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https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/A/lesson/A.1.9/problem/A-96 | 1,723,060,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00198.warc.gz | 238,510,854 | 15,953 | ### Home > INT1 > Chapter A > Lesson A.1.9 > ProblemA-96
A-96.
1. The expression for the left side is $1 − 3 − \left(2x\right)$. The expression for the right side is $−x − 3 −\left(2 + x\right)$. Simplify each side by flipping tiles from the negative to the positive region and removing zero pairs. Record your work.
$2x − 2 = −2x − 5$
$4x = −3$
$x\ =\ -\frac{3}{4}$
The simplified expression for the left side is$y + 2$, for the right side is $2y + 2$. | 157 | 459 | {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.6545 |
www.home-cost.com | 1,722,892,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00805.warc.gz | 651,524,188 | 58,206 | Edit Content
# How to Calculate Overhead Costs in Construction Projects
Learn how to accurately calculate overhead costs in construction projects with our comprehensive guide.
# How to Calculate Overhead Costs in Construction Projects
Construction companies face numerous costs when executing a construction project. One of the most significant expenses is overhead costs, which can impact the profitability of a project. As such, understanding how to calculate overhead costs is vital to ensure that construction projects remain on track financially. This article will highlight the key subtopics that will help readers understand how to calculate overhead costs in construction projects.
## Understanding Overhead Costs in Construction
Overhead costs are a crucial aspect of any construction project. They are indirect expenses that a business incurs when executing construction projects. Unlike direct costs, which are easily traceable to a specific project, overhead costs apply to all projects a construction company undertakes.
It is important to note that although overhead costs are essential to business operations, they do not form part of the final cost of construction projects. Instead, they cover expenditures such as rent, salaries, and insurance, among others.
Overhead costs are indirect expenses that a company incurs when executing construction projects. They cannot be traced back to a specific project and apply to all projects a company undertakes. Overhead costs include expenses such as rent, salaries, and utilities, among others.
Overhead costs in construction fall under two primary categories. These are:
• Administrative Overhead Costs – includes expenses such as office rent, salaries of administrative staff, utilities expense, office supplies, and other general administration expenses.
• Project Overhead Costs – includes expenses directly related to a particular project. These costs can include wages and expenses related to supervising a project, construction equipment rental, safety expenses, and tool expenses, among others.
It is important to note that both types of overhead costs are essential to the successful completion of any construction project. Administrative overhead costs ensure that the business runs smoothly, while project overhead costs ensure that the project is completed efficiently and effectively.
### Importance of Accurate Overhead Cost Calculation
Accurate cost calculation is essential in project management as it helps in budget planning and control. Overhead costs represent a significant portion of indirect costs incurred from construction projects. Calculating these costs accurately will help construction businesses make better decisions about project pricing and profitability, ultimately ensuring the success of the business.
Furthermore, accurate overhead cost calculation can help construction businesses identify areas where they can reduce costs and increase profitability. For example, if a business determines that their administrative overhead costs are too high, they can explore options such as downsizing their office space or reducing their administrative staff to cut costs.
In conclusion, understanding overhead costs in construction is critical to the success of any construction business. By accurately calculating overhead costs and identifying areas for improvement, construction businesses can increase profitability and ensure the success of their projects.
## Steps to Calculate Overhead Costs
Calculating overhead costs can be a daunting task for many construction companies. However, with a clear guideline, the process can become a lot easier. Here are some key steps to calculating overhead costs in construction:
### Identify Direct and Indirect Costs
The first step in calculating overhead costs is to distinguish between direct and indirect costs. Direct costs are costs that can be tied to specific projects, while indirect costs are those that cannot be traced back to a particular project. Examples of direct costs include materials, labor, and equipment costs, while indirect costs include rent, utilities, and administrative expenses.
It is important to note that indirect costs are necessary to keep a business running, but they cannot be charged directly to a specific project. Therefore, identifying indirect costs is crucial in calculating overhead costs because they will help you determine what expenses to allocate to the project’s overhead costs.
### Allocate Indirect Costs to Overhead
Once you have identified indirect costs, the next step is to allocate them to the project’s overhead costs. This can be done by categorizing indirect costs and creating an overhead cost pool. You can then divide this pool by the total direct labor hours or direct costs involved in the project to determine the overhead cost rate per unit.
For example, if your overhead cost pool is \$100,000 and your total direct labor hours for a project are 1,000, your overhead cost rate per unit would be \$100 (\$100,000 ÷ 1,000 hours).
### Determine the Allocation Base
The allocation base refers to the basis used to allocate overhead costs to a project. In construction, the allocation base can either be labor hours or material costs. By determining the allocation base, you can accurately allocate overhead costs to the appropriate project and maintain profitability.
For instance, if you decide to use labor hours as your allocation base, you would allocate overhead costs based on the number of hours worked on a project. On the other hand, if you use material costs as your allocation base, you would allocate overhead costs based on the total cost of materials used on a project.
Calculating the overhead rate enables businesses to determine the overhead costs applied to different jobs or projects. To calculate the overhead rate, divide the total overhead costs by the total allocation base. The result will give you an overhead rate per unit, which can be used to allocate indirect costs accurately.
For example, if your total overhead costs are \$100,000 and your allocation base is 1,000 labor hours, your overhead rate per unit would be \$100 (\$100,000 ÷ 1,000 hours).
### Apply the Overhead Rate to Construction Projects
Once you have established the overhead rate, the next step is to apply it to construction projects. Multiply the overhead rate for your project by the number of direct labor hours or direct costs to generate your overhead cost for each job or project.
For instance, if your overhead rate per unit is \$100 and your project requires 500 labor hours, your overhead cost for the project would be \$50,000 (\$100 × 500 hours).
By following these steps, construction companies can accurately calculate their overhead costs and ensure that they are allocating expenses correctly. This will help them maintain profitability and make informed business decisions.
## Common Overhead Cost Components in Construction
Overhead costs in construction can vary from one business to another. However, some common components cut across all construction companies. Some of the most common overhead cost components include:
### Labor and Salaries
Construction companies have both direct and indirect labor costs that need to be included in the calculation of overhead costs. Direct labor refers to the personnel involved in the actual construction of a project, while indirect labor comprises support staff such as administrative staff and management.
Direct labor costs are relatively straightforward to calculate, as they are based on the number of hours worked by each employee multiplied by their hourly wage. However, indirect labor costs can be more challenging to calculate, as they are not always directly related to the construction project. For example, administrative staff may spend some of their time on tasks that are not directly related to the project, such as answering phone calls or responding to emails.
It is essential to accurately calculate both direct and indirect labor costs to ensure that the construction company can cover its expenses and make a profit.
### Equipment and Maintenance
Construction projects rely heavily on equipment such as cranes, bulldozers, and other specialized machinery. These machines require regular maintenance to keep them operational, thereby contributing to overhead costs.
In addition to maintenance costs, equipment also has a limited lifespan and will eventually need to be replaced. Construction companies must factor in the cost of purchasing new equipment when calculating overhead costs.
It is essential to keep equipment in good working condition to prevent delays and ensure that projects are completed on time. Regular maintenance and replacement of equipment are necessary to achieve this goal.
### Insurance and Taxes
Construction businesses must pay taxes, including income tax, property tax, payroll tax, and sales tax. Insurance is also a significant overhead cost component in construction. Insurance premiums can be extensive, depending on the level and scope of coverage required.
Construction companies must have insurance coverage to protect themselves against potential losses and liabilities. Insurance coverage may include general liability insurance, workers’ compensation insurance, and property insurance.
Taxes and insurance premiums can be significant expenses for construction companies, and it is essential to accurately calculate these costs to ensure that the company can cover its expenses and remain profitable.
### Office Expenses
Office expenses include various costs such as rent, utilities, office supplies, and communication expenses. These expenses are necessary to keep the administrative side of the construction business running smoothly.
Office expenses can vary depending on the size of the construction company and the scope of its operations. Larger companies may have more significant office expenses, as they may need to rent larger office spaces and employ more administrative staff.
It is essential to accurately calculate office expenses to ensure that the construction company can cover its expenses and operate efficiently.
### Professional Services
Professional services such as legal expenses, accounting, and marketing expenses also form part of overhead costs in construction. Although professional services can be expensive, they are essential for the smooth running of a construction business.
Legal expenses may include fees for drafting contracts, obtaining permits, and resolving disputes. Accounting expenses may include fees for bookkeeping, tax preparation, and financial analysis. Marketing expenses may include fees for advertising, website development, and social media management.
Professional services can be significant expenses for construction companies, and it is essential to accurately calculate these costs to ensure that the company can cover its expenses and remain profitable.
## Conclusion
Accurately calculating overhead costs is an essential part of tracking and forecasting project costs in construction. Taking the time to understand your business’s overhead costs and implementing accurate cost allocation strategies can help ensure project profitability. By utilizing these strategies, businesses can keep their revenues and profits high for years to come.
### How Home-Cost can help professionals with better construction cost estimates
Home-Cost PRO is designed to cater to the unique needs of contractors and construction companies. With our advanced technology and industry expertise, we provide you with accurate and detailed home construction cost estimates, helping you make informed decisions and optimize your project planning.
By utilizing our software, you can eliminate the hassle of manually estimating costs, which often leads to inaccuracies and time-consuming revisions. Our platform allows you to input project-specific details, such as materials, labor, permits, and fees, and generates a precise cost breakdown in a fraction of the time it would take to do it manually.
In addition, our service offers customization options to reflect your preferred construction methods and regional factors, ensuring that the estimates align with your specific requirements. You’ll have the ability to adjust variables and see how changes impact the overall cost, empowering you to make strategic decisions that align with your budget and project goals.
Furthermore, our detailed cost reports provide you with comprehensive insights, presenting more lines of detail than competitors’. This level of transparency enables you to have a clear understanding of cost allocation, aiding in accurate budgeting, resource allocation, and decision-making throughout the construction process.
Our PRO plan lets you work on as many projects as you like and lets you access your estimates from anywhere, through any device. In addition, you have total control over unit prices and descriptions, which you can edit, and export to your preferred format for your convenience.
By leveraging our service, contractors and construction companies can improve accuracy, save time, reduce costly errors, and optimize resource allocation. Experience the benefits of streamlined cost estimation and take your construction projects to new heights. Contact us today to explore how our service can transform your construction cost estimation process.
Optimized by Seraphinite Accelerator
Turns on site high speed to be attractive for people and search engines. | 2,339 | 13,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.946042 |
https://alexwlchan.net/2020/11/maths-is-about-facing-ambiguity-not-avoiding-it/ | 1,657,179,412,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00119.warc.gz | 137,567,482 | 3,404 | # Maths is about facing ambiguity, not avoiding it
Every so often, an expression like this goes viral on Twitter:
What’s the result? It’s ambiguous.
There are two interpretations, depending on which order of operations you use. Whenever these expressions goes viral, there are heated arguments about which interpretation is correct.
Seeing these arguments reminds me of the school stereotype “maths always has one right answer!” – and how unhelpful it is. People want to find the singular answer, but there isn’t always one to be found. Like everything else, maths has plenty of ambiguous problems, and learning to deal with this sort of ambiguity is a key part of a mathematical, problem-solving skillset.
When faced with an ambiguous problem, you can do one of two things:
1. Guess between the available options, or
2. Try to find more information to aid your decision.
Both can be appropriate responses, depending on the context, and learning to choose between them (and how to go about them) is a skill.
For example, when I’m writing software, there are often aspects of the design which haven’t been explicitly defined by the person who wrote the spec. I have to decide how to how to approach those gaps in the spec, and how that affects the software I’m writing. I might guess if it’s a cosmetic detail, but ask for more information if it’s something security-critical.
There are lots of things that might affect whether I guess or ask for more detail, including:
• If I guess, how easy is it to change my mind later?
• What are the consequences if I make a bad guess?
• How difficult is it to get more information?
• Do I have a good track record of guessing this sort of thing?
If I guess, I should be able to explain and justify whatever option I end up choosing. If I try to find more information, I should be able to ask the right questions to resolve the ambiguity, and make sure I’m getting the details I need. These are both skills in their own right.
If we only ever look for a singular answer, we’re just avoiding ambiugity – we’re not addressing it. Problem solving often involves working with incomplete information, and we’re better off practicing this skill, not pretending problems will always be neatly specified. “There’s always one right answer” might be true in the artifical problems of maths textbooks, but it breaks down in the real world.
This post was originally a thread on Twitter. | 509 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | longest | en | 0.93144 |
https://www.geogebra.org/m/HWQURaTZ | 1,721,054,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00602.warc.gz | 686,868,221 | 27,674 | GeoGebra Classroom
# number machine
Try the following task with this number machine.
• Think of a series of basic operations. It can be adding, subtracting, multiplying or dividing by a number. There can be two or more operations in a series and more than one operation of the same kind.
• You will apply the same series of operations to 4 or more numbers of your choice and calculate the results.
• Enter the first two chosen numbers and their results in the first two pairs of boxes (blue and green).
• Enter two more chosen numbers in the yellow and orange boxes. This time, the machine will calculate the results for you. Click the check boxes to see the results. Are they the same with what you have got?
• Enter one more number in the pink box and check the steps for generating the result. Are the steps the same with what you have done? Does the machine give the same result even the steps may be different from yours?
What questions or discussions can be considered based on this task? | 209 | 996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-30 | latest | en | 0.918856 |
https://timecalculator.info/how-many-days-until-june-14/ | 1,716,654,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00007.warc.gz | 505,436,346 | 2,772 | How Many Days Until June 14 | Time Calculator
How many days until June 14
There are 19 days until June 14.
This year and other years :
From :To :How Many Days ?
25 May 2024June 1419
25 May 2024June 14 2025384
25 May 2024June 14 2026749
25 May 2024June 14 20271114
How many days until exact date?
Calculating the number of days between two dates can be useful in a variety of situations, such as planning a trip or project, tracking a deadline, or simply counting down to an upcoming event.
You can use our day-to-day calculator or if you prefer to calculate the number of days manually, you can use the following formula:
Days until target date = (Target date - Current date)
Here's an example of how to use this formula:
1. Determine the current date. Let's say today's date is March 23, 2023.
2. Determine the target date. Let's say you want to find out how many days until your birthday, which is on May 1, 2023.
3. Subtract the current date from the target date:
Days until birthday = (May 1, 2023 - March 23, 2023) = 39 days
So there are 39 days until your birthday.
Time units calculation tool. | 297 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-22 | latest | en | 0.935855 |
https://math.stackexchange.com/questions/3143105/when-does-the-limit-not-exist | 1,695,874,609,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00391.warc.gz | 413,530,240 | 33,877 | # When does the limit not exist
I have two functions f(x) and g(x), and I am trying to take the limit of f(x)/g(x):
$$\lim_{x\to ∞} f(x)/g(x)$$
The value of f(x) is a constant (greater than 0) and after substituting infinity into g(x), I got 0. Since the denominator is 0, does this mean that the limit does not exist?
• Not necessarily: $\lim_{x\to\infty}\frac{1}{e^{-x}}=\infty$. Mar 11, 2019 at 0:12 | 137 | 406 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.862975 |
https://www.salonbellisimo.net/barbershop/how-much-should-you-tip-our-barber.html | 1,653,736,544,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00288.warc.gz | 1,133,597,317 | 11,985 | # How Much Should You Tip Our Barber?
## How much do you tip for a \$20 haircut?
To answer ‘how much do you tip for a \$20 haircut’ you should tip between \$3 and \$4 on a \$20 haircut, depending on how good your haircut was and how much tip you’d like to leave. \$3 is a 15% tip and \$4 is a 20% tip.
## How much do you tip on a \$30 haircut?
Generally speaking, for good service, you should tip 15 to 20 percent of the entire bill.
## How much do you tip for a \$40 haircut?
Remember the golden rule: “You should tip 20 percent on the entire service cost, not per individual,” says Schweitzer. So if your haircut and blow-dry cost \$40 total, and your color was \$60, your total service cost comes to \$100. That means you should tip \$20 divided between the colorist and stylist.
You might be interested: Question: Tiki Barber Works Where?
## How much do you tip for a \$25 haircut?
For good service, barbers should be tipped 15 to 20 percent of the total amount of the hair cut. Tipping your barber is considered common courtesy and makes up a large part of their income. The tip amount can change if your cut was exceptional (or none if it was bad).
Even though tipping is completely personal, it doesn’t have to be weird or awkward. General rule: Tip your hairdresser 20 percent, but if you can/want to go higher, by all means, do. Just remember that no one is expecting anything—how you tip is up to you and your relationship with the stylist.
## Do you tip a barber if he owns the shop?
Tipping the Owner Most owners agree that while they don’t expect tips, they always appreciate them. “If the salon owner is cutting or coloring your hair, it is customary to tip them 15 to 20%, just as you would any other stylist—they are still providing a service even if they own the business,” Abramite notes.
## How much do you tip on a 50 haircut?
\$50 Haircut Tip Calculator To answer ‘how much do you tip for a \$50 haircut’ you should tip between \$7.50 and \$10 on a \$50 haircut, depending on how good your haircut was and how much tip you’d like to leave. \$7.50 is a 15% tip and \$10 is a 20% tip.
## Is \$5 a good tip for haircut?
A: Depending on the price point of your salon, you’ll want to give your stylist \$5 to \$20 for a complimentary bang trim. If you’re unsure, play it safe by tipping your stylist the equivalent of 20 percent of a full haircut. Q: If the service is a gift, do I still need to tip?
You might be interested: FAQ: How To Start A Spa Or Barber Shop Henderspn?
## How much is a good haircut?
If your mane is getting a bit long, you might be wondering, “How much does a haircut cost?” We surveyed salons across the nation to find the average prices for men’s and women’s haircuts. Overall, we found that the nationwide average price range for a haircut is \$40-\$66. Most haircuts cost around \$53 on average.
## How much should I tip on a \$200 hair service?
According to the unspoken industry standard, if your haircut or dyeing session cost you around one hundred bucks, it will be acceptable to give from eighteen to twenty percent tips if the service was excellent.
## Should I tip my lawn guy?
Do tip lawn-mowing crews, snowplow drivers, oil-truck drivers, and sprinkler servicers—but only if you’re dealing with employees, not the business owner, and only if you see the same guys come around every time. Don’t tip at the time of service, however.
## How much do you tip on a 45 haircut?
Square says the average woman’s haircut costs \$45 and a 20% tip is left. If you had coloring, highlights, extensions, etc that will add to the cost so just put in what you’ve paid. A tip of 15 to 20% is considered appropriate if you’re happy with the job your hairdresser’s done.
## How much do you tip for a \$25 pedicure?
What customers do: The standard tip is anywhere from 15 to 20 percent of the cost of services, before taxes. Some customers tip more than 20 percent if it’s a low-cost service where the tip would otherwise be less than \$2, or if they’re especially thrilled with the service, say our experts.
You might be interested: FAQ: How To Legaly Be A Barber With No Liscense?
## What is the average price for a men’s haircut?
Men’s Haircut Prices – How Much Does A Haircut Cost? (2021 Guide) Ever wonder about the average price for a men’s haircut? While haircut prices can range from \$10 to \$100, the average cost of a haircut in the United States is \$28.
## How much do barbers make?
The mean annual wage of a barber is \$30,480. Barbers in the 75-90th percentile can make between \$37,490-\$48,480 annual salary. The average hourly and annual wages of barbering vary, however, according to your location and position. | 1,138 | 4,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.960295 |
https://www.pyimagedata.com/how-to-build-an-artificial-neural-network-with-python/ | 1,642,977,067,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00137.warc.gz | 1,012,265,973 | 57,263 | In the previous tutorial, we build an artificial neural network from scratch using only matrices. In this tutorial, we’ll build an artificial neural network with python just using the NumPy library. While we create this neural network we will move on step by step. But you can use any programming language to create this neural network too.
## Describe The Network Structure
The artificial neural network that we will build consists of three inputs and eight rows. But we will use only six-row and the rest of the rows will be test data. We will build an artificial neural network that has a hidden layer, an output layer. The Hidden layer will consist of five neurons. Weights and bias of the neural network will be created randomly.
### Define the Variables
In this part, we will define the variables that we use. The variables will consist of the matrices. Therefore we can simply define these matrices using the python NumPy library.
Firstly we need to install the NumPy library and we can install the NumPy library using the following command.
``pip install numpy``
Defining inputs and output:
``````import numpy as np
inputs = np.array([[0,0,0],[0,0,1],[0,1,0],[0,1,1],[1,0,0],[1,0,1]])
output = np.array([[0],[1],[0],[1],[0],[1]])
print("inputs : ")
print(inputs)
print(".....................")
print("output : ")
print(output)``````
Defining weights
``````w1 = np.random.randn(inputs.shape[1],5)
w2 = np.random.randn(5,output.shape[1])
print("w1")
print(w1)
print(".......................")
print("w2")
print(w2)``````
Defining biases:
``````b1 = 1
b2 = 1``````
### Activation Function
The activation functions allow us to bring our output values in each layer to the 0-1 range. If we do our operations without using the activation function, the output values will increase exponentially after each process. This will cause us to train our artificial neural network for very long periods of time. There are many activation functions to eliminate this. These are sigmoid function, tanh function, relu function etc. We will use the sigmoid function in this tutorial.
``````def sigmoid(x):
return 1/(1 + np.exp(-x))``````
## Buid the Artificial Neural Network
Artificial neural network training consists of two main parts.
• Calculating the predicted output ŷ, known as feedforward
• Updating the weights and biases, known as backpropagation
The following image neural network shows training.
#### Feedforward
z = x * w + b
a = 1/(1 + e^-z )
y_head = a2
error = loss function
error = (1/2) * (output – y_head)^2
``````z1 = np.dot(inputs,w1) + b1
a1 = sigmoid(z1)
z2 = np.dot(a1, w2) + b2
a2 = sigmoid(z2)
error = np.sum((1/2)*(output - a2)**2)``````
``````print("z1 : ",z1.shape)
print("a1 : ",a1.shape)
print("z2 : ",z2.shape)
print("a2 : ",a2.shape)
print("error : ", error)``````
z1 : (6, 5)
a1 : (6, 5)
z2 : (6, 1)
a2 : (6, 1)
error : 0.8829125977000097
#### BackPropagation
So far, we found the error value of our prediction. But we need to update our weights and bias values. So we need to know the derivative of the loss function with respect to the weights and biases. We know that derivative of the function is the slope of the function. If we can calculate the derivative of the function we can simply update weights and bias values by increasing/reducing. This called gradient descent.
However, we can’t directly calculate the derivative of the loss function with respect to the weights and biases. Because the derivative of the loss function doesn’t contain weights and bias. So we need to use the chain rule to calculate it. Each part of the chain rule is called the partial derivative.
Step 1: Find derivatives of layer one.
``````error_d_a2 = (a2 - output)
# derivative of the a2 with respect to the z2
a2_d_z2 = a2*(1 - a2)
# derivative of the z2 with respect to the w2
z2_d_w2 = a1
# derivative of the z2 with respect to the b2_w
z2_d_b2 = b2</pre></div>``````
``````print("error_d_a2 : ", error_d_a2.shape)
print("a2_d_z2 : ", a2_d_z2.shape)
print("z2_d_w2 : ", z2_d_w2.shape)
print("z2_d_b2 : ", z2_d_b2)``````
error_d_a2 : (6, 1)
a2_d_z2 : (6, 1)
z2_d_w2 : (6, 5)
z2_d_b2 : 1
Step 2: Calculate the derivative of the error with respect to the w2.
``````delta_w2 = error_d_a2 * a2_d_z2
delta_w2 = np.dot(z2_d_w2.T, delta_w2)
print("delta_w2 : ", delta_w2.shape)``````
delta_w2 : (5, 1)
Step 3: Calculate the derivative of the error with respect to the b2.
``````delta_b2 = error_d_a2 * a2_d_z2
delta_b2 = delta_b2 * z2_d_b2
delta_b2 = np.sum(delta_b2)
print("delta_b2 : ", delta_b2)``````
delta_b2 : 0.34414836643934316
Step 4: Update w2 and b2_w.
``````w2 = w2 - delta_w2
b2 = b2 - delta_b2
print("new w2 : ")
print(w2)
print(".....................")
print("new b2_w : ")
print(b2)``````
Step 5: Find derivatives of layer two.
``````# derivative of the z2 with respect to the a1
z2_d_a1 = w2
# derivative of the a1 with respect to the z1
a1_d_z1 = a1*(1 - a1)
# derivative of the z1 with respect to the w1
z1_d_w1 = inputs
# derivative of the z1 with respect to the b1_w
z1_d_b1_w = b1``````
``````print("z2_d_a1 : ", z2_d_a1.shape)
print("a1_d_z1 : ", a1_d_z1.shape)
print("z1_d_w1 : ", z1_d_w1.shape)
print("z1_d_b1_w : ", z1_d_b1_w)``````
z2_d_a1 : (5, 1)
a1_d_z1 : (6, 5)
z1_d_w1 : (6, 3)
z1_d_b1_w : 1
Step 6: Calculate the derivative of the error with respect to the w2.
``````delta_w1 = error_d_a2 * a2_d_z2
delta_w1 = np.dot(delta_w1,z2_d_a1.T)
delta_w1 = delta_w1 * a1_d_z1
delta_w1 = np.dot(inputs.T,delta_w1)
print("delta_w1 : ", delta_w1.shape)``````
delta_w1 : (3, 5)
Step 7: Calculate the derivative of the error with respect to the b1.
``````delta_b1 = error_d_a2 * a2_d_z2
delta_b1 = np.dot(delta_b1,z2_d_a1.T)
delta_b1 = delta_b1 * a1_d_z1
delta_b1 = delta_b1 * z1_d_b1_w
delta_b1 = np.sum(delta_b1)
print("delta_b1: ", delta_b1)``````
delta_b1: -0.0020394367989461973
Step 8: Update w1 and b1.
``````w1 = w1 - delta_w1
b1 = b1 - delta_b1
print("new w1 : ")
print(w1)
print(".....................")
print("new b1 : ")
print(b1)``````
#### Training the neural network
So far, we calculated all parameters and now we will train our neural network.
``````# Defining all variables
import numpy as np
error_list = list()
inputs = np.array([[0,0,0],[0,0,1],[0,1,0],[0,1,1],[1,0,0],[1,0,1]])
output = np.array([[0],[1],[0],[1],[0],[1]])
# Weights
w1 = np.random.randn(inputs.shape[1],5)
w2 = np.random.randn(5,output.shape[1])
# Biases
b1 = 1
b2 = 1
# Sigmoid Function
def sigmoid(x):
return 1/(1 + np.exp(-x))
# Update the weights times 100.
for i in range(100):
# Feedforward
z1 = np.dot(inputs,w1) + b1
a1 = sigmoid(z1)
z2 = np.dot(a1, w2) + b2
a2 = sigmoid(z2)
error = np.sum((1/2)*(output - a2)**2)
# Backpropagation
## LAYER 2
### derivative of the error with respect to the a2
error_d_a2 = (a2 - output)
### derivative of the a2 with respect to the z2
a2_d_z2 = a2*(1 - a2)
### derivative of the z2 with respect to the w2
z2_d_w2 = a1
### derivative of the z2 with respect to the b2_w
z2_d_b2 = b2
### delta weights 2
delta_w2 = error_d_a2 * a2_d_z2
delta_w2 = np.dot(z2_d_w2.T, delta_w2)
### delta biases
delta_b2 = error_d_a2 * a2_d_z2
delta_b2 = delta_b2 * z2_d_b2
delta_b2 = np.sum(delta_b2)
### Update weights and bias
w2 = w2 - delta_w2
b2 = b2 - delta_b2
## LAYER 1
### derivative of the z2 with respect to the a1
z2_d_a1 = w2
### derivative of the a1 with respect to the z1
a1_d_z1 = a1*(1 - a1)
### derivative of the z1 with respect to the w1
z1_d_w1 = inputs
### derivative of the z1 with respect to the b1_w
z1_d_b1_w = b1
### delta weights 1
delta_w1 = error_d_a2 * a2_d_z2
delta_w1 = np.dot(delta_w1,z2_d_a1.T)
delta_w1 = delta_w1 * a1_d_z1
delta_w1 = np.dot(inputs.T,delta_w1)
### delta bias 1
delta_b1 = error_d_a2 * a2_d_z2
delta_b1 = np.dot(delta_b1,z2_d_a1.T)
delta_b1 = delta_b1 * a1_d_z1
delta_b1 = delta_b1 * z1_d_b1_w
delta_b1 = np.sum(delta_b1)
### update w1 and b1
w1 = w1 - delta_w1
b1 = b1 - delta_b1
error_list.append(error)
print("error : ", error)``````
#### Show the Trained Data With Matplotlib Library
We trained our artificial neural network times 100. Now let’s show trained data on the screen using the python matplotlib library.
