked by Sara on May 3, 2010
96. US history
how did the Virginia colony survive with the help of the English colonists?
asked by brian on September 19, 2016
97. spanish
what is the english translation of all fill del compare?
asked by vicente on March 2, 2012
98. English
List and explain 10 functions of english language
asked by Anonymous on January 17, 2019
99. Social Studies
Why do so many Asian countries speak English or French?
asked by Scott on May 18, 2011
100. History
why did the English settle in Virginia in the 17th century ?
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Weegy: What is?
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User: Find the slope of the line that contains the points named. G(3, -2), H(7, -2)
Question
Asked 6/19/2012 11:40:29 AM
Updated 7/31/2014 2:38:49 PM
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The slope of the line that contains the points named G(3, -2), and H(7, -2) is 0. m = [-2-(-2)]/(7-3); m = 0/4 = 0
Added 7/31/2014 2:38:48 PM
This answer has been confirmed as correct and helpful.
Confirmed by yumdrea [7/31/2014 2:45:30 PM]
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Questions asked by the same visitor
Find the coordinates for the midpoint of the segment with endpoints given. (3, 5) and (-2, 0) (1/2, 5/2) (5/2, 1/2) (1, 5/2)
Weegy: midpoint is 4/2,-2/2,So (2,-1) User: Find the coordinates for the midpoint of the segment with endpoints given. (3, 5) and (-2, 0) Weegy: midpoint is 4/2,-2/2,So (2,-1) User: A circle is inscribed in a square whose vertices have coordinates R(0, 4), S(6, 2), T(4, -4), and U(-2, -2). Find the equation of the circle. (x - 2)² + y² = 68 (x + 2)² + y² = 20 (x - 2)² + y² = 10 (More)
Question
Updated 7/15/2014 8:57:08 AM
The coordinates for the midpoint of the segment with endpoints (3, 5) and (-2, 0) is (1/2, 5/2).
Added 7/15/2014 8:45:52 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [7/15/2014 8:50:22 AM]
A circle is inscribed in a square whose vertices have coordinates R(0, 4), S(6, 2), T(4, -4), and U(-2, -2). The equation of the circle is (x - 2)² + y² = 10
Added 7/15/2014 8:57:08 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [7/15/2014 9:00:13 AM]
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Question
The diagram shows the sketch of a curve and the tangent to the curve at P.
The curve has equation and the point P(-2,24) lies on the curve. The tangent at P crosses the x-axis at Q.
a.
i. Find the equation of the tangent to the curve at the point P, giving your answer in the form .
ii. Hence find the x-coordinate of Q.
b.
i. Find
ii. The point R(1,0) lies on the curve. Calculate the area of the shaded region bounded by the curve and the lines PQ and QR.
Solution
a.
i.
We are required to find the equation of the tangent to the curve at the point P(-2,24).
To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
We already have coordinates of a point on the tangent P(-2,24). Therefore, we need slope of the tangent to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we have gradient for the curve at point P, we can find the slope of tangent to the curve at point P.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;
We are given the equation of the curve as;
Therefore;
Rule for differentiation is of is:
Rule for differentiation is of is:
Rule for differentiation is of is:
Now we need gradient of the curve at point P(-2,24). Therefore, we substitute ;
Hence,
Now we can find slope of tangent to the curve at point P(-2,24).
Now we can write equation of the tangent.
Point-Slope form of the equation of the line is;
ii.
We are required to find coordinates of x-intercept (point Q) of the tangent to the curve at point P.
The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
From (a:i) we know the equation of the tangent to the curve at point P as;
Therefore;
Hence coordinates of x-intercept (point Q) of the tangent to the curve at point P are;
b.
i.
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
ii.
Consider the diagram below.
It is evident from the diagram that;
First we find area under curve.
To find the area of region under the curve , we need to integrate the curve from point to along x-axis.
We are given equation of the curve as;
It is evident that shaded region extends from point P (x=-2) to point R (x=1) along x-axis.
Therefore;
We have found in (b:i) that;
Therefore;
Next we find area of triangle PQS.
Expression for the area of the triangle is;
For triangle PQS;
It is evident that SQ and PS are distances between point S&Q and P&S, respectively.
To find these distance we need coordinates of all three points. We are already given coordinates of point P(-2,24) and we have found coordinates of in (a:ii).
It is evident from the diagram above that x-coordinate of S is -2 (as that of point P) and y-coordinate 0 (as it is x-intercept of PS). Therefore, S(-2,0).
Expression to find distance between two given points and is:
Therefore, for PS;
Hence;
Next we find QS.
Hence;
Now we can find area of triangle PSQ.
Finally; | 981 | 3,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.795138 |
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# if cotA+1/cotA=2 then find the value of cot^2A+1/cot^2A
SHAIK AASIF AHAMED
7 years ago
Hello student,
givencotA+1/cotA=2
let x=cotA
then x+1/x=2
squaring on both sides we get
x2+1/x2+2=4
so x2+1/x2=2
substituting x=cotA in above expression we get
cot2A+1/cot2A=2
rahul
16 Points
7 years ago | 238 | 642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.711057 |
http://slideplayer.com/slide/1423456/ | 1,511,158,168,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805914.1/warc/CC-MAIN-20171120052322-20171120072322-00183.warc.gz | 282,258,018 | 20,504 | # Physics Wheeled and levitating vehicles STEM Box Challenge Launch, 22.10.13 Dr Stephen Gibson, Lecturer in Physics.
## Presentation on theme: "Physics Wheeled and levitating vehicles STEM Box Challenge Launch, 22.10.13 Dr Stephen Gibson, Lecturer in Physics."— Presentation transcript:
Physics Wheeled and levitating vehicles STEM Box Challenge Launch, 22.10.13 Dr Stephen Gibson, Lecturer in Physics
Physics challenge overview: Phase one, Wheeled Vehicles -Farthest distance travelled. -Fastest vehicle over a set distance. -Fattest (heaviest) load pulled over a set distance. 2 Challenge your students: » Using only the items in the box and your ingenuity, design a self- powered, wheeled vehicle that is powered by an elastic band or a balloon. » Experimentally test which power source to choose and optimise the vehicle differently to compete in three challenges:
Physics challenge overview: Phase one, Wheeled Vehicles Mechanical piece for vehicle construction: 4 wheels, chassis pieces,/ straws. axles, bearings. Power sources; elastic bands of different strengths / sizes, including spares; balloons + straws. Small weights for testing elastic band extension. Document showing outline of vehicle construction Worksheet with ideas for experiment test and vehicle optimisation 3 Whats in the box? » Box include the essentials to construct your vehicles:
Physics challenge overview: Phase one, Wheeled Vehicles 4 Some vehicle ideas: » The design is up to your students, but must use one of the power sources provided. » The challenge is how to optimize the design for the three different tasks: » Think about vehicle mass, acceleration, stored energy, speed of release…
Physics challenge overview: Phase II, compete at RHUL + levitation! 5 » On completing the challenges at your school, enter your vehicle in a summer competition at Royal Holloway, University of London. » In the final challenge, make your vehicles levitate with a little help from our liquid nitrogen cooled superconductors and watch it take off and soar along a magnetic track.
Curriculum links 6 stored (potential) energy in the elastic band or balloon and conversion of energy to other forms (kinetic + heat). kinematics: time, distance, speed and acceleration. forces, motion and reducing friction. » Students learning outcomes are linked to the most recent Key Stage 3 curriculum: » Aim to bring the concepts to life through the practical demonstrations! » No knowledge of superconductors is needed…
Added value: 7 » Royal Holloway will support your school with two contact/interface sessions: Interface 1: Initial skype meeting with teacher / pupils regarding the wheeled vehicle challenges in the competition. Explain challenge and give pointers on optimising the models for the three challenges. Interface 2: School visit to include demonstration with liquid nitrogen, to test static levitation with a short test track.
Added value: 8 http://www.youtube.com/watch?v=DJV5PSpYPL0#t=183 » Superconducting / low temperature technology support through Royal Holloway Physics Department: Superconductors manufactured via the crystal growth facility in the RHUL Physics. Diamagnetic (pyrolytic graphite) can be used for static levitation demonstration. Well also provide a magnetically levitating racetrack for the competition at RHUL. A glimpse of whats possible:
Questions?
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http://www.howeverythingworks.org/print1.php?QNum=76 | 1,532,366,255,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00488.warc.gz | 464,992,738 | 1,650 | MLA Citation: Bloomfield, Louis A. "Question 76"How Everything Works 23 Jul 2018. 23 Jul 2018 .
76. When you transfer momentum between two objects, why is it that the change in total momentum is 0?
Suppose you are standing motionless on extremely slippery ice. If you now take off your shoe and throw it northward as hard as you can, you will transfer momentum to it. Since you and your shoe were initially motionless, your combined momentum was 0. Neither of you nor the shoe was moving, so the product of mass times velocity was 0. But after you throw the shoe, both you and the shoe have momentum. Your momentum is equal to your mass times your velocity, so your momentum points in the direction you are going. The shoe also has momentum, equal to its mass times its velocity. But since it is heading in the opposite direction from you, it has the opposite momentum from you. Together, your combined momentum remains exactly 0—it didn't change. In general, momentum is transferred from one object to another so that any change in momentum in one object is always compensated for by an opposite change in momentum in the other object. | 249 | 1,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-30 | latest | en | 0.975725 |
https://aviation.stackexchange.com/questions/70792/can-surface-roughing-effects-on-an-airfoil-reynolds-number-be-quantified | 1,719,106,241,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00472.warc.gz | 102,961,908 | 36,969 | # Can surface roughing effects on an airfoil Reynolds number be quantified?
I am building balsa free flight gliders with a chord of 0.10 meters and a velocity of around 3 meters per second. This works out to a Reynolds number of around 15,000.
The wings are around 1 meter in span, and are made with the wood grain running lengthwise. They are thin undercambered (no bottom covering) with maximum thickness of 7 mm at 30% of chord from leading edge.
Because of the roughness of the unfinished grain, is my Reynolds more like 50,000, or am I still down in the range where a flat plate might be better?
Roughness has no influence on the Reynolds number. It will, however, help to trigger a laminar-turbulent transition. That is the reason for dimples in golf balls. Also, roughness will increase the friction drag in turbulent boundary layers.
Flat plate friction diagram (picture source). Note the horizontal lines: They show that drag does not drop any more with increasing Reynolds number once the surface roughness exceeds a certain value. Roughness is given here relative to the chord length as $$\frac{\epsilon}{l}$$, with $$\epsilon$$ being the mean height of the roughness and $$l$$ the chord length.
The diagram only starts at Re = 100,000, so it is clearly outside the range of your application. If you sand your wooden surface with fine grit sandpaper, I don't expect that the remaining roughness will increase drag. At those Reynolds numbers it might be advantageous to add a trip wire in order to induce a transition to turbulent flow – see the golf ball example.
• Frisbee grooves may be a bit closer. Roughness to chord length ratio interesting way of trying it, as well as location on the airfoil. Commented Oct 18, 2019 at 16:08
• What's causing the increasing in drag with Re for the low surface roughness lines?
– JZYL
Commented Oct 18, 2019 at 23:08
• @Jimmy Don't confuse "increasing drag" and "increasing the drag coefficient". For a given structure, lower Re means lower velocity and the drag coefficient is multiplied by $v^2$ to get the drag force. But for laminar flow the drag force is proportional to $v$ not $v^2$, so the drag coefficient used in the conventional formula appears to be bigger at low Re even though the drag force is less. Commented Oct 19, 2019 at 0:01
• @alephzero I didn't confuse. 1e6 Re is a standard wind tunnel Re for engineering design. No one will say the drag is linear to V.
– JZYL
Commented Oct 19, 2019 at 0:39
• What I'm getting is that larger roughness to chord ratio is "completely turbulent" at lower Reynolds, therefor "transitional turbulent" at a lower airspeed, where drag reduction occurs. This may have been what Kline-Fogelman were after, but their "disrupter" steps may have been too large. Commented Oct 19, 2019 at 11:53 | 685 | 2,797 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.904393 |
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# 700-800 Level Quant Problem Collection + Detailed Solutions
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Math Expert
Joined: 02 Sep 2009
Posts: 52390
Re: 700-800 Level Quant Problem Collection + Detailed Solutions [#permalink]
### Show Tags
07 Mar 2017, 23:09
muthappashivani wrote:
Q33. A, B, C and D are all different digits between 0 and 9. If AB + DC = 7B (AB, DC and 7B
are two digit numbers), what is the value of C?
A) 0
B) 1
C) 2
D) 3
E) 5
Can someone help me with this. Thank you
Please follow the rules when posting a question. Kindly re-post ion the PS forum. Thank you.
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Re: 700-800 Level Quant Problem Collection + Detailed Solutions [#permalink]
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27 Mar 2017, 09:43
I Guess there is a mistake in the answers.
In the Geometry, 2nd part for the triangles, the 4th question - both of the statements alone are sufficient to answer the question, while in the answers the 2nd one is ignored. If someone is interested I may provide the explanation.
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Re: 700-800 Level Quant Problem Collection + Detailed Solutions [#permalink]
### Show Tags
02 Jun 2017, 18:13
By medium/hard I mean 600/700 level questions.[/quote]
So the questions are 600-700 or 700+ ?
In my opinion questions are 600-700 range.
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Re: 700-800 Level Quant Problem Collection + Detailed Solutions [#permalink]
### Show Tags
24 Jan 2018, 11:13
I could not find solutions for Calculations, Exponents, Basic Algebra
Re: 700-800 Level Quant Problem Collection + Detailed Solutions &nbs [#permalink] 24 Jan 2018, 11:13
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Display posts from previous: Sort by | 878 | 3,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-04 | latest | en | 0.879587 |
https://tiffytaffy.com/riddle-thats-playing-with-so-many-peoples-minds/ | 1,632,630,181,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057796.87/warc/CC-MAIN-20210926022920-20210926052920-00562.warc.gz | 585,624,772 | 10,096 | Here’s A Riddle That’s Playing With So Many People’s Minds! Can You Answer Correctly?
When was the last time you tried your hand at a little math riddle? One that not only challenges your math skills, but that also forces you to pay attention to detail and do some critical thinking? Well, if it’s been a while, it’s time to dust off that part of your brain and use it!
Can Your Solve This Math Riddle Without Any Help?
This riddle was allegedly created by a third-grader, but none of their peers could get the right answer. The next step was for the adults to tackle the problem, but it stumped many of them, too. Now it’s your turn to see if you can solve it or not. Check out the problem below.
Think you have the correct answer? Before you jump to say “Yes, that was so easy!” maybe give the problem another read-over before being so bold. Are you sure you read it correctly?
70!
If you’re thinking right now “What?! How did you get 70?” allow me to explain. The riddle says divide 30 by half, not in half. When you divide by half, you actually end up multiplying by two. This means that 30 divided by half is 60. Add 10, and you’ve got 70. | 278 | 1,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.961903 |
http://en.wikipedia.org/wiki/Gear_ratios | 1,429,972,007,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246649738.26/warc/CC-MAIN-20150417045729-00278-ip-10-235-10-82.ec2.internal.warc.gz | 86,626,127 | 19,164 | # Gear ratio
(Redirected from Gear ratios)
The gear ratio of a gear train, also known as its speed ratio, is the ratio of the angular velocity of the input gear to the angular velocity of the output gear.[1] The gear ratio can be calculated directly from the numbers of teeth on the gears in the gear train. The torque ratio of the gear train, also known as its mechanical advantage, is determined by the gear ratio.[2] The speed ratio and mechanical advantage are defined so they yield the same number in an ideal linkage.
Illustration of gears of an automotive transmission
## Gear trains with two gears
The simplest example of a gear train has two gears. The "input gear" (also known as drive gear) transmits power to the "output gear" (also known as driven gear). The input gear will typically be connected to a power source, such as a motor or engine. In such an example, the power output of the output (driven) gear depends on the ratio of the dimensions of the two gears.
Two meshing gears transmit rotational motion.
### Formula
The teeth on gears are designed so that the gears can roll on each other smoothly (without slipping or jamming). In order for two gears to roll on each other smoothly, they must be designed so that the velocity at the point of contact of the two pitch circles (represented by v) is the same for each gear.
Mathematically, if the input gear GA has the radius rA and angular velocity $\omega_A \!$, and meshes with output gear GB of radius rB and angular velocity $\omega_B \!$, then:
$v = r_A \omega_A = r_B \omega_B, \!$
The number of teeth on a gear is proportional to the radius of its pitch circle, which means that the ratios of the gears' angular velocities, radii, and number of teeth are equal. Where NA is the number of teeth on the input gear and NB is the number of teeth on the output gear, the following equation is formed:
$\frac{\omega_A}{\omega_B} = \frac{r_B}{r_A} = \frac{N_B}{N_A}.$
This shows that a simple gear train with two gears has the gear ratio R given by
$R = \frac{\omega_A}{\omega_B} = \frac{N_B}{N_A}.$
This equation shows that if the number of teeth on the output gear GB is larger than the number of teeth on the input gear GA, then the input gear GA must rotate faster than the output gear GB.
## Speed ratio
Gear teeth are distributed along the circumference of the pitch circle so that the thickness t of each tooth and the space between neighboring teeth are the same. The pitch p of a gear, which is the distance between equivalent points on neighboring teeth along the pitch circle, is equal to twice the thickness of a tooth,
$p=2t.\!$
The pitch of a gear GA can be computed from the number of teeth NA and the radius rA of its pitch circle
$p = \frac{2\pi r_A}{N_A}.$
In order to mesh smoothly two gears GA and GB must have the same sized teeth and therefore they must have the same pitch p, which means
$p = \frac{2\pi r_A}{N_A} = \frac{2\pi r_B}{N_B}.$
This equation shows that the ratio of the circumference, the diameters and the radii of two meshing gears is equal to the ratio of their number of teeth,
$\frac{r_B}{r_A} = \frac{N_B}{N_A}.$
The speed ratio of two gears rolling without slipping on their pitch circles is given by,
$R =\frac{\omega_A}{\omega_B}=\frac{r_B}{r_A},$
therefore
$R = \frac{\omega_A}{\omega_B} = \frac{N_B}{N_A}.$
In other words, the gear ratio, or speed ratio, is inversely proportional to the radius of the pitch circle and the number of teeth of the input gear.
## Torque ratio
A gear train can be analyzed using the principle of virtual work to show that its torque ratio, which is the ratio of its output torque to its input torque, is equal to the gear ratio, or speed ratio, of the gear train.
This means that the input torque ΤA applied to the input gear GA and the output torque ΤB on the output gear GB are related by the ratio
$R = \frac{T_B}{T_A},$
where R is the gear ratio of the gear train.
The torque ratio of a gear train is also known as its mechanical advantage
$\mathit{MA} = \frac{T_B}{T_A}.$
## Idler gears
In a sequence of gears chained together, the ratio depends only on the number of teeth on the first and last gear. The intermediate gears, regardless of their size, do not alter the overall gear ratio of the chain. However, the addition of each intermediate gear reverses the direction of rotation of the final gear.
An intermediate gear which does not drive a shaft to perform any work is called an idler gear. Sometimes, a single idler gear is used to reverse the direction, in which case it may be referred to as a reverse idler. For instance, the typical automobile manual transmission engages reverse gear by means of inserting a reverse idler between two gears.
Idler gears can also transmit rotation among distant shafts in situations where it would be impractical to simply make the distant gears larger to bring them together. Not only do larger gears occupy more space, the mass and rotational inertia (moment of inertia) of a gear is proportional to the square of its radius. Instead of idler gears, a toothed belt or chain can be used to transmit torque over distance.
### Formula
If a simple gear train has three gears, such that the input gear GA meshes with an intermediate gear GI which in turn meshes with the output gear GB, then the pitch circle of the intermediate gear rolls without slipping on both the pitch circles of the input and output gears. This yields the two relations
$\frac{\omega_A}{\omega_I} = \frac{N_I}{N_A}, \quad \frac{\omega_I}{\omega_B} = \frac{N_B}{N_I}.$
The speed ratio of this gear train is obtained by multiplying these two equations to obtain
$R = \frac{\omega_A}{\omega_B} = \frac{N_B}{N_A}.$
Notice that this gear ratio is exactly the same as for the case when the gears GA and GB engaged directly. The intermediate gear provides spacing but does not affect the gear ratio. For this reason it is called an idler gear. The same gear ratio is obtained for a sequence of idler gears and hence an idler gear is used to provide the same direction to rotate the driver and driven gear, if the driver gear moves in clockwise direction, then the driven gear also moves in the clockwise direction with the help of the idler gear.
### Example
2 gears and an idler gear on a piece of farm equipment, with a ratio of 42/13 = 3.23:1
In the photo, assuming that the smallest gear is connected to the motor, it is called the drive gear or input gear. The somewhat larger gear in the middle is called an idler gear. It is not connected directly to either the motor or the output shaft and only transmits power between the input and output gears. There is a third gear in the upper-right corner of the photo. Assuming that that gear is connected to the machine's output shaft, it is the output or driven gear.
The input gear in this gear train has 13 teeth and the idler gear has 21 teeth. Considering only these gears, the gear ratio between the idler and the input gear can be calculated as if the idler gear was the output gear. Therefore, the gear ratio is driven/drive = 21/13 ≈1.62 or 1.62:1.
At this ratio it means that the drive gear must make 1.62 revolutions to turn the driven gear once. It also means that for every one revolution of the driver, the driven gear has made 1/1.62, or 0.62, revolutions. Essentially, the larger gear turns slower.
The third gear in the picture has 42 teeth. The gear ratio between the idler and third gear is thus 42/21, or 2:1, and hence the final gear ratio is 1.62x2≈3.23. For every 3.23 revolutions of the smallest gear, the largest gear turns one revolution, or for every one revolution of the smallest gear, the largest gear turns 0.31 (1/3.23) revolution, a total reduction of about 1:3.23 (Gear Reduction Ratio (GRR) = 1/Gear Ratio (GR)).
Since the idler gear contacts directly both the smaller and the larger gear, it can be removed from the calculation, also giving a ratio of 42/13≈3.23. The idler gear serves to make both the drive gear and the driven gear rotate in the same direction, but confers no mechanical advantage.
## Belt drives
Belts can have teeth in them also and be coupled to gear-like pulleys. Special gears called sprockets can be coupled together with chains, as on bicycles and some motorcycles. Again, exact accounting of teeth and revolutions can be applied with these machines.
Valve timing gears on a Ford Taunus V4 engine — the small gear is on the crankshaft, the larger gear is on the camshaft. The crankshaft gear has 34 teeth, the camshaft gear has 68 teeth and runs at half the crankshaft RPM.
(The small gear in the lower left is on the balance shaft.)
For example, a belt with teeth, called the timing belt, is used in some internal combustion engines to synchronize the movement of the camshaft with that of the crankshaft, so that the valves open and close at the top of each cylinder at exactly the right time relative to the movement of each piston. A chain, called a timing chain, is used on some automobiles for this purpose, while in others, the camshaft and crankshaft are coupled directly together through meshed gears. Regardless of which form of drive is employed, the crankshaft to camshaft gear ratio is always 2:1 on four-stroke engines, which means that for every two revolutions of the crankshaft the camshaft will rotate once.
## Automotive applications
Automobile drivetrains generally have two or more major areas where gearing is used. Gearing is employed in the transmission, which contains a number of different sets of gears that can be changed to allow a wide range of vehicle speeds, and also in the differential, which contains the final drive to provide further speed reduction at the wheels. In addition, the differential contains further gearing that splits torque equally between the two wheels while permitting them to have different speeds when travelling in a curved path. The transmission and final drive might be separate and connected by a driveshaft, or they might be combined into one unit called a transaxle. The gear ratios in transmission and final drive are important because different gear ratios will change the characteristics of a vehicle's performance.[citation needed]
### Example
A 2004 Chevrolet Corvette C5 Z06 with a six-speed manual transmission has the following gear ratios in the transmission:
Gear Ratio
1st gear 2.97:1
2nd gear 2.07:1
3rd gear 1.43:1
4th gear 1.00:1
5th gear 0.84:1
6th gear 0.56:1
reverse -3.38:1
In 1st gear, the engine makes 2.97 revolutions for every revolution of the transmission’s output. In 4th gear, the gear ratio of 1:1 means that the engine and the transmission's output rotate at the same speed. 5th and 6th gears are known as overdrive gears, in which the output of the transmission is revolving faster than the engine's output.
The Corvette above has an axle ratio of 3.42:1, meaning that for every 3.42 revolutions of the transmission’s output, the wheels make one revolution. The differential ratio multiplies with the transmission ratio, so in 1st gear, the engine makes 10.16 revolutions for every revolution of the wheels.
The car’s tires can almost be thought of as a third type of gearing. This car is equipped with 295/35-18 tires, which have a circumference of 82.1 inches. This means that for every complete revolution of the wheel, the car travels 82.1 inches (209 cm). If the Corvette had larger tires, it would travel farther with each revolution of the wheel, which would be like a higher gear. If the car had smaller tires, it would be like a lower gear.
With the gear ratios of the transmission and differential, and the size of the tires, it becomes possible to calculate the speed of the car for a particular gear at a particular engine RPM.
For example, it is possible to determine the distance the car will travel for one revolution of the engine by dividing the circumference of the tire by the combined gear ratio of the transmission and differential.
$d = \frac{c_t}{gr_t \times gr_d}$
It is also possible to determine a car's speed from the engine speed by multiplying the circumference of the tire by the engine speed and dividing by the combined gear ratio.
$v_c = \frac{c_t \times v_e}{gr_t \times gr_d}$
Gear Distance per engine revolution Speed per 1000 RPM
1st gear 8.1 in (210 mm) 7.7 mph (12.4 km/h)
2nd gear 11.6 in (290 mm) 11.0 mph (17.7 km/h)
3rd gear 16.8 in (430 mm) 15.9 mph (25.6 km/h)
4th gear 24.0 in (610 mm) 22.7 mph (36.5 km/h)
5th gear 28.6 in (730 mm) 27.1 mph (43.6 km/h)
6th gear 42.9 in (1,090 mm) 40.6 mph (65.3 km/h)
### Wide-ratio vs. close-ratio transmission
A close-ratio transmission is a transmission in which there is a relatively little difference between the gear ratios of the gears. For example, a transmission with an engine shaft to drive shaft ratio of 4:1 in first gear and 2:1 in second gear would be considered wide-ratio when compared to another transmission with a ratio of 4:1 in first and 3:1 in second. This is because the close-ratio transmission has less of a progression between gears. For the wide-ratio transmission, the first gear ratio is 4:1 or 4, and in second gear it is 2:1 or 2, so the progression is equal to 4/2 = 2 (or 200%). For the close-ratio transmission, first gear has a 4:1 ratio or 4, and second gear has a ratio of 3:1 or 3, so the progression between gears is 4/3, or 133%. Since 133% is less than 200%, the transmission with the smaller progression between gears is considered close-ratio. However, the difference between a close-ratio and wide-ratio transmission is subjective and relative.[3]
Close-ratio transmissions are generally offered in sports cars, sport bikes, and especially in race vehicles, where the engine is tuned for maximum power in a narrow range of operating speeds, and the driver or rider can be expected to shift often to keep the engine in its power band.
Factory 4-speed or 5-speed transmission ratios generally have a greater difference between gear ratios and tend to be effective for ordinary driving and moderate performance use. Wider gaps between ratios allow a higher 1st gear ratio for better manners in traffic, but cause engine speed to decrease more when shifting. Narrowing the gaps will increase acceleration at speed, and potentially improve top speed under certain conditions, but acceleration from a stopped position and operation in daily driving will suffer.
Range is the torque multiplication difference between 1st and 4th gears; wider-ratio gear-sets have more, typically between 2.8 and 3.2. This is the single most important determinant of low-speed acceleration from stopped.
Progression is the reduction or decay in the percentage drop in engine speed in the next gear, for example after shifting from 1st to 2nd gear. Most transmissions have some degree of progression in that the RPM drop on the 1-2 shift is larger than the RPM drop on the 2-3 shift, which is in turn larger than the RPM drop on the 3-4 shift. The progression may not be linear (continuously reduced) or done in proportionate stages for various reasons, including a special need for a gear to reach a specific speed or RPM for passing, racing and so on, or simply economic necessity that the parts were available.
Range and progression are not mutually exclusive, but each limits the number of options for the other. A wide range, which gives a strong torque multiplication in 1st gear for excellent manners in low-speed traffic, especially with a smaller motor, heavy vehicle, or numerically low axle ratio such as 2.50, means that the progression percentages must be high. The amount of engine speed, and therefore power, lost on each up-shift is greater than would be the case in a transmission with less range, but less power in 1st gear. A numerically low 1st gear, such as 2:1, reduces available torque in 1st gear, but allows more choices of progression.
There is no optimal choice of transmission gear ratios or a final drive ratio for best performance at all speeds, as gear ratios are compromises, and not necessarily better than the original ratios for certain purposes. | 3,818 | 16,124 | {"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} | 3.953125 | 4 | CC-MAIN-2015-18 | latest | en | 0.930563 |
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## Presentation on theme: "Hypothesis Testing with One Sample"— Presentation transcript:
Hypothesis Testing with One Sample
Chapter 7 Hypothesis Testing with One Sample Larson/Farber 4th ed.
Chapter Outline 7.1 Introduction to Hypothesis Testing
7.2 Hypothesis Testing for the Mean (Large Samples) 7.3 Hypothesis Testing for the Mean (Small Samples) 7.4 Hypothesis Testing for Proportions 7.5 Hypothesis Testing for Variance and Standard Deviation Larson/Farber 4th ed.
Introduction to Hypothesis Testing
Section 7.1 Introduction to Hypothesis Testing Larson/Farber 4th ed.
Section 7.1 Objectives State a null hypothesis and an alternative hypothesis Identify type I and type I errors and interpret the level of significance Determine whether to use a one-tailed or two-tailed statistical test and find a p-value Make and interpret a decision based on the results of a statistical test Write a claim for a hypothesis test Larson/Farber 4th ed.
Hypothesis Tests Hypothesis test
A process that uses sample statistics to test a claim about the value of a population parameter. For example: An automobile manufacturer advertises that its new hybrid car has a mean mileage of 50 miles per gallon. To test this claim, a sample would be taken. If the sample mean differs enough from the advertised mean, you can decide the advertisement is wrong. Larson/Farber 4th ed.
Hypothesis Tests Statistical hypothesis
A statement, or claim, about a population parameter. Need a pair of hypotheses one that represents the claim the other, its complement When one of these hypotheses is false, the other must be true. Larson/Farber 4th ed.
complementary statements
Stating a Hypothesis Null hypothesis A statistical hypothesis that contains a statement of equality such as , =, or . Denoted H0 read “H subzero” or “H naught.” Alternative hypothesis A statement of inequality such as >, , or <. Must be true if H0 is false. Denoted Ha read “H sub-a.” complementary statements Larson/Farber 4th ed.
Stating a Hypothesis To write the null and alternative hypotheses, translate the claim made about the population parameter from a verbal statement to a mathematical statement. Then write its complement. H0: μ ≤ k Ha: μ > k H0: μ ≥ k Ha: μ < k H0: μ = k Ha: μ ≠ k Regardless of which pair of hypotheses you use, you always assume μ = k and examine the sampling distribution on the basis of this assumption. Larson/Farber 4th ed.
Example: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A university publicizes that the proportion of its students who graduate in 4 years is 82%. Solution: H0: Ha: p = 0.82 Equality condition (Claim) p ≠ 0.82 Complement of H0 Larson/Farber 4th ed.
Example: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is less than 2.5 gallons per minute. Solution: H0: Ha: μ ≥ 2.5 gallons per minute Complement of Ha Inequality condition μ < 2.5 gallons per minute (Claim) Larson/Farber 4th ed.
Example: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim. A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is more than 20 ounces. Solution: H0: Ha: μ ≤ 20 ounces Complement of Ha Inequality condition μ > 20 ounces (Claim) Larson/Farber 4th ed.
Types of Errors No matter which hypothesis represents the claim, always begin the hypothesis test assuming that the equality condition in the null hypothesis is true. At the end of the test, one of two decisions will be made: reject the null hypothesis fail to reject the null hypothesis Because your decision is based on a sample, there is the possibility of making the wrong decision. Larson/Farber 4th ed.
Types of Errors Actual Truth of H0 Decision H0 is true H0 is false Do not reject H0 Correct Decision Type II Error Reject H0 Type I Error A type I error occurs if the null hypothesis is rejected when it is true. A type II error occurs if the null hypothesis is not rejected when it is false. Larson/Farber 4th ed.
Example: Identifying Type I and Type II Errors
The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether the meat inspector’s claim is true. When will a type I or type II error occur? Which is more serious? (Source: United States Department of Agriculture) Larson/Farber 4th ed.
Solution: Identifying Type I and Type II Errors
Let p represent the proportion of chicken that is contaminated. H0: Ha: Hypotheses: p ≤ 0.2 p > 0.2 (Claim) 0.16 0.18 0.20 0.22 0.24 p H0: p ≤ 0.20 H0: p > 0.20 Chicken meets USDA limits. Chicken exceeds USDA limits. Larson/Farber 4th ed.
Solution: Identifying Type I and Type II Errors
Hypotheses: H0: Ha: p ≤ 0.2 p > 0.2 (Claim) A type I error is rejecting H0 when it is true. The actual proportion of contaminated chicken is less than or equal to 0.2, but you decide to reject H0. A type II error is failing to reject H0 when it is false. The actual proportion of contaminated chicken is greater than 0.2, but you do not reject H0. Larson/Farber 4th ed.
Solution: Identifying Type I and Type II Errors
Hypotheses: H0: Ha: p ≤ 0.2 p > 0.2 (Claim) With a type I error, you might create a health scare and hurt the sales of chicken producers who were actually meeting the USDA limits. With a type II error, you could be allowing chicken that exceeded the USDA contamination limit to be sold to consumers. A type II error could result in sickness or even death. Larson/Farber 4th ed.
Level of Significance Level of significance
Your maximum allowable probability of making a type I error. Denoted by , the lowercase Greek letter alpha. By setting the level of significance at a small value, you are saying that you want the probability of rejecting a true null hypothesis to be small. Commonly used levels of significance: = 0.10 = 0.05 = 0.01 P(type II error) = β (beta) Larson/Farber 4th ed.
Standardized test statistic
Statistical Tests After stating the null and alternative hypotheses and specifying the level of significance, a random sample is taken from the population and sample statistics are calculated. The statistic that is compared with the parameter in the null hypothesis is called the test statistic. σ2 χ2 (Section 7.5) s2 z (Section 7.4) p t (Section 7.3 n < 30) z (Section 7.2 n 30) μ Standardized test statistic Test statistic Population parameter Larson/Farber 4th ed.
P-values P-value (or probability value)
The probability, if the null hypothesis is true, of obtaining a sample statistic with a value as extreme or more extreme than the one determined from the sample data. Depends on the nature of the test. Larson/Farber 4th ed.
Nature of the Test Three types of hypothesis tests left-tailed test
right-tailed test two-tailed test The type of test depends on the region of the sampling distribution that favors a rejection of H0. This region is indicated by the alternative hypothesis. Larson/Farber 4th ed.
Left-tailed Test The alternative hypothesis Ha contains the less-than inequality symbol (<). H0: μ k Ha: μ < k z 1 2 3 -3 -2 -1 P is the area to the left of the test statistic. Test statistic Larson/Farber 4th ed.
Right-tailed Test The alternative hypothesis Ha contains the greater-than inequality symbol (>). H0: μ ≤ k Ha: μ > k P is the area to the right of the test statistic. z 1 2 3 -3 -2 -1 Test statistic Larson/Farber 4th ed.
Two-tailed Test The alternative hypothesis Ha contains the not equal inequality symbol (≠). Each tail has an area of ½P. H0: μ = k Ha: μ k P is twice the area to the right of the positive test statistic. P is twice the area to the left of the negative test statistic. z 1 2 3 -3 -2 -1 Test statistic Test statistic Larson/Farber 4th ed.
Example: Identifying The Nature of a Test
For each claim, state H0 and Ha. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the P-value. A university publicizes that the proportion of its students who graduate in 4 years is 82%. Solution: H0: Ha: p = 0.82 z -z ½ P-value area p ≠ 0.82 Two-tailed test Larson/Farber 4th ed.
Example: Identifying The Nature of a Test
For each claim, state H0 and Ha. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the P-value. A water faucet manufacturer announces that the mean flow rate of a certain type of faucet is less than 2.5 gallons per minute. Solution: z -z P-value area H0: Ha: μ ≥ 2.5 gpm μ < 2.5 gpm Left-tailed test Larson/Farber 4th ed.
Example: Identifying The Nature of a Test
For each claim, state H0 and Ha. Then determine whether the hypothesis test is a left-tailed, right-tailed, or two-tailed test. Sketch a normal sampling distribution and shade the area for the P-value. A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is more than 20 ounces. Solution: z P-value area H0: Ha: μ ≤ 20 oz μ > 20 oz Right-tailed test Larson/Farber 4th ed.
Making a Decision Decision Rule Based on P-value
Compare the P-value with . If P , then reject H0. If P > , then fail to reject H0. Claim Decision Claim is H0 Claim is Ha Fail to reject H0 Reject H0 There is enough evidence to reject the claim There is enough evidence to support the claim There is not enough evidence to reject the claim There is not enough evidence to support the claim Larson/Farber 4th ed.
Example: Interpreting a Decision
You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0? H0 (Claim): A university publicizes that the proportion of its students who graduate in 4 years is 82%. Larson/Farber 4th ed.
Solution: Interpreting a Decision
The claim is represented by H0. If you reject H0 you should conclude “there is sufficient evidence to indicate that the university’s claim is false.” If you fail to reject H0, you should conclude “there is insufficient evidence to indicate that the university’s claim (of a four-year graduation rate of 82%) is false.” Larson/Farber 4th ed.
Example: Interpreting a Decision
You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0? Ha (Claim): Consumer Reports states that the mean stopping distance (on a dry surface) for a Honda Civic is less than 136 feet. Solution: The claim is represented by Ha. H0 is “the mean stopping distance…is greater than or equal to 136 feet.” Larson/Farber 4th ed.
Solution: Interpreting a Decision
If you reject H0 you should conclude “there is enough evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is less than 136 feet.” If you fail to reject H0, you should conclude “there is not enough evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is less than 136 feet.” Larson/Farber 4th ed.
Steps for Hypothesis Testing
State the claim mathematically and verbally. Identify the null and alternative hypotheses. H0: ? Ha: ? Specify the level of significance. α = ? Determine the standardized sampling distribution and draw its graph. Calculate the test statistic and its standardized value. Add it to your sketch. This sampling distribution is based on the assumption that H0 is true. z z Test statistic Larson/Farber 4th ed.
Steps for Hypothesis Testing
Find the P-value. Use the following decision rule. Write a statement to interpret the decision in the context of the original claim. Is the P-value less than or equal to the level of significance? No Fail to reject H0. Yes Reject H0. Larson/Farber 4th ed.
Section 7.1 Summary Stated a null hypothesis and an alternative hypothesis Identified type I and type I errors and interpreted the level of significance Determined whether to use a one-tailed or two-tailed statistical test and found a p-value Made and interpreted a decision based on the results of a statistical test Wrote a claim for a hypothesis test Larson/Farber 4th ed.
Hypothesis Testing for the Mean (Large Samples)
Section 7.2 Hypothesis Testing for the Mean (Large Samples) Larson/Farber 4th ed.
Section 7.2 Objectives Find P-values and use them to test a mean μ
Use P-values for a z-test Find critical values and rejection regions in a normal distribution Use rejection regions for a z-test Larson/Farber 4th ed.
Using P-values to Make a Decision
Decision Rule Based on P-value To use a P-value to make a conclusion in a hypothesis test, compare the P-value with . If P , then reject H0. If P > , then fail to reject H0. Larson/Farber 4th ed.
Example: Interpreting a P-value
The P-value for a hypothesis test is P = What is your decision if the level of significance is 0.05? 0.01? Solution: Because < 0.05, you should reject the null hypothesis. Solution: Because > 0.01, you should fail to reject the null hypothesis. Larson/Farber 4th ed.
Finding the P-value After determining the hypothesis test’s standardized test statistic and the test statistic’s corresponding area, do one of the following to find the P-value. For a left-tailed test, P = (Area in left tail). For a right-tailed test, P = (Area in right tail). For a two-tailed test, P = 2(Area in tail of test statistic). Larson/Farber 4th ed.
Example: Finding the P-value
Find the P-value for a left-tailed hypothesis test with a test statistic of z = Decide whether to reject H0 if the level of significance is α = 0.01. Solution: For a left-tailed test, P = (Area in left tail) z -2.23 P = Because > 0.01, you should fail to reject H0 Larson/Farber 4th ed.
Example: Finding the P-value
Find the P-value for a two-tailed hypothesis test with a test statistic of z = Decide whether to reject H0 if the level of significance is α = 0.05. Solution: For a two-tailed test, P = 2(Area in tail of test statistic) z 2.14 1 – = P = 2(0.0162) = 0.9838 Because < 0.05, you should reject H0 Larson/Farber 4th ed.
Z-Test for a Mean μ Can be used when the population is normal and is known, or for any population when the sample size n is at least 30. The test statistic is the sample mean The standardized test statistic is z When n 30, the sample standard deviation s can be substituted for . Larson/Farber 4th ed.
Using P-values for a z-Test for Mean μ
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the standardized test statistic. Find the area that corresponds to z. State H0 and Ha. Identify . Use Table 4 in Appendix B. Larson/Farber 4th ed.
Using P-values for a z-Test for Mean μ
In Words In Symbols Find the P-value. For a left-tailed test, P = (Area in left tail). For a right-tailed test, P = (Area in right tail). For a two-tailed test, P = 2(Area in tail of test statistic). Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. Reject H0 if P-value is less than or equal to . Otherwise, fail to reject H0. Larson/Farber 4th ed.
Example: Hypothesis Testing Using P-values
In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at = 0.01? Use a P-value. Larson/Farber 4th ed.
Solution: Hypothesis Testing Using P-values
Ha: = Test Statistic: μ ≥ 30 min μ < 30 min P-value z -2.57 0.0051 0.01 < 0.01 Reject H0 Decision: At the 1% level of significance, you have sufficient evidence to conclude the mean delivery time is less than 30 minutes. Larson/Farber 4th ed.
Example: Hypothesis Testing Using P-values
You think that the average franchise investment information shown in the graph is incorrect, so you randomly select 30 franchises and determine the necessary investment for each. The sample mean investment is \$135,000 with a standard deviation of \$30,000. Is there enough evidence to support your claim at = 0.05? Use a P-value. Larson/Farber 4th ed.
Solution: Hypothesis Testing Using P-values
Ha: = Test Statistic: μ = \$143,260 μ ≠ \$143,260 P-value P = 2(0.0655) = z -1.51 0.0655 0.05 Decision: > 0.05 Fail to reject H0 At the 5% level of significance, there is not sufficient evidence to conclude the mean franchise investment is different from \$143,260. Larson/Farber 4th ed.
Rejection Regions and Critical Values
Rejection region (or critical region) The range of values for which the null hypothesis is not probable. If a test statistic falls in this region, the null hypothesis is rejected. A critical value z0 separates the rejection region from the nonrejection region. Larson/Farber 4th ed.
Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution Specify the level of significance . Decide whether the test is left-, right-, or two-tailed. Find the critical value(s) z0. If the hypothesis test is left-tailed, find the z-score that corresponds to an area of , right-tailed, find the z-score that corresponds to an area of 1 – , two-tailed, find the z-score that corresponds to ½ and 1 – ½. Sketch the standard normal distribution. Draw a vertical line at each critical value and shade the rejection region(s). Larson/Farber 4th ed.
Example: Finding Critical Values
Find the critical value and rejection region for a two-tailed test with = 0.05. Solution: 1 – α = 0.95 z z0 ½α = 0.025 ½α = 0.025 -z0 = -1.96 z0 = 1.96 The rejection regions are to the left of -z0 = and to the right of z0 = 1.96. Larson/Farber 4th ed.
Decision Rule Based on Rejection Region
To use a rejection region to conduct a hypothesis test, calculate the standardized test statistic, z. If the standardized test statistic is in the rejection region, then reject H0. is not in the rejection region, then fail to reject H0. z z0 Fail to reject H0. Reject H0. Left-Tailed Test z < z0 z z0 Reject Ho. Fail to reject Ho. z > z0 Right-Tailed Test z z0 Two-Tailed Test z0 z < -z0 z > z0 Reject H0 Fail to reject H0 Larson/Farber 4th ed.
Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Sketch the sampling distribution. Determine the critical value(s). Determine the rejection region(s). State H0 and Ha. Identify . Use Table 4 in Appendix B. Larson/Farber 4th ed.
Using Rejection Regions for a z-Test for a Mean μ
In Words In Symbols Find the standardized test statistic. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If z is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed.
Example: Testing with Rejection Regions
Employees in a large accounting firm claim that the mean salary of the firm’s accountants is less than that of its competitor’s, which is \$45,000. A random sample of 30 of the firm’s accountants has a mean salary of \$43,500 with a standard deviation of \$5200. At α = 0.05, test the employees’ claim. Larson/Farber 4th ed.
Solution: Testing with Rejection Regions
Ha: = Rejection Region: μ ≥ \$45,000 μ < \$45,000 Test Statistic 0.05 Decision: Fail to reject H0 z -1.645 0.05 At the 5% level of significance, there is not sufficient evidence to support the employees’ claim that the mean salary is less than \$45,000. -1.645 -1.58 Larson/Farber 4th ed.
Example: Testing with Rejection Regions
The U.S. Department of Agriculture reports that the mean cost of raising a child from birth to age 2 in a rural area is \$10,460. You believe this value is incorrect, so you select a random sample of 900 children (age 2) and find that the mean cost is \$10,345 with a standard deviation of \$1540. At α = 0.05, is there enough evidence to conclude that the mean cost is different from \$10,460? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion) Larson/Farber 4th ed.
Solution: Testing with Rejection Regions
Ha: = Rejection Region: μ = \$10,460 μ ≠ \$10,460 Test Statistic 0.05 Decision: Reject H0 z -1.96 0.025 1.96 At the 5% level of significance, you have enough evidence to conclude the mean cost of raising a child from birth to age 2 in a rural area is significantly different from \$10,460. -1.96 1.96 -2.24 Larson/Farber 4th ed.
Section 7.2 Summary Found P-values and used them to test a mean μ
Used P-values for a z-test Found critical values and rejection regions in a normal distribution Used rejection regions for a z-test Larson/Farber 4th ed.
Hypothesis Testing for the Mean (Small Samples)
Section 7.3 Hypothesis Testing for the Mean (Small Samples) Larson/Farber 4th ed.
Section 7.3 Objectives Find critical values in a t-distribution
Use the t-test to test a mean μ Use technology to find P-values and use them with a t-test to test a mean μ Larson/Farber 4th ed.
Finding Critical Values in a t-Distribution
Identify the level of significance . Identify the degrees of freedom d.f. = n – 1. Find the critical value(s) using Table 5 in Appendix B in the row with n – 1 degrees of freedom. If the hypothesis test is left-tailed, use “One Tail, ” column with a negative sign, right-tailed, use “One Tail, ” column with a positive sign, two-tailed, use “Two Tails, ” column with a negative and a positive sign. Larson/Farber 4th ed.
Example: Finding Critical Values for t
Find the critical value t0 for a left-tailed test given = 0.05 and n = 21. Solution: The degrees of freedom are d.f. = n – 1 = 21 – 1 = 20. Look at α = 0.05 in the “One Tail, ” column. Because the test is left-tailed, the critical value is negative. t -1.725 0.05 Larson/Farber 4th ed.
Example: Finding Critical Values for t
Find the critical values t0 and -t0 for a two-tailed test given = 0.05 and n = 26. Solution: The degrees of freedom are d.f. = n – 1 = 26 – 1 = 25. Look at α = 0.05 in the “Two Tail, ” column. Because the test is two-tailed, one critical value is negative and one is positive. t -2.060 0.025 2.060 Larson/Farber 4th ed.
t-Test for a Mean μ (n < 30, Unknown)
A statistical test for a population mean. The t-test can be used when the population is normal or nearly normal, is unknown, and n < 30. The test statistic is the sample mean The standardized test statistic is t. The degrees of freedom are d.f. = n – 1. Larson/Farber 4th ed.
Using the t-Test for a Mean μ (Small Sample)
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Identify the degrees of freedom and sketch the sampling distribution. Determine any critical value(s). State H0 and Ha. Identify . d.f. = n – 1. Use Table 5 in Appendix B. Larson/Farber 4th ed.
Using the t-Test for a Mean μ (Small Sample)
In Words In Symbols Determine any rejection region(s). Find the standardized test statistic. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If t is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed.
Example: Testing μ with a Small Sample
A used car dealer says that the mean price of a 2005 Honda Pilot LX is at least \$23,900. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of \$23,000 and a standard deviation of \$1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book) Larson/Farber 4th ed.
Solution: Testing μ with a Small Sample
α = df = Rejection Region: μ ≥ \$23,900 μ < \$23,900 Test Statistic: Decision: 0.05 14 – 1 = 13 Reject H0 At the 0.05 level of significance, there is enough evidence to reject the claim that the mean price of a 2005 Honda Pilot LX is at least \$23,900 t -1.771 0.05 -1.771 -3.026 Larson/Farber 4th ed.
Example: Testing μ with a Small Sample
An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at α = 0.05? Assume the population is normally distributed. Larson/Farber 4th ed.
Solution: Testing μ with a Small Sample
α = df = Rejection Region: μ = 6.8 μ ≠ 6.8 Test Statistic: Decision: 0.05 19 – 1 = 18 Fail to reject H0 At the 0.05 level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8. t -2.101 0.025 2.101 -2.101 2.101 -1.816 Larson/Farber 4th ed.
Example: Using P-values with t-Tests
The American Automobile Association claims that the mean daily meal cost for a family of four traveling on vacation in Florida is \$118. A random sample of 11 such families has a mean daily meal cost of \$128 with a standard deviation of \$20. Is there enough evidence to reject the claim at α = 0.10? Assume the population is normally distributed. (Adapted from American Automobile Association) Larson/Farber 4th ed.
Solution: Using P-values with t-Tests
Ha: μ = \$118 μ ≠ \$118 TI-83/84set up: Calculate: Draw: Decision: > 0.10 Fail to reject H0. At the 0.10 level of significance, there is not enough evidence to reject the claim that the mean daily meal cost for a family of four traveling on vacation in Florida is \$118. Larson/Farber 4th ed.
Section 7.3 Summary Found critical values in a t-distribution
Used the t-test to test a mean μ Used technology to find P-values and used them with a t-test to test a mean μ Larson/Farber 4th ed.
Hypothesis Testing for Proportions
Section 7.4 Hypothesis Testing for Proportions Larson/Farber 4th ed.
Section 7.4 Objectives Use the z-test to test a population proportion p Larson/Farber 4th ed.
z-Test for a Population Proportion
A statistical test for a population proportion. Can be used when a binomial distribution is given such that np 5 and nq 5. The test statistic is the sample proportion . The standardized test statistic is z. Larson/Farber 4th ed.
Using a z-Test for a Proportion p
Verify that np ≥ 5 and nq ≥ 5 In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Sketch the sampling distribution. Determine any critical value(s). State H0 and Ha. Identify . Use Table 5 in Appendix B. Larson/Farber 4th ed.
Using a z-Test for a Proportion p
In Words In Symbols Determine any rejection region(s). Find the standardized test statistic. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If z is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed.
Example: Hypothesis Test for Proportions
Zogby International claims that 45% of people in the United States support making cigarettes illegal within the next 5 to 10 years. You decide to test this claim and ask a random sample of 200 people in the United States whether they support making cigarettes illegal within the next 5 to 10 years. Of the 200 people, 49% support this law. At α = 0.05 is there enough evidence to reject the claim? Solution: Verify that np ≥ 5 and nq ≥ 5. np = 200(0.45) = 90 and nq = 200(0.55) = 110 Larson/Farber 4th ed.
Solution: Hypothesis Test for Proportions
Ha: = Rejection Region: p = 0.45 p ≠ 0.45 Test Statistic 0.05 Decision: Fail to reject H0 z -1.96 0.025 1.96 At the 5% level of significance, there is not enough evidence to reject the claim that 45% of people in the U.S. support making cigarettes illegal within the next 5 to 10 years. -1.96 1.96 1.14 Larson/Farber 4th ed.
Example: Hypothesis Test for Proportions
The Pew Research Center claims that more than 55% of U.S. adults regularly watch their local television news. You decide to test this claim and ask a random sample of 425 adults in the United States whether they regularly watch their local television news. Of the 425 adults, 255 respond yes. At α = 0.05 is there enough evidence to support the claim? Solution: Verify that np ≥ 5 and nq ≥ 5. np = 425(0.55) ≈ 234 and nq = 425 (0.45) ≈ 191 Larson/Farber 4th ed.
Solution: Hypothesis Test for Proportions
Ha: = Rejection Region: p ≤ 0.55 p > 0.55 Test Statistic 0.05 Decision: Reject H0 z 1.645 0.05 At the 5% level of significance, there is enough evidence to support the claim that more than 55% of U.S. adults regularly watch their local television news. 1.645 2.07 Larson/Farber 4th ed.
Section 7.4 Summary Used the z-test to test a population proportion p
Larson/Farber 4th ed.
Hypothesis Testing for Variance and Standard Deviation
Section 7.5 Hypothesis Testing for Variance and Standard Deviation Larson/Farber 4th ed.
Section 7.5 Objectives Find critical values for a χ2-test
Use the χ2-test to test a variance or a standard deviation Larson/Farber 4th ed.
Finding Critical Values for the χ2-Test
Specify the level of significance . Determine the degrees of freedom d.f. = n – 1. The critical values for the χ2-distribution are found in Table 6 of Appendix B. To find the critical value(s) for a right-tailed test, use the value that corresponds to d.f. and . left-tailed test, use the value that corresponds to d.f. and 1 – . two-tailed test, use the values that corresponds to d.f. and ½ and d.f. and 1 – ½. Larson/Farber 4th ed.
Finding Critical Values for the χ2-Test
Right-tailed Left-tailed χ2 1 – α χ2 1 – α Two-tailed χ2 1 – α Larson/Farber 4th ed.
Example: Finding Critical Values for χ2
Find the critical χ2-value for a left-tailed test when n = 11 and = 0.01. Solution: Degrees of freedom: n – 1 = 11 – 1 = 10 d.f. The area to the right of the critical value is 1 – = 1 – 0.01 = 0.99. χ2 From Table 6, the critical value is Larson/Farber 4th ed.
Example: Finding Critical Values for χ2
Find the critical χ2-value for a two-tailed test when n = 13 and = 0.01. Solution: Degrees of freedom: n – 1 = 13 – 1 = 12 d.f. The areas to the right of the critical values are χ2 From Table 6, the critical values are and Larson/Farber 4th ed.
The Chi-Square Test χ2-Test for a Variance or Standard Deviation
A statistical test for a population variance or standard deviation. Can be used when the population is normal. The test statistic is s2. The standardized test statistic follows a chi-square distribution with degrees of freedom d.f. = n – 1. Larson/Farber 4th ed.
Using the χ2-Test for a Variance or Standard Deviation
In Words In Symbols State the claim mathematically and verbally. Identify the null and alternative hypotheses. Specify the level of significance. Determine the degrees of freedom and sketch the sampling distribution. Determine any critical value(s). State H0 and Ha. Identify . d.f. = n – 1 Use Table 6 in Appendix B. Larson/Farber 4th ed.
Using the χ2-Test for a Variance or Standard Deviation
In Words In Symbols Determine any rejection region(s). Find the standardized test statistic. Make a decision to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim. If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0. Larson/Farber 4th ed.
Example: Hypothesis Test for the Population Variance
A dairy processing company claims that the variance of the amount of fat in the whole milk processed by the company is no more than You suspect this is wrong and find that a random sample of 41 milk containers has a variance of At α = 0.05, is there enough evidence to reject the company’s claim? Assume the population is normally distributed. Larson/Farber 4th ed.
Solution: Hypothesis Test for the Population Variance
Ha: α = df = Rejection Region: σ2 ≤ 0.25 σ2 > 0.25 Test Statistic: Decision: 0.05 41 – 1 = 40 Fail to Reject H0 χ2 55.758 At the 5% level of significance, there is not enough evidence to reject the company’s claim that the variance of the amount of fat in the whole milk is no more than 0.25. 55.758 43.2 Larson/Farber 4th ed.
Example: Hypothesis Test for the Standard Deviation
A restaurant claims that the standard deviation in the length of serving times is less than 2.9 minutes. A random sample of 23 serving times has a standard deviation of 2.1 minutes. At α = 0.10, is there enough evidence to support the restaurant’s claim? Assume the population is normally distributed. Larson/Farber 4th ed.
Solution: Hypothesis Test for the Standard Deviation
Ha: α = df = Rejection Region: σ ≥ 2.9 min. σ < 2.9 min. Test Statistic: Decision: 0.10 23 – 1 = 22 Reject H0 14.042 χ2 At the 10% level of significance, there is enough evidence to support the claim that the standard deviation for the length of serving times is less than 2.9 minutes. 14.042 11.536 Larson/Farber 4th ed.
Example: Hypothesis Test for the Population Variance
A sporting goods manufacturer claims that the variance of the strength in a certain fishing line is A random sample of 15 fishing line spools has a variance of At α = 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed. Larson/Farber 4th ed.
Solution: Hypothesis Test for the Population Variance
Ha: α = df = Rejection Region: σ2 = 15.9 σ2 ≠ 15.9 Test Statistic: Decision: 0.05 15 – 1 = 14 Fail to Reject H0 5.629 χ2 26.119 At the 5% level of significance, there is not enough evidence to reject the claim that the variance in the strength of the fishing line is 15.9. 5.629 26.119 19.194 Larson/Farber 4th ed.
Section 7.5 Summary Found critical values for a χ2-test
Used the χ2-test to test a variance or a standard deviation Larson/Farber 4th ed. | 9,011 | 34,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.87949 |
https://differencebetween.io/gauge-pressure-and-atmospheric-pressure/ | 1,701,492,512,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00810.warc.gz | 255,520,516 | 18,200 | Difference Between Gauge Pressure and Atmospheric Pressure
• Post category:Science
• Post author:
Definition of Gauge Pressure and Atmospheric Pressure
Atmospheric Pressure: Atmospheric pressure is the pressure exerted by the Earth’s atmosphere on its surface. It is caused by the weight of the air above the surface and is measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa). The average atmospheric pressure at sea level is around 101.3 kilopascals (kPa), which is also known as one standard atmosphere (atm).
Atmospheric pressure is typically measured using a barometer, which consists of a glass tube filled with mercury or other liquid, with one end sealed and the other end open to the atmosphere. The pressure of the atmosphere pushes down on the surface of the mercury, causing it to rise up the tube. The height of the mercury in the tube is proportional to the atmospheric pressure.
The atmospheric pressure decreases with increasing altitude, as there is less air above the surface exerting pressure. This is why the air pressure at the top of a mountain is lower than at sea level. In addition, changes in atmospheric pressure can be caused by weather systems, such as high and low pressure systems, which can affect the weather conditions in a particular region.
Atmospheric pressure has many practical applications, including in meteorology, aviation, and the operation of vacuum cleaners. Understanding atmospheric pressure is also important in the study of the Earth’s atmosphere and climate, as well as in the design and operation of buildings and other structures that are exposed to the elements.
Gauge Pressure: Gauge pressure is the pressure measured relative to the local atmospheric pressure. It is typically measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa). Gauge pressure is used to measure pressure differences within a system, such as the pressure difference between two points in a pipe or the pressure inside a vessel relative to the ambient pressure.
Gauge pressure is measured using instruments called pressure gauges, which can be mechanical, electrical or digital. A common example of a mechanical pressure gauge is a Bourdon tube gauge, which consists of a curved tube that is connected to the system being measured. As the pressure in the system increases, the tube straightens out, which is then translated into a reading on a dial. Electrical and digital pressure gauges use electronic sensors to measure the pressure and display the readings.
One of the most common uses of gauge pressure is in hydraulic and pneumatic systems. In hydraulic systems, the pressure generated by a pump is used to generate force to move objects or perform work. In pneumatic systems, compressed air is used to power machines and tools. In both cases, gauge pressure is used to monitor and control the pressure within the system to ensure that it is operating within safe limits.
Gauge pressure is also used in many industrial and scientific applications, including in the measurement of fluid flow rates, the monitoring of gas and liquid pipelines, and the control of industrial processes. Understanding gauge pressure is important for engineers and technicians who work with these systems, as it allows them to ensure that the systems are operating safely and efficiently.
Differences between Gauge Pressure and Atmospheric Pressure
The key differences between gauge pressure and atmospheric pressure are as follows:
1. Definition: Gauge pressure is the pressure measured relative to the local atmospheric pressure, while atmospheric pressure is the pressure exerted by the Earth’s atmosphere on its surface.
2. Measurement: Gauge pressure is typically measured using pressure gauges, while atmospheric pressure is typically measured using barometers.
3. Units: Gauge pressure is typically measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa), while atmospheric pressure is also measured in units of force per unit area, such as kilopascals (kPa) or millibars (mb).
4. Reference Point: Gauge pressure is measured relative to the local atmospheric pressure, while atmospheric pressure is the reference point for gauge pressure measurements.
5. Application: Gauge pressure is used to measure pressure differences within a system, such as the pressure difference between two points in a pipe or the pressure inside a vessel relative to the ambient pressure. Atmospheric pressure is important in the study of the Earth’s atmosphere and climate, as well as in the design and operation of buildings and other structures that are exposed to the elements.
6. Range: Gauge pressure can be positive or negative, depending on whether the pressure in the system being measured is greater or less than the local atmospheric pressure. Atmospheric pressure is always positive and typically ranges from around 950 mb to 1050 mb.
Understanding the differences between gauge pressure and atmospheric pressure is important in many industrial and scientific applications, as it allows engineers and technicians to ensure that systems are operating within safe and efficient limits.
Applications of Gauge Pressure and Atmospheric Pressure
Applications of Gauge Pressure:
1. Hydraulic systems: Gauge pressure is used to monitor and control the pressure within hydraulic systems, which are used to generate force to move objects or perform work.
2. Pneumatic systems: Gauge pressure is used to monitor and control the pressure within pneumatic systems, which are used to power machines and tools with compressed air.
3. Industrial processes: Gauge pressure is used to monitor and control pressure in various industrial processes, such as chemical reactions, oil refining, and power generation.
4. HVAC systems: Gauge pressure is used to monitor and control the pressure within heating, ventilation, and air conditioning (HVAC) systems, which regulate the temperature and air quality in buildings.
5. Automotive systems: Gauge pressure is used to monitor and control the pressure within automotive systems, such as engine oil pressure, fuel pressure, and tire pressure.
Applications of Atmospheric Pressure:
1. Weather forecasting: Changes in atmospheric pressure can indicate the arrival of weather systems, such as high and low pressure systems, which can affect weather conditions in a particular region.
2. Aviation: Atmospheric pressure is used in the operation of aircraft, as changes in pressure can affect altitude and the performance of aircraft systems.
3. Building design and safety: Understanding atmospheric pressure is important in the design and operation of buildings and other structures that are exposed to the elements.
4. Oceanography: Atmospheric pressure can affect ocean currents and tides, and is therefore important in the study of oceanography.
5. Climate science: Atmospheric pressure is important in the study of the Earth’s atmosphere and climate, as it can affect temperature, humidity, and other environmental factors.
Conclusion
Gauge pressure and atmospheric pressure are both important concepts in the fields of engineering, science, and technology. Gauge pressure is used to measure pressure differences within a system, while atmospheric pressure is the pressure exerted by the Earth’s atmosphere on its surface.
Understanding the differences between these two types of pressure, as well as their applications, is crucial for engineers and technicians who work with hydraulic and pneumatic systems, industrial processes, HVAC systems, automotive systems, weather forecasting, aviation, building design and safety, oceanography, and climate science.
The study and application of gauge pressure and atmospheric pressure are essential in ensuring the safe and efficient operation of various systems and processes in our daily lives.
References Website
Here are some references that you can use to learn more about gauge pressure and atmospheric pressure:
1. “Gauge Pressure vs. Absolute Pressure: What’s the Difference?” – Omega Engineering. Available at: https://www.omega.com/en-us/resources/gauge-pressure-vs-absolute-pressure
2. “Atmospheric Pressure” – National Weather Service. Available at: https://www.weather.gov/jetstream/atmosphere_pressure
3. “What is Gauge Pressure?” – WIKA Instrument Corporation. Available at: https://www.wika.us/knowledge-center/what-is-gauge-pressure-10968_en_us.WIKA
4. “Atmospheric Pressure” – Encyclopedia Britannica. Available at: https://www.britannica.com/science/atmospheric-pressure
5. “What Is Gauge Pressure and Absolute Pressure?” – ThoughtCo. Available at: https://www.thoughtco.com/gauge-pressure-and-absolute-pressure-373353. | 1,655 | 8,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.949207 |
http://gundam.aeug.org/archives/2000/05/1288.html | 1,600,414,025,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00367.warc.gz | 76,711,769 | 2,326 | Vince Leon (vleon@u.arizona.edu)
Thu, 25 May 2000 11:37:27 -0700
Hey -Z-
Maybe you should add some info about a colony drop to your High Frontier web page?
--- --- --- --- --- --- --- --- --- --- --- ---
Vince Leon
vleon@u.arizona.edu
-----Original Message-----
From: D Fink [SMTP:lioconvoy@home.com]
Sent: Thursday, May 25, 2000 11:30 AM
To: gundam@aeug.org
Subject: Re: [gundam] Colony Drop was Re: New Gundam timeline
I know this all just anime physics, but would a colony even survive
re-entry? Colonies strike me as being rather fragile. Over half of their
surface area is windows, which I cant image were ever designed to stand much
heat. Just my 200 yen.
Dave
----- Original Message -----
From: "Lim Jyue" <lim_jyue@pacific.net.sg>
To: <gundam@aeug.org>; <gundam@aeug.org>
Sent: Wednesday, May 24, 2000 12:49 PM
Subject: Re: [gundam] Colony Drop was Re: New Gundam timeline
> At 20:21 05/23/2000 -0400, core@gundam.com wrote:
> >Can't comment on the 500km crater size, but the problem of crater size as
> >a function of the dropped body has been solved. For example, you can
look
> >at a crater on the Moon or Mars or Earth and give a very good estimate of
> >the meteorite. So given the approximate mass of a colony, the crater
size
> >should be known. Is there a planetary astronomer in the house?
>
> Rule of thumb from my astronomy course -- the size of the crater
is
> 10x that of the object's diameter, and the depth 1.5-2x. That means the
> impacting object of a 500km object is about 50km in diameter, and will
cause
> a 75km deep crater -- won't that let the magma out?
>
> Can't tell how the shape of a colony will affect this though..
-
Gundam Mailing List Archives are available at http://gundam.aeug.org/
This archive was generated by hypermail 2.0b3 on Fri May 26 2000 - 03:30:50 JST | 546 | 1,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | latest | en | 0.898996 |
http://greylabyrinth.com/solution/213 | 1,527,406,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00427.warc.gz | 115,761,141 | 3,264 | Logain Potpourri II
VI) Roll Of The Dice
You can determine the die’s opposite sides from the rolls, but not the arrangement of them. Specifically, one set of opposite sides can be the mirror image of the other.
The opposite sides are
2&8, 7&4, 5&11
VII) DA Next Puzzle (D’n’A)
D and A stand for DOWN and ACROSS. The series is a simple number series 1,2,3,4,5,…
The number of Ds and As are the number of down and across lines needed to make numbers on a digital readout such as a clock.
1) 2D
2) 2D3A
3) 2D3A
4) 3D1A
5) 2D3A
6) 3D3A
7) 2D1A
8) 4D3A
9) 3D3A
10) 6D2A
11) ? = 4D
12) ? = 4D3A
13) ? = 4D3A
14) ? = 5D1A
VIII) The Birds
The grandfather bird sits on top.
The sentence is referencing each bird’s status in a “family” tree.
Each word represents a common family name that goes with it, and the grandfather is the oldest family member and hence sits on top.
Farmer’s daughter
Undercover brother
Fortunate son
Mother goose
Grandfather clause
Twisted sister
Uncle Buck
Father Time
IX) String Order
Each word can unscramble into a number after removing one letter:
1-8 TWO -- N
2-3 ONE -- N
3-7 EIGHT -- O
4-1 FOUR -- M
6-2 NINE -- L
8-5 TEN -- D
9-6 SIX -- A
10-4 THREE -- I
Match the left number with the appropriate “spelled” word, and the right number is the order of the omitted letter from it. Once unscrambled you get MANDOLIN…which is no guitar of course.
3.18 stars. Votes are updated daily.
On a scale of 1 to 5, 1 being among your least favorite, 5 being among your most favorite, how would you rate this puzzle?
1 2 3 4 5 | 516 | 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} | 2.765625 | 3 | CC-MAIN-2018-22 | latest | en | 0.875801 |
https://forum.allaboutcircuits.com/threads/help-with-simple-circuit-opamp-project.81276/ | 1,571,608,118,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00107.warc.gz | 494,783,139 | 17,799 | # Help with simple circuit/opamp project
Discussion in 'The Projects Forum' started by bears13, Feb 19, 2013.
1. ### bears13 Thread Starter New Member
Feb 19, 2013
1
0
I am working on a project in which I have to attach a hydrocarbon sensor to a microcontroller in order to read in the voltage and know if the sensor has detected something. I have attached the circuit for the hydrocarbon sensor. As you can see there are two resistors in parallel so the voltage will be constant. We need to figure out what supporting circuitry will be needed in order to read a voltage and amplify it so that our microcontroller can read it. We were thinking of connecting an opamp or two to amplify our signal, but we are not sure how to connect them to the sensor.
We are really confused and need a good, easy-to-understand explanation. Any help would be appreciated.
Thanks!
http://s.eeweb.com/members/laura_webb/discussions/2013/02/19/COBRA-MODEL-100-HC-schematic-1361297798.pdf
2. ### tindel Well-Known Member
Sep 16, 2012
659
224
It looks like the HC sensor resistor will go as high as open circuit and as low as 10kohm. put this in parallel with 330kohm resistor and you get an equivalent resistance between 10k (no hydrocarbons) and 330k (lots of hydrocarbons). You can supply the equivalent resistance with less than a 12V source and sense the current and that will tell you how many hydrocarbons you have.
That's what I get out of that spec sheet anyway. It's not very descriptive - maybe a phone call to the manufacture would help
3. ### wayneh Expert
Sep 9, 2010
16,102
6,220
I don't think you need anything more than a resistor voltage divider. Check the input impedance of your µC. If it's >10X higher than the typical resistance of the sensor, you should be able to measure voltages with little disturbance of the reading by the µC itself. If not, then a voltage follower op-amp circuit may be called for. There are op-amps available that have very high input impedance. You may not need anything exceptional, but that is an important factor in selecting the right amp.
Your µC needs to see a voltage, so you need to convert the changing resistance of the sensor to a changing voltage. Putting a known-value resistor in series with the "unknown" resistor, and measuring the voltage drop across that known resistor, allows you calculate the current by Ohm's law and then the unknown resistance. (You need to know the total voltage drop, also, either by measuring the 12V supply or by measuring the drop across the sensor as well as the known resistor.) You'll then want to apply a formula to relate the measured resistance back to hydrocarbon level. It may be best to establish this by experimentation, rather than theory.
The manufacturer can likely supply you a typical application circuit. | 657 | 2,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.947109 |
https://www.mathworks.com/matlabcentral/answers/491935-how-to-check-whether-any-two-elements-are-equal-or-not-in-a-nx1-array-in-matlab | 1,579,857,174,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250616186.38/warc/CC-MAIN-20200124070934-20200124095934-00259.warc.gz | 970,939,770 | 20,611 | # How to check whether any two elements are equal or not in a Nx1 array in matlab?
1 view (last 30 days)
Dear all,
I have one column matrix with N array, and I want to check whether any two elements are equal or not? And if two or more numbers are equal, how to find their position on the array? How can I do that by coding?
Thanks
Guillaume on 19 Nov 2019
x = [1 2 3 4 2 5 6 3 7 8 9 10]; %demo data. Works with column or row vectors
[loc1, loc2] = find(tril(x == x.', -1)); %location of duplicates
dupvalues = x(loc1); %same as x(loc2) obviously
table(dupvalues(:), loc1, loc2, 'VariableNames', {'DuplicateValue', 'FirstIndex', 'SecondIndex'}) %for pretty display
#### 1 Comment
Swarnajyoti Mukherjee on 20 Nov 2019
It is working perfectly, thanks.
Vladimir Sovkov on 19 Nov 2019
%%
z=randi(5,10,1); % forms a sample array; replace it with an array of your interest
%%
[~,indx]=sort(z);
k0=1;
while k0<=numel(z)
if k0<numel(z)
k1=find(z(indx(k0+1:end))>z(indx(k0)),1); % if you do not need the exact coincidence but a tolerant closeness, change it to "k1=find(z(indx(k0+1:end))-z(indx(k0))>epss,1)" with the epss being the satisfying tolerance
else
k1=[];
end
if isempty(k1)
k1=numel(z);
else
k1=k1+k0-1;
end
disp(strcat('for z=',num2str(z(indx(k0)))));
disp(sort(indx(k0:k1))');
k0=k1+1;
end
#### 1 Comment
Swarnajyoti Mukherjee on 20 Nov 2019
Perfect, Thanks. | 461 | 1,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-05 | latest | en | 0.663604 |
https://millionstories.net/draw-a-picture-strategy-for-math-problem-solving.php | 1,600,595,406,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400197946.27/warc/CC-MAIN-20200920094130-20200920124130-00470.warc.gz | 522,890,693 | 8,810 | Sunday, September 20
draw a picture strategy for math problem solving - millionstories.net
# Interesting №24335: draw a picture strategy for math problem solving
## Contents
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## draw a picture to solve a math problem?
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I am sorry, that has interfered... I understand this question. I invite to discussion. Write here or in PM. | 1,529 | 6,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.9883 |
https://pl.tradingview.com/script/VocSxIq4-excellent-adx/ | 1,610,842,675,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00565.warc.gz | 522,911,142 | 77,936 | 2298 obejrzeń
The Average Directional movement indeX ( ADX ) is an indicator that helps you determine the trend direction, pivot points , and much more else! But it looks not so easy as other famous indicators. It seems strange or even terrible, but don't be afraid. Let's understand how it works and get its power into your analysis tactics.
In the beginning, imagine a drunk man goes through a ladder: step by step. Up, up, down, up, down, down, up...
How can we understand which direction he goes? Exactly! We can count the number of steps in each direction. In the above example, in the upward – 4, in the downward – 3. So, it looks like he goes in an upward direction.
The ADX indicator counts the same steps, but for price. The size of each step equals 1 ATR for "DI Length" candles. On the indicator chart, we have the green and red lines. The green line represents a number of steps upward. The red line shows one downward. When the red line upper green, then the price goes below, then the trend is directed down. Later the green line comes above the red one, and then the trend changes the direction to upward. Wow? After that, you can easy detect the trend direction on the market!
But it is still not the end. On the chart, we also have the fat blue line. This is the ADX line, and it represents the power of the trend. It is calculated from a distance between the green and red curves. The ADX line value grows if the distance is increased. If the movement is really powerful, then a number of steps into a direction much more prominent than one in an opposed direction. Then the blue line grows faster. But if the growth has stopped and the blue line turns back or already had changed self-direction, then it is a signal that the trend has ended too. It's an excellent sign to close the position (but not always). Easy? Not quite. Thresholds help you there. The indicator has two additional parameters: upper and lower thresholds to evaluate the trend-over signal strength. An u-turn of the ADX line above the upper threshold sends a strong signal. If one occurs between both thresholds, it is a bit weak signal. But if the blue line goes below the lower threshold, it looks like there is no trend, and the price goes side. We can also say that the price goes side when the ADX value gradually falls down.
The Excellent ADX indicator helps you catch pivot /pullback signals based on green, red, and blue lines. Each such signal is highlighted as a green (buy) or red (sell) dot on the plot. The size of the dot represents the strength of the signal. You can also check the position of green and red lines from each other to determine the trend direction and the place where it has been changed. The Excellent ADX indicator helps you there too. It highlights the trend direction by the background-color, so you'll never miss it! The Excellent ADX good compliance with the Price Channel indicator built for the same length. You can use them together to be on a trend wave always!
Informacje o Wersji: I've added a additional ADX line (light yellow color). This line much faster than the original blue ADX line. The faster line can be helfpul to be ready to close your position.
Informacje o Wersji: I've fixed names of ADX lines
Informacje o Wersji: Fix the title screen
## Komentarze
Odpowiedz
works this indicator for heikin ashi and cryptocurrency? or do i need to change settings for crypto?
Odpowiedz
The ADX close position signal also has a good compliance with MACD divergence signal. So, now you have more reason to think go out.
Odpowiedz
newmen41
I mean the case, when you see that trend has ended, and then, after some side move, you see an other peak of the blue line and, at the same time, MACD show divergence.
Odpowiedz
Very good job !
Odpowiedz
You need to hold in mind that if the trend is really downwards, then the ADX line must grow. It grows maybe not immediately after the trend direction has changed, but it begins to grow if the trend power enough.
Odpowiedz
newmen41
@newmen41, Hey can you explain more about the DOTS on the indicator. Got the detail explanation about DMI+ and DMI- as well as ADX line
But what those Red and Green Dots denote on respective lines.
1. Does the RED Dot on ADX means a downward correction in uptrend so a sell signal?
2. Does the GREEN DOT on ADX means a upward correction in downtrend so a buy signal?
3. And similarly the Green/Red Dot on DMI+ line and Green/Red DOT on DMI- line?
Thanks. The indicator looks great just a little more simple explanation for the addons will be really helpful and amazing :)
Odpowiedz
In general, each dot has the same meaning as another one. If the dot has a red color, then you need to sell. If the green then buy. But I know that there are many different buy/sell cases, like open/increase/close/close-half position, and I see that you want more details. :)
1. The red dot on the ADX line means that the upward trend has been over. In this case, better to close your short position. If the signal dot is not so large (when the ADX line has turned under the high threshold), it may be better to close just 1/2 of the position. But anyway, be ready to close it totally if you feel that the trend is really finished. If the price movement has a correction character, it mustn't affect the ADX line significantly. But you know that it depends on the correction force. :) Must you open a long position after you see the red dot on the ADX line? I feel no. Better to open one after the red line appears above the green line, and the background color on the chart changed to red.
2. And the same about the green dot on the ADX line. The green dot on the ADX line means that the downward trend has finished. In this case, better to close your long position or it half. Better don't open a short position just after you've seen the green dot on the ADX line. Wait and get a confirmation that the trend direction has changed. For example, you can wait for a case when the green line goes upper the red line, and the background color on the chart changed to green.
3. Small and little dots on DMI+ DMI- lines highlight pullback signals. You can skip them if you want. But there are times when you want to jump into the last car of a departing train... If your case like this, watch pullback signals at the beginning of a new trend movement and open or increase your positions corresponding with the dot color.
Odpowiedz
newmen41
But I feel you have mixed up between Short & Longs :)
1. The red dot on the ADX line means that the upward trend has been over. In this case, better to close your short position.
-- Here if the upward trend is over I guess the long entry should be closed.
2. Open a long position after you see the red dot on the ADX line? I feel no. Better to open one after the red line appears above the green line, and the background color on the chart changed to red.
-- I guess once the Red Dot on ADX line appears we need to be ready to open short position and for extra confirmation we can check the DMI red line to appear above green line and back ground goes RED
Similar confusion I can see in your next two points.
I feel it's because of some translation issue from your native language to English. But overall I got what you want to convey and really impressed with this indicator.
Thank you so much for sharing it with the TradingView Community :)
Odpowiedz | 1,698 | 7,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.94393 |
http://community.wolfram.com/groups/-/m/t/1224634 | 1,534,503,405,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212040.27/warc/CC-MAIN-20180817104238-20180817124238-00480.warc.gz | 78,755,443 | 21,362 | # [✓] Implement an external function to a discretionary notebook?
Posted 9 months ago
893 Views
|
3 Replies
|
3 Total Likes
|
Hey guys, I've got a function which computes the propagation of uncertainty for a given function. It is called "fehler" and an example would be: fehler[Sin[2*a], a] which gives me 2 Sqrt[\[CapitalDelta]a^2 Cos[2 a]^2] So basically the syntax is fehler[function, variable1, variable2,...]. The whole file is attached to the post, the credit goes to some random guy from my university, unluckily I could not figure out who made this function.Now I'd like to make this function usable inside every notebook, because right now I always use it externally which is not really nice. So for example if i have a random notebook it should work like this: (*random notebook*) fehler= (*this is the code i need*) a = 1; b=2; const=9; fehler[a*const+b*2, a, b] (*the output should like this*) Sqrt[const^2 \[CapitalDelta]a^2 + 4 \[CapitalDelta]b^2] So that in every notebook I use i can simply use the "fehler"-function to calculate to propagation of uncertainty without having to use it globally. I tried to make this possible for a long time, but the main problem is that the existing fehler-function requires to clear all Variables with ClearAll["Global*"]; before executing fehler. It is important that the output can be further used as a function, and not is simply a (numerical) result.Do you guys have an idea how I can solve this problem? It would help me (and other students) a lot, if we could the function inside our notebooks and not externally.Thanks you and best regards, Tobias Attachments:
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Posted 9 months ago
Tobias,you can do a NotebookEvaluate["Fehlerrechnung-gauss.nb"] if you delete the ClearAll and I suggest removing all the other extraneous evaluatable cells. This is simplest.The "better" way to do this is to make a package out of it --Delete the ClearAll at the beginning and create a package:You can load the package with Needs["yourpackageName"] (note the prime at the end of the name)Regards,Neil
Tobias,Also, you can simplify your function quite a bit and make it not use any global variables with: fehler2[exp_, vars_List] := Sqrt[Plus @@ Map[(D[exp, #]*ToExpression["\[CapitalDelta]" <> ToString[#]])^2 &, vars]] which is used like this fehler2[a*const + b*2, {a, b}] Note that the variables are given as a list and not inlined in the function. You can make them inlined as you originally had but I think that is less conforming to Mathematica standards for functions: fehler2[exp_, vars__] := Sqrt[Plus @@ Map[(D[exp, #]* ToExpression["\[CapitalDelta]" <> ToString[#]])^2 &, {vars}]] which is called your old way: fehler2[a*const + b*2, a, b] You also have a big mistake in the way you use the function -- you must have the a,b,etc NOT defined as numbers when you call the function (in any of the versions). My suggestion is to do the following: error = fehler[a*const + b*2, a, b] error /. {a -> 1, b -> 2, const -> 9} The same applies for fehler2 in either format. This approach will not define the a,b,const so you can change things without clearing anything in the kernel. Hope this helps.Regards,Neil | 831 | 3,175 | {"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} | 2.984375 | 3 | CC-MAIN-2018-34 | latest | en | 0.860286 |
https://math.stackexchange.com/questions/2436333/problem-on-conditional-expected-value-with-to-a-random-variable | 1,558,481,383,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256586.62/warc/CC-MAIN-20190521222812-20190522004812-00171.warc.gz | 544,604,186 | 33,257 | # Problem on conditional expected value with to a random variable
I have a problem with the following exercise.
Let $(\Omega, \mathcal{H}, \mathbb{P})$ be the probability space defined by $\Omega = (0,1)$, $\mathcal{H} = \mathcal{B}(0,1)$ and $\mathbb{P}$ is the Lebesgue measure. Consider the following random variables: $X(\omega)=\omega^{\,2}$, $Y(\omega)=\omega(1-\omega)$ for each $\omega \in \Omega$. I have to calculate $\mathbb{E}[X|Y]$ but I do not know how to proceed.
Every suggestion or help is appreciated.
• just a hunch, not 100% sure -- if you know $y = w(1-w)$ you can find $w=w(y)$ and from it compute $w^2$? – gt6989b Sep 19 '17 at 17:52
For every $B\in\mathcal B$ we have the equality:$$\int^{0.5}_{-0.5}z1_B\left(0.25-z^2\right)dz=0\tag1$$
Substituting $\omega=0.5-z$ leads to:$$\int^1_0\left(\omega-0.5\right)1_B\left(\omega\left(1-\omega\right)\right)d\omega=0\tag2$$Of course we have: $$\omega-0.5=\omega^2-\left[0.5-\omega(1-\omega)\right]=X(\omega)-\left[0.5-Y(\omega)\right]\tag3$$ so $(2)$ is equivalent with:$$\int^1_0X\left(\omega\right)1_B\left(Y\left(\omega\right)\right)d\omega=\int^1_0\left[0.5-Y(\omega)\right]1_B(Y(\omega))d\omega\tag4$$
We can write this also as:$$\int_{\{Y\in B\}}X(\omega)d\omega=\int_{\{Y\in B\}}0.5-Y(\omega)d\omega\tag5$$
This being true for every $B\in\mathcal H$ states exactly that: $$\mathbb E(X\mid Y)=0.5-Y$$
• Which of the steps $1\to2$,$2\to3$,$3\to4$, $4\to5$ (or something in between) needs further explaining? – drhab Sep 20 '17 at 8:18
• I reread it and now I think I understand why you replace $X(\omega)$ with $0.5-Y(\omega)$ in (4). I guess (1) comes from your idea/observation. How did you write it? – LJG Sep 20 '17 at 22:05
• I started with finding an expression for $X$ on base of $Y=y$. Actually there are two expressions and taking their average gives $X=0.5-y$. That was a hint. It is hard to describe how I arrived at $(1)$ and I can imagine that in my answer it seems as if it is some gift from heaven. For a big deal it is a matter of intuiton and mathematical maturity. I cannot hand over a recipe. Sorry for that. – drhab Sep 21 '17 at 7:48 | 760 | 2,131 | {"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} | 3.984375 | 4 | CC-MAIN-2019-22 | latest | en | 0.800426 |
https://tensorflow.google.cn/versions/r2.2/api_docs/python/tf/keras/metrics/BinaryAccuracy?hl=pl | 1,653,065,131,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00413.warc.gz | 632,568,692 | 58,146 | Google I/O is a wrap! Catch up on TensorFlow sessions
# tf.keras.metrics.BinaryAccuracy
Calculates how often predictions matches binary labels.
This metric creates two local variables, `total` and `count` that are used to compute the frequency with which `y_pred` matches `y_true`. This frequency is ultimately returned as `binary accuracy`: an idempotent operation that simply divides `total` by `count`.
If `sample_weight` is `None`, weights default to 1. Use `sample_weight` of 0 to mask values.
#### Usage:
````m = tf.keras.metrics.BinaryAccuracy()`
`_ = m.update_state([[1], [1], [0], [0]], [[0.98], [1], [0], [0.6]])`
`m.result().numpy()`
`0.75`
```
````m.reset_states()`
`_ = m.update_state([[1], [1], [0], [0]], [[0.98], [1], [0], [0.6]],`
` sample_weight=[1, 0, 0, 1])`
`m.result().numpy()`
`0.5`
```
Usage with tf.keras API:
``````model = tf.keras.Model(inputs, outputs)
model.compile('sgd', loss='mse', metrics=[tf.keras.metrics.BinaryAccuracy()])
``````
`name` (Optional) string name of the metric instance.
`dtype` (Optional) data type of the metric result.
`threshold` (Optional) Float representing the threshold for deciding whether prediction values are 1 or 0.
## Methods
### `reset_states`
View source
Resets all of the metric state variables.
This function is called between epochs/steps, when a metric is evaluated during training.
### `result`
View source
Computes and returns the metric value tensor.
Result computation is an idempotent operation that simply calculates the metric value using the state variables.
### `update_state`
View source
Accumulates metric statistics.
`y_true` and `y_pred` should have the same shape.
Args
`y_true` Ground truth values. shape = `[batch_size, d0, .. dN]`.
`y_pred` The predicted values. shape = `[batch_size, d0, .. dN]`.
`sample_weight` Optional `sample_weight` acts as a coefficient for the metric. If a scalar is provided, then the metric is simply scaled by the given value. If `sample_weight` is a tensor of size `[batch_size]`, then the metric for each sample of the batch is rescaled by the corresponding element in the `sample_weight` vector. If the shape of `sample_weight` is `[batch_size, d0, .. dN-1]` (or can be broadcasted to this shape), then each metric element of `y_pred` is scaled by the corresponding value of `sample_weight`. (Note on `dN-1`: all metric functions reduce by 1 dimension, usually the last axis (-1)).
Returns
Update op.
[]
[] | 644 | 2,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.680793 |
https://www.w3resource.com/c-programming-exercises/tree/c-tree-exercises-10.php | 1,718,932,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00260.warc.gz | 927,292,892 | 29,593 | Implementing AVL Tree in C: Insertion and Deletion
# Implementing AVL Tree in C: Insertion and Deletion Operations
## C Program to implement Tree Structure: Exercise-10 with Solution
Write a C program that implements an AVL tree in C. Include functions for insertion and deletion while maintaining the AVL tree's balance property.
In computer science, an AVL tree (named after inventors Adelson-Velsky and Landis) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations.
Sample Solution:
C Code:
``````// Including necessary header files
#include <stdio.h>
#include <stdlib.h>
// Structure for a tree node
struct TreeNode {
int data;
struct TreeNode* left;
struct TreeNode* right;
int height; // Height of the node
};
// Function to get the height of a node
int height(struct TreeNode* node) {
if (node == NULL)
return 0;
return node->height;
}
// Function to get the maximum of two integers
int max(int a, int b) {
return (a > b) ? a : b;
}
// Function to create a new node with a given key
struct TreeNode* createNode(int key) {
struct TreeNode* newNode = (struct TreeNode*)malloc(sizeof(struct TreeNode));
if (newNode != NULL) {
newNode->data = key;
newNode->left = NULL;
newNode->right = NULL;
newNode->height = 1; // New node is initially at height 1
}
return newNode;
}
// Function to right rotate subtree rooted with y
struct TreeNode* rightRotate(struct TreeNode* y) {
struct TreeNode* x = y->left;
struct TreeNode* T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right)) + 1;
x->height = max(height(x->left), height(x->right)) + 1;
// Return new root
return x;
}
// Function to left rotate subtree rooted with x
struct TreeNode* leftRotate(struct TreeNode* x) {
struct TreeNode* y = x->right;
struct TreeNode* T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right)) + 1;
y->height = max(height(y->left), height(y->right)) + 1;
// Return new root
return y;
}
// Function to get the balance factor of a node
int getBalance(struct TreeNode* node) {
if (node == NULL)
return 0;
return height(node->left) - height(node->right);
}
// Function to insert a key into the AVL tree
struct TreeNode* insert(struct TreeNode* root, int key) {
// Perform standard BST insert
if (root == NULL)
return createNode(key);
if (key < root->data)
root->left = insert(root->left, key);
else if (key > root->data)
root->right = insert(root->right, key);
else // Duplicate keys not allowed
return root;
// Update height of the current node
root->height = 1 + max(height(root->left), height(root->right));
// Get the balance factor to check whether this node became unbalanced
int balance = getBalance(root);
// Left Left Case
if (balance > 1 && key < root->left->data)
return rightRotate(root);
// Right Right Case
if (balance < -1 && key > root->right->data)
return leftRotate(root);
// Left Right Case
if (balance > 1 && key > root->left->data) {
root->left = leftRotate(root->left);
return rightRotate(root);
}
// Right Left Case
if (balance < -1 && key < root->right->data) {
root->right = rightRotate(root->right);
return leftRotate(root);
}
// Return the unchanged node pointer
return root;
}
// Function to find the node with the minimum value
struct TreeNode* minValueNode(struct TreeNode* node) {
struct TreeNode* current = node;
while (current->left != NULL)
current = current->left;
return current;
}
// Function to delete a key from the AVL tree
struct TreeNode* deleteNode(struct TreeNode* root, int key) {
if (root == NULL)
return root;
// Perform standard BST delete
if (key < root->data)
root->left = deleteNode(root->left, key);
else if (key > root->data)
root->right = deleteNode(root->right, key);
else {
// Node with only one child or no child
if ((root->left == NULL) || (root->right == NULL)) {
struct TreeNode* temp = root->left ? root->left : root->right;
// No child case
if (temp == NULL) {
temp = root;
root = NULL;
} else // One child case
*root = *temp; // Copy the contents of the non-empty child
free(temp);
} else {
// Node with two children, get the inorder successor
struct TreeNode* temp = minValueNode(root->right);
// Copy the inorder successor's data to this node
root->data = temp->data;
// Delete the inorder successor
root->right = deleteNode(root->right, temp->data);
}
}
// If the tree had only one node, then return
if (root == NULL)
return root;
// Update height of the current node
root->height = 1 + max(height(root->left), height(root->right));
// Get the balance factor to check whether this node became unbalanced
int balance = getBalance(root);
// Left Left Case
if (balance > 1 && getBalance(root->left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root->left) < 0) {
root->left = leftRotate(root->left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root->right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root->right) > 0) {
root->right = rightRotate(root->right);
return leftRotate(root);
}
return root;
}
// Function to perform in-order traversal of the AVL tree
void inOrderTraversal(struct TreeNode* root) {
if (root != NULL) {
inOrderTraversal(root->left);
printf("%d ", root->data);
inOrderTraversal(root->right);
}
}
// Function to free the memory allocated for the AVL tree
void freeAVLTree(struct TreeNode* root) {
if (root != NULL) {
freeAVLTree(root->left);
freeAVLTree(root->right);
free(root);
}
}
int main() {
struct TreeNode* root = NULL;
int choice, key;
do {
printf("\nAVL Tree Operations:\n");
printf("1. Insert a node\n");
printf("2. Delete a node\n");
printf("3. In-order Traversal\n");
printf("4. Exit\n");
scanf("%d", &choice);
switch (choice) {
case 1:
printf("Enter the key to insert: ");
scanf("%d", &key);
root = insert(root, key);
break;
case 2:
printf("Enter the key to delete: ");
scanf("%d", &key);
root = deleteNode(root, key);
break;
case 3:
printf("In-order Traversal: ");
inOrderTraversal(root);
printf("\n");
break;
case 4:
// Free allocated memory
freeAVLTree(root);
printf("Exiting...\n");
break;
default:
printf("Invalid choice! Please enter a valid option.\n");
}
} while (choice != 4);
return 0;
}
```
```
Output:
```AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
Enter the key to insert: 87
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
In-order Traversal: 87
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
Enter the key to insert: 45
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
Enter the key to insert: 34
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
In-order Traversal: 34 45 87
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
Enter the key to delete: 87
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
In-order Traversal: 34 45
AVL Tree Operations:
1. Insert a node
2. Delete a node
3. In-order Traversal
4. Exit
Exiting...
```
Explanation:
In the exercise above,
Here's a brief explanation:
• Node Structure (struct TreeNode):
• Represents a node in the AVL tree.
• Contains data, left child, right child, and the height of the node.
• Utility Functions:
• height: Calculates the height of a node.
• max: Returns the maximum of two integers.
• createNode: Creates a new node with a given key.
• rightRotate and leftRotate: Perform right and left rotations, respectively, during balancing.
• Insertion (insert function):
• Inserts a new key into the AVL tree while maintaining the balance property.
• Utilizes rotations to ensure the tree remains balanced.
• Deletion (deleteNode function):
• Deletes a node with a given key while maintaining the balance property.
• Performs rotations to rebalance the tree as needed.
• In-order Traversal (inOrderTraversal function):
• Prints the elements of the AVL tree in sorted order (in-order traversal).
• Main Function:
• Interactively allows the user to perform AVL tree operations:
• Insert a node
• Delete a node
• In-order traversal
• Exit
• Freeing Memory (freeAVLTree function):
• Frees the memory allocated for the entire AVL tree.
Flowchart:
C Programming Code Editor:
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
| 2,326 | 8,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.761493 |
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## Recommended Posts
Here goes with a quick go at the mystical numbers on your telescopes and how to decode them.
I'll start off with some definitions in case some of these are non-obvious, the whys and wherefores of the f/ number will have to wait until I can condense the f/ concept better as I've just tried and messed up.
Aperture
This is the diameter of the hole that the light enters the telescope through. A 100mm refractor will have a front lens of roughly 100mm. A reflector will have a tube slightly larger than the aperture as the actual aperture is the primary mirror (the tube is merely a way of holding all the bits together).
Focal length
This is the distance from the primary (front lens in a refractor or the big mirror at the bottom of the tube in a Newtonian) to the focal point. The focal point is where you put the eyepiece or camera.
f/ ratio
This is the relationship between the aperture and the focal length.
So what does all this malarky mean?
The significance of these three numbers is, in simple terms:-
Aperture
More aperture means that you are collecting more light and more light makes the image appear brighter. Bright things are easier to see and to photograph. More is better.
Focal length
This has a direct bearing on the magnification. No other number is involved as far as the telescope is concerned. The magnification is calculated by dividing this number by the focal length of the eyepiece. Using the above example and a 10mm eyepice gives a magnification of 1000/10 = 100. If you use a 20mm eyepiece the magnification goes down to 1000/20 = 50.
For imagers that doesn't help much so there's another way of looking at it. If you could look through a hole the size of your camera's sensor you would be able, by moving your eye from side to side, to see a section of the sky. Bigger sensor, more sky. To work it out more exactly you need to imagine that all the light comes through a tiny section in the centre of the lens or primary mirror so you can use the focal length of the mirror and the sensor size to work out how much sky you can "see" using that sensor. If you sketch an isosceles triangle with a base dmension equal to your sensor width and a height equal to the focal length of the 'scope, the angle at the pointy end is the angle which also covers the piece of sky that you can image widthwise. Do the same for the height and you get the angle corresponding to the amount of sky you can image in the other direction.
To be continued...
Captain Chaos
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1 The net force on a moving object is suddenly reduced to zero. As a consequence, the object (A) stops abruptly (B) stops during a short time interval (C) changes direction (D) continues at a constant velocity (E) changes velocity in an unknown manner 5 m m A 5 m uniform plank of mass 00 kg rests on the top of a building with m extended over the edge as shown above. How far can a 50-kg person venture past the edge of the building on the plank before the plank just begins to tip? (A) 0.5 m (B).0 m (C).5 m (D).0 m (E) It is impossible to make the plank tip since the person would have to be more than meters from the edge of the building. An astronomer estimates the mass of galaxy to be 0 solar masses. Galaxy has a mass one-third that of galaxy. Which of the following is the best order of magnitude estimate for the mass of the galaxy? (A) 0 3 solar masses (B) 0 5 solar masses (C) 0 8 solar masses (D) 0 solar masses (E) 0 4 solar masses
2 5 N N kg A block of mass kg slides along a horizontal tabletop. A horizontal applied force of N and a vertical applied force of 5 N act on the block, as shown above. If the coefficient of kinetic friction between the block and the table is 0., the frictional force exerted on the block is most nearly (A) N (B) 3 N (C) 4 N (D) 5 N (E) 7 N X Y Consider the figure above which shows a ball that is projected and follows a parabolic path. Point Y is the highest point on the path, and air resistance is negligible. Which of the following best shows the direction of the acceleration of the ball at point X? (A) (D) (B) (E) (C)
3 x A B C D E The motion of a particle along a straight line is represented by the position, x, versus time, t, graph above. At which of the labeled points on the graph is the magnitude of the acceleration of the particle greatest? (A) A (B) B (C) C (D) D (E) E t v = C x In the above equation, the distance x is measured in meters, the time t is in seconds, and the velocity v is in meters per second. What are the SI units of the constant C? (A) m (B) m s (C) m/s (D) m (E) m A massless rigid rod of length 3d is pivoted at a fixed point W, and two forces each of magnitude F are applied vertically upward as shown above. A third vertical force of magnitude F may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude F is applied at point (A) W only (B) Y only (C) V or X only (D) V or Y only (E) V, W, or X
4 An airplane heads due east at 600 km/hr. The wind blows from the north at 50 km/hr. The cardinal directions (N, S, E, W) are represented as indicated below. With respect to the earth, the plane s velocity is best represented by which vector? Three forces act on an object. If the object is in translational equilibrium, which of the following must be true? I. The vector sum of the three forces must equal zero. II. The magnitudes of the three forces must be equal III. All three forces must be parallel (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III velocity (m/s) time (s) An object of mass kg moves according to the velocity-time graph pictured above. The net force on the object from 0 to 4 seconds is (A).7 N (B).0 N (C) 0 (D) +.0 N (E) +.7 N
5 5 m 3 m 37 o 4 m A kg block sits at rest on the incline shown above. The magnitude of the normal force exerted on the block by the plane is most nearly (A) 0 N (B) N (C) 6 N (D) 0 N (E) 33 N Questions - relate to five particles that start at position x = 0 at time t = 0 and move in one dimension independently of one another. Graphs of the velocity of each particle versus time are shown below. v (m/s) v (m/s) 0 time (s) 0 time (s) - Particle A - Particle B v (m/s) v (m/s) 0 time (s) 0 time (s) - Particle C - Particle D v (m/s) 0 time (s) - Particle E Which particle is farthest from the origin at t = seconds? (A) A (B) B (C) C (D) D (E) E Which particle moves with a constant nonzero acceleration? (A) A (B) B (C) C (D) D (E) E
6 The figure above shows a cart of mass M accelerating to the right with a block of mass m held to the front surface only by friction. The coefficient of friction between the surfaces is µ. What is the minimum acceleration a of the cart such that the block will not fall? A stone of mass kg and weight 0 N falls through the air. The air resistance acting on the stone is N. What is the acceleration of the stone? (A) m/s (B) 9.8 m/s (C) 8 m/s (D) 5 m/s (E) m/s A massless rod is supported at point P as shown above. A block weighing 40 N is attached to the rod 0. m from P. How far from point P must a block weighing 80 N be attached in order to balance the rod? (A) 0. m (B) 0. m (C) 0.4 m (D) 0.5 m (E) 0.8 m
7 A horizontal force F is used to pull a 5-kg block across a floor at a constant speed of 3 m/s. The frictional force between the block and the floor is 0 N. What is the value of the coefficient of kinetic friction, µ k? (A) 0.0 (B) 0.5 (C) 0.0 (D) 0.5 (E) 0.50 The one of the following that is not a vector is (A) acceleration (B) force (C) speed (D) velocity (E) momentum If an object experiences no net force, which of the following must be true about its motion? I. The object has an average velocity of zero. II. The object's velocity does not change. III. The object cannot move. (A) I only (B) II only (C) III only (D) I and II only (E) I and III only A car rounds a curve at a steady 40 m/s. Which of the following is happening to the car s velocity vector? (A) It is increasing in magnitude. (B) It is decreasing in magnitude. (C) It does not change. (D) Its magnitude remains constant but its direction changes. (E) Its magnitude is zero. Position X Y T t Two trains, X and Y, begin moving on parallel tracks from the same starting position at a train station. The graph above shows their respective distances from the starting point as functions of time. At time T, which of the following is true? (A) Train X is moving faster than train Y. (B) Train Y is moving faster than train X. (C) Trains X and Y are moving at the same speed. (D) Train X is ahead of train Y. (E) Train Y is ahead of train X.