``````import matplotlib.pyplot as plt
x = np.arange(len(error_list))
y = error_list
plt.figure(figsize=(10,8))
plt.plot(x,y)
plt.xlabel("iteration")
plt.ylabel("error")
plt.title("Artificial Neural Networks Training")
plt.show()`````` | 2,742 | 8,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.755722 |
https://tolstoy.newcastle.edu.au/R/help/06/02/20385.html | 1,591,226,731,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00481.warc.gz | 572,524,387 | 4,958 | # Re: [R] Difficulty with qqline in logarithmic context
From: Prof Brian Ripley <ripley_at_stats.ox.ac.uk>
Date: Thu 02 Feb 2006 - 03:06:16 EST
Is there a good reason to use qqnorm in a single-log context? Should one not rather use
> qqnorm(log(freq))
> qqline(log(freq))
since you are (I guess) looking at log-normality of freq? Another way to look at that is
> qqplot(qlnorm(ppoints(length(freq))), freq, log="xy")
the same plot, different scales. (I believe a QQ plot should always have comparable scales on the two axes.)
The point is that qqline is tied to normality, not to log-normality.
On Wed, 1 Feb 2006, François Pinard wrote:
> Hi, R friends. I had some difficulty with the following code:
>
> qqnorm(freq, log='y')
> qqline(freq)
>
> as the line drawn was seemingly random. The exact data I used appears
> below. After wandering a bit within the source code for "abline",
> I figured out I should rather write:
>
> qqnorm(freq, log='y')
> par(ylog=FALSE)
> qqline(log10(freq))
> par(ylog=TRUE)
>
> I'm proposing that this little stunt be rather be hidden and
> automatically effected within "qqline" proper, whenever par('ylog') is
> TRUE. I thought about providing a patch, as "qqline" is so small. Yet
> it would be more noise than useful, as I'm not familiar with the "datax"
> argument usage, which should probably be addressed as well.
>
>
>
> Here is the data, in case useful:
>
> freq <-
> as.integer(c(33, 79, 21, 436, 58, 18, 1106, 498, 1567, 393, 2,
> 104, 50, 67, 113, 76, 327, 331, 196, 145, 86, 59, 12, 215, 293,
> 154, 500, 314, 246, 587, 85, 23, 323, 3, 13, 576, 29, 37, 24,
> 21, 1230, 137, 13, 93, 3, 101, 72, 218, 59, 17, 2, 8, 86, 143,
> 150, 22, 19, 234, 119, 157, 4, 255, 146, 126, 76, 15, 271, 170,
> 4, 6, 16, 3048, 2175, 3350, 5017, 5706, 1610, 665, 322, 1, 16,
> 47, 51, 168, 94, 66, 154, 99, 11, 547, 953, 1, 1071, 80, 184,
> 168, 52, 187, 103, 187, 361, 46, 85, 135, 597, 121, 283, 26,
> 12, 20, 169, 9, 79, 15, 114, 75, 30, 111, 556, 173, 32, 99, 438,
> 2, 2, 1, 117, 5, 3, 51, 8, 41, 12, 23, 2, 13, 5, 1, 9, 4, 1,
> 7, 15, 5, 48, 16, 112, 6, 1, 39, 60, 5, 23, 5, 19, 1, 8, 32,
> 4, 13, 1, 14, 71, 5, 1, 35, 30, 100, 389, 22, 8, 1, 192, 40,
> 6, 3, 17, 2, 14, 71, 14, 1, 5, 4, 32, 21, 18, 13, 2, 2, 45, 342,
> 46, 144, 18, 131, 188, 112, 37, 85, 90, 8, 195, 173, 5, 53, 96,
> 37, 16, 16, 281, 64, 50, 92, 336, 31, 744, 4, 134, 74, 1, 227,
> 6, 48, 418, 64, 66, 59, 20, 45, 20, 370, 148, 22, 7, 30, 601,
> 29, 82, 113, 938, 252, 65, 137, 72, 22, 98, 12, 152, 212, 13,
> 8, 35, 3, 77))
>
> Yet this really is the value of "courriel\$freq" after "data(courriel)",
> with a file ".../R/data/courriel.R" here, holding:
>
> courriel <- read.table(pipe('grep -c \'^From \' ../courriel/*'),
> sep=':', as.is=T, row.names=1,
> col.names=c('fichier', 'freq'))
>
> My goal, which is nothing serious, was merely to toy with the number of
> messages per folder, for folders massaged out of R archives.
>
>
>
> Version:
> platform = i686-pc-linux-gnu
> arch = i686
> os = linux-gnu
> system = i686, linux-gnu
> status =
> major = 2
> minor = 2.1
> year = 2005
> month = 12
> day = 20
> svn rev = 36812
> language = R
>
> Locale:
>
> Search Path:
> .GlobalEnv, package:methods, package:stats, package:graphics, package:grDevices, package:utils, package:datasets, fp.etc, Autoloads, package:base
>
>
> --
> François Pinard http://pinard.progiciels-bpi.ca
>
> ______________________________________________
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
>
```--
Brian D. Ripley, ripley@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
______________________________________________
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help | 1,611 | 3,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.859958 |
https://pythonawesome.com/aero-421-spacecraft-attitude-dynamics-and-control-final-project/ | 1,660,505,640,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00283.warc.gz | 448,479,384 | 10,651 | AERO – 421 Final Project Redevelopment
Spacecraft Attitude, Dynamics, and Control: Simulation to determine and control a satellite’s attitude in LEO.
Background
AERO-421, or Spacecraft Attitude, Dynamics, and Controls, is a class taught at the California Polytechnic State University at San Luis Obispo (Cal Poly SLO) which serves as an…
“Introduction to spacecraft attitude dynamics and control… [and] fundamentals of guidance and navigation systems… [with emphasis in] analysis and design of control systems for aerospace vehicles.” – Cal Poly Aerospace Engineering Course Catalog
The final project in the course was to develop a simulation to determine a satellite’s attitude in Low Earth Orbit (LEO), consider and model detumbling from a launch vehicle, consider and model disturbances due to external forces i.e., Solar Radiation Pressure (SRP), and to consider and model control via onboard reaction wheels.
The initial project was developed in MATLAB, however, the project will be completely redeveloped in Python to showcase controls and software development skillsets.
Part 0: Context and Given Data
The project will explore modeling and simulation of the various stages of a spacecraft mission, specifically simulating the attitude dynamics from initial spacecraft deployment to operation. In this simulation, the spacecraft is an Earth observing satellite and an attitude determination and control system must be designed using reaction wheels to ensure the spacecraft maintains pointing nadir.
Orbital Data:
• h (angular momentum) = 53335.2 km^2/s
• e (eccentricity) = 0
• Ω (Right Ascension of Ascending Node) = 0 deg
• i (inclination) = 98.43 deg
• ω (Argument of Perigee) = 0 deg
• θ (True Anomaly) = 0 deg
• ϵ-LVLH (Initial Quaternion relating the body to the LVLH frame) = [0, 0, 0]; η = 1
Detumble Phase:
• Spacecraft is a square box with 2 meters on each edge with total mass of 640 kg
• Initial Angular Velocity of [-0.05, 0.03, 0.2] rad/s relating the body to the ECI frame
Normal Operations:
• Spacecraft bus is a 2 meter cube with mass of 500 kg. The center of mass of the spacecraft is located at the geometric center of the bus.
• A rectangular sensor is attached to the face pointing towards the Earth (+Z-axis) and is 1 meter long and 0.25 meters square. The sensor has a mass of 100 kg.
• Two solar panels are deployed along the +/- Y-axis and are constrained to rotate about the +/- Y-axis. The solar panels are 3 meters long (in the Y-axis), 2 meters wide, and 0.05 meters thick. Each panel has a mass of 20 kg and the center of mass is located at the geometric center of the panel. The solar panels do not rotate relative to the spacecraft bus.
• Assume all spacecraft components have uniform density with centers of mass located at the geometric centers of each component
• Magnetically, the spacecraft residual magnetic dipole moment can be modeled as pointing in the -Z direction with magnitude 0.5 A-m^2
• See the figure below for the spacecraft schematic
• Because the thrusters are not actually fully-modulated thrusters, the spacecraft will have a residual angular velocity of [0.001, -0.001, 0.002] rad/s relating the body to the ECI frame after the detumble phase.
• During operation the spacecraft is required to point at the target on the ground to within 0.001 degrees 3-sigma using the reaction wheels used in the reaction wheels part.
Part 1: Mass Properties
Determine the mass and inertial properties of the spacecraft for both the detumble and the normal operations phases.
Outputs:
• Total mass of the spacecraft
• Center of mass relative to the spacecraft bus center of mass. The body frame will be located at the center of mass of the whole spacecraft
• Intertia matrix of the whole spacecraft about the center of mass of the spacecraft
Part 2: Torque Free Motion
Model the torque free orbital and attitude motion of the spacecraft
Outputs: Plots for…
• Euler angles and quaternions relating body to ECI reference frames
• Angular velocity of the spacecraft in body components for one orbit of the normal operations phase
Part 3: Detumble
Simulate the motion of the satellite during the detumble phase. Assume fully modulated thrusters and use direct velocity feedback
Outputs: Plots for…
• Euler angles and quaternions relating body to ECI reference frames
• Angular velocity of the spacecraft in body components for the detumble phase
• Torque components in the body frame
Part 4: Disturbance Simulation
Add the four disturbance models to the simulation:
• Atmospheric Drag
• Solar Pressure
• Earth Magnetic Field
Use the following model for the atmospheric density. Notice that h is the height above the Earth’s surface in kilometers where R_Earth equals 6378km
Consider the simulation epoch to be March 20, 2021. Disregard any variations of the ECI representation of the sunlight direction during the simulation.
Outputs: Plots for…
• Euler angles and quaternions relating the body to the ECI reference frame
• Euler angles and quaternions relating the body to the LVLH reference frame
• Angular velocity of the spacecraft relative to the ECI frame expressed in body components
• Angular velocity of the spacecraft relative to the LVLH frame expressed in body components
• Torque components for atmospheric drag, solar radiation pressure, gravity gradient, and earth magnetic field
Part 5: Reaction Wheel Control
Determine the control gains for a full state feedback 3-axis reaction wheel control system. Use the requirements of ζ = 0.65 and t_s = 30 sec
The positions of the 3 reaction wheels are [1, 0, 0], [0, 1, 0], and [0, 0, 1]. Each reaction wheel can be modeled as a simple cylinder with radius of 0.3 m and a height of 0.02 m
Outputs: Plots for…
• Euler angles and quaternions relating the body to ECI reference frame
• Euler angles and quaternions relating the body to LVLH reference frame
• Angular velocity of the spacecraft relative to the ECI reference frame expressed in body components
• Angular velocity of the spacecraft relative to LVLH frame expressed in body components
• Commanded moment from the determined control law
• Wheel speed of each reaction wheel
Part 6: Visualization
Determine and animate the quanterions of the spacecraft, from ECI to body frame, for the duration of 1-5 revolutions.
Output:
• Table of quaternion and time data
• Video or other animation file to show the revolution of the spacecraft
View Github | 1,457 | 6,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.863549 |
https://solvedlib.com/n/financial-algebra | 1,721,834,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00559.warc.gz | 444,336,979 | 10,578 | # Financial Algebra
All new and solved questions in Financial Algebra category
##### Jimmy invest 4000 in an account
Jimmy invest 4000 in an account...
...
...
##### Identify the source of finance that are available to a business and assess the implication of different sources
identify the source of finance that are available to a business and assess the implication of different sources...
##### Chase Bank started its first day of operations with $40 million in capital. A total of$190 million in checkable deposits is received. The bank makes a $25 million commercial loan and lends another$120 million in mortgage loans. The bank also purchases 30-day T-Bills for 20 million. If required reserves are 10%, what does the bank balance sheet look like? (5 marks)
Chase Bank started its first day of operations with $40 million in capital. A total of$190 million in checkable deposits is received. The bank makes a $25 million commercial loan and lends another$120 million in mortgage loans. The bank also purchases 30-day T-Bills for 20 million. If required re...
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##### There is an inverse floater with a current annual coupon of $35 and a face of$500. Its companion floater is selling at par, $500, and has a current annual coupon$5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and what is the price of the inverse floater? What is the effective coupon rate ceiling (the highest the coupon rate could be and still get Pfloater = $500) on the floater? There is an inverse floater with a current annual coupon of$35 and a face of $500. Its companion floater is selling at par,$500, and has a current annual coupon $5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and wh... 5 answers ##### There is an inverse floater with a current annual coupon of$35 and a face of $500. Its companion floater is selling at par,$500, and has a current annual coupon $5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and what is the price of the inverse floater? What is the effective coupon rate ceiling (the highest the coupon rate could be and still get Pfloater =$500) on the floater?
There is an inverse floater with a current annual coupon of $35 and a face of$500. Its companion floater is selling at par, $500, and has a current annual coupon$5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and wh...
##### There is an inverse floater with a current annual coupon of $35 and a face of$500. Its companion floater is selling at par, $500, and has a current annual coupon$5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and what is the price of the inverse floater? What is the effective coupon rate ceiling (the highest the coupon rate could be and still get Pfloater = $500) on the floater? There is an inverse floater with a current annual coupon of$35 and a face of $500. Its companion floater is selling at par,$500, and has a current annual coupon $5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and wh... 5 answers ##### There is an inverse floater with a current annual coupon of$35 and a face of $500. Its companion floater is selling at par,$500, and has a current annual coupon $5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and what is the price of the inverse floater? What is the effective coupon rate ceiling (the highest the coupon rate could be and still get Pfloater =$500) on the floater?
There is an inverse floater with a current annual coupon of $35 and a face of$500. Its companion floater is selling at par, $500, and has a current annual coupon$5. Both of these instruments mature in 5 years. What is the current price of the bond that was used to generate these instruments and wh...
##### Valerie is saving money for her college education. She deposited some money in a savings account paying 5% and $1200 less than that amount in a second account paying 4%. The two accounts produced a total of$141 interest in 1 yr. How much did she invest at each rate?
Valerie is saving money for her college education. She deposited some money in a savings account paying 5% and $1200 less than that amount in a second account paying 4%. The two accounts produced a total of$141 interest in 1 yr. How much did she invest at each rate?... | 1,131 | 5,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.937151 |
http://davion.com/index.php/tutorial/risks/risk-probability/cost-weighted-probability | 1,519,098,192,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00116.warc.gz | 91,545,866 | 4,458 | # Cost-Weighted Probability
Cost-weighted probability is used when calculating the overall probability for a risk when cost allocation is in force and different probabilities have been assigned to a risk for each allocation block.
The obvious way to calculate overall probability in such situations is simply to average the probabilities over all the allocation blocks. However, this can give unrealistic results.
In the example shown below, no costs will ever be incurred, because in years 1 and 3 there are no costs, and in year 2, while there is a \$1,000,000 cost, there is zero probability of it being incurred. However, if we were to average the probabilities over all three years, we would arrive at the result that there is a 67% probability of a \$1,000,000 cost.
Allocation Block Year 1 Year 2 Year 3 Estimated Cost \$0 \$1,000,000 \$0 Probability 100% 0% 100%
To get around this problem Mandrel uses cost-weighted probability, which is the sum of (probability x cost) for each allocation block divided by the sum of the costs:
Applying this to the example above gives an overall probability of 0%, which is the correct answer in this case.
Zero-Cost Case
Cost-weighted probability breaks down if there are no costs, in which case Mandrel calculates overall probability as a simple average. This situation might arise, for example, where you have one or more non-cost risk impacts, and a particular risk has these impacts but no cost impact. | 317 | 1,460 | {"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} | 3.46875 | 3 | CC-MAIN-2018-09 | latest | en | 0.952654 |
http://en.wikipedia.org/wiki/Blotto_games | 1,438,170,412,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986423.95/warc/CC-MAIN-20150728002306-00253-ip-10-236-191-2.ec2.internal.warc.gz | 83,209,405 | 14,221 | # Blotto games
Blotto games (or Colonel Blotto games, or "Divide a Dollar" games) constitute a class of two-person zero-sum games in which the players are tasked to simultaneously distribute limited resources over several objects (or battlefields). In the classic version of the game, the player devoting the most resources to a battlefield wins that battlefield, and the gain (or payoff) is then equal to the total number of battlefields won.
The Colonel Blotto game was first proposed and solved by Émile Borel[1] in 1921, as an example of a game in which "the psychology of the players matters". It was studied after the Second World War by scholars in Operation Research, and became a classic in Game Theory.[2]
The game is named after the fictional Colonel Blotto from Gross and Wagner's 1950[3] paper. The Colonel was tasked with finding the optimum distribution of his soldiers over N battlefields knowing that:
1. on each battlefield the party that has allocated the most soldiers will win, but
2. both parties do not know how many soldiers the opposing party will allocate to each battlefield, and:
3. both parties seek to maximize the number of battlefields they expect to win.
## Example
As an example Blotto game, consider the game in which two players each write down three positive integers in non-decreasing order and such that they add up to a pre-specified number S. Subsequently, the two players show each other their writings, and compare corresponding numbers. The player who has two numbers higher than the corresponding ones of the opponent wins the game.
For S = 6 only three choices of numbers are possible: (2, 2, 2), (1, 2, 3) and (1, 1, 4). It is easy to see that:
Any triplet against itself is a draw
(1, 1, 4) against (1, 2, 3) is a draw
(1, 2, 3) against (2, 2, 2) is a draw
(2, 2, 2) beats (1, 1, 4)
It follows that the optimum strategy is (2, 2, 2) as it does not do worse than breaking even against any other strategy while beating one other strategy. There are however several Nash equilibria. If both players choose the strategy (2, 2, 2) or (1, 2, 3), then none of them can beat the other one by changing strategies, so every such strategy pair is a Nash equilibrium.
For larger S the game becomes progressively more difficult to analyse. For S = 12, it can be shown that (2, 4, 6) represents the optimal strategy, while for S > 12, deterministic strategies fail to be optimal. For S = 13, choosing (3, 5, 5), (3, 3, 7) and (1, 5, 7) with probability 1/3 each can be shown to be the optimal probabilistic strategy.
Borel's game is similar to the above example for very large S, but the players are not limited to round integers. They thus have an infinite number of available pure strategies, indeed a continuum.
This concept is also implemented in a story of Sun Tzu when watching a chariot race with three different races running concurrently. In the races each party had the option to have one chariot team in each race, and each chose to use a strategy of 1, 2, 3 (with 3 being the fastest chariot and 1 being the slowest) to deploy their chariots between the three races creating close wins in each race and few sure outcomes on the winners. When asked how to win Sun Tzu advised the chariot owner to change his deployment to that of 2, 3, 1. Though he would be sure to lose the race against the fastest chariots (the 3 chariots); he would win each of the other races, with his 3 chariot easily beating 2 chariots and his 2 chariot beating the 1 chariots.
## Application
This game is commonly used as a metaphor for electoral competition, with two political parties devoting money or resources to attract the support of a fixed number of voters.[4][5] Each voter is a "battlefield" that can be won by one or the other party. The same game also finds application in auction theory where bidders must make simultaneous bids.[6]
Several variations on the original game have been solved by Laslier,[7] Roberson,[8] Kvasov.[9] | 982 | 3,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-32 | latest | en | 0.969518 |
https://blog.benoitvallon.com/sorting-algorithms-in-javascript/the-shellsort-algorithm/ | 1,675,886,167,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00286.warc.gz | 153,549,585 | 6,514 | # The Shellsort algorithm
## #sorting-algorithms series
Posted on March 16, 2016
The #sorting-algorithms series is a collection of posts about reimplemented sorting algorithms in JavaScript.
If you are not familiar with sorting algorithms, a quick introduction and the full list of reimplemented sorting algorithms can be found in the introduction post of the series on sorting algorithms in JavaScript.
If you feel comfortable with the concept of each sorting algorithm and only want to see the code, have a look at the summary post of the series. It removes all explanations and contains only the JavaScript code for all sorting algorithms discussed in the series.
## Get the code on Github
Of course, all the code can also be found on Github in the repository sorting-algorithms-in-javascript.
## A good way to compare all of them
Unlike the data structures, all sorting algorithms have the same goal and they can all take the same input data. So, for every sorting algorithms of the series, we are going sort an `array` of 10 numbers from 1 to 10.
By doing so we will be able to compare the different sorting algorithms more easily. Sorting algorithms are very sensitive to the input data so we will also try different input data to see how they affect the performances.
# The Shellsort algorithm
## Definition
Shellsort is an in-place comparison sort which can be seen as either a generalization of sorting by exchange (bubble sort) or sorting by insertion (insertion sort). The method starts by sorting pairs of elements far apart from each other, then progressively reducing the gap between elements to be compared. Starting with far apart elements can move some out-of-place elements into position faster than a simple nearest neighbor exchange. From Wikipedia
Shellsort is a generalization of insertion sort that allows the exchange of items that are far apart. It is worth noting that for a value of `gap` equals to 1, this algorithm is equal to insertion sort. Have a look at the code below and you will quickly notice the similarities.
## Visualization
If you want to have a nice visualization of the algorithm, the visualgo.net website is a nice resource. You can play with many parameters and see which part of the algorithm is doing what.
## Complexity
Time complexity
Best Average Worst
? ? ?
Note that the complexity of the Shellsort algorithm depends of the sequence of number that you use as gaps and there are a lot of different possible sequences. I think that it is nice to have a look at this table to get an overview of those possible sequences and see the resulting complexity.
I didn’t mention any specific complexity above because the sequence that I am using in my example (the last one of the previous table) makes calculating the complexity impossible (it is an empirically sequence of numbers unrelated to n).
Anyway, you can still have an overview of the time and space complexity of the Shellsort algorithm with this excellent Big O cheat sheet. Be careful because they look wrong to me at the time of writing this post, but hopefully they will be fixed in the future.
## The code
For each sorting algorithm, we are going to look at 2 versions of the code. The first one is the final/clean version, the one that you should remember. The second one implements some counters in order to demonstrate the different time complexities depending of the inputs.
### Clean version
``````// array to sort
var array = [9, 2, 5, 6, 4, 3, 7, 10, 1, 8];
// gaps
var gaps = [701, 301, 132, 57, 23, 10, 4, 1];
function shellsort(array) {
for(var g = 0; g < gaps.length; g++) {
var gap = gaps[g];
for(var i = gap; i < array.length; i++) {
var temp = array[i];
for(var j = i; j >= gap && array[j - gap] > temp; j -= gap) {
array[j] = array[j - gap];
}
array[j] = temp;
}
}
return array;
}
console.log(shellsort(array)); // => [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
``````
### Version with counters
``````// sample of arrays to sort
var arrayRandom = [9, 2, 5, 6, 4, 3, 7, 10, 1, 8];
var arrayOrdered = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
var arrayReversed = [10, 9, 8, 7, 6, 5, 4, 3, 2, 1];
// gaps
var gaps = [701, 301, 132, 57, 23, 10, 4, 1];
function shellsort(array) {
var countOuter = 0;
var countInner = 0;
var countSwap = 0;
for(var g = 0; g < gaps.length; g++) {
var gap = gaps[g];
for(var i = gap; i < array.length; i++) {
countOuter++;
var temp = array[i];
for(var j = i; j >= gap && array[j - gap] > temp; j -= gap) {
countInner++;
countSwap++;
array[j] = array[j - gap];
}
array[j] = temp;
}
}
console.log('outer:', countOuter, 'inner:', countInner, 'swap:', countSwap);
return array;
}
shellsort(arrayRandom.slice()); // => outer: 15 inner: 11 swap: 11
shellsort(arrayOrdered.slice()); // => outer: 15 inner: 0 swap: 0
shellsort(arrayReversed.slice()); // => outer: 15 inner: 13 swap: 13
`````` | 1,258 | 4,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-06 | latest | en | 0.8988 |
https://math.answers.com/questions/What_is_34.70_divided_by_15 | 1,716,110,896,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00010.warc.gz | 344,254,949 | 47,582 | 0
# What is 34.70 divided by 15?
Updated: 9/25/2023
Wiki User
8y ago
2.3133
Lenore Murphy
Lvl 10
3y ago
Wiki User
8y ago
34.7 ÷ 15 = 2 47/150 = 2.31333... ≈ 2.313
Wiki User
8y ago
2.313
Earn +20 pts
Q: What is 34.70 divided by 15?
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3470 meters = ~11,384.5 feet.
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18
3,470
1.4667
9.6667
### What is 15 divided by one?
15 divided by one is 15. Anything divided by one is that same thing. One is the identity.
### What is the sum of 3470 and 4587?
I recommend you use a calculator - if you need it.
### What divided by 2.5 15?
37.5 divided by 2.5 = 15
64
10.6667 | 348 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.936218 |
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# Use the following to answer the questions below:Fast food restaurants are been required to publish ...
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Use the following to answer the questions below:
Fast food restaurants are been required to publish nutrition information about the foods they serve. Nutrition information about a random sample of McDonald's lunch/dinner menu items (excluding sides and drinks) was obtained from their website. We wish to use the sodium content (in milligrams) to better understand the number of calories in the lunch/dinner menu items at McDonald's. Some summary statistics, partial computer output from a regression analysis, and a scatterplot (with regression line) of the data are provided.
Use two decimal places when reporting the results from any calculations, unless otherwise specified.
Variable Mean StDev Calories 477.3 164.6 Sodium (mg) 1021.3 373.8
The regression equation is Calories = 99.69 + 0.3698 Sodium (mg)
Source DF SS MS F P Regression 1 267501 267501 31.11 0.000 Error 13 111793 8599 Total 14 379293
What is the estimated slope in this regression model? Interpret the slope in context.
▸ 0.3698
For each additional mg of sodium in McDonald's lunch/dinner menu items, we predict the number of calories to increase by 0.3698 calories.
▸ 0.3698
For each additional calorie in McDonald's lunch/dinner menu items, we predict the grams of sodium to increase by 0.3698 mg.
▸ 99.69
For each additional calorie in McDonald's lunch/dinner menu items, we predict the grams of sodium to increase by 99.69 mg.
▸ 99.69
For each additional mg of sodium in McDonald's lunch/dinner menu items, we predict the number of calories to increase by 99.69 calories.
Textbook
## Statistics: Unlocking the Power of Data
Edition: 3rd
Authors:
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Relations and Functions (Ten Test Papers)
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/* ]]> */ | 1,539 | 6,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | longest | en | 0.821144 |
https://basescripts.com/calculating-the-total-cost-using-google-apps-script | 1,726,444,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00071.warc.gz | 113,082,668 | 26,448 | # Calculating the Total Cost Using Google Apps Script
Google Apps Script is a powerful tool that allows you to automate tasks and enhance your Google Sheets capabilities. In this blog post, we will demonstrate how to use Google Apps Script to calculate the total cost from a sample data table. This script can be especially useful for financial calculations, budgeting, or any scenario where you need to sum values across multiple rows and columns.
### Sample Data Table
Let’s start with a sample data table in a Google Sheet named `CostsSheet`. This table has the following columns:
### Goals
1. Calculate the total cost for each item.
2. Calculate the grand total cost of all items.
``function calculateTotalCost() { // Open the spreadsheet and the CostsSheet var ss = SpreadsheetApp.getActiveSpreadsheet(); var sheet = ss.getSheetByName('CostsSheet'); // Get the data from the sheet var data = sheet.getDataRange().getValues(); // Initialize the total cost variable var totalCost = 0; // Loop through the data to calculate the total cost for each item for (var i = 1; i < data.length; i++) { // Skip the header row var quantity = data[i][1]; var unitPrice = data[i][2]; var itemTotal = quantity * unitPrice; // Write the total cost for each item in the next column sheet.getRange(i + 1, 4).setValue(itemTotal); // Add the item total to the grand total cost totalCost += itemTotal; } // Write the grand total cost at the bottom of the total cost column sheet.getRange(data.length + 1, 4).setValue('Grand Total:'); sheet.getRange(data.length + 1, 5).setValue(totalCost); // Set the header for the total cost column sheet.getRange(1, 4).setValue('Total Cost');}``
### Explanation
1. Open Spreadsheet and Sheet:`var ss = SpreadsheetApp.getActiveSpreadsheet(); var sheet = ss.getSheetByName('CostsSheet'); `This opens the active spreadsheet and the specific sheet we are working with: `CostsSheet`.
2. Get Data from Sheet:`var data = sheet.getDataRange().getValues(); `This retrieves all the data from the `CostsSheet`.
3. Initialize the Total Cost Variable:`var totalCost = 0; `This initializes a variable to store the grand total cost.
4. Loop Through the Data to Calculate Total Cost for Each Item:`for (var i = 1; i < data.length; i++) { // Skip the header row var quantity = data[i][1]; var unitPrice = data[i][2]; var itemTotal = quantity * unitPrice; // Write the total cost for each item in the next column sheet.getRange(i + 1, 4).setValue(itemTotal); // Add the item total to the grand total cost totalCost += itemTotal; } `This loop calculates the total cost for each item by multiplying the quantity by the unit price. It then writes the total cost for each item in the next column and adds it to the grand total cost.