8 Consider a system consisting only of the earth and a bowling ball, which moves upward in a parabola 0 m above earth s surface. The downward force of earth s gravity on the ball, and the upward force of the ball s gravity on the earth, form a Newton s third law force pair. Which of the following statements about the ball is correct? (A) The ball must be in equilibrium since the upward forces must cancel downward forces. (B) The ball accelerates toward the earth because the force of gravity on the ball is greater than the force of the ball on the earth. (C) The ball accelerates toward the earth because the force of gravity on the ball is the only force acting on the ball. (D) The ball accelerates away from earth because the force causing the ball to move upward is greater than the force of gravity on the ball. (E) The ball accelerates away from earth because the force causing the ball to move upward plus the force of the ball on the earth are together greater than the force of gravity on the ball. A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling m. The experiment is repeated on the same track, but now the cart is initially moving at m/s. How far does the cart travel before coming to rest? (A) m (B) m (C) 3 m (D) 4 m (E) 8 m A wheel of radius R and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses m, M, and M, respectively, are mounted on the rim of the wheel, as shown above. If the system is in static equilibrium, what is the value of m in terms of M? (A) (B) (C) (D) (E) M M 3M M 5m
9 A rubber ball bounces on the ground as shown. After each bounce, the ball reaches one-half the height of the bounce before it. If the time the ball was in the air between the first and second bounce was second, what would be the time between the second and third bounce? (A) 0.50 s (B) 0.7 s (C).0 s (D).4 s (E).0 s A rock of mass m is thrown horizontally off a building from a height h, as shown above. The speed of the rock as it leaves the thrower's hand at the edge of the building is υ 0. How much time does it take the rock to travel from the edge of the building to the ground? (A) hυ 0 (B) h υ 0 (C) hυ 0 g (D) (E) h g h g 4. A 0 kg boulder rests on the surface of the Earth, which has a mass of 5.98 x 0 4 kg. What is the magnitude of the gravitational force that the boulder exerts on Earth? (A) 5.98 x 0 5 N (B) 5.98 x 0 4 N (C) 00 N (D) 0 N (E) The boulder exerts no force on the earth.
10 Which of the above graphs could correctly represent the motion of an object moving with a constant speed in a straight line? (F) graph I only (G) graph II only (H) graphs I and IV only (I) graphs II and V only (J) All of the above graphs represent constant speed in a straight line. Three stones of different mass (m, m, and 3m) are thrown vertically upward with different velocities (v, v, and 3v, respectively). The diagram indicates the mass and velocity of each stone. Rank from high to low the maximum height of each stone. Assume air resistance is negligible. (K) [high] I, II, III [low] (L) [high] II, I, III [low] (M) [high] III, II, I [low] (N) [high] I, III, II [low] (O) All reach the same height. Is it possible for an object s velocity to increase while its acceleration decreases? (P) No, this is impossible because of the way in which acceleration is defined. (Q) No, because if acceleration is decreasing the object will be slowing down. (R) No, because velocity and acceleration must always be in the same direction. (S) Yes, an example would be an object falling near the surface of the moon. (T) Yes, an example would be an object falling in the presence of air resistance.
11 Above is velocity-time graph representing a duck flying south for the winter. At what point did the duck stop its forward motion? (A) A (B) B (C) C (D) D (E) Between A and B A block rests on an incline that makes an angle θ with the horizontal. The block remains at rest as θ is slowly increased. What happens to the magnitudes of the normal force and the static friction force of the incline on the block? a. Both increase b. Both decrease c. Both remain the same d. The normal force increases while the friction force decreases. e. The friction force increases while the normal force decreases. Two boxes are accelerated to the right on a frictionless horizontal surface as shown above. The larger box has a mass of 9 kg, and the smaller box has a mass of 3 kg. If a 4 N horizontal force pulls on the larger box, with what force does the larger box pull on the smaller box? (A) 3 N (B) 6 N (C) 8 N (D) 8 N (E) 4 N
12 A toy car moves along the x-axis according to the velocity-time graph shown above. When does the car have zero acceleration? f. At.0 and 4.0 seconds g. At 3.0 seconds h. At 3.3 and 5. seconds i. The acceleration is always zero. j. The acceleration is never zero. A ball of mass m is thrown horizontally at a vertical wall with a speed v and bounces off elastically and horizontally. What is the magnitude of the change in momentum of the ball? (A) mv (B) mv (C) zero mv (D) 4 mv (E) A solid metal ball and a hollow plastic ball of the same external radius are released from rest in a large vacuum chamber. When each has fallen m, they both have the same (A) inertia (B) speed (C) momentum (D) kinetic energy (E) change in potential energy If a ball is thrown directly upwards with twice the initial speed of another, how much higher will it be at its apex? (A) 8 times (B) 4 times (C) times (D) times (E) times
13 . Which one of the following quantities can have its unit expressed as kg m s (A) force (B) power (C) kinetic energy (D) pressure (E) linear momentum? 5. The free fall trajectory of an object thrown from the top of a building is shown as the dashed line in the figure above. Which sets of arrows best correspond to the directions of the velocity and of the acceleration for the object at the point labeled P on the diagram?. Consider the motion of an object described by the position-time graph above. For what time(s) is the speed of the object the greatest? (A) At all times from t = 0 s to t =.0 s (B) At time t = 3.0 s (C) At time 4.0 s (D) At all times from t = 5.0 s to 7.0 s (E) At time t = 8.5 s 3. A car with mass M initially travels to the east with speed 4v. A truck initially travels to the west with speed 3v. After the vehicles collide, they move together to the west with common speed v. What is the mass of the truck? (A) M (B) 3M (C) 4M (D) 5M (E) 6M 4. A 5.0 kg solid sphere falls at more than its terminal velocity near the surface of the earth. The mass of the earth is 6.0 x 0 4 kg. What is the magnitude of the gravitational force exerted by the solid sphere on the earth? 6. In the velocity-time graph above, the positive direction is east. Which of the following is correct? (A) The object moves to the west the entire time. (B) The object moves to the east the entire time. (C) The object moves to the west, and then to the east. (D) The object moves to the east, and then to the west. (E) The object changes its direction of motion more than once. (A) 0 N (B) 5 N (C) 50 N (D) 6 x 0 5 N (E) It is immeasurably small, but not zero.
14 7. A 0 kg box remains at rest while a person pushes directly to the right on the box with a force of 60 N. The coefficient of kinetic friction between the box and the surface is µ k = 0.0. The coefficient of static friction between the box and the surface is µ s = What is the magnitude of the force of friction acting on the box during the push? (A) 00 N (B) 0 N (C) 60 N (D) 40 N (E) 0 N 8. A point object is connected to the end of a long string of negligible mass and the system swings as a simple pendulum with period T. What is the period of the pendulum if the string is made to have onequarter of its original length? (A) 4T (B) T (C) T (D) T / (E) T / 4 9. A mass m is pulled outward until the string of length L to which it is attached makes a 90 o angle with the vertical. The mass is released from rest and swings through a circular arc. What is the tension in the string when the mass swings through the bottom of the arc? (A) 0 (B) ½mg (C) mg (D) mg (E) 3mg 0. Two boxes of different masses in an orbiting space station appear to float at rest one above the other with respect to the station. An astronaut applies the same force to both boxes. Can the boxes have the same acceleration with respect to the space station? (A) No, because the boxes are moving in orbits of different radius. (B) No, because the box of greater mass requires more force to reach the same acceleration. (C) Yes, because both boxes appear weightless. (D) Yes, because both boxes are accelerating toward earth at the same rate. (E) It cannot be determined without knowing whether the boxes are being pushed parallel or perpendicular to earth s gravity.. An object is dropped from rest from a certain height. Air resistance is negligible. After falling a distance d, the object s kinetic energy is proportional to which of the following? (A) (B) (C) (D) d (E) d d d d. An object is projected vertically upward from ground level. It rises to a maximum height H. If air resistance is negligible, which of the following must be true for the object when it is at height H/? (A) Its speed is half of its initial speed. (B) Its kinetic energy is half of its initial kinetic energy. (C) Its potential energy is half of its initial potential energy. (D) Its total mechanical energy is half of its initial value. (E) Its total mechanical energy is half of its value at the highest point.
15 3. A boy of mass m and a girl of mass m are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed v across the frictionless ice surface and he slides to the right as shown above. What is the total work done by the children? (A) Zero (B) mv (C) mv (D) mv (E) 3mv 6. A kg object initially moving with a constant velocity is subjected to a force of magnitude F in the direction of motion. A graph of F as a function of time t is shown above. What is the increase, if any, in the velocity of the object during the time the force is applied? (A) 0 m/s (B).0 m/s (C) 3.0 m/s (D) 4.0 m/s (E) 6.0 m/s 4. An object of mass M travels along a horizontal air track at a constant speed v and collides elastically with an object of identical mass that is initially at rest on the track. Which of the following statements is true for the two objects after the impact? (A) The total momentum is Mv and the total kinetic energy is ½Mv. (B) The total momentum is Mv and the total kinetic energy is less than ½Mv. (C) The total momentum is less than Mv and the total kinetic energy is ½Mv. (D) The momentum of each object is ½Mv. (E) The kinetic energy of each object is ¼Mv. 5. A spherical planet has mass greater than that of earth, but its density is unknown. The weight of an object on that planet compared with its weight on earth is which of the following? 7. A particle P moves around the circle of radius R shown above under the influence of a radial force of magnitude F. What is the work done by the radial force as the particle moves from position to position halfway around the circle? (A) Zero (B) RF (C) RF (D) πrf (E) πrf (A) Larger (B) The same (C) Smaller (D) It cannot be determined without information about the planet s size. (E) It cannot be determined without information about the planet s atmosphere.
16 Questions 8 and 9 Two blocks of wood, each of mass kg, are suspended from the ceiling by strings of negligible mass, as shown above.. Three students were arguing about the height of a parking garage. One student suggested that to determine the height of the garage, they simply had to drop tennis balls from the top and time the fall of the tennis balls. If the time for the ball to fall was.4 seconds, approximately how tall is the parking garage? (A) 5 m (B) 7 m (C) 0 m (D) 4 m (E) 0 m 8. What is the tension in the upper string? (A) 0 N (B) 0 N (C) 40 N (D) 50 N (E) 60 N 9. What is the force exerted on the upper block by the lower string? (A) Zero (B) 0 N upward (C) 0 N downward (D) 0 N upward (E) 0 N downward 0. The period of a mass-spring system undergoing simple harmonic oscillation is T. If the amplitude of the mass-spring s motion is doubled, what will be the new period? (A) ¼T (B) ½T (C) T (D) T (E) 4T. A deliveryman moves 0 cartons from the sidewalk, along a 0-meter ramp to a loading dock, which is.5 meters above the sidewalk. If each carton has a mass of 5 kg, what is the total work done by the deliveryman on the cartons to move them to the loading dock? (A) 500 J (B) 3750 J (C) J (D) J (E) J 3. An astronaut on the Moon simultaneously drops a feather and a hammer. The fact that they reach the surface at the same instant shows that (A) No gravity forces act on a body in a vacuum. (B) The acceleration due to gravity on the Moon is less than the acceleration due to gravity on the Earth. (C) The gravitational force from the Moon on heavier objects (the hammer) is equal to the gravitational force on lighter objects (the feather). (D) A hammer and feather have less mass on the Moon than on Earth. (E) In the absence of air resistance all bodies at a given location fall with the same acceleration.
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### PRACTICE TEST for Midterm Exam
South Pasadena AP Physics PRACTICE TEST for Midterm Exam FORMULAS Name Period Date / / d = vt d = v ot + ½ at d = v o + v t v = v o + at v = v o + ad v = v x + v y = tan 1 v y v v x = v cos v y = v sin
### Physics Exam 2 October 11, 2007
INSTRUCTIONS: Write your NAME on the front of the blue exam booklet. The exam is closed book, and you may have only pens/pencils and a calculator (no stored equations or programs and no graphing). Show
### Physics Semester 1 Review
Physics Semester 1 Review Name: 1. Define: Speed Velocity Acceleration Use the graph to the right to answer questions 2-4. 2. How far did the object travel in 3 seconds? 3. How long did it take for the
### Name: Class: 903 Active Physics Winter Break Regents Prep December 2014
In this section use the following equations for velocity and displacement to solve: 1. In a drill during basketball practice, a player runs the length of the 30.meter court and back. The player does this | 12,582 | 47,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.911541 |
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JEE Main 2022 (Online) 27th July Evening Shift
+4
-1
Let S be the set of all $$(\alpha, \beta), \pi<\alpha, \beta<2 \pi$$, for which the complex number $$\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$$ is purely imaginary and $$\frac{1+i \cos \beta}{1-2 i \cos \beta}$$ is purely real. Let $$Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S$$. Then $$\sum\limits_{(\alpha, \beta) \in S}\left(i Z_{\alpha \beta}+\frac{1}{i \bar{Z}_{\alpha \beta}}\right)$$ is equal to :
A
3
B
3 i
C
1
D
2 $$-$$ i
2
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1
Let the minimum value $$v_{0}$$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$$ is attained at $${ }{z}=z_{0}$$. Then $$\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$$ is equal to :
A
1000
B
1024
C
1105
D
1196
3
JEE Main 2022 (Online) 26th July Evening Shift
+4
-1
If $$z=x+i y$$ satisfies $$|z|-2=0$$ and $$|z-i|-|z+5 i|=0$$, then :
A
$$x+2 y-4=0$$
B
$$x^{2}+y-4=0$$
C
$$x+2 y+4=0$$
D
$$x^{2}-y+3=0$$
4
JEE Main 2022 (Online) 26th July Morning Shift
+4
-1
Let O be the origin and A be the point $${z_1} = 1 + 2i$$. If B is the point $${z_2}$$, $${\mathop{\rm Re}\nolimits} ({z_2}) < 0$$, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?
A
$$\arg {z_2} = \pi - {\tan ^{ - 1}}3$$
B
$$\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}$$
C
$$|{z_2}| = \sqrt {10}$$
D
$$|2{z_1} - {z_2}| = 5$$
EXAM MAP
Medical
NEET | 697 | 1,473 | {"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} | 3.203125 | 3 | CC-MAIN-2024-18 | latest | en | 0.582367 |
https://www.chalet-ars-vivendi.de/4895/what-39-s-the-molar-mass-of-iron-iii-acetate.html | 1,611,671,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704800238.80/warc/CC-MAIN-20210126135838-20210126165838-00462.warc.gz | 723,909,221 | 6,891 | 1. Home
2. What 39 S The Molar Mass Of Iron Iii Acetate
# What 39 S The Molar Mass Of Iron Iii Acetate
April 16,2020
The molar mass of k2cr2o7 is 294.What is the mass of 4.80 mole of barium hydride, bah2 669.Using the stock system of nomenclature, cr2so4 is named.A compounds molar mass is numerically equal to.What is the correct molar mass of ag3po4.What is the percentage of iron in ironiii chloride.
### If The Molar Mass Of Asa Is 180Gmol How Many
The problem ask what is the mass of ironiii chloride contains 2.35 x 1023 chloride ions.I understand the molar mass is 162.35 x 1023 chloride ions would be around 0.- the work that my teacher gave me is 0.Asked by shadow on january 29, 2013 chloral hydrate.
### Gl To Moll Conversion Endmemo
Molar mass of frequently calculated chemicals c2h52o ether nh42c2o4 ammonium oxalate nh42co3.Ethyl acetate c4h9oh butyl alcohol c5h10 cyclopentane c5h10o5 ribose c5h12 pentane c5h12o methyl tert-butyl ether.Ironiii bromide fecl2 ironii chloride fecl3 ironiii chloride fef3 ironiii fluoride feo.
### What Is The Molar Solubility Of Iron Iii Hydroxide Ksp
Youll need to set up an ice table.Feoh3 - fe3 3oh- initial for fe3 0.25 and 0 for oh- change for fe3 x and for oh- 3x equilibrium for fe3 0.25 x ignore x here too small and for oh- 3x so your equation is ksp fe3o.
### Stoichiometry Mrraswells Chemistry Help
A sample of any element with a mass equal to that elements atomic weight in grams will contain precisely one mole of atoms 6.02 x 10 23 atoms.For example, helium has an atomic weight of 4.00 grams of helium will contain one mole of helium atoms.You can also work with fractions or multiples of moles.
### Wwwerthelpm Molar Mass Of 2 23326
Molar mass of cuso 4 63.5 g of cuso 4 contains 63.100 g of cuso 4 will contain of copper.Amount of copper that can be obtained from 100 g cuso 4 39.8 determine the molecular formula of an oxide of iron in which the mass per cent of iron.
### Cbse Ncert Solutions For Class 11 Chemistry Chapter 1
Iii ch s solution i the molecular mass of water h.Molar mass of sodium acetate is.We know the atomic mass of iron 55.85 the atomic mass of oxygen 16 number of moles of iron present in the oxide.
### Molar Mass Table Of 465 Common Chemical Compounds
The molar mass of a substance, also often called molecular mass or molecular weight although the definitions are not strictly identical, but it is only sensitive in very defined areas, is the weight of a defined amount of molecules of the substance a mole and is expressed in gmol.
### Naming Ionic Compounds Chemunlimitedm
Give the name and molar mass of the following ionic compounds name molar mass 1 na 2 co 3 sodium carbonate 129 gramsmole 2 naoh sodium hydroxide 40 gramsmole 3 mgbr 2 magnesium bromide 184.1 gramsmole 4 kcl potassium chloride 74.6 gramsmole 5 fecl 2 iron ii chloride 126.8 gramsmole 6 fecl 3.
### Ironiii Ion Fe Chemspider
Bioaccumulation estimates from log kow bcfwin v2.17 log bcf from regression-based method 0.162 log kow used -0.77 estimated volatilization from water henry lc 0.0245 atm-m3mole estimated by bond sar method half-life from model river 0.83 min half-life from model lake 71.966 days removal.
### Learn How To Calculate Molarity Of A Solution
Molarity is a unit of concentration, measuring the number of moles of a solute per liter of solution.The strategy for solving molarity problems is fairly simple.This outlines a straightforward method to calculate the molarity of a solution.
### Find The Molar Mass Of Iron Sulfate Yahoo Answers
For instance iron iii sulfate in your case has 2 iron atoms each of mass 55.845 gm, 3 sulfur atoms each weighs 32.065 gm and 12 4x3 oxygen atoms of mass 15.So the molar mass which is the molecular mass expressed in grams of iron iii sulfate would be 2x55.
### What Is The Molar Mass Of Argon Studym
The molar mass of argon is equal to 39.948 grams-per-mole, or 39.There are three major pieces of information in an elements box on the.
### Equilibrium Calculating Solubility Of Ironiii
The following example in skoogs analytical chemistry uses mass-balance and charge-balance equations to calculate the solubility of ironiii hydroxide in aqueous solution.
### Molar Mass Of Kc6h5co2 Webqcg
Examples of molecular weight computations c14o162, s34o162.Definitions of molecular mass, molecular weight, molar mass and molar weight.Molecular mass molecular weight is the mass of one molecule of a substance and is expressed in the unified atomic mass units u.1 u is equal to 112 the mass of one atom of carbon-12.
### Some Basic Concepts 1alculate The Molecular Mass Of
Molar mass of the compound 100 now, molar mass of na 2 so 4 2 23.0 142 gmol 100mass percent of sodium 46 142 32.39 mass percent of sulphur 32 142 100 22.54 mass percent of oxygen 64 142 100 45.Determine the empirical formula of an oxide of iron which has 69.
### Chapter 3 Molar Mass Calculation Of Molar Masses
Molar mass molar ratio 180.1 g mol c6h12 o6 2.33 mol o strategy once the number of moles of a substance present is known, we can use molar mass to find the number of grams avogadros number to find the number of atoms, ions, or molecules moles b grams b atoms b molar mass na moles a molar ratio molar conversions chapter 3 h grams.
### Molecular Weight Of Alanine Convert Units
More information on molar mass and molecular weight.In chemistry, the formula weight is a quantity computed by multiplying the atomic weight in atomic mass units of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
### Student Worksheet 4 Extra Practice Questions Molar
Able on molar mass and conversions and the factor-label method in the enrichment exercises.Calculate the mass of ironiii oxide rust produced by the reaction of 500 g of iron with oxygen from the air.What mass of precipitate should form if 2.00 g of silver nitrate in solution is reacted with excess.Acetate solution.
### Copperii Acetate C4h6cuo4 Chemspider
S22,s2425,s26,s363739,s45 synquest 57389, m029-1-01 tbc synquest m029-1-01 warning causes gi injury, skin and eye irritation alfa aesar 19417, 44355 warning irritates skin and eyes, harmful if swallowed alfa aesar b23615, 44355 xn,n abblis chemicals ab1003371.
### Ironiii Chloride Cas 7705 08 0 803945
Ironiii chloride anhydrous for synthesis.Cas , ec number 231-729-4, chemical formula fecl.- find msds or sds, a coa, data sheets and more information.
### Chromiumiii Acetate C6h9cro6 Pubchem
Combinations of basic chromiumiii acetate and basic chromiumiii formate also find application in the textile industry as mordants.In addition, chromiumiii acetate is used as a starting compound in the production of organic chromium dyes.
### Molecular Weight Of Ironiii Oxide Convert Units
Ironiii oxide molecular weight.Molar mass of fe2o3 159.Convert grams ironiii oxide to moles or moles ironiii oxide to grams.Molecular weight calculation 55.99943 percent composition by element.
### What 39 S The Molar Mass Of Iron Iii Acetate
What 39 s the molar mass of iron iii acetate.Sch 4c1 unit 2 problem set upper grand district school board.A typical bottle of nail polish remover contains 2.5 mol of ethyl acetate c 4 h 8 o 2. | 1,952 | 7,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-04 | latest | en | 0.786415 |
https://quizzes.studymoose.com/2-7-terms-quiz/ | 1,719,212,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00368.warc.gz | 424,650,283 | 13,043 | # 2.7 Terms
question
Standard Deviation
A number that measures how far data values are from their mean. Provides a numerical measure of the overall amount of variation in a data set, and Can be used to determine whether a particular data value is close to or far from the mean.
question
The standard deviation is _____ when the data are all concentrated close to the mean, exhibiting little variation or spread.
Small.
question
The standard deviation is ______ when the data values are more spread out from the mean, exhibiting more variation.
Larger.
question
value = mean + (#ofSTDEV)(standard deviation)
(#STDEV) does not need to be an integer.
question
sample: x = x¯ + (#ofSTDEV)(s) Population: x=?+(#ofSTDEV)(?)
The lower case letter s represents the sample standard deviation and the Greek letter ? (sigma, lower case) represents the population standard deviation. The symbol x¯ is the sample mean and the Greek symbol ? is the population mean.
question
If x is a number, then the difference "x - mean" is called its ___.
Deviation.
question
Variance
The average of the squares of the deviations (the x - x¯ values for a sample, or the x - ? values for a population).
question
How much the statistic varies from one sample to another is known as the ____ _____ of a statistic.
Sampling variability.
question
The standard error of the mean.
Measures the sampling variability of a statistic. An example of standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean.
question
The notation for the standard error of the mean is ?/?n
Where ? is the standard deviation of the population and n is the size of the sample.
question
If you add the deviations, the sum is always ____.
Zero.
question
By squaring the deviations, you make them positive numbers, and the sum will also be ____.
Positive.
question
#ofSTDEVs is often called a "___-_______"; we can use the symbol z.
z-score.
question
For ANY data set, no matter what the distribution of the data is:
At least 75% of the data is within two standard deviations of the mean. At least 89% of the data is within three standard deviations of the mean. At least 95% of the data is within 4.5 standard deviations of the mean. This is known as Chebyshev's Rule.
question
For data having a distribution that is BELL-SHAPED and SYMMETRIC:
Approximately 68% of the data is within one standard deviation of the mean. Approximately 95% of the data is within two standard deviations of the mean. More than 99% of the data is within three standard deviations of the mean. This is known as the Empirical Rule. It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.
1 of 15
## Unlock all answers in this set
question
Standard Deviation
A number that measures how far data values are from their mean. Provides a numerical measure of the overall amount of variation in a data set, and Can be used to determine whether a particular data value is close to or far from the mean.
question
The standard deviation is _____ when the data are all concentrated close to the mean, exhibiting little variation or spread.
Small.
question
The standard deviation is ______ when the data values are more spread out from the mean, exhibiting more variation.
Larger.
question
value = mean + (#ofSTDEV)(standard deviation)
(#STDEV) does not need to be an integer.
question
sample: x = x¯ + (#ofSTDEV)(s) Population: x=?+(#ofSTDEV)(?)
The lower case letter s represents the sample standard deviation and the Greek letter ? (sigma, lower case) represents the population standard deviation. The symbol x¯ is the sample mean and the Greek symbol ? is the population mean.
question
If x is a number, then the difference "x - mean" is called its ___.
Deviation.
question
Variance
The average of the squares of the deviations (the x - x¯ values for a sample, or the x - ? values for a population).
question
How much the statistic varies from one sample to another is known as the ____ _____ of a statistic.
Sampling variability.
question
The standard error of the mean.
Measures the sampling variability of a statistic. An example of standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean.
question
The notation for the standard error of the mean is ?/?n
Where ? is the standard deviation of the population and n is the size of the sample.
question
If you add the deviations, the sum is always ____.
Zero.
question
By squaring the deviations, you make them positive numbers, and the sum will also be ____. | 1,074 | 4,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-26 | latest | en | 0.902924 |
https://www.storyboardthat.com/storyboards/62b293b5/statsss | 1,726,536,016,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00363.warc.gz | 930,083,989 | 78,762 | • My Storyboards
# STATSSS
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#### Storyboard Text
• Hello! My name is John and I am standing here to report on how to use the appropriate statistic tool on different situations.
• HOW TO DETERMINE THE APPROPRIATE TOOL WHEN:ON DIFFERENT SITUATIONS
• A Report of Group ATEEEEE......
• First, when the population variance is unknown, the appropriate test statistic to use is the z-test.
• First, when the population variance is known and you are dealing with a normally distributed population, the appropriate test statistic is using z-test. This test is appropriate when you have a large enough sample size.
• HOW TO DETERMINE THE APPROPRIATE TOOL WHEN:VARIANCE IS KNOWN
• A Report of Group ATEEEEE......
• REMEMBER:Z-test is the appropriate tool to used in central limit theorem when:If the population is normally distributed or the sample size is large and the true population means μ = μo, the z has a standard normal distribution.The sample size is extremely large and the variance (σ2) is known or unknown.The sample standard deviation (s) may be used as an estimate of the population standard deviation (σ) when the value of σ is unknown.
• HOW TO DETERMINE THE APPROPRIATE TOOL WHEN:CENTRAL LIMIT THEOREM
• A Report of Group ATEEEEE......
• What is a t-test?A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups, which may be related in certain features.
• HOW TO DETERMINE THE APPROPRIATE TOOL WHEN:VARIANCE IS UNKNOWN
• A Report of Group ATEEEEE......
• REMEMBER:The T-test is the appropriate tool to used when:The population standard deviation (σ) is unknown.The sample standard deviation (s) is known.The population is normal or nearly normally distributed.The sample size is less than 30 (n≥30).
• HOW TO DETERMINE THE APPROPRIATE TOOL WHEN:VARIANCE IS UNKNOWN
• A Report of Group ATEEEEE......
• Z-testn ≥ 30, σ is knownn ≥ 30, σ is unknown (Central Limit Theorem)n 30, σ is known
• So, that is all for our reporting today. On behalf of Group ATEEEEE...... My name is John and Thank you for listening.
• T-testn 30, σ is not knownn ≥ 30, s is known
Over 30 Million Storyboards Created | 544 | 2,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-38 | latest | en | 0.858171 |
guzitoqiby.cerrajeriahnosestrada.com | 1,601,350,292,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00349.warc.gz | 371,857,884 | 4,212 | We know that linear hyphens graph a straight line, so I liberal what a quadratic function is incomplete to look like. The vertex of f x is 0, 0while the least of p x is 4, 0.
Use the website of values to write the following function: Complete cursor problems and get more feedback. How to Leave Quadratic Equations:.
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Let's graph all three parts:. Improve your math knowledge with free questions in "Complete a function table: quadratic functions" and thousands of other math skills.
Relating to a mathematical expression containing a term of the second degree, such as x 2 + 2. ♦ A quadratic equation is an equation having the general form ax 2 + bx + c = 0, where a, b, and c are constants. ♦ The quadratic formula is x = - b ± √(b 2 - 4 ac)/2 a. It is used in algebra to calculate the roots of quadratic equations.
Quadratic Functions-Worksheet Find the vertex and “a” and then use to sketch the graph of each function. Find the intercepts, axis of symmetry, and range of each function. Here is a set of practice problems to accompany the Quadratic Equations - Part I section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.
CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams.
In order to get functions, we'll graph quadratic equations of the form. y = ax 2 + bx + c. This will ensure that we only have one value of y for each value of x.
Sample Problem. Graph the equation y = x 2. This is the simplest quadratic equation there is. It's also a relation, so we already know how to graph it. | 898 | 3,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.932777 |
https://www.pinterest.co.uk/andy2695/number-bonds/ | 1,502,899,509,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102307.32/warc/CC-MAIN-20170816144701-20170816164701-00392.warc.gz | 973,709,537 | 100,878 | ### Smack the Number Counting Game
Preschool kids will have a blast while playing this simple Smack the Number counting game! Work on number recognition and one to one correspondence in a fun way!
Common core place value K-1, numbers 11 to 19, number chains and what to do with them!
Used for subtraction from ten with the Early Years - this looks like a great way to keep maths fun! This can also be adapted by labelling each pin from 1-10. Whichever ones have been knocked down, the child must add them up e.g. knocks down number 4, 7 and 9. The child must then add these up to find the total.
Counting, maths ideas in school, preschool, nursery
You Might be a First Grader...: Number Bond. This is for Susan, Ashlee, Nancy and Shannon.
Using number bonds to teach students how to decompose numbers #addition #subtraction
Part Part Whole practice. Simply brilliant! Need to spend some time searching through her blog! There's great stuff there!
True or false Number Sorts last week... writing their own calculations this week... brilliant maths Class1!
number bonds to 10 - Cuisenaire rods , Write fractions to show what color each part is.
### Number Match Slap with Playing Cards
A number match game with playing cards
##### More ideas
Making 10! great idea! First Grade Smiles Done as partners ~ Each student had ten cubes of one color and a blank copy of the chart. As a class, traded one color cube with our partner and colored in our chart to match. Fantastic to reach all learners: oral, visual, kinethetic, tactile.
Rainbow Number Combinations to 10 Freebie {from Cupcake for the Teacher} - Awesome!
Freebie! Combinations of ten. Enjoy! (My sister just downloaded and used with some of her K kids...and didn't know it was mine!)
Using number bonds to teach students how to decompose numbers #addition #subtraction
Cool Teach - Adventures in Teaching: Numbers
### Favorite Math Pins
Flip TEN-Fun addition card game from Guided Math! Kids line up cards in four rows of five. Then, they flip two cards over. If the sum of the two cards equals they keep the cards and replace the cards with two more from the deck. The game ends when ther
Number Bonds Worksheets. Great for teachers using Singapore Math. Really helps students understand number relationships.
This shows how students can show their work on the B & W version. As they progress, they could write the actual number sentence instead of using "and" and "make".
Pinterest | 550 | 2,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-34 | longest | en | 0.895824 |
http://www.numbersaplenty.com/554 | 1,590,918,128,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00577.warc.gz | 193,680,346 | 3,274 | Search a number
554 = 2277
BaseRepresentation
bin1000101010
3202112
420222
54204
62322
71421
oct1052
9675
10554
11464
123a2
13338
142b8
1526e
hex22a
554 has 4 divisors (see below), whose sum is σ = 834. Its totient is φ = 276.
The previous prime is 547. The next prime is 557. The reversal of 554 is 455.
Subtracting from 554 its product of digits (100), we obtain a palindrome (454).
Subtracting from 554 its reverse (455), we obtain a palindrome (99).
554 is nontrivially palindromic in base 11.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 529 + 25 = 23^2 + 5^2 .
It is a magnanimous number.
554 is an undulating number in base 11.
It is a Curzon number.
It is a plaindrome in base 13, base 15 and base 16.
It is a nialpdrome in base 10.
It is not an unprimeable number, because it can be changed into a prime (557) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 137 + ... + 140.
554 is a deficient number, since it is larger than the sum of its proper divisors (280).
554 is a wasteful number, since it uses less digits than its factorization.
554 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 279.
The product of its digits is 100, while the sum is 14.
The square root of 554 is about 23.5372045919. The cubic root of 554 is about 8.2130270823.
The spelling of 554 in words is "five hundred fifty-four", and thus it is an aban number.
Divisors: 1 2 277 554 | 478 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-24 | longest | en | 0.913187 |
https://techgrabyte.com/facebook-ai-mathematician-solve-university-calculus-problems/ | 1,713,768,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818081.81/warc/CC-MAIN-20240422051258-20240422081258-00811.warc.gz | 503,798,738 | 18,107 | # Facebook’s AI Mathematician Can Solve University Calculus Problems
Maths is a highly complex and challenging subject for most of the human beings, even for Artificial Neural Networks. Most neural networks are only capable of performing simple addition and multiplication problem.
But in Paris, at Facebook AI Research, François Charton and Guillaume Lample have developed an algorithm that has learned to solve university-level calculus problems in seconds.
Both have trained an AI on tens of millions of calculus problems randomly generated by a computer.
The Facebook team of researcher have trained a neural network to perform the necessary symbolic reasoning in order to differentiate and integrate mathematical expressions for the very first time.
Neural networks are highly capable of recognizing patterns, but when it comes to symbolic reasoning they struggle, at best neural networks were able to do addition and multiplication of whole numbers.
Facebook’s neural network is a significant step toward more powerful mathematical reasoning and a new way of applying neural networks beyond traditional pattern-recognition tasks.
In order to teach AI about the Mathematical problem and finding its solutions, the team trained the AI by using natural language processing (NLP), a computational tool commonly used to analyze language.
François Charton and Guillaume Lample teach a neural network to recognize the patterns of mathematical manipulation that are equivalent to integration and differentiation.
Finally, they let the neural network loose on expressions it has never seen and compares the results with the answers derived by conventional solvers like Mathematica and Matlab.
Lample and Charton break the mathematical expressions into their component parts by representing expressions as tree-like structures. The leaves on these trees are numbers, constants, and variables like x.
This works because the mathematics in each problem can be thought of as a sentence, with variables, normally denoted x, playing role of nouns & operations, such as finding the square root, playing role of verbs.
The AI then “translates” the problem into a solution.
François Charton and Guillaume Lample create this database by randomly assembling mathematical expressions from a library of binary operators such as addition, multiplication, and so on; unary operators such as cos, sin, and exp; and a set of variables, integers, and constants, such as π and e.
They also limit the number of internal nodes to keep the equations from becoming too big.
By this, the team generates a massive training data set consisting, for example, of 80 million examples of first- and second-order differential equations and 20 million examples of expressions integrated by parts.
And finally, François Charton and Guillaume Lample fed the neural network with 5,000 expressions it has never seen before and comparing the results it produces in 500 cases with those from commercially available solvers, such as Maple, Matlab, and Mathematica.
When the pair tested AI on 500 calculus problems, it found a solution with an accuracy of 98%. A comparable standard program for solving maths problems had only an accuracy of 85% on the same problems.
In many cases, the neural network was able to find solutions within the 30s.
One interesting observation was that the neural network was able to find more than one solution for a single problem and that because a math problem can be solved in many ways.
“On all tasks, we observe that our model significantly outperforms Mathematica,” say the researchers.
“On function integration, our model obtains close to 100% accuracy, while Mathematica barely reaches 85%.” And the Maple and Matlab packages perform less well than Mathematica on average.
Despite this, it could still correctly answer questions that confounded other maths programs.
“The ability of the model to recover equivalent expressions, without having been trained to do so, is very intriguing,” say Lample and Charton.
That’s a significant breakthrough. “To the best of our knowledge, no study has investigated the ability of neural networks to detect patterns in mathematical expressions,” says the pair.
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The DeepMind CEO & Co-Founder Is Joining Google As The AI Lab Positions | 862 | 4,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | longest | en | 0.939476 |
https://math.stackexchange.com/questions/2847994/proof-verification-extension-lemma-for-functions-on-submanifolds | 1,563,712,628,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00316.warc.gz | 470,889,859 | 37,252 | # (Proof verification) Extension Lemma for Functions on Submanifolds
There is a similar question about this lemma in Lee's book, but I wasn't really satisfied with the detail in the proposed solution. In particular, the OP of the linked question doesn't describe how to get the smooth maps which I am calling $$f_p$$ below. I am hoping someone can give critical feedback on this proof and point out where, if anywhere, I have erred.
Lemma 5.34 (Extension Lemma for Functions on Submanifolds). Suppose $$M$$ is a smooth manifold, $$S\subset M$$ is a smooth submanifold, and $$f\in C^\infty(S)$$.
1. If $$S$$ is embedded, then there exist a neighborhood $$U$$ of $$S$$ in $$M$$ and a smooth function $$\widetilde f\in C^\infty(U)$$ such that $$\widetilde f|_S = f$$.
2. If $$S$$ is properly embedded, then the neighborhood $$U$$ in part (a) can be taken to be all of $$M$$.
Proof of 1:
Let $$n = \dim S$$ and $$m = \dim M$$. Since $$S$$ is embedded, each $$p\in S$$ is in the domain of a slice chart $$(U_p,\varphi_p)$$ in $$M$$ such that $$S\cap U_p$$ is a single slice in $$U$$. Since $$f$$ is smooth, at each $$p\in S$$ we can find a slice chart so that $$\mathbf{f}_p:= f\circ\varphi_p^{-1}\colon \varphi_p(S\cap U_p)\to\Bbb R$$ is smooth. Since $$\varphi_p(S\cap U_p) = \{(x^1,\dots,x^n,0,\dots,0)\in \varphi_p(U_p)\}$$, and $$\mathbf{f}_p$$ is smooth on this set, there is a smooth extension $$\widehat f_p\colon V_p\subset\Bbb R^m\to \Bbb R$$ of $$\mathbf{f}_p$$, where $$V_p$$ is an open set in $$\Bbb R^m$$ containing $$\varphi(p)$$.
(Here is where I see a potential "bug" in this proof. By the definition of a smooth map on $$S$$, the map $$\mathbf{f}_p$$ is "smooth." However, since $$\mathbf{f}_p$$ is defined on a set that is not an open subset of $$\Bbb R^n$$, I believe, and please do address this point, "smooth" means that there is the extension I am calling $$\widehat f_p$$.)
The restriction of $$\widehat f_p$$ to the open set $$W_p := \varphi_p(U_p)\cap V_p$$ is still smooth and gives us a map $$f_p\colon \varphi_p^{-1}(W_p)\to\Bbb R$$ defined by $$f_p(x) = \big(\widehat f_p|_{W_p}\big)\circ\varphi_p(x)$$.
The collection of open sets $$\varphi_p^{-1}(W_p)$$ is an open cover of $$S$$, so let $$(\psi_p)_{p\in S}$$ be a partition of unity subordinate to this cover. For any $$x\in U:=\bigcup_{p\in S}\varphi^{-1}(W_p)$$, define $$\widetilde f(x) = \sum_{p\in S}\psi_p(x)f_p(x)$$. If $$x\in S$$, then $$\widetilde f(x) = \sum_{p\in S}\psi_p(x) f(x) = f(x).$$ Hence $$\widetilde f$$ is an extension of $$f$$. Since the collection of supports of the $$\psi_p$$ is locally finite, each point in $$U$$ has a neighborhood in which the sum is finite, so this defines a smooth map since every point has a neighborhood such that $$\widetilde f$$ restricted to that neighborhood is smooth. Hence $$\widetilde f\in C^\infty(U)$$ is the desired extension. $$\square$$
Proof of 2:
In this case, the only thing that changes is that $$S$$ is actually closed in $$M$$ by a lemma in Lee's book. By appending $$M\smallsetminus S$$ to the collection $$\{\varphi_p^{-1}(W_p)\}$$ and taking a partition of unity $$\{\psi_p: p\in S\}\cup\{\psi_0\}$$ subordinate to this open cover of $$M$$ with $$\operatorname{supp}\psi_0\subset M\smallsetminus S$$, define $$\widetilde f(x) = \psi_0(x) + \sum_{p\in S}\psi_p(x)f_p(x)$$, where $$f_p$$ is defined in the first part. $$\square$$
I heartily welcome feedback, but I believe the proof in my question works. The first section where I thought the proof was lacking can indeed be improved. I got the idea for the improvement from the hint to Exercise 2.3 in Lee's Riemannian Manifolds, restated here for interest:
Exercise 2.3. Suppose $M\subset \widetilde M$ is an embedded submanifold.
(a) If $f$ is any smooth function on $M$, show that $f$ can be extended to a smooth function on $\widetilde M$ whose restriction to $M$ is $f$. [Hint: Extend $f$ locally in slice coordinates by letting it be independent of $(x^{n+1},\dots,x^m)$, and patch that together using a partition of unity.]
Here is the improved proof of 1:
Let $n = \dim S$. Since $S$ is embedded, each $p\in S$ is in the domain of a slice chart $(U_p,\varphi_p)$ in $M$ such that $f|_{S\cap U_p} = f(x^1,\dots,x^n)$ and the $x^i$ are slice coordinates. Extend $f|_{S\cap U_p}$ to $U_p$ by setting $f_p(x^1,\dots,x^n,x^{n+1},\dots,x^m) = f|_{S\cap U_p}(x^1,\dots,x^n)$. The map $f_p\colon U_p\to\Bbb R$ is smooth because it is independent of its last $m-n$ coordinates and it is smooth with respect to its first $n$ coordinates.
The collection of open sets $U_p$ is an open cover of $S$, so let $(\psi_p)_{p\in S}$ be a partition of unity subordinate to this cover, and define $\widetilde f(x) = \sum_{p\in S} \psi_p(x)f_p(x)$. If $x\in S$, then $$\widetilde f(x) = \sum_{p\in S}\psi_p(x) f(x) = f(x).$$ Hence $\widetilde f$ is an extension of $f$.
Since the collection of supports of the $\psi_p$ is locally finite, each point in $U$ has a neighborhood in which the sum is finite, so this defines a smooth map since every point has a neighborhood such that $\widetilde f$ restricted to that neighborhood is smooth. Hence $\widetilde f\in C^\infty(U)$ is the desired extension. $\square$
The proof of 2 is modified by replacing $\varphi_p^{-1}(W_p)$ with $U_p$ defined in the modified proof of 1 above. $\square$
The reason that the original proof still works is that the extensions I describe in the first proof of 1 can be taken to be the $f_p$ explicitly described in the proof in this answer. The approach taken here simplifies some of the notation and construction of the extension $f_p$ by blurring the distinction between points $p$ in the manifold and their local coordinate representations $(x^1,x^2,x^3,\dots)$. Smoothness of the coordinate representation of $f$ is equivalent to smoothness of the actual map defined on the manifold. | 1,816 | 5,869 | {"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": 64, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-30 | latest | en | 0.849218 |
https://math.answers.com/Q/What_is_25_percent_of_17_pounds | 1,717,060,353,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00305.warc.gz | 326,012,557 | 47,007 | 0
# What is 25 percent of 17 pounds?
Updated: 9/16/2023
Wiki User
7y ago
4.25 pounds or 4 and 1/4 pounds
25% = 0.25
0.25 x 17 = 4.25
Or you can use the fraction 1/4, which equals 0.25
17/4 = 4 1/4 = 4.25
Wiki User
7y ago | 105 | 231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-22 | latest | en | 0.840507 |
https://www.convertit.com/Go/SmartPages/Measurement/Converter.ASP?From=li&To=length | 1,620,407,002,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00126.warc.gz | 742,491,457 | 3,740 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Chinese li = 644.652 length (length) ``` Related Measurements: Try converting from "li" to agate (typography agate), arpentcan, astronomical unit, barleycorn, bottom measure, cable length, city block (informal), cloth finger, digitus (Roman digitus), engineers chain, fathom, furlong (surveyors furlong), inch, Israeli cubit, pace, palm, Roman mile, soccer field, sun (Japanese sun), verst (Russian verst), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: li = 18.17 actus (Roman actus), .01455553 arpentcan, 76,140 barleycorn, 2,538,000 caliber (gun barrel caliber), 352.5 fathom, 6.45E+17 fermi, 29,005.71 finger, 1,933,956 French, 2,785.95 Greek span, 6,345 hand, 25,380 inch, .13352273 league, 3,204.55 link (surveyors link), 644,652,000 micron, .11444805 parasang, 2.09E-14 parsec, 1,834,212.62 point (typography point), .16416729 ri (Japanese ri), 2,820 span (cloth span), 705 yard.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 452 | 1,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.679868 |
https://albertteen.com/uk/gcse/mathematics/probability-and-statistics/the-mode | 1,718,876,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00130.warc.gz | 77,857,269 | 39,743 | Albert Teen
YOU ARE LEARNING:
The Mode
# The Mode
### The Mode
The mode is a method of finding the average, and represents the most commonly found number.
One way of analyzing and describing the average of a set of data is by looking at the mode. The mode or modal group is the data that appears the most frequently, or, in other words, is most popular.
Here is a set of data collected on the shoe sizes of 10 students. $5, \space 5, \space 6, \space 6, \space 7, \space 7, \space 7, \space 7, \space 8, \space 8$
Which is the most common number here? $5, \space 5, \space 6, \space 6, \space 7, \space 7, \space 7, \space 7, \space 8, \space 8$
$7$ is the most common number in the set of data. Therefore, $7$ can be said to be the mode.
Here is another set of data collected on the height of 10 students. What is the mode? $1.67, \space 1.35 ,\space 1.49 ,\space 1.56 ,\space 1.78 ,\space 1.7 , \space 1.62 ,\space 1.69 ,\space 1.68 ,\space 1.58$
1
We can put the heights into a grouped frequency table
This gives intervals within which individual values can fall.
2
Instead of a mode, we now have a modal class interval
The modal class interval is the group with the most frequency, which in this case is: $1.6 < x \leq 1.7$ | 387 | 1,243 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.837916 |
http://jitu368.com/essay/Calorie-Management/F3YJDQ3XJ | 1,544,723,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825029.40/warc/CC-MAIN-20181213171808-20181213193308-00600.warc.gz | 146,495,547 | 11,499 | # Calorie Management Essay example
2328 Words Sep 15th, 2013 10 Pages
Calorie Management Over the course of this class and the few weeks that encompassed this learning process, we have learned several different conceptual applications for the integration of algorithms within the infrastructure of a company. The idea of using an algorithm or a program to influence or benefit our daily lives is an obvious choice for an application. We cannot use a program to cure cancer, diabetes or other wide ranging diseases, but we can use it to help with weight management, the mastery of which can lead to several different areas of improvement. By creating a program that will aid in the eradication of obesity, we can go a long way towards improving the mass health of the worldwide human population. The obesity …show more content…
Problem Statement A team of software developers has been awarded a contract to develop a program to determine if a person is balancing calorie intake. The challenge to the developers is to determine what tasks need to be completed so that the end result shows the person an accurate result as to whether or not they are losing or gaining weight. To determine these tasks the developers need to write a top notch algorithm in pseudo code that helps to lay out what needs to be accomplished. Once the tasks are identified the program can be written.
Program Solution The solution the developers came up with is to create a calorie calculator. The calorie calculator will ask for inputs from the person and those inputs will be processed by the program to show the person what their basal metabolic rate is and whether they are gaining or losing weight. The results will allow the person to evaluate their diet and decide whether or not what they are doing is helping or hampering their health.
Inputs and Outputs The input of the program is the variables that are inputted to produce a result. When the inputs are entered into the program the program takes the information and makes calculations based on the information it is given. The program then performs an output that shows the results of the calculations that were performed. The calorie calculator will take the variables that are inputted by an
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https://www.studioarmata.com/print-all-fibonacci-numbers-python/ | 1,568,807,363,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00096.warc.gz | 1,048,693,325 | 11,895 | Print All Fibonacci Numbers Python
As observed, the reaction to the headline NFP print, in case of a relative deviation. released by the US Department of.
Meta Analysis Psychology Strengths Weaknesses Meta-Analysis of Early Intensive Behavioral Intervention for Children with Autism. Journal of Clinical Child & Adolescent Psychology, 38 (3), 439-450. Reviewed by Amy Hansford, MS Carl Sagan Demon Haunted World
. en.wikipedia.org/wiki/Fibonacci_number) example to show how to use ipdb debugger. The following is a simple program ([a very inefficient.
If so, it might be time for you to learn Python. An initial investment. groupby([‘Page Type’]).agg(‘sum’) print(aggregated_df) After we create the Page Type column, we group all pages by type and.
C program to print fibonacci series using recursion In this program, we will read value of N (N for number of terms) and then print fibonacci series till N terms using recursion. Fibonacii series: Is a series of number in which each number is the sum of preceding two numbers.
Unlike other popular operating systems, where the command line is a scary proposition for all but the most experienced. contains duplicate lines, it will print out each line and the number of.
Nov 22, 2016 · If you have written some code in Python, something more than the simple “Hello World” program, you have probably used iterable objects. Iterable objects are objects that conform to the “Iteration Protocol” and can hence be used in a loop.
Each function can also have a return statement that returns a computed result. For example, here is a simple Python program from lab 4 that illustrates the calling. def main(): # compute the 6th Fibonacci number recursively print fib(6) def.
This flowchart answers the question "How do you draw a flowchart to calculate the first N Fibonacci numbers?" If N is 20, then you get Fibonacci numbers 0 through 19. The Fibonacci series are the numbers:. The output statements in the flowchart show the value of i and the Fibonacci number fib. You would print or display one line and then go.
Fibonacci series is defined as a sequence of numbers in which the first two numbers are 1 and 1, or 0 and 1, depending on the selected beginning point of the sequence, and each subsequent number is the sum of the previous two. So, in this series, the n th term is the sum of (n-1) th term and (n-2) th term. In this tutorial, we’re going to.
def print_square(x): print(square(x)). At its highest level, In general, Python code is a sequence of statements. A simple. Consider the sequence of Fibonacci numbers, in which each number is the sum of the preceding two: 0, 1, 1 , 2, 3, 5, 8,
So I can run the above on a list, tuple or set of numbers. I can even run it on a dictionary whose keys are all numbers. x = 1 >>> y = ‘1’ >>> print(x+y) That code will result in an error, because.
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I wrote a fibonacci. big numbers took less than 1 second. The code was written in C as C++ didn’t yet exist. You have been told before to before to watch your coding style. Stop using global. There.
Code Python When you think about this query, the first thing that comes to your mind is to run a loop and check if all elements in list are same. Below is a simple program.
A number of new Python features have been added to the version 3.8. Some of the features are simply awesome. This article will outline the new features of Python 3.8 which you should be familiar with.
recur_fibonacci(41) will take more than twice as long. However, contrary to what some people think recursion is not the problem here. Rather, the problem is algorithmic: For every Fibonacci number you calculate, you first calculate all previous Fibonacci numbers, and you do this again for each previous number, without remembering intermediate.
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Given all. the number of iterations or the length of a matrix): name = raw_input(‘Type your name and press enter: ‘) print ‘Hi ‘ + name num = raw_input(‘How many times do you want to print your.
Apr 19, 2019 · All other terms are obtained by adding the preceding two terms. This means to say the nth term is the sum of (n-1) th and (n-2) th term. Python Program to Display Fibonacci Sequence Using Recursive Function
We’ve all heard of 3D printing by now – from prosthetic limbs to car parts, 3D printers are creating a growing number of products. “You could print a house in the shape of a Fibonacci spiral if you.
Feb 12, 2018 · Fibonacci trading tools are used for determining support/resistance levels or to identify price targets. It is the presence of Fibonacci series in nature which attracted technical analysts’ attention to use Fibonacci for trading. Fibonacci numbers work like magic in.
Answer to Write a short Python program using a for loop to print the Fibonacci series up to 10 terms. The Fibonacci series is desc.
Apr 13, 2018. The Fibonacci numbers are a sequence of integers in which every. We present each algorithm as implemented in the Python programming.
This simple isprime(number) function checks if the given integer number is a prime number and returns. of n # for all odd numbers for x in range(3, int(n**0.5)+1, 2): if n % x == 0: return False.
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Python. Exercise 1 Experiment with Python: try some computations using it as a cal-culator, then enter the code for factorand experiment with it. Use control-C to abort a computation if it takes too long. Exercise 2 Devise and test a function fib(n)which returns the n-th Fibonacci number Fn. Fn is de ned by the initial conditions F0 = 1, F1 = 1.
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Print all numbers from A to B inclusively. In all the problems input the data using input() and print the result using print(). Maintainer: Vitaly Pavlenko ([email protected]) Credits to: Denis Kirienko, Daria Kolodzey, Alex Garkoosha, Vlad Sterzhanov, Andrey Tkachev, Tamerlan Tabolov, Anthony Baryshnikov, Denis Kalinochkin, Vanya Klimenko.
So, let’s start with the root module from which almost all other scientific modules are built, NumPY. Out of the box, Python. numbers: import time a = range(100000) c = [] starttime = time.clock().
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Jan 30, 2019 · A simple verification in your Python interpreter would show that factorial(5) gives you 120. Fibonacci Sequence. A Fibonacci sequence is one where each number is the sum of the proceeding two numbers. This sequence makes an assumption that Fibonacci numbers for 0 and 1 are also 0 and 1. The Fibonacci equivalent for 2 would therefore be 1.
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Python Program for Fibonacci Series – Example python program to find the fibonacci series.
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I do not expect you to do it all for me!! I have got this far on my own. I have used a for loop for the fibonacci numbers and it is ok. But I am really stuck on the ratio and displaying it. Again,
The Fibonacci. playing around with Python ‘for loops’ # for large ranges replace range() with the more efficient xrange() print "iterate over 0 to <6:" for k in range(6): print k print ‘-‘*30 #.
Using Python’s os.walk. This allows you to perform all sorts of reports and interesting tasks. For example, the above code will print all of the subdirectories under the current directory, ".". Let.
Last time, we wrote our Python commands directly into the shell, by typing at the prompt ( >>> ). This is. In the new window, type print "Hello everyone!. Note that all Python files end with the extension.py.
An example how to find a file (Windows7): # search for a filename in a given directory # and all. print("Success") # do something here else: print("Failure") # do something else here test() If you.
Read writing about Programming in Coding Puzzles. A collection of programming interview questions.
Here is how you can find a Fibonacci number in a Fibonacci pattern when you want to find a number in that pattern. Check it out! Here is how you can find a Fibonacci number in a Fibonacci pattern when you want to find a number in that pattern. Check it out!. Python 2.7. Ruby. Roy. Python. Nodejs.
Jun 21, 2017. Code Challenge #11: June 21, 2017 Every week, we feature the type of. Output: an integer, the 50th number in the Fibonacci Sequence. Here's my quick resolution using Python for the Basic and Intermediate difficulties.
Sep 07, 2014 · Computing Fibonacci sequence with list comprehension I’m trying to learn list comprehensions, but the example I decided to use requires some additional python features I haven’t learned yet. I’ve searched and found not the solution.
Dec 19, 2018. Trying to make an efficient Fibonacci sequencer in Python. added_to iterator+= 1 return(added_to) print(fibonacci(200000)). Oh man, every time I see the Fibonacci sequence I remember one of my interviews when I was to.
When python. Fibonacci number. And is the same function in Python C-API. You may yield: ”What the heck, why it’s so big”, and here comes the explanation: In C code on the very first line, we have.
We are looking for a developer (or small team) to develop a python / django based maintenance tracking system. The ideal candidate can do both front end and back end with multi-user, multi-role, and scaleability in mind. Here are the requirements of the project: We need a system that tracks maintenance on any type of machine (automobile, drone, engine, refrigerator etc).
If invoked with no options, it executes all the files listed in sequence and drops you into the interpreter while still acknowledging any options you may have set in your ipython_config.py. This behavior is different from standard Python, which when called as python -i will only execute one file and. | 2,728 | 11,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-39 | latest | en | 0.845815 |
http://www.adras.com/Hlookup-with-Merged-cell-as-look-up-value.t80958-13.html | 1,638,505,725,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00371.warc.gz | 92,105,953 | 3,015 | From: srs on 18 Mar 2010 16:59 Hopefully someone can help me out on this one. I have a spreadsheet that looks similar to this: 2/15/2010 3/15/2010 . . . Site# INV VALUE INV VALUE 1 34 332 40 440 2 47 250 47 250 3 55 476 36 420 Total 136 1058 123 1100 I am trying to use a Hlookup to return both the Totals for both the INV and Value in a compainon sheet. The date cells are merged over the INV and VAL Colums so my Hlookup formula only returns the INV value. Anyone know of a way to have a Hlookup return the value in the right column or another work around for this problem. From: T. Valko on 18 Mar 2010 17:51 Kind of hard to figure out where your stuff is so you'll have to go by my sample... This is in the range B6:E6 >136,1058,123,1100 This in the merged cells B1:C1 2/15/2010 This in the merged cells D1:E1 3/15/2010 A10 = 2/15/2010 B10 = INV C10 = VALUE Enter this formula in B11: =INDEX(B6:E6,MATCH(A10,B1:E1,0)) Enter this formula in C11: =INDEX(B6:E6,MATCH(A10,B1:E1,0)+1) -- Biff Microsoft Excel MVP "srs" wrote in message news:0D1837F9-4306-4FDB-A1AB-08F670BFBD1B(a)microsoft.com...> Hopefully someone can help me out on this one. I have a spreadsheet that > looks similar to this: > > 2/15/2010 3/15/2010 . . . > Site# INV VALUE INV VALUE > 1 34 332 40 440 > 2 47 250 47 250 > 3 55 476 36 420 > Total 136 1058 123 1100 > > > I am trying to use a Hlookup to return both the Totals for both the INV > and > Value in a compainon sheet. The date cells are merged over the INV and > VAL > Colums so my Hlookup formula only returns the INV value. Anyone know of a > way to have a Hlookup return the value in the right column or another work > around for this problem. | | 555 | 1,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-49 | latest | en | 0.824995 |
http://www.mathworks.co.jp/jp/help/physmod/elec/examples/nonlinear-inductor-characteristics.html?action=changeCountry | 1,386,334,840,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163051590/warc/CC-MAIN-20131204131731-00092-ip-10-33-133-15.ec2.internal.warc.gz | 429,080,372 | 8,652 |
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## Nonlinear Inductor Characteristics
This example shows a comparison of nonlinear inductor behavior for different parameterizations. Starting with fundamental parameter values, the parameters for linear and nonlinear representations are derived. These parameters are then used in a Simscape™ model and the simulation outputs compared.
Specification of Parameters
Fundamental parameter values used as the basis for subsequent calculations:
• Permeability of free space,
• Relative permeability of core,
• Number of winding turns,
• Effective magnetic core length,
• Effective magnetic core cross-sectional area,
• Core saturation begins,
• Core fully saturated,
mu_0 = pi*4e-7;
mu_r = 3000;
Nw = 10;
le = 0.032;
Ae = 1.6e-5;
B_sat_begin = 0.75;
B_sat = 1.5;
Calculate Magnetic Flux Density and Magnetic Field Strength Data
Where:
• Magnetic flux density,
• Magnetic field strength,
Linear representation:
Nonlinear representation (including coefficient, a):
% Use linear representation to find value of H corresponding to B_sat_begin
H_sat_begin = B_sat_begin/(mu_0*mu_r);
% Rearrange nonlinear representation to calculate coefficient, a
a = atanh( B_sat_begin/B_sat )/H_sat_begin;
% Linear representation
H_linear = [-500 500];
B_linear = mu_0*mu_r*H_linear;
% Nonlinear representation
H_nonlinear = -5*H_sat_begin:H_sat_begin:5*H_sat_begin;
B_nonlinear = B_sat*tanh(a*H_nonlinear);
Display Magnetic Flux Density Versus Magnetic Field Strength
The linear and nonlinear representations can be overlaid.
figure,plot( H_linear, B_linear, H_nonlinear, B_nonlinear );
grid( 'on' );
title( 'Magnetic flux density, B, versus Magnetic field strength, H' );
xlabel( 'Magnetic field strength, H (A/m)' );
ylabel( 'Magnetic flux density, B (T)' );
legend( 'B=\mu_0 \mu_r H', 'B-H curve', 'Location', 'NorthWest' )
Calculate Magnetic Flux and Current Data
Where:
• Magnetic flux,
• Current,
Linear representation:
Nonlinear representation:
% Linear inductance
L_linear = mu_0*mu_r*Ae*Nw^2/le;
% Linear representation
I_linear = [-1.5 1.5];
phi_linear = I_linear.*L_linear/Nw;
% Nonlinear representation
I_nonlinear = le.*H_nonlinear./Nw;
phi_nonlinear = B_nonlinear.*Ae;
Display Magnetic Flux Versus Current
The linear and nonlinear representations can be overlaid.
figure, plot( I_linear, phi_linear, I_nonlinear, phi_nonlinear );
grid( 'on' );
title( 'Magnetic flux, \phi, versus current, I' );
xlabel( 'Current, I (A)' );
ylabel( 'Magnetic flux, \phi (Wb)' );
legend( '\phi=I L/N_w', '\phi-I curve', 'Location', 'NorthWest' );
Use Parameters in Simscape Model
The parameters calculated can now be used in a Simscape model. Once simulated, the model is set to create a Simscape logging variable, simlog.
modelName = 'elec_nonlinear_inductor';
open_system( modelName );
sim( modelName );
Conclusion
The state variable for all representations is magnetic flux, . Current, I, and magnetic flux, , can be obtained from the Simscape logging variable, simlog, for each representation. Overlaying the simulation results from the representations permits direct comparison.
figure, plot( ...
simlog.Linear_Inductor.inductor.i.series.values,...
simlog.Linear_Inductor.inductor.phi.series.values,...
simlog.B_vs_H.inductor.i.series.values,...
simlog.B_vs_H.inductor.phi.series.values,...
simlog.phi_vs_I.inductor.i.series.values,...
simlog.phi_vs_I.inductor.phi.series.values,...
'o'...
);
grid( 'on' );
title( 'Magnetic flux, \phi, versus current, I' );
xlabel( 'Current, I (A)' );
ylabel( 'Magnetic flux, \phi (Wb)' );
legend( 'Linear (single inductance)', 'B-H characteristic', '\phi-I characteristic', 'Location', 'NorthWest' ); | 938 | 3,754 | {"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} | 2.640625 | 3 | CC-MAIN-2013-48 | latest | en | 0.600311 |
https://www.magicalapparatus.com/pencil-reading/ri-t-i-r-i-r.html | 1,561,521,120,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00128.warc.gz | 804,542,575 | 12,127 | ## Ri t i r i r
After that—turn the card and wave it casually so that all may see it has SOMETHING written on it—but no one will be able to see what!
Finally, tell the spectator to remove his card and call out the name—which he then shows to the audience. As a last word, should the spectator choose the card which corresponds to your prediction—it is obvious that you allow him to open the envelope and read the card aloud.
If you haven't got the nerve to try this trick the way I suggest, you can always end clean by having nineteen predictions in a Pocket Index ready to switch the card after it has been mis-read—so that a card bearing the correct prediction can be handed to the spectator afterwards—I'm glad you thought of that! *
(15) "The Informative Joker" by Corinda *
This is a quick card trick which ends on a light hearted note. It is a novel idea for close up card workers—not suitable for performance as an effect for large audiences.
The Effect and The Method. You force the three of clubs and have it returned to the pack. I said this effect was for Card Workers—so you do not need me to tell you how to force a card ! You now explain that the joker is the one card that sees all, hears all and tells everything ! You take out the joker and pretend to listen to him— suddenly your eyes light up— "He tells me you took the Three of Clubs". You look at the spectator and wave the joker at him,44 is that Correct ?" He will probably smile and say yes—and you then look very serious and say—44 You don't believe it I can tell that, but let me tell you that he told me and showed me—look, he knew what card you would choose ! You drop the joker on the table and it is seen that in one hand he holds a card—it is the Three of Clubs !
Some Jokers are pictured as a clowned man standing with one arm in the air. Draw with ink a small card in this hand and let it resemble the Three of Clubs. When you wave the Joker to show it during the routine, put your thumb over the drawing to conceal the card in his hand.
(16) "Double-Impact Prediction" by Corinda
The Effect. "I want you to imagine that I have a pack of playing cards and that I am showing them to you now. (Make an imaginary fan and show). You look along the cards and you see one that you fancy—what do you see? Good, now would you count along from the start here (indicate an imaginary starting place) and tell me the position of your card? Fourteen? Thank you, and the card was the Eight of Spades—correct? That is a very odd thing— because here I have an envelope and on it, as you will see, is the number fourteen—have a look and see what's inside". He takes out the eight of spades! Please don't rush to tell me it's a good effect, I've been doing it for years—and I know it's good!
The Method. Use a Pocket Index and a Swami Gimmick. All this I have discussed before!
(17) "The Matchbox Mystery" by Corinda
The Effect. The performer writes a prediction on a small slip of paper which he folds and drops into an empty cup. He then throws a box of matches on to the table and invites anybody to remove as many as they like. This done, the remainder are counted and then the prediction is read. It tells the exact number of matches that will be left in the box—and is correct.
The Method. Write a prediction—fold it twice and appear to drop it into a cup. In actual fact you palm it out leaving the cup empty. Have a box of matches with no more than 52 matches in it and drop it on the table. When the total of the matches left is declared—reach into a Billet Index where you have in readiness 52 predictions and palm out the appropriate one. Pick up the cup and drop the billet in as you do so—passing the cup to a spectator to read the billet inside. Effect No. 24 (Step One) is a variation of this trick.
(18) "The Mystery of the Chest" by Corinda
The Effect. On a table stands a nicely decorated chest and beside it stands a drinking glass. To the left of these are two or three items which include seven keys and seven numbered tags. A committee of seven people take part in the proceedings and they are invited to come on to the stage or to one end of the room. Now this is very important so please read carefully. This is one of those rare effects where most of the work is done by the assistants and practically none by the performer. In this case, our performer stands well to one side and calls out his instructions to the committee.
To start the routine, the Mentalist explains to the audience and assistants that the box on the table is a chest—containing something which will surprise everybody (you do not say what it is). You explain further—that on the table are seven keys and that only one of them will open the chest—which at the moment is locked. Next you tell the audience that the committee must be numbered and then must each select a key. You instruct each member of the committee to:—
"Go to the table, select any .number tag you like (they run from 1 to 7) and pin it on your lapel. Next choose any key and test it—should it be the one that unlocks the chest—relock the chest after testing. Whether the key does or does not fit—when it has been tested it must be dropped into the glass".
This is done up to the point when the last person comes to the table. After he has dropped his key into the glass—you ask him to shake them all up and then take them himself along the row of assistants—and have each person remove one. He takes the last one—the one that is left after everybody has had a choice.
Once more the committeemen are called to try their keys and see who has the one that fits the lock. One at a time they come to the table and see if their key fits. Since all the keys look alike—there is no way of telling until they are tested in the lock. No matter who has the key that fits—every key is tried to prove that six do not unlock the chest. The assistant who takes the right key stands aside—the others return to the side of the stage. Now
you approach the box for the FIRST TIME and ask that it be unlocked. From inside the chest you remove an envelope and tearing it open, ask the assistant left with you to remove a card from inside—and read it aloud. It reads:—
"It is my Prediction that whoever chooses to call himself number three —will also choose the only key to open this chest—Corinda". Everybody can see that the assistant standing beside you wears a tag No. 3 and you have not influenced their choice in any way. This is my effect, which although somewhat long (as with most versions of the "Seven Keys") is well worth the preliminaries. I would suggest it is more suitable for the drawing room where the use of seven spectators draws a good number of the audience into the effect—thus giving the trick a personal interest.
The Method. In the Chest is an envelope. It is a window envelope as described on page 12 of "Step One" and it is prepared with a pre-worded prediction as told to you on Page 14 of the same book. The trick is painfully simple. When you see who gets the key that opens the lock you look at his number. When you take the envelope out of the chest, with a Swami Gimmick—fill in his number so that the full prediction reads accordingly! To do this, even though you are not proficient with a Swami Gimmick—is childs play—you must be able to write any one of the numbers from one to seven—is that hard? In view of the method you will understand that the chest is unfaked, the lock ordinary, the keys ordinary, there is no force of the number tags and it does not matter who takes part—if anyone criticises this I will send them a pint of my blood!
The Effect. The Mentalist writes a Prediction naming a card. He folds it ' and gives it to the spectator—telling him to drop it in his pocket. A pack of cards is taken and shuffled. It is explained that one must be chosen—and that the choice must not only be free of any influence—but also, no one must know what is selected. To achieve this—you tell the spectator to hold the cards behind his back and to mark any one on the face with a cross. You give him the cards and a pencil. After that, the cards are examined and the marked one is found—then the prediction is opened by the spectator and read—it names what card has been chosen.
The Method. This is an adaption of a trick given in Thurston's Book of Magic and there are many uses for the principle. The secret is very simple— it is the pencil. Although it looks absolutely normal—it will not write. It is just a dowel rod painted up to look like a pencil—or a real pencil with a gimmicked tip. On the sly, you mark one card on the face with a cross. That card you predict. You mix the cards and give them to the spectator with suitable instructions so that he will apparently draw on the face of a card. Give him the fake pencil—and be sure to recover it afterwards. As an added touch, you could switch the fake pencil for a real one—just in case!
(20) "Nicely Suited"
The Effect. The performer shuffles a pack of cards and places them on the table. He declares that he knows what card is most likely to be chosen— before it is chosen! This he proves to be true—a card is chosen and in a striking manner it is revealed.
The Method. I have found this of great use to me with card effects. It is a definite improvement on the old verbal force which was used to make the spectator take any card you wanted him to take. Prepare by putting the Three of Clubs on top of the pack, the Three of Hearts you reverse in the middle, the Three of Spades goes to the bottom of the deck and the Three of Diamonds you put into your pocket. By this arrangement you will observe that whatever suit is chosen—you have what is called a "get-out"—or a suitable card. The deck is placed on the table in this condition and you must now force (verbally) the number three. To do this quickly—suggest Court Cards are ignored—deal first with "odds or evens"—force the odds and eliminate to the three. At this point you stop to explain that there may be some doubt as to the method of selection—so you will give the spectator a chance to call out any suit he likes—and whatever suit is called will be accepted . . . If it's Hearts—Spread the deck face down and show that the only card reversed is the Three of Hearts. For Clubs—simply turn over the top card; for Spades pick up the deck and show the bottom card—for—have the spectator reach into your pocket and remove one card—his card! Sometimes the little things make a lot of difference and somebody once said 'Long live the little difference'(misquote).
## Practical Mental Influence
Unlock the Powers of Mental Concentration to Influence Other People and to Change Situations. Learn How to mold the mind one-pointed, until you have focused and directed a mighty degree of Mental Influence toward the desired object.
Get My Free Ebook | 2,437 | 10,837 | {"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} | 2.828125 | 3 | CC-MAIN-2019-26 | latest | en | 0.972131 |
https://blogs.sas.com/content/iml/2012/10/04/dice-probabilities-and-the-game-of-craps.html | 1,722,805,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00055.warc.gz | 117,911,975 | 11,953 | Gambling games that use dice, such as the game of "craps," are often used to demonstrate the laws of probability. For two dice, the possible rolls and probability of each roll are usually represented by a matrix. Consequently, the SAS/IML language makes it easy to compute the probabilities of various events.
For two fair dice, the probability of any side appearing is 1/6. Because the rolls are independent, the joint probability is computed as the product of the marginal probabilities. You can also create a matrix that summarizes the various ways that each particular number can occur:
```proc iml; dieProb = j(1, 6, 1/6); /* each die is fair: Prob(any face)=1/6 */ /* joint prob is indep = product of marginals */ probs = dieProb` * dieProb; /* 6x6 matrix; all elements = 1/36 */ events = repeat(1:6, 6) + T(1:6); print events[c=("1":"6") r=("1":"6")];```
You can use the event and probs matrices to compute the probabilities of each possible sum of two dice. There are several ways to do this, but I like to use the LOC function to find the elements of the event matrix that correspond to each roll, and then add the associated probabilities:
```P = j(1, 12, 0); /* probability of each event */ do i = 2 to ncol(P); idx = loc( events=i ); P[i] = sum( probs[idx] ); end; print P[format=6.4 c=("P1":"P12")];```
The preceding table shows the probabilities of each roll. (Note that the probability of rolling a 1 with 2 dice is zero; the P[1] element is not used for any computations.) You can use the table to compute the probability of winning at craps. If you roll a 7 or 11 on the first roll, you win. If you roll a 2, 3, or 12, you lose. If you roll a 4, 5, 6, 8, 9, or 10 on the first roll (called "the point"), you continue rolling. You win if you can successfully re-roll your "point" before you roll a 7. It is an exercise in conditional probability that you can compute the probability of making a "point" in craps by summing a ratio of probabilities, as follows:
```win = {7 11}; lose = {2 3 12}; point = (4:6) || (8:10); Pwin1 = sum( P[win] ); /* Prob of winning on first roll */ PLose1 = sum( P[lose] ); /* Prob of losing on first roll */ /* Prob winning = Pr(7 or 11) + Prob(Making Point) */ Pwin = P[7] + P[11] + sum(P[point]##2 / (P[point] + P[7])); print Pwin1 PLose1 Pwin;```
According to the computation, the probability of winning at craps is almost—but not quite!—50%.
This exercise shows that having a matrix and vector language is useful for computing with probabilities. More importantly, however, this post sets the foundations for looking at the interesting case where the two dice are not fair. If you change the definition of dieProb (on the first line of the program) so that it is no longer a constant vector, all of the computations are still valid!
Next week I will post an article that shows how the odds change if some sides of the dice are more likely to appear than others. Feel free to conduct your own investigation of unfair dice before then.
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Distinguished Researcher in Computational Statistics
Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of SAS/IML software. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
1. >>>If you roll a 7 or 11 on the first roll, you win. If you roll a 2, 3, or 11, you lose.
Ths should be "...If you roll a 2, 3, or 12, you lose."
• Thanks. Corrected.
2. David Bruckman on
While the odds shown above are fairly generic, the game really rests on chip management, balancing a Pass/Don't Pass game, and sizing up the rollers.
Craps players know that the come out (first) roll favors them, but many play a "Don't Pass" strategy when a point has to be made, especially with longer odds (a point of 4 or 10).
Even Don't Pass players will hedge with early bets on 6 or 8, then then turn off these bets after the first point rolls. Hedging shrinks an overall gain, but who wouldn't want to play a positive margin?
Study up, or give it up.
• Of course there is much more to playing craps, than just knowing the base probabilities, primarily because the house payouts are less than the event probabilities. This post does not discuss betting strategies or the expected return of various bets. For people who actually play craps, there are many Web sites that discuss these issues.
3. "If you roll a 2, 3, or 11, you lose."
Should be 2, 3, or 12. More importantly, the code is correct :) | 1,183 | 4,639 | {"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} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.905605 |
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# Heirloom tomatoes, grown from seeds saved from the previous
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28 Aug 2012, 20:33
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Heirloom tomatoes, grown from seeds saved from the previous year, only look less appetizing than their round and red supermarket cousins, often green and striped, or have plenty of bumps and bruises, but are more flavorful.
(A) cousins, often green and striped, or have plenty of bumps and bruises, but are
(B) cousins, often green and striped, or with plenty of bumps and bruises, although
(C) cousins, often green and striped, or they have plenty of bumps and bruises, although they are
(D) cousins; they are often green and striped, or with plenty of bumps and bruises, although
(E) cousins; they are often green and striped, or have plenty of bumps and bruises, but they are
[Reveal] Spoiler: OA
If you have any questions
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28 Aug 2012, 20:35
Above question has already been discussed in this forum.
But I need more explanation in reference to the usage of although in D.
Someone wrote in the previous three that Although don't have a subject in D, hence option D is wrong.
Please explain the above concept in detail.
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28 Aug 2012, 22:18
Heirloom tomatoes, grown from seeds saved from the previous year, only look less appetizing than their round and red supermarket cousins, often green and striped, or have plenty of bumps and bruises, but are more flavorful.
The 2 bold parts are Subject and Verb of the sentence. Therefore, often green and striped part constitutes a run-on sentence. -> eliminate A, B, C.
(D) cousins; they are often green and striped, or with plenty of bumps and bruises, although more flavorful
(E) cousins; they are often green and striped, or have plenty of bumps and bruises, but they are more flavorful
To my view, the 2 above sentences successfully deliver the meaning of the original sentence excepting the bold part in D.
Although is used to convey a contrast meaning with the previous part. D should be correct if the previous part told somethings not flavorful -> E will be wrong.
But in the previous part of the sentence, it deals with other issues. Therefore, 'but' is more suitable
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vivekdixit07 wrote:
Above question has already been discussed in this forum.
But I need more explanation in reference to the usage of although in D.
Someone wrote in the previous three that Although don't have a subject in D, hence option D is wrong.
Please explain the above concept in detail.
The word "although" kicks of a dependent CLAUSE and needs a cleear subject.
Examples:
Although they had no umbrellas, the excited children ran outside to enjoy the summer rainstorm. -- Logical and Correct.
Although had no umbrellas, the excited children ran outside to enjoy the summer rainstorm - Illogical without the subject "they".
In option D [although are more flavorful], "although" does not have an explicit (or even clearly implied) subject, therefore it is incorrect.
KW
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29 Aug 2012, 10:35
KyleWiddison wrote:
vivekdixit07 wrote:
Above question has already been discussed in this forum.
But I need more explanation in reference to the usage of although in D.
Someone wrote in the previous three that Although don't have a subject in D, hence option D is wrong.
Please explain the above concept in detail.
The word "although" kicks of a dependent CLAUSE and needs a cleear subject.
Examples:
Although they had no umbrellas, the excited children ran outside to enjoy the summer rainstorm. -- Logical and Correct.
Although had no umbrellas, the excited children ran outside to enjoy the summer rainstorm - Illogical without the subject "they".
In option D [although are more flavorful], "although" does not have an explicit (or even clearly implied) subject, therefore it is incorrect.
KW
Thanks you very much for the explanation.
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Re: Heirloom tomatoes, grown from seeds saved from the previous [#permalink]
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07 Sep 2016, 00:36
The incorrect modifier is placed after the red supermarket cousins. E is the correct option
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07 Sep 2016, 04:25
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View Full Version : Additive Morphing
spider85304-09-2012, 12:04 PMHi, I need some hints, ideas or guides on how to make an additive morphing. By additive morphing I mean the fallowing: Having a mesh A. Two duplicates of mesh A with some deformation B and C. For example I need to deform the mesh A as A -> B + C That mean that I need to deform A to B (this is easy its just the same B) Then I need to add the C - A difference deformation to B. In other words I need to make deformation C in B space (relative deformation of C to A, applied to B). The first idea that I had is to somehow make a vertex space, and store the deformation B and C relative to same vertex index space. then modifying to B, I recalculate the vertex space in B, and apply the difference vector (C - A) to B. Sounds good but 3ds max doesn't have vertex space... I saw that its have some pseudovertex space by using an Edge and a Normal to compute the vertex space. But because its depends on one edge, it may be modified in B deformation and resulting deformation may look incorrect. Because a small rotation on that edge will result in rotation of all vertex space without counting the other edges. Another idea could be to use sum of all edges and normalize to get the vector for computing the vertex space.. but then another problem may arrive, 2 or more vectors can cancel each other resulting in a null vector or vector that doesn't count other edges, and by this making an undefined state. So does anybody has some idea about this? Thanks
denisT
04-09-2012, 01:53 PM
the morphing is always additive. your questions is not clear for me. try to explain it better...
when you set 100% for target B it means you add (B-A) to A
when you set 100% for target C after you set B that it means you add (C-A) to B.
the order doesn't matter. It will be any way A + (B-A) + (C-A).
denisT
04-09-2012, 02:09 PM
The first idea that I had is to somehow make a vertex space, and store the deformation B and C relative to same vertex index space.
what is vertex space? is it vertex normal based or deformation A->B direction?
spider853
04-09-2012, 02:14 PM
the morphing is always additive. your questions is not clear for me. try to explain it better...
when you set 100% for target B it means you add (B-A) to A
when you set 100% for target C after you set B that it means you add (C-A) to B.
the order doesn't matter. It will be any way A + (B-A) + (C-A).
Sorry, the example with B + (C-A) is not good, as it looks like you just add the difference.
To imagine easier, I will bring a 1D example along the normal.
Imagine you have a plane with normals pointing up. (obj A)
You clone that plane (obj C) and deform the vertecies along the normals to form a X bump.
You clone the obj A and apply a bend modifier with full 360* rotation. now the normals are pointing from the center out circular.
When you make de additive deformation they go in this order.
For each Ai, Ci vertex you get de distance along the Ai normal of the Ci vertex and store it (for example Li).
now you modify the A to B. It will be the same object as B.
Now when you add the C deformation, for each of the Bi vertex, vertex gets pushed towards normal at the distance Li.
Now you have a cilinder with a bumped X on it
So this is in 1D only around normal.
but I want to save the 3D relative position to the vertex space.
So bassicaly I have normal for up vector, and I need to find another vector to compute the vertex space. That vector needs to be consistent along deformations to perserve the right logical direction
denisT
04-09-2012, 02:39 PM
following this scenario it's not morphing. it sounds like displacement for me. maybe the right way will be use Projection modifier and its Project Mapping as a source data for Displace Modifier.
spider853
04-09-2012, 02:45 PM
No, displacement are along normals like the 1D example I wrote above,
More like vector displacement.
Btw, does vector displacement occurs in tangent space?
denisT
04-09-2012, 03:21 PM
No, displacement are along normals like the 1D example I wrote above,
More like vector displacement.
Btw, does vector displacement occurs in tangent space?
displacement in the mapping space. you can generate any mapping including the one based on direction of deformation.
spider853
04-09-2012, 04:53 PM
To be clear, I don't need/want to do any maps.
The thing that I need is somewhere similar to vector displacement map 3D (but not displacement map because its only on the normal of the vertex 1D) but without doing any mapping.
Also I didn't found any example of implementation of vector displacement on google and the question is still open.
What I need is a good way to know where is right-left, front-back of the vertex (up-down is normal) so I can save the relative position of deformation and restore it an a different vertex positions.
The way I'm doing it now is by using
MNMesh::GetVertexSpace ()
but I get wrong results on big deformations on the base object
denisT
04-09-2012, 05:03 PM
The thing that I need is somewhere similar to vector displacement map 3D (but not displacement map because its only on the normal of the vertex 1D) but without doing any mapping.
displacement is NOT along vertex normal. The direction of displacement depends on UV mapping. You can make any mapping to give the direction.
spider853
04-12-2012, 12:51 PM
I think I found a similar problem with solution here:
MatanH
04-15-2012, 05:33 AM
Hi, I don't know if you found a solution but here is an idea.
Let's mark the edges arround vertex i as ei1, ... , ein.
For every j between 1 and n lets calculate the following:
dij <- arccos(dot normalized(cross normi ei1) normalized(eij))
Now let's categorize the edges in the following 4 groups:
Group 1: -45 < dij <=45
Group 2: 45 < dij <= 135
Group 3: 135 < dij <= 180 or -180 <= dij <= -135
Group 4: -135 < dij <= -45
Now you can avarrage the vectors in group 1 - group 3,
to create v1, and avarrage the vectors in group 2 - group 4 to create v2.
define the normal as v3 and you have a base for the vertex i's space.
spider853
04-15-2012, 09:41 AM
Hi, I don't know if you found a solution but here is an idea.
Let's mark the edges arround vertex i as ei1, ... , ein.
For every j between 1 and n lets calculate the following:
dij <- arccos(dot normalized(cross normi ei1) normalized(eij))
Now let's categorize the edges in the following 4 groups:
Group 1: -45 < dij <=45
Group 2: 45 < dij <= 135
Group 3: 135 < dij <= 180 or -180 <= dij <= -135
Group 4: -135 < dij <= -45
Now you can avarrage the vectors in group 1 - group 3,
to create v1, and avarrage the vectors in group 2 - group 4 to create v2.
define the normal as v3 and you have a base for the vertex i's space.
That's interesting,
Thanks, I'll have a look
Looks like a nice solution as its using all the edges. One thing I see here is that the matrix will not be ortogonal. From some point of view this looks like a feature because when the geometry will be bended in, the deformation will be scaled down.
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04-15-2012, 09:41 AM
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Journal Article
# Small-Sample Confidence Intervals for p1 - p2 and p1/p2 in 2 × 2 Contingency Tables
Thomas J. Santner and Mark K. Snell
Journal of the American Statistical Association
Vol. 75, No. 370 (Jun., 1980), pp. 386-394
DOI: 10.2307/2287464
Stable URL: http://www.jstor.org/stable/2287464
Page Count: 9
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## Abstract
Consider two binomial populations Π1 and Π2 having "success" probabilities p1 in (0, 1) and p2 in (0, 1), respectively. This article studies the problem of constructing small-sample confidence intervals for the difference of the success probabilities, $\Delta \equiv p_1 - p_2$ and their ratio (the "relative risk"), $\rho \equiv p_1/p_2$ based on independent random samples of sizes n1 and n2 from Π1 and Π2, respectively. These are nuisance parameter problems; hence the proposed intervals achieve coverage probabilities greater than or equal to their nominal (1 - α) levels. Three methods of constructing intervals are proposed. The first one is based on the well -known conditional intervals for the odds ratio $\psi \equiv p_1(1 - p_2)/p_2(1 - p_1)$. It yields easily computable Δ and ρ intervals. The second method directly generates unconditional intervals of the desired size. An algorithm is given for producing the intervals for arbitrary n1 and n2. The 2 × 2 case is given as an illustrative example. The third method constructs unconditional intervals based on a generalization of the classical (Fisher) tail method. Some comparisons are made.
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Solving Equations With Angle Relationships Worksheets Pdf – The aim of Expressions and Equations Worksheets is for your child to be able to learn more effectively and efficiently. The worksheets include interactive exercises as well as problems based on the sequence in which operations are performed. These worksheets are designed to make it easier for children to grasp complex concepts as well as simple concepts in a short time. The PDF worksheets are free to download and can be utilized by your child to test math problems. These resources are useful for students from 5th-8th grade.
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There are plenty of worksheets that teach quadratic equations. Each level has its own worksheet. These worksheets are a great way to practice solving problems up to the fourth degree. Once you’ve finished a level, you can begin to work on solving other types of equations. You can then take on the same problems. You could, for instance solve the same problem as an elongated one. | 623 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-14 | longest | en | 0.958051 |
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« on: March 14, 2013, 04:09:01 PM »
Ok guys, bear with me, trying to learn as much as I can here. How do I decide the amount of tankage( ferms, bright, serving) that I am going to need for a particular system? Does it come down to bbls brewed, how many beers on rotation, etc. I can't really seem to find a routine formula or thought, is there such a thing?
#### anthony
• Brewer
• Posts: 263
• Hoppy to help!
##### Re: Dumb question about tanks
« Reply #1 on: March 14, 2013, 04:48:46 PM »
On the production side, you figure out what you are projecting to produce, you figure out the average time the beer will spend in the fermenter, you figure out what your batch size will be, and you solve for x.
For example:
You project that you're most likely to produce and sell 1000 BBL a year. You're brewing 7 BBL batches. Your beer spends, on average, 14 days in the fermenter. That means, at its maximum, every fermenter can ferment 26 batches.
26 cycles * 7 BBL = 182 BBL per fermenter, so you would need at least 6 fermenters to produce 1000 BBL a year in that setup.
The other tanks depend on what your overall business plan is. Is it a brew pub, is it a tap room, is it a distributing brewery, etc. And the fermenters can vary based on the same things. Maybe you project that you are going to make and sell 1000 BBL a year but that 750 BBL of that will be of the same beer. Then you would want to double-batch beers, maybe you would get 2 15 BBL fermenters and 2 7 BBL fermenters.
#### hopfenundmalz
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##### Re: Dumb question about tanks
« Reply #2 on: March 14, 2013, 04:52:33 PM »
Are you going to do lagers? That means more tank time, and more tanks. A local place is mainly doing lagers, and could make 3 times the beer if doing ales.
Jeff Rankert
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Ann Arbor Brewers Guild
Home-brewing, not just a hobby, it is a lifestyle!
#### majorvices
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« Reply #3 on: March 14, 2013, 08:07:42 PM »
Pretty much what Anthony says. You can move some batches out faster than others. The log jam often happens at the BBT. And, of course, iof you are planning on serving clear beer you will need a little extra conditioning time or filter.
I think the double batch day makes a lot of sense. You sanitize one tank, you sanitize your hoses and heat exchanger once, you MT is already pre heated, you only add an extra 4 hours to your day to brew twice the amount. | 710 | 2,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.933763 |
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## manometer
(mənŏm`ĭtər): see pressurepressure,
in mechanics, ratio of the force acting on a surface to the area of the surface; it is thus distinct from the total force acting on a surface. A force can be applied to and sustained by a single point on a solid.
.
## Manometer
A double-leg liquid-column gage used to measure the difference between two fluid pressures. Micromanometers are precision instruments which typically measure from very low pressures to 50 mm of mercury (6.7 kilopascals). The barometer is a special case of manometer with one pressure at zero absolute. See Barometer
The various types of manometers have much in common with the U-tube manometer, which consists of a hollow tube, usually glass, a liquid partially filling the tube, and a scale to measure the height of one liquid surface with respect to the other (see illustration). If the legs of this manometer are connected to separate sources of pressure, the liquid will rise in the leg with the lower pressure and drop in the other leg. The difference between the levels is a function of the applied pressure and the specific gravity of the pressurizing and fill fluids.
U-tube manometer
A well-type manometer has one leg with a relatively small diameter, and the second leg is a reservoir. The cross-sectional area of the reservoir may be as much as 1500 times that of the vertical leg, so that the level of the reservoir does not change appreciably with a change of pressure. Mercurial barometers are commonly made as well-type manometers.
The inclined-tube manometer is used for gage pressures below 10 in. (250 mm) of water differential. The leg of the well-type manometer is inclined from the vertical to elongate the scale. Inclined double-leg U-tube manometers are also used to measure very low differential pressures. See Pressure measurement
## Manometer
an instrument for measuring the pressure of liquids and gases. A distinction is made among manometers for the measurement of absolute pressure, which is measured from zero (full vacuum); for the measurement of excess pressure, or the difference between absolute and atmospheric pressure, when the absolute pressure is greater; and differential manometers, for the measurement of differences between two pressures, each of which is usually different from atmospheric pressure. Pressure measurements corresponding to atmospheric pressure are made with barometers, whereas measurements of the pressure of rarefied gases are made with vacuum gauges (mainly in vacuum technology). Manometers with scales graduated in various units are used in the measurement of pressure.
The basis of the measurement system of a manometer is a sensing element, which is a primary pressure transducer. A distinction is made among liquid, piston, and spring manometers, depending on the principle of operation and the design of the sensing element. In addition, instruments are used in which the principle of operation is based on measurement of the changes in the physical properties of various materials caused by pressure.
Figure 1. Areas of use of various types of manometers
Scaleless manometers, which produce standard pneumatic or electric output signals that are then fed into control devices for automatic regulation of various industrial processes, are used in addition to manometers with direct readout or recording of the readings. The areas of use of various types of manometers are shown in Figure 1.
The sensing element of liquid manometers is a column of liquid that balances the pressure being measured. The idea of using a liquid for the measurement of pressure was first advanced by the Italian scientist E. Torricelli (1640). The first mercury manometers were produced by the Italian mechanical engineer V. Viviani (1642) and the French scientist B. Pascal (1646). The structural design of liquid manometers is extremely varied. The main types are U-tube, well-type, and double-well manometers. Modern liquid manometers have ranges of measurement from 0.1 newton per sq m (N/m2) to 0.25 meganewton per sq m (MN/ m2), or about 0.01 mm of water (mm H2O) to 1900 mm of mercury (mm Hg); such instruments are used mainly for high-precision measurements in the laboratory. Liquid manometers used for measurement of small excess pressures and rarefactions below 5 kN/m2 (37.5 mm Hg) are called micromanometers. Manometers used for measurements within narrow ranges are filled with light liquids (water, alcohol, toluene, or silicone oils), instruments for wider measurement ranges are filled with mercury. During the measurement of pressure with an inclined-tube micromanometer, the liquid filling the reservoir is displaced into the tube and the change in the level of the liquid is compared with the scale, which is calibrated in units of pressure. The upper limit of measurement with such an instrument is no higher than 2 kN/m2 (about 200 mm of water) at the greatest angle of inclination. Accurate measurements and testing of micromanometers require the use of double-well manometers of the compensation type, in which one of the vessels is fixed, but the second may be displaced vertically to generate the column of liquid required to balance the pressure. The displacement, determined by means of a precise scale with a vernier or from gauge blocks, defines the pressure being measured. Micromanometers of the compensation type may be used for pressure measurements to 5 kN/m2 (about 500 mm H2O). In this case, the error does not exceed (2-5) × 10-3 N/m2, or (2-5) × 10-2 mm H2O.
The upper measurement limit of liquid manometers may be raised by increasing the height of the liquid column and by using a liquid of higher density. However, even the use of mercury in manometers rarely raises its upper limit above 0.25 MN/m2 (about 1900 mm Hg)—for example, in well-type manometers, in which the wide vessel is connected to a vertical tube. Liquid manometers designed for high-precision measurements are equipped with electrical or optical readout devices, and their structural design features permit the elimination of various sources of error (the effect of temperature, vibrations, capillary force, and so on). High accuracy is achieved by a double-well mercury manometer for absolute pressure measurement with a volumetric readout (Figure 2), which is used for the determination of temperature in standard gas thermometers (D. I. Mendeleev All-Union Institute of Metrology). The measurement limits of this manometer are 0-0.13 MN/m2 (0-1000 mm Hg).
Figure 2. Diagram of an absolute pressure manometer with volumetric readout: (1) vessels, (2) metal plates, (3) mercury, (4) glass connecting tubes, (5) readout microscope, (6) scale
To improve the performance of manometers (mainly accuracy of readings), monitoring systems are used for automatic determination of the height of the column of liquid.
The sensing element in piston-type manometers is a piston or other body that is used to balance the pressure by means of a weight or some other device for the measurement of force. A widely used type of piston manometer has an unsealed piston, which is ground in to fit into the cylinder with a small clearance and is displaced axially. The original instrument of this type was designed by the Russian scientists E. I. Parrot and H. F. E. Lenz in 1833. Piston manometers found numerous uses in the second half of the 19th century because of the studies of E. Ruchholz (Germany) and A. Amag (France), who independently proposed the use of the unsealed piston.
The basic advantage of piston manometers over liquid manometers is the ability to measure high pressures with high accuracy. A relatively small piston-type manometer (height, about 0.5 m) has a wider range of measurement and greater accuracy than the 300-m mercury manometer designed by the French scientist L. Cailleté (1891). The manometer was mounted on the Eiffel Tower in Paris. The upper limit of measurement of piston manometers is about 3.5 GN/m2 (3.5 × 108 mm H2O). In this case, the height of the apparatus is no more than 2.5 m. Measurement of such pressures with a mercury manometer would necessitate an increase in its height to 26.5 km.
The most widespread piston-type manometers have a plain unsealed piston (Figure 3). The space below the piston is filled with oil, which flows under pressure into the clearance between the piston and the cylinder, which lubricates the sliding surfaces. Rotation of the piston relative to the cylinder prevents contact friction. The pressure is determined from the weights used to balance the pressure and from the cross section of the piston. The range of measurement may be varied within wide limits by changing the weights and the cross section of the piston; the range is 0.04-10.0 MN/m2, or 0.4-100.0 kilograms-force per sq cm (kgf/cm2), for piston manometers of this type. In this case, the errors of the most accurate standard manometers are not greater than 0.002-0.005 percent. A further increase in the upper limit of measurement makes the piston area so small that it becomes necessary to design special supports for the weights (support rods and lever devices). For example, to decrease the total weights used in the manometer based on the system of M. K. Zhokhovskii (USSR), the balancing force is generated by a hydraulic booster. In this case, even at the high pressures of 2.5 GN/m2 (2.5 × 104 kgf/cm2) the measuring device is of maximum compactness and does not require a large number of weights.
Figure 3. MP-60 weighted-piston manometer with simple unsealed piston: (1) weights, (2) weight pan, (3) stop, (4) funnel, (5) piston, (6) cylinder
Piston manometers of special types are also used to measure small excess pressures and rarefactions, as well as absolute and atmospheric pressure. The piston systems of such manometers are usually given preliminary balancing by a special device, which makes possible a decrease in the lower limit of measurement virtually to zero. For example, the piston may be balanced by a spring mechanism. The piston is rotated by an electric motor. When a rarefaction is generated in the space above the upper part of the piston, the excess of atmospheric pressure balances the weights placed on its lower part.
Spherical and conical pistons are used in addition to cylindrical pistons. In bell manometers, the function of the piston is fulfilled by a bell, and in manometers of the “ring balance” type, by a flat partition within a hollow ring.
Piston-type manometers are used for calibrating and testing other types of manometers and for accurate measurement and monitoring of pressure, with input of readings to a numerical readout device or transmission over a distance.
The sensing element in spring (elastic) manometers is an elastic envelope that absorbs the pressure. The deformation of the envelope is a measure of the pressure that causes it. A distinction is made among tube, diaphragm, and bellows types, depending on the design of the sensing element. The principle of determining pressure from the elastic deformation of a thin envelope was proposed in 1846 by the German scientist R. Schintz; a specific aspect of this method, consisting in the determination of pressure from the deformation of a hollow tubular spring, was proposed in 1848 by the French scientist E. Bourdon, after whom the tubular spring has been named the Bourdon tube. The measurement limits of spring manometers include a wide range of pressures (from 10 N/m2 to 1000 MN/m2, or 1-108 mm H2O).
Simplicity of operation, compactness of design, and convenience of use have led to industrial use of manometers of the elastic type. The simplest tubular manometer (Figure 4) contains a hollow tube bent into an arc, one end of which is attached to the volume whose pressure is to be measured, and the other, which is sealed, is connected to the lever of the transmission mechanism. A change in pressure causes deformation of the tube, and the displacement of its end is transmitted to a pointer, which indicates the pressure on a scale. In addition to tubular springs, a diaphragm or bellows may be used in elastic manometers. In addition to mechanical conversion of the deformation of the sensing element into manometer readings, electrical and optical methods of transformation are used, including those involving the transmission of the results of measurements over a distance.