5. Write the Grand Total Cost:`sheet.getRange(data.length + 1, 4).setValue('Grand Total:'); sheet.getRange(data.length + 1, 5).setValue(totalCost); `This writes the grand total cost at the bottom of the total cost column.
6. Set the Header for the Total Cost Column:`sheet.getRange(1, 4).setValue('Total Cost'); `This sets the header for the total cost column.
### Conclusion
This Google Apps Script efficiently calculates the total cost for each item and the grand total cost from a sample data table. By following the steps outlined in this blog post, you can automate and streamline your financial calculations in Google Sheets. Happy scripting! | 803 | 3,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.522834 |
https://se.mathworks.com/matlabcentral/profile/authors/3558229 | 1,717,107,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00618.warc.gz | 423,348,797 | 22,051 | # F.
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nästan 12 år ago | 1,523 | 5,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.732313 |
https://www.techwhiff.com/issue/if-f-x-3x-10-and-g-x-4x-2-find-f-g-x--150384 | 1,674,829,743,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00738.warc.gz | 1,020,029,245 | 12,431 | # If f(x)=3x + 10 and g(x) = 4x - 2 find (f - g)(x)
###### Question:
if f(x)=3x + 10 and g(x) = 4x - 2 find (f - g)(x)
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### The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $400 and standard deviation$20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $400 and standard deviation$20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted a... | 1,035 | 4,572 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.927989 |
https://forum.qorvo.com/t/decaranging-pc-channel-response-log-interpretation/752 | 1,660,065,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00663.warc.gz | 270,117,512 | 5,836 | Hello all,
I’ve recently started working with the EVK-1000 kit doing some simple point-to-point, single Anchor, single Tag, range testing to familiarize with the system. I’m testing with the Tag standalone running the DecaRanging ARM application and powered by a USB battery pack and the Anchor powered by PC USB and controlled by the DecaRanging PC application. I’ve been utilizing the “Channel Respose Log” functionality to have a record of the distances as I conduct testing. That aspect has worked well, but I’m trying to better understand the Rx and Tx timestamps (reference DecaRanging PC User’s Guide sections 4.2 and 4.3).
I see in the log file various Tx and Rx timestamps. I understand the nature of the communication protocol with the Poll/Response/Final framework, but it is the actual timestamps here expressed in scientific notation that I don’t understand. What is this time in reference to? What unit is the time in?
When I see a series of Rx/Tx/Rx such as:
21 74 Rx time = 1.608443040713579e+001 EF4B0FA59F
90 Tx time = 1.623443065844414e+001
29 74 Rx time = 1.643443053953513e+001 F48011C6AB
It seems to make sense because each timestamp is a higher number than the previous one. However, when I see a series such as:
21 76 Rx time = 1.708640614285983e+001 FE332DFD42
92 Tx time = 2.900536998259716e-002
29 76 Rx time = 2.290052341715495e-001 0368301A73
I don’t really know what to make of it.
Please note that I’m using Notepad++ to open the log file and then using the Find function with Regex to search for the string: ^((.21 … RX time =.)|(.29 … RX time =.)|(.TX time =.)|(.Anchor.))\$ effectively condensing the log file into timestamps and distance calculations.
Any help would be greatly appreciated.
Thanks!
Nick
Rx time is the time of reception of a frame - decimal is the DW1000 time converted to seconds, hex is the DW1000 time (40 bit number)
Rx time(un) is the raw time stamp before any DW1000 time adjustments after first path calculation in LDE
RXDATA: these are the received bytes
txdly and rxdly are the TX and RX antenna delays as programmed
Accum Len 1016 - these are the real and imaginary parts of the accumulator CIR for the received frame
Tx time is the time of the frame transmission (has TX antenna delay added)
RX OK - this signifies good reception
HLP - this is first path index in the accumulator
PSC - number of accumulated preamble symbols
NTH - noise threshold
T - temperature and voltage - read from DW1000 on frame reception
RSL - received signal level (dBm) - calculated as given by the formula in User Manual
FSL - first path signal level (dBm) - calculated as given by the formula in User Manual
/Leo
Thank you for the reply. So if a timestamp is in seconds, what is the reference point? Is it time since the test started or is there just some running timer?
What’s going on in the example I posted where you have a poll at 1.70e+1, response at 2.90e-2, and final at 2.29e-1? The timestamps don’t seem to line up correctly.
21 76 Rx time = 1.708640614285983e+001 FE332DFD42
92 Tx time = 2.900536998259716e-002
29 76 Rx time = 2.290052341715495e-001 0368301A73
Please read the DW1000 UM. And DecaRanging Application user manual where the log format is described.
If you still can’t figure it out let me know.
Z
Hi Nick,
According to the user’s manual, page 73:
The timestamp register is 40-bit, counting at 128499.2 MHz (which is approximately 64Hz or a 15ps period).
Then after (2^40)/(128
499.2*10^6) = 17 . 207 401 025 641 seconds it will wrap around.
Best regards,
Fellipe Saldanha Garcia
Hi,
Just to correct the previous message: (which is approximately 64GHz or a 15ps period).
Best regards,
Fellipe Saldanha Garcia
Felipe,
Thank you for the response. I was missing the connection with the System Time Counter. It makes sense now.
Regards,
Nick | 1,026 | 3,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.901339 |
https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&oldid=103089 | 1,695,668,620,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00480.warc.gz | 124,403,415 | 12,491 | # 2011 AIME II Problems/Problem 6
## Problem 6
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a, and $a+d>b+c$. How many interesting ordered quadruples are there?
## Solution 1
Rearranging the inequality we get $d-c > b-a$. Let $e = 11$, then $(a, b-a, c-b, d-c, e-d)$ is a partition of 11 into 5 positive integers or equivalently: $(a-1, b-a-1, c-b-1, d-c-1, e-d-1)$ is a partition of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is $\binom{6+4}4 = \binom{10}4 = 210$. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if $N$ is the number of partitions where the second element is equal to the fourth, our answer is $(210-N)/2$.
We find $N$ as a sum of 4 cases:
• two parts equal to zero, $\binom82 = 28$ ways,
• two parts equal to one, $\binom62 = 15$ ways,
• two parts equal to two, $\binom42 = 6$ ways,
• two parts equal to three, $\binom22 = 1$ way.
Therefore, $N = 28 + 15 + 6 + 1 = 50$ and our answer is $(210 - 50)/2 = \fbox{080}$
## Solution 2
Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding $\binom83 = 56$ ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding $\binom 63 = 20$ ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding $\binom43 = 4$ ways and there cannot be 3 holders between a and b so our total is 56+20+4=$\fbox{080}$.
## Solution 3 (Slightly bashy)
We first start out when the value of $a=1$.
Doing casework, we discover that $d=5,6,7,8,9,10$. We quickly find a pattern.
Now, doing this for the rest of the values of $a$ and $d$, we see that the answer is simply:
$(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}$
## Solution 4 (quick)
Notice that f $a+d>b+c$, then $(11-a)+(11-d)<(11-b)+(11-c)$, so there is a 1-to-1 correspondence between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d.
Quick counting gives that the number of ordered quadruples such that $a+d=b+c$ is 50.
Thus the answer is $\frac{\binom{10}{4}-50}{2} = \boxed{80}.$
## See also
2011 AIME II (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Login to AoPS | 966 | 3,024 | {"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": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-40 | latest | en | 0.914478 |
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To solve this eq we want to have only one x on the LHS. First we have to multiply 2 by argument from the parenthesis.
So:
4+2(3x+5)=11-x
4+6x+10=11-x (now i simplify 4 and 10
6x+14=11-x /+x (im taking x to LHS)
7x+14=11 /-14 (im taking 14 to the RHS
7x=-3 :7 (im isolating x)
x= - its the result. | 203 | 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} | 3.859375 | 4 | CC-MAIN-2017-04 | latest | en | 0.86341 |
https://quantumcomputing.stackexchange.com/questions/tagged/bloch-sphere | 1,726,807,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00539.warc.gz | 430,815,242 | 52,891 | # Questions tagged [bloch-sphere]
For questions related to the Bloch sphere. In quantum mechanics, the Bloch sphere is a geometrical representation of the pure state space of a two-level quantum mechanical system (qubit), named after the physicist Felix Bloch. (Wikipedia)
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### Bloch representation of a quantum channel acting on a 2-qubit density-matrix
A previous answer nicely show the relationship between the Pauli transfer matrix (PTM) and the Bloch representation of a quantum channel that acts on single-qubit density matrix. In short, given a ...
1 vote
18 views
### Visualizing the Effect of a Parameterized Gate on a Qubit Using the Bloch Sphere in Qiskit 1.0
I am seeking assistance with visualizing the effect of a parameterized gate on a qubit using the Bloch Sphere representation. I have a preliminary code snippet as follows: ...
• 293
1 vote
82 views
### Can we find the quantum state using the Bloch vector in the Bloch sphere?
I had a question about qft. When we apply QFT in a circuit, we can display the qubits separately in the bloch sphere with the plot_bloch_multivector function. I realized that the bloch vectors do not ...
67 views
• 161
72 views
### What do the angles in a Poincaré sphere represent?
I understand the principle of the Bloch sphere. You write your state in the following way: $$|\Psi\rangle = \cos(\theta/2)|0\rangle+e^{i\varphi}\sin(\theta/2)|1\rangle.$$ The angles $\varphi,\theta$ ...
• 2,366
1 vote
67 views
### Converting $H$ gate to $R_x$ and $R_z$
EDIT: My solution is supposed to work for $|1\rangle$ state too. See https://imgur.com/a/7F1cHu4 Right of the bat the answer is $$H=R_z(\pi/2)R_x(\pi/2)R_z(\pi/2)\,.$$ My question is, I cannot reach ...
• 123
184 views
### Affine transformation of the Bloch sphere to Kraus representation of qubit channels
It is known that qubit channels can be written in the form: \begin{align} \Phi(\rho) = \frac{1}{2}\left(I+(T\vec{r}+\vec{t})\cdot\sigma\right)\ \end{align} where $\vec{r}$ is the Bloch vector ...
• 31
1 vote
101 views
### In the phase flip action on standard basis, why do we consider the $-1$ phase only for the $|1\rangle$?
Prof. Watrous in the first lecture of Qiskit summer school 2023, mentions: "....the significance of putting a minus sign in front of the $|1\rangle$ basis vector and not $|0\rangle$ will be more ...
• 19
27 views
28 views
### Is a qubit passing through a gate a 3D rotation of the vector on the Bloch sphere?
When a qubit passes through a gate isnt it a 3D rotation of the vector on the Bloch sphere?
• 151
475 views
### Closeness between two unitaries on the Bloch sphere
The fidelity between two (single-qubit) quantum states can be easily translated into the euclidean distance between the two states on the Bloch sphere (hilbert-schidmit distance). I'm curious if this ...
• 169
307 views
### Sufficient conditions for a single-qubit unitary to be the identity
Say I have a unitary $U = e^{-iHt}$ where $H = \alpha X + Z$. First, suppose $U = I$. Then it rotates a set of initial states to themselves. Say I'm working on a computational basis, then on the Bloch ...
• 169
167 views
• 143
1 vote
94 views
### Representing networks with qubits as edges
I am looking to take a classical non-negative real valued network and generalize it to the quantum case for processing. A network is given by an adjacency matrix, essentially edge weights $e_{ij}$ for ...
112 views
### How to represent an entangle state on Bloch Sphere [duplicate]
If states are not entangle, both bloch sphere represent the state of qubit but if they are entangle, why bloch sphere show nothing??
610 views
### Plotting the Bell state on a Bloch sphere
How can you plot the bell state on a Bloch sphere? bell = QuantumCircuit(2) bell.h(0) bell.x(1) bell.cx(0,1) Is there any good reference for understanding how ...
102 views
### Is the Bloch sphere the same as the electron spin?
A single qubit is represented with a bloch sphere and implemented on an electron with only two energy values from lots of them. So, I am confused about why an energy range is represented as a ...
• 285
162 views
### Show that for pure states the description of the Bloch vector we have given coincides with that in section 1.2
$\newcommand\bra[1]{\left\langle#1\right|}\newcommand\ket[1]{\left|#1\right\rangle}$ I am having a little bit of difficulty with part (4) of Exercises 2.72 from Nielsen and Chuang's "Quantum ...
• 705
1k views
104 views
### Why are probabilities represented with alpha^2 and beta^2?
To preserve the Complementary Rule of probability, the sum of the probabilities of the outcomes (measured |0> or measured |1>) must equal 1 or 100%. That's why alpha^2+beta^2=1. However, why the ...
1 vote
138 views
### Why is the conjugate being used when rewriting a qubit state in another basis?
I'm currently going through Introduction to Classical and Quantum Computing, by Thomas Wong, and I'm struggling with an example given on page 88 (point 4). The author gives the two following states, ...
108 views
### Is there a criteria to ensure a one-qubit operator is exactly of the form $R_n(\theta)$ (i.e without a global phase $e^{i\alpha}$)?
Reading the Nielsen and Chuang, I saw that every unitary operator $U$ can be written as $e^{i\alpha} R_n(\theta)$ for some well chosen $n \in \mathbb{R}^3$ and $0 \leq \theta < 2\pi$. I would like ...
• 101
107 views
### Is the function PU(2) and SO(3) induced by the Bloch sphere bijective?
I have difficulty understanding the fact that, as written in this reference, every single-qubit unitary corresponds to a unique rotation of R3 and vice versa. If I understand well, this means there ...
• 101
399 views
### Finding the polar and azimuthal angles of the Bloch vector corresponding to $\frac1{\sqrt2}(1-i)|0\rangle-\frac i{\sqrt2}|1\rangle$ [duplicate]
I need to find the polar angles and azimuthal angles of the following bloch vector: $$\frac{1-i}{2}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle$$ I just couldn't figure it out, and I could also not find ...
382 views
### How many parameters do we need to characterize a pure state?
Suppose I have a pure qubit. I can think of starting with the state $\vert 0\rangle$ and apply some unitary to it. Such a unitary has three parameters according to this link. In $d$ dimensions, the ...
• 23
104 views
### Geometric representation of rotation operator on Bloch sphere
I'm studying Nielsen&Chuang Book and need some clarification on mapping arbitrary rotation operator onto geometric tranformation of vector on Bloch sphere. I thought that $R_{\vec{n}}(\theta)$ ...
1 vote
369 views
### Where does 1/sqrt(2) come from in the state of i
I’m trying to learn about calculating coordinates for $\theta$ and $\varphi$ in a Bloch-sphere. I came accross this book about it, including example questions. At question 2.12b, they ask to give the ...
So far I learned that a qubit can be written as $| \psi \rangle = \alpha | 0 \rangle + \beta | 1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$ and reparametrized as $| \psi \rangle = cos( \theta / 2) + e^{... • 143 2 votes 2 answers 367 views ### What is the Bloch sphere representation of$\rho\to\mathcal{E}(\rho) = |+\rangle\langle+|ρ|+\rangle\langle+| + |−\rangle\langle−|ρ|−\rangle\langle−|$? Suppose a projective measurement is performed on a single qubit in the basis$|+\rangle, |−\rangle$, where$|±\rangle \equiv (|0\rangle\pm |1\rangle)/\sqrt{2}$. In the event that we are ignorant of ... • 831 2 votes 0 answers 39 views ### What do the values of θ, ɸ, and λ represent visually in the Bloch Sphere when defining a unitary gate? [duplicate] In IBM Quantum Docs, it is stated that a unitary matrix can be defined as$U = \begin{bmatrix} \cos(\theta/2) & -e^{j\lambda}\sin(\theta/2) \\ e^{j\phi}\sin(\theta/2) & e^{j\lambda+j\phi}\cos(\...
$$\begin{array}{l} |\psi\rangle=\frac{\sqrt{3 i}}{2}|0\rangle-\frac{1}{2}|1\rangle \\ |\psi\rangle=0.924|0\rangle-0.382 i |1\rangle \end{array}$$ Basically I’m trying to convert these to a standard ... | 2,228 | 8,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.830215 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/95/11/n/ | 1,695,562,871,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00214.warc.gz | 942,183,836 | 61,720 | # Properties
Label 95.11.n Level $95$ Weight $11$ Character orbit 95.n Rep. character $\chi_{95}(21,\cdot)$ Character field $\Q(\zeta_{18})$ Dimension $396$ Newform subspaces $1$ Sturm bound $110$ Trace bound $0$
# Learn more
## Defining parameters
Level: $$N$$ $$=$$ $$95 = 5 \cdot 19$$ Weight: $$k$$ $$=$$ $$11$$ Character orbit: $$[\chi]$$ $$=$$ 95.n (of order $$18$$ and degree $$6$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$19$$ Character field: $$\Q(\zeta_{18})$$ Newform subspaces: $$1$$ Sturm bound: $$110$$ Trace bound: $$0$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{11}(95, [\chi])$$.
Total New Old
Modular forms 612 396 216
Cusp forms 588 396 192
Eisenstein series 24 0 24
## Trace form
$$396 q + 132 q^{3} - 1830 q^{4} + 3066 q^{6} - 95436 q^{9} + O(q^{10})$$ $$396 q + 132 q^{3} - 1830 q^{4} + 3066 q^{6} - 95436 q^{9} - 93750 q^{10} + 2985984 q^{12} - 2110026 q^{13} + 1224960 q^{14} - 6809430 q^{16} + 532380 q^{17} - 7959948 q^{19} - 9580518 q^{21} - 31211136 q^{22} - 13532700 q^{23} + 21417984 q^{24} - 159720 q^{26} - 183408156 q^{27} + 144654336 q^{28} - 140983194 q^{29} - 119625000 q^{30} + 122776776 q^{31} + 393148500 q^{32} - 359004150 q^{33} - 578370492 q^{34} - 59812500 q^{35} - 640932912 q^{36} + 503327622 q^{38} + 1302975096 q^{39} - 96000000 q^{40} - 1194423528 q^{41} - 3390185778 q^{42} - 670104582 q^{43} + 2150443914 q^{44} + 3016431090 q^{46} - 151507770 q^{47} + 364629606 q^{48} - 8225668806 q^{49} + 3009006312 q^{51} - 2363609406 q^{52} - 4515798744 q^{53} - 9311985042 q^{54} + 952636938 q^{57} + 4039140864 q^{58} + 5041664862 q^{59} - 2779312500 q^{60} + 5128963260 q^{61} - 4033454760 q^{62} - 12114094326 q^{63} + 11990621196 q^{64} - 3916687500 q^{65} - 1806217476 q^{66} + 6379678362 q^{67} + 18479546670 q^{68} + 14567082498 q^{69} - 1621125000 q^{70} - 1788156144 q^{71} - 16923552852 q^{72} - 26786299530 q^{73} + 15012453948 q^{74} + 18346685868 q^{76} + 6003303804 q^{77} + 69668310660 q^{78} - 9538312266 q^{79} + 5565750000 q^{80} + 2012656020 q^{81} + 30701977302 q^{82} - 21873008760 q^{83} - 51655671540 q^{84} - 14398500000 q^{85} - 14081837694 q^{86} - 17926323216 q^{87} + 110136441168 q^{88} + 77332013844 q^{89} + 18945093750 q^{90} - 85587979728 q^{91} + 50548839756 q^{92} - 13898204208 q^{93} + 262589518236 q^{96} + 29548158042 q^{97} - 235808629536 q^{98} - 91369111794 q^{99} + O(q^{100})$$
## Decomposition of $$S_{11}^{\mathrm{new}}(95, [\chi])$$ into newform subspaces
Label Dim $A$ Field CM Traces $q$-expansion
$a_{2}$ $a_{3}$ $a_{5}$ $a_{7}$
95.11.n.a $396$ $60.359$ None $$0$$ $$132$$ $$0$$ $$0$$
## Decomposition of $$S_{11}^{\mathrm{old}}(95, [\chi])$$ into lower level spaces
$$S_{11}^{\mathrm{old}}(95, [\chi]) \cong$$ $$S_{11}^{\mathrm{new}}(19, [\chi])$$$$^{\oplus 2}$$ | 1,239 | 2,829 | {"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-40 | latest | en | 0.244923 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-common-core/chapter-1-expressions-equations-and-inequalities-1-4-solving-equations-practice-and-problem-solving-exercises-page-31/55 | 1,534,890,282,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219109.94/warc/CC-MAIN-20180821210655-20180821230655-00509.warc.gz | 863,731,556 | 13,972 | ## Algebra 2 Common Core
$x=\frac{-3b-2c+5}{-b+c}$ or $x=\frac{3b+2c-5}{b-c}$
We start by distributing. We have $cx+2c-5=bx-3b$. We want the terms with a $x$ to be together, and all the others to be on the other side of the equation. We add $-2c$, $5$, and $-bx$ to both sides of the equation to get $cx-bx=-2c-3b-5$. Now, we factor out $x$ on the left side to get $x(c-b)=-2c-3b+5$. Finally, we divide both sides by $c-b$ to isolate x. This gives us $x=\frac{-2c-3b+5}{c-b}$ but usually we put variables in alphabetical order. $x=\frac{-3b-2c+5}{-b+c}$ | 213 | 554 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-34 | longest | en | 0.741886 |
https://askdev.io/questions/4780/characterizing-dense-subgroups-of-the-reals | 1,618,689,701,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00516.warc.gz | 224,223,419 | 6,902 | # Characterizing Dense Subgroups of the Reals
Possible Duplicate :
Subgroup of $\mathbb{R}$ either dense or has a least positive element?
Let $(\mathbb{R},+)$ be the team of Real Numbers under enhancement. Allow $H$ be a correct subgroup of $\mathbb{R}$. Confirm that either $H$ is thick in $\mathbb{R}$ or there is an $a \in \mathbb{R}$ such that $H=\{ na : n=0, \pm{1},\pm{2},\dots\}$.
I am unable to continue.
0
2019-05-07 15:05:07
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If there is a tiniest favorable component, after that we are done, given that any kind of favorable component has to be an integer multiple of it, or otherwise we can make use of a euclidean - type - algorithm to get a favorable component with smaller sized value. (I.e., intend $a$ is the tiniest favorable component, and also $b$ a favorable component which is not an integer multiple of $a$ - - - maintain deducting duplicates of $a$ till you get something that is purely in between $0$ and also $a$.)
So think there is a series $a_n$ had in the team that contains favorable numbers often tending to absolutely no. After that the team has each ${\mathbb{Z} a_n}$. This suggests that for each and every $n$, any kind of number in $\mathbb{R}$ is within $|a_n|$ of a component of the team. Given that the $|a_n|$ can be tiny, we locate that the team is thick. | 364 | 1,312 | {"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} | 3.3125 | 3 | CC-MAIN-2021-17 | latest | en | 0.899337 |
http://cboard.cprogramming.com/cplusplus-programming/2068-need-some-help.html | 1,475,330,717,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662882.88/warc/CC-MAIN-20160924173742-00289-ip-10-143-35-109.ec2.internal.warc.gz | 45,808,298 | 11,177 | 1. ## Need some help
I need a little help trying to figure out why my program is not doing what I need it to do. Here is the program:
#include <iostream.h>
#include <iomanip.h>
using std::cout;
using std::cin;
using std::endl;
int main()
{
int choice1;
float area1, radi1, leng1, heig1, heig2, bas1;
do
{
cout << "The Geometry Calculator" << endl;
cout << endl;
cout << "1. Claculate the area of a circle." << endl;
cout << "2. Claculate the area of a rectangle." << endl;
cout << "3. Calculate the area of a triangle." << endl;
cout << "4. Quit." << endl;
cout << endl;
cout << "Enter your choice (1-4) : " << flush;
cin >> choice1;
if ( choice1 = 1 )
{
cout << endl;
cout << "The area of the circle with radius " << radi1 << " is " << setprecision ( 4 )
<< setiosflags (ios::fixed | ios::showpoint ) << area1 << endl;
}
else if ( choice1 = 2 )
{
cout << "Please enter the length of the rectangle: ";
cin >> leng1;
cout << "Please enter the height of the rectangle: ";
cin >> heig1;
area1 = leng1 * heig1;
cout << endl;
cout << "The area of your rectangle with length " << leng1 << " and height " << heig1 << " is "
<< setprecision ( 2 ) << setiosflags ( ios::fixed | ios::showpoint ) << area1 << endl;
}
else if ( choice1 = 3 )
{
cout << "Please enter the height of the triangle: ";
cin >> heig2;
cout << "Please enter the base of the triangle: ";
cin >> bas1;
area1 = bas1 * ( 0.5 * heig2 );
cout << endl;
cout << "The area of your triangle with height " << heig2 << " and base " << bas1 << " is "
<< setprecision ( 2 ) << setiosflags ( ios::fixed | ios::showpoint ) << area1 << endl;
}
else if ( choice1 = 4 )
{
cout << endl;
cout << "Thanks for making me think! Please Come Again!!!" << endl;
break;
}
}
while (( choice1 >=1 ) && ( choice1 <= 4 ));
return 0;
}
Now what it is supposed to do is run a continuous loop till option 4 is entered. Everytime I run it, I will only go to option 1 and no other ones. It also makes it tough to escape the loop. What am I doing wrong?
2. what is it doing wrong?
3. When I run the program it will ask for an option and if you choose any other option it still feels that I entered 1 as the option and does the work for "if ( choice = 1 )" If I take out the elses then it will go through each if action like the question is not even there.
Hopefully this will help.
4. Got it figured out!
I am a dummy!
I am new to C++ programming.
I forgot you need to use double equals (==)to do the comparison for an if statement rather then single (=) which means assign
The Ski
5. ah a good one...hehe....old comp teacher swore he'd thouroghly embarrass any student who did that....of course that didn't stop me from doing it at least once a month. | 766 | 2,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | latest | en | 0.71221 |
https://en.wikipedia.org/wiki/Talk:Upsampling | 1,501,268,135,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550977093.96/warc/CC-MAIN-20170728183650-20170728203650-00184.warc.gz | 628,186,019 | 13,510 | # Talk:Upsampling
## Bandwidth versus sampling rate
Something seems wrong here. The page says
"Upsampling is the process of increasing the sampling rate of a signal. This is usually done to increase the bandwidth of a signal."
I think the second sentence is not correct. Ideally, upsampling should not increase bandwidth, only sample rate.
Quoting further:
"The upsampling factor (commonly denoted by L) is usually an integer or a rational fraction greater than unity. This factor multiplies the sampling rate or, equivalently, divides the sampling period. For example, if compact disc audio was upsampled by a factor of 5/4 then the resulting sampling rate goes from 44,100 Hz to 55,125 Hz, ... The range of valid frequencies (i.e., those that satisfy the Nyquist-Shannon sampling theorem) has gone from 22,050 Hz to 27,562.5 (an increase in 5,512.5 Hz."
While it is true that at the 55,125 Hz sampling rate, frequencies up to 27,562 Hz could be represented, producing such frequency content is not the purpose of the upsampling process. In fact, any frequencies that are above 22,050 Hz after the upsampling must be undesirable artifacts, as they were not represented in the original (lower sample rate) signal - and the purpose of the low-pass filter is to eliminate any such increase in bandwidth. After applying the low-pass interpolation filter, the bandwidth of the original signal and the bandwidth of the upsampled signal should be approximately the same if the filter has done a good job. Thus the purpose of upsampling is not to increase the bandwidth of the signal. --- SudoMonas
Note, please sign your posts with ~~~~
Certainly upsampling by 16 will increase your bandwidth by 16, but your original signal won't fill the extra bandwidth. Regardless, increasing your sampling rate gives you more bandwidth. Cburnett 05:34, 9 Mar 2005 (UTC)
This seems to be a matter of confusion over the meaning of the term bandwidth. To me, bandwidth is the width of the band of frequencies that is necessary to represent the content of the signal in a frequency-based analysis (e.g., a Fourier transform), not the number of bits used to represent the signal in the time (or space) domain. After the low-pass interpolation filtering, the frequency content of an upsampled signal should not have been substantially altered. (Thanks for the signing tip - I was wondering how that was done.) SudoMonas 06:37, 9 Mar 2005 (UTC)
The frequency content of the signal has not changed, but the bandwidth increases. For example, if my original signal is 1 khz and the sampling rate is 2 khz and I upsample by 16 (so sampling rate is now 32 khz). After filtering, the signal still has a frequency content of 1 khz but the bandwidth of the signal is now 16 khz. I could, for example, add in a 10 khz sine wave and be perfectly ok. Or I could add in another single if it's less than 16 khz.
Ultimately, it would appear you're not divorcing the bandwidth of the signal from that of frequency content of the signal. Cburnett 07:09, 9 Mar 2005 (UTC)
I removed "This is usually done to increase the bandwidth of a signal, which provides room for adding more information. Upsampling by itself does not add information."
Bandwidth can refer to both "the actual width of the frequency information carried by a signal", and "the width of frequency information which can be carried by a certain sample/data rate".
if the first definition is used, upsampling does NOT increasing the bandwith. if the latter definition is used, it's not possible to upsample without increasing the bandwith.
So it's incorrect either way. (even without "This is usually done to", I don't see how the context "upsampling" would totally "rule out" the first definition)
Upsampling in practice add some degree of aliasing, which perhaps could be considered "information"?