Figure 4. MM-40 tubular manometer: (1) tube, (2) lever of transmission mechanism, (3) transmission mechanism, (4) pointer
Systems of automatic control and monitoring of industrial processes use elastic manometers with force compensation (in terms of the method of measurement). In this case, the manometer consists of a measuring unit and a standard electric or pneumatic force transducer. The measured pressure is transformed by the sensing element into a force, which is balanced by a force generated by a feedback mechanism rather than by the deformation of the sensor. A standard electric or pneumatic signal proportional to the pressure being measured is generated at the output of the transducer. This system makes it possible to use the same transducer in manometers for determining absolute or excess pressure, as well as vacuum, pressure differences, and parameters of heat and mass transfer, such as the temperature, level, density, and feed rate. In this case, the measurement limits may be changed over a wide range by varying the ratios of the lever arms of the transducer and the area of the bellows. The absolute pressure measurement unit consists of two bellows (Figure 5) attached to a T-shaped lever of the transducer. A vacuum is generated in one of the bellows and the second bellows communicates with the volume in which the pressure is to be measured. The action of the pressure presses the restrictor (stop) on the T-shaped lever against the nozzle, which leads to an increase of pressure in the feedback bellows and the generation of a balancing force. The transducer is supplied with compressed air from an external source. The output pressure is transmitted by means of a pneumatic amplifier to the instrumentation, which records the results of measurement.
The use of manometers of the above types is difficult or impossible
Figure 5. Diagram of an MASP-1 scaleless absolute pressure manometer: (1) reference bellows, (2) measuring bel-lows, (3) nozzle, (4) stop, (5) feedback bellows, (6) pneumatic amplifier
when the pressures to be measured are either very high (above 2.5 MN/m2) or close to zero (below 10 N/m2). These cases require the use of manometers based on the measurement of some parameter related to pressure by a fixed relationship. lonization, thermal, viscosity, and radiometric manometers are used for measurement of low absolute pressures. An example of manometers used for the measurement of high pressures are Manganin manometers, in which the electric resistance of a thin Manganin wire is measured as a function of pressure. Manometers based on the magnetostrictive effect and the speed of sound are also used.
Manometers based on the change in the melting point of mercury with pressure are distinguished by high precision. The transition of mercury from the solid to liquid state is accompanied by a sharp change in the volume, which makes possible accurate recording of the temperature and pressure that correspond to the instant of melting and provides good reproducibility of results. A measuring unit with a manometer of this type makes it possible to determine pressures of up to 4 GN/m2 (approximately 4 × 108 mm H2O) with an error not exceeding 1 percent and is used as a standard for superhigh pressure in calibrating and testing manometers for use up to 4 GN/m2.
Further improvement of manometers includes an increase in their accuracy, extension of the range of measurement, and an increase in reliability, durability, and convenience of use. An increase in accuracy is facilitated by the use of such materials as age-hardenable alloys, quartz (for example, in making the sensing elements of spring manometers) and of elastic supports and optical and electrical methods of transmitting and recording readings. Various means for the transmission of the results of measurements to devices with numerical readout, as well as recorders and printers, which may be located at considerable distances from the measurement sites (for example, transmission of the results of atmospheric pressure measurements from Mars and Venus during flybys of artificial satellites), are used in automatic measurement.
### REFERENCES
Zhokhovskii, M. K. Tekhnika izmereniia davleniia i razrezheniia, 2nd ed. Moscow, 1952.
Zhokhovskii, M. K. Teoriia i raschetpriborov s neuplotnennym porshnem, 2nd ed. Moscow, 1966.
Andriukhina, O. V., and V. N. Gramenitskii. Obraztsovye gruzoporshnevyepribory dlia izmereniia davleniia, sily i massy [survey]. Moscow, 1969.
Khansuvarov, K. I. Tochnye pribory dlia izmereniia absoliutnogo davleniia. Moscow, 1971.
K. I. KHANSUVAROV
## manometer
[mə′näm·əd·ər]
(engineering)
A double-leg liquid-column gage used to measure the difference between two fluid pressures.
## manometer
An instrument for the measurement of pressure; a U-shaped glass tube partially filled with water or mercury, one side of which is connected to the source of pressure. The amount of displacement of the liquid is a measure of the magnitude of the pressure.
## manometer
Any pressure-measuring instrument. Typically, it consists of a calibrated glass tube filled with liquid and linked to a remote pressure source. Air pressure forces the liquid up in the glass tube and its magnitude is measured against a calibrated scale.
References in periodicals archive ?
The current situation in SA is suboptimal and must be addressed at a national level so that manometers are available at all levels of care.
The purchase of aneroid or other reliable manometers that can be calibrated and serviced on a regular basis, together with strict measurement protocols can offer the safety and efficacy needed for clinical dental hygiene treatment.
The actual intracuff pressure was then measured using a handheld manometer and the readings were recorded.
Using proprietary bladder technology, on which a patent is pending, the PULMANEX(R) Disposable Airway Pressure Manometer measures airway pressure utilizing a multi-positional adaptor which allows the clinician the flexibility to position the monitor to provide a view of both the patient and the pressure readings simultaneously.
METHODS * We randomly selected 25 pharmacies and compared blood pressure readings obtained from their automated machines with from a mercury manometer. We used 3 volunteers with arm circumferences at the low, medium, and high ends of the acceptable range of a normal adult cuff size.
The box contained a very sensitive anaeroid capsule which communicated with a rubber arm cuff enclosed in canvas; the apparatus also had a pump, an escape valve and an anaeroid manometer to indicate the pressure.
Yarows and Brook (2000) found little difference between BPs taken by 12 automated devices on 2 normo-tensive subjects when compared with those of mercury or aneroid and validated automated manometers. Also found were similar ranges of deviation between traditional and automated methods of measurement, indicating operator interpretation during traditional methods are as variable as output by automated devices with digital displays.
The proposal for a Directive (to amend the existing Directive 76/769/EEC to either restrict or prevent the marketing of substances or products which may adversely affect public health) - bans the marketing of mercury in new fever and room thermometers, barometers, blood pressure gauges and manometers and sphygmomanometers.
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F, which is the common vertex of the triangles: that is », together with four right angles. Therefore all the angles of the figure, together with four right angles are equal to twice as many right angles as the figure has sides.
The Elements of Euclid, the parts read in the University of Cambridge [book ... - Side 33
av Euclides - 1846
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## The Young Surveyor's Guide: Or, A New Introduction to the Whole Art of ...
Edward Laurence - 1716 - 375 sider
...external < a, is equal to t wo right Angles (by tbe^tb /)confequently all the internal and external Angles are equal to twice as many right Angles as the Figure has fides. But all its internal Angles are equal to twiee as many right Angles eicept 4 as it has fides...
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## The Young Gentleman's Arithmetick, and Geometry: Containing Such Elements of ...
Edward Wells - 1723 - 294 sider
...the Sum of all the Angles in all the Tri~ angles, into which the Figure is divided, will together be equal to twice as ma-ny right Angles, as the Figure has Sides. But the Angles about P, the inward Point of each Figure, wherein all the Triangles concur, are (by...
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## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - 1762 - 466 sider
...is the common Vertex of the triangles ; that is ", toM. i. gether with four right angles. Therefore all the angles of the figure^ together with four right...equal to twice as many right angles as the figure has fides. C o R. 2 . All the exterior angles of any rectilineal figure are together equal to four right...
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## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books
Euclid - 1765 - 464 sider
...(for every whole ig equal to all its parts taken together) therefore .all the angles of a right-lined figure, together with four right angles, are equal to twice as many right angles as the figure has fides. And taking away four right angles from each, there will remain all the angles of the figure...
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## A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the ...
Thomas Malton - 1774 - 440 sider
...by the Sides. ie equal to four Right Angles. And, all the internal Angles of any Right-lined Figure are equal to twice as many Right Angles as the Figure has Sides, wanting four, (Th. i. i0. i.) confequently, the external Angles being equal to thofe four (Th. 2. of...
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## The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ...
Robert Simson - 1775 - 520 sider
...F, which is the common vertex of the triangles; that is", together with four right angles. Therefore all the angles of the figure, together with four right...equal to twice as many right angles as the figure has fides. CoR. 2. All the exterior angles of any re&ilineal figure, are together equal to four right angles....
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## The Geometrician: Containing Essays on Plane Geometry, and Trigonometry ...
Benjamin Donne - 1775 - 159 sider
...; that is, ь together with four Rt. L. s. Therefore this Corollary is true. »68. 88. COROLL. 10. All the exterior Angles of any rectilineal Figure are together equal to four Right Angles. Pl.iF24. Becaufe every interior ¿.ABC, with its adjacent „ 6 exterior ¿_ ABD, is a — 2 Rt. ¿.s...
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## The First Six Books: Together with the Eleventh and Twelfth
Euclid - 1781 - 520 sider
...F, which is the common vertex of the triangles; that isa, together with four right angles. Thprefpre all the angles of the figure, together with four right...equal to twice as many right angles as the figure has fides. CoR. 2. All the exterior angles of any rectilineal figure, arc together equal to four right...
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## A Complete Treatise on Practical Mathematics: Including the Nature and Use ...
John McGregor (teacher of mathematics.) - 1792 - 431 sider
...angles of any regular polygon* 3By cor. i ft, I. 32. Euclid. All the anterior angles of any reoilineal figure, together with four right angles, are equal to twice as many right angles as the figure has fides. Hence the following rule. RULÉ. From double thé number of fides f übt vail: 4, and the remainder...
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## Elements of Geometry;: Containing the First Six Books of Euclid, with Two ...
John Playfair - 1795 - 400 sider
...which B the common vertex of the triangles : that is a, together witi four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many rigit angles as the fi rurehas fides. CoR. 2. All the exterior angles of any reftilineal figure an...
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The gear is a toothed wheel, and its meshing action transmits the rotation of one shaft to other shafts, and transmits motion or torque between the two shafts. There are many types of gears. According to the axiality of the gears, there are parallel axis, intersecting axis, and non-parallel and non-intersecting axis.
Modulus: a pointer indicating the size of a gear. The modulus of a pair of engaged gears must be the same. Otherwise, the two gears have different tooth specifications and cannot run smoothly. The modulus can be defined as: the ratio of the pitch diameter and the number of teeth, the unit is usually (mm), the modulus is usually used in the metric system. and in the inch system, the same position is the diameter pitch, that is, the number of gear teeth and pitch and diameter The ratio of is in the sense the reciprocal of the modulus. The unit is (teeth / inch). Modulus: m = D / N Diameter section: Pd = N / D
Pitch circle: It is a theoretical circle. In a pair of meshed gears, the pitch circle must be tangent to each other
Pitch diameter: is the diameter of the pitch circle, D
Pressure line: When the two gears are in contact, the direction perpendicular to the contact surface is that, usually the forward load is transmitted along the direction of the pressure line.
Pressure angle PA: This is the angle between the common tangent of the two gear pitch circles and the pressure line. The general pressure angles are 14.5 degrees and 20 degrees. Today, more than 20 pressure angles are used to prevent interference.
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https://www.doubtnut.com/qna/96593107 | 1,716,251,955,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00353.warc.gz | 642,019,116 | 36,217 | # Find the general solution of : cosx+sinx=1
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Step by step video, text & image solution for Find the general solution of : cosx+sinx=1 by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Similar Practice Problems
• Question 1 - Select One
## General solution of cosx=1 is
Anπ,nZ
B2nπ,nZ
C3nπ,nZ
D4nπ,nZ
• Question 1 - Select One
## General solution of cosx−sinx=1 is
A2nπ,2nππ2,nZ
B2nπ,2nπ+π2,nZ
C2nπ,2nππ4,nZ
D2nπ,2nπ+π4,nZ
• Question 1 - Select One
## Find the general solution of each of the following differential equations: cosx(1+cosy)dx−siny(1+sinx)dy=0
A(1+sinx)(1+cosy)=C
B(1sinx)(1+cosy)=C
C(1sinx)(1cosy)=C
D(1+2sinx)(3+cosy)=C
• Question 1 - Select One
## If n is any integer, then the general solution of the equation cosx−sinx=1√2 is
Ax=2nππ12orx=2nπ+7π12
Bx=nπ±π12
Cx=2nπ+π12orx=2nπ7π12
Dx=nπ+π12orx=nπ7π12
### Similar Questions
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# Why are so people so concerned about the number of miles your food travel to reach your plate?
Updated: 9/21/2023
Wiki User
12y ago
because they dont want it to be cold i think :)
Wiki User
12y ago
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Q: Why are so people so concerned about the number of miles your food travel to reach your plate?
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Houses got bombed, and millions of people had to travel miles for work.
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### If you travel at a constant speed and the distance you travel varies directly with your travel timeSuppose in 1.5 hours you can walk 5 miles What is the costant of variation and how did you find this?
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### How fast did caravans travel?
The speed of travel depended much on the terrain and the number in the caravan. They averaged eight miles per day but could travel as fast as twelve miles in a day or be stalled a day without motion. The speed of travel depended much on the terrain and the number in the caravan. They averaged eight miles per day but could travel as fast as twelve miles in a day or be stalled a day without motion.
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110
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### How long does it take to travel 39.20 miles?
The number of hours it takes is39.2/your speed, in miles per hour . | 558 | 2,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.9477 |
https://stacks.math.columbia.edu/tag/03ER | 1,718,628,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00855.warc.gz | 480,048,942 | 6,000 | Definition 18.28.1. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings.
1. A presheaf $\mathcal{F}$ of $\mathcal{O}$-modules is called flat if the functor
$\textit{PMod}(\mathcal{O}) \longrightarrow \textit{PMod}(\mathcal{O}), \quad \mathcal{G} \mapsto \mathcal{G} \otimes _{p, \mathcal{O}} \mathcal{F}$
is exact.
2. A map $\mathcal{O} \to \mathcal{O}'$ of presheaves of rings is called flat if $\mathcal{O}'$ is flat as a presheaf of $\mathcal{O}$-modules.
3. If $\mathcal{C}$ is a site, $\mathcal{O}$ is a sheaf of rings and $\mathcal{F}$ is a sheaf of $\mathcal{O}$-modules, then we say $\mathcal{F}$ is flat if the functor
$\textit{Mod}(\mathcal{O}) \longrightarrow \textit{Mod}(\mathcal{O}), \quad \mathcal{G} \mapsto \mathcal{G} \otimes _\mathcal {O} \mathcal{F}$
is exact.
4. A map $\mathcal{O} \to \mathcal{O}'$ of sheaves of rings on a site is called flat if $\mathcal{O}'$ is flat as a sheaf of $\mathcal{O}$-modules.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 405 | 1,155 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.581785 |
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Solving Linear Equations In One Variable Involving Fractions Worksheet. Solve linear equations with rational coefficients by multi plying by the least common denominator to clear the fractions. Steps for solving a linear equation in one variable:
Solving a system of linear equations in one variable the linear equations in one variable are represented in the form ax+b = 0 where x is a variable, a is a coefficient and b is a constant. The easy teacher worksheets will allow you to practice and learn linear equations. Solving systems of equations word problems worksheet pdf
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3 ways to solve one step equations solving involving with fractions algebra linear math. If a line goes through a number and a letter put the letter in the numbered box below and watch the answer to the riddle appear. Solving multi step equations worksheet answers pre.
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Algebra solving linear equations in one variable lesson 1 3 of 4 fraction you. Solving linear equations with fractions worksheet pdf. Count on our printable 6th grade math worksheets with answer keys for a thorough practice. | 427 | 2,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.894565 |
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# Convolution of Bitmaps
, 29 Mar 2006
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Article discussing how to convolve images, specificially the convolution of bitmaps.
## Introduction
Image convolution plays an important role in computer graphics applications. Convolution of images allows for image alterations that enable the creation of new images from old ones. Convolution also allows for important features such as edge detection, with many widespread uses. The convolution of an image is a simple process by which the pixel of an image is multiplied by a kernel, or masked, to create a new pixel value. Convolution is commonly referred to as filtering.
## Details
First, for a given pixel (x,y), we give a weight to each of the surrounding pixels. It may be thought of as giving a number telling how important the pixel is to us. The number may be any integer or floating point number, though I usually stick to floating point since floats will accept integers as well. The kernel or mask that contains the filter may actually be any size (3x3, 5x5, 7x7); however, 3x3 is very common. Since the process for each is the same, I will concentrate only on the 3x3 kernels.
Second, the actual process of convolution involves getting each pixel near a given pixel (x,y), and multiplying each of the pixel's channels by the weighted kernel value. This means that for a 3x3 kernel, we would multiply the pixels like so:
```(x-1,y-1) * kernel_value[row0][col0]
(x ,y-1) * kernel_value[row0][col1]
(x+1,y-1) * kernel_value[row0][col2]
(x-1,y ) * kernel_value[row1][col0]
(x ,y ) * kernel_value[row1][col1]
(x+1,y ) * kernel_value[row1][col2]
(x-1,y+1) * kernel_value[row2][col0]
(x ,y+1) * kernel_value[row2][col1]
(x+1,y+1) * kernel_value[row2][col2]```
The process is repeated for each channel of the image. This means that the red, green, and blue color channels (if working in RGB color space) must each be multiplied by the kernel values. The kernel position is related to the pixel position it is multiplied by. Simply put, the kernel is allocated in `kernel[rows][cols]`, which would be `kernel[3][3]` in this case. The 3x3 (5x5 or 7x7, if using a larger kernel) area around the pixel (x,y) is then multiplied by the kernel to get the total sum. If we were working with a 100x100 image, allocated as `image[100][100]`, and we wanted the value for pixel (10,10), the process for each channel would look like:
```float fTotalSum =
Pixel(10-1,10-1) * kernel_value[row0][col0] +
Pixel(10 ,10-1) * kernel_value[row0][col1] +
Pixel(10+1,10-1) * kernel_value[row0][col2] +
Pixel(10-1,10 ) * kernel_value[row1][col0] +
Pixel(10 ,10 ) * kernel_value[row1][col1] +
Pixel(10+1,10 ) * kernel_value[row1][col2] +
Pixel(10-1,10+1) * kernel_value[row2][col0] +
Pixel(10 ,10+1) * kernel_value[row2][col1] +
Pixel(10+1,10+1) * kernel_value[row2][col2] +```
Finally, each value is added to the total sum, which is then divided by the total weight of the kernel. The kernel's weight is given by adding each value contained in the kernel. If the value is zero or less, then a weight of 1 is given to avoid a division by zero.
The actual code to convolve an image is:
```for (int i=0; i <= 2; i++)//loop through rows
{
for (int j=0; j <= 2; j++)//loop through columns
{
//get pixel near source pixel
/*
if x,y is source pixel then we loop through and
get pixels at coordinates:
x-1,y-1
x-1,y
x-1,y+1
x,y-1
x,y
x,y+1
x+1,y-1
x+1,y
x+1,y+1
*/
COLORREF tmpPixel = pDC->GetPixel(sourcex+(i-(2>>1)),
sourcey+(j-(2>>1)));
//get kernel value
float fKernel = kernel[i][j];
//multiply each channel by kernel value, and add to sum
//notice how each channel is treated separately
rSum += (GetRValue(tmpPixel)*fKernel);
gSum += (GetGValue(tmpPixel)*fKernel);
bSum += (GetBValue(tmpPixel)*fKernel);
//add the kernel value to the kernel sum
kSum += fKernel;
}
}
//if kernel sum is less than 0, reset to 1 to avoid divide by zero
if (kSum <= 0)
kSum = 1;
//divide each channel by kernel sum
rSum/=kSum;
gSum/=kSum;
bSum/=kSum;```
The source code included performs some common image convolutions. Also included is a Convolve Image menu option that allows users to enter their own kernel. Common 3x3 kernels include:
```gaussianBlur[3][3] = {0.045, 0.122, 0.045, 0.122,
0.332, 0.122, 0.045, 0.122, 0.045};
gaussianBlur2[3][3] = {1, 2, 1, 2, 4, 2, 1, 2, 1};
gaussianBlur3[3][3] = {0, 1, 0, 1, 1, 1, 0, 1, 0};
unsharpen[3][3] = {-1, -1, -1, -1, 9, -1, -1, -1, -1};
sharpness[3][3] = {0,-1,0,-1,5,-1,0,-1,0};
sharpen[3][3] = {-1, -1, -1, -1, 16, -1, -1, -1, -1};
edgeDetect[3][3] = {-0.125, -0.125, -0.125, -0.125,
1, -0.125, -0.125, -0.125, -0.125};
edgeDetect2[3][3] = {-1, -1, -1, -1, 8, -1, -1, -1, -1};
edgeDetect3[3][3] = {-5, 0, 0, 0, 0, 0, 0, 0, 5};
edgeDetect4[3][3] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
edgeDetect5[3][3] = {-1, -1, -1, 2, 2, 2, -1, -1, -1};
edgeDetect6[3][3] = {-5, -5, -5, -5, 39, -5, -5, -5, -5};
sobelHorizontal[3][3] = {1, 2, 1, 0, 0, 0, -1, -2, -1 };
sobelVertical[3][3] = {1, 0, -1, 2, 0, -2, 1, 0, -1 };
previtHorizontal[3][3] = {1, 1, 1, 0, 0, 0, -1, -1, -1 };
previtVertical[3][3] = {1, 0, -1, 1, 0, -1, 1, 0, -1};
boxBlur[3][3] = {0.111f, 0.111f, 0.111f, 0.111f,
0.111f, 0.111f, 0.111f, 0.111f, 0.111f};
triangleBlur[3][3] = { 0.0625, 0.125, 0.0625,
0.125, 0.25, 0.125, 0.0625, 0.125, 0.0625};```
Last but not least is the ability to show a convoluted image as a grayscale result. In order to display a filtered image as grayscale, we just add a couple lines to the bottom of the `Convolve` function:
```//return new pixel value
if (bGrayscale)
{
int grayscale=0.299*rSum + 0.587*gSum + 0.114*bSum;
rSum=grayscale;
gSum=grayscale;
bSum=grayscale;
}
clrReturn = RGB(rSum,gSum,bSum);```
This means that the entire `Convolve` function now looks like:
```COLORREF CImageConvolutionView::Convolve(CDC* pDC, int sourcex,
int sourcey, float kernel[3][3], int nBias,BOOL bGrayscale)
{
float rSum = 0, gSum = 0, bSum = 0, kSum = 0;
COLORREF clrReturn = RGB(0,0,0);
for (int i=0; i <= 2; i++)//loop through rows
{
for (int j=0; j <= 2; j++)//loop through columns
{
//get pixel near source pixel
/*
if x,y is source pixel then we loop
through and get pixels at coordinates:
x-1,y-1
x-1,y
x-1,y+1
x,y-1
x,y
x,y+1
x+1,y-1
x+1,y
x+1,y+1
*/
COLORREF tmpPixel = pDC->GetPixel(sourcex+(i-(2>>1)),
sourcey+(j-(2>>1)));
//get kernel value
float fKernel = kernel[i][j];
//multiply each channel by kernel value, and add to sum
//notice how each channel is treated separately
rSum += (GetRValue(tmpPixel)*fKernel);
gSum += (GetGValue(tmpPixel)*fKernel);
bSum += (GetBValue(tmpPixel)*fKernel);
//add the kernel value to the kernel sum
kSum += fKernel;
}
}
//if kernel sum is less than 0, reset to 1 to avoid divide by zero
if (kSum <= 0)
kSum = 1;
//divide each channel by kernel sum
rSum/=kSum;
gSum/=kSum;
bSum/=kSum;
//add bias if desired
rSum += nBias;
gSum += nBias;
bSum += nBias;
//prevent channel overflow by clamping to 0..255
if (rSum > 255)
rSum = 255;
else if (rSum < 0)
rSum = 0;
if (gSum > 255)
gSum = 255;
else if (gSum < 0)
gSum = 0;
if (bSum > 255)
bSum = 255;
else if (bSum < 0)
bSum = 0;
//return new pixel value
if (bGrayscale)
{
int grayscale=0.299*rSum + 0.587*gSum + 0.114*bSum;
rSum=grayscale;
gSum=grayscale;
bSum=grayscale;
}
clrReturn = RGB(rSum,gSum,bSum);
return clrReturn;
}```
Last but not least, I did a little tweaking to get the program to load a default image from a resource (`IDB_BITMAP1`). Then, I added the ability to convolve this default image. The program will still load image from a file, the only difference is that it will now show a default image at startup.
Please note that this article is, by no means, an example of fast processing of pixels. It is merely meant to show how convolution can be done on images. If you would like a more advanced image processor, then feel free to email me with the subject "WANT CODE:ImageEdit Please". That is an unreleased image processor I have done, though parts are not implemented yet due to lack of time, that contains much more functionality, using the CxImage library as its basis for reading and saving images.
This article has no explicit license attached to it but may contain usage terms in the article text or the download files themselves. If in doubt please contact the author via the discussion board below.
A list of licenses authors might use can be found here
## About the Author
Web Developer United States
Programming using MFC and ATL for almost 12 years now. Currently studying Operating System implementation as well as Image processing. Previously worked on DSP and the use of FFT for audio application. Programmed using ADO, ODBC, ATL, COM, MFC for shell interfacing, databasing tasks, Internet items, and customization programs.
## Comments and Discussions
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Does not compile in MSVC 2003 Dark Alchemist6-Jan-06 9:27 Dark Alchemist 6-Jan-06 9:27
```Compiling... StdAfx.cpp WINVER not defined. Defaulting to 0x0501 (Windows XP and Windows .NET Server) Compiling... MainFrm.cpp MainFrm.cpp(110) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(111) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(112) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(113) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(114) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(115) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(116) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(117) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data MainFrm.cpp(118) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data ImageConvolutionView.cpp ImageConvolutionView.cpp(248) : warning C4244: 'initializing' : conversion from 'double' to 'int', possible loss of data ImageConvolutionView.cpp(249) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data ImageConvolutionView.cpp(250) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data ImageConvolutionView.cpp(251) : warning C4244: '=' : conversion from 'int' to 'float', possible loss of data ImageConvolutionDoc.cpp ImageConvolution.cpp ImageConvolution.cpp(58) : warning C4996: 'CWinApp::Enable3dControls' was declared deprecated C:\Program Files\Microsoft Visual Studio .NET 2003\Vc7\atlmfc\include\afxwin.h(4369) : see declaration of 'CWinApp::Enable3dControls' ConvolveDlg.cpp ConvolveDlg.cpp(18) : error C2143: syntax error : missing ')' before ',' ConvolveDlg.cpp(18) : error C2365: 'IDD' : redefinition; previous definition was a 'enumerator' c:\unzipped\ImageConvolution\ConvolveDlg.h(21) : see declaration of 'IDD' ConvolveDlg.cpp(18) : error C2244: 'IDD' : unable to match function definition to an existing declaration c:\unzipped\ImageConvolution\ConvolveDlg.h(21) : see declaration of 'IDD' definition 'IDD' existing declarations ConvolveDlg.cpp(18) : error C2059: syntax error : ')' ConvolveDlg.cpp(20) : error C2143: syntax error : missing ';' before '{' ConvolveDlg.cpp(20) : error C2447: '{' : missing function header (old-style formal list?) Generating Code... Build log was saved at "file://c:\unzipped\ImageConvolution\Release\BuildLog.htm" ImageConvolution - 6 error(s), 14 warning(s)```That says it all for MSVC .NET 2003 (7.1). So, how do I get this thing compiled?
Re: Does not compile in MSVC 2003 Aqiruse29-Mar-06 10:08 Aqiruse 29-Mar-06 10:08
Last Visit: 31-Dec-99 19:00 Last Update: 19-Feb-17 11:37 Refresh 1
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# Specific Object Recognition: Matching in 2D - PowerPoint PPT Presentation
Specific Object Recognition: Matching in 2D. engine model. image containing an instance of the model. Is there an engine in the image? If so, where is it located?. Alignment. Use a geometric feature-based model of the object. Match features of the object to features in the image.
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Specific Object Recognition: Matching in 2D
engine model
image containing an
instance of the model
Is there an engine in the image?
If so, where is it located?
Alignment
• Use a geometric feature-based model of the object.
• Match features of the object to features in the image.
• Produce a hypothesis h (matching features)
• Compute an affine transformationT from h
• Apply T to the features of the model to map the model features to the image.
• Use a verification procedure to decide how well the model features line up with actual image features
Alignment
model
image
Most often used:
2D Affine Transformations
• translation
• rotation
• scale
• skew
Point Representation and Transformations (review)
Normal Coordinates for a 2D Point
P = [x, y] =
Homogeneous Coordinates
P = [sx, sy, s] where s is a scale factor
x
y
t
t
Scaling
x´ cx 0 x cx * x
y´ 0 cy y cy * y
=
=
scaling by
a factor of 2
Rotation
x´ cos -sin x x cos - y sin
y´ sin cos y x sin + y cos
=
=
rotate point rotate axes
Translation
2 X 2 matrix doesn’t work for translation!
Here’s where we need homogeneous coordinates.
x´ 1 0 x0 x x + x0
y´ 0 1 y0 y y + y0
1 0 0 1 1 1
=
=
(x+x0, y + y0)
(x,y)
Rotation, Scaling and Translation
xw 1 0 x0 s 0 0 cos -sin 0 xi
yw = 0 1 y0 0 s 0 sin cos 0 yi
1 0 0 1 0 0 1 0 0 1 1
T S R
TR
P2=(x2,y2)
P3´=(u3,v3)
P2´=(u2,v2)
P1´=(u1,v1)
P1=(x1,y1)
P3=(x3,y3)
Given 3 matching pairs of points, the affine transformation
can be computed through solving a simple matrix equation.
u1 u2 u3a11 a12 a13 x1 x2 x3
v1 v2 v3 = a21 a22 a23 y1 y2 y3
1 1 1 0 0 1 1 1 1
A More Robust Approach
Using only 3 points is dangerous, because if even one is
off, the transformation can be far from correct.
Instead, use many (n =10 or more) pairs of matching
control points to determine a least squares estimate of
the six parameters of the affine transformation.
Error(a11, a12, a13, a21, a22, a23) =
((a11*xj + a12*yj + a13 - uj)2 +
(a21*xj + a22*yj + a23 - vj)2 )
j=1,n
What is this for?
• Many 2D matching techniques use it.
• 1. Local-Feature Focus Method
• 2. Pose Clustering
• 3. Geometric Hashing
Local-Feature-Focus Method
• Each model has a set of features (interesting points).
• -The focus features are the particularly detectable features,
• usually representing several different areas of the model.
• - Each focus feature has a set of nearby features that
• can be used, along with the focus feature, to compute
• the transformation.
focus feature
LFF Algorithm
Let G be the set of detected image features.
Let Fm be focus features of the model.
Let S(f) be the nearby features for feature f.
for each focus feature Fm
for each image feature Gi of the same type as Fm
1. find the maximal subgraph Sm of S(Fm) that matches a subgraph Si of S(Gi).
2. Compute transformation T that maps the points
of each feature of Sm to the corresponding one of Si.
3. Apply T to the line segments of the model.
4. If enough transformed segments find evidence in the image, return(T)
Pose Clustering
Let T be a transformation aligning model M with image object O
The pose of object O is its location and orientation, defined byT.
The idea of pose clustering is to compute lots of possible pose
transformations, each based on 2 points from the model and
2 hypothesized corresponding points from the image.*
Then cluster all the transformations in pose space and try
to verify the large clusters.
* This is not a full affine transformation, just RST.
Pose Clustering
Model
Image
Pose Clustering
Model
Image
Geometric Hashing
• This method was developed for the case where there is
• a whole database of models to try to find in an image.
• a large amount of offline preprocessing and
• a large amount of space
• for potentially fast online
• object recognition
• pose detection
Theory Behind Geometric Hashing
• A model M is a an ordered set of feature points.
3
1
2
M =
8
4
6
7
5
• An affine basis is any subset E={e00,e01,e10}
• of noncollinear points of M.
• For basis E, any point x M can be represented in
• affine coordinates(,).
• x = (e10 – e00) + (e01-e00) + e00
e01
x = (,)
e10
e00
Affine Transform
If x is represented in affine coordinates (,).
x = (e10 – e00) + (e01- e00) + e00
and we apply affine transform T to point x, we get
Tx =(Te10 – Te00) + (Te01-Te00) + Te00
In both cases, x has the same coordinates (,).
Example
transformed object
original object
Offline Preprocessing
For each model M
{
Extract feature point set FM
for each noncollinear triple E of FM (basis)
for each other point x of FM
{
calculate (,) for x with respect to E
store (M,E) in hash table H at index (,)
}
}
Hash Table
list of model /
basis pairs
M1, E1
M2, E2
.
.
Mn, En
E1 … EmOnline Recognition
M1
.
.
initialize accumulator A to all zero
extract feature points from image
for each basis triple F/* one basis */
for each other point v /* each image point */
{
calculate (,) for v with respect to F
retrieve list L from hash table at index (,)
for each pair (M,E) of L
A[M,E] = A[M,E] + 1
}
find peaks in accumulator array A
for each peak (M,E) in A
calculate and try to verify T : F = TE
Mk
(M,E)->T
Verification
How well does the transformed model line up
with the image.
Whole segments work better, allow less halucination,
but there’s a higher cost in execution time.
• compare positions of feature points
• compare full line or curve segments
2D Matching Mechanisms
• We can formalize the recognition problem as finding
• a mapping from model structures to image structures.
• Then we can look at different paradigms for solving it.
- interpretation tree search
- discrete relaxation
- relational distance
- continuous relaxation
Formalism
• A part (unit) is a structure in the scene,
• such as a region or segment or corner.
• A label is a symbol assigned to identify the part.
• An N-ary relation is a set of N-tuples defined over a
• set of parts or a set of labels.
• An assignment is a mapping from parts to labels.
Example
image
model
circle4
circle5
arc1
eye1
eye2
circle1
circle2
smile
circle3
What are the relationships?
What is the best assignment
of model labels to image features?
Consistent Labeling Definition
Given:
1. a set of units P
2. a set of labels for those units L
3. a relation RP over set P
4. a relation RL over set L
A consistent labeling f is a mapping f: P -> L satisfying
if (pi, pj) RP, then (f(pi), f(pj)) RL
which means that a consistent labeling preserves relationships.
Abstract Example
binary relation RL
b
binary relation RP
c
1
a
2
3
d
e
P = {1,2,3}
L = {a,b,c,d,e}
RP={(1,2),(2,1),(2,3)}
RL = {(a,c),(c,a),(c,b),
(c,d),(e,c),(e,d)}
One consistent labeling is {(1,a),(2,c),(3,d)
House Example
P
L
RP and RL are
connection relations.
f(S1)=Sj
f(S2)=Sa
f(S3)=Sb
f(S4)=Sn
f(S5)=Si
f(S6)=Sk
f(S10)=Sf
f(S11)=Sh
f(S7)=Sg
f(S8) = Sl
f(S9)=Sd
1. Interpretation Tree
• An interpretation tree is a tree that represents all
• assignments of labels to parts.
• Each path from the root node to a leaf represents
• a (partial) assignment of labels to parts.
• Every path terminates as either
• 1. a complete consistent labeling
• 2. a failed partial assignment
Interpretation Tree Example
b
RP
1
2
RL
a
c
3
root
e
d
(1,a)
(2,b) (2,c)
X
(3,b) (3,d) (3,e)
(1,2) RP
(a,b) RL
X
OK
OK
2. Discrete Relaxation
• Discrete relaxation is an alternative to (or addition to)
• the interpretation tree search.
• Relaxation is an iterative technique with polynomial
• time complexity.
• Relaxation uses local constraints at each iteration.
• It can be implemented on parallel machines.
How Discrete Relaxation Works
1. Each unit is assigned a set of initial possible labels.
2. All relations are checked to see if some pairs of labels
are impossible for certain pairs of units.
3. Inconsistent labels are removed from the label sets.
4. If any labels have been filtered out
then another pass is executed
else the relaxation part is done.
5. If there is more than one labeling left, a tree search
can be used to find each of them.
Example of Discrete Relaxation
RP
RL
L1
L2
L3
Pi
L1 L2 L3
X
L8
L6
Pj
L6 L8
There is no label in Pj’s label set that is connected to
L2 in Pi’s label set. L2 is inconsistent and filtered out.
3. Relational Distance Matching
• A fully consistent labeling is unrealistic.
• An image may have missing and extra features;
• required relationships may not always hold.
• Instead of looking for a consistent labeling,
• we can look for the best mapping from P to L,
• the one that preserves the most relationships.
b
1
2
a
c
3
Preliminary Definitions
Def: A relational description DP is a sequence of
relations over a set of primitives P.
• Let DA = {R1,…,RI} be a relational description over A.
• Let DB = {S1,…,SI} be a relational description over B.
• Let f be a 1-1, onto mapping from A to B.
• For any relation R, the composition Rf is given by
• Rf = {(b1,…,bn) | (a1,…,an) is in R and f(ai)=(bi), i=1,n}
Example of Composition
Rf = {(b1,…,bn) | (a1,…,an) is in R and f(ai)=(bi), i=1,n}
b
1
2
1 a
2 b
3 c
a
c
3
f
R
Rf
Rf is an isomorphic copy of R with nodes renamed by f.
Relational Distance Definition
Let DA be a relational description over set A,
DB be a relational description over set B,
and f : A -> B.
• The structural error of f for Ri in DA and Si in DB is
• E (f) = | Ri f - Si | + | Si f - Ri |
• The total error of f with respect to DA and DB is
• E(f) = E (f)
• The relational distance GD(DA,DB) is given by
• GD(DA,DB) = min E(f)
-1
i
S
I
i
S
i=1
f : A B, f is 1-1 and onto
Example
a
b
1
2
3
4
c
d
What is the best mapping?
What is the error of the best mapping?
ExampleLet f = {(1,a),(2,b),(3,c),(4,d)}
a
b
1
2
S
R
3
4
c
d
| Rf - S | = |{(a,b)(b,c)(c,d)(d,b)} - {(a,b)(b,c)(c,b)(d,b)} |
= |{(c,d)}| = 1
-1
|S f - R | = |{(1,2)(2,3)(3,2)(4,2)} - {(1,2)(2,3)(3,4)(4,2)}|
= |{(3,2)}| = 1
Is there a better mapping?
E(f) = 1+1 = 2
Variations
• Different weights on different relations
• Normalize error by dividing by total possible
• Attributed relational distance for attributed relations
• Penalizing for NIL mappings
Implementation
• Relational distance requires finding the lowest cost mapping from object features to image features.
• It is typically done using a branch and bound tree search.
• The search keeps track of the error after each object part is assigned an image feature.
• When the error becomes higher than the best mapping found so far, it backs up.
• It can also use discrete relaxation or forward checking (see Russel AI book on Constraint Satisfaction) to prune the search.
4. Continuous Relaxation
• In discrete relaxation, a label for a unit is either possible or not.
• In continuous relaxation, each (unit, label) pair has a probability.
• Every label for unit i has a prior probability.
• A set of compatibility coefficients C = {cij} gives the influence
• that the label of unit i has on the label of unit j.
• The relationship R is replaced by a set of unit/label compatibilities
• where rij(l,l´) is the compatibility of label l for part i with
• label l´ for part j.
• An iterative process updates the probability of each label for
• each unit in terms of its previous probability and the compatibilities
• of its current labels and those of other units that influence it.
Initialize probability of label l for part i to the a priori probability.
At stepk, compute a multiplicative factor based on looking at every
other part j, how muchi and jconstrain one another, the possible labels
for part j, their current probabilities and their compatability with label l.
Recognition by Appearance
• Appearance-based recognition is a competing paradigm to
• features and alignment.
• No features are extracted.
• Images are represented by basis functions (eigenvectors)
• and their coefficients.
• Matching is performed on this compressed image
• representation.
Eigenvectors and Eigenvalues
Consider the sum squared distance of a point x to all of the orange points:
What unit vector v minimizes SSD?
What unit vector v maximizes SSD?
Solution: v1 is eigenvector of A with largest eigenvalue
v2 is eigenvector of A with smallest eigenvalue
Principle component analysis
• Suppose each data point is N-dimensional
• Same procedure applies:
• The eigenvectors of A define a new coordinate system
• eigenvector with largest eigenvalue captures the most variation among training vectors x
• eigenvector with smallest eigenvalue has least variation
• We can compress the data by only using the top few eigenvectors
The space of faces
• An image is a point in a high-dimensional space
• An N x M image is a point in RNM
• We can define vectors in this space
Dimensionality reduction
The set of faces is a “subspace” of the set of images.
• We can find the best subspace using PCA
• Suppose it is K dimensional
• This is like fitting a “hyper-plane” to the set
• of faces
• spanned by vectors v1, v2, ..., vK
• any face x a1v1 + a2v2 + , ..., + aKvK
Turk and Pentland’s Eigenfaces: Training
• Let F1, F2,…, FM be a set of training face images.
• Let F be their mean and i = Fi – F.
• Use principal components to compute the eigenvectors
• and eigenvalues of the covariance matrix.
• M
• C = (1/M) ∑iiT
• i=1
• Choose the vector u of most significant M eigenvectors
• to use as the basis.
• Each face is represented as a linear combination of eigenfaces
• u = (u1, u2, u3, u4, u5); F27 = a1*u1 + a2*u2 + … + a5*u5
Matching
unknown
face image
I
convert to its
eigenface
representation
= (1, 2, …, m)
Find the face class k that minimizes
k = || - k ||
training
images
3 eigen-
images
mean
image
linear
approxi-
mations
Extension to 3D Objects
• Murase and Nayar (1994, 1995) extended this idea to 3D
• objects.
• The training set had multiple views of each object, on a
• dark background.
• The views included multiple (discrete) rotations of the object on
• a turntable and also multiple (discrete) illuminations.
• The system could be used first to identify the object and then to
• determine its (approximate)pose and illumination.
Significance of this work
• The extension to 3D objects was an important contribution.
• Instead of using brute force search, the authors observed that
• All the views of a single object, when transformed into the
• eigenvector space became points on a manifold in that space.
• Using this, they developed fast algorithms to find the closest
• object manifold to an unknown input image.
• Recognition with pose finding took less than a second.
Appearance-Based Recognition
• Training images must be representative of the instances
• of objects to be recognized.
• The object must be well-framed.
• Positions and sizes must be controlled.
• Dimensionality reduction is needed.
• It is not powerful enough to handle general scenes
• without prior segmentation into relevant objects.
• * The newer systems that use “parts” from interest operators
• are an answer to these restrictions.
Summary
• 2D object recognition for specific objects (usually industrial) is done by alignment.
-Affine transformations are usually powerful enough to handle objects that are mainly two-dimensional.
-Matching can be performed by many different “graph-matching” methodoloties.
-Verification is a necessary part of the procedure.
• Appearance-based recognition is another 2D recognition paradigm that came from the need for recognizing specific instances of a general object class, motivated by face recognition.
• Parts-based recognition combines appearance-based recognition with interest-region detection. It has been used to recognize multiple generic object classes. | 4,768 | 16,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-47 | latest | en | 0.853727 |
http://www.intmath.com/quadratic-equations/ans-1.php?a=2 | 1,369,425,991,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705020058/warc/CC-MAIN-20130516115020-00069-ip-10-60-113-184.ec2.internal.warc.gz | 459,212,625 | 1,585 | `2-1/x=3/(x+2)`
Multiply throughout by `x(x+2)` to remove the denominators (bottoms) of the fractions:
`2x(x+2)-(x(x+2))/x=(3(x)(x+2))/(x+2)`
Cancelling gives:
`2x(x+2)-(x+2)=3x`
Expanding the brackets:
`2x^2+4x-x-2=3x`
`2x^2-2=0`
`x^2-1=0`
Factoring gives:
`(x+1)(x-1)=0`
So `x = -1` or `x = 1`.
CHECK: Substituting `x = -1` into both the left hand side and right hand side of the question gives:
`{: ("LHS",=2-1/x),(,=2-1/-1), (,=3) :}`
`{: ("RHS",=3/(x+2)),(,=3/(-1+2)), (,=3),(,="LHS") :} `
Likewise, for `x = +1`,
LHS `= 2 - 1 = 1`
RHS `= 3/3 = 1 =` LHS | 279 | 575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.618817 |
https://www.playtaptales.com/numbers/octooctogintaquadringentillion/ | 1,555,773,264,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529839.0/warc/CC-MAIN-20190420140859-20190420162859-00443.warc.gz | 776,429,532 | 4,222 | A Octooctogintaquadringentillion (1 Octooctogintaquadringentillion) is 10 to the power of 1467 (10^1467). This is a tremendously astronomical number!
## How many zeros in a Octooctogintaquadringentillion?
There are 1,467 zeros in a Octooctogintaquadringentillion.
A Octooctogintaquadringentillionaire is someone whos assets, net worth or wealth is 1 or more Octooctogintaquadringentillion. It is unlikely anyone will ever be a true Octooctogintaquadringentillionaire. If you want to be a Octooctogintaquadringentillionaire, play Tap Tales!
## Is Octooctogintaquadringentillion the largest number?
Octooctogintaquadringentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
## Big Numbers
This is just one of many really big numbers! | 1,206 | 2,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.66757 |
http://caml.inria.fr/pub/ml-archives/caml-list/2002/07/e5ad30ad3d085e195c5e4d8b7ad94acd.en.html | 1,529,556,429,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00289.warc.gz | 52,512,064 | 5,065 | This site is updated infrequently. For up-to-date information, please visit the new OCaml website at ocaml.org.
[Caml-list] Caml productivity.
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Date: 2002-07-30 (16:50) From: eijiro_sumii@a... Subject: Re: Games (Re: [Caml-list] Caml productivity.)
```[Excuse me for citing a personal message without prior permission.
Also, excuse me for repeating all the same points as in my previous
messages...]
> I think you'll find that the emphasis on symbolic programming in ICFP was
> typical of ICFP rather than the supposed application domain.
I very much expected this response.:-) Indeed ICFP must have put
_more_ emphasis, I suppose, but it is of course just one example.
Also in some other CG programs involving Bezier curves, hidden
line/surface removal, radiosity, polygons, meta balls, etc. etc. that
I wrote, most non-trivial parts were symbolic ones. The numeric parts
were "mathematics" rather than "programming" - once I solved some
equations, the rest of the work was routine: it was just writing down
let x = (-. b +. sqrt(b *. b -. 4. *. a *. c)) /. (2. *. a)
or things like that.
Chatting with game/CG professionals around me _strengthened_ this
impression: the symbolic parts seem even more non-trivial in realistic
applications. For instance, just imagine implementing various kinds
of graphics editors or scripting languages for animation description!
Unfortunately, all of these are mere personal experience and
impression: just I believe that CG programs in the real world are more
symbolic than classroom exercises or laboratory experiments, for the
reasons that I wrote above and in my previous messages. If somebody
doesn't believe this, I have no more means (and time) to convince
him/her - this is just "FYI".
--
Eijiro Sumii (http://www.yl.is.s.u-tokyo.ac.jp/~sumii/)
Research Associate, Department of Computer Science, University of Tokyo
-------------------
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``` | 565 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-26 | latest | en | 0.928413 |
https://gateoverflow.in/1110/gate-cse-2003-question-23?show=362752 | 1,627,201,158,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00661.warc.gz | 292,768,437 | 32,183 | GATE CSE 2003 | Question: 23
19.9k views
In a min-heap with $n$ elements with the smallest element at the root, the $7^{th}$ smallest element can be found in time
1. $\Theta (n \log n)$
2. $\Theta (n)$
3. $\Theta(\log n)$
4. $\Theta(1)$
in DS
edited
3
But exact question was "in a heap of size n the 7th smallest element can be found in ?" It is not mentioned anywhere it is min heap or max heap so it should be Θ(n).
4
so just extract root element no do it 7 times
and after each extract take log n so total 7 * log n = O(log n )
0
If it is extracted n times and each time it takes logn time, then why not its nlogn
0
Digvijay Pandey "In a heap with n elements with the smallest element at root"
0
It is given smallest element at the root so it might be minheap
0
Why the complexity is not O(log n).
Because after finding the every smallest element from the min heap we have to heapify the tree too.
0
https://stackoverflow.com/questions/50940998/finding-the-7th-smallest-element-in-min-heap
@Digvijay Pandey Why cant't we use Heapsort to find time complexity?
0
Heapsort takes $O(n log n)$ time, which is inefficient for this purpose
Time to find the smallest element on a min-heap- one retrieve operation - $\Theta(1)$
Time to find the second smallest element on a min-heap- requires $2^2 - 1 = 3$ check operations to find the second smallest element out of $3$ elements - $\Theta(1)$
Time to find the $7^{th}$ smallest element - requires $O(2^7-1) = O(127)$ check operations to find the seventh smallest element out of $127$ possible ones - $\Theta(1)$
In short if the number of required operations is independent of the input size $n$, then it is always $\Theta(1)$.
(Here, we are doing a level order traversal of the heap and checking the elements)
If we are not allowed to traverse the heap and allowed only default heap-operations, we will be restricted with doing Extract-min $7$ times which would be $O(\log n)$.
Correct Answer: $D$
edited
23
If elements are repeated, then also this holds true. Consider the heap array
1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 4 5 6 7
The given procedure returns "1" and that is the correct answer. "7" must be returned only if the questions asks "7th" distinct smallest element which might require entire scanning of the heap and needs a hashtable to be run in $O(n)$ time.
3
how will it be constant time? If 7th smallest element is in last level then we need to compare all elements in heap. Then there will be O(n) time required.
If we delete root element (smallest element) and do heapify (requires log n time).
Do it for 7 time to find 7th smallest element. Total time= 7 * log n = log n .
0
How 7th smallest element can be in the last level of a min-heap?
0
7th smallest element can be the largest one in heap. For example 1 3 9 5 13 11 12 is a list and 13 is 7th smallest element. If we construct heap then 13 will be in last level.
1
It can go only till the 7th level. So, the number of checks is bounded by 2= 127.
2
Why is this 2 -1?
Suppose in a min heap of height 2,we can find the second min in max 2 comparisons.Plz explain anyone
0
5
How is it 3 check operations for finding 2nd minimum element , for finding 1st minimum , we don't require any comparison since we can retrieve it immediately , for retrieving the 2nd minimum element , we require 1 comparison , which is between the two child nodes of the root node and for computing the 3rd minimum we require 2 comparisons since either it will be one among the sibling of 2nd minimum or any child of 2nd minimum element , so likewise series will be formed for finding kth minimum element like
1+2+3+...+k-1=k(k-1)/2 comparisons , although it is also a constant but still value must be considered , so please clarify how u came up with ur logic and what is wrong in logic stated by me .
1
I think ans is o(1), correct but following logic u came up with is wrong. We can find second smallest in only 1 comparision, 3rd smallest requires 2 more comparisions.
20
The answer to this question completely depends on how we phrase out our query. If we were to find the 99th element in a min-heap of 100 elements, it will take $O(1)$. However, if were asked to find the $(n-1)^{th}$ smallest element in that same min-heap, will take $O(\log{n})$.
16
^Well yes, thats how theory is defined. And in your n case, if its added that n is constant (a fixed number), it becomes $O(1)$. The complexity is defined in terms of input- whatever is not part of input can be taken as constant. If $n$ is an input- it becomes $O(\log n)$.
47
we have to find the 7th smallest element. Heap is min heap so smallest is root itself.
For 2nd smallest compare both children of the root. whichever is smallest that will be the 2nd smallest element. Let's say Left child is smaller than the right child so, 2nd smallest is the Left child of the root.
for 3rd smallest compare left and right child of 2nd smallest and sibling of 2nd smallest. total 3 elements. Find the minimum of these three and declare it as 3rd min. let sibling of 2nd min is the smallest so it will be 3rd min.
for 4th minimum compare left and right child 2nd minimum and left and right child of the 3rd minimum. total of four elements. find minimum. declare it as the 4th minimum.
up to the 7th minimum.
complexity = 0 (for minimum no comparison) + 1 (for 2nd min) + 2 (for 3rd minimum) + 3 (for 4th minimum) ..........
= 0+1+2+3.......+6 = 21 comparison. which is constant.
O(1) time only.
0
@Arjun Sir what does it mean by
"If we are not allowed to traverse the heap and allowed only default heap-operations, we will be restricted with doing Extract-min 7 times which would be O(logn). " in the best answer given above..
0
In order to find the 7 th element one must remove elements one by one from the heap, until the 7 th smallest element is found.
As a heap may contain duplicate values, we may need to remove more then 7 elements. At the worst case we will have to remove
all n elements.
1 deletion takes O(log(n)). Hence ‘n’ deletes will take O(nlog(n)).
0
There is no question 36 or I cannot see Questions after 27.
And see this http://algs4.cs.princeton.edu/24pq/
It says priority queue (which indeed min/max heap ) with duplicate gives any one of largest value for delete_max () operation
So similarly for min heap with duplicate elements it will perform deletemin operation 7 times and each time within logN time it will give minmum element of the min heap. LogN will be taken for heapify.
Taking into consideration what you said then there should be a method to compare previously deleted element with newly deleted element to be sure that both are not identical. And you get truly 7th smallest element (if exists)
0
can you please explain waht 3 check operations will be performed ?
1
@Arjun Sir
I have a doubt sir if say in our array there is no 7th smallest element say our array contain [1,1,1,1,1,1,1,1,1,1]
now in this case we need to do extract min operation n number of times which would give nlogn time?
Plz Correct me Sir
0
But How do you know it is a binary heap..??
2
according to question it is min-heap so always the smallest among all the elements is in the root after applying heapify.
Now 1st delete the 1st smallest element which will result an exchange of the last element of the heap(here, think as element is stored in an array) with the root, now this element can't stay in the root because of heapify which will bring 2nd smallest elements up in the root.It takes O(logn) time.
this procedure will continue upto deletion of 6th smallest element.after that heapify will bring up the 7th smallest element in the root.Now we can find the 7th smallest element by in the root at O(1) time.
So, total time complexity = 6 * O(logn) + O(1) = O(logn)
0
0
@Digvijay Pandey
since I'm an avg student this is the best explanation that i have found but i think 2^7-1 = 127 will be the all possible possibilities for finding the 7th smallest element in heap.(I'm considering all elements are distinct in heap)
but what if if they give kth smallest element :-
then there will be two cases : k is independent to n i.e. k is constant hence again theta(1).
second k is O(n) then time complexity would be O(logn)
and one more thing since they just asked the "Found element or search it " but what if they ask extract it hence kth smallest element extraction operation will take (if k is O(n) ) will be k*log n or (n logn )
isn't it ?
0
sir, why did you take "if we are allowed to traverse root" as a default case? Shouldn't the default be the extract_min() one since that's the default-allowed-operation for a heap?
I mean, how to know if we're allowed to traverse children in a heap or not?
In worst case, we will not be right? So, shouldn't the extract_min() be the default one?
I'm so confused :(
Edit -
There's $O(k)$ algorithm for finding $k^{th}$ smallest element in a min-heap that uses array representation, which makes option (D) the correct choice.
0
@Arjun $sir$,
In a min-heap with n elements
1) The $7^{th}$ smallest element can be found in time, if duplicates are allowed: $O(1)$
2) The $7^{th}$ distinct smallest element can be found in time, If duplicates are allowed: $O(n)$.
Why we are getting $O(n)$ for the second case. Please do clarify.
1
@Kushagra गुप्ता In the second case if 122345 is in the heap and they want to know 3rd distinct smallest element then it will be 3 not 2. Similarly in worst case heap can be 1222223, in this case to find 3rd smallest element you have to traverse whole heap.
0
When comparisions are 21 then what is 127?The number of positions we come across? @Digvijay Pandey
0
Could you please explain how you are getting O(log n ) for (n-1)th element?
By using default heap operations only, we get O(log n), but if we are doing heap traversal then, searching elements should take O(n) time?
0
duplicates are not allowed, What will be the answer if we were asked to find the
1. $\\ (n-7)^{th} \\$ smallest element
2. $\\ \frac{n}{7}^{th}\\$ smallest element
1
The 7th most element is always within the 7 levels from the root so we have constant number of nodes in worst condition .we need to check only constant number of nodes to find the 7th smallest number so constant time
41
Key point: The kth smallest element cannot be deeper than the kth level of the heap, since the path from it to the root must go through elements of decreasing value.
2
@Sachin Mittal. Bro from where did u read this fact? Plz mention the reference :)
0
@Sachin Mittal 1 If input is like 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 3 4, Now we want to find 3rd smllest then your statement will become False. Please correct me if i am wrong?
1
even this is an old thread but i want clarification for concept building
since kth smallest element never be beyond to kth level it is true but i think we have to add one more restriction to it, like there are all elements are distinct isn't it ?
The best known algorithm to find Kth smallest element in Min Heap takes - O(K*logK) time and here K = 7th element.
Thus we get O(7*log7) = O(1) as Answer.
0
But it required you to maintain a heap of size k itself but in the given question they have mentioned that the heap size is n and not 7.
1 vote
3 approaches....
1st- Selection Sort
7 passes to find 7th minimum i.e. 7 × n = O(n)
2nd- Deletion from heap
Deletion of 1 element is logn
Deletion of 7 elements is 7 ×logn = O(logn)
3rd- sum of 1st 6 natural number in heap
O(1)
Among all 3 approaches 3rd one is best so option d is the answer.
0
What do you mean by 1st 6 natural numbers in heap? Can u elaborate?
The question is quite ambiguous.
If the question means
i) it's a min-heap then answer would be O(1), because at worst case we need to check the roots at depth 7, and as we know asymptotic complexity tells the order of growth of time wrt 'n' or input size, here irrespective of n, if it's greater than 7 we only need to check at depth 7 at worst case as I said, even if n is 10, 100, 1000 and so on. So, it's not depending on 'n'.
ii) it's just a binary tree with the least element at the root of the tree then T would be O(nlogn) because at worst case we need to search the entire tree.
Sine the 7th smallest element will be from ciel( a[7/2] to a[7] hence only “k” comparison are required thus 0(1) is the answer
ago
0
@rish1602
We have to find the 7th smallest element.
Heap is min heap so smallest is root itself.
For 2nd smallest compare both children of the root. whichever is smallest that will be the 2nd smallest element.
Let's say Left child is smaller than the right child so, 2nd smallest is the Left child of the root. for 3rd smallest compare left and right child of 2nd smallest and sibling of 2nd smallest. total 3 elements.
Find the minimum of these three and declare it as 3rd min.
let sibling of 2nd min is the smallest so it will be 3rd min.
for 4th minimum compare left and right child 2nd minimum and left and right child of the 3rd minimum.
total of four elements.
O(1) time only....
The search of 7th smallest element is independent on the input size n so it takes constant time for every input n O(n)
0
As heap is minheap already indicating the question as smallest element at the root so to find the 7th smallest from that minheap it will take O(1) i.e. theta(1) time so (d) is correct.
0
if question is about to find nth smallest element then it will be O(n)???????
1
lg(n)
Related questions
1
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Consider the function $f$ defined below. struct item { int data; struct item * next; }; int f(struct item *p) { return ((p == NULL) || (p->next == NULL)|| ((p->data <= p ->next -> data) && f(p->next))); } For a given linked list ... in non-decreasing order of data value the elements in the list are sorted in non-increasing order of data value not all elements in the list have the same data value
2
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Let S be a stack of size $n \geq1$. Starting with the empty stack, suppose we push the first n natural numbers in sequence, and then perform $n$ pop operations. Assume that Push and Pop operations take $X$ seconds each, and $Y$ seconds elapse between the end of one such stack operation and the ... $n(X+Y)$ $3Y+2X$ $n(X+Y)-X$ $Y+2X$
3
11.8k views
A data structure is required for storing a set of integers such that each of the following operations can be done in $O(\log n)$ time, where $n$ is the number of elements in the set. Deletion of the smallest element Insertion of an element if it is not already ... tree can be used but not a heap Both balanced binary search tree and heap can be used Neither balanced search tree nor heap can be used | 3,965 | 14,643 | {"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} | 3.890625 | 4 | CC-MAIN-2021-31 | latest | en | 0.817612 |
https://www.unitconverters.net/pressure/break-to-picopascal.htm | 1,669,934,793,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00672.warc.gz | 1,115,810,179 | 3,325 | Home / Pressure Conversion / Convert Break to Picopascal
# Convert Break to Picopascal
Please provide values below to convert break to picopascal [pPa], or vice versa.
From: break To: picopascal
### Break to Picopascal Conversion Table
BreakPicopascal [pPa]
0.01 break1.0E+28 pPa
0.1 break1.0E+29 pPa
1 break1.0E+30 pPa
2 break2.0E+30 pPa
3 break3.0E+30 pPa
5 break5.0E+30 pPa
10 break1.0E+31 pPa
20 break2.0E+31 pPa
50 break5.0E+31 pPa
100 break1.0E+32 pPa
1000 break1.0E+33 pPa
### How to Convert Break to Picopascal
1 break = 1.0E+30 pPa
1 pPa = 1.0E-30 break
Example: convert 15 break to pPa:
15 break = 15 × 1.0E+30 pPa = 1.5E+31 pPa | 263 | 647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-49 | latest | en | 0.448501 |
https://www.physicsforums.com/threads/sliding-block-moving-up-incline.341359/ | 1,600,962,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00362.warc.gz | 998,870,794 | 19,341 | # Sliding Block moving up incline
## Homework Statement
A block with mass m = 5.0 kg slides down an inclined plane of slope angle 36.5o with a constant velocity. It is then projected up the same plane with an initial speed 1.05 m/s. How far up the incline will the block move before coming to rest?
## The Attempt at a Solution
$$a_x = -g sin \theta$$
$$\Delta v^2 = 2a \Delta x$$
$$\frac{-v_o^2}{2a} = \Delta x$$
$$\frac{-1.05^2}{-2gsin \theta} = \Delta x$$
$$\frac{-1.05^2}{-11.658} = \Delta x$$
But this wasn't right.
Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
As it slides down it's moving at constant velocity. So what force must be acting on the block besides gravity?
I would say friction but there is no mention of it in the problem.
Doc Al
Mentor
I would say friction but there is no mention of it in the problem.
Good. They expect you to conclude that on your own, based on the fact that it's not accelerating as it slides down.
So what must that friction force equal?
Good. They expect you to conclude that on your own, based on the fact that it's not accelerating as it slides down.
So what must that friction force equal?
I'm getting
$$f = -mg sin \theta$$
Doc Al
Mentor
I'm getting
$$f = -mg sin \theta$$
Good. The magnitude is mgsinθ, but the direction depends on which way the block moves.
Good. The magnitude is mgsinθ, but the direction depends on which way the block moves.
I don't get how the friction helps any.
Doc Al
Mentor
I don't get how the friction helps any.
To determine the acceleration, you need the net force. Friction is one of the forces acting on the block.
You'd only do that when the block is projected back up the incline right? Then use the kinematic equation I had in OP?
Edit: So I was thinking about this more. When the block is projected back up the incline, you'd have friction and gravity in the X direction. Since I found the magnitude of the friction above, the acceleration would be:
$$+mg sin \theta + mg sin \theta = ma_x$$ masses would cancel, so:
$$2 g sin \theta = a_x$$
Does this look right so far? (I set the positive X direction going down the incline)
Edit 2: Yes, I was right, thanks Doc :P
Last edited:
Doc Al
Mentor
You'd only do that when the block is projected back up the incline right?
Friction acts whether the block is sliding up or down the incline. The problem you need to solve involves the block sliding up, so you need to find the block's acceleration as it slides up.
Then use the kinematic equation I had in OP?
Yes.
Okay I have a new question that is similar to this.
A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is μs= 0.44, and the coefficient of kinetic friction is μk = 0.17. The angle of elevation θ is increased slowly from the horizontal. At what value of θ does the block begin to slide (in degrees)?
So I started out with:
Y-Direction
$$F_N - mg cos \theta = ma_y$$
$$F_N = mg cos \theta$$
X-Direction
$$-f + mg sin \theta = ma_x$$
$$-(\mu_s mg cos \theta) + mg sin \theta = ma_x$$
$$-\mu_s g cos \theta + g sin \theta = a_x$$
$$g(-\mu_s cos \theta + sin \theta) = a_x$$
But here is where I'm stuck and dunno what to do next.
Doc Al
Mentor
Two hints:
- What does ax equal?
- When does static friction = μN?
- When does static friction = μN?
I don't understand what you're asking, when there's no sliding?
- What does ax equal?
Since you're giving this as a hint, I'm guessing you mean it's equal to zero. But I thought that since the block is moving (question asks for kinetic friction) you don't know if ax is zero or not.
Doc Al
Mentor
I don't understand what you're asking, when there's no sliding?
No, my point was that static friction does not always equal μN. μN represents the maximum value of static friction between two surfaces--the actual friction may well be less than that, depending on the situation. (For example: A book rests on a table. What's the static friction acting on it?)
Since you're giving this as a hint, I'm guessing you mean it's equal to zero.
Right.
But I thought that since the block is moving (question asks for kinetic friction) you don't know if ax is zero or not.
As the surface is tilted up, the acceleration remains zero until you get to the point where it just begins to start moving. That's the point you want, so it's only static friction that you care about. What's the maximum angle you can have and still not have the block slide? (Use the first hint, of course.)
Yes, I knew that about static friction, however I think I'm stuck more mathematically, because if you simplify what I had up above when ax = 0 you get:
$$-\mu_s cos \theta + sin \theta = \frac{1}{g}$$ What next?
Doc Al
Mentor
...if you simplify what I had up above when ax = 0 you get:
$$-\mu_s cos \theta + sin \theta = \frac{1}{g}$$
That's not quite correct. Take another look.
Oops. I still don't see how to solve for the angle.
$$-\mu_s cos \theta + sin \theta = 0$$
Doc Al
Mentor
That's better. Play with it a bit. Move terms around.
Just a question, is this going to end up being an identity, because I don't know those very well. It'd help if I knew there was an identity involved.
Doc Al
Mentor
No fancy trig identities involved. Hint: What trig function can be defined in terms of sine and cosine?
Was in class. So:
$$\theta = tan^{-1} \mu_s$$
Now, is this always true? Like, if you're given a friction problem, and you need to find the angle, you can do this? Or vice versa, say you're given the angle and are asked to find the coefficient of static friction, you can immediately do this?
Last edited:
Doc Al
Mentor
Was in class. So:
$$\theta = tan^{-1} \mu_s$$
Right.
Now, is this always true? Like, if you're given a friction problem, and you need to find the angle, you can do this? Or vice versa, say you're given the angle and are asked to find the coefficient of static friction, you can immediately do this?
What you found is the angle at which something just starts to slip as a function of the coefficient of friction. That's a specific answer to a specific question. Don't go blindly using tanθ = μ, just because a problem has friction and an incline.
Note that equations work two ways. Given the angle at which slipping starts, you can find the coefficient of friction.
What you found is the angle at which something just starts to slip as a function of the coefficient of friction. That's a specific answer to a specific question. Don't go blindly using tanθ = μ, just because a problem has friction and an incline.
Note that equations work two ways. Given the angle at which slipping starts, you can find the coefficient of friction.
Hm, I didn't say what I wanted to say. What I wanted to ask is, when a mass starts slipping down an incline, will this always work? Or, will it only work when the angle of the incline is being increased?
Doc Al
Mentor
What I wanted to ask is, when a mass starts slipping down an incline, will this always work? Or, will it only work when the angle of the incline is being increased?
If you mean anytime you have a mass starting to slip down an incline does tanθ = μ? No, not at all. | 1,853 | 7,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.859375 | 4 | CC-MAIN-2020-40 | longest | en | 0.942624 |
https://www.wyzant.com/resources/answers/division?f=active&pagesize=20&pagenum=1 | 1,527,302,686,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00111.warc.gz | 868,449,944 | 13,948 | Simple mathematics
Buttercup, has had 15 kittens, born in two litters. Buttercup had 7 more kittens in her second litter than her first. How many kittens did she have in each litter?
Molly uses 192 beads to make a bracelet and a necklace. It takes 5 times as many beads to make a necklace than it does to make a bracelet. How many beads are used to make the necklace?
i can't do 63 divided by 17 or 138 divided by 42 and i can't think of any other division problems with a quotient of 12 and a remainder of 3
if you divide it by 2, 3, 4, 5, or 6 you will geta remainder of 1. If you divide it by 11, the remainder is 0. What number is this? How do I figure that out?
Its not a trick question, I am sincerely struggling to find the answer. List of prime numbers says that is it not prime, so it HAS to divide by SOMETHING. OCD has led me to this.
How do I solve this question it's hard
5,6/5 = 1,12, but when i set it up i get 1,11. i think 5 goes into 5 one time ( 1,) 5 goes into 6 one time. leaving 1 in rest. ending up with 1,11. ? | 301 | 1,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-22 | latest | en | 0.979725 |
https://www.geeksforgeeks.org/get-level-node-binary-tree-iterative-approach/?ref=lbp | 1,656,919,160,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00373.warc.gz | 823,008,122 | 31,488 | Get level of a node in binary tree | iterative approach
• Difficulty Level : Easy
• Last Updated : 31 Mar, 2022
Given a Binary Tree and a key, write a function that returns level of the key.
For example, consider the following tree. If the input key is 3, then your function should return 1. If the input key is 4, then your function should return 3. And for key which is not present in key, then your function should return 0.
Recursive approach to this problem is discussed here
https://www.geeksforgeeks.org/get-level-of-a-node-in-a-binary-tree/
The iterative approach is discussed below :
The iterative approach is modified version of Level Order Tree Traversal
Algorithm
```create a empty queue q
push root and then NULL to q
loop till q is not empty
get the front node into temp node
if it is NULL, it means all nodes of
one level are traversed, so increment
level
else
check if temp data is equal to data
to be searched
if yes then return level
if temp->left is not NULL,
enqueue temp->left
if temp->right is not NULL,
enqueue temp->right
C++
`// CPP program to print level of given node``// in binary tree iterative approach``/* Example binary tree``root is at level 1` ` ``20`` ``/ \`` ``10 30`` ``/ \ / \`` ``5 15 25 40`` ``/`` ``12 */``#include ``using` `namespace` `std;` `// node of binary tree``struct` `node {`` ``int` `data;`` ``node* left;`` ``node* right;``};` `// utility function to create``// a new node``node* getnode(``int` `data)``{`` ``node* newnode = ``new` `node();`` ``newnode->data = data;`` ``newnode->left = NULL;`` ``newnode->right = NULL;``}` `// utility function to return level of given node``int` `getlevel(node* root, ``int` `data)``{`` ``queue q;`` ``int` `level = 1;`` ``q.push(root);` ` ``// extra NULL is pushed to keep track`` ``// of all the nodes to be pushed before`` ``// level is incremented by 1`` ``q.push(NULL);`` ``while` `(!q.empty()) {`` ``node* temp = q.front();`` ``q.pop();`` ``if` `(temp == NULL) {`` ``if` `(q.front() != NULL) {`` ``q.push(NULL);`` ``}`` ``level += 1;`` ``} ``else` `{`` ``if` `(temp->data == data) {`` ``return` `level;`` ``}`` ``if` `(temp->left) {`` ``q.push(temp->left);`` ``}`` ``if` `(temp->right) {`` ``q.push(temp->right);`` ``}`` ``}`` ``}`` ``return` `0;``}` `int` `main()``{`` ``// create a binary tree`` ``node* root = getnode(20);`` ``root->left = getnode(10);`` ``root->right = getnode(30);`` ``root->left->left = getnode(5);`` ``root->left->right = getnode(15);`` ``root->left->right->left = getnode(12);`` ``root->right->left = getnode(25);`` ``root->right->right = getnode(40);` ` ``// return level of node`` ``int` `level = getlevel(root, 30);`` ``(level != 0) ? (cout << ``"level of node 30 is "` `<< level << endl) :`` ``(cout << ``"node 30 not found"` `<< endl);` ` ``level = getlevel(root, 12);`` ``(level != 0) ? (cout << ``"level of node 12 is "` `<< level << endl) :`` ``(cout << ``"node 12 not found"` `<< endl);` ` ``level = getlevel(root, 25);`` ``(level != 0) ? (cout << ``"level of node 25 is "` `<< level << endl) :`` ``(cout << ``"node 25 not found"` `<< endl);` ` ``level = getlevel(root, 27);`` ``(level != 0) ? (cout << ``"level of node 27 is "` `<< level << endl) :`` ``(cout << ``"node 27 not found"` `<< endl);`` ``return` `0;``}`
Java
`// Java program to print level of given node``// in binary tree iterative approach``/* Example binary tree``root is at level 1` ` ``20`` ``/ \`` ``10 30`` ``/ \ / \`` ``5 15 25 40`` ``/`` ``12 */``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{` ` ``// node of binary tree`` ``static` `class` `node`` ``{`` ``int` `data;`` ``node left, right;` ` ``node(``int` `data)`` ``{`` ``this``.data = data;`` ``this``.left = ``this``.right = ``null``;`` ``}`` ``}` ` ``// utility function to return level of given node`` ``static` `int` `getLevel(node root, ``int` `data)`` ``{`` ``Queue q = ``new` `LinkedList<>();`` ``int` `level = ``1``;`` ``q.add(root);` ` ``// extra NULL is pushed to keep track`` ``// of all the nodes to be pushed before`` ``// level is incremented by 1`` ``q.add(``null``);`` ``while` `(!q.isEmpty())`` ``{`` ``node temp = q.poll();`` ``if` `(temp == ``null``)`` ``{`` ``if` `(q.peek() != ``null``)`` ``{`` ``q.add(``null``);`` ``}`` ``level += ``1``;`` ``}`` ``else`` ``{`` ``if` `(temp.data == data)`` ``{`` ``return` `level;`` ``}`` ``if` `(temp.left != ``null``)`` ``{`` ``q.add(temp.left);`` ``}`` ``if` `(temp.right != ``null``)`` ``{`` ``q.add(temp.right);`` ``}`` ``}`` ``}`` ``return` `0``;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{` ` ``// create a binary tree`` ``node root = ``new` `node(``20``);`` ``root.left = ``new` `node(``10``);`` ``root.right = ``new` `node(``30``);`` ``root.left.left = ``new` `node(``5``);`` ``root.left.right = ``new` `node(``15``);`` ``root.left.right.left = ``new` `node(``12``);`` ``root.right.left = ``new` `node(``25``);`` ``root.right.right = ``new` `node(``40``);` ` ``// return level of node`` ``int` `level = getLevel(root, ``30``);`` ``if` `(level != ``0``)`` ``System.out.println(``"level of node 30 is "` `+ level);`` ``else`` ``System.out.println(``"node 30 not found"``);` ` ``level = getLevel(root, ``12``);`` ``if` `(level != ``0``)`` ``System.out.println(``"level of node 12 is "` `+ level);`` ``else`` ``System.out.println(``"node 12 not found"``);` ` ``level = getLevel(root, ``25``);`` ``if` `(level != ``0``)`` ``System.out.println(``"level of node 25 is "` `+ level);`` ``else`` ``System.out.println(``"node 25 not found"``);` ` ``level = getLevel(root, ``27``);`` ``if` `(level != ``0``)`` ``System.out.println(``"level of node 27 is "` `+ level);`` ``else`` ``System.out.println(``"node 27 not found"``);`` ``}``}` `// This code is contributed by``// sanjeev2552`
Python3
`# Python3 program to find closest``# value in Binary search Tree` `_MIN ``=` `-``2147483648``_MAX ``=` `2147483648` `# Helper function that allocates a new``# node with the given data and None``# left and right pointers. ``class` `getnode:` ` ``# Constructor to create a new node`` ``def` `__init__(``self``, data):`` ``self``.data ``=` `data`` ``self``.left ``=` `None`` ``self``.right ``=` `None` `# utility function to return level``# of given node``def` `getlevel(root, data):` ` ``q ``=` `[]`` ``level ``=` `1`` ``q.append(root)` ` ``# extra None is appended to keep track`` ``# of all the nodes to be appended`` ``# before level is incremented by 1`` ``q.append(``None``)`` ``while` `(``len``(q)):`` ``temp ``=` `q[``0``]`` ``q.pop(``0``)`` ``if` `(temp ``=``=` `None``) :`` ``if` `len``(q) ``=``=` `0``:`` ``return` `0`` ``if` `(q[``0``] !``=` `None``):`` ``q.append(``None``)`` ``level ``+``=` `1`` ``else` `:`` ``if` `(temp.data ``=``=` `data) :`` ``return` `level`` ``if` `(temp.left):`` ``q.append(temp.left)`` ``if` `(temp.right) :`` ``q.append(temp.right) `` ``return` `0` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``# create a binary tree`` ``root ``=` `getnode(``20``)`` ``root.left ``=` `getnode(``10``)`` ``root.right ``=` `getnode(``30``)`` ``root.left.left ``=` `getnode(``5``)`` ``root.left.right ``=` `getnode(``15``)`` ``root.left.right.left ``=` `getnode(``12``)`` ``root.right.left ``=` `getnode(``25``)`` ``root.right.right ``=` `getnode(``40``)` ` ``# return level of node`` ``level ``=` `getlevel(root, ``30``)`` ``if` `level !``=` `0``:`` ``print``(``"level of node 30 is"``, level)`` ``else``:`` ``print``(``"node 30 not found"``)` ` ``level ``=` `getlevel(root, ``12``)`` ``if` `level !``=` `0``:`` ``print``(``"level of node 12 is"``, level)`` ``else``:`` ``print``(``"node 12 not found"``)`` ` ` ``level ``=` `getlevel(root, ``25``)`` ``if` `level !``=` `0``:`` ``print``(``"level of node 25 is"``, level)`` ``else``:`` ``print``(``"node 25 not found"``)` ` ``level ``=` `getlevel(root, ``27``)`` ``if` `level !``=` `0``:`` ``print``(``"level of node 27 is"``, level)`` ``else``:`` ``print``(``"node 27 not found"``)` `# This code is contributed by``# Shubham Singh(SHUBHAMSINGH10)`
C#
`// C# program to print level of given node``// in binary tree iterative approach``/* Example binary tree``root is at level 1` ` ``20`` ``/ \`` ``10 30`` ``/ \ / \`` ``5 15 25 40`` ``/`` ``12 */``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `GFG``{` ` ``// node of binary tree`` ``public` `class` `node`` ``{`` ``public` `int` `data;`` ``public` `node left, right;` ` ``public` `node(``int` `data)`` ``{`` ``this``.data = data;`` ``this``.left = ``this``.right = ``null``;`` ``}`` ``}` ` ``// utility function to return level of given node`` ``static` `int` `getLevel(node root, ``int` `data)`` ``{`` ``Queue q = ``new` `Queue();`` ``int` `level = 1;`` ``q.Enqueue(root);` ` ``// extra NULL is pushed to keep track`` ``// of all the nodes to be pushed before`` ``// level is incremented by 1`` ``q.Enqueue(``null``);`` ``while` `(q.Count > 0)`` ``{`` ``node temp = q.Dequeue();`` ` ` ``if` `(temp == ``null``)`` ``{`` ``if` `(q.Count > 0)`` ``{`` ``q.Enqueue(``null``);`` ``}`` ``level += 1;`` ``}`` ``else`` ``{`` ``if` `(temp.data == data)`` ``{`` ``return` `level;`` ``}`` ``if` `(temp.left != ``null``)`` ``{`` ``q.Enqueue(temp.left);`` ``}`` ``if` `(temp.right != ``null``)`` ``{`` ``q.Enqueue(temp.right);`` ``}`` ``}`` ``}`` ``return` `0;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main(String []args)`` ``{` ` ``// create a binary tree`` ``node root = ``new` `node(20);`` ``root.left = ``new` `node(10);`` ``root.right = ``new` `node(30);`` ``root.left.left = ``new` `node(5);`` ``root.left.right = ``new` `node(15);`` ``root.left.right.left = ``new` `node(12);`` ``root.right.left = ``new` `node(25);`` ``root.right.right = ``new` `node(40);` ` ``// return level of node`` ``int` `level = getLevel(root, 30);`` ``if` `(level != 0)`` ``Console.WriteLine(``"level of node 30 is "` `+ level);`` ``else`` ``Console.WriteLine(``"node 30 not found"``);` ` ``level = getLevel(root, 12);`` ``if` `(level != 0)`` ``Console.WriteLine(``"level of node 12 is "` `+ level);`` ``else`` ``Console.WriteLine(``"node 12 not found"``);` ` ``level = getLevel(root, 25);`` ``if` `(level != 0)`` ``Console.WriteLine(``"level of node 25 is "` `+ level);`` ``else`` ``Console.WriteLine(``"node 25 not found"``);` ` ``level = getLevel(root, 27);`` ``if` `(level != 0)`` ``Console.WriteLine(``"level of node 27 is "` `+ level);`` ``else`` ``Console.WriteLine(``"node 27 not found"``);`` ``}``}` `// This code is contributed by Arnab Kundu`
Javascript
``
Output:
```level of node 30 is 2
level of node 12 is 4
level of node 25 is 3 | 4,350 | 13,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.61939 |
https://rational-equations.com/in-rational-equations/multiplying-fractions/algebra-2-explorations-and.html | 1,527,071,606,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00111.warc.gz | 640,603,956 | 11,764 | Algebra Tutorials!
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algebra 2: explorations and applications
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Author Message
Trawemf
Registered: 02.08.2003
From: USA
Posted: Thursday 28th of Dec 13:28 1. Hello Everyone Can someone out there show me a way out? My algebra teacher gave us algebra 2: explorations and applications problem today. Normally I am good at decimals but somehow I am just stuck on this one assignment. I have to turn it in by this weekend but it looks like I will not be able to complete it in time. So I thought of coming online to find assistance. I will really be grateful if a math master can help me work this (topicKwds) out in time.
Jahm Xjardx
Registered: 07.08.2005
From: Odense, Denmark, EU
Posted: Friday 29th of Dec 08:45 I find these routine problems on almost every forum I visit. Please don’t misunderstand me. It’s just as we advance to high school , things change in a flash. Studies become complex all of a sudden. As a result, students encounter trouble in completing their homework. algebra 2: explorations and applications in itself is a quite complex subject. There is a program named as Algebrator which can assist you in this situation.
Troigonis
Registered: 22.04.2002
From: Kvlt of Ø
Posted: Saturday 30th of Dec 08:18 Yeah! I agree with you! The refund guarantee that comes with the purchase of Algebrator is one of the attractive options. In case you are not happy with the assistance offered to you on any math topic, you can get a refund of the payment you made towards the purchase of Algebrator within the number of days specified on the label. Do have a look at http://www.rational-equations.com/solving-linear-and-quadratic-equations.html before you place the order since that gives a lot of information about the areas on which you can expect to get assisted.
Trawemf
Registered: 02.08.2003
From: USA
Posted: Sunday 31st of Dec 09:36 I just hope this tool isn’t very complex . I am not so good with the technical stuff. Can I get the product description, so I know what it has to offer?
TC
Registered: 25.09.2001
From: Kµlt °ƒ Ø, working on my time machine
Posted: Tuesday 02nd of Jan 10:00 I am a regular user of Algebrator. It not only helps me complete my homework faster, the detailed explanations given makes understanding the concepts easier. I strongly advise using it to help improve problem solving skills.
cufBlui
Registered: 26.07.2001
From: Scotland
Posted: Thursday 04th of Jan 10:15 It can be ordered right here – http://www.rational-equations.com/linear-equations-in-two-variables.html. A friend told me that they even offer a ‘no strings attached’ money back guarantee, so go ahead and order a copy, I’m sure you’ll love it . | 1,208 | 5,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.822796 |
https://www.tutorialspoint.com/program-to-sort-given-set-of-cartesian-points-based-on-polar-angles-in-python | 1,686,202,453,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654097.42/warc/CC-MAIN-20230608035801-20230608065801-00112.warc.gz | 1,136,042,859 | 10,544 | # Program to sort given set of Cartesian points based on polar angles in Python
Suppose we have a set of Cartesian points in a list called points. We have to sort them based on their polar angles. The polar angles vary in range 0 and 2*PI. If some points have same polar angles, then arrange them based on the distance of that point from the origin.
So, if the input is like points = [(1,1), (1,-2),(-2,2),(5,4),(4,5),(2,3),(-3,4)],
then the output will be [(5, 4), (1, 1), (4, 5), (2, 3), (-3, 4), (-2, 2), (1, -2)]
To solve this, we will follow these steps −
• Define a comparator function key() . This will take x
• atan := tan-inverse of x[1]/x[0]
• return pair (atan, x[1]^2+x[0]^2) if atan >= 0 otherwise (2*pi + atan, x[0]^2+x[1]^2)
• then sort points using comparator function key()
## Example
Let us see the following implementation to get better understanding −
import math
def solve(points):
def key(x):
atan = math.atan2(x[1], x[0])
return (atan, x[1]**2+x[0]**2) if atan >= 0 else (2*math.pi + atan, x[0]**2+x[1]**2)
return sorted(points, key=key)
points = [(1,1), (1,-2),(-2,2),(5,4),(4,5),(2,3),(-3,4)]
print(solve(points))
## Input
[(1,1), (1,-2),(-2,2),(5,4),(4,5),(2,3),(-3,4)]
## Output
[(5, 4), (1, 1), (4, 5), (2, 3), (-3, 4), (-2, 2), (1, -2)]
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https://myengineeringworld.net/2012/05/optimal-speed-for-minimum-fuel.html | 1,726,501,874,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00740.warc.gz | 370,270,521 | 69,703 | Optimal Car Average Speed For Minimum Fuel Consumption
May 20, 2012
As the title implies, in this post I will try to answer to the following question: Which is the average car speed that minimizes the fuel consumption? Well, the straight answer is that there is not only one speed, but, for each car there is an optimum velocity range.
The chart above depicts the relationship between the fuel consumption and the average speed of a typical small gasoline (< 1400 cm³) Euro 4 passenger car. The chart can be divided into four areas, corresponding to four velocity ranges:
0 – 30 km/h (0 – 18.64 mph): For velocities below 30 km/h the fuel consumption is quite high (above 8 l/100km – 29.34 mpg). Unfortunately this speed range is typical for cars circulating in a city – urban environment. Cars are subjected to a continuous start and stop motion due to junctions and traffic lights. Furthermore, at the main city roads (at the center of the city for example) there is very often traffic congestion. These two parameters – as well as some others – lead to a very low average speed, and, consequently, to high fuel consumption and emissions.
30 – 55 km/h (18.64 – 34.18 mph): Until 55 km/h, as the average speed increases the fuel consumption drops. This velocity range is quite common in sub-urban or rural areas, where there is less traffic congestion and vehicles are able to move with higher speeds.
55 – 80 km/h (34.18 – 49.71 mph): For the particular vehicle, this is the optimum velocity range that minimizes the fuel consumption (around 6 l/100 km – 39.15 mpg). This range corresponds to rural or even highway driving conditions. Fuel consumption is almost steady in this range. Let’s look an example with numbers: The flat line in this range suggesting that the car A, which moves with an average speed of 55 km/h will consume almost the same fuel with a similar car B that moves with 80 km/h. So, if these two cars have to cross a distance of 100 km, then car B will reach to the destination 34 minutes earlier than car A, having consumed almost the same fuel (6 l). So, if the driving conditions are suitable and the speed limits are high, within this velocity range, driving in higher speeds will save travel time!
80 – 120 km/h (49.71 – 74.56 mph): Within this range as the average speed increases the fuel consumption augments too. This velocity range is quite common in highways. Contrary to the previous example, car C that moves in a highway with an average speed of 120 km/h will reach to the destination 25 minutes earlier than car B, which moves with 80 km/h, BUT, car C will have consumed almost one liter more fuel. The fuel consumptions below were calculated with the fuel consumption calculator.
The previous analysis is quite idealized. There are many parameters that are involved in fuel consumption, like engine’s operational conditions, air resistance, driver behavior, road and ambient conditions, tire pressure and many others. However, the average car speed is a useful indicator of fuel consumption. So, the next time that you will have to drive for a long distance try to keep your instantaneous car speed within the range 55 – 80 km/h as much as possible, in order to minimize the fuel consumption of your car. | 721 | 3,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.946223 |
https://justbeginningsflowers.com/what-ratio-gas-to-oil-do-i-mix/ | 1,695,769,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00409.warc.gz | 362,409,119 | 21,313 | Poulan/Weed Eater #Wt200 711724 24cc Gas Blower/Vac
What ratio gas to 2-cycle oil do I mix for a Poulin Weedeater? My husband used to use the machine and now he has Alzheimer’s and can no longer do or remember anything about it, so I have to learn.Liz Hazell
Experienced gardeners share their insights in answering this question :
Use 3.2 oz of 2-cycle oil to one gallon of gasoline. If you use 3.3 up to 4 oz you would be okay nut it may smoke a little. Sorry to hear about your husband. Good luck…Lee
How to Identify and Fix Common Gardening Problems ?
We provide a variety of viewpoints on how to identify and fix common gardening problems. Our sources include academic articles, blog posts, and personal essays from experienced gardeners :
A 40:1 ratio is recommended for this particular model. To obtain the correct mixture, pour 3.2 ounces (95 ml) of 2-cycle synthetic oil into one gallon (4 liters) of fresh unleaded gasoline.
Most STIHL gasoline-powered equipment runs on a 50:1 mixture of gasoline and 2-cycle engine oil.
Fuel Mix. According to Stihl, gas and oil for leaf blowers should be mixed at a ratio of 50 parts gas to 1 part oil. This amounts to about 2.6 ounces of oil to each gallon of gas.
A: The fuel-oil mixture ratio is 40:1. You can obtain this ratio by mixing 3.2 oz. of two cycle air cooled engine oil with one gallon of regular gas. Over a period of time oil will separate from gasoline.
If your manufacturer recommends a 50:1 fuel/oil mix, it means you need 50 parts of gas to one part two-stroke oil. To mix one gallon of fuel at 50:1, add 2.6 ounces of two-stroke oil to one gallon of gas, as shown in the chart below.
The stoichiometric mixture for a gasoline engine is the ideal ratio of air to fuel that burns all fuel with no excess air. For gasoline fuel, the stoichiometric air–fuel mixture is about 14.7:1 i.e. for every one gram of fuel, 14.7 grams of air are required.
This ratio indicates how much fresh, regular unleaded gasoline (containing no more than 10% ethanol) to mix with how much oil. For all Remington 2-cycle products, the ratio is 40:1. This means you`ll want to add 3.2 oz of oil to every gallon of gasoline.
For a 50:1 ratio of gas to oil, use 2.6 fluid ounces of oil per gallon of gas. For a 40:1 mixture, use 3.2 fluid ounces of oil per gallon of gas.
100:1 – Mix 50ml of oil per 5ltrs of fuel.
Recommended by Yamaha and Suzuki for most of their small two stroke engines up to about 30hp, this ratio requires the least amount of TCW3 two stroke oil. The result of using this ratio (less oil) is reduced spark plug fouling and less smoke.
It`ll probably work in a pinch but I wouldn`t do it long term. You`re putting less oil in a motor that requires more, in turn providing less lubrication than is required.
40 to 1 ratio
This equals 25mls of two stroke oil to 1 litre of petrol.
For a mixing ratio of 1 : 50 you need 5 litres of high-octane gas and 0.10 litres (100 ml) of STIHL two-stroke oil.
Q: How do I mix gas and oil? A: Mix 3.2 ounces of Poulan 2-cycle air cooled engine oil to one gallon of fresh unleaded gasoline to obtain the recommended 40:1 fuel to oil ratio.
For a 50:1 ratio of gas to oil, use 2.6 fluid ounces of oil per gallon of gas. For a 40:1 mixture, use 3.2 fluid ounces of oil per gallon of gas.
Discover Relevant Questions and Answers for Your Specific Issue
the most relevant questions and answers related to your specific issue
Do I need gas and oil mix – Poulan Super 250a Chainsaw Plate
ANSWER : I think Poulan uses a 40:1 ratio to one gallon of gas, however I have always used 50:1 the same as my sthl saws with no issue. I would rather have more oil mix that to little.
Do you mix gas and oil in this trimmer – Ryobi 30 Cubic Centimeter Curved Shaft Gas Trimmer
ANSWER : Yes. Verifythe ratio, either 40:1 or 50:1
What is ment by 50 to 1 mix
ANSWER : 50 to one is a ratio of fuel to lubricating oil. Typically 50 parts petrol/gasoline to one part two stroke oil.
Oil typically comes in a special plastic bottle that incoporates a measure. Simply squeeze the right ammount of oil into the little chamber and add to 5 litres of gas.
I need the gas mix oil ratio for the Poluan Pro 42…
ANSWER : 40:1 gas to oil ratio.
DCS520 PETROL/OIL RATIO
ANSWER : 40 to one is a good mix
How do u mix the 50/1 ratio oil and gas mixture
ANSWER : Mix 2 1/2 oz of 2 cycle oil into 1 gal of gas
What is the oil/gas mix ratio? – Husqvarna , No.455 Rancher, 20" Gas – Powered Chainsaw
ANSWER : 50:1 ratio 2.6 fl oz of 2 stroke oil to 1 gal of fuel
What is the oil mix for 2 gallons of gas for a 2 cycle gas leaf blower
ANSWER : Only use 50 to 1 as that is what is specified.
To get 50 to one ratio with the 6.4 oz bottle of 2 stroke oil just add 2 1/2 U.S.. gallons and that will be 50 to 1 ratio so you will have the correct ratio . | 1,328 | 4,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.918793 |
http://de.metamath.org/mpegif/funimass3.html | 1,618,680,759,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00193.warc.gz | 26,690,853 | 5,962 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > funimass3 Structured version Visualization version Unicode version
Theorem funimass3 6013
Description: A kind of contraposition law that infers an image subclass from a subclass of a preimage. Raph Levien remarks: "Likely this could be proved directly, and fvimacnv 6012 would be the special case of being a singleton, but it works this way round too." (Contributed by Raph Levien, 20-Nov-2006.)
Assertion
Ref Expression
funimass3
Proof of Theorem funimass3
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 funimass4 5930 . . 3
2 ssel 3412 . . . . . 6
3 fvimacnv 6012 . . . . . . 7
43ex 441 . . . . . 6
52, 4syl9r 73 . . . . 5
65imp31 439 . . . 4
76ralbidva 2828 . . 3
81, 7bitrd 261 . 2
9 dfss3 3408 . 2
108, 9syl6bbr 271 1
Colors of variables: wff setvar class Syntax hints: wi 4 wb 189 wa 376 wcel 1904 wral 2756 wss 3390 ccnv 4838 cdm 4839 cima 4842 wfun 5583 cfv 5589 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1677 ax-4 1690 ax-5 1766 ax-6 1813 ax-7 1859 ax-9 1913 ax-10 1932 ax-11 1937 ax-12 1950 ax-13 2104 ax-ext 2451 ax-sep 4518 ax-nul 4527 ax-pr 4639 This theorem depends on definitions: df-bi 190 df-or 377 df-an 378 df-3an 1009 df-tru 1455 df-ex 1672 df-nf 1676 df-sb 1806 df-eu 2323 df-mo 2324 df-clab 2458 df-cleq 2464 df-clel 2467 df-nfc 2601 df-ne 2643 df-ral 2761 df-rex 2762 df-rab 2765 df-v 3033 df-sbc 3256 df-dif 3393 df-un 3395 df-in 3397 df-ss 3404 df-nul 3723 df-if 3873 df-sn 3960 df-pr 3962 df-op 3966 df-uni 4191 df-br 4396 df-opab 4455 df-id 4754 df-xp 4845 df-rel 4846 df-cnv 4847 df-co 4848 df-dm 4849 df-rn 4850 df-res 4851 df-ima 4852 df-iota 5553 df-fun 5591 df-fn 5592 df-fv 5597 This theorem is referenced by: funimass5 6014 funconstss 6015 fvimacnvALT 6016 fimacnv 6027 r0weon 8461 iscnp3 20337 cnpnei 20357 cnclsi 20365 cncls 20367 cncnp 20373 1stccnp 20554 txcnpi 20700 xkoco2cn 20750 xkococnlem 20751 basqtop 20803 kqnrmlem1 20835 kqnrmlem2 20836 reghmph 20885 nrmhmph 20886 elfm3 21043 rnelfm 21046 symgtgp 21194 tgpconcompeqg 21204 eltsms 21225 ucnprima 21375 plyco0 23225 plyeq0 23244 xrlimcnp 23973 rinvf1o 28306 xppreima 28324 cvmliftmolem1 30076 cvmlift2lem9 30106 cvmlift3lem6 30119 mclsppslem 30293
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https://realmathinaminute.wordpress.com/2009/04/26/problems-20-24/ | 1,529,370,262,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00326.warc.gz | 697,836,072 | 18,120 | # Problems #20-24
20. A building that is 50 feet tall casts a shadow 30 feet long. Nearby, a tree casts a 75 foot long shadow. How tall is the tree? (Hint: Use similar triangle ratios) Read it twice. Wording is tricky.
a. 95 ft
b. 110 ft
c. 125 ft
d. 140 ft
Answers: 50/30 =x/75 so x = 125
21. Which unit is correct for describing the surface area of a hexagonal pyramid?
a. cm
b. cm squared
c. cm cubed
Answer: (b) Area is always “units squared”.
22. If r>0 and s <0, in which quadrant of the xy-plane does the point (r,s) lie?
23. Which set of points is equidistant from the rays that form an angle? (Hint: Draw it)
a. Perpendicular bisector
b. Skew Line
c. Angle Bisector
d. Central Angle | 220 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-26 | latest | en | 0.780424 |
https://byjus.com/questions/due-to-internal-combustion-inside-a-rocket-gases-are-ejected-at-the-rate-of-2-4-kgminute-the-velocity-of-the-gases-with-respect-to-the-rocket-is-400-ms-calculate-force-exerted-on-the-rocket/ | 1,627,237,968,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00348.warc.gz | 177,360,765 | 37,894 | # Due to internal combustion inside a rocket, gases are ejected at the rate of 2.4 kg/minute. The velocity of the gases with respect to the rocket is 400 m/s. Calculate force exerted on the rocket
Rate of change of mass = (dm/dt) = 2.4kg/minute = (2.4/60) kg/sec = 0.04 kg/sec
Force= Rate of change of momentum =v(dm/dt) + m(dv/dt).
= 400(0.04) + Mass of rocket(0) = 16N
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# Given 15 cot A = 8, find sin A and sec A.
Given,
$$15 cot A \ = \ 8$$
$$\Rightarrow cotA \ = \ \frac{8}{15} \ = \ \frac{B}{P} \ = \ \frac{AB}{BC}$$
Let AB = 8k and BC = 15k.
Then, by Pythagoras theorem
$$AC \ = \ \sqrt{AB^2 + BC^2} \$$
$$= \ \sqrt{(8k)^2 + (15k)^2} \$$
$$= \ \sqrt{ 64k^2 + 225k^2} \$$
$$= \ \sqrt{289k^2} \ = \ 17k$$
$$sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{15k}{17k} \ = \ \frac{15}{17}$$
and, $$secA \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \ = \ \frac{17k}{8k} \ = \ \frac{17}{8}$$ | 275 | 564 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-40 | latest | en | 0.473638 |
https://chemistry.stackexchange.com/questions/49236/how-do-i-determine-the-average-carbon-oxidation-numbers-in-ethanol-and-ethanal | 1,721,782,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00729.warc.gz | 139,056,837 | 42,435 | # How do I determine the average carbon oxidation numbers in ethanol and ethanal?
I have an assignment (I quote it word by word):
Determine the oxidation number of carbon in (A) ethanol and (B) ethanal.
Answers are (A) -2 and (B) -1.
I reviewed similar question at SE: Oxidation of Carbons and I found first answer to be clear and links given in this answer to be useful.
I also found a good Khan Academy video, which basically confirms all in this answer
Now, here is my problem: There are two carbons in ethanol. Oxidation number of carbon which is connected to three hydrogens is $-3$; oxidation number of carbon which is connected to two hydrogens and one oxygen is $-1$. How do we get answer $-2$ out of it then? (Do we add $-3 + (-1)$ and then divide by 2?)
And again there are two carbons in ethanal. Oxidation number of carbon which is connected to three hydrogens is $-3$; oxidation number of carbon which is connected to one hydrogen and one oxygen (but with double bond) is $+1$. How do we get answer $-1$ out of it? (Do we add $-3 + 1$ and divide it by 2?)
Overall, my question does not seem to concentrate on one of the carbons present in ethanol (or ethanal). And answers given look like there is oxidation number of first carbon added to oxidation number of second carbon and then divided by 2. This situation seems to be true for ethanol, and again the same happens for ethanal. Where can I read about adding two oxidation states and then calculating average (I have never heard of it)? Or maybe these calculations done in some other way? Thank you very much in advance for answering.
• This question is really hard to read. If you would, please check out the information here and here about formatting posts with HTML and/or MathJax, and rewrite your question using these tools. Thanks! Commented Apr 9, 2016 at 14:57
• Thank you, Brian! I made paragraphs and hid link for Khan Academy video. I would make some other changes, if I would have a clue what exactly I should change . Your advice on this matter will be very much appreciated. Commented Apr 9, 2016 at 15:14
• Yes, it's simply average oxidation number. You were unnecessarily verbose here IMO. Commented Apr 9, 2016 at 15:18
• Thank you, Mithoron! I asked about link for any source which would confirm that idea. If you know one, please post it. I would be most grateful. I also have a reason to write a detailed question, but thank you for sharing your opinion. Commented Apr 9, 2016 at 15:43
• @SleepyHollow Those edits are great! Much easier to read. Thanks! Commented Apr 9, 2016 at 16:46
What you have been asked to calculate is what is called an average oxidation number. This is often used in inorganic chemistry (not because it is rather powerful, but because you can easily get away with it) but lesser in organic chemistry.