Also, I don't understand the "Unlike in downsampling which uses a low-pass filter as an anti-aliasing filter, upsampling uses an interpolation filter, which also is a low-pass filter."
Both upsampling and downsampling refer to the sinc filter, which "strips high-frequency data from a signal", Could it be better explained what the difference is between "using sinc as an interpolation filter" and "using sinc as an anti-aliasing filter", I just find it highly confusing.
(was I too bold removing?) teadrinker 14:44, 31 March 2006 (UTC)
The filter is the same, only the purpose is different. With upsampling, there is no problem with aliasing. Mirror Vax 15:36, 31 March 2006 (UTC)
## Additional text to upsampling
Towards the end of the section on 'Upsampling by integer factor', I would think that the following text might be useful
--- start text--
When zeros are inserted into the signal for upsampling, the spectrum of the upsampled sequence will have aliases at integer multiples of the original sampling frequency. These aliases can be removed to a reasonable extent by a finite impulse response low pass filter. The presence of zeros in the sequence which is passed through the filter can be used to reduce the complexity of the filter implementation. The original filter can be split to ${\displaystyle L}$ subfilters and the output of each of these subfilters is sequentially tapped to obtain the filtered output sequence [1].
1. ^ Polyphase filters for interpolation [1]
---stop text----
What say?
Beetelbug 14:54, 12 May 2007 (UTC)
## an interesting point of understanding upsampling
Maybe we can understand upsampling like this: First the sampling rate is Fs, and we want to upsampling the rate to positive infinite. (also it seems the situation is going to change from Digital world to Analog world~. ) Note that when the signal is presented by the rate of Fs, in fact the signal is periodic in spectrum whose period is Fs, the reason we can't see it is just because the sampling rate Fs gives no ability to handle. So when the sampling rate comes to really big, there will be a lot of signals repeated on the spectrum with period of Fs. Now abviously we need a interpolation filter after the upsampling process.
Terrence007 (talk) 03:47, 6 November 2011 (UTC)
interesting. Dicklyon (talk) 03:55, 6 November 2011 (UTC) | 1,400 | 6,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-30 | latest | en | 0.950068 |
http://www.velocityreviews.com/forums/t746156-mathematical-operations-on-array.html | 1,397,761,683,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00443-ip-10-147-4-33.ec2.internal.warc.gz | 762,698,607 | 8,361 | Velocity Reviews > Mathematical Operations on Array
# Mathematical Operations on Array
Bryan.Fodness@gmail.com
Guest
Posts: n/a
04-01-2011
I am loading text into an array and would like to convert the values.
from math import *
from numpy import *
from pylab import *
mincos=degrees(acos(data[:,0]))
minazi=degrees(data[:,1])
minthick=data[:,2]/0.006858
I am not sure why degrees() works, but acos() does not.
I receive the following
Traceback (most recent call last):
File "C:\ test.py", line 6, in ?
mincos=degrees(acos(float(data[:,0])))
TypeError: only length-1 arrays can be converted to Python scalars
Can anyone tell me what I am doing wrong?
Peter Otten
Guest
Posts: n/a
04-01-2011
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> I am loading text into an array and would like to convert the values.
>
> from math import *
> from numpy import *
> from pylab import *
>
> mincos=degrees(acos(data[:,0]))
> minazi=degrees(data[:,1])
> minthick=data[:,2]/0.006858
>
> I am not sure why degrees() works, but acos() does not.
>
> I receive the following
>
> Traceback (most recent call last):
> File "C:\ test.py", line 6, in ?
> mincos=degrees(acos(float(data[:,0])))
> TypeError: only length-1 arrays can be converted to Python scalars
>
> Can anyone tell me what I am doing wrong?
Using star-imports.
Among other things it makes it hard to keep track of where things are coming
from:
>>> from math import *
>>> from numpy import *
>>> degrees
<ufunc 'degrees'>
>>> acos
<built-in function acos>
>>> acos.__module__
'math'
>>> arccos
<ufunc 'arccos'>
Bryan.Fodness@gmail.com
Guest
Posts: n/a
04-01-2011
On Apr 1, 9:52*am, Peter Otten <(E-Mail Removed)> wrote:
> (E-Mail Removed) wrote:
> > I am loading text into an array and would like to convert the values.
>
> > from math import *
> > from numpy import *
> > from pylab import *
>
> > mincos=degrees(acos(data[:,0]))
> > minazi=degrees(data[:,1])
> > minthick=data[:,2]/0.006858
>
> > I am not sure why degrees() *works, but acos() does not.
>
> > I receive the following
>
> > Traceback (most recent call last):
> > * File "C:\ test.py", line 6, in ?
> > * * mincos=degrees(acos(float(data[:,0])))
> > TypeError: only length-1 arrays can be converted to Python scalars
>
> > Can anyone tell me what I am doing wrong?
>
> Using star-imports.
>
> Among other things it makes it hard to keep track of where things are coming
> from:
>
> >>> from math import *
> >>> from numpy import *
> >>> degrees
> <ufunc 'degrees'>
> >>> acos
>
> <built-in function acos>>>> acos.__module__
> 'math'
> >>> arccos
>
> <ufunc 'arccos'>- Hide quoted text -
>
> - Show quoted text -
Thank you.
Terry Reedy
Guest
Posts: n/a
04-01-2011
On 4/1/2011 9:35 AM, (E-Mail Removed) wrote:
> Can anyone tell me what I am doing wrong?
Posting the same question twice is a bad idea, as it splits answers and
may lead to duplication. I answered your first post without seeing
Peter's response to you second post, which is further down the list.
--
Terry Jan Reedy | 863 | 3,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-15 | latest | en | 0.742643 |
https://cboard.cprogramming.com/cplusplus-programming/130158-quadratic-gaussian-blur-printable-thread.html | 1,508,243,145,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821088.8/warc/CC-MAIN-20171017110249-20171017130249-00704.warc.gz | 686,812,449 | 3,252 | • 09-19-2010
TIMBERings
This is a paint like program, where on a click or drag, a circle with certain fills will appear.
I'm having issues with the Quadratic and Gaussian Brushes. Each provide me with a small dot, the QuadraticBush is about 20 pixels in diameter and the GaussianBrush is 1 pixel. Each start with a radius and flow of 128. These are the two functions that create the mask which stores the flow.
They are wrapped in a double for loop holding i as each row and j as each column.
Size is the width and height of the array.
Distance is the distance from the click location to each pixel in the mask.
Flow can be from 0 to 255 and represents the amount of color applied.
Rho is some constant for the Gaussian equation.
Code:
```if( distance <= (double)radius ) { if (distance < 1) { m_mask[i*(size-1)+j] = m_flow; } else { m_mask[i*(size-1)+j] = m_flow / (distance * distance); if (m_mask[i*(size-1)+j] > m_flow) { m_mask[i*(size-1)+j] = m_flow; } } } else { m_mask[i*(size-1)+j] = 0; }```
Gaussian:
Code:
```if( distance <= (double)radius ) { float base = 1 /(2 * PI * rho * rho); float exponent = -1.0 * ((distance * distance) / (2.0 * rho * rho)); m_mask[i*(size-1)+j] = m_flow * base * exp(exponent); if (m_mask[i*(size-1)+j] > m_flow) { m_mask[i*(size-1)+j] = m_flow; } } else { m_mask[i*(size-1)+j] = 0; }```
This is a linear blur which works just fine. It goes from 100% inside and 0% outside.
Linear:
Code:
``` int radius = m_radius; int size = radius*2 +1; int count=0; int arraysize = size * size; m_mask = new float[arraysize]; double distance; for(int i=0; i<size; i++) { for(int j=0; j<size; j++) { distance = sqrt((double)((radius-i)*(radius-i) + (radius-j)*(radius-j))); if( distance <= (double)radius ) { m_mask[i*(size-1)+j] = ((radius-distance)/(double)radius) * m_flow; } else { m_mask[i*(size-1)+j] = 0; } } }```
Any help would be great. | 621 | 2,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.499173 |
https://carlesto.com/past-questions/967/mechanics-and-properties-of-matter-tutorial-question-by-sure-excellence-tutorial-2003-past-question | 1,708,896,877,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474643.29/warc/CC-MAIN-20240225203035-20240225233035-00206.warc.gz | 163,128,635 | 12,298 | # Download Mechanics and Properties of Matter tutorial question by sure excellence tutorial - PHY114 Past Question PDF
You will find Mechanics and Properties of Matter tutorial question by sure excellence tutorial past question PDF which can be downloaded for FREE on this page. Mechanics and Properties of Matter tutorial question by sure excellence tutorial is useful when preparing for PHY114 course exams.
Mechanics and Properties of Matter tutorial question by sure excellence tutorial past question for the year 2003 examines 100-level Science and Technology students of UI, offering PHY114 course on their knowledge of force, vector, projectile
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### Tests related to Mechanics and Properties of Matter tutorial question by sure excellence tutorial
School: WAEC, JAMB & POST UTME
Department:
Course Code: JAMB
Topics: Physics, JAMB, Friction, work, force, motion, speed, velocity, energy, hydraulic press, relative density, hydrometer, gas law, sound wave, wave, light, mirror,capacitor, electricity, pressure | 3,553 | 16,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.824429 |
https://softuni.bg/forum/34858/snow-white-more-exercise-associative-arrays | 1,611,307,054,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00044.warc.gz | 564,552,066 | 17,705 | ## Snow White - More Exercise: Associative Arrays
Здравейте, имам нужда от помощ при сортировката на следната задача. Искам удобен начин за сортиране според броя на джуджетата с еднаква шапка приложимо към моето решение.
Моето решение: https://pastebin.com/77KGd8TP
You will be receiving several input lines which contain data about dwarfs in the following format:
{dwarfName} <:> {dwarfHatColor} <:> {dwarfPhysics}
The dwarfName and the dwarfHatColor are strings. The dwarfPhysics is an integer.
You must store the dwarfs in your program. There are several rules though:
• If 2 dwarfs have the same name but different color, they should be considered different dwarfs, and you should store both of them.
• If 2 dwarfs have the same name and the same color, store the one with the higher physics.
When you receive the command “Once upon a time”, the input ends. You must order the dwarfs by physics in descending order and then by total count of dwarfs with the same hat color in descending order.
Then you must print them all.
### Input
• The input will consists of several input lines, containing dwarf data in the format, specified above.
• The input ends when you receive the command “Once upon a time”.
### Output
• As output you must print the dwarfs, ordered in the way , specified above.
• The output format is: ({hatColor}) {name} <-> {physics}
### Constraints
• The dwarfName will be a string which may contain any ASCII character except ‘ ’ (space), ‘<’, ‘:’, ‘>’.
• The dwarfHatColor will be a string which may contain any ASCII character except ‘ ’ (space), ‘<’, ‘:’, ‘>’.
• The dwarfPhysics will be an integer in range [0, 231 – 1].
• There will be no invalid input lines.
• If all sorting criteria fail, the order should be by order of input.
• Allowed working time / memory: 100ms / 16MB.
### Examples
Input Output Pesho <:> Red <:> 2000 Tosho <:> Blue <:> 1000 Gosho <:> Green <:> 1000 Sasho <:> Yellow <:> 4500 Prakasho <:> Stamat <:> 1000 Once upon a time (Yellow) Sasho <-> 4500 (Red) Pesho <-> 2000 (Blue) Tosho <-> 1000 (Green) Gosho <-> 1000 (Stamat) Prakasho <-> 1000 Pesho <:> Red <:> 5000 Pesho <:> Blue <:> 10000 Pesho <:> Red <:> 10000 Gosho <:> Blue <:> 10000 Once upon a time (Blue) Pesho <-> 10000 (Blue) Gosho <-> 10000 (Red) Pesho <-> 10000 | 659 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.689774 |
https://stacks.math.columbia.edu/tag/0BP9 | 1,660,023,521,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00386.warc.gz | 485,136,566 | 6,117 | Remark 15.90.18. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module that is not necessarily glueable for $(R \to R', f)$. Setting $M' = M \otimes _ R R'$ and $M_1 = M_ f$ we obtain the glueing datum $\text{Can}(M) = (M', M_1, \text{can})$. Then $\tilde M = H^0(M', M_1, \text{can})$ is an $R$-module that is glueable for $(R \to R', f)$ and the canonical map $M \to \tilde M$ gives isomorphisms $M \otimes _ R R' \to \tilde M \otimes _ R R'$ and $M_ f \to \tilde M_ f$, see Theorem 15.90.17. From the exactness of the sequences
$M \to (M \otimes _ R R' )\oplus M_ f \to M \otimes _ R (R')_ f \to 0$
and
$0 \to \tilde M \to (\tilde M \otimes _ R R') \oplus \tilde M_ f \to \tilde M \otimes _ R (R')_ f \to 0$
we conclude that the map $M \to \tilde M$ is surjective.
There are also:
• 4 comment(s) on Section 15.90: The Beauville-Laszlo theorem
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). | 392 | 1,058 | {"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} | 2.796875 | 3 | CC-MAIN-2022-33 | latest | en | 0.686367 |
http://www.appszoom.com/android_applications/tools/scientific-calculator-plus_cccek.html | 1,386,795,622,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164047228/warc/CC-MAIN-20131204133407-00065-ip-10-33-133-15.ec2.internal.warc.gz | 299,538,439 | 13,630 | Close
Scientific Calculator Plus
Scientific Calculator Plus is a powerful mathematical tool to do mathematical analysis and evaluate complicated mathematical expressions similar to Matlab. More than Matlab, it has a mathematical equation(s) solver which helps user to solve mathematical problems. It supports complex number, array (matrix), (higher level) integration, 2D, polar and 3D chart, string, programming (using a language similar to Basic and Matlab) and unit conversion. It can run in both Android phones and tablets, PC and MAC with JAVA support enabled. Scientific Calculator Plus for Android includes a Scientific Calculator Plus for PC and Mac. User can connect mobile device to any PC or MAC via a USB cable, copy SDcard\AnMath\ folder to his/her own directory and launch the JMathCmd.jar in it.
This calculator has provided more than 100 diffused functions covering trigonometric calculation, complex number, matrix, integration, string and chart plotting. It is also able to handle mathematical operators like plus, division, power, percent, etc. And these operators fully support matrix and complex numbers. Thus user is able to evaluate complicated expressions, which are either not supported or hard to input in most traditional calculators. Moreover, the calculator is actually based on an easy-to-use programming language called MFP. Users can develop their own functions using this tool by a PC editor, store them in mobile SD-card and use them in calculator or command line. As a programmable tool, it can do everything.
The calculator can also think for user. It is able to solve unknown variables from complicated expressions and show the result to user. To calculate result of an expression, user input an expression like 3 + log(4.1 / avg(1,5,-3)). To solve unknown variables, user input single expression like log(x) - 7== 3, or a group of expressions like
y1/3+y2==7
y2/2-y3+y1==9
y3+y1+y2==2.4
. Then simply press start button, user will see value of 3 + log(4.1 / avg(1,5,-3)) or solved x or y1, y2 and y3 value.
This calculator has two basic user interfaces: smart calculator interface and command line interface. In the smart calculator interface, user is able to define variables, solve equation(s) and input expression(s) from historical records, using keyboard or clicking function buttons. Function buttons are configurable so that user is able to define his/her input keypads to input user-defined functions quickly.
User can also input expressions and plot 2D, polar or 3D charts in smart calculator. The number of unknown variables and variable name determine whether 2D, polar or 3D chart is plotted. After chart is plotted, smart calculator will show chart snapshot in the output box and history record. User can review the chart by simply clicking the chart snapshot
In the command-line interface (which works like Matlab), user uses keyboard to input expressions and sees the print output (which is not shown in the calculator interface) and return value. Command-line interface is best tool for developers to output intermediate results in their functions.
In order to help user input and evaluate (higher level) integration and plot charts, this calculator includes built-in integration and chart-plotting utilities. User no longer needs to type long and complicated expressions. Instead, by simply filling the text fields and pressing buttons, user will see the results or the charts. All the charts are automatically saved in SD-card which can be viewed and managed by file manager of the calculator later on.
Like Matlab, the calculator also provides a small program editor which gives developers much convenience to build their own functions. Nevertheless, they can also user a PC editor to do this job and save the program files in SD-card for future use.
Tags: scientific calculator phones, scientific calculator, 程式開發科學計算器, 科學計算器, scientific calculator plus, matlab phones, scientific calculator phone, scientific calculator code, mfp programming language, free scientific calculator plus.
Recently changed in this version
* Be able to plot polar graph. In Smart Calculator a polar graph is generated if the expressions include only 2 variables and at least one is Greek letter α, β, γ or θ. The input pad of Smart Calculator includes a θ button to assist user to input polar expressions;
* Add input function to input values from command line while program is running;
* Support multiple SD cards. User is able to select the storage to place app's data and SCP for JAVA;
* Functionality optimized and bugs fixed;
Last activity on Scientific Calculator Plus
Comments and ratings for Scientific Calculator Plus
• (72 stars)
Great calculator with a lot of functions and capabilities, but the interface is not very user friendly
• (72 stars)
Excellent
• (72 stars)
In the market no other app has so powerful 3 d functionality
• (72 stars)
This app is great. Pls keep up the gd wrk
• (72 stars)
So convenient
• (72 stars)
Wow...
• (72 stars)
I definitely erect my thumb | 1,058 | 5,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-48 | latest | en | 0.885942 |
http://www.cfd-online.com/Forums/cfx/85585-reasonable-order-domain-imbalances-print.html | 1,477,194,737,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00154-ip-10-171-6-4.ec2.internal.warc.gz | 352,859,582 | 2,259 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- CFX (http://www.cfd-online.com/Forums/cfx/)
- - Reasonable order of Domain imbalances (http://www.cfd-online.com/Forums/cfx/85585-reasonable-order-domain-imbalances.html)
saisanthoshm88 March 2, 2011 02:04
Reasonable order of Domain imbalances
CFX has got two forms of convergence criteria:
1. The residuals of continuity, momentum and energy equations.
2. The Domain Imbalances.
It is possible for the residuals to have converged well but the imbalances to not converge; and it is also possible for the residuals to not have converged but the imbalances have converged.
So to consider a solution to have good convergence what should be the percentage of domain imbalance in general?
(For a Steady state simulation and a Transient simulation)
I know that some quantities of interest should also be monitored to judge the convergence.
But I’m particularly interested in knowing how the domain imbalances are indeed monitored.
I mean reasonably of what order they should seem to be to say that the solution had converged well in terms of the domain imbalances.
Is it always required to have them converged by 1%.
ghorrocks March 3, 2011 18:41
Quote:
It is possible for the residuals to have converged well but the imbalances to not converge;
Yes
Quote:
and it is also possible for the residuals to not have converged but the imbalances have converged.
Yes
Different simulations need different convergence. Read the CFX documentation for a guide, but ultimately you need to do a sensitivity analysis to check you have sufficiently converged for you specific case.
All times are GMT -4. The time now is 23:52. | 380 | 1,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-44 | latest | en | 0.934955 |
http://math.stackexchange.com/questions/53161/how-do-i-show-that-e1-x-x2x2-5x-4-x12-2x-1-has-only-one-po | 1,466,850,598,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393093.59/warc/CC-MAIN-20160624154953-00075-ip-10-164-35-72.ec2.internal.warc.gz | 196,217,232 | 17,843 | # How do I show that $e^{1/x} = x(2x^2 + 5x + 4)/[(x+1)^2 (2x-1)]$ has only one positive real solution?
I need to show that, for the function $f(x) = x^2 (e^{1/x}-1) - x^2/(x+1)$, there is some $x_0 > 0$ such that $f(x)$ is decreasing on $(0,x_0)$ and increasing on $(x_0,\infty)$.
By setting $f'(x) = 0$ I can see that $f(x)$ is decreasing on $(0,1/2)$, and taking the series expansion of $f(x)$ at $\infty$ tells me that $f(x) \to 3/2$ as $x \to \infty$, but this isn't enough. The problem boils down to the question in the title.
I just can't seem to make it give up the goods.
-
I could simplify the problem quite a bit but left a simpler problem for you to solve. We have
$$f(x) = \left(e^{\frac{1}{x}}-1\right) x^2-\frac{x^2}{x+1}$$
using some basic algebra you can factor its derivate to:
$$f'(x) = \frac{2 e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}}{(x+1)^2}$$
So what is left to show is that
$$h(x)= e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}$$
has exactly one real root in $(0, \infty)$, but this is not hard because if we look at the graph of h we see that
And therefore assume that $$h'(x)=(x+1) \left(e^{\frac{1}{x}} \left(\frac{1}{x^2}+6 x-\frac{1}{x}-2\right)-6 x-4\right)>0$$
Where you can also use the notation
$$h'(t)=h'(1/x)=\frac{(t+1) \left(-4 t+e^t ((t-2) t (t+1)+6)-6\right)}{t^2}$$
So it remains again to show that
$$(t+1) \left(t^3-t^2-2 t+6\right) e^t\geq 2 \left(2 t^2+5 t+3\right)$$
Then we are done if we apply the mean value theorem. It should be doable to show this.
-
Setting $t= 1/x$ and working with $t$ should simplify things a bit... – Aryabhata Jul 22 '11 at 20:29
@Aryabhata: Thanks! I used your suggestion to reduce it even further. – Listing Jul 22 '11 at 20:41
Thanks Listing, following in this manner did eventually yield the solution. Also, @Aryabhata: yes, I had the same idea in the shower :) – Antonio Vargas Jul 23 '11 at 0:42
No problem, good to see I could help. – Listing Jul 23 '11 at 9:19 | 786 | 2,020 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-26 | latest | en | 0.87601 |
https://www.physicsforums.com/threads/parallel-path-interaction.226715/ | 1,575,837,488,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00204.warc.gz | 830,048,973 | 19,694 | # Parallel path interaction
## Main Question or Discussion Point
Two masses A and B are accelerated to a velocity V on close “parallel” paths.
They will experience gravity or free-fall toward one another.
If I take the limit as mass A and B are reduced to zero and the velocity A and B is accelerated to approach the speed of light, ( I am approaching light photon conditions for A and B ), will the objects ( photons ) A and B attract one another / will their paths bend toward one another / will their paths cross one another
Related Special and General Relativity News on Phys.org
You don’t need to take limits of mass somehow converting into photons. Photons and mass are different. You can just consider individual photons or particles of mass they will behave the same.
The problem you will have is when you define how the “bend” in your straight lines need to be it will likely be much smaller than your ability to define straight or parallel lines.
It would be like expecting a beam of collimated light to remain the same width forever, light doesn’t work that way.
pervect
Staff Emeritus
Two masses A and B are accelerated to a velocity V on close “parallel” paths.
They will experience gravity or free-fall toward one another.
If I take the limit as mass A and B are reduced to zero and the velocity A and B is accelerated to approach the speed of light, ( I am approaching light photon conditions for A and B ), will the objects ( photons ) A and B attract one another / will their paths bend toward one another / will their paths cross one another
No, they will not. (I used to have some more specific references on this point, which can also be stated by saying that parallel light beams won't attract each other gravitationally. Note that anit-parallel light beams will attract each other).
Am I off base yet?
This makes sense to me when I consider that light would only have “relativistic mass” that only has an affect on an object in the direction of motion. So photons traveling in a perpendicular path would not gravitationally affect one another as they have no velocity perpendicular to the path of travel.
Am I correct in assuming that two photons on parallel paths passing one another from opposite directions would experience a gravitational interaction?
I would expect this because they do have a velocity component towards or away from each other at some point during the passing process.
How is this for a definition.
If P is a plane through photon A and perpendicular to the velocity vector of photon A then all photons traveling perpendicular to plane P and contain in plane P will not have a gravitational interaction with photon A.
This makes sense to me
What does?
My comment that they will, but you will not be able to show it.
Or that it they will not.
You are working on individual particles (photons) not beams of particles.
(Also, I don't know what "anit-parallel" means, I assume it does not just mean perpendicular)
I was responding to Pervects' post.
Light which is traveling in Parallel and adjacent paths do not gravitationally attract because relativistic mass is directional. That is to say it requires a velocity with respect to an observer to exit. Light has no velocity perpendicular to the direction of travel and therefore has no relativistic mass in the direction perpendicular to travel.
I was responding to Pervects' post.
Light which is traveling in Parallel and adjacent paths do not gravitationally attract because relativistic mass is directional. That is to say it requires a velocity with respect to an observer to exit. Light has no velocity perpendicular to the direction of travel and therefore has no relativistic mass in the direction perpendicular to travel.
Isn’t this contrary to standard GR assumptions that massless photon particles of energy should follow the same paths as traditional particles of mass would?
Thus, light bends around the sun just as a particle of mass would do.
I assume you expect two particles of mass starting off on your trajectory would not remain parallel as they curve together. Yet here the rule of photons behaving gravitationally the same as regular particles is not followed for some reason.
The reason for not retaining this GR assumption is not well explained IMO.
Two mass particles would follow a similar path as the photon described.
If two masses are traveling together they have no relative velocity with respect to one another and would therefore only observe the rest mass of the adjacent particle.
If the particles are small then the rest mass gravitational attraction will be small.
If however the same two masses pass each other at a velocity close to the speed of light then either of the particles will observe the other particle as having a large velocity and attribute relative mass with this velocity.
If the differential velocity approaches the speed of light and the particle mass quantities approach zero then the paths will approach a path identical to a path photons would follow.
A lot of odd things happen when the speed of light is approached.
JesseM
Isn’t this contrary to standard GR assumptions that massless photon particles of energy should follow the same paths as traditional particles of mass would?
Thus, light bends around the sun just as a particle of mass would do.
Are you sure this is a "standard GR assumption"? Particles of mass certainly don't follow the same geodesics through spacetime as photons, so I'm not so sure their paths through curved space would look the same either, although they would both obviously be deflected in some way by the Sun.
Are you sure this is a "standard GR assumption"? Particles of mass certainly don't follow the same geodesics through spacetime as photons, so I'm not so sure their paths through curved space would look the same either, although they would both obviously be deflected in some way by the Sun.
Of the two potential concepts:
1) “Particles of mass certainly don't follow the same geodesics through spacetime as photons” - - should mean GR does not define the same GR curved space path for them.
2) “they would both obviously be deflected in some way by the Sun” - - They being photons vs, particles of mass; plus they would not be deflected just “in some way” but in exactly the same way, just as various values of small mass would all be deflected in exactly the same way by the Sun.
The only accepted and tested elements of GR assumptions I’m aware of confirm item #2.
And IMO if #2 is true it implies that #1 is false.
I’m not aware of any well founded and tested assumptions based on GR that indicate #1 could be true. If you have references to such I’ll look.
I suspect the reason is “space-time” was never and is not a good foundation from which the complete GR Theory was or could be built upon.
E.G. Einstein never gave “space-time” any real credit beyond being able to excite the interest of the general public that could not understand the details of something like GR. But he did feel that being able to involve the public at level they could appreciate was a good thing for science.
JesseM
1) “Particles of mass certainly don't follow the same geodesics through spacetime as photons” - - should mean GR does not define the same GR curved space path for them.
I don't understand what you mean by "GR does not define the same GR curved space path for them". Are you disputing the statement of mine that you quoted, or not?
RandallB said:
I’m not aware of any well founded and tested assumptions based on GR that indicate #1 could be true. If you have references to such I’ll look.
You mean, references for the fact that different particles follow different geodesics depending on their velocity, and in particular that the geodesic path of light is different from the geodesic path of a massive particle? You might try looking up the difference between "timelike geodesics" and "null geodesics" (or 'lightlike geodesics') for starters. But are you doubting that two test particles can cross each other's paths at a single point in spacetime and yet follow different geodesics because their velocity at that point is different? If two particles are in different orbits around a planet but the orbits are elliptical so they cross paths at some point, do you doubt that both particles are both following geodesics even though the two paths go in different directions from the crossing-point?
RandallB said:
I suspect the reason is “space-time” was never and is not a good foundation from which the complete GR Theory was or could be built upon.
What are you talking about? The whole point of GR is that it tells you the curvature of spacetime based on the distribution of mass and energy, and test particles follow geodesic paths in spacetime which maximize the proper time along that path (for timelike geodesics), just like how the inertial twin in the flat-spacetime twin paradox follows a geodesic path between two events on his worldline which causes him to age more (larger proper time) than his twin who follows a non-inertial path between the same two events.
RandallB said:
E.G. Einstein never gave “space-time” any real credit beyond being able to excite the interest of the general public that could not understand the details of something like GR.
This is a very ignorant statement. Again, the basic tensor equations of GR (i.e. the Einstein field equations) are all about how matter and energy curve the 4D spacetime manifold, not how they curve space. For example, see this page from an introduction to Einstein's GR equation which says:
Similarly, in general relativity gravity is not really a `force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial. If you toss a ball, it follows a parabolic path. This is far from being a geodesic in space: space is curved by the Earth's gravitational field, but it is certainly not so curved as all that! The point is that while the ball moves a short distance in space, it moves an enormous distance in time, since one second equals about 300,000 kilometers in units where c=1. This allows a slight amount of spacetime curvature to have a noticeable effect.