To properly determine oxidation states, you should do the calculation on a per-atom basis. And doing that would give you $\mathrm{-III}$ and $\mathrm{-I}$ for ethanol’s two carbons and $\mathrm{-III}$ and $\mathrm{+I}$ for ethanal’s. This method has a number of advantages:
• you are always dealing with integers; no need for any fractional states
• it is easy to see exactly which carbon’s oxidation state changed.
The downside of this method is that you cannot use it if you don’t know the structural formula. Consider if, for example, you only saw $\ce{C2H6O}$ or $\ce{C2H4O}$ — especially the first one could represent two different structures with different properties. This is were average oxidation states come in handy: you calculate an average of all the atoms of a single element. You may remember that oxygen is typically given $\mathrm{-II}$, hydrogen $\mathrm{+I}$ (except in peroxides and hydrides, respectively; although oxofluorides should get a special mention). You may also remember that the sum of oxidation states of a neutral compound must be $\pm 0$ that allows you to calculate the average oxidation states as follows:
\begin{align}\text{For }\ce{C2H6O}\text{:}&\\ 0 &= 2 x + 6 \times (+1) + (-2)\\ 0 &= 2x + 6 - 2\\ 0 &= 2x + 4\\ -4 &= 2x\\ x &= -2\\ \\ \text{For }\ce{C2H4O}\text{:}&\\ 0 &= 2 x + 4 \times (+1) + (-2)\\ 0 &= 2x + 4 - 2\\ 0 &= 2x + 2\\ -2 &= 2x\\ x &= -1\end{align}
Since these are averages, you can also arrive at the same conclusion by taking the arithmetic mean of all carbon atoms’ oxidation state; you will obviously arrive at the same result:
\begin{align}\ce{C2H6O}\text{:}&\\ \bar{x} &= \frac{-3 + (-1)}{2}\\ \bar{x} &= \frac{-4}{2}\\ \bar{x} &= -2\\ \\ \ce{C2H4O}\text{:}&\\ \bar{x} &= \frac{-3 + 1}{2}\\ \bar{x} &= \frac{-2}{2}\\ \bar{x} &= -1\end{align} | 1,241 | 4,633 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.954214 |
https://bitcoin.stackexchange.com/questions/114601/how-throughput-latency-finality-waiting-time-are-defined | 1,701,734,586,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00541.warc.gz | 156,491,405 | 40,880 | # How throughput, latency, finality, waiting time are defined?
I am confused with the below terms. Could someone explain them in simple terms and confirm that my rationale below is correct?
Waiting Time - Throughput - Finality - Latency - Confirmation Time
I have created the following graph.
I guess when a node propagates a transaction until this is included in a block, this is called waiting time (the period A below). The number of transactions that are included in a block is called throughput. Finality is defined as the period (the period B below) since the tx was included in the Block X0 until the Block X6 was created (i.e., 6 confirmations). Latency (or confirmation time), is called the period from when the node propagated the transaction until the Block X6 was created (the period C below : A + B).
Could someone confirm if this is the real meaning of the terms waiting time, throughput, finality, latency, and confirmation time? | 203 | 949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.977331 |
https://socialsci.libretexts.org/Under_Construction/Purgatory/Book%3A_Principles_of_Microeconomics_(Casolari)/04%3A_Elasticity/4.01%3A_Prelude_to_Elasticity | 1,726,867,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00746.warc.gz | 499,120,728 | 31,511 | # 4.1: Prelude to Elasticity
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Skills to Develop
• Price Elasticity of Demand and Price Elasticity of Supply
• Polar Cases of Elasticity and Constant Elasticity
• Elasticity and Pricing
• Elasticity in Areas Other Than Price
Figure $$\PageIndex{1}$$: Netflix, Inc. is an American provider of on-demand Internet streaming media to many countries around the world, including the United States, and of flat rate DVD-by-mail in the United States. (Credit: modification of work by Traci Lawson/Flickr Creative Commons)
## That Will Be How Much?
Imagine going to your favorite coffee shop and having the waiter inform you the pricing has changed. Instead of $$\3$$ for a cup of coffee, you will now be charged $$\2$$ for coffee, $$\1$$ for creamer, and $$\1$$ for your choice of sweetener. If you pay your usual $$\3$$ for a cup of coffee, you must choose between creamer and sweetener. If you want both, you now face an extra charge of $$\1$$. Sound absurd? Well, that is the situation Netflix customers found themselves in—a $$60\%$$ price hike to retain the same service in 2011.
In early 2011, Netflix consumers paid about $$\10$$ a month for a package consisting of streaming video and DVD rentals. In July 2011, the company announced a packaging change. Customers wishing to retain both streaming video and DVD rental would be charged $$\15.98$$ per month, a price increase of about $$60\%$$. In 2014, Netflix also raised its streaming video subscription price from $$\7.99$$ to $$\8.99$$ per month for new U.S. customers. The company also changed its policy of 4K streaming content from $$\9.00$$ to $$\12.00$$ per month that year.
How would customers of the $$18$$-year-old firm react? Would they abandon Netflix? Would the ease of access to other venues make a difference in how consumers responded to the Netflix price change? The answers to those questions will be explored in this chapter: the change in quantity with respect to a change in price, a concept economists call elasticity.
Anyone who has studied economics knows the law of demand: a higher price will lead to a lower quantity demanded. What you may not know is how much lower the quantity demanded will be. Similarly, the law of supply shows that a higher price will lead to a higher quantity supplied. The question is: How much higher? This chapter will explain how to answer these questions and why they are critically important in the real world.
To find answers to these questions, we need to understand the concept of elasticity. Elasticity is an economics concept that measures responsiveness of one variable to changes in another variable. Suppose you drop two items from a second-floor balcony. The first item is a tennis ball. The second item is a brick. Which will bounce higher? Obviously, the tennis ball. We would say that the tennis ball has greater elasticity.
Consider an economic example. Cigarette taxes are an example of a “sin tax,” a tax on something that is bad for you, like alcohol. Cigarettes are taxed at the state and national levels. State taxes range from a low of $$17$$ cents per pack in Missouri to $$\4.35$$ per pack in New York. The average state cigarette tax is $$\1.51$$ per pack. The 2014 federal tax rate on cigarettes was $$\1.01$$ per pack, but in 2015 the Obama Administration proposed raising the federal tax nearly a dollar to $$\1.95$$ per pack. The key question is: How much would cigarette purchases decline?
Taxes on cigarettes serve two purposes: to raise tax revenue for government and to discourage consumption of cigarettes. However, if a higher cigarette tax discourages consumption by quite a lot, meaning a greatly reduced quantity of cigarettes is sold, then the cigarette tax on each pack will not raise much revenue for the government. Alternatively, a higher cigarette tax that does not discourage consumption by much will actually raise more tax revenue for the government. Thus, when a government agency tries to calculate the effects of altering its cigarette tax, it must analyze how much the tax affects the quantity of cigarettes consumed. This issue reaches beyond governments and taxes; every firm faces a similar issue. Every time a firm considers raising the price that it charges, it must consider how much a price increase will reduce the quantity demanded of what it sells. Conversely, when a firm puts its products on sale, it must expect (or hope) that the lower price will lead to a significantly higher quantity demanded.
## Contributor
This page titled 4.1: Prelude to Elasticity is shared under a not declared 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. | 2,778 | 9,072 | {"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} | 3.28125 | 3 | CC-MAIN-2024-38 | latest | en | 0.19987 |
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Question
# If 97 and a63 are equivalent fractions, then the value of a is ___ .
A
18
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B
81
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C
30
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D
60
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Solution
## The correct option is B 81 97=a63 are equivalent fractions. Equivalent fractions are obtained when you divide or multiply both numerator and denominator with the same number. Here in the denominator, 7 ×9=63 So, 9 (numerator) has to be multiplied by 9 to get a. Hence, a=9×9=81 Therefore, 9×97×9=8163.
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Join BYJU'S Learning Program | 267 | 924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.828472 |
www.automeasure.com | 1,623,749,040,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00298.warc.gz | 54,172,237 | 111,994 | ## Sunday, May 30, 2021
### Pulse Response Examiner - a Python Jupyter Notebook
This notebook simulates the pulse response of a basic RLC circuit. You can use this notebook to help you choose a resistor R that you solder in series at your pulse-generating source, to minimize ringing overshoot (due to inductance) while not rounding the pulse edges too much (due to capacitance). In the last plot below, we find a good resistance value for R.
### Define the function that performs an inverse Laplace transform¶
# PULSE_RESPONSE_EXAMINER PROGRAM
# Version 0.1, 30-May-2021
# Contact: xejgyczlionoldmailservertigernet replace lion with
# an at sign and tiger with a dot.
#
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#
# Unless required by applicable law or agreed to in writing, software
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
import math, cmath
import numpy as np
from scipy.fft import fft
import matplotlib.pyplot as plt
#FUNCTION inv_laplace(): employs the 'fourier method' for inversion of the Laplace
#transform. The algorithm here is an adaptation by Joe Czapski of a Mathcad
#worksheet by Sven Lindhardt. It's a practical approximation of inverse Laplace for
#systems close to stability, whether a little overstable or understable, when the
#rightmost pole is not greater than zero. SciPy's fast complex FFT is used in this
#application to generate a fast inverse Laplace transform.
def inv_laplace(n, R, L, C, t1, t2, tspan):
#2.5 is an error correction factor to reduce artifacts when both L and C are tiny
sigma = 2.5 / tspan
#calculate delta omega (angular frequency sample interval)
domega = math.pi / tspan
#convert impedance and time values to complex numbers for transfer function
Rc = R + 0.0j; Lc = L + 0.0j; Cc = C + 0.0j; t1c = t1 + 0.0j; t2c = t2 + 0.0j
Vouts = np.zeros(n+1, dtype=np.cdouble)
for i in range(0, n+1):
omega = i * domega
s = sigma - omega * 1.0j
sL = s * Lc
sCinv = (1.0 + 0.0j) / (s * Cc)
#Calculate H(s) for the desired circuit.
#For an RLC circuit H(s) = (1/sC) / (R + sL + 1/sC)
Hs = sCinv / (Rc + sL + sCinv)
#Calculate Vin(s) for the input waveform.
#For an ideal pulse Vin(s) = 100 * (e^-t1s - e^-t2s) / s
Vins = (100.0 + 0.0j) * (cmath.exp(-t1c*s) - cmath.exp(-t2c*s)) / s
#Calculate Vout(s) = H(s) * Vin(s)
Vouts[i] = Hs * Vins
#Perform the complex fourier transform. Before calling fft(),
#construct a symmetric version of complex array Vouts.
Vouts[n] = Vouts[n].real + 0.0j
Vouts_symmetric = np.concatenate([Vouts, (Vouts[1:n])[::-1]])
Vout_unscaled = fft(Vouts_symmetric)
#Take the real part of Vout_unscaled, first half of the symmetric array,
#and scale for inverse Laplace using time span, sigma, and exp function.
#The sample interval in seconds = tspan / n
sigdx = sigma * tspan / n
Vout = np.zeros(n)
for i in range(0, n):
Vout[i] = math.exp(i*sigdx) * Vout_unscaled[i].real / tspan
return Vout
### Set the circuit component values, pulse width, and time scale, and generate the plot¶
You can use this handy inductance calculator for your cabling or traces situation. In the formulas on that webpage, use µr = 1 if the intervening material is air or a dielectric insulator.
R = 10.0 #ohms
L = 4.0e-9 #henries
t1 = 1.0e-8 #pulse rising edge seconds from t0
t2 = 2.0e-8 #pulse falling edge seconds from t0
#Set tspan which is the time span of the analysis in seconds from t0.
#Artifacts can arise when Vout doesn't settle to zero before the end of the
#analysis time. You may need to adjust tspan to increase the analysis time
#following the end of the pulse to allow settling to zero.
tspan = 4.0e-8
#Set n which is the number of points to use for the analysis. n must be a power of 2.
#Set n=2048 for most cases. Set n=8192 to reduce artifacts when both L and C are tiny.
n = 2048
ns_per_s = 1.0e9 #time display unit correction
xdata = np.linspace(0.0, tspan * ns_per_s * (n-1)/n, n)
ydata = inv_laplace(n, R, L, C, t1, t2, tspan)
plt.rcParams['figure.figsize'] = [12, 6]
plt.title('Output Voltage Waveform, R = %d Ω' % (R))
plt.xlabel('Time ns')
plt.ylabel('Magnitude %')
plt.grid(True)
plt.plot(xdata, ydata, color='firebrick')
plt.show()
### Try different values of R, regenerating the plot for each value¶
R = 200.0 #ohms
ydata = inv_laplace(n, R, L, C, t1, t2, tspan)
plt.title('Output Voltage Waveform, R = %d Ω' % (R))
plt.xlabel('Time ns')
plt.ylabel('Magnitude %')
plt.grid(True)
plt.plot(xdata, ydata, color='firebrick')
plt.show()
R = 50.0 #ohms
ydata = inv_laplace(n, R, L, C, t1, t2, tspan)
plt.title('Output Voltage Waveform, R = %d Ω' % (R))
plt.xlabel('Time ns')
plt.ylabel('Magnitude %')
plt.grid(True)
plt.plot(xdata, ydata, color='firebrick')
plt.show() | 1,491 | 4,829 | {"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} | 2.921875 | 3 | CC-MAIN-2021-25 | longest | en | 0.783953 |
https://intercom.help/fotop/english/support-photographer/f26-how-does-your-sales-commission-work | 1,555,748,972,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529472.24/warc/CC-MAIN-20190420080927-20190420102927-00219.warc.gz | 463,021,002 | 4,541 | Como regra geral, a Fotop. Obtém 25% das vendas de todas as fotos para cobrir os custos de operação da plataforma.
When you confirm your participation in an event, your moderator may choose to collect a percentage of your sales of your photos, for example, 10% of each order made. Therefore, this moderator can offer you 65% commission
Example, if the event moderator is offering 65% commission:
• Example, if the event moderator is offering 65% commission:
• Total sale value of the photographer: R\$ 300,00
• Fotop Commission: 25% = R\$ 75,00
• Moderator's Committee: 10% = R\$ 30.00
• Photographer's Committee: 65% = R\$ 195,00
• At the end of the day you will receive the net value of R\$ 195.00
• Example, if the event moderator is offering 70% commission:
• Total sale value of the photographer: R\$ 300,00
• Fotop Commission: 25% = R\$ 75,00
• Moderator's Committee: 5% = R\$ 15,00
• Photographer's Committee: 70% = R\$ 210,00
• At the end of the day you will receive the net amount of R\$ 210,00
It's cool to see what the moderator of the event is offering in return for commissions, as there is often a marketing effort on the part of the moderator, and perhaps for this reason the commission may be a bit smaller. If you have this kind of effort, it pays to invest your time in photographing such proof.
Encontrou sua resposta? | 361 | 1,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-18 | latest | en | 0.871455 |
https://in.mathworks.com/matlabcentral/answers/1943474-i-have-data-for-each-milli-second-how-can-i-average-the-1000-samples-and-convert-in-to-1-sec-i-got | 1,716,581,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00580.warc.gz | 266,284,160 | 25,560 | # I have data for each milli second. How can I average the 1000 samples and convert in to 1 sec? I got 322 sec data. 1K samples for each second.
2 views (last 30 days)
srinivas chekuri on 7 Apr 2023
Answered: Walter Roberson on 8 Apr 2023
I am having 322759 rows. It means 322 seconds. How can I average the mili seconds data and convert in to seconds. I just want 322 rows(322 seconds). Accordingly the next column should also average up according to the time. Attaching a sample
Fangjun Jiang on 7 Apr 2023
Data=(1:25)';
Ten=10;
NofData=floor(length(Data)/Ten)*Ten;
temp=reshape(Data(1:NofData),Ten,[])
temp = 10×2
1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20
av=mean(temp)
av = 1×2
5.5000 15.5000
You may want to use imresize：
Table=imresize(Table,[322,2]);
Table(:,1)=1:322;
Walter Roberson on 8 Apr 2023
### Categories
Find more on Tables in Help Center and File Exchange
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Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 325 | 1,016 | {"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} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.771573 |
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# South Valley University Theory of Machine
## Faculty of Engineering Code: ENM222
Dept. of Mechanical Engineering 2013/2014
1
Final Report
Question One:
1. What is the main purpose of gears?
2. Explain the terms: Module, Pitch circle, Circular pitch, Clearance, Face of a tooth,
Flank of a tooth, Backlash, Dedendum and Addendum.
3. What is the main advantage and disadvantage of: spur, helical, double helical, bevel,
worm, Rack and pin Gears.
4. What are the main factors considered in gear selection?
5. What do you understand by gear train? Discuss the various types of gear trains.
6. Explain briefly the differences between simple, compound, and epicyclic gear trains.
7. What are the special advantages of epicyclic gear trains?
8. Explain the procedure adopted for designing the spur wheels.
9. How the velocity ratio of epicyclic gear train is obtained by tabular method?
10. Explain with a neat sketch the sun and planet wheel.
11. What are the various types of the torques in an epicyclic gear train?
12. Explain the difference between helical gears and spur gears?
13. What are the different types of bevel gears?
14. State the purpose of: bevel gear, helical gears, spur gears, rack and pinion gears and
worm gears.
15. Identify the advantages and disadvantages of: bevel gear, helical gears, spur gears
Question Two
:
1. Explain with a neat sketch the main types of gear trains?
2. State the different methods of Gear Manufacture
3. Figure below, illustrates an automotive sliding-mesh transmission gearbox.
The numbers of teeth on the various gear wheels are as follows:
South Valley University Theory of Machine
Faculty of Engineering Code: ENM222
Dept. of Mechanical Engineering 2013/2014
2
Gear wheel 1 is keyed to the input shaft and cannot slide along it. Gear wheels 2, 3,
and 4 form a compound cluster that is keyed to the output shaft but can be slid along
it. Gear wheel 10 is a reverse idler. Determine the gear ratios for each speed selection,
starting from the lowest forward selection to the highest and then to the reverse
4. A compound train consists of six gears. The number of teeth on the gears are as
follows :
A B C D E F
60 40 50 25 30 24
The gears B and C are on one shaft while the gears D and E are on another shaft. The
gear A drives gear B, gear C drives gear D and gear E drives gear F. If the gear A
transmits 1.5 kW at 100 r.p.m. and the gear train has an efficiency of 80 per cent, find
the torque on gear F.
Question Three:
1. In a compound epicyclic gear train as shown in the figure, has gears A and an annular
gears D & E free to rotate on the axis P. B and C is a compound gear rotate about axis
Q. Gear A rotates at 90 rpm CCW and gear D rotates at 450 rpm CW. Find the speed
and direction of rotation of arm F and gear E. Gears A,B and C are having 18, 45 and
21 teeth respectively. All gears having same module and pitch.
South Valley University Theory of Machine
Faculty of Engineering Code: ENM222
Dept. of Mechanical Engineering 2013/2014
3
2. An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal wheel
C with 82 teeth is concentric with B. A Compound gears DE meshed with the two internal
wheels. D has 28 teeth and meshes with internal gear C while E meshes with B. The
compound wheels revolve freely on pin which projects from a arm keyed to a shaft A co-
axial with F. if the wheels have the same pitch and the shaft A makes 800 rpm, what is the
speed of the shaft F? Sketch the arrangement.
3. The fig shows an Epicyclic gear train. Wheel E is fixed and wheels C and D are
integrally cast and mounted on the same pin. If arm A makes one revolution per sec
(Counter clockwise) determine the speed and direction of rotation of the wheels B
and F.
Question Four:
1. What is the function of a governor ? How does it differ from that of a flywheel ?
2. State the different types of governors. What is the difference between centrifugal and
inertia type governors ? Why is the former preferred to the latter ?
3. Explain the term height of the governor. Derive an expression for the height in the
case of a Watt governor. What are the limitations of a Watt governor ?
4. What are the effects of friction and of adding a central weight to the sleeve of a Watt
governor ?
5. Discuss the controlling force and stability of a governor and show that the stability of
a governor depends on the slope of the curve connecting the controlling force (FC)
and radius of rotation (r) and the value (FC /r).
South Valley University Theory of Machine
Faculty of Engineering Code: ENM222
Dept. of Mechanical Engineering 2013/2014
4
6. The length of the upper arm of a Watt governor is 400 mm and its inclination to the
vertical is 30. Find the percentage increase in speed, if the balls rise by 20 mm.
7. All the arms of a Porter governor are 178 mm long and are hinged at a distance of 38
mm from the axis of rotation. The mass of each ball is 1.15 kg and mass of the sleeve
is 20 kg. The governor sleeve begins to rise at 280 r.p.m. when the links are at an
angle of 30 to the vertical. Assuming the friction force to be constant, determine the
minimum and maximum speed of rotation when the inclination of the arms to the
vertical is 45.
8. A spring controlled governor of the Hartnell type has the following data : Mass of the
ball = 1.8 kg ; Mass of the sleeve = 6 kg ; Ball and sleeve arms of the bell crank lever
= 150 mm and 120 mm respectively. The equilibrium speed and radius of rotation for
the lowest position of the sleeve are 400 r.p.m. and 150 mm respectively. The sleeve
lift is 10 mm and the change in speed for full sleeve lift is 5%. During an overhaul,
the spring was compressed 2 mm more than the correct compression for the initial
setting. Determine the stiffness of the spring and the new equilibrium speed for the
lowest position of the sleeve.
9. A Porter governor has all four arms 200 mm long. The upper arms are pivoted on the
axis of rotation and the lower arms are attached to a sleeve at a distance of 25 mm
from the axis. Each ball has a mass of 2 kg and the mass of the load on the sleeve is
20 kg. If the radius of rotation of the balls at a speed of 250 r.p.m. is 100 mm, find the
speed of the governor after the sleeve has lifted 50 mm. Also, determine the effort
and power of the governor.
10. A governor of the Proell type has each arm 250 mm long. The pivots of the upper and
lower arms are 25 mm from the axis. The central load acting on the sleeve has a mass
of 25 kg and the each rotating ball has a mass of 3.2 kg. When the governor sleeve is
in mid-position, the extension link of the lower arm is vertical and the radius of the
path of rotation of the masses is 175 mm. The vertical height of the governor is 200
mm. If the governor speed is 160 r.p.m. when in mid-position, find : 1. length of the
extension link; and 2. tension in the upper arm.
Question Five:
1. What do you mean by unbalance and why it is so important?
2. How is unbalanced force due to single rotating mass balanced.
3. How is unbalanced force due to several rotating masses in the same plane
determined?
4. Discuss causes of unbalance?
5. Explain the method of balancing of different masses revolving in the same plane.
6. How the different masses rotating in different planes are balanced ?
7. Why is balancing of rotating parts necessary for high speed engines ?
8. Explain clearly the terms static balancing and dynamic balancing. State the
necessary conditions to achieve them.
9. Discuss how a single revolving mass is balanced by two masses revolving in different
planes.
10. Why all the rotating systems are not balanced?
11. What do you understand by balancing of revolving masses?
South Valley University Theory of Machine
Faculty of Engineering Code: ENM222
Dept. of Mechanical Engineering 2013/2014
5
12. If not balanced what effects are induced on shaft bearing system due to unbalanced
rotating masses.
13. Five masses A, B, C, D and E revolve in the same plane at equal radii. A, B and C are
respectively 10, 5 and 8 kg in mass. The angular direction from A are 60
o
, 135
o
, 210
o
and 270
o
. Find the masses D and E for complete balance.
14. A shaft carries three pulleys A, B and C at distance apart of 600 mm and 1200 mm.
The pulleys are out of balance to the extent of 25, 20 and 30 N at a radius of 25 mm.
The angular position of out of balance masses in pulleys B and C with respect to that
in pulley A are 90
o
and 210
o
respectively. It is required that the pulleys be completely
balanced by providing balancing masses revolving about axis of the shaft at radius of
125 mm. The two masses are to be placed in two transverse planes midway between
the pulleys.
15. The four masses A, B, C and D are 100 kg, 150 kg, 120 kg and 130 kg attached to a
shaft and revolve in the same plane. The corresponding radii of rotations are 22.5 cm,
17.5 cm, 25 cm and 30 cm and the angles measured from A are 45
o
, 120
o
and 255
o
.
Find the position and magnitude of the balancing mass, if the radius of rotation is 60
cm.
With my best wishes
Dr. Nouby M. Ghazaly | 2,283 | 9,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.901838 |
https://www.nichesblog.com/search/law-of-cosines-calculator-sss | 1,643,460,128,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00468.warc.gz | 929,522,874 | 9,001 | # Keyword Analysis & Research: law of cosines calculator sss
## Keyword Research: People who searched law of cosines calculator sss also searched
What is the formula for cosine law?
Cosine Rule a2 = b2 + c2 - 2bc cos ∠x b2 = a2 + c2 - 2ac cos ∠y c2 = a2 + b2 - 2ab cos ∠z
When to use law of sines vs law of cosines?
Depending on the information we have available, we can use the law of sines or the law of cosines. The law of sines relates the length of one side to the sine of its angle and the law of cosines relates the length of two sides of the triangle to their intermediate angle.
When to use cosine rule?
The cosine rule is a formula commonly used in trigonometry to determine certain aspects of a non-right triangle when other key parts of that triangle are known or can otherwise be determined.
What is the cosine rule?
The cosine rule, also known as the law of cosines, relates all 3 sides of a triangle with an angle of a triangle. It is most useful for solving for missing information in a triangle. For example, if all three sides of the triangle are known, the cosine rule allows one to find any of the angle measures. | 280 | 1,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-05 | latest | en | 0.902097 |
http://nrich.maths.org/public/leg.php?code=-262&cl=4&cldcmpid=1393 | 1,506,281,939,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690203.42/warc/CC-MAIN-20170924190521-20170924210521-00220.warc.gz | 257,210,336 | 5,468 | # Search by Topic
#### Resources tagged with Geodesics similar to Impossible Polyhedra:
Filter by: Content type:
Stage:
Challenge level:
### There are 8 results
Broad Topics > 3D Geometry, Shape and Space > Geodesics
### Weekly Challenge 47: Weird Universes
##### Stage: 5 Challenge Level:
Consider these weird universes and ways in which the stick man can shoot the robot in the back.
### Geometry and Gravity 1
##### Stage: 3, 4 and 5
This article (the first of two) contains ideas for investigations. Space-time, the curvature of space and topology are introduced with some fascinating problems to explore.
### How Many Geometries Are There?
##### Stage: 5
An account of how axioms underpin geometry and how by changing one axiom we get an entirely different geometry.
### When the Angles of a Triangle Don't Add up to 180 Degrees
##### Stage: 4 and 5
This article outlines the underlying axioms of spherical geometry giving a simple proof that the sum of the angles of a triangle on the surface of a unit sphere is equal to pi plus the area of the. . . .
### Flight Path
##### Stage: 5 Challenge Level:
Use simple trigonometry to calculate the distance along the flight path from London to Sydney.
### Spherical Triangles on Very Big Spheres
##### Stage: 5 Challenge Level:
Shows that Pythagoras for Spherical Triangles reduces to Pythagoras's Theorem in the plane when the triangles are small relative to the radius of the sphere.
### Over the Pole
##### Stage: 5 Challenge Level:
Two places are diametrically opposite each other on the same line of latitude. Compare the distances between them travelling along the line of latitude and travelling over the nearest pole.
### Pythagoras on a Sphere
##### Stage: 5 Challenge Level:
Prove Pythagoras' Theorem for right-angled spherical triangles. | 409 | 1,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-39 | latest | en | 0.840078 |
https://industrial-electronics.com/industrial-electricity-com/dste5_11.html | 1,679,558,169,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00570.warc.gz | 365,801,412 | 4,847 | # Using Wire Tables and Determining Conductor Sizes--part 1
TOPICS:
• Intro; The American Wire Gauge (AWG)
• Using the NEC Charts
• Factors that Determine Ampacity
• Correction Factors
• Calculating Conductor Sizes and Resistance
• Calculating Voltage Drop
• Parallel Conductors
• Testing Wire Installations
• Summary/Quiz
TERMINOLOGY:
• Ambient air temperature
• American Wire
• Gauge (AWG)
• Ampacity (current-carrying ability)
• Circular mil
• Correction factor
• Damp locations
• Dry locations
• Insulation
• Maximum operating temperature
• MEGGER
• Mil-foot
• National Electrical Code (NEC)
• Parallel conductors
• Wet locations
INTRO: Being able to determine the amount of current a conductor is permit ted to carry or the size wire need for an installation is essential to any electrician, whether he or she works as an installation electrician or as a maintenance electrician. This guide:
• explains how the amount of current a conductor is permitted to carry is not the same as selecting the proper wire for an installation and describes the differences.
• differentiates the different types of wire insulation and the appropriate use of each based on ambient temperatures.
• explains the method for using tools such as a MEGGER when deter mining the resistance of wire insulation.
• discusses how conductor length and size impact resistance and determine the required conductor size.
• provides the tools for determining ampacity rating of conductors when applying correction factors for wiring in a raceway.
• explains that, as a general rule, electricians select wire sizes from the NEC.
However, there are instances where the wire run is too long or some special type of wire is being employed. In those cases, wire size and type are chosen by determining the maximum voltage drop and calculating the resistance of the wire.
This guide explains how to determine wire size using the NEC and calculating wire resistance.
Learning goals:
• select a conductor from the proper wire table.
• discuss the different types of wire insulation.
• determine insulation characteristics.
• use correction factors to determine the proper ampacity rating of conductors.
• determine the resistance of long lengths of conductors.
• determine the proper wire size for loads located long distances from the power source.
• list the requirements for using parallel conductors.
• discuss the use of a MEGGER for testing insulation.
FUNDAMENTALS:
The size of the conductor needed for a particular application can be determined by several methods. The National Electrical Code (NEC) is used throughout industry to determine the conductor size for most applications. It’s imperative that an electrician be familiar with Code tables and correction factors.
In some circumstances, however, wire tables cannot be used, as in the case of extremely long wire runs or for windings of a transformer or motor. In these instances, the electrician should know how to determine the conductor size needed by calculating maximum voltage drop and resistance of the conductor.
### The American Wire Gauge (AWG)
The American Wire Gauge was standardized in 1857 and is used mainly in the United States for the diameters of round, solid, nonferrous electrical wire. The gauge size is important for determining the current-carrying capacity of a conductor. Gauge sizes are determined by the number of draws necessary to produce a given diameter or wire. Electrical wire is made by drawing it through a succession of dies.
+++++One side of the wire gauge is marked with the AWG size.
+++++The other side of the wire gauge lists the diameter of the wire in thousandths of an inch.
+++++The slot, not the hole, determines the wire size.
+++++Wire is drawn through a succession of dies to produce the desired diameter. Wire Die Draw block
+++++Wire gauge
Each time a wire passes through a die, it’s wrapped around a draw block several times. The draw block provides the pulling force necessary to draw the wire through the die. A 24 AWG wire would be drawn through 24 dies, each having a smaller diameter. In the field, wire size can be determined with a wire gauge. One side of the wire gauge lists the AWG size of the wire. The opposite side of the wire gauge indicates the diameter of the wire in thousandths of an inch. When determining wire size, first remove the insulation from around the conductor. The slots in the wire gauge, not the holes behind the slots, are used to determine the size.
The largest AWG size is 4/0, which has an area of 211,600 circular mills (CM). Conductors with a larger area are measured in thousand circular mills.
The next largest conductor past 4/0 is 250 thousand circular mills (250 kcmil). Conductors can be obtained up to 2000 kcmil. In practice, large conductors are difficult to pull through conduit. It’s sometimes desirable to use parallel conductors instead of extremely large conductors.
### Using the NEC Charts
NEC 310 deals with conductors for general wiring. Table 310.15(B)(16) through Table 310.15(B)(19) are generally used to select a wire size according to the requirements of the circuit. Each of these tables lists different conditions. The table used is determined by the wiring conditions. Table 310.15(B)(16) lists ampacities (current-carrying ability) of not more than three single
insulated conductors in raceway or cable or buried in the earth based on an ambient (surrounding) air temperature of 30 degree C (868F). Table 310.15(B)(17) lists ampacities of single insulated conductors in free air based on an ambient temperature of 3 C. Table 310.15(B)(18) lists the ampacities of three single insulated conductors in raceway or cable based on an ambient temperature of 40 degree C (1048F). The conductors listed in Table 310.15(B)(18) and Table 310.15(B)(19) are generally used for high-temperature locations. The heading at the top of each table lists a different set of conditions.
+++++ Insulation around conductor. Insulation Conductor
### Factors That Determine Ampacity
Conductor Material: One of the factors that determines the resistivity of wire is the material from which the wire is made. The wire tables list the current-carrying capacity of both copper and aluminum or copper-clad aluminum conductors. The currents listed in the left-hand half of Table 310.15(B)(16), e.g., are for copper wire. The currents listed in the right-hand half of the table are for aluminum or copper-clad aluminum. The table indicates that a copper conductor is permit ted to carry more current than an aluminum conductor of the same size and insulation type. An 8 American Wire Gauge (AWG) copper conductor with Type TW insulation is rated to carry a maximum of 40 amperes. An 8 AWG aluminum conductor with Type TW insulation is rated to carry only 35 amperes.
One of the columns of Table 310.15(B)(18) and Table 310.15(B)(19) gives the ampacity rating of nickel or nickel-coated copper conductors.
====
EXAMPLE: Find the maximum operating temperature of Type RHW insulation. (Note: Refer to the NEC.)
Solution: Find Type RHW in the second column of Table 310.104(A). The third column lists a maximum operating temperature of 758C, or 1678F.
Can Type THHN insulation be used in wet locations?
Solution:
Locate Type THHN insulation in the second column. The fourth column indicates that this insulation can be used in dry and damp locations. This type of insulation cannot be used in wet locations. For an explanation of the difference between damp and wet locations, consult "locations" in Article 100 of the NEC.
====
Insulation Type: Another factor that determines the amount of current a conductor is permitted to carry is the type of insulation used. This is due to the fact that different types of insulation can withstand more heat than others. The insulation is the non conductive covering around the wire). The voltage rating of the conductor is also determined by the type of insulation. The amount of voltage a particular type of insulation can withstand without breaking down is determined by the type of material it’s made of and its thickness. NEC Table 310.104(A) lists information concerning different types of insulation. The table is divided into columns that list the trade name; identification letters; maximum operating temperature; whether the insulation can be used in a wet, damp, or dry location; material; thickness; and outer covering.
A good thing to remember is that insulation materials that contain the letter W, such as RHW, THW, THWN, and so on may be used in wet locations.
Top of Page PREV: Measuring Instruments--part 6: Summary/Quiz NEXT: part 2 Index
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Could someone help me with this Physics question please? Watch
1. I think the value of 'h' is 0.1m from using sin(78)*2. I got 1.9 and I took this away from 2m.
I don't know how to do the second part, any help would be appreciated.
2. use mgh = 0.5mv^2 and rearrange for v
3. (Original post by DenizS)
I think the value of 'h' is 0.1m from using sin(78)*2. I got 1.9 and I took this away from 2m.
I don't know how to do the second part, any help would be appreciated.
You are on the right track, but need to be more accurate and round to three decimal places.
Once you have the height the pendulum falls through, plug that into mgh to get the initial potential energy.
Them rearrange the kinetic energy formula with v as the subject and using the value of mgh you calculated.
4. (Original post by uberteknik)
You are on the right track, but need to be more accurate and round to three decimal places.
Once you have the height the pendulum falls through, plug that into mgh to get the initial potential energy.
Them rearrange the kinetic energy formula with v as the subject and using the value of mgh you calculated.
I have rearranged the formula so it is v= the square root of 2mgh/m
I then substitute in the square root, 2(0.2*9.81*0.1)/0.2. This gives me 1.4.
Is 1.4 my initial velocity?
5. (Original post by DenizS)
I have rearranged the formula so it is v= the square root of 2mgh/m
I then substitute in the square root, 2(0.2*9.81*0.1)/0.2. This gives me 1.4.
Is 1.4 my initial velocity?
rearrange this to get v on it's own:
NB because m is on both sides of the equation it cancels out to leave:
now rearrange:
Your value of h using trigonometry is not accurate enough. Round this value to 2 decimal places (the question gave g = 9.81ms-2 so you need to be consistent with this.
6. (Original post by uberteknik)
rearrange this to get v on it's own:
NB because m is on both sides of the equation it cancels out to leave:
now rearrange:
Your value of h using trigonometry is not accurate enough. Round this value to 2 decimal places (the question gave g = 9.81ms-2 so you need to be consistent with this.
Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
7. (Original post by DenizS)
Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
You want to check your calculation again.
8. (Original post by DenizS)
Okay, I substituted 0.04 as the change in height into the equation to get 0.7848. Is 0.78 my initial velocity?
9. (Original post by uberteknik)
0.93
10. (Original post by DenizS)
0.93
Yes.
Maximum v is when all of the potential energy is converted to kinetic energy at the bottom of the swing.
v = 0.93 ms-1 (2 d.p.)
11. (Original post by uberteknik)
Yes.
Maximum v is when all of the potential energy is converted to kinetic energy at the bottom of the swing.
v = 0.93 ms-1 (2 d.p.)
Thank you
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# Assume that Japan and Korea each has 2400 hours available. Originally, each country divided its time equally between the production of cars and airplanes. Now, each country spends all its time producing the good in which it has a comparative advantage. As a result, the total output of cars increased by a. 80. b. 40. c. 16. d. 64.
+1
1. 14 July, 14:56
0
Explanation:
Since 60minutes = 1 hour
2400 : 60
= 40
The total output of cars increased by 40
The law of comparative advantage states that a country will specialise in producing commodities in which the country has greatest comparative advantage. Therefore a country will produce for export those commodities she can produce cheaply than other countries. | 178 | 734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-31 | latest | en | 0.93528 |
https://manoxblog.com/2016/11/06/how-many-visitors-per-day-does-a-site-need-to-draw-any-significant-ad-revenue/ | 1,721,570,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00181.warc.gz | 319,268,922 | 19,293 | # How many visitors per day does a site need to draw any significant ad revenue?
The text that follows is owned by the site above referred.
There are already some good answers on this topic. I didn’t see any that took the \$100 a week number in the question details into consideration. So let’s sharpen our pencils, bust out the calculators, and figure out how many visitors per day are needed to get to \$100 a week.
First off, let’s get the numbers on the same time frame.
\$100/week / 7 days/week = \$14.29/day
Now let’s figure how many clicks we’ll need to get to \$14.29.
This depends on the cost per click and the publisher/network revenue share. Google AdSense has a revenue share of 68% to the publisher in most cases. That means for every \$1 click, the publisher makes \$0.68 and Google makes \$0.32.
This is where your niche really comes into play. If your site is about financial services or legal, the cost per click is usually pretty high. If your site is about some consumer products (retail, car dealers, real estate, travel) the cost per click will probably be pretty low. We can chart this out by cost per click though:
\$21 CPC * 68% = \$14.28
\$14.29/day / \$14.29/click = ~1 click per day
\$1 CPC * 68% = \$0.68
\$14.29/day / \$0.68/click = ~21 clicks per day
\$0.10 CPC * 68% = \$0.068
\$14.29/day / \$0.068/click = ~210 clicks per day
Now let’s figure how many impressions are needed to get that number of clicks. To get this we’ll need to know the click through rate (CTR) for ads on the site. Generally on display advertising you can expect a CTR of about 0.1%. Niche sites with good ad placement may drive that to 0.5%, but display advertising rarely, if ever, has CTR anywhere near what you would see in search (1%+). Let’s figure both the high and low end of CTR for both the high and low end of clicks needed.
1 click / 0.1% = 1,000 impressions (makes for easy math with CPM)
1 click / 0.5% = 500 impressions
210 clicks / 0.1% = 210,000 impressions
210 clicks / 0.5% = 105,000 impressions
And finally, we can figure how many visitors are needed to get to the required number of impressions. This will depend on the number of impressions per visit. If there is one ad on a page and a visitor tends to visit one page, the impressions per visit would be one. As the number of impressions per visit go up, the CTR will tend to go down, so everything is a balancing act of sorts to optimize the site revenue.
impressions needed / impressions/visit = visits needed
210,000 impressions / 1 impression/visit = 210,000 daily visits
500 impressions / 10 impressions/visit = 50 daily visits
Based on those numbers, the range is somewhere between 50 and 210,000 daily visits.
To sum up…
1. There are a lot of factors at play to determine how many visitors per day are needed to reach \$100/week in ad revenue.
2. \$100/week is probably not going to be enough to attract many (if any) direct advertising buys. This puts you at the mercy of the networks like Google AdSense.
3. Depending on the niche, content, ad placements, number of ads per page, number of pages visitors view, reason for the visit, and more, the actual answer to this question can vary from probably something on the order of very few to quite a lot.