Last edited: | 2,124 | 10,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-51 | latest | en | 0.945394 |
https://uk.mathworks.com/matlabcentral/cody/problems/2880-matlab-basics-ii-determine-if-an-array-has-a-3rd-dimension/solutions/1036552 | 1,579,421,015,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00297.warc.gz | 721,114,235 | 15,516 | Cody
# Problem 2880. Matlab Basics II - Determine if an array has a 3rd dimension
Solution 1036552
Submitted on 29 Oct 2016 by Gareth Lee
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x(:,:,1) = [1 2 3]; x(:,:,2) = [4 5 6]; x(:,:,3) = [7 8 9]; y_correct = 1; assert(isequal(three_d(x),y_correct))
2 Pass
x(:,1) = [4 7 38]; x(:,2) = [42 45 6]; y_correct = 0; assert(isequal(three_d(x),y_correct)) | 181 | 523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | latest | en | 0.737592 |
https://mathexamination.com/test/exponential-hierarchy.php | 1,627,316,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00333.warc.gz | 398,466,091 | 8,060 | ## Take My Exponential Hierarchy Test
If you need to study for a Exponential Hierarchy Test, there are numerous tools out there that you can utilize. Nevertheless, if you want to find out more about the methods of teaching, there are also many resources offered. This post takes a look at a few of these resources and their usefulness.
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It is specifically important to study for a Exponential Hierarchy Test if you understand how particular questions are structured. The majority of people discover that they have a respectable idea of what is anticipated of them by the instructor, however the specific details are quite difficult to pin down. Getting specific suggestions on what is expected from you by utilizing a guide will help you prepare your finest.
The theory of test preparation, particularly for Exponential Hierarchy, is nothing new. As long as there has been assessments, individuals have been preparing for them. They are often used by schools and companies also. In fact, many universities and colleges now need their trainees to take tests.
Typically, independent students will take a test just before they graduate. This is the time when they are trying to comprise any lost ground that they may have made in class. If you are going to take an examination soon after graduating, make certain that you are prepared well beforehand.
Prior to you go to take your Test, you ought to check out various sources of information. Start with Google and search for the Test topics you are most interested in. You need to likewise check out magazines and books on the subject to get a good photo of what the Test resembles.
In order to prepare for your tests, it is a good idea to check out great deals of product on the subject. This will help you think rationally about the problems that you might be provided with. Understanding the test layout is also useful. You should be familiar with the design of each section.
Lots of people are intimidated by their examinations. Sadly, this is something that impacts most trainees. Even those who feel confident are still most likely to have doubts about the test. If you feel that you do not actually understand the concern, you might want to request help before the start of the examination.
Selecting the right timing to check out the questions will help you get to grips with the product. An example of this is a typical issue for trainees. They may know the answer to a question, however they still have problem with other parts of the Test. For this reason, it is typically better to invest a little time at the start of the evaluation and ensure that you totally comprehend the question prior to proceeding.
To help you with your Exponential Hierarchy tests, I have actually consisted of a quick guide to assist you prepare. This is based upon reading books on the subject. It is a really succinct list of resources that you can use in order to get the most out of your evaluations.
Many students do not like the tests. This is probably since they think that they are tests that are hard to get through. On the contrary, if you have actually prepared effectively, then you must have the ability to pass the test with flying colours.
Lastly, you need to never let the Test distract you. Make certain that you take your time to check out the test content completely. Thiswill help you get the most out of your Test.
## Pay Me To Do Your Exponential Hierarchy Test
There are a great deal of offered online options when it comes to online mathematics test aid. You just require to pick the one that is most appropriate for your requirements.
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If you are searching for mathematics test assistance, you can pick from a number of websites which provide comprehensive responses and sample questions. Some sites also give specific directions on how to address the mathematics problems correctly. These websites are a lot more popular than those who use basic aid.
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Your understanding of the subject is essential if you want to study well in today times and depend on date with existing patterns. That is why it is important to look for online tips that you can utilize to get the very best out of your mathematics test.
You must first ensure that you have actually made the time to take the test. That way, you will have the ability to answer the questions at your own rate and you will not have to hurry through the whole process.
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It is necessary to think about the quantity of time that you can spare for studying. Whether you are a trainee or a parent, you should be extremely clear on your time and how much assistance you can pay for. Check out what the typical time is prior to you select which choice is best for you.
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When you believe you have picked a site that provides the very best online choices for your math test, you can proceed and check out the different type of aid that are offered. Understanding the type of aid that is offered can provide you comfort and will assist you discover better and faster than ever before.
## Hire Someone To Take My Exponential Hierarchy Test
When you are participating in a Mathematics class, you are bound to come across the requirement to discover how to best study for your mathematics test. It is necessary that you comprehend that no matter how many math books you may own, there will constantly be a scenario when you require to discover Mathematics Test Aid. It is much better to understand how to study for your Exponential Hierarchy test early than not, and without knowing the exact situation, you might wind up failing this test or even worse.
You will first need to understand what your Exponential Hierarchy Test is going to be prior to you can know how to study for it. Some mathematics classes have a specific examination which is among the requirements for the test to be provided. If this holds true, then you will need to learn if this Test will be covering what you require to cover in your real test. It will likewise help if you discover exactly what questions you will need to respond to on the Exponential Hierarchy test before you go to the Test.
You will then need to identify the kind of Mathematics Test Help you will need to study for your Exponential Hierarchy test. There are various resources that can be utilized for studying for any Exponential Hierarchy test. The most typical ones include online math practice tests, along with complete practice tests. These resources work because they assist you find out everything you need to understand about the subject in order to pass your actual test.
The more you study for your particular mathematics test, the better prepared you will be for the Test. You will also find out a lot about the test itself, and this will assist you be more prepared for your real examination. It is best to take as many tests as possible, as it will assist you to learn how the test is set up, as well as what you will require to know.
Take a look at a few various tests and see which ones you would choose to take for a comprehensive evaluation of the subject. This is very crucial, because it can avoid you from getting a low rating on your Exponential Hierarchy test. It can likewise help you much better prepare for your real Test.
It is likewise crucial to choose where you are going to take the Test. This is very important due to the fact that it can identify what type of materials you will need to bring with you. For instance, you might require flashcards or pencils and paper to assist you with your test.
Many tests have concerns that ask you to look into the correct products on a graph, matrix, or chart. If you do not know how to do this, then it will assist to bring a set of reference charts with you so that you can have an idea of what each item is.
You will also require to choose which type of calculator is going to be right for your Mathematics Evaluate Help. The majority of tests will have a number of various types of calculators offered to you, and you will require to understand which one is going to be right for your particular mathematics test.
There are some that will have extra features like moving number keys, however these might not be something you will need. Most of calculators are going to have a small number of pre-programmed settings. It is necessary to determine what these are prior to you begin to attempt to determine the remainder of the calculator.
When you are attempting to determine a mathematical idea, it can help to utilize a calculator in order to get a feel for how it is done. There are great deals of them to select from, and each of them will differ in how easy it is to utilize. For instance, some of them will be hand held, while others are computerized.
A lot of the first timers that get in a class might not have much basic mathematics skills. This is why it is best to use a calculator throughout an exam. A number of these will have different interactive alternatives which enable you to get a feel for the idea and after that carry on to practice the genuine product on the calculator.
Practice using the various tools and general rules on the calculator, and then study the issue on the calculator and the equation it produces. This is the very best way to get a feel for using a calculator to fix for a number. | 2,216 | 11,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.970203 |
https://www.enotes.com/homework-help/infinity-larger-than-largest-number-imaginable-592984 | 1,521,739,424,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647901.79/warc/CC-MAIN-20180322170754-20180322190754-00265.warc.gz | 807,665,533 | 10,485 | # If Infinity is larger than the largest number imaginable, then how do you square infinity? Because if Infinity is the largest number then even infinity squared could still be smaller.
Borys Shumyatskiy | Certified Educator
Hello!
Infinity is not an ordinary number, and its square isn't a number also. Actually, the square of infinity is also infinity.
Consider a model where ordinary numbers and infinity are represented as a limits of sequences.
If a sequence `{a_n}` has a limit `a` (a number), i.e.
`AA` e>0 `EE` N(e) | `AA` n>N(e) `|a_n-a|lte,`
then it is considered as a representative of a number `a.`
If a sequence `{a_n}` has an infinite limit, i.e.
`AA` E>0 `EE` N(E) | `AA` n>N(E) `|a_n|gtE,`
then it is considered as a representative of the infinity.
In this model we can add and multiply numbers AND infinity (with some restrictions). In particular, infinity squared is also infinity.
There is another model, where infinity is a cardinality of an infinite set. There are many different infinities in this model, some of them are greater than another:)
But "infinity squared" (the cardinality of the Cartesian product of the corresponding infinite set) is the same infinity.
cmsmichael | Student
Infinity is a term used to describe anything that is endless. That can apply to a scientific principle, mathematics, spiritual belief, or a philosphical idea. Something that is endless has no beginning or end and cannot be measured or managed using traditional knowledge. Therefore, attempting to apply known principles or formulas to the concept of endlessness, or infinity in this case, is not possible.
I thought of the line from the song in "The Sound of Music" that goes "How do you hold a moonbeam in your hand?" when I read your question. There are some things that we intuitively know are real but can never fully explain or manage with our limited human resources. That's what keeps life interesting and drives us toward greater understanding of our world and the universe that surrounds us. | 459 | 2,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.937896 |
https://cris.openu.ac.il/iw/publications/approximating-connectivity-augmentation-problems-2 | 1,716,930,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00335.warc.gz | 151,132,913 | 10,553 | # Approximating connectivity augmentation problems
פרסום מחקרי: פרסום בכתב עתמאמרביקורת עמיתים
## תקציר
Let G = (V,E) be an undirected graph and let S V. The S-connectivity ΛSG(u,v) of a node pair (u,v) in G is the maximum number of uv-paths that no two of them have an edge or a node in S - {u,v} in common. The corresponding Connectivity Augmentation (CA) problem is: given a graph G = (V,E), a node subset S V, and a nonnegative integer requirement function r(u,v) on V × V, add a minimum size set F of new edges to G so that ΛSG+F(u,v) ≥ r(u,v) for all (u,v) ε V × V. Three extensively studied particular cases are: the Edge-CA (S = Ø), the Node-CA (S = V), and the Element-CA (r(u,v)= 0 whenever u S or v S). A polynomial-time algorithm for Edge-CA was developed by Frank. In this article we consider the Element-CA and the Node-CA, that are NP-hard even for r(u,v) ε {0,2}. The best known ratios for these problems were: 2 for Element-CA and O(rmax · ln n) for Node-CA, where r max = maxu,vεV r(u,v) and n = V. Our main result is a 7/4-approximation algorithm for the Element-CA, improving the previously best known 2-approximation. For Element-CA with r(u,v) ε {0,1,2} we give a 3/2-approximation algorithm. These approximation ratios are based on a new splitting-off theorem, which implies an improved lower bound on the number of edges needed to cover a skew-supermodular set function. For Node-CA we establish the following approximation threshold: Node-CA with r(u,v) ε {0,k} cannot be approximated within O(2log1-εn) for any fixed ε > 0, unless NP ⊆ DTIME(npolylog(n)).
שפה מקורית אנגלית 5 ACM Transactions on Algorithms 6 1 https://doi.org/10.1145/1644015.1644020 פורסם - 1 דצמ׳ 2009
## טביעת אצבע
להלן מוצגים תחומי המחקר של הפרסום 'Approximating connectivity augmentation problems'. יחד הם יוצרים טביעת אצבע ייחודית. | 648 | 1,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.638653 |
https://www.slideserve.com/adamma/compare-and-order-non-rational-numbers | 1,513,337,033,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00029.warc.gz | 813,966,868 | 12,963 | 1 / 13
# Compare And Order Non-rational numbers - PowerPoint PPT Presentation
Compare And Order Non-rational numbers. Wilma L. Collins Cobb 6 th Grade Campus GPISD. TEKS Objective 6.1A. (1) Number, operation, and quantitative reasoning. The student represents and uses rational numbers in a variety of equivalent forms. The student is expected to:
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### Compare And Order Non-rational numbers
Wilma L. Collins
GPISD
• (1) Number, operation, and quantitative reasoning. The student represents and uses rational numbers in a variety of equivalent forms. The student is expected to:
• (A) compare and order non-negative rational numbers;
• The purpose of the lesson is to determine students understanding of number concepts
• Students will successfully read, compare, and order whole numbers and decimals.
Materials
• 2 sets of Paper Plate Presentation plates
• Transparencies or list of Paper Plate Presentation numbers
• two teams with a set of plates.
• Each team member should have a plate.
• This is a silent game, if a student talks during play, a point will be deducted from the team’s score.
• Show the first problem.
• Once the problem is uncovered, the team members with the appropriate plates should arrange themselves at the front of the class to represent the number/problem displayed from the transparency.
• The students have 30 seconds to form the number/problem at the front of the class. After 30 seconds, each team is given a point for a correct answer.
7. An additional point is given to the first team to “present” the correct answer.
8. Play continues until all of the numbers/problems have been displayed.
9. The winning team is the one with the most points at the end of the game.
Presentation Numbers “present” the correct answer.
1. five thousand, two and one tenth
2. Seven hundred twenty-three and eight hundredths
3. Eighty and four thousandths
4. Nine thousandths
5. One and fifty-three hundredths
6. Two hundred one thousand, thirty-six
Presentation numbers continue “present” the correct answer.
7. Fifteen and two tenths.
8. One hundred twenty thousand, three hundred seven and four tenths.
9. Four hundred twenty-five thousand, three hundred seventeen and eight thousandths.
10.one hundred forty-six thousand, three hundred ninety-seven
Guided Practice……….. “present” the correct answer.
Place the following numbers in order from greatest to least.
0.75, 0.615, 0.58, 0.195
1. Line-up numbers and add zero(s)
0.750
0.615
0.580
0.195
2. Look at number in the tenths place, all the numbers are different, so arrange that number in order from greatest to least.
.615 ,.195, .580, .750, than reorder all
.750
.615
.580
.195
### How do you compare and order non-negative rational numbers? different, so arrange that number in order from greatest to least.
To answer this question click on the webpage. Do the practice exercise of your choice, complete exercise according to web directions.
http://www.aaamath.com
Conclusion different, so arrange that number in order from greatest to least. | 853 | 3,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-51 | latest | en | 0.904608 |
https://forum.smartpls.com/viewtopic.php?t=30555&sid=6e218a6eb3849a25facf40e1e5d138a7 | 1,701,815,842,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00454.warc.gz | 302,748,954 | 7,048 | ## sample size estimation for multi group analysis
Questions about the implementation and application of the PLS-SEM method, that are not related to the usage of the SmartPLS software.
kini
PLS Junior User
Posts: 2
Joined: Fri Sep 08, 2023 12:44 pm
Real name and title: Kini.S PhD student
### sample size estimation for multi group analysis
Dear All,
I want to run a multi-group analysis in smart pls. For this, I would like to do an apriori estimation of the minimum sample size requirement for each group using G*power3.1. Which test should I use in G*power? Any guidance on this would be appreciated.
gertrudecolclough
PLS Junior User
Posts: 1
Joined: Fri Nov 03, 2023 7:32 pm
Real name and title: smartpls
### Re: sample size estimation for multi group analysis
kini wrote: Fri Sep 08, 2023 1:41 pm Dear Alluno online,
I want to run a multi-group analysis in smart pls. For this, I would like to do an apriori estimation of the minimum sample size requirement for each group using G*power3.1. Which test should I use in G*power? Any guidance on this would be appreciated.
According to my research, there are different methods and tests for conducting a multi-group analysis in PLS-SEM, such as the confidence intervals, the partial least squares multigroup analysis (PLS-MGA), the parametric test, and the Welch-Satterthwait test. Each method has its own assumptions and advantages, and you may need to choose the one that best suits your research question and data characteristics.
However, a general approach for estimating the minimum sample size requirement for each group is to use the one-way independent samples ANOVA test in G*power3.1. This test allows you to compare the means of two or more groups on a continuous outcome variable. To use this test, you need to specify the following parameters:
The type of power analysis: You should choose “A priori: Compute required sample size - given alpha, power, and effect size”.
The statistical test: You should choose “F tests: ANOVA: Fixed effects, omnibus, one-way”.
The type of effect size: You should choose “f”, which is the standardized mean difference between groups.
The input parameters: You should enter the values for alpha (the significance level, usually 0.05), power (the probability of detecting an effect, usually 0.80 or higher), and f (the effect size, which can be small, medium, or large, according to Cohen’s benchmarks5).
The output parameters: You should enter the number of groups that you want to compare, and the software will calculate the total sample size and the sample size per group.
You can find some examples and tutorials on how to use G*power3.1 for one-way independent samples ANOVA here, here, and here. | 635 | 2,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.88259 |
https://qanda.ai/en/solutions/zu56HtwCnm-1-Draw-the-graph-of-yx2-5x-6-and-hence-solve-x2-5x-140- | 1,618,262,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069267.22/warc/CC-MAIN-20210412210312-20210413000312-00011.warc.gz | 579,782,273 | 13,791 | Symbol
Problem
$1$ Draw the graph of $y=x^{2}-5x-6$ and hence solve $x^{2}-5x-14=0$ | 38 | 83 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-17 | latest | en | 0.737851 |
https://solvedlib.com/n/solve-each-second-order-differential-equation-with-trigonometric,12021109 | 1,685,385,056,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00741.warc.gz | 607,038,373 | 17,015 | # Solve each second-order differential equation. With Trigonometric Expressions. $y^{\prime \prime}+4 y=\sin 2 x$
###### Question:
Solve each second-order differential equation. With Trigonometric Expressions. $y^{\prime \prime}+4 y=\sin 2 x$
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##### QUESTION 263.36 m long bar moves on parallel rails that are connected tnrough 28.4 chm resistor. The circuit is in a uniform magnetic field 0.68 T tnat is directed out of the plane of the paper. At an instant when the induced current in the circuit is clockwise and equalto 6.12 A, what is the magnitude of the velocity of the oar in units?
QUESTION 2 63.36 m long bar moves on parallel rails that are connected tnrough 28.4 chm resistor. The circuit is in a uniform magnetic field 0.68 T tnat is directed out of the plane of the paper. At an instant when the induced current in the circuit is clockwise and equalto 6.12 A, what is the magni...
Problem 0.1. Given the stochastic differential equation for the stock price in the risk neutral measure dS(t) = rS(t) dt + oS(t) dW(t) 1) Derive the value of binary option which pays 1 $if the Stock Price at time T is greater than the strike K, namely Lsr>K 2) Derive the value of & capped pu... 1 answer ##### (4pts) 25) Patients not only has rights but also D must recognize he or she has... (4pts) 25) Patients not only has rights but also D must recognize he or she has no responsibilities because of his or her medical condition l must recognize the effects of his or her lifestyle habits on his or her health I work periodically, as desired. collaboratively with healthcare providers D ne... 1 answer ##### Y Mode: Connect HW 21 Help Save& Exit Submit 0 Problem 02.003.b- Resultant of forces applied... y Mode: Connect HW 21 Help Save& Exit Submit 0 Problem 02.003.b- Resultant of forces applied to a hook using the triangle rule Knowing that P 75 N and 215 N determine graphically the magnitude and direction of their resultant using the triangle rule The magnitude of the resultant is The directio... 5 answers ##### Let g and h be differentiable functions defined OII and f(,y) be differentiable function defined on R?. Say that we know the following values for these functions 9(2) ~3, h(2) =4 9(2) =5, "'(2) = Ald J(2,2) =-3,f(-3,4) =9 J.(2,2) =4,6.(-3,4) =-7 fv(2,2) = 6,5v(-3.4) = 8 Find the derivative of f(g(t). h(t)) at + = 2 Let g and h be differentiable functions defined OII and f(,y) be differentiable function defined on R?. Say that we know the following values for these functions 9(2) ~3, h(2) =4 9(2) =5, "'(2) = Ald J(2,2) =-3,f(-3,4) =9 J.(2,2) =4,6.(-3,4) =-7 fv(2,2) = 6,5v(-3.4) = 8 Find the derivative... 1 answer ##### Year Alt. 1 -$1000 + 1500 Alt.2 -$2000 + 2800 Incremental Alt.2 - Alt. 1 -$1000...
Year Alt. 1 -$1000 + 1500 Alt.2 -$2000 + 2800 Incremental Alt.2 - Alt. 1 -$1000 + 1300 7. Excel Problems: Which alternative you choose based on the incremental analysis? Pls do this in excel. Machine Machine X$700 Initial Cost \$200 Uniform annual benefit Salvage value Useful life, in years 8. A 401...
##### Question 8Give the form of particular solution ofy+ _6 y 110 > -6 19y=4e ~cos(x) +given that r1 li is & foot of the characteristic equation1B cos(x) + Cx sin(x) - D=Axe< + B cos(r) + € sin(x) + D:=Axe 1B_ cos(x) + Cx sin(x) - D=Ae 1B cos(x) + € sin(x) - D2=4 e + B cos(x) + Cx sin(x) - DVone of the above:742
Question 8 Give the form of particular solution of y+ _6 y 110 > -6 19y=4e ~cos(x) + given that r1 li is & foot of the characteristic equation 1B cos(x) + Cx sin(x) - D =Axe< + B cos(r) + € sin(x) + D :=Axe 1B_ cos(x) + Cx sin(x) - D =Ae 1B cos(x) + € sin(x) - D 2=4 e + B cos(...
##### Which of the following is a reason for organizations to store data in separate databases? Question...
Which of the following is a reason for organizations to store data in separate databases? Question 15 options: A) Storing data in multiple databases completely eliminates information silos. B) Duplicated data becomes inconsistent when changes are made to one set of data. C) Each department in an org...
##### Suppose the random varieble folloxs= uniforn isbibution gver thg interva (3.1]. If five (5) independent %alues of thg rardumn variable 32 gelected what is the Probability they all lie in thg middle half of the interval; that is in the interyz [0 25,0 75]
Suppose the random varieble folloxs= uniforn isbibution gver thg interva (3.1]. If five (5) independent %alues of thg rardumn variable 32 gelected what is the Probability they all lie in thg middle half of the interval; that is in the interyz [0 25,0 75]...
##### Queston 330l33 Step5) + 25 thousand people: How fast will thc of J clty wll nurnber F() = (0.30 3)(0.21 Itts estlmated that (years (rom now the population . Round your answer tO two decimal places; population (In thougsands) be groring In 1Jyears? =Keypad Keyboard ShortcutsRntr 3.03 PointsWhalannd peoplc per ycat
Queston 33 0l33 Step 5) + 25 thousand people: How fast will thc of J clty wll nurnber F() = (0.30 3)(0.21 Itts estlmated that (years (rom now the population . Round your answer tO two decimal places; population (In thougsands) be groring In 1Jyears? = Keypad Keyboard Shortcuts Rntr 3.03 Points Whala...
##### Choose the major product of the following reacton;CHQ DMSONo Reaction
Choose the major product of the following reacton; CHQ DMSO No Reaction...
##### 1) Calculate 4G rx (350 K) for the reaction: H2O(g) +0.502(g) → H2O2(1) using the Gibbs-Helmholtz equation and (46°...
1) Calculate 4G rx (350 K) for the reaction: H2O(g) +0.502(g) → H2O2(1) using the Gibbs-Helmholtz equation and (46°F 2120-228.57 kJ/mol; 4G°F (H202)-120.35 kJ/mol; Hºr ano)-241.82 kJ/mol; AHF (H202)=-198.78 kJ/mol). You may assume that the difference between 4GºrxN and AH&osla...
##### Consider a spherical mirror and lens separated by 45 cm. The mirror is on the left...
Consider a spherical mirror and lens separated by 45 cm. The mirror is on the left with a focal length of 100 cm. The lens is on the right with a focal length of −20 cm. A 5 cm tall object is placed 20 cm to the left of the lens. a) If you only consider the rays that move to the right from the...
##### I 2) For the chemical reaction CO2(g) + H2O(0) S H2CO3(aq) give the correct expression for...
I 2) For the chemical reaction CO2(g) + H2O(0) S H2CO3(aq) give the correct expression for Kc, Kn, and K, or explain why such an expression cannot be given. [12 points)...
##### Hour mst 5umciorNour ScoieWlnoingsPigwdus AnswetsOSCALCI 4.1,005-006.WA.Tut:FagnaA5tour TeacherHndcetreters Icng rests nouontz Gronnden lerns Juant dcrticzwdi Ine loct top*ano doentnt Keten fojt of the Edocr Iror Ieall?the Iadott ntenAvan Inom FrMAtre ra020.3 msec: How fast 5meccTulcrlae KnttntLabel the dutarce from the 6ase I Uduct tho wull *und {ho hoight Jon) (F" All from tha groind t0 the top racu OiLn Aa prcblem mathenutical eertOck Uno imoliot a ffcrentiaton, solve tor Inc Jpe opriat
Hour mst 5umcior Nour Scoie Wlnoings Pigwdus Answets OSCALCI 4.1,005-006.WA.Tut: Fagna A5tour Teacher Hndcet reters Icng rests nouontz Gronnden lerns Juant dcrticzwdi Ine loct top*ano doentnt Keten fojt of the Edocr Iror Ieall? the Iadott ntenAvan Inom FrMA tre ra02 0.3 msec: How fast 5 mecc Tulcrla...
##### 1(IO pls ) Draw @x hbekdr"tion protike for single step reaetion: ~B with the following data. Idlentify hether thee raction enelothennic Or exolheric, and calculate the: enthalpy of Taction und the activalion energy Fuergy of renctants S0. kJmol; Eueryy of produets 7u. kJhl: Energy of trunsition stale 80. kJhol.
1(IO pls ) Draw @x hbekdr"tion protike for single step reaetion: ~B with the following data. Idlentify hether thee raction enelothennic Or exolheric, and calculate the: enthalpy of Taction und the activalion energy Fuergy of renctants S0. kJmol; Eueryy of produets 7u. kJhl: Energy of trunsi...
##### Domains of functions of three or more variables Find the domain of the following functions. If possible, give a description of the domains (for example, all points outside a sphere of radius 1 centered at the origin). $p(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}-9}$
Domains of functions of three or more variables Find the domain of the following functions. If possible, give a description of the domains (for example, all points outside a sphere of radius 1 centered at the origin). $p(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}-9}$...
##### Question: The composite bar ABC is made of same material having E- 100 GPa. When the...
Question: The composite bar ABC is made of same material having E- 100 GPa. When the load was applied, the diameter for segment AB decreased by 0.025mm, calculate the final diameter of segment BC 20 mm 15 mm 100 kN 100 kN B 200 mm C A 350 mm...
##### Chapter 9B Practice ProblemsFrom the following heats ofreaction: 2 Hz (g) 0z (g) 2 H,o (g) 3 Oz (g) 2 0, (g) Calculate the heat of reaction for:AH = -483.6 k]AH =+284.6 kJ3 Hz (g)0, (g) 7 3 H,O (g)
Chapter 9B Practice Problems From the following heats ofreaction: 2 Hz (g) 0z (g) 2 H,o (g) 3 Oz (g) 2 0, (g) Calculate the heat of reaction for: AH = -483.6 k] AH =+284.6 kJ 3 Hz (g) 0, (g) 7 3 H,O (g)...
##### (14 points) Solve the following y" + 2y IVP by byysin '1-81 € (0) =4 the method of Laplace transforms: y' (0) = 2
(14 points) Solve the following y" + 2y IVP by byysin '1-81 € (0) =4 the method of Laplace transforms: y' (0) = 2...
-- 0.065297-- | 4,002 | 12,811 | {"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": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.839583 |
https://www.coursehero.com/tutors-problems/Statistics-and-Probability/20199584-Given-a-normal-distribution-determine-the-following-using-the-0-Z-tab/ | 1,582,303,096,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00160.warc.gz | 685,197,255 | 27,163 | View the step-by-step solution to:
Question
# Given a normal distribution, determine the following using the 0-Z table which can be found in
Content/Distribution Tables. Show table values used in each part. See Content/Videos-Topics In Stat 200/Normal Distribution & Z values for examples.
a) P(Z<1.8) Note: area of lower half of curve is .5 must be added to value for Z from 0 to 1.8.
b) P(Z>1.8) Note: table value of Z =1.8 is subtracted from .5
c) P(Z<-1.8) Note: answer should be the same as for b).
d) P(-.04<Z<1.4) Note: total area under curve between -.04 and 1.4
e) P(0.6<Z<1.8) Note: total area under curve between .6 and 1.8.
a) P ( Z < 1 , 8 ) = F( 1,8)= 0,9641 b) P ( Z > 1 , 8 ) = 1 P ( Z 1 , 8... View the full answer
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Browse Documents | 332 | 1,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.793112 |
https://rdrr.io/cran/etm/man/plot.etmCIF.html | 1,506,411,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695113.88/warc/CC-MAIN-20170926070351-20170926090351-00326.warc.gz | 720,599,550 | 15,426 | # plot.etmCIF: Plot cumulative incidence functions In etm: Empirical Transition Matrix
## Description
Plot function for `etmCIF` objects. The function plots cumulative incidence curves, possibly with pointwise confidence intervals.
## Usage
```1 2 3 4 5 6``` ```## S3 method for class 'etmCIF' plot(x, which.cif, xlim, ylim, ylab = "Cumulative Incidence", xlab = "Time", col = 1, lty, lwd = 1, ci.type = c("none", "bars", "pointwise"), ci.fun = "cloglog", ci.col = col, ci.lty = 3, legend = TRUE, legend.pos, curvlab, legend.bty = "n", pos.ci = 27, ci.lwd = 3, ...) ```
## Arguments
`x` A `etmCIF` object `which.cif` A numeric vector indicating which CIFs should be plotted. When missing, only the CIF of interest is plotted (determined through the `failcode` argument in `etmCIF`.) `xlim` x-axis limits for the plot. By default, `c(0, max(time))` `ylim` y-axis limits. Default is `c(0, 1)` `ylab` Label for y-axis. Default is `"Cumulative Incidence"` `xlab` Label for x-axis. Default is "Time" `col` Vector describing colours used for the CIF curves. Default is black `lty` Vector of line type `lwd` Thickness of the lines `ci.type` One of ```c("none", "bars", "pointwise")```. `none` plots no confidence interval, `bars` plots the confidence intervals in the form of a segment for one time point, and `pointwise` draws pointwise confidence intervals for the whole follow-up period. `ci.fun` Transformation used for the confidence intervals. Default is "clolog", and is a better choice for cumulative incidences. Other choices are "log" and "log-log" `ci.col` Colour for the pointwise confidence interval curves. Default is same as the CIF curves `ci.lty` Line type for the confidence intervals. Default is 3 `legend` Logical. Whether to draw a legend. Default is `TRUE` `legend.pos` A vector giving the legend's position. See `legend` for further details `curvlab` A character or expression vector to appear in the legend. Default is CIF + event label `legend.bty` Box type for the legend. Default is none ("n") `pos.ci` If `ci.type = "bars"`, vector of integers indicating at which time point to put the confidence interval bars. Default is 27 `ci.lwd` Thickness of the confidence interval segment (for `ci.type = "bars"`) `...` Further graphical arguments
## Details
The function relies on `plot.etm` and `lines.etm` with more or less the same options. Exception is the drawing of the confidence intervals, for which several displays are possible.
## Value
No value returned
## Author(s)
Arthur Allignol [email protected]
`etmCIF`, `plot.etm`, `lines.etm`
``` 1 2 3 4 5 6 7 8 9 10 11``` ```data(abortion) cif.ab <- etmCIF(Surv(entry, exit, cause != 0) ~ group, abortion, etype = cause, failcode = 3) cif.ab plot(cif.ab, ci.type = "bars", pos.ci = 24, col = c(1, 2), lty = 1, curvlab = c("Control", "Exposed")) plot(cif.ab, which = c(1, 2)) ``` | 815 | 2,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-39 | longest | en | 0.717992 |
https://rdoc.info/gems/powerpack/Enumerable | 1,657,068,351,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00187.warc.gz | 524,100,520 | 5,979 | # Module: Enumerable
Defined in:
lib/powerpack/enumerable/sum.rb,
lib/powerpack/enumerable/average.rb,
lib/powerpack/enumerable/exactly.rb,
lib/powerpack/enumerable/several.rb,
lib/powerpack/enumerable/drop_last.rb,
lib/powerpack/enumerable/take_last.rb,
lib/powerpack/enumerable/frequencies.rb,
lib/powerpack/enumerable/drop_last_while.rb,
lib/powerpack/enumerable/take_last_while.rb
## Instance Method Summary collapse
• Calculates the average of a numeric collection.
• Drops the last n elements of an enumerable.
• Drops the last elements of an enumerable meeting a predicate.
• Checks if exactly n elements meet a certain predicate.
• Counts the number of occurrence of items in the enumerable.
• Checks if two or more elements meet a certain predicate.
• Sums up elements of a collection by invoking their `+` method.
• Take the last n elements of an enumerable.
• Take the last n elements of an enumerable meeting a certain predicate.
## Instance Method Details
### #average(default = nil) ⇒ Object
Calculates the average of a numeric collection.
Examples:
``````[1, 2, 3].average #=> 2
[1, 2, 3, 4].average #=> 2.5
[].average #=> nil
[].average(0) #=> 0``````
Parameters:
• default (Object) (defaults to: nil)
an optional default return value if there are no elements. It's nil by default.
Returns:
• The average of the elements or the default value if there are no elements.
``` 15 16 17 18``` ```# File 'lib/powerpack/enumerable/average.rb', line 15 def average(default = nil) coll_size = to_a.size coll_size > 0 ? reduce(&:+) / coll_size.to_f : default end```
### #drop_last(n) ⇒ Array
Drops the last n elements of an enumerable.
Examples:
``````[1, 2, 3].drop_last(1) #=> [1, 2]
[].drop_last(5) #=> []``````
Parameters:
• n (Fixnum)
the number of elements to drop
Returns:
• (Array)
an array containing the remaining elements
``` 11 12 13 14 15 16 17 18``` ```# File 'lib/powerpack/enumerable/drop_last.rb', line 11 def drop_last(n) fail ArgumentError, 'attempt to drop negative size' if n < 0 ary = to_a return [] if n > ary.size ary[0...(ary.size - n)] end```
### #drop_last_while ⇒ Array
Drops the last elements of an enumerable meeting a predicate.
Examples:
``[1, 2, 3].drop_last_while(&:odd?) #=> [1, 2]``
Returns:
• (Array)
an array containing the remaining elements
``` 9 10 11 12 13 14 15 16 17 18 19``` ```# File 'lib/powerpack/enumerable/drop_last_while.rb', line 9 def drop_last_while return to_enum(:drop_last_while) unless block_given? result = [] dropping = true reverse_each do |obj| result.unshift(obj) unless dropping &&= yield(obj) end result end```
### #exactly?(n) ⇒ Boolean
Checks if exactly n elements meet a certain predicate.
Without a block uses the identify of the elements as default predicate. This means that nil and false elements will be ignored.
Examples:
``````[1, 2, 3, 4].exactly?(1) { |n| n > 3 } #=> true
[1, 2, 3, 4].exactly?(2, &:even?) #=> true
[1, 1, 3, 3].exactly?(2, &:even?) #=> false``````
``````[1, false, nil].exactly?(3) #=> false
[1, false, nil].exactly?(1) #=> true
[false, nil].exactly?(0) #=> true
[1, 2, 3].exactly?(3) #=>true``````
Parameters:
• n (Fixnum)
the number of matches required
Returns:
• (Boolean)
true if we get exactly n matches, false otherwise
``` 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41``` ```# File 'lib/powerpack/enumerable/exactly.rb', line 21 def exactly?(n) found_count = 0 if block_given? each do |*o| if yield(*o) found_count += 1 return false if found_count > n end end else each do |o| if o found_count += 1 return false if found_count > n end end end n == found_count end```
### #frequencies ⇒ Hash
Counts the number of occurrence of items in the enumerable.
Examples:
``````[].frequencies # => {}
[1, :symbol, 'string', 3, :symbol, 1].frequencies
#=> { 1 => 2, :symbol => 2, 'string' => 1, 3 => 1 }``````
Returns:
• (Hash)
in the format value => count
``` 13 14 15``` ```# File 'lib/powerpack/enumerable/frequencies.rb', line 13 def frequencies each_with_object(Hash.new(0)) { |e, a| a[e] += 1 } end```
### #several? ⇒ Boolean
Checks if two or more elements meet a certain predicate.
Without a block uses the identify of the elements as default predicate. This means that nil and false elements will be ignored.
Examples:
``````[1, 2, 3, 4].several?(&:even?) #=> true
[1, 1, 3, 3].several?(&:even?) #=> false``````
``````[1, false, nil].several? #=> false
[1, 2, 3].several? #=>true``````
Returns:
• (Boolean)
``` 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35``` ```# File 'lib/powerpack/enumerable/several.rb', line 15 def several? found_count = 0 if block_given? each do |*o| if yield(*o) found_count += 1 return true if found_count > 1 end end else each do |o| if o found_count += 1 return true if found_count > 1 end end end false end```
### #sum(initial = nil) ⇒ Object
Sums up elements of a collection by invoking their `+` method. Most useful for summing up numbers.
Examples:
``````[1, 2, 3].sum #=> 6
[[1], [2], [3]].sum #=> [1, 2, 3]
[].sum #=> 0
["a"].sum #=> "a"
["b", "c"].sum("a") #=> "abc"``````
Parameters:
• initial (Object) (defaults to: nil)
an optional initial value. It defaults to 0 for an empty collection.
Returns:
• The sum of the elements, or the initial value if there are no elements.
``` 17 18 19 20 21 22 23``` ```# File 'lib/powerpack/enumerable/sum.rb', line 17 def sum(initial = nil) if initial reduce(initial, &:+) else reduce(&:+) || 0 end end```
### #take_last(n) ⇒ Array
Take the last n elements of an enumerable.
Examples:
``````[1, 2, 3].take_last(2) #=> [2, 3]
[].take_last(5) #=> []``````
Parameters:
• n (Fixnum)
the number of elements to take
Returns:
• (Array)
an array containing the requested elements
``` 11 12 13 14 15 16 17 18``` ```# File 'lib/powerpack/enumerable/take_last.rb', line 11 def take_last(n) fail ArgumentError, 'attempt to take negative size' if n < 0 ary = to_a return ary if n > ary.size ary[(ary.size - n)..-1] end```
### #take_last_while ⇒ Array
Take the last n elements of an enumerable meeting a certain predicate.
Examples:
``[1, 2, 3, 5].take_last_while(&:odd?) #=> [3, 5]``
Returns:
• (Array)
an array containing the matching elements
``` 9 10 11 12 13 14 15``` ```# File 'lib/powerpack/enumerable/take_last_while.rb', line 9 def take_last_while return to_enum(:take_last_while) unless block_given? result = [] reverse_each { |elem| yield(elem) ? result.unshift(elem) : break } result end``` | 1,985 | 6,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.440057 |
https://forum.arduino.cc/t/arduino-uno-led-control-array-problem/197833 | 1,620,437,292,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00546.warc.gz | 279,549,902 | 5,148 | # Arduino Uno - Led Control - Array Problem
Hallo,
ich habe an einen Uno Rev3, um 19 LEDs zu kontrollieren, 3 74HC595 Schieberegister angeschlossen.
Folgendes Problem ist:
Das Programm läuft stabil für eine Größe der Arrays von 18. Für die eigentliche Anwendung brauch ich aber 234 Werte.
int latchPin = 8; //Shifregister Latch
int dataPin = 11; //Shiftregister Data
int clockPin = 12; //Shiftregister Clock
int pushButton = 7; //Push Button -> Start Sequence
int ledShift [234] = {128,64,32,16,8,4,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,4,
8,16,32,64,128,0,128,64,32,16,8,4,2,1,0,0,0,0,0,0,0,0,0,0,0,128,64,32,16,8,4,2,1,0,0,0,0,0,
0,0,0,0,0,128,192,160,80,40,20,10,5,2,1,0,0,0,0,0,0,0,0,128,64,32,16,8,4,2,1,0,0,0,0,0,0,0,
0,4,8,16,32,64,128,64,32,16,8,4,2,1,0,0,0,0,0,0,36,34,33,32,32,32,32,32,16,8,4,2,1,0,0,0,0,
16,0,16,8,4,2,1,0,0,0,0,24,24,88,72,72,72,72,0,0,4,2,1,0,0,0,128,64,32,16,8,4,2,1,0,0,128,64,
32,16,8,4,2,1,0,128,64,32,16,8,4,2,1,3,5,9,17,33,65,129,1,129,65,33,17,9,5,3,131,67,35,19,11,
7,15,31,127,255,127,63,31,31,15,7,3,1,0,0,0}; //Sequence LED 1-8
int ledShift2 [234] = {0,0,0,0,0,0,0,0,128,64,32,16,8,4,2,1,0,0,0,0,0,1,2,4,8,16,32,64,128,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,128,64,32,16,8,4,2,1,0,0,0,0,0,0,0,0,0,0,0,128,64,32,16,8,
4,2,1,0,0,0,0,0,0,0,0,0,0,128,64,160,80,40,20,10,5,2,1,1,1,1,1,1,1,1,1,1,129,65,33,17,9,5,3,3,
3,3,3,3,3,3,3,3,3,3,3,3,131,67,35,19,11,7,7,7,7,135,71,39,23,15,15,15,15,15,15,143,79,47,31,31,
29,31,31,31,31,31,159,95,63,43,43,41,41,41,45,13,31,31,31,31,31,31,159,95,63,63,63,63,63,63,63,
63,63,191,127,127,127,127,127,127,127,127,127,255,255,255,255,255,255,255,255,255,255,255,255,
255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,255,
255,254,252,248,240,224,192,64,0}; //Sequence LED 9-16
int ledShift3 [234] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,128,64,32,64,128,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,128,64,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,
160,96,96,96,96,96,96,96,96,96,96,96,96,96,96,96,96,96,224,224,224,224,224,224,224,224,224,224,224,
224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,
224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,
224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,
224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,
224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,224,192,128,0,0,0,0,
0,0,0,0,0}; //Sequence LED 17-19
long time [234] = {443,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,
130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,5600,400,130,130,130,130,130,130,130,
130,130,130,130,130,130,130,130,130,130,4460,400,130,130,130,130,130,130,130,130,130,130,130,130,130,130,
130,130,2719,5558,1447,130,130,130,130,130,130,130,130,130,130,130,130,130,130,130,1964,1834,130,130,130,
130,130,130,130,130,130,130,130,130,130,130,1748,1825,130,130,130,130,430,100,110,110,110,110,110,110,110,
110,110,110,110,2587,1831,130,130,130,130,130,130,258,130,130,130,130,130,130,130,130,2580,3668,3710,1835,
130,130,130,130,130,130,4478,1800,1840,640,1170,1820,750,1240,1660,1150,760,670,130,130,130,130,4800,400,
130,130,130,130,130,130,130,130,130,400,130,130,130,130,130,130,130,130,400,130,130,130,130,130,130,1440,200,
200,200,200,200,200,1240,4100,400,130,130,130,130,130,1240,400,130,130,130,130,1550,100,3662,3674,3,250,250,250,
250,250,250,250,250,250,250,5000}; //Delay between LEDs in millis
int value = 0; //shiftout counter
int value2 = 0;
int value3 = 0;
void setup ()
{
pinMode (latchPin, OUTPUT);
pinMode (dataPin, OUTPUT);
pinMode (clockPin, OUTPUT);
pinMode (statusLed, OUTPUT);
pinMode (pushButton, INPUT);
Serial.begin (9600);
}
void loop ()
{
for (int counter=0; counter < 233; counter ++)
{
value = ledShift[counter];
value2 = ledShift2[counter];
value3 = ledShift3[counter];
writeOutput();
delay (time[counter]);
}
}
void writeOutput() //Shiftregister Output
{
digitalWrite (latchPin, LOW); //Latch LOW -> Send Data
shiftOut (dataPin, clockPin, MSBFIRST, value3); //Send Data To Shiftreg 3
shiftOut (dataPin, clockPin, MSBFIRST, value2); //Send Data To Shiftreg 2
shiftOut (dataPin, clockPin, MSBFIRST, value); //Send Data To Shiftreg 1
digitalWrite (latchPin, HIGH); //Latch HIGH -> Display Data
Serial.write ("Ding Dong");
}
setze ich nun für das delay in der loop einen festen Wert ein, dann läuft das Ganze, aber für den “time” Wert nicht. Dann kommt nicht mal was am Serial Monitor raus. ich versteh nicht was genau das Problem ist :~ habt ihr da ne Idee?
Auf dem UNO ist hierfür bei weitem nicht genug Speicher. Der hat nur 2kB RAM! Ein int benötigt 2 Byte. Ein long hat 4 Byte!
Du musst den Speicher optimieren. Wenn du nur Werte von 0-255 brauchst, byte statt int verwenden. Und generell für alle Arrays PROGMEM verwenden: http://arduino.cc/en/Reference/PROGMEM Dann landet das alles im Flash.
Time muss hier auch gar nicht long sein. In einen unsigned int gehen Werte bis 65535
Juhuuu :) es läuft:) danke schön:) | 2,573 | 5,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-21 | latest | en | 0.543687 |
https://gitlab.utc.fr/meurouth/nf26-project/-/commit/f0029c0c97539f7cc76c375b5acaf5181b2deb33?view=inline | 1,656,650,946,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00155.warc.gz | 340,266,141 | 34,447 | Commit f0029c0c by Unknown
### Little change
parent a929da85
... ... @@ -2,167 +2,9 @@ import numpy as np import matplotlib.pyplot as plt from functools import reduce from database_pre2 import connection import matplotlib.pyplot as plt import re import folium def add (x,y): return x+y def abs_diff(x,y): return abs(x-y) def diff(x,y): return x-y #caculate mean reduce #input [count,mean] def reduceFonction (x,y): result = [] for i in range(2): result.append(reduce(add,[x[i],y[i]])) return result #input [valeur] -> [count,mean] def mapFonction1 (x): return [1,x] #input [count,mean] -> [mean] def mapFonction2 (x): return x[1]/x[0] def testNan (x): test = x != x return test def mapReduce_kmeans(data,targetNB): results = dict() for row in data.result(): data_target = row[targetNB] if testNan(data_target): continue data_espace = (row[1],row[2],row[3]) if results.get(data_espace) is None: results[data_espace] = mapFonction1(data_target) else: mapresult = mapFonction1(data_target) results[data_espace] = reduceFonction(mapresult,results[data_espace]) for eachEspace in results: results[eachEspace] = mapFonction2(results[eachEspace]) return results def cluster_nb_diff(centre_new,centre): sum = 0 for i in range(3): sum += abs(centre_new[i][0]-centre[i][0]) return sum/3 #input [tmpt] -> [tmpt,tmpt,tmpt,tmpt] def map1_kmeans(x): return [x,x,x,x] def mapCentre(x): return [x[0],x[1],x[2],0] #input [tmpt,tmpt,tmpt,tmpt] and [c1,c2,c3,0] -> [|tmpt - c1|,|tmpt - c2|,|tmpt - c3|,tmpt] def reduceKmeans (x,y): result = [] for i in range(4): result.append(reduce(abs_diff,[x[i],y[i]])) return result #input [|tmpt - c1|,|tmpt - c2|,|tmpt - c3|,tmpt] -> [cluster number, min(|tmpt - c|), tmpt] def map2_kmeans(x): min_value = 10000000000000 index = 0 for each in range(3): if min_value > x[each]: min_value = x[each] index = each return [index,min_value,x[3]] def MapnewCentre(x): return x[1]/x[0] def kmeans (data,targetNB): #3centre with [point count, temprature centre] centre = {0:[1,0],1:[1,0],2:[1,0]} #cluster est pour stocler lat, lon de chaque point de chaque cluster cluster = [[],[],[]] result = mapReduce_kmeans(data,targetNB) #mettre le premier 3 point comme le centres init init_point_values = [result[i] for i in result.keys()][:3] init_point_keys = [i for i in result.keys()][:3] for key in centre.keys(): centre[key] = [1,init_point_values[key]] cluster[key].append(init_point_keys[key]) #init the centre new and result new for mapreduce centre_new = {0:[0,0],1:[0,0],2:[0,0]} result_new = dict() #When the number of point of cluster don't change,stop while True: for eachkey in result: if eachkey in cluster[0] or eachkey in cluster[1] or eachkey in cluster[2]: continue #caculate the distance between the data of this lingne and the centre #Map1_kemeans result_new[eachkey] = map1_kmeans(result[eachkey]) centre_values = [] for each in centre: centre_values.append(centre[each][1]) centre_values = mapCentre(centre_values) #Reduce result_new[eachkey] = reduceKmeans(result_new[eachkey],centre_values) #Map2_kmeans result_new[eachkey] = map2_kmeans(result_new[eachkey]) #Put all the distance and points into the clusters #Result format [cluster number, min(|tmpt - c|),tmpt - c] for eachpoint in result_new: clusterNB = result_new[eachpoint][0] centre_new[clusterNB][0] += 1 centre_new[clusterNB][1] += result_new[eachpoint][2] cluster[clusterNB].append(eachpoint) #compare centre_new and centre, if if not cluster_nb_diff(centre_new,centre) > 1: break else: #caculate the new centre print ('jasdlkjalsdkjalskd ',cluster_nb_diff(centre_new,centre)) for eachculster in centre_new: centre_new[eachculster][1] = MapnewCentre(centre_new[eachculster]) centre = centre_new centre_new = {0:[0,0],1:[0,0],2:[0,0]} result_new = dict() cluster = [[],[],[]] createMap(cluster) def createMap(data): mean_lat = 0 ... ... @@ -179,27 +21,30 @@ def createMap(data): mean_lat = mean_lat/count mean_lon = mean_lon/count m = folium.Map(location=[mean_lon,mean_lat], zoom_start=6) m = folium.Map(location=[mean_lon,mean_lat],zoom_start=6) color = {0:'blue',1:'red',2:'green'} # Attributes names which will be displayed on the map attributes = ["alti", "drct", "dwpf", "feel", "gust", "ice_accretion_1hr", "ice_accretion_3hr", "ice_accretion_6hr", "metar", "mslp", "p01i", "peak_wind_drct", "peak_wind_gust", "peak_wind_time", "relh", "sknt", "skyc1", "skyc2", "skyc3", "skyc4", "skyl1", "skyl2", "skyl3", "skyl4", "tmpf", "vsby", "wxcodes"] for each in data.result(): # print(each) # Here we choose not to display the "nan" values and the METAR ID l = [attributes[i] + ":" + str(each[i + 4]) for i in range(len(attributes)) if str(each[i + 4]) != 'nan' and attributes[i] != "metar"] string='\n'.join(l) folium.Marker([each[2],each[1]], popup=string, icon=folium.Icon(color='red')).add_to(m) popup=string, icon=folium.Icon(color='red')).add_to(m) m.save("Projet-NF26/map.html") if __name__ == "__main__": session = connection() timestamp = '2017-12-02 00:30:00' # Timestamp user wants to search date = input("Please enter the date you want to search (format 'yyyy-MM-dd'):") time = input("Please enter the time you want to search (format 'hh:MM:ss'):") #timestamp = '2017-12-02 00:30:00' timestamp = date + ' ' + time data = session.execute_async("select * from meurouth_cql.database_time where date = '%s' ALLOW FILTERING"%(timestamp)) createMap(data)
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Finish editing this message first! | 1,717 | 5,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.382753 |
https://www.thedailycafe.com/discussion-board/adding-bit-fun | 1,723,171,212,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00189.warc.gz | 796,291,533 | 28,121 | The April 3 tip of the week is about adding a bit of fun to our time in the classroom. We want to hear how you add a bit of fun to your day. Let’s learn from each other!
03.12.2015
9y
5
2.7k
3
10y
Everyone in first grade has a job to do at the end of the day. I turn on some peppy, fun music, and as chores are finished, everyone meets in the gathering space for a dance party. Students get their chores done quickly so they can participate. It ends the day with fun, laughter, and a little exercise.
9y
Kim Jorgensen
We do something similar for transitions. The students have selected a favourite Go Noodle clip - Pop Se Ko. I start the clip minimised so we can only hear the music. When this begins, we start packing up - singing as we go - ready to gather on the carpet for our next mini-lesson. Once everyone is ready, I maximise the clip and we all finish dancing together. The quicker we pack up, the more dancing we get!
9y
That sounds like fun! I’m going to try it!
9y
Michelle Jenkins
There are several fun things we do that come to mind in first grade:
• Playing various Dr. Jean’s songs during brain breaks and mini lessons to enhance learning skills and promote activity and motion with learning
• On April Fool’s Day I asked the students if they wanted to have a “brownie” and they soon found out I had offered a “brown E” (a di-cut of the letter E that is brown)! The next day I offered them a true brownie
• Cooking and baking together is always fun. Blends math with language arts. One time I had parents bring in food to cook from their country of origin. This was a great success!
• Last week we took our math test on Time on the outside steps with our clipboards and pencils.
• Play all sorts of chimes, maracas, bells, etc, as consistent messages throughout the day to mark transition times.
• We created kites with fractions on them and wrote about the fractions we made. We then got to fly our own kites outside for a 5 minute break!
• Playing classical music from all over the world during Writer’s Workshop time.
• For our poem of the week, one day every week we read the poem using silly voices of our choice (slow motion, country style, British style, super fast, monster style, baby, etc). This is a big hit!
These are just a few ideas!
9y
You are a master at fun! They are going to remember that brownie joke forever!
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Resources for At-Home Learning Miscellaneous 6 1.9k 3 years 11 months | 709 | 2,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.955641 |
https://www.mvtec.com/doc/halcon/1911/en/difference.html | 1,695,623,528,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00674.warc.gz | 1,001,222,106 | 6,050 | # difference (Operator)
## Name
`difference` — Calculate the difference of two regions.
## Signature
`difference(Region, Sub : RegionDifference : : )`
## Description
`difference` calculates the set-theoretic difference of two regions: (Regions in Region) - (Regions in Sub) The resulting region is defined as the input region (`Region`) with all points from `Sub` removed. Note that, internally, all regions of `Sub` are united to a single region before the differences between the individual regions of `Region` and the united region are calculated.
## Attention
Empty regions are valid for both parameters. On output, empty regions may result. The value of the system flag 'store_empty_region' determines the behavior in this case.
## Execution Information
• Multithreading type: reentrant (runs in parallel with non-exclusive operators).
• Processed without parallelization.
## Parameters
`Region` (input_object) region(-array) `→` object
Regions to be processed.
`Sub` (input_object) region(-array) `→` object
The union of these regions is subtracted from Region.
`RegionDifference` (output_object) region(-array) `→` object
Resulting region.
## Example (HDevelop)
```* provides the region X without the points in Y
difference(X,Y,RegionDifference)
```
## Complexity
Let N be the number of regions, F_1 be their average area, and F_2 be the total area of all regions in `Sub`. Then the runtime complexity is O(F_1 * log(F_1) + N * (sqrt(F_1) + sqrt(F_2))).
## Result
`difference` always returns the value 2 (H_MSG_TRUE). The behavior in case of empty input (no regions given) can be set via `set_system('no_object_result',<Result>)` and the behavior in case of an empty input region via `set_system('empty_region_result',<Result>)`. If necessary, an exception is raised.
## Possible Predecessors
`threshold`, `connection`, `regiongrowing`, `pouring`, `class_ndim_norm`
## Possible Successors
`select_shape`, `disp_region`
`intersection`, `union1`, `union2`, `complement`, `symm_difference` | 491 | 2,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.649185 |
https://mathworld.wolfram.com/TotalDominationNumber.html | 1,701,555,924,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00078.warc.gz | 455,543,629 | 25,968 | TOPICS
# Total Domination Number
The total domination number of a graph is the size of a smallest total dominating set, where a total dominating set is a set of vertices of the graph such that all vertices (including those in the set itself) have a neighbor in the set. Total dominating numbers are defined only for graphs having no isolated vertex (plus the trivial case of the singleton graph ).
For example, in the Petersen graph illustrated above, since the set is a minimum dominating set (left figure), while since is a minimum total dominating set (right figure).
For any simple graph with no isolated points, the total domination number and ordinary domination number satisfy
(1)
(Henning and Yeo 2013, p. 17). In addition, if is a bipartite graph, then
(2)
(Azarija et al. 2017), where denotes the graph Cartesian product.
For a connected graph with vertex count ,
(3)
(Cockayne et al. 1980, Henning and Yeo 2013, p. 11).
Dominating Set, Domination Number
## Explore with Wolfram|Alpha
More things to try:
## References
Azarija, J.; Henning, M. A.; and Klavžar, S. "(Total) Domination in Prisms." Electron. J. Combin. 24, No. 1, paper 1.19, 2017. http://www.combinatorics.org/ojs/index.php/eljc/article/view/v24i1p19.Cockayne, E. J., Dawes, R. M., and Hedetniemi, S. T. "Total Domination in Graphs." Networks 10, 211-219, 1980.Henning, M. A. and Yeo, A. Total Domination in Graphs. New York: Springer, 2013.
## Cite this as:
Weisstein, Eric W. "Total Domination Number." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/TotalDominationNumber.html | 442 | 1,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | longest | en | 0.72901 |
https://dev.betterlesson.com/lesson/522186/mentally-speaking?from=mtp_lesson | 1,607,161,177,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00320.warc.gz | 268,775,825 | 23,748 | # Mentally Speaking!
38 teachers like this lesson
Print Lesson
## Objective
Students will use base ten blocks to enhance their conceptual understanding of addition and subtraction procedures.