4. Using the methodology outlined above, you can calculate the number of visitors needed in your unique situation. | 844 | 3,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.923787 |
https://religiondocbox.com/Atheism_and_Agnosticism/72972318-On-truth-at-jeffrey-c-king-rutgers-university.html | 1,623,850,130,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00272.warc.gz | 452,339,403 | 30,890 | # On Truth At Jeffrey C. King Rutgers University
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1 On Truth At Jeffrey C. King Rutgers University I. Introduction A. At least some propositions exist contingently (Fine 1977, 1985) B. Given this, motivations for a notion of truth on which propositions can be true at worlds according to which they don t exist. 1. (Fine 1977) Some propositions seem to correctly (partly) characterize worlds in which they don t exist. a. The proposition that Will Ferrell doesn t exist and a deprived Ferrell-less world. 2. (Adams 1981, King 2007) The following seems true to most of us: 1. It is possible that Will Ferrell didn t exist. a. But for 1 to be true, the proposition that Will Ferrell doesn t exist must be true according to some possible world w. 3. (Stalnaker 2009) Possible worlds are maximal, which requires that for any world w and any proposition p, either p or ~p is true according to w. a. Consider a proposition p that fails to exist at some world w. b. Plausible to think ~p doesn t exist at w either. c. But since w is maximal, either p or ~p is true according to w. C. I ll call the notion of truth according to which propositions can be true according to worlds where they don t exist truth at; I ll use true in for the notion of truth according to which propositions must exist at worlds to be true there. D. In King (2007) I defended a novel account of propositions. 1. On the view defended there, it seems likely that all propositions exist contingently. E. Other theoretical commitments 1. Actualism the actual world exhausts all that there is. 2. Serious actualism: an entity possesses a property at a world w only if it exists at w F. The Plan 1
2 1. State my account of propositions. 2. Explain why propositions exist contingently on that view. 3. Discuss some features we want truth at to have. 4. Formulate my account of truth at and discuss some of its consequences. II. A Theory of Propositions A. Motivation for the theory 1. Perhaps the most common view of propositions is that they are eternal, abstract entities that by their very nature and independently of all minds and languages represent the world as being a certain way and so have truth conditions. 2. I cannot accept this view. 3. Having decided that propositions can t be the sorts of things that represent/have truth conditions by their very natures and independently of minds and languages, one can either reject propositions altogether or construct an account of propositions on which they were somehow endowed with their representational capacities. a. Given the many jobs propositions perform in philosophy, I set out to do the latter. b. I call the resulting account an account of naturalized propositions. B. Statement of the theory 1. An object possessing a property, n objects standing in an n-place relation, n properties standing in an n-place relation and so on are all facts. a. If an object o possesses the property P, there is a fact of o possessing P. 1 If not, there is no such fact. 2. I claim propositions are certain kinds of facts. a. If a proposition is true, it is made true by a fact (or facts) distinct from it. b. So, for example, the fact that is the proposition that Rebecca swims is distinct from the fact of Rebecca swimming and the latter makes the former true. 1 I am purposely using this rather unwieldy expression a/the fact of o possessing P, as I also did in King (2007), to avoid saying the fact that o possesses P. The latter expression is an expression of ordinary English and it is a substantive claim that the English expressions beginning the fact that designate what I am calling facts here. Hence I don t want to use these expressions. 2
4 a. First, in so far as one finds serious actualism intuitively compelling, one finds it intuitively plausible to think that it is correct to deny that an object possesses a property or stands in a relation at a world according to which it doesn t exist. i. This, in turn, suggests that a proposition that denies that an object possesses a property or stands in a relation ought to be true at a world according to which the object doesn t exist. ii. And it is plausible to think that Not[Ra 1,,a n ] denies that a 1,,a n (in that order) stand in R. ii. But then it should be true at a world according to which at least one of a 1,,a n fails to exist. b. Second, intuitively, it just seems as though the propositions that Socrates doesn t exist, that Socrates isn t tall and so on are true at worlds where Socrates doesn t exist. c. Third, and related to this, counterfactuals such as the following intuitively seem true: 2. If I hadn t existed, I wouldn t have been six feet tall. i. Assuming that the truth of such counterfactuals requires the consequent to be true at (some range of ) worlds where the antecedent is, the truth of such counterfactuals seems to require that the proposition that I am not six feet tall is true at worlds according to which I don t exist. 3 IV. Adams on truth at and propositions of the form applep or p, where p is a singular proposition about o. A. Adams claims that any proposition of the form applep or p, where p is a singular proposition about o, is false at any world w according to which o doesn t exist. 1. As Adams himself recognizes, this has many unfortunate consequences. a. For example, consider the following proposition: (3) Not[Bush exists] Not[Bush exists] 2. So why does Adams adopt the view that any proposition of the form p or applep, where p is a singular proposition about o, is false at worlds where o doesn t exist? 3 Not to mention requiring that the proposition that I don t exist is true at worlds according to which I don t exist, (the counterfactual hardly seems vacuously true!). Note that if this is correct, counterfactuals invoke true at in the sense that for a counterfactual to be true, the proposition expressed by its consequent must be true at (some range of) worlds at which the proposition expressed by its antecedent is true. 4
6 world where it doesn t exist; but a proposition of the form of (5) where Ψ is a property (and so Ψ(o) is atomic) can be true at a world where o doesn t exit. iii. So by Adams lights (5) asserts that o lacks a property rather than ascribing the property of non-ψ-ing to o. But then Adams need to explain why (4), unlike (5), does attribute a property (again, presumably the property of possibly Ψ-ing) to o. c. Second problem (related to previous point): it is at least a little odd to claim that while propositions of the form (5) need not ascribe properties to o, the results of embedding them under a modal element must: (6) Not[Ψ(o)] i. Surely we should be told why the modal element has this effect. ii. After all, in the case of negation, we do have an explanation of why a complex proposition consisting of negation embedding a singular proposition about o need not ascribe a property to o. Such a proposition may require for its truth at a world merely that o fails to possess a property. iii. But what is the account of why embedding a singular proposition that doesn t ascribe a property to o under a modal element yields a proposition that does ascribe a property to o? Some explanation of this should be given. C. Adams second reason for thinking that propositions of the form p and applep, where p is a singular proposition about o, are false at worlds where o doesn t exist. 1. Adams thinks that such propositions ascribe properties to the proposition that p. 2. I assume he means that a proposition of the form (4) Ψ(o) where Ψ(o) is a singular proposition about o, attributes to Ψ(o) the property of being possible. 3. If that s right, the truth of the proposition Ψ(o) at a world w requires Ψ(o) to possess this property there, in violation of serious actualism. 4. Problems with Adams second reason for thinking that propositions of the form applep or p, where p is a singular proposition about o, are false at any world w according to which o doesn t exist. a. Consider again the case of negation and the proposition that Adams doesn t exist: 6
8 3. Third, we took the intuitive truth of counterfactuals like If Socrates hadn t existed, he wouldn t have been tall to be evidence that the proposition that Socrates isn t tall is true at worlds where Socrates doesn t exist. B. The same three reasons can be given in the present case. 1. First, for the proposition that possibly Socrates exists to be true at w the proposition that Socrates exists must be true at some w. This doesn t seem to require Socrates to exist or possess any properties at w at all! Hence, we have an intuitive account of how the proposition that possibly Socrates exists can be true at a world w where Socrates doesn t exist that is consistent with serious actualism. 2. Second, when we ask whether the proposition that it is possible that Socrates exists is true at worlds where he doesn t exist, again intuitively the answer is yes. But then, just as in the case of the proposition that Socrates doesn t exist, this is at least some reason to think that the proposition really is true at such a world. 3. Third, consider the following counterfactual: 9. If Socrates hadn t existed, it would have (still) been possible that he existed. Such counterfactuals again strike people as true. But their truth presumably requires the proposition that it is possible that Socrates existed to be true at worlds where Socrates and the proposition that Socrates exists don t exist. So here again the truth of 9 provides some evidence that the proposition that it is possible that Socrates exists is true at worlds according to which he doesn t exist. C. To square the conclusion that the proposition that possibly Socrates exists can be true at a world w where Socrates doesn t exist with serious actualism, we have to say that this proposition being true at such a world doesn t require Socrates or the proposition that Socrates exists to possess an properties at w. 1. In particular, it doesn t require Socrates to possess the property of possibly existing at w and it doesn t require the proposition to possess the property of being possible at w. 2. But something similar has to be said in the case of the proposition that Socrates doesn t exist. a. It can be true at a world w without Socrates possessing the property of nonexistence at w and without the proposition that Socrates exists possessing the property of not being the case at w. 3. Finally, it is worth noting that our intuitions regarding the proposition that Socrates doesn t exist and the proposition that possibly Socrates exists contrast sharply with our intuitions regarding propositions that do require Socrates or the proposition that Socrates exists to possess properties in order to be true at a world. 8
9 a. Intuitively, we think the proposition that Socrates doesn t exist and the proposition that possibly Socrates exists are true at a world where Socrates doesn t exist. b. Intuitively, we do not think the proposition that Socrates is snubnosed or the proposition that Plato believes Socrates exists are true at such a world. D. In short, the serious actualist who thinks that singular propositions about Socrates don t exist at worlds where Socrates doesn t exist should hold that the propositions that Socrates doesn t exist and that possibly Socrates exists are on par in relevant respects. To the extent that you, like me, think the former should be true at worlds where Socrates doesn t exist, you should think the latter should be as well. VI. Preliminaries to the definition of truth at A. Possible worlds 1. I take merely possible worlds to be uninstantiated, maximal properties that the universe could have had. a. The modality here is primitive. b. As to maximality, the natural idea is to begin with all properties the universe could have had. i. Call these world properties. ii. Say that a world property p necessitates a world property q iff it is impossible for p to be instantiated and q not be instantiated. A world property p antinecessitates a world property q iff it is impossible for p to be instantiated and for q to be instantiated. 5 Again here, the modality is primitive. iii. A maximal world property p is one such that for any world property q, p necessitates or anti-necessitates q. 2. The properties that I take to be possible worlds exist uninstantiated in the actual world. a. Hence, the existence of possible worlds is compatible with actualism: the uninstantiated properties that are possible worlds exist in the actual world. 5 The characterizations of world properties necessitating and anti-necessitating other world properties bear an obvious resemblance to Plantinga s (1976) characterizations of states of affairs including and precluding each other. 9
10 b. Of course, propositions exist in the actual world as well. c. In attempting to characterize a relation between propositions and possible worlds---the truth at relation we are attempting to characterize a relation that obtains between propositions and worlds in the actual world. B. I ll adopt the view that a proposition of the form [Ra 1,,a n ] (where R is an n- place relation) is false, and its negation is true, at a world according to which at least one of a 1,,a n doesn t exist. C. As I said in discussing Adams, I ll also hold that propositions of the form p and applep, where p is a singular proposition about o, can be true at worlds where o doesn t exist. D. Worlds w according to which something is F, though there is no (actual) individual o such that according to w o is F. 1. Though my father Richard has no brother, he might have had one. So according to some possible world w, Richard has a brother. 2. Assume again that nothing in the actual world could have been Richard s brother. 3. Then the fact that Richard has a brother according to w cannot be the result of some actual thing having the property of being Richard s brother according to w. 4. So how is it that Richard has a bother according to w? a. On views like Adams (1981) or Stalnaker s (2008a, 2008b), where possible worlds are sets of propositions, it is easy for w to do this. b. But how is it that worlds construed as I construe them as maximal properties the world might have had are such that according to them Richard has a brother? c. Let s say in such a case that the world depicts Richard having a brother. d. Presumably, worlds depict Richard having a brother by depicting a bunch of properties being jointly instantiated and depicting the thing that instantiates these properties as bearing relations to other things. i. On this picture, worlds depict actual individuals as possessing properties and standing in relations and also depict bunches of properties being jointly instantiated (and represent the joint instantiater as bearing relations to other things) without depicting any (actual) object as instantiating these properties (and bearing relations to other things). ii. For example, a world might depict the properties P, Q and the relational property Ra (where R is a two-place relation) as standing in the relation of joint instantiation. 10
11 5. On my view of properties and possible worlds as properties the universe might have had, the properties that are worlds are quite complex and have lots of parts. 6 a. For example, any world that depicts humans existing has the property of being human as a part; and any world that depicts me as existing has me as a part. Similarly, when a world depicts a nonactual individual as existing by depicting properties as jointly instantiated and relations as obtaining, there is a part of the world that does this depicting (which itself is complex and has the properties it depicts as instantiated and the relations it depicts as obtaining as parts). b. Call these parts of worlds, whereby nonactual individuals are depicted by depicting properties as jointly instanced, relations as obtaining, etc. faux individuals. c. When a world depicts a faux individual by depicting properties being jointly instantiated and relations obtaining, I ll say the world depicts the faux individual as possessing the properties and standing in the relations. d. I assume faux individuals are world bound; no faux individual in one world is the same faux individual as a faux individual in another world. VII. A definition of truth at A. Suppose our language contains n-place predicates ('A', 'B', with or without numerical subscripts) for all n>0; individual variables ('x','y', with or without numerical subscripts) and names ('a','b','c', with or without numerical subscripts). 1. Assume our languages also contains the one-place sentential connective ~ ;the twoplace sentential connective & ; the determiners 'every' and 'some'; and the operator possibly. 2. The syntax is as follows: 1. If δ is a determiner, α is a variable and Σ is a formula containing free occurrences of α [δασ] is a quantifier phrase. 2. If Π is an n-place predicate and α 1,...,α n are names or variables, [Πα 1,...,α n ] is a formula. 3. If Ω is a quantifier phrase and Σ is a formula, then [ΩΣ] is a formula. 4. If Ψ and Φ are formulas, so are [Φ] and [Φ&Ψ]. 6 This view of some properties and relations being complex and having other properties and relations as parts is discussed in King (2007), especially Chapter 7. 11
12 5. If Φ is a formula, so is possibly[φ]. Sentences are formulae with no free variables. B. Formulae express propositional frames as follows: 1. The propositional frame expressed by [Πα 1,...,α n ] is [Π * α 1 *,...,α n * ], where Π * is the n-place relation expressed by Π; α i * (1<i<n) is the bearer of α i if α i is a name, and α i itself if α i is a variable. 2. The propositional frames expressed by ~Σ and [Σ&Ψ] are NOT[Σ ] and [Σ AND Ψ ], respectively, where NOT and AND are the truth functions expressed by ~ and &, respectively; and Σ, Ψ are the propositional frames expressed by Σ and Ψ, respectively. 3. The propositional frame expressed by [[δξσ]ψ] (where ξ is a variable) is [[δ ξσ ]Ψ ], where Σ, Ψ are the propositional frames expressed by Σ, Ψ respectively; and δ * is the semantic value of δ. 4. The propositional frame expressed by possibly[σ] is POSSIBLY[Σ ], where Σ is the propositional frame expressed by Σ and POSSIBLY is a function that maps a proposition S and world w to true iff for some w, S is true at w. 5. I call these propositional frames because they include things expressed by formulae containing free variables. Propositions are propositional frames containing no free variable. D. Assume we have a set W of possible worlds and for every w ε W, we have the domain D w of (non-faux) individuals that exist according to w. 1. Further, assume that for each world w, we have functions that assign individuals in D w or faux individuals that exist according to w to variables and assign individuals to themselves For a world w, let g w be such a function. Also, let Θ, Ξ, Ω be propositional frames, R be an n-place relation (1< n), e 1,,e n be individuals or variables, ξ be a variable, EVERY and SOME be the semantic values of every and some, and POSSIBLY as above. 3. Then g w satisfies a propositional frame Ξ relative to w iff 7 Since faux individuals are complex parts of worlds, there is nothing odd about functions assigning them to variables. However, note that g w assigns to variables only faux individuals that exist according to w. As I said above, I view faux individuals as world bound and so I don t try to say under what conditions a faux individual in w is the same faux individual as a faux individual in another world w. 12
13 1. Ξ = [R e 1,,e n ]: w depicts g w (e 1 ), g w (e n ) in that order standing in R. 2. Ξ = NOT[Ω]: g w fails to satisfy Ω relative to w. 3. Ξ = [Θ AND Ω]: g w satisfies both Θ and Ω relative to w. 4. Ξ = [[SOME ξ Ω] Θ]: some g w that differs from g w at most on what it assigns to ξ satisfies Ω and Θ relative to w. (similar clause for [[EVERY ξ Ω] Θ]) 5. Ξ = POSSIBLY[Ω]: for some w and some g w that agrees with g w on all the free variables in Ω, g w satisfies Ω relative to w. 8 Finally, a proposition X is true at w iff every function g w satisfies X relative to w; otherwise, it is false at w. VIII. But what about McMichael (1983)? A. Consider the following sentence: 10. It is possible that Richard King should have had a brother who was a lawyer but might not have been a lawyer. 10a. x(rx & Lx & ~Lx) 1. On standard semantic approaches, the truth of 10 in the actual world requires there to be a possible world w containing an individual o who is Richard s brother and a lawyer and for there to be another world w in which o fails to be a lawyer. 2. The problem is that for the actualist, there are no merely possible (i.e. possible and non-actual) individuals. 3. Hence, assuming, as before, that no actual person could have been my father s brother, there is no w that contains an individual o that is a lawyer and Richard s brother. 4. Of course, on my view, there will be a world that depicts the properties of being Richard King s brother and being a lawyer as jointly instantiated. a. And that there is such a world containing such a faux individual makes the following sentence true: 11. It is possible that Richard King should have had a brother who was a lawyer. 8 See previous note. If g w assigns a faux individual in w to some variable in Ω, then for w w, g w cannot agree with g w on all variables in Ω. 13
14 5. But for 10 to be true we would have to be able to make sense of the faux individual in w that makes11 true existing in some w ( w) and not being a lawyer there. 6. Nothing in the machinery I have sketched does this. Hence, the proposition expressed by 10 is false for all I ve said to this point. B. I am inclined to think that the actualist should accept this conclusion. 1. That 10 is false is something we learn when we learn that actualism is true. 2. Though there might have been things that aren t actual in the sense that merely possible worlds depict properties being jointly instantiated without depicting any actual object instantiating them, it is just false to say that such things might have been different from the way such worlds depict them as being. C. However, one might also try to explain why 10 seems true by claiming that it aims at a true claim in the vicinity. 1. What s true is that if certain merely possible worlds, say w, had been actual, there would have been individuals who aren t in fact actual; and had that been so, these individuals might have been different from the way there are in w. 14
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### Bob Hale: Necessary Beings
Bob Hale: Necessary Beings Nils Kürbis In Necessary Beings, Bob Hale brings together his views on the source and explanation of necessity. It is a very thorough book and Hale covers a lot of ground. It | 11,258 | 48,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-25 | longest | en | 0.901062 |
https://discuss.leetcode.com/topic/87705/simple-and-fast-c-solution-using-vector | 1,516,462,097,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00795.warc.gz | 658,117,406 | 8,397 | # Simple and fast C++ solution using vector
• A lot of the simpler C++ solutions suffer from using string concatenation on every iteration, which makes them take order n^2 steps. This simple solution computes the solution in linear time by building the result backwards in a vector, then reversing it. This allows it to avoid having to pre-compute the size of the result, making it simpler. It's also easier and faster to simply add the values and then extract the two bits of the sum, rather than compute each bit separately.
``````class Solution {
public:
string addBinary(string a, string b) {
vector<char> result;
int carry = 0;
for (int ai = a.length() - 1, bi = b.length() - 1; ai >= 0 || bi >= 0; ai--, bi--) {
int a_val = ai < 0 ? 0 : (a[ai] - '0');
int b_val = bi < 0 ? 0 : (b[bi] - '0');
int sum = a_val + b_val + carry;
result.push_back((sum & 1) + '0');
carry = sum >> 1;
}
if (carry) result.push_back('1');
reverse(result.begin(), result.end());
return string(result.begin(), result.end());
}
};
``````
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# hw4solns - CMPS 101 Summer 2009 Homework Assignment 4...
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1 CMPS 101 Summer 2009 Homework Assignment 4 Solutions 1. (3 Points) Consider the function ) ( n T defined by the recurrence formula + < = 3 ) 3 / ( 2 3 1 6 ) ( n n n T n n T a. (1 Points) Use the iteration method to write a summation formula for ) ( n T . Solution: ) 3 / ( 2 ) ( n T n n T + = ) ) 3 / 3 / ( 2 3 / ( 2 n T n n + + = ) 3 / ( 2 3 / 2 2 2 n T n n + + = ) 3 / ( 2 3 / 2 3 / 2 3 3 2 2 n T n n n + + + = etc.. After substituting the recurrence into itself k times, we get ) 3 / ( 2 3 2 ) ( 1 0 k k k i i i n T n n T + = - = . This process terminates when the recursion depth k is chosen so that 3 3 / 1 < k n , which is equivalent to 3 3 / 1 < k n , whence 1 3 3 + < k k n , so 1 ) ( log 3 + < k n k , and hence ) ( log 3 n k = . With this value of k we have 6 ) 2 or 1 ( ) 3 / ( = = T n T k . Therefore ) ( log 1 ) ( log 0 3 3 2 6 3 2 ) ( n n i i i n n T + = - = . b. (1 Points) Use the summation in (a) to show that ) ( ) ( n O n T = Solution: Using the above summation, we have ) ( log 1 ) ( log 0 3 3 2 6 ) 3 / 2 ( ) ( n n i i n n T + - = since x x for any x ) 2 ( log 0 3 6 ) 3 / 2 ( n n i i + = adding -many positive terms ) 2 ( log 3 6 ) 3 / 2 ( 1 1 n n + - = by a well known formula ) ( 6 3 ) 2 ( log 3 n O n n = + = ) ( 1 ) 2 ( log 3 2 ) 2 ( log 3 3 n o n = < < Therefore ) ( ) ( n O n T = .
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2 c. (1 Points) Use the Master Theorem to show that ) ( ) ( n n T Θ = Solution: Let 0 ) 2 ( log 1 3 > - = ε . Then 1 ) 2 ( log 3 = + ε , and ) ( ) 2 ( log ) 2 ( log 3 3 ε ε + + Ω = = n n n . Also for any c in the range 1 3 / 2 < c , and any positive n , we have cn n n = ) 3 / 2 ( ) 3 / ( 2 , so the regularity condition holds. By case (3) of the Master Theorem ) ( ) ( n n T Θ = .
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https://practicaldev-herokuapp-com.global.ssl.fastly.net/rmion/smoke-basin-20h2 | 1,726,715,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00436.warc.gz | 429,145,591 | 48,519 | Robert Mion
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# Smoke Basin
## The task at hand
### Solve for X where
`X = a combination of individual counts`
1. Sum of risk levels of each low point
2. Product of the size of the three largest basins
### Input is
• A multi-line string
### It represents
• A 100x100 square area
• Each location is a number between `0-9`
• Each number represents an altitude
## Part 1: Rounds 1 and 2
I solved this part of the challenge on the day it was released.
### In 50 lines of code, my original algorithm looked something like this:
``````Setup:
Split the file's contents into an array of strings
Split each string into an array of characters
Convert each character into a number 0-9
Create an array to collect low points
Sub-routine to record a low point:
Add the current number to the array of low points, only if:
The minimum number among a list of numbers...
...is equal to the number at the current location...
...and the number in the last location...
...of the sorted and reversed list of numbers...
...is the number at the current location
For each row in the grid
For each cell in the row
Through a series of nested if..else if..else clauses...
...call the sub-routine, passing in the appropriate amount of arguments, each referencing the correct adjacent cell values
For each value in the collection of low points
Update the value to itself plus one
For each value in the incremented collection of low points
Add the value to an accumulating sum of all previous values
Return the final sum
``````
### Round 2: Now only 10 lines of code!
``````Same setup: process input file into a 2D array of numbers
Set risk level to 0
For each row in the grid
For each cell in the row
If the value in the cell is less than...
...the minimum number in a list of 2-4 numbers...
...resulting from a starting array of four relative coordinates (up one, left one, right one, down one)...
...updated to contain the values in each of the cells referenced by those relative coordinates...
...filtered to remove undefined values attained by references to cells that are out of boundary
Increment risk level by the current cell's value plus one
Return risk level
``````
## Feeling invigorated after refactoring Part 1's code
• I put into practice a slew of new coding tricks I learned from studying other solvers' code in prior challenges
• I resolved a silly mistake in my code that failed to omit cell values that were less than their adjacent cells' values
• I felt proud of the conciseness and eloquence of my code, especially as compared to Round 1
## Part 2: Round 1
When it was originally released, I didn't even attempt to solve part 2.
That's because I was unable to understand what was happening in the example diagrams.
This time around - with careful analysis, and perhaps an improved eye for interpreting puzzle instructions - I understood how the size of each basin was determined.
### Two ways to solve
1. Use `Find` in my browser with the query `9` and my eyes to identify the largest basins and tally the sizes manually
2. Write an algorithm that tallies each basin size after processing the entire grid area
Obviously, I felt motivated to solve it via #2.
### Defining what the algorithm should return
``````21 AA
39 A
``````
Given the input on the left, the algorithm should find a basin of size 3
``````219 AA
399 A
996 B
``````
Given the input above, the algorithm should two basins: one of size 3, another of size 1
``````89876995 A BBB C
93987896 D BBB C
54598939 DDD B E
95999123 D EEE
89891019 F G EEE
``````
Given the input above, the algorithm should seven basins with sizes: 1,7,2,5,7,1,1
``````9299999 A
9193939 A A A
9012349 AAAAA
9101999 AAA
9912329 AAAA
9999939 A
``````
Given the input above, the algorithm should find a basin of size 17
And, of course, the example provided:
``````2199943210 AA BBBBB
3987894921 A CCC B BB
9856789892 CCCCC D B
8767896789 CCCCC DDD
9899965678 C DDDDD
``````
Given the input above, the algorithm should four basins with sizes: 3,9,9,14
### Describing how this algorithm should work
``````Create a unique set to track visited locations
Create an empty list to track basin sizes
For each row in the grid
For each cell in the row
If the cell's value is not 9, and its coordinates are not in the list of traversed cells:
* For each of 2-4 cells above, left, right and below it:
If the cell's value is 9:
Return the locations visited along the way to this cell
Else if the cell's value is 1 higher or lower than the source cell's value:
Continue at * with the current list of visited locations
Add to the list of basins:
The size of the unique set of visited locations in the basin that this cell resides
For each size in basins
Sort the list of basins from largest to smallest
Extract the largest three values
Return the product of all three values
``````
### Delight, then disappointment
• Running my algorithm on each of the samples above produces the correct output: it found each basin and determined the size correctly
• Sadly, when I ran my algorithm on my input, the answer it generated was too low
• To diagnose, I searched for a large-looking basin in my output. I found one at the bottom. I counted the values. I ran my algorithm. I found a problem.
### UPDATED: Then extreme delight!
I solved it later the same day! See my revised notes below to learn how I stumbled upon - and resolved - my algorithm's error.
### A summary of my journey through this challenge
• I solved Part 1 a while ago
• At that time, I didn't understand Part 2 enough to even attempt it
• Now, I solved Part 1 with an algorithm 1/5 the original's size!
• I actually understand what's happening in Part 2
• I wrote an algorithm that generates the correct answer on several small example data sets
• I discovered an example basin that my algorithm doesn't properly compute, in case I ever want to come back and meticulously debug it
My reflection from before solving Part 2:
``````It's a bummer that I couldn't solve Part 2.
I'm definitely proud of how much progress and practice this challenge offered me.
Now, it's time to move on.
``````
## Part 2: Round 2
• I came back later the same day to investigate how my algorithm processed that basin
• I discovered a new truth about the numbers in the grid: they may not always change by 1 from cell to cell
• I removed parts of my code that were overly conditional
• I generated the correct answer for Part 2!
• Bonus: I made another simulator to show the low points and basins for any valid grid
My reflection from after solving Part 2:
``````Just kidding!
It turns out, I couldn't let this puzzle go unsolved.
Luckily, with a simple logging statement, I could see which locations my algorithm started from.
Upon reviewing the first unexpected starting location, I noticed it was a difference of 2 from each of its adjacent cells.
I assumed there was only ever a difference of 1.
I removed that condition from my code.
Viola! My algorithm now calculated the correct - larger - sizes of each basin...and thus determined the correct product of the three largest basins
``````
What a huge relief, fantastic sense of closure, and an incredibly rewarding personal achievement!
Bonus for code golf fans: my final code for both solutions is only 40 lines.
Next! | 1,798 | 7,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.827108 |
https://www.akaqa.com/question/q19192030740-If-1liter8kilometer520kilometerliter?page=1 | 1,695,516,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00725.warc.gz | 714,973,424 | 12,621 | if 1liter=8kilometer,520kilometer=liter?
1l=8k, 520k=l?
0 Views: 701 Answers: 3 Posted: 11 years ago
This is impossible to answer. A kilometer is a measure of distance, while a liter is a measure of volume.
donxeon
buh still there is an answer cuz he asked if it was possible
8 kilo= 1liter
520 kilo= (520 X 1)/8
= 520/8=65
520 kilos= 65 liters....dats the answer to yur question
64lts
ROMOS
Karma: 2940
Bob/PKB
Karma: 1830
Colleen
Karma: 1760
Benthere | 172 | 468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-40 | latest | en | 0.730778 |
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Straight Dope Message Board Remember Me?
#1
09-13-2017, 01:27 PM
HeyHomie Charter Member Join Date: Sep 1999 Location: Viburnum, MO Posts: 9,291
Space Elevator Question: If You're At The "Top," Are You In Zero-G?
Let's say the prefect the Space Elevator. You get on it and ride it up to the top, 400km above the surface of the Earth. Will you be "weightless" in the sense that astronauts aboard the ISS are?
__________________
If you see "Sent from my phone blah blah blah" in my post, please understand that this is automatic when I post from my phone, and I don't know how to disable it. Sorry.
#2
09-13-2017, 01:50 PM
Weisshund Guest Join Date: Dec 2016 Posts: 1,658
Neat question.
I have no answer, but it definitely made me wonder, since you would be technically spinning pretty fast, would you experience false gravity and be drawn outwards?
#3
09-13-2017, 01:51 PM
Darren Garrison Guest Join Date: Oct 2016 Posts: 5,987
No. You would be weightless only at the height of geosynchronous orbit.
#4
09-13-2017, 01:53 PM
scr4 Member Join Date: Aug 1999 Location: Alabama Posts: 14,118
You can't have a 400 km tall space elevator. A space elevator is basically a ridiculously long satellite in geostationary orbit. One end of the elevator reaches all the way to the ground, but the center of mass of the elevator is at 35,800 km altitude.
#5
09-13-2017, 01:59 PM
scr4 Member Join Date: Aug 1999 Location: Alabama Posts: 14,118
P.s. if you are just talking about a 400km tall tower, then no, you wouldn't be weightless at the top. If you step off the tower at the top, you'll fall straight down to the ground. Which is why it's not really a space elevator. A satellite at that altitude is "weightless" only because it's moving very fast, about 7 km/s. If you had such a tower and wanted to launch a satellite from the top, you'd still have to attach a rocket to it and accelerate it to 7 km/s.
Last edited by scr4; 09-13-2017 at 02:01 PM.
#6
09-13-2017, 02:04 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Actually the center of mass would have to be well above the geostationary altitude, and the actual top with the counterweight much higher still, to counteract the weight of the cable and its cargo. Everything above geostationary altitude of course wants to have a longer orbital period than geostationary, so at geostationary velocity it tries to pull away to a higher altitude, producing tension on the cable.
#7
09-13-2017, 02:06 PM
DPRK Guest Join Date: May 2016 Posts: 1,160
You would be weightless after you stepped off the tower, just like the satellite, until you hit the atmosphere. The weightless satellite is falling as well, only it keeps missing the Earth.
#8
09-13-2017, 02:11 PM
eburacum45 Guest Join Date: Feb 2003 Location: Old York Posts: 2,602
Quote:
Originally Posted by wolfpup Actually the center of mass would have to be well above the geostationary altitude, and the actual top with the counterweight much higher still, to counteract the weight of the cable and its cargo.
If you count the mass of the cable and its cargo together, the centre of mass needs to be at the geostationary point, or only a tiny bit above to cause some tension in the cable. If you put the centre of gravity too high the cable will fall upwards away from the Earth.
#9
09-13-2017, 02:14 PM
carnivorousplant KB not found. Press any key Charter Member Join Date: Apr 2000 Location: Central Arkansas Posts: 55,073
Quote:
Originally Posted by eburacum45 If you put the centre of gravity too high the cable will fall upwards away from the Earth.
I thought you tied the Earth end to a very large rock or something.
Was Clarke's tethered in the Fountains of Paradise?
__________________
You callous bastard!
More of my illusions have just been shattered!!
-G0sp3l
Last edited by carnivorousplant; 09-13-2017 at 02:16 PM.
#10
09-13-2017, 02:22 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Quote:
Originally Posted by eburacum45 If you count the mass of the cable and its cargo together, the centre of mass needs to be at the geostationary point, or only a tiny bit above to cause some tension in the cable. If you put the centre of gravity too high the cable will fall upwards away from the Earth.
Well, the cable is anchored at the equator, it's not going anywhere! If you make the center of mass too high you're going to needless expense and putting needless extra tension on the cable, but it does have to be above the geosynchronous point to a sufficient extent to support the payload, whose maximum weight will be at the surface.
#11
09-13-2017, 02:28 PM
scr4 Member Join Date: Aug 1999 Location: Alabama Posts: 14,118
Quote:
Originally Posted by carnivorousplant I thought you tied the Earth end to a very large rock or something. Was Clarke's tethered in the Fountains of Paradise?
The earth end of Clarke's space elevator was indeed anchored, with some tension, but just enough to keep the elevator stable. If I remember correctly, in the book, someone asks what happens if the cable were severed, and the engineer replies it'll just hang there - and someone else says that's not literally true, there's some tension to keep it stable.
#12
09-13-2017, 02:47 PM
carnivorousplant KB not found. Press any key Charter Member Join Date: Apr 2000 Location: Central Arkansas Posts: 55,073
Quote:
Originally Posted by scr4 someone asks what happens if the cable were severed, and the engineer replies it'll just hang there - and someone else says that's not literally true, there's some tension to keep it stable.
I remembered that, but I thought it was narration. Thanks!
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#13
09-13-2017, 02:55 PM
dstarfire Member Join Date: Oct 2009 Location: Tacoma, WA; USA Posts: 1,374
If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth's surface) is in a stable orbit. So, at the top, you'd also be in a stable orbit and, therefore, weightless.
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#14
09-13-2017, 03:10 PM
iamthewalrus(:3= Guest Join Date: Jul 2000 Location: Santa Barbara, CA Posts: 10,757
Quote:
Originally Posted by scr4 P.s. if you are just talking about a 400km tall tower, then no, you wouldn't be weightless at the top. If you step off the tower at the top, you'll fall straight down to the ground. Which is why it's not really a space elevator. A satellite at that altitude is "weightless" only because it's moving very fast, about 7 km/s. If you had such a tower and wanted to launch a satellite from the top, you'd still have to attach a rocket to it and accelerate it to 7 km/s.
This is wrong in many ways.
If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn't hold together at all.
Note that they don't have to put rockets at the top of skyscrapers to accelerate them to the appropriate speed after they're built, even though the tops are moving faster than the foundations.
#15
09-13-2017, 03:16 PM
sbunny8 Guest Join Date: Nov 2009 Location: Eugene, Oregon Posts: 1,087
Geosynchronous orbit is about 35,800 km above Earth's surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit. At every point, from the bottom to the middle to the top, your perceived gravity would be the difference between how fast Earth's gravity is pulling you downward and how fast the floor under your feet is moving away from you centripetally due to rotating once every 24 hours. At the equator, you're 6,371 km from Earth's center, rotating 40,030 km in 24 hours, which is 1668 kph or 463.3 meters per second, hence Earth is moving away from your feet at about .03 m/s2. Actual gravity there is 9.81 m/s2. Subtract the two and your apparent gravity is 9.78 m/s2 downward.
Now step in to the elevator and to up to geosynchronous orbit, 35,800 km up. Now you're 42,164 km from Earth's center. At that point, you're moving sideways at 3,066 m/s, so your centripetal acceleration is downward at 0.22 m/s2. You're nearly seven times farther away from Earth's center where actual gravity is also 0.22 m/s2. The difference between the two is zero, and you feel weightless. But you're only halfway up the elevator.
Go all the way to the far end of the elevator and you're 71,600 km from Earth's surface (78,000 km from the center). You're moving sideways at 5,672 m/s, so your centripetal acceleration is 0.41 m/s2 but your actual gravity is down to just 0.05 m/s2. The difference is 0.36 m/s2 NEGATIVE. If you step out of the elevator and let go of the hand rail, two seconds later you'll find your feet have drifted 36 cm off the floor. In ten seconds, you'll be 18 meters up and drifting further away at 3.6 m/s. You aren't really drifting upwards. Both you and the platform are moving downwards; it is moving faster than you are. The other end of the elevator is attached to Earth's surface and Earth is rotating, pulling the platform away from you, while gravity is pulling you downward at a much smaller magnitude. Sixty seconds after you let go, you'll be 648 meters away (nearly half a mile).
So, no, you aren't weightless at the top. You weight is NEGATIVE.
#16
09-13-2017, 03:20 PM
scr4 Member Join Date: Aug 1999 Location: Alabama Posts: 14,118
Quote:
Originally Posted by iamthewalrus(:3= This is wrong in many ways. If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn't hold together at all. Note that they don't have to put rockets at the top of skyscrapers to accelerate them to the appropriate speed after they're built, even though the tops are moving faster than the foundations.
In that post, I was talking about a 400 km tall tower, in case what the OP meant by "space elevator" was just a tall tower, tall enough that the top of it was in space.
A 400 km tall tower is just a tower. It's just supported by compressive strength of the material. The top of the tower is nowhere near orbital speed for that height. If you dropped something from it, it would drop almost straight down.
#17
09-13-2017, 03:24 PM
scr4 Member Join Date: Aug 1999 Location: Alabama Posts: 14,118
Quote:
Originally Posted by carnivorousplant I remembered that, but I thought it was narration. Thanks!
Actually you're right:
Quote:
[The visitor] reached out a cautious hand and stroked the narrow ribbon linking the planet with its new moon. "What would happen," he asked, "if it broke?" That was an old question. Most people were surprised at the answer. [Morgan answered,] "Very little. At this point, it's under practically no tension. If you cut the tape, it would just hang there, waving in the breeze." Kingsley made an expression of distaste; both knew that this was a considerable oversimplification. At the moment, each of the four tapes was stressed at about a hundred tons, but that was negligible compared to the design loads they would be handling when the system was in operation and they had been integrated into the structure of the Tower. There was no point, however, in confusing the boy with such details.
#18
09-13-2017, 03:26 PM
carnivorousplant KB not found. Press any key Charter Member Join Date: Apr 2000 Location: Central Arkansas Posts: 55,073
Quote:
Originally Posted by sbunny8 Geosynchronous orbit is about 35,800 km above Earth's surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit.
Place a massive object at the top; would that not move the center of mass upwards? We move an asteroid to synchronous orbit, and lower a cable from it.
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Last edited by carnivorousplant; 09-13-2017 at 03:26 PM.
#19
09-13-2017, 03:27 PM
Darren Garrison Guest Join Date: Oct 2016 Posts: 5,987
Quote:
Originally Posted by iamthewalrus(:3= If you somehow built a tower that tall on the Earth, which is rotating, then the top would have to already be moving that fast. Each piece of the tower would be accelerated to the appropriate orbital velocity as it was built, or the tower wouldn't hold together at all.
Nope, a rigid tower can't be at proper orbital velocity at every height. Look at this calculator. At a point 50 km up, orbital period is 1.42 hours. At 100 km, 1.44 hours. At 200 km, 1.47 hours. At 400 km, 1.54 hours. (All figures rounded.) But if it is a rigid tower rising off of Earth, each section would have to have an orbital period of 24 hours.
#20
09-13-2017, 03:35 PM
eburacum45 Guest Join Date: Feb 2003 Location: Old York Posts: 2,602
There is a way to put a 400km high space elevator up that would just hang there; build an orbital ring first. This link explains how that could work.
https://en.wikipedia.org/wiki/Orbital_ring
Note that an orbital ring is even more speculative than a geostationary space elevator. If you stepped off a 400km orbital ring you'd just fall towards the Earth, so I wouldn't advise it.
#21
09-13-2017, 03:35 PM
Robot Arm Guest Join Date: Jun 2000 Location: Medford, MA Posts: 21,751
Quote:
Originally Posted by dstarfire If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth's surface) is in a stable orbit. So, at the top, you'd also be in a stable orbit and, therefore, weightless.
Not correct.
A space elevator, as a whole, is in a stable orbit, but not every part of it is. The center of mass is at 22,200 miles high (or close to it). And because there's 22,200 miles of cable below that point, there needs to be a hell of a lot of mass above that point, too. If you were in an elevator climbing up this cable, at 22,200 miles you'd be weightless. Your height and speed would be just right to stay in orbit, so it doesn't matter if the elevator cab is around you or not. But below that point you'd be going slower than a satellite, so you'd still feel some sense of gravity inside the elevator. Above that, you'd be going faster than a satellite and you'd be pressed, slightly, against the ceiling of the elevator.
#22
09-13-2017, 03:37 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Quote:
Originally Posted by dstarfire If we ignore the height you specified, then yes, you would be weightless. The principle of a space elevator is that the top is far enough out, and massive enough that center of mass of the entire structure (from satellite to elevator shaft to anchor point on earth's surface) is in a stable orbit. So, at the top, you'd also be in a stable orbit and, therefore, weightless.
I'm not sure what you mean by "stable orbit" but that's not correct by most definitions, and you certainly would not be weightless at the top of a space elevator if by "top" you mean the maximum extent of the cable, where the counterweight is. The counterweight is in an artificially constrained orbit, orbiting with a geosychronous period but at a higher altitude where its velocity is higher than it should be for a stable orbit. In effect everything at that speed and altitude has negative weight; if the counterweight were released, it would climb to a higher orbit, just as a satellite would do if it got a rocket boost. That's what gives the cable its tension. It would keep climbing until the earth starting pulling it back, and that point would become the apogee of a new stable orbit.
The only point on a space elevator where you would be weightless would be at the geosynchronous point. You would get lighter and lighter until you weighed zero at that point. Climbing further, toward the center of mass and the counterweight, your weight would become increasingly negative, pointing away from the earth. If you jumped out into space you would of course immediately become weightless since you would be on a ballistic trajectory, but you'd end up rising higher and then settling into a stable higher orbit.
#23
09-13-2017, 04:08 PM
RedSwinglineOne Guest Join Date: Jan 2007 Location: Silicon valley Posts: 1,307
So if I understand correctly, a space elevator would be of little use getting to low earth orbit because at that altitude, whatever you are lifting is not moving fast enough to maintain orbit.
#24
09-13-2017, 04:10 PM
Marvin the Martian Member Join Date: Jun 2015 Location: Phoenix, AZ, USA Posts: 732
Quote:
Originally Posted by wolfpup If you jumped out into space you would of course immediately become weightless since you would be on a ballistic trajectory, but you'd end up rising higher and then settling into a stable higher orbit.
Plus, if my quick calculations are correct, if you went past about 11km above the geosynchronous point and let go you would be moving faster than escape velocity - you would go not just a higher orbit but out of the earth's gravitational field completely. I believe Clarke made this point also in Fountains of Paradise - the space elevator could also be used as a slingshot to launch interstellar payloads.
Last edited by Marvin the Martian; 09-13-2017 at 04:10 PM.
#25
09-13-2017, 04:19 PM
Darren Garrison Guest Join Date: Oct 2016 Posts: 5,987
Quote:
Originally Posted by eburacum45 There is a way to put a 400km high space elevator up that would just hang there; build an orbital ring first. This link explains how that could work. https://en.wikipedia.org/wiki/Orbital_ring Note that an orbital ring is even more speculative than a geostationary space elevator.
Also touched on in this thread.
#26
09-13-2017, 04:23 PM
carnivorousplant KB not found. Press any key Charter Member Join Date: Apr 2000 Location: Central Arkansas Posts: 55,073
Quote:
Originally Posted by Marvin the Martian I believe Clarke made this point also in Fountains of Paradise - the space elevator could also be used as a slingshot to launch interstellar payloads.
Interstellar?
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#27
09-13-2017, 04:28 PM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 74,386
With a space elevator, getting to LEO would be cheaper than it is now, but it'd still be awkward. But that's OK, because there's very little value in LEO in its own right. The only reason we launch so much stuff to LEO right now is because it's the cheapest orbit (with our current technology). A space elevator would make GEO orbits and highly-eccentric orbits much cheaper, so most of what we launch to LEO now would just be launched to some other orbit instead.
EDIT: And yes, you could do interstellar launches from a space elevator if you wanted. But you'd still have to solve all of the other problems with an interstellar mission, like providing power for thousands of years. Much more practical would be interplanetary missions.
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Last edited by Chronos; 09-13-2017 at 04:30 PM.
#28
09-13-2017, 04:33 PM
Marvin the Martian Member Join Date: Jun 2015 Location: Phoenix, AZ, USA Posts: 732
Quote:
Originally Posted by carnivorousplant Interstellar?
Meant interplanetary.
#29
09-13-2017, 04:56 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Quote:
Originally Posted by RedSwinglineOne So if I understand correctly, a space elevator would be of little use getting to low earth orbit because at that altitude, whatever you are lifting is not moving fast enough to maintain orbit.
Except that (as Chronos already said) getting up to an orbital altitude and with a good fraction of the needed orbital speed (even if insufficient) already saves you a very, very high percentage of the cost of trying to lift it from the ground in a rocket at standstill! Plus, if you go higher than you need to be, you have free energy to play with, though I confess I don't have a sufficient intuitive sense of the orbital mechanics to know how you would effectively use that -- I believe you would inevitably need some amount of rocket thrust to maneuver into the desired orbit. The international consortium working on this proposes a "LEO gate" for low earth orbits that would release the payload at around 24,000 km, IIRC, as opposed to the GEO (geosynchronous point) at 35,786 km. They also propose a "lunar gate" and a "Mars gate" at higher altitudes above GEO.
Quote:
Originally Posted by Marvin the Martian Plus, if my quick calculations are correct, if you went past about 11km above the geosynchronous point and let go you would be moving faster than escape velocity - you would go not just a higher orbit but out of the earth's gravitational field completely. I believe Clarke made this point also in Fountains of Paradise - the space elevator could also be used as a slingshot to launch interstellar payloads.
That intuitively seems far too little. The Wikipedia article on space elevators states that escape velocity is reached at 53,100 km, as compared to GEO at 35,786.
#30
09-13-2017, 05:17 PM
Marvin the Martian Member Join Date: Jun 2015 Location: Phoenix, AZ, USA Posts: 732
Quote:
Originally Posted by wolfpup That intuitively seems far too little. The Wikipedia article on space elevators states that escape velocity is reached at 53,100 km, as compared to GEO at 35,786.
The Wikipedia article says, "An object attached to a space elevator at a radius of approximately 53,100 km would be at escape velocity when released." Radius, not altitude. The radius of a geosynchronous orbit is about 42,000 km.
#31
09-13-2017, 05:29 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Quote:
Originally Posted by Marvin the Martian The Wikipedia article says, "An object attached to a space elevator at a radius of approximately 53,100 km would be at escape velocity when released." Radius, not altitude. The radius of a geosynchronous orbit is about 42,000 km.
OK, thanks, but what caught my attention was that you wrote "11km above the geosynchronous point", which seemed wrong. I wasn't trying to be snarky, it actually didn't occur to me that this was probably a typo and that you probably meant 11K km! If so, then apologies and all is well -- you were indeed in the ballbark.
#32
09-13-2017, 05:38 PM
Marvin the Martian Member Join Date: Jun 2015 Location: Phoenix, AZ, USA Posts: 732
Quote:
Originally Posted by wolfpup OK, thanks, but what caught my attention was that you wrote "11km above the geosynchronous point", which seemed wrong. I wasn't trying to be snarky, it actually didn't occur to me that this was probably a typo and that you probably meant 11K km! If so, then apologies and all is well -- you were indeed in the ballbark.
Yes, another typo . Meant 11,000 km. Really wish Mm (megameter) were in common usage.
#33
09-13-2017, 05:48 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Quote:
Originally Posted by Marvin the Martian Yes, another typo . Meant 11,000 km. Really wish Mm (megameter) were in common usage.
From your username, I should have realized that you'd be closely familiar with the velocities necessary to return home!
#34
09-13-2017, 06:18 PM
RedSwinglineOne Guest Join Date: Jan 2007 Location: Silicon valley Posts: 1,307
Quote:
Originally Posted by wolfpup Except that (as Chronos already said) getting up to an orbital altitude and with a good fraction of the needed orbital speed (even if insufficient) already saves you a very, very high percentage of the cost of trying to lift it from the ground in a rocket at standstill! ...
But on the surface of the earth, you are not at a standstill. At the equator you are moving east at over 1000mph. 300 miles up you are only going very slightly faster. 1100mph or so if my math is correct.
It is a great help to be above the atmosphere of course, but speed seems still to be an issue.
You are probably right that something may be gained by going higher (and therefore faster) and changing orbit later on.
#35
09-13-2017, 07:11 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
I'm certainly not an expert on orbital mechanics, but I think that's the general idea. Tremendous cost and energy is saved by gaining altitude, because a rocket has to lift not only the payload, but also itself and all its fuel, so it all snowballs exponentially. Plus, you're not talking about hundreds of miles in altitude but potentially thousands -- as mentioned, the LEO gate proposed by the elevator consortium to the best of my knowledge is at 24,000 km, about two-thirds of the way to GEO -- which gives the payload lots of orbital speed at release. I would imagine relatively minimal thrusters could put it into a nice circular LEO.
#36
09-13-2017, 08:58 PM
gazpacho Guest Join Date: Oct 1999 Posts: 5,634
Quote:
Originally Posted by wolfpup I'm certainly not an expert on orbital mechanics, but I think that's the general idea. Tremendous cost and energy is saved by gaining altitude, because a rocket has to lift not only the payload, but also itself and all its fuel, so it all snowballs exponentially. Plus, you're not talking about hundreds of miles in altitude but potentially thousands -- as mentioned, the LEO gate proposed by the elevator consortium to the best of my knowledge is at 24,000 km, about two-thirds of the way to GEO -- which gives the payload lots of orbital speed at release. I would imagine relatively minimal thrusters could put it into a nice circular LEO.
This is not correct. It takes much more energy to achieve orbital speed than it does to achieve orbital height.
https://what-if.xkcd.com/58/
#37
09-13-2017, 09:22 PM
enipla Member Join Date: Jul 2001 Location: Colorado Rockies. Posts: 11,860
Ummm.... Perhaps this is too simple of a view. But thinking about the OP's question. - Wouldn't the anchor in space have to have 'negative' gravity to hold up the climbing ribbon/cable to keep it taught? Not even considering if the payload climbs or is pushed up with lasers.
So, if you 'stepped off', you would fly into space away from earth.
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#38
09-13-2017, 09:30 PM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 74,386
Well, getting above the atmosphere does still save you something. Still, I imagine that if, for some reason, you really must have LEO, then the most efficient way to get there is probably to lift to a higher height and release, to go into an orbit with its perigee at LEO height. Then, when you're down there, you do a rocket burn to circularize. Or maybe a combination of rocket burns and aerobraking. I haven't done the calculations on how high you'd need to go for your original release, but the 24 Mm cited by Wikipedia seems plausible.
EDIT: enipla, the top point must be at least some amount above GEO in order to balance it, so if you went to the top and let go, you would certainly "fall" upwards, at least initially. But depending on the design of the elevator, you might end up still in an orbit around the Earth and come back down to that same height again a little over a day later (and then up again and so on), or you might end up escaping completely. How high up the end needs to be depends on how massive the counterweight is, and with a really big counterweight, it might be only a little above GEO height. Still, it'd be really nice to be able to launch things on escape trajectories for free, so in practice, I expect that any space elevator would be at least tall enough for that.
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Last edited by Chronos; 09-13-2017 at 09:33 PM.
#39
09-13-2017, 10:09 PM
wolfpup Guest Join Date: Jan 2014 Posts: 7,923
Quote:
Originally Posted by gazpacho This is not correct. It takes much more energy to achieve orbital speed than it does to achieve orbital height. XKCD has an article about this. https://what-if.xkcd.com/58/
That's quite true, and worth pointing out. My comment was badly worded. Satellite launches tend to go fairly straight up only to clear the worst of the atmosphere, and then they heel over and go for speed. The thing is, an enormous proportion of the fuel is consumed in those early seconds and minutes, which is why experimental airlifted spacecraft ("air launch to orbit") have had some success, even though the predominant advantage offered by air launch is high altitude and thin atmosphere rather than any substantial velocity. But certainly your statement about where most of the energy goes is correct. And the beauty of space elevators is that enormous altitude and huge orbital velocity go hand in hand.
#40
09-13-2017, 11:27 PM
JWT Kottekoe Guest Join Date: Apr 2003 Location: California Posts: 933
nm
Last edited by JWT Kottekoe; 09-13-2017 at 11:29 PM.
#41
09-14-2017, 01:28 PM
iamthewalrus(:3= Guest Join Date: Jul 2000 Location: Santa Barbara, CA Posts: 10,757
Quote:
Originally Posted by scr4 In that post, I was talking about a 400 km tall tower, in case what the OP meant by "space elevator" was just a tall tower, tall enough that the top of it was in space. A 400 km tall tower is just a tower. It's just supported by compressive strength of the material. The top of the tower is nowhere near orbital speed for that height. If you dropped something from it, it would drop almost straight down.
My mistake. I took the OP's "400km" as a wild-ass guess about the proper height for an actual geosynchronous elevator, and was responding as such, but you're correct, a 400km tower is just a tall tower.
Quote:
Originally Posted by Darren Garrison Nope, a rigid tower can't be at proper orbital velocity at every height. Look at this calculator. At a point 50 km up, orbital period is 1.42 hours. At 100 km, 1.44 hours. At 200 km, 1.47 hours. At 400 km, 1.54 hours. (All figures rounded.) But if it is a rigid tower rising off of Earth, each section would have to have an orbital period of 24 hours.
Sorry, I meant "rotational velocity", not orbital velocity. It sounded like scr4 was saying that the top of the tower wouldn't already be rotating at the proper speed (to remain a tower, not to be in orbit).
My point was that if you are building a rigid tower, at no point do you have to do something special to accelerate the higher bits, because they are accelerated as you raise them up to build the tower. But I used some pretty imprecise language to get there.
#42
09-14-2017, 02:03 PM
Elendil's Heir SDSAB Join Date: Jun 2004 Location: my Herkimer Battle Jitney Posts: 73,042
Quote:
Originally Posted by carnivorousplant ...Was Clarke's tethered in the Fountains of Paradise?
Fascinating book - I just reread it a year or so ago. Star Trek: Voyager had a space-elevator episode, too: http://memory-alpha.wikia.com/wiki/Rise_(episode)
#43
09-14-2017, 04:55 PM
HopDavid Guest Join Date: Jul 2011 Posts: 17
Quote:
Originally Posted by HeyHomie Let's say the prefect the Space Elevator. You get on it and ride it up to the top, 400km above the surface of the Earth. Will you be "weightless" in the sense that astronauts aboard the ISS are?
Gravity scales with 1/r^2 and so called centrifugal force scales with r. Earth's radius is about 6378 kilometers.
If you're a 150 lb guy, climbing to the top of a 400 km tower on the equator would lose you about 3 lbs.
Centrifugal force doesn't cancel gravity until you reach an altitude of about 36,000 km. This is the altitude of geosynchronous satellites.
To reach an orbit with a perigee above earth's surface you need to climb to an altitude of around 30,000 kilometers.
Here I have some drawings of the paths payloads would follow if released from different parts of an elevator.
#44
09-14-2017, 05:03 PM
k9bfriender Guest Join Date: Jul 2013 Posts: 6,056
Quote:
Originally Posted by HopDavid If you're a 150 lb guy, climbing to the top of a 400 km tower on the equator would lose you about 3 lbs.
I think I'd lose more than just 3 lbs if I climbed a 400km tower.
#45
09-14-2017, 05:04 PM
HopDavid Guest Join Date: Jul 2011 Posts: 17
Quote:
Originally Posted by sbunny8 Geosynchronous orbit is about 35,800 km above Earth's surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit.
As you go outward gravity falls faster than centrifugal force climbs. So the acceleration gradient isn't symmetrical about geosynchronous orbit.
To balance and maintain tension, the tether above geosynchronous would need to be around twice as long as the tether below geosynchronous orbit.
#46
09-14-2017, 05:20 PM
carnivorousplant KB not found. Press any key Charter Member Join Date: Apr 2000 Location: Central Arkansas Posts: 55,073
Quote:
Originally Posted by k9bfriender I think I'd lose more than just 3 lbs if I climbed a 400km tower.
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#47
09-14-2017, 07:26 PM
K364 Member Join Date: Nov 2001 Location: Edmonton, Alberta Posts: 2,290
Quote:
Originally Posted by HopDavid As you go outward gravity falls faster than centrifugal force climbs. So the acceleration gradient isn't symmetrical about geosynchronous orbit. To balance and maintain tension, the tether above geosynchronous would need to be around twice as long as the tether below geosynchronous orbit.
I'm thinking that below geosynchronous the tower wants to drop, and above geosynchronous it wants to fly away to a higher orbit.
So, centripetal force depends on the speed and mass. In an unconstrained orbit these are equal to gravity. In the tower above geosynchronous it is more than gravity and pulls the tower up. So, you don't need any special length for the upper tether, just more mass.
BTW, in the category of "Possible, just needs more technology than we have today" I think this engineering will never happen. The scale of this thing is immense and any defect or mishap will be catastrophic.
Last edited by K364; 09-14-2017 at 07:26 PM.
#48
09-14-2017, 09:57 PM
HopDavid Guest Join Date: Jul 2011 Posts: 17
Quote:
Originally Posted by K364 I'm thinking that below geosynchronous the tower wants to drop, and above geosynchronous it wants to fly away to a higher orbit.
Correct. Gravity is GM/r^2 . And so called centrifugal force is ω^2 r. ω is angular speed in radians.
Net acceleration is GM/r^2 - ω^2 r.
If r > than geosynchronous, centrifugal force is greater and net acceleration is up. If r is below geosynchronous, gravity is greater and net acceleration is down.
Quote:
Originally Posted by K364 So, you don't need any special length for the upper tether, just more mass.
Net acceleration is small if you're close to geosynchronous orbit. So a counterweight near geosynchronous would provide only a small amount of upward newtons unless it was many tonnes.
Having the tether length extend to an altitude of about 144,000 kilometers would be the minimum mass way to counterbalance the downward newtons.
And if your tower is that tall, it could fling payloads most the way to Neptune.
#49
09-15-2017, 08:41 AM
eburacum45 Guest Join Date: Feb 2003 Location: Old York Posts: 2,602
I am pretty sure that you could use a space elevator to leave the Solar System if you aimed the payload just right- using a flyby manoeuvre at Jupiter, for instance, like Voyager.
It is a remarkable thing that once you get past geostationary orbit all that acceleration would come from the rotation of the Earth. But launching payloads towards the outer solar system would have an effect - there ain't no such thing as a free lunch. It would cause a drag on the Earth, and slow it down, That's why you need to attach the elevator firmly at the bottom - otherwise the bottom end of elevator would career off through the atmosphere, and extract angular momentum by friction.
#50
09-15-2017, 09:06 AM
Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 74,386
On the other hand, you could also set up an exchange of material: Say, put another elevator on Mars, and ship equal tonnages of Earthly seawater to Mars, and Martian iron ore to Earth. In that case, you'd have no net effect on the rotation or orbit of Earth or of Mars. And you could set it up so the energy cost per ton (both ways) was arbitrarily small.
(nitpicking myself: You couldn't actually do arbitrarily small, since you'd need computers to calculate the proper trajectories, and it costs energy to do those calculations. But that's so small compared to the energies involved in the movement of the matter that it might as well be zero.)
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