#### Big Idea
Using base ten blocks and place value, students explore how to become more fluent in adding and subtracting two-digit numbers without regrouping.
## Opening Questions!
15 minutes
I am going to write a two-digit addition problem on the board (56 + 98). I want you guys to align the numbers so that the tens and ones are in the same column. Now, turn and tell your neighbor how you aligned your numbers and why. Make sure to use place value when you are telling your neighbor.
As students are discussing, I circulate and ask students questions. What number is in the ones place? What number is in the tens place? How do you know? Which number is greater? What number do you place on top when aligning your numbers?
MP8- Look for repeated reasoning
After students have talked with their partner for 3 minutes or so, I ask student volunteers to stand up and share their answers. As students share I make sure to high-five students who are using place value vocabulary to discuss their numbers. Some students respond by saying, "I know 98 is larger than 56 because one number has 9 tens and the other number only has 5 tens." Another student points out how he aligns his numbers by labeling the value above each number. "I place a T over the 9 and an O over the 8, and I place a T over the 5 and an O over the 6."
I repeat this task using a two-digit subtraction problem without regrouping.
Example:
98 - 27 = ?
Solution:
Subtract the digits in the ones column. 8 - 7 = 1. Now, we need to subtract the digits in the tens column.
90 - 20 = 70. So, 98 - 27 = 71
I encourage students to use their place value mats and base-ten materials to explain their answers.
## Applying the Knowledge!
10 minutes
In this portion of the lesson I want my students to spend some time working with this concept on their own. I ask students to move to their assigned partner. I give them a set of base ten blocks, a place value chart, and four two-digit addition and subtraction problems. I hope that my students will use this exploration activity to discover place value concepts and learn addition and subtraction. The blocks can be set to represent ones, tens, and hundreds. Hopefully it will help students visualize the numbers with which they are working.
MP2 - Reason abstractly and quantitatively.
Materials: Place Value Mat, Base-Tens Cut-Outs
As students are working, I circulate the room to check for understanding. For instance, some students organize their base-tens where the numbers in the ones and tens place align. Some students organize one place at a time.
I ask, why did you decide to organize the problem like that? One student explains that he already knows how to align his numbers, so he decided to represent numbers first to show he understood the task. So, I decide to probe him a bit more. Is there a more efficient strategy? He explains it would be easier for him to go ahead and add. However, he did say working with the base ten blocks helps him become more efficient.
On the other hand, the student who represented his work problem by problem needed the visual representation to support his learning. What did you notice? Do you think this may work with other numbers? How can you be sure? He hesitates for a moment, but realizes that the strategy he is using, to add or subtract the digits in each place, can be used to help him solve the other three problems.
After about ten minutes or so, I ask students to turn and discuss their work with their partner. I want to see what my students are thinking, or if they need additional support. Most importantly I want my student talking and speaking mathematically to learn how to communicate better when they are explaining their own problem solving techniques.
## Testing the Skills
15 minutes
The purpose of this lesson was to allow my students an opportunity to work with base ten blocks and place value concepts to help them develop mental imagery when adding and subtracting numbers.
In this activity, I want to check students' growth. To do this I invite students to the carpet. I ask a couple of probing questions. How did base-tens help you become better at adding and subtracting two-digit numbers? Some students say they normally get confused when problems are written as number sentences. Applying a visual helped them align their numbers correctly according to place value.
I write 76, 21, and 11 on the board. I ask: What digits are in the ones and tens places in each of these numbers?
Example:
In the number 76, 7 is in the tens place and 6 is in the ones place. I go on to ask students to represent the numbers using base-tens. I do this to determine if students know the value of each digit, and not just its place. I continue until we have discussed all three numbers.
I ask students what will happen if we add two of the numbers together. I ask students to choose two numbers. They decide to add 76 + 11. I write the problem in number sentence form to see if my students can align the numbers correctly according to their place value. I encourage students to use base ten blocks.
MP8 - Look for and express regularity in repeated reasoning. This practice helps students to reason mathematically.
Additional questions to support reasoning skills:
1. How would you prove that?
2. What assumptions are you making? | 1,190 | 5,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2020-50 | latest | en | 0.930609 |
https://www.women.com/1230821/best-common-sense-questions-and-answers/ | 1,721,682,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00467.warc.gz | 920,688,593 | 27,486 | # 51 Of The Best Common Sense Questions And Their Answers
Common sense riddles will stump even the smartest person but they are so fun to twist your mind around. The following 51 best common sense brain teasers are great one-liners at a cocktail party or making even the grumpiest frump smile. Let us know which brain teaser stumped you in the comments below!
## 1. A farmer had 752 sheep and took one shot that got them all. How did he do it?
Answer: He took a panoramic view.
## 2. How many times you can subtract the number 5 from 25?
Answer: Once. After the first calculation, you will be subtracting 5 from 20, then 5 from 15, and so on.
## 3. What is the easiest way to throw a ball, have it stop, and completely reverse direction after traveling a short distance?
Answer: Throw the ball straight up.
## 4. There are five oranges in a basket. How will distribute the oranges to five people, such that one orange is still in the basket?
Answer: Distribute 1 orange each to the four people. And then give the fifth orange together with the basket to the last person.
## 5. A man buys a new car and goes home to tell his girlfriend. He goes the wrong way up a one way street, bumps into seven people, goes on the pavement and takes a shortcut through a public park. He is seen by a policeman but is not arrested. Why?
Answer: He is walking.
## 6. A guy is condemned to death. He has to choose a room. Room #1: A fiery inferno. Room #2: 50 Assassins with loaded guns. Room #3: A room full of hungry lions that didn't eat in 3 months. Which room is the safest?
Answer: Room #3 because the lions would be dead if they didn't eat in 3 months.
## 8. Some months have 30 days, and some have 31. How many have 28?
Answer: All of them.
## 10. Two fathers and two sons went fishing one day. They were there the whole day and only caught 3 fish. One father said, that is enough for all of us, we will have one each. How can this be possible?
Answer: There was the father, his son, and his son's son. This equals 2 fathers and 2 sons for a total of 3!
## 11. The peacock is a bird that does not lay eggs. How do they get baby peacocks?
Answer: The Peahen lays eggs!
## 13. A little girl kicks a soccer ball. It goes 10 feet and comes back to her. How is this possible?
Answer: Ever heard of gravity? She kicked it up.
## 14. A 10 foot rope ladder hangs over the side of a boat with the bottom rung on the surface of the water. The rungs are one foot apart, and the tide goes up at the rate of 6 inches per hour. How long will it be until three rungs are covered?
Answer: Never. The boat rises as the tide goes up.
## 15. A is the father of B. But B is not the son of A. How's that possible?
Answer: B is the daughter you!
## 16. A man dressed in all black is walking down a country lane. Suddenly, a large black car without any lights on comes round the corner and screeches to a halt. How did the car know he was there?
Answer: It was day time.
## 17. A rooster laid an egg on top of the barn roof. Which way did it roll?
Answer: It didn't roll – since when did roosters start laying eggs?
## 18. A truck driver is going down a one way street the wrong way, and passes at least ten cops. Why is he not caught?
Answer: Because he was not driving! He's walking on the sidewalk.
## 19. An electric train is moving north at 100mph and a wind is blowing to the west at 10mph. Which way does the smoke blow?
Answer: There is no smoke with an electric train.
## 20. How can a man go eight days without sleep?
Answer: By sleeping during the night time
## 21. How can you drop a raw egg onto a concrete floor without cracking it?
Answer: The Egg won't crack the concrete floor!
## 22. How can you lift an elephant with one hand?
Answer: It is not a problem, since you will never find an elephant with one hand.
## 23. How much dirt is there in a hole 3 feet deep, 6 ft long and 4 ft wide?
Answer: None, or else it wouldn't be a hole.
## 24. If a doctor gives you 3 pills and tells you to take one pill every half hour, how long would it take before all the pills had been taken?
Answer: 1 hour! Take the 1st pill right away, half an hour later take the 2nd and half an hour after that the 3rd. Total time spent: 1 hour!
## 25. If it took eight men ten hours to build a wall, how long would it take four men to build it?
Answer: No time at all it is already built.
## 26. If Mr Smith's peacock lays an egg in Mr Jones' yard, who owns the egg?
Answer: Peacocks don't lay eggs, just peahens.
## 27. If there are 6 apples and you take away 4, how many do you have?
Answer: The 4 you took.
## 29. Is it legal for a man to marry his widow's sister?
Answer: No, but since he's dead it would be kind of difficult.
## 30. Some months have 31 days, others have 30 days. How many have 28 days?
Answer: All months have 28 days.
## 31. Larry's father has five sons named Ten, Twenty, Thirty, Forty...Guess what would be the name of the fifth?
Answer: Larry! He would be the fifth son.
## 32. There was an airplane crash, every single person on board died, but yet two people survived. How is this possible?
Answer: The two were married.
## 34. How far can you walk into the woods?
Answer: Half way. After that you are walking out of the woods.
## 35. If you throw a red stone into the blue sea what it will become?
Answer: It will become Wet.
## 38. Name the most recent year in which New Year's came before Christmas.
Answer: This year. New Year's always comes before Christmas of the same year.
## 40. Which is heavier, 100 pounds of rocks or 100 pounds of feathers?
Answer: They both weigh the same – 100 pounds.
## 42. If there are 12 fish and half of them drown, how many are there?
Answer: 12, fish don't drown!
## 43. How many times can you subtract 10 from 100?
Answer: Once. Next time you would be subtracting 10 from 90.
## 45. Before the Mount Everest was discovered, what was the highest mountain in the world?
Answer: It was still Mount Everest
## 47. Do they have a 4th of July in England?
Answer: Yes, just like they have a 3rd of July and a 5th of July.
## 48. How many birthdays does the average man have?
Answer: Just one – his actual date of birth.
## 49. If you were to put a coin into an empty bottle and then insert a cork into the neck, how could you remove the coin without taking out the cork or breaking the bottle?
Answer: Simply push the cork into the bottle and shake the coin out. | 1,674 | 6,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-30 | latest | en | 0.97503 |
http://maths.crusoecollege.vic.edu.au/s1t8h/ | 1,685,920,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00256.warc.gz | 33,347,718 | 7,870 | # Semester 2 Topic 1 Level H
Task Name : Number stories Task Level : H (Year 7) Semester : 1 Topic : Introducing Algebra VC Strand : Patterns and Algebra Web Address : http://maths.crusoecollege.vic.edu.au/paH Equipment Needed : nil Victorian Curriculum outcome : Introduce the concept of variables as a way of representing numbers using letters. Extend and apply the laws and properties of arithmetic to algebraic terms and expressions. Task description : Students create eight number stories, two involving addition, two involving subtraction, two involving multiplication and two involving division. Four of these stories must involve the use of brackets. For each story, the student must write the story, draw a picture and write a number sentence with a pronumeral representing the unknown number. They must then solve each story using inverse operations (you will need to explain what this is). Assessment options : Photographs of drawings; digital equivalents using Paint or similar. Teacher notes : Students can really struggle conceptually with this task: many don’t seem to understand the difference where the unknown is at the end of the story compared. For students lacking imagination, provide a scaffold to help them create a story: the video provides models which can be easily adapted to involve addition, subtraction, multiplication and division. Be careful with the subtraction stories: if the student has the unknown at the beginning of the story, things can get confusing. Extension task: Solve equations by balancing (see-saw analogy interactive) | 306 | 1,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.889014 |
https://socratic.org/questions/how-do-you-graph-y-2-by-plotting-points-4 | 1,695,627,442,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00792.warc.gz | 579,217,260 | 5,795 | # How do you graph y= -2 by plotting points?
y = 0 ____
y = -1
___
y = -2
____
#### Explanation:
Just plot $\left(0 , - 2\right) , \left(1 , - 2\right) , \left(2 , - 2\right)$, etc.
and you will see that it's a straight line parallel to x-axis. | 92 | 247 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.76225 |
http://dharmath.blogspot.com/2014/02/random-walk-sequences.html | 1,531,876,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589980.6/warc/CC-MAIN-20180718002426-20180718022426-00088.warc.gz | 101,507,305 | 15,423 | ## Sunday, February 2, 2014
### Random walk sequences
Let $n$ be a natural number. Suppose $a_1, a_2, \dots$ are sequences of +1s and -1s. Also define $s_k = a_1 + a_2 + \dots + a_k$, and define $s_0 = 0$. Such sequence is called "complete" if for each $i \in {0, \dots, n-1}$ there exists $k$ such that $s_k \equiv i \mod n$.
What is the largest number $S$ such that the following statement is true: For every complete sequence, there exists $j,k \geq 0$ such that $|s_j - s_k| = S$.
#### Answer:
$n-1$
We can construct $a_i$ as follows: $n-1$ +1s, followed by $n-1$ -1s, and followed by $n-1$ +1s again, alternatingly. It's easy to see that this sequence is complete. Moreover, the sum of any contiguous block ranges from $-(n-1)$ to $(n-1)$. So clearly for this particular sequence, $S \leq n-1$ fulfills the statement, and $S > n-1$ is not satisfied by this sequence.
Now we have to prove that for $S=n-1$ and any given complete sequence, the problem statement is fulfilled.
#### First Solution
The sequence is infinite, but suppose we can truncate it to the smallest sequence so that it's still complete. In other words, suppose the sequence $a_1, \dots, a_N$ is complete but $a_1, \dots, a_{N-1}$ is not. Let $s_N = x$. Because $N$ is the smallest index such that it's still complete, that means $s_i \equiv x \mod n$ does not happen before $N$. Consider $a_N$. It can be either +1 or -1. Suppose it's +1 (the case of -1 is similar). So that means $s_{N-1} = x-1$.
Let $p$ be the an index such that $s_p \equiv x+1 \mod n$. We know $p$ exists because the sequence is still complete. Also that $p < N-1$. That means $s_{N-1} - s_p \equiv (x-1) - (x+1) \equiv -2 \mod n$. But from $i = p$ to $i = N-1$ we can't have $s_i = x$, so that means $s_{N-1} - s_p = tn - 2$ for some $t > 0$. This means that $s_N - s_p = tn-1$.
But if $t > 1$, then we must have some $r$ such that $p < r < N$ and $s_N - s_r = n, s_r - s_p = (t-1)n-1$. But this yields a contradiction, because then $s_r \equiv s_N \equiv x \mod n$ and thus $a_1, \dot, a_r$ is complete, violating our condition that $N$ is the smallest complete sequence. Thus we must have that $t=1$ and consequently $s_N - s_p = n-1$
#### Second Solution
Suppose that there is no contiguous block whose sum is $\pm(n-1)$. Each contiguous block must have sum of at most $(n-2)$ and at least $-(n-2)$. Now, if there is $k$ such that $s_k = n-1$ then we are done, because it would imply a contiguous block from $0$ to $k$ with sum $n-1$. So let $x$ be the maximum value of $s_i$, where $x \leq n-2$. Likewise, let $y$ be the minimum value of $s_i$ where $y \geq -(n-2)$.
Because the sequence is complete, then we must have that $x-y \geq n-1$, otherwise some values won't be "reached" by $s_i$ at all. But this means that there is a contiguous block from the maximum to the minimum or vice versa, and their sum is $\pm(n-1)$ | 943 | 2,882 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-30 | latest | en | 0.909374 |
http://stackoverflow.com/questions/11144666/why-is-number-0-and-number-nan-in-javascript/11144726 | 1,462,576,207,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461864121714.34/warc/CC-MAIN-20160428172201-00042-ip-10-239-7-51.ec2.internal.warc.gz | 269,091,804 | 21,352 | Why is Number([]) === 0 and Number({}) === NaN in Javascript?
I was looking at the first table on http://zero.milosz.ca/, and wanted to understand why, for example, `0 == []` and `0 != {}`. I'm assuming it's because `Number([]) == 0` and `Number({}) == NaN`. However, that part seems arbitrary. Why is an empty list `0` and empty object a `NaN`?
-
Arrays are weird like that. Like `[[[[[[[[123]]]]]]]] == 123`. – Niet the Dark Absol Jun 21 '12 at 18:42
@Kolink ...but what causes this to work? (What rule of `==`, not found in `===`, is being applied?) – user166390 Jun 21 '12 at 18:44
@pst: The strict comparison returns false if the operands are not of the same type (as you probably know). But `==` will convert both operands to numbers in this case. – Felix Kling Jun 21 '12 at 18:56
Why does the title refer to `===` but the body use `==`? Which one(s) is the question actually asking about? – Lawrence Johnston Jun 22 '12 at 18:24
Using `Number(some_object)` will use the string representation of the given object. For your examples the string representations are:
``````js> ({}).toString();
[object Object]
js> [].toString();
js>
``````
The string `'[object Object]'` cannot be converted to a number but the empty string `''` can.
-
But why is the empty string `''` converted to 0? – tskuzzy Jun 21 '12 at 18:47
It is pretty common that falsy values are converted to 0 when used in a numeric context. Empty objects are not falsy by the way. – ThiefMaster Jun 21 '12 at 18:48
@tskuzzy: Because it is defined this way: "The MV of StringNumericLiteral ::: [empty] is 0.". See es5.github.com/#x9.3.1 – Felix Kling Jun 21 '12 at 18:49
To elaborate a bit on ThiefMaster's answer, I've taken a look into ECMAScript's specifications:
When converting a string into a number, a grammar is used for the conversion. In particular, the mathematical value of `StringNumericLiteral ::: [empty]` is defined as 0. In fact, it's 0 for any whitespace.
-
For those who are too lazy to search for this in the PDF, es5.github.com/#x9.3.1 is the relevant section of the spec. – ThiefMaster Jun 21 '12 at 18:52
Thanks, I updated the link with the official HTML version. – tskuzzy Jun 21 '12 at 18:52
When one value is an object ([],{}) and the other is a number or string, operator == converts the object to a primitive value (a number in this case) using the built-in conversion methods which all objects in Javascript inherit: toString() and valueOf().
For generic objects like {}, valueOf is used, and by default it returns the object itself, which is != 0.
For built-in arrays, toString is used. This method applied to an array returns a string containing all the elements joined by commas. For the empty array, it returns an empty string, ''.
Then the interpreter applies valueOf to that string; the return value of this method for an empty string is 0, so [] == 0.
- | 782 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | latest | en | 0.848787 |
https://qanda.typicalstudent.org/category/mathematics?page=11&%3Bper-page=50&per-page=50 | 1,563,617,249,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526506.44/warc/CC-MAIN-20190720091347-20190720113347-00549.warc.gz | 523,335,094 | 27,115 | Hot Student Stories
Mathematics
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Which shows the expressions in the order they would appear on a number line from least to greatest? a.17 over 9, square root of 6, square root of 15, square root of 30, 3 to the power of 3 b.3 to the power of 3, square root of 30, square root of 15, square root of 6, 17 over 9 c.3 to the power of 3, square root of 6, square root of 15, 17 over 9, square root of 30 d.17 over 9, square root of 30, square root of 15, 3 to the power of 3, square root of 6
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Which of the following expressions is the inverse of the function y=3x+4
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1.Is the simplified form of 2square root of 3 • 2square root of 6 rational? Yes No 2.Is the simplified form of 2square root of 3 + 3square root of 3 rational? Yes No
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Zach Chandler
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followin | 789 | 2,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-30 | longest | en | 0.846999 |
https://mom6.readthedocs.io/en/main/api/generated/pages/Discrete_PG.html | 1,718,908,288,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00054.warc.gz | 343,595,523 | 7,773 | Following [3], the horizontal momentum equation in the general coordinate $$r$$ can be written as:
$\frac{\partial \vec{u}}{\partial t} + \nabla_r \Phi + \alpha \nabla_r p = \cal{F}$
where the vector $$\cal{F}$$ represents all the forcing terms other than the pressure gradient. Here, $$\vec{u}$$ is the horizontal component of the velocity, $$\Phi$$ is the geopotential:
$\Phi = gz$
$$\alpha = 1/\rho$$ is the specific volume and $$p$$ is the pressure. The gradient operator is a gradient along the coordinate surface $$r$$.
MOM6 offers two options, an older one using a Montgomery potential as described in [26] and [62]. However, it can have the instability described in [29]. The version described here is that in [3] and is the recommended option (ANALYTIC_FV_PGF = True). The paper describes the Boussinesq form while the code supports that and also a non-Boussinesq form.
In two dimensions ( $$x$$ and $$p$$), we can integrate the zonal component of the momentum equation above over a finite volume:
(8)$\begin{split}\begin{eqnarray} - \int dx \int dp \frac{\partial u}{\partial t} &= \int dx \int dp \left. \frac{\partial \Phi}{\partial x}\right|_p \\ &= \int_{p_{br}}^{p_{tr}} \Phi dp + \int_{p_{tr}}^{p_{tl}} \Phi dp + \int_{p_{tl}}^{p_{bl}} \Phi dp &+ \int_{p_{bl}}^{p_{br}} \Phi dp \label{eq:PG_loop} \end{eqnarray}\end{split}$
We convert to line integrals thanks to the Leibniz rule. See the figure for the location of the line integral ranges:
The only approximations made are (i) that the potential temperature $$\theta$$ and the salinity $$s$$ can be represented continuously in the vertical within each layer although discontinuities between layers are allowed and (ii) that $$\theta$$ and $$s$$ can be represented continuously along each layer. MOM6 has options for piecewise constant (PCM), piecewise linear (PLM), and piecewise parabolic (PPM) in the vertical.
If we use the Wright equation of state ([70]), we can integrate the above integrals analytically. This equation of state can be written as:
$\alpha(s, \theta, p) = A(s, \theta) + \frac{\lambda(s, \theta)}{P(s, \theta) + p}$
where $$A, \lambda$$ and $$P$$ are functions only of $$s$$ and $$\theta$$. The integral form of hydrostatic balance is:
$\Phi(p_t) - \Phi(p_b) = \int_{p_t}^{p_b} \alpha(s, \theta, p) dp$
Assuming piecewise constant values for $$\theta$$ and $$s$$ and the above equation of state, we get:
(9)$\begin{split}\begin{eqnarray} \Phi(p_t) - \Phi(p_b) &= \int_{p_t}^{p_b} \alpha(s, \theta, p) dp \\ &= (p_b - p_t) A + \lambda \ln \left| \frac{P + p_b}{P + p_t} \right| \\ &= \Delta p \left( A + \frac{\lambda}{P + \overline{p}} \frac{1}{2 \epsilon} \ln \left| \frac{1 + \epsilon}{1 - \epsilon} \right| \right) \label{eq:PG_vert} \end{eqnarray}\end{split}$
which is the exact solution for the continuum only if $$\theta$$ and $$s$$ are uniform in the interval $$p_t$$ to $$p_b$$. Here, we have introduced the variables:
$\Delta p = p_b - p_t$
$\overline{p} = \frac{1}{2}(p_t + p_b)$
and
$\epsilon = \frac{\Delta p}{2 (P + \overline{p})}$
We will show later that $$\epsilon \ll 1$$. Note the series expansion:
$\frac{1}{2 \epsilon} \ln \left| \frac{1 + \epsilon}{1 - \epsilon} \right| = \sum_{n=1}^\infty \frac{\epsilon^{2n-2}}{2n - 1} = 1 + \frac{\epsilon^2}{3} + \frac{\epsilon^4}{5} + \cdots \forall |\epsilon | \leq 1$
Typical values for the deep ocean with 100 m layer thickness are $$6 \times 10^8$$ Pa and $$10^6$$ Pa, respectively, yielding $$\epsilon \sim 8 \times 10^{-4}$$ and a corresponding accuracy in the geopotential height calculation of $$\frac{\lambda \epsilon^3}{g} \sim 10^{-5}$$ m. For this value of $$\epsilon$$, the series converges with just three terms. In MOM6, we use series rather than the intrinsic log function , since the log is machine dependent and insufficiently accurate. In extreme circumstances, $$\Delta p \sim 6 \times 10^7$$ Pa (limited by the depth of the ocean) for which $$\epsilon \sim 0.04$$ with geopotential height errors of order 1 m. In this case, the series converges to machine precision with six terms.
The finite volume acceleration is expression terms of four integrals around the volume, $$\int \Phi dp$$. The side integrals can be calculated by direct integration of (9), which gives:
$\begin{split}\begin{eqnarray} \int_{p_t}^{p_b} \Phi dp &= \Delta p \left( \Phi_b + \frac{1}{2} A \Delta p + \lambda \left( 1 - \frac{1 - \epsilon}{2 \epsilon} \ln \left| \frac{1 + \epsilon}{1 - \epsilon} \right| \right) \right) \\ &= \Delta p \left( \Phi_b + \frac{1}{2} A \Delta p + \lambda \left( 1 - (1 - \epsilon) \left( 1 + \frac{\epsilon^2}{3} + \frac{\epsilon^4}{5} + \cdots \right) \right) \right) \\ &= \Delta p \left( \Phi_b + \frac{1}{2} A \Delta p + \lambda \left( \epsilon - (1 - \epsilon) \epsilon^2 \left( \frac{1}{3} + \frac{\epsilon^2}{5} + \cdots \right) \right) \right) \end{eqnarray}\end{split}$
where $$\Phi, \Delta p, P, A$$ and $$\lambda$$ are each evaluated on the left or right side of the volume.
The top and bottom integrals in (8) must allow for the effect of varying $$\theta$$ and $$s$$ on $$A, \lambda$$ and $$P$$. We evaluate these integrals numerically using sixth-order quadrature; Boole’s rule requires evaluating the coefficients in the equation of state at five points, two of which have already been evaluated for the side integrals. For efficiency, we linearly interpolate the coefficients $$A, P$$ and $$\lambda$$ between the end points, which seems to make very little difference to the solution. We also verified that tenth-order quadrature makes little difference to the solution. The values of the top and bottom integrals are carried upward in a hydrostatic-like integration, obtained as follows:
$\begin{split}\begin{eqnarray} \int_{p_{tl}}^{p_{tr}} \Phi_t dp &= (p_{tr} - p_{tl}) \int_0^1 \Phi_t dx \\ &= (p_{tr} - p_{tl}) \int_0^1 \left( \Phi_b + A(x) \Delta p(x) + \lambda (x) \ln \left| \frac{1 + \epsilon (x)}{1 - \epsilon (x)} \right| \right) dx \\ &= (p_{tr} - p_{tl}) \int_0^1 \Phi_b dx \\ &+ \int_0^1 \Delta p(x) \left( A(x) + \frac{\lambda (x)}{P(x) + \overline{p} (x)} \sum_{n=1}^\infty \frac{\epsilon^{2n-2}}{2n-1} \right) dx \end{eqnarray}\end{split}$
The first integral is either known from the top integral of the layer below or the boundary condition at the ocean bottom. The second integral is evaluated numerically.
All the above definite integrals are specific to the Wright equation of state; the use of a different equation of state requires analytic integration of the appropriate equations. We have found, however, that high-order numerical integration appears to be sufficient. Although the numerical implementation is more general (allowing the use of arbitrary equations of state), it is significantly more expensive and so we advocate the analytic implementation for efficiency. | 2,061 | 6,818 | {"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} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.853064 |
https://encyclopedia2.thefreedictionary.com/eigenfrequency | 1,571,027,544,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00197.warc.gz | 573,240,526 | 12,347 | # eigenfrequency
Also found in: Dictionary, Wikipedia.
## eigenfrequency
[′ī·gən‚frē·kwən·sē]
(physics)
One of the frequencies at which an oscillatory system can vibrate.
References in periodicals archive ?
The increase in the absolute power of the first eigenfrequency while that of the other eigenfrequencies decreases may be due to damping of the oscillations.
The theoretical value of the eigenfrequency can be obtained using an analytical model based on (8).
A damping factor of 1% has been set for a frequency about 7 Hz, which is close to the first eigenfrequency of the undamaged plate.
Constraints are the first Eigenfrequency, which is required to be above 950 Hz, and a draw constraint to assert manufacturability.
The eigenfrequency for this fourth mode in the Gemona region is evaluated to be about 3.8 Hz and it compares well with one of the dominant frequencies (96) for the spectral response estimated from the Friuli 1976-1977 earthquake sequence at the Tolmezzo-Ambiesta dam (TLM1) accelerograph site.
A typical example of a physical constraint is the relationship between eigenfrequency of a valve plate and rotation speed in a compression chamber:
This range results from the parameters of used seismometers (the eigenfrequency of seismometer Lennartz LE3D/5s is 0.2 Hz) and from the parameters of recording (sampling frequency of the data recorded by the seismometer Guralp CMG-3ESP is 20 Hz).
This model only contains two parameters, the eigenfrequency [[omega].sub.0] = [square root of k+c/m+[m.sub.a]] of buoy/alternator, and the relative damping [zeta] = [gamma]/2 [square root of (m+[m.sub.a])(k+c).
In addition, [[eta].sub.d] = [square root of [[mu].sub.d]/[[epsilon].sub.d] and [[eta].sub.c] = [square root of [[mu].sub.c]/[[epsilon].sub.c] are the intrinsic impedances of the spherical cavity's internal dielectric filler and infinitely thick conducting wall materials, respectively, [mathematical expression not reproducible] and the complex angular resonance frequency (or eigenfrequency) is given by [[??].sub.0] = [[omega]'.sub.0] + j[[omega]".sub.0].
where [[omega].sub.0] = [square root of ([C.sub.Eq] / [J.sub.0])] and [[xi].sub.0]--an angular eigenfrequency of oscillations and a coefficient of the relative damping of the mechanical subsystem, while [[xi].sub.0] = 0.5 [T.sub.E][[omega].sub.0].
They include calculation models for revealing the relation between the eigenfrequency spectrum of anisotropic plates and the piezoelectric effect in the case of different boundary-value problems (Narita 2003).
3D Structural Mechanics Module (eigenfrequency analysis solver) of Comsol Multiphysics package was used to calculate the mode shapes and the undamped natural frequencies.
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Open / Close | 674 | 2,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-43 | longest | en | 0.867739 |
http://forums.parallax.com/archive/index.php/t-103745.html | 1,406,051,721,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997862121.60/warc/CC-MAIN-20140722025742-00086-ip-10-33-131-23.ec2.internal.warc.gz | 71,131,853 | 2,626 | PDA
View Full Version : hell of a problem using data types or adding an array
grasshopper
05-22-2008, 11:25 PM
Basically i am trying to take an array that a user can change by storing a baud rate in EEPROM. My problem is that if i add up the string that the user inputs i can seem to get it to be a correct value.
here is an example if a user inputs 9600
FLASH[ 2 ] := ((RxString[ 9 ] * 1000) + (RxString[ 10 ] * 100) + ( RxString[ 11 ] * 10) + (RxString[ 12 ]))
In the code the array Flash[ 2 ] is where i would expect to find 9600 logicaly i thought adding RxString [ 9 - 12] but it is not the case
9600 where the 9 is RxString[ 9 ] and the 6 is RxString[ 10 ] and the 0 is RxString [ 11 ] and the 0 is RxString [ 12 ]
My other foreseeable problem is that if a user inputs a number 19200 then the code will have to be modified
FLASH[ 2 ] := ((RxString[ 9 ] * 10000) + (RxString[ 10 ] * 1000) + ( RxString[ 11 ] * 100) + (RxString[ 12 ] *10 ) + (RxString[ 12 ] ))
I could address this with an If statement but some help would be nice
grasshopper
05-23-2008, 01:59 AM
Any one ?
Did i explain it enough?
I really need help on this one
Mike Green
05-23-2008, 02:22 AM
You basically want to convert a digit string into a binary value. There is already a routine (rxDec) in the Extended FullDuplexSerial object that does this. You could also look at the "getAnyNumber" method near the end of FemtoBasic.spin which allows for decimal, hexadecimal, and binary numbers.
allanlane5
05-23-2008, 03:00 AM
{
You've forgotten that the "Character" recieved will be in ASCII.· The ASCII character for "0" (zero) is 30.
}
PUB GetNum(InStrAddr) : Result | I, MyLen
· I := 0
· MyLen := STRSIZE(InStrAddr)
· repeat I from MyLen TO 0 STEP -1
··· Result := Result * 10
··· Result := Result + InStrAddr[I] - 30
· RETURN Result
grasshopper
05-23-2008, 03:16 AM
allanlane5 said...
PUB GetNum(InStrAddr) : Result | I, MyLen
I := 0
repeat I from MyLen TO 0 STEP -1
Result := Result * 10
Result := Result + InStrAddr - 30
RETURN Result
Nice work thanks a million!!
This is what I was looking for .
http://forums.parallax.com/images/smilies/tongue.gif
StefanL38
05-23-2008, 04:06 AM
Hello,
the ASCII-Code for "0" zero is DECIMAL 48 and hexadecimal this is 30
a number without a leading "\$" is interpreted as DECIMAL
so code has to be
PUB GetNum(InStrAddr) : Result | I, MyLen
I := 0
repeat I from MyLen TO 0 STEP -1
Result := Result * 10
Result := Result + InStrAddr - 48 'intead of 30
RETURN Result
best regards
Stefan
Post Edited (StefanL38) : 5/22/2008 9:29:47 PM GMT | 831 | 2,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.779424 |
http://www.ncsu.edu/human_resources/payroll/fntax/subprestest.php | 1,412,217,347,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663637.20/warc/CC-MAIN-20140930004103-00141-ip-10-234-18-248.ec2.internal.warc.gz | 790,531,248 | 6,607 | Human Resources > Payroll > Fntax > Current Page
# Substantial Presence Test
## Ask a question
Name: Your email: Question:
## What is the substantial presence test? (SPT)
This is a test defined by the IRS to determine if a foreign national qualifies as a resident alien for tax purposes.
• An individual who arrives on an F-1 or J-1 student visa will be exempt from counting days towards SPT for a period of five years.
Note:The five year exemption is measured from the year the individual arrives, not from the length of time the individual is in the United States. For example, if the individual arrives in the United States on December 15, 2006, she will be out of her exemption on January 1, 2011, not December 15, 2011.
• An individual, who arrives on a non-student J or Q visa, may also be exempt from counting days towards SPT for a period of time depending upon the length of time in the United States, the purpose of the current visit, and information concerning prior visits.
• An individual, who may arrive as a nonresident alien, can have that status changed for a number of different reasons.
• Always make an appointment with the foreign national tax specialist in the University Payroll Department to accurately determine your correct tax status.
How can I determine if I meet the qualifications for the substantial presence test?
The substantial presence test counts the number of days over the last three years that you are physically present in the United States, using the following formula:
1. You have 31 days at least during the current year, PLUS
2. You have 183 days total, if you add the current year and the two preceding years, calculated as follows:
1. all the days that you were present in this current year, PLUS
2. 1/3 of the days you were present in the one year before this current year, PLUS
3. 1/6 of the days you were present two years before the current year.
Example1: You are physically present in the United States for 120 days in the current year, 120 days in the one year past, and 120 days in the two years past.
120 + 1/3(120) + 1/6(120) =
120 + 40 + 20 = 180 days
Was substantial presence met?
1. Were you here 31 days during current year? YES
2. Were you here 183 days total in the last three years? NO
Since both statements must be answered YES to meet the substantial presence test; the test WAS NOT met.
Example 2: You are physically present in the United States for 180 days in the current year, and for 180 days in the one year past, but only 120 days in the two years past.
180 +1/3 (180) + 1/6 (180) =
180 + 60 + 30 = 270 days
Was substantial presence met?
1. Were you here 31 days during the current year? YES
2. Were you here 183 days total in the last three years? YES
Since both statements must be answered YES to meet the substantial presence test; the test WAS met.
## Are there exceptions to these rules?
Yes! Each exception depends on the purpose of your visit, your visa type, and prior visits. The foreign national tax specialist will help you assess your status.
Last Edited: December 3rd, 2010 | 723 | 3,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | longest | en | 0.943871 |
http://www.evi.com/q/how_many_ounces_is_30_grams | 1,394,397,029,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010305626/warc/CC-MAIN-20140305090505-00072-ip-10-183-142-35.ec2.internal.warc.gz | 328,203,302 | 13,932 | # How many ounces is 30 grams?
• 1.0582188594206 ounces
the mass 1.0582188594206 ounces
• tk10npubl tk10ncanl
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• 30grams into ouncezs | 588 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2014-10 | longest | en | 0.877919 |
https://www.scribd.com/document/135465028/CEamp-Theory | 1,537,716,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159470.54/warc/CC-MAIN-20180923134314-20180923154714-00065.warc.gz | 838,109,251 | 34,531 | # Source: www.mwftr.com/MobileLab.
html
Dr. Charles Kim
Common-Emitter Amplifier
A. Before We Start As the title of this lab says, this lab is about designing a Common-Emitter Amplifier, and this in this stage of the lab course is premature, in my opinion, of course. How can one design a BJT amplifier only after one simple characteristic experiment? Maybe students are all brilliant or this subject is already covered in a class, extensively. Even with that assumption, experiment with a BJT amplifier should come before asking for designing such circuit. Even before that, a much simpler circuit investigation would be more beneficial to understand the Common-Emitter Amplifier.
B. Common-Emitter Amplifier Experiment B.1 Theory
Let’s start our discussion on Common Emitter Amplifier (CE Amp), rather from the first BJT Lab, in which we discussed about Base voltage and Collector voltage in the operating region. To revive your memory, here I bring the CE circuit configuration. This DC voltage application into BJT is usually called “DC Biasing”
In CE circuit, however, more popular biasing method is to supply single DC voltage (instead of two: VBB and VCC), also with a resistor at the Emitter. From this circuit, let’s calculate the voltage corresponding to VBB and the resistance corresponding to RB. From the upper circuit. Between B and GND, the voltage is VBB and equivalent resistance is RB.
1
Opening the terminals means we cut the wire between the Base and the junction of R1 and R2. the terminal voltage is nothing but the voltage across R2. Here we apply Thevenin theorem. The thevenin voltage can be acquired by finding the terminal voltage (at B and GND) after opening the terminal.Dr. and R1 and R2 are in series with voltage VCC across the series resistors. So we can apply R2 “voltage divider” to find the terminal voltage: Vth = VCC ⋅ R1 + R2 2 . we also try to find the equivalent voltage and resistance seen at the terminals B and GND. on the single voltage biased circuit. Charles Kim Similarly. When you open the circuit (See below).
In other words. we have the two parallel resistors R1 and R2 at the terminals of B and GND. OK. AC wants to be riding over the DC and gets some boost. we usually mean by amplifying AC signal. Deactivation of a voltage source means shorting the voltage source. At the same time. we can picture this way. we apply the “Input resistance method” which gets the equivalent circuit at the terminals after deactivating the voltage source. Now it’s time to consider a CE Amp circuit. when we say amplifier.Dr. the DC bias voltage should not interrupt the input AC signals and the amplified output AC signal. Also. By the way. Charles Kim How about Thevenin resistor? Since we have only independent voltage source VCC. is governed by a DC system. Therefore. Whatever entered through the 3 . Thevenin R ⋅R resistance is Rth = 1 2 R1 + R2 Finally we have the following circuit which exactly corresponds to the initial biasing circuit we studied in the BJT 1 Lab. with entrance and exit gates. This means that the biased voltages are DC values and they are not to be disturbed. A castle surrounded by circling high wall.
In the circuit above. and CE is the bypass capacitor. all other elements are exactly the same as the circuit we first discussed. Charles Kim entrance gate is exiting the exit gate with amplification. As you can see. The circuit below is one of the popular AC Amp circuits. C1 and C2 are coupling capacitors. the DC system inside the castle cannot be leaked through the gates. and its impact on frequency response of the whole circuit. This in circuit formation is a bypass capacitor at the Emitter. C2. of C1. In DC analysis.Dr. It’s a tightly controlled system. The castle here is the base voltage divider biased circuit. when we assume that Vin is a DC source. The size of the capacitor corresponds to the size of the gate. Therefore the gates’ other function is to block any leakage from the castle to outside world. Actually there is one more device inside the castle to nullify any effect of those foreign objects entering the castle. while larger gate passes through larger object. Why? Capacitor stores DC energy. That means the DC-only circuit would look like this: 4 . In other words capacitor works as a open circuit for DC source. Smaller gates pass only smaller object. while keeping the DC system stable and intact. and CE will be discussed shortly. DC Analysis DC analysis of the circuit is very important in that (i) it makes sure DC biasing is all right and (ii) it finds IE which (indirectly) is used for voltage gain of the circuit. that DC source cannot cross coupling capacitor C1. And the gates are realized by coupling capacitors. we assume that there are only DC sources. What a great castle it must be! However. The size.
That means the DC source VCC will be grounded. then. Remember this. then V the Emitter current IE is determined by I E = E . If we assume that the AC signal we provide is high frequency signal. we can redraw the circuit as follows: 5 . Why? The impedance of capacitor is reverse proportional to the capacitance and the frequency of AC signal.Dr. R2. This we will discuss in the AC Analysis of the circuit. we assume there is no DC sources. we can safely say the impedance of capacitor is almost zero.7 V lower than VB (typical silicon diode voltage drop). the capacitors are shorted. If we assume that Base current is negligible. All DC sources are deactivated. Charles Kim R2 R1 + R2 Then the voltage at the Emitter is 0. And zero impedance means short circuit. The AC analysis circuit then looks like this: Since all three CE Amp resistors (R1. and RC) are all connected to the GND. In AC voltage gain (amplification). By the voltage divider rule. RE V then I C = I E = E RE This IC is very important element in determining the AC voltage gain. the value of IC is an important factor to be included. Moreover. the Base voltage is VB = VCC ⋅ AC Analysis In AC analysis.
Rescue: Have you heard about Ebers-Moll model of BJT? In the model. then.Dr. Charles Kim As we know the voltage gain is the ratio output voltage Vout and the input voltage Vin. Then our final good equivalent circuit looks like: 6 .) Now we can apply this Emitter resistance re (with simpler form of 25/IC[mA]) into the circuit. How do we get the ratio? If we blindly apply that the Emitter current is close to Collector current. we would end up at the following “equivalent” circuit: As you see the above approach has problem: the input voltage is tied to the ground. (Here we learn that BJT performance is dependent upon temperature. the small-signal VT impedance looking into the Emitter is found by the equation: re = . where VT=25. If you disconnect input circuit from Base (since Base current is zero). then there is no way to connect input and output.3[mV] I c [mA] at room temperature. at the same time Base current is zero.
Design steps and consideration are discussed in the next Common Emitter Amplifier. 7 . (3) Since the load resistor is also in the equation. swamped Common Emitter Amplifier. Remember this when you design your Common Emitter Amplifier. the voltage gain of the circuit can be unstable. and (c) the Emitter resistance. we have the following conclusion: (1) The popular CE Amp circuit’s voltage gain is dependent upon (a) RC (b) Load resistor . Here I intentionally summarize the way we design for a common emitter amplifier because the above amplifier is not the best one I suggest. (2) Since Emitter resistance is dependent on temperature. so-called. load affects the voltage gain of the circuit.Dr. Here the idea is to add some bypassed emitter resistance for stable biasing with no change in gain at signal frequencies. Charles Kim Whew! After a long road to the final circuit.
DC ANALYSIS: 8 . part of the Emitter resistance is bypassed by the capacitor CE.Dr. Charles Kim Swamped Common Emitter Amplifier The only change here in the most popular Common Emitter Amplifier is that we increase the AC resistance of the Emitter circuit to reduce variations in voltage gain. The Common Emitter Amplifier circuit is shown below: As you see above.
Then how “high” is high enough? This will answer the bandwidth of our Common Emitter Amplifier. the capacitor cannot be removed from the consideration. we can see the circuit from Vin to the Base of the BJT (which is the input coupling circuit) forms a simple typical passive high pass filter circuit. Charles Kim AC ANALYSIS: FREQUENCY ANALYSIS: Well. Let’s get the AC analysis circuit with capacitors are not shorted out. It’s cutoff frequency is determined by the capacitor and resistor. 9 . Let’s look at the left circled circuit part. we’ve run far and. This means that when AC signal frequency is low.Dr. we short out capacitors when the AC signal frequency is high. If we reduce the circuit A then. A. in AC analysis. this is the last subject of the discussion. thankfully. Before we said.
10 . that means you change the frequency response of the circuit. B. the circuit part at the bottom for bypassing. Charles Kim From here we can guess (and use it in the design) what the lowest frequency it can pass without any reduction in the promised gain. with its cutoff frequency controlled by. we can see this part is a simple typical passive low pass filter. again. On the other hand. capacitor and resistor. What about the output side capacitor? That part is again another high pass filter circuit.Dr. If you change the load.
Right? (d) Why those values of resistors? Try to answer by DC analysis. (c) The load resistance in the circuit is chosen 100kΩ. Circuit Formation in PSPice Note that: (a) Input signal is VAC (instead of usual VSIN) in the circuit since we need frequency response of the output voltage. (b) To ease the limitation of power supply by IOBoard. (e) Why those values of capacitors? Perform AC analysis and find expected low and high cutoff frequencies. 11 . I mean. Find expected voltage gain from the DC analysis. Charles Kim B. take the circuit presented here as a starter. AC Sweep is done only with VAC with only amplitude specified.Dr. homework. This value may be significantly different from your assigned work. Since I cannot satisfy everybody.2 Simulation Since we spent enough time in Theory on Common Emitter Amplifier. The circuit presented here may somewhat disappoint you mainly because this circuit does not satisfy the assignment. we are now eager to design a circuit. VCC is supplied by 5V source. or project of your class. Different frequency will be applied by the simulator. and I do not intend to do so by the way. but very important starter. Let's do some simulation first. But wait for a second. You. you need to do some work too.
Dr. From here you can get a lot of information. Charles Kim Getting Voltage and Current (all DC of course) gives you the pseudo-DC analysis of the circuit. In the AC Sweep we have to assign our frequency band of interest. your AC source must be VAC. and I picked 0.5 in the circuit above. AC Sweep To do AC Sweep. 12 . You decide only ACMAG (AC magnitude). Here I set from 0 to 100MHz.
the input voltage. so called "load effect"? 13 .Dr. Charles Kim Then we place two voltage probes for frequency analysis: Simulated Results (Frequency Response) The green tracing is the voltage at the load and red. at each of the frequency at the range of 0 . What is the low cutoff frequency? What is the high cutoff frequency? What is the mid-band voltage gain? How can you change the cutoff frequencies by changing capacitor values? How do you change the gain by changing resistor values? Do you see the influence of load.100MHz. | 2,616 | 11,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-39 | latest | en | 0.917823 |
http://www.haskell.org/pipermail/haskell-cafe/2006-April/015342.html | 1,411,405,380,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657137108.99/warc/CC-MAIN-20140914011217-00263-ip-10-234-18-248.ec2.internal.warc.gz | 542,226,955 | 2,098 | # [Haskell-cafe] The Marriage of Heaven and Hell: Type Classes and Bit-Twiddling
David F. Place d at vidplace.com
Tue Apr 11 18:41:11 EDT 2006
```Hi:
Looking around further for a charming way to count bits, I found a
method that is completely inscrutable might be very fast.
http://graphics.stanford.edu/~seander/bithacks.html
Counting bits set in 12, 24, or 32-bit words using 64-bit instructions
I thought it would be neat to have size function constrained to
members of Bounded that would automatically choose which method
depending on the extent of the domain of the set. So, a small set
could be computed by the very fast method and the compiler would be
able to do the dispatch as part of constant folding.
This is what I came up with. It works, but it seems kind of forced.
I wonder if there is a better way.
sizeB :: (Bounded a,Enum a) => a -> Set a -> Int
sizeB e =
case fromEnum \$ maxBound `asTypeOf` e of
x | x <= 12 -> \(Set w) -> fromIntegral \$ c12 \$ fromIntegral w
x | x <= 24 -> \(Set w) -> fromIntegral \$ c24 \$ fromIntegral w
x | x <= 32 -> \(Set w) -> fromIntegral \$ c32 \$ fromIntegral w
_ -> \(Set w) -> bitcount 0 w
c12 :: Word64 -> Word64
c12 v = (v * 0x1001001001001 .&. 0x84210842108421) `rem` 0x1f
c24' :: Word64 -> Word64
c24' v = ((v .&. 0xfff) * 0x1001001001001 .&. 0x84210842108421) `rem`
0x1f
c24 :: Word64 -> Word64
c24 v = (c24' v) + ((((v .&. 0xfff000) `shiftR` 12) *
0x1001001001001 .&. 0x84210842108421) `rem` 0x1f)
c32 :: Word64 -> Word64
c32 v = (c24 v) + (((v `shiftR` 24) * 0x1001001001001 .&.
0x84210842108421) `rem` 0x1f)
For example:
data Test1 = Foo | Bar | Baaz | Quux deriving (Enum, Bounded)
sizeTest1 :: (Set Test1) -> Int
sizeTest1 = sizeB Foo
Cheers, David
--------------------------------
David F. Place
mailto:d at vidplace.com
``` | 609 | 1,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2014-41 | latest | en | 0.748963 |
https://wiki.kerbalspaceprogram.com/wiki/Geosynchronous_Orbit_(Math) | 1,713,060,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00693.warc.gz | 585,448,600 | 8,849 | # Geosynchronous Orbit (Math)
A geosynchronous or, more specifically, geostationary orbit is an orbit where your orbital period is equal to that of the gravitational body's "day" (specifically the sidereal time or sidereal rotation period), so you remain in the same spot over the planet consistently. Also the gravitational force and the centripetal force needs to be equal, which is the case for any circular orbit. The difference between geosynchronous and geostationary orbits is that geosynchronous orbits have an orbital period of 1 day, but geostationary orbits are equatorial as well.
In order to calculate the geostationary orbit around any given body, we must first create an equation with the force of gravity and the centripetal force. The force of gravity is equal to:
${\displaystyle F_{G}={\frac {G\cdot M_{1}\cdot M_{2}}{r^{2}}}}$
where G is the gravitational constant (${\displaystyle 6.67384\cdot 10^{-11}{\frac {m^{3}}{kg\cdot s^{2}}}}$), M1 is the mass of the body, M2 is the mass of the satellite, and r is the distance between the center of mass of the planet and that of the satellite.
Because centripetal force is the same as gravitational force in a geostationary orbit, we can put them in opposite sides of the equation:
${\displaystyle F_{G}={\frac {G\cdot M_{1}\cdot M_{2}}{r^{2}}}=M_{2}\cdot {\frac {v^{2}}{r}}=F_{C}}$
The masses of the satellite cancel out, so we are left with:
${\displaystyle {\frac {G\cdot M_{1}}{r^{2}}}={\frac {v^{2}}{r}}}$
So, no matter how large your satellite is, the geostationary altitude will be the same.
We know that velocity is equal to distance divided by time. In this case, the distance is the circumference of your orbit, which is 2πr, and time in this instance is your orbital period, or one day for the planetary body. So, we now have:
${\displaystyle v={\frac {2\cdot \pi \cdot r}{t}}}$
But we are dealing with velocity squared, so we square our equation and get:
${\displaystyle v^{2}={\frac {4\cdot \pi ^{2}\cdot r^{2}}{t^{2}}}}$
Plugging that back into the original equation, we get:
${\displaystyle {\frac {G\cdot M_{1}}{r^{2}}}={\frac {4\cdot \pi ^{2}\cdot r^{2}}{t^{2}\cdot r}}}$
The r2 cancels out the r in the denominator and becomes a plain old r. Thus, we have:
${\displaystyle {\frac {G\cdot M_{1}}{r^{2}}}={\frac {4\cdot \pi ^{2}\cdot r}{t^{2}}}}$
We now multiply by r2:
${\displaystyle G\cdot M_{1}={\frac {4\cdot \pi ^{2}\cdot r^{3}}{t^{2}}}}$
And multiply by t2/4π2:
${\displaystyle r^{3}={\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}}$
And by taking the cube root of that, we arrive at our answer (well, sort of):
${\displaystyle r={\sqrt[{3}]{\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}}}$
But remember how r is the distance from the center of mass of the planet to that of the satellite? Well, it turns out that, unlike satellites, the center of mass of a planet is rather far away from its surface:
${\displaystyle a={\sqrt[{3}]{\frac {G\cdot M_{1}\cdot t^{2}}{4\cdot \pi ^{2}}}}-R_{p}}$
Where ${\displaystyle a}$ is the altitude from the sea level of the planet, and ${\displaystyle R_{p}}$ is the radius of the planet. This formula calculates the altitude above sea level for any given orbital period ${\displaystyle t}$. To get the height for a stationary orbit, the orbital period must be as long as the sidereal rotation period, the time of a full revolution of the planet relative to the sky. | 1,001 | 3,422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.876061 |
http://stackoverflow.com/questions/tagged/equations | 1,461,954,680,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111392.88/warc/CC-MAIN-20160428161511-00063-ip-10-239-7-51.ec2.internal.warc.gz | 266,200,381 | 29,100 | # Tagged Questions
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### Solve Systems of Linear Equations in MATLAB
I have the following Equations T2+T4-3.615*T1=0; T1+10+2*T5-5.752*T2=0; T1+38+2*T5-4*T4=0; 83+T4+T2+10-4*T5=0; i've tried the following to find the values of T1,T2,T4,T5 syms T1 T3 T4 T2 T5 ...
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### Find a vector components value in matlab
I have thw following problem Considering the matrix M = [1 2 3; 4 5 6; 7 8 10] And the vector E = [Ex; Ey; Ez] And B = [1; 4; 10] Does anyone knows how to find the values of E components in ...
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### I want to solve a system of equations containing a*x
Is it possible to solve a system of three equations in Excel, that contain x*y?? Let's suppose that my unknowns are a,b,x The equations are a + b = 1 a * x - 20y = 0 10x * a - 20a + b = 0 Is ... | 3,422 | 12,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.893752 |
www.flem-ath.com | 1,524,740,489,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948126.97/warc/CC-MAIN-20180426105552-20180426125552-00349.warc.gz | 403,730,273 | 8,898 | # The Speed of Light Code
The Speed of Light Code (2017)
Rand Flem-Ath
Scientists at the Search for Extraterrestrial Intelligence (SETI) comb the heavens in a quest to find universal constants like ф (1.618…) and π (3.1415…). These ‘Civilization Signature Constants’ (CSC) are fundamental to SETI because such numbers are highly unlikely to occur at random. If a message came from a distant star system printing out ф, or π digits, then SETI would cry Eureka! If that message contained two universal constants, then SETI might even declare contact!
But what if we look at the problem from another angle? What if we back-engineer SETI’s strategy? Rather than pointing our instruments at the sky maybe we should be looking at our own planet to determine if a Super-Intelligent-Entity-or-Entities (SIEE) has/have already left ф (1.618…) and π (3.1415…) footprints upon Earth.
The idea that the speed of light might contain a code seems odd at first. But that’s only because we are accustomed to measuring c (the speed of light) using the metric system. In 1975, the speed of light was determined at 299,792,458 meters per second with a measurement uncertainty of 4 parts per billion. 1983, the measurement uncertainty was corrected by redefining the meter as being 1/299,792,458 of the distance light travels in one second. The number 299,792,458 has nothing special about it. It seems random, but look what happens when we measure c (speed of light) with a different measuring tool, nautical miles (M). The speed of light travels at 161,874.977 nautical miles per second. We call this c/M.
The first four digits of c/M are 1618 which is ф (phi) packed right into c. Two universal constants in one set of numbers. Coincidence? Maybe? But what if this same message (ф combined with c) was coming from outer space? How would we react?
The ф in c is not the only part of this ancient cryptogram. Let’s look at the next three digits (749). When we divide 21600 nautical miles (the circumference of the Earth) into c/M we get 7.494211898148148. In other words, in one second light travels approximately 7.49 times the circumference of the Earth. So, the 749 in c/M maybe a confirmation (redundancy) that we’re dealing with the speed of light as measured by our planet’s circumference.
So, if 1618 is ф and 7.49 are consistent with a cryptogram, what then of the 77? I suggest it is a check digit, something that has only recently become a part of 21st century civilization. We use check digits because we use computers. For inventory purposes, numbers like 4977, suggest a confirmation that c/M is not random. The 4977 in c/M is a check digit because 7 times 7 is 49. I suggest that the 77 in c/M is a kind of exclamation mark meaning: it’s real. You’re on the right track to cracking the cryptogram.
c/M is an ancient cryptogram c = the speed of light M = nautical miles /s = per second c = 161874.977M/s 1.618 = ф 7.49 = the number of earth circumferences light travels in one second 77 = check digit (49 = 7 x 7)
So how did this come about? I suggest that SIEE has/have tuned the velocity of the Earth’s spin to print out an ancient cryptogram using nautical miles as the key. SIEE left us a message in the very spin of our planet. SIEE then set about the task of insuring that humanity would use a specific time system otherwise the code would never be broken. SIEE did this, I suggest, by imparting to the Sumerians a time system (24 hours, 60 minutes and 60 seconds) that ultimately would provide the suitable measurement of one second.
10ф can also be used for the velocity of the Earth’s orbit when measured in nautical miles.
Earth’s Orbital Velocity M/s = nautical miles per second 15.85M/s = minimum 16.07M/s = mean 16.10M/s = midpoint 16.18M/s = 10фM/s 16.1874977M/s = 1/10,000th the speed of light 16.35M/s = maximum
Each year, the Earth’s orbital velocity varies from a low of 15.85M/s to a high of 16.35M/s. There are days when our orbital velocity produces the universal constant ф (10фM/s). And other days when our planet’s speed passes through 1/10,000th of c (16.1875977M/s).
An Earth that has been “tuned” to produce universal constants challenges the prevailing paradigm that our planet’s spin and orbital velocity are simply the residue effects of the creation of our solar system. In that model, the Earth’s orbital and spin velocity should be a random number, not ordered, and especially not ordered by universal constants like c and ф.
~~~~~~~
The Speed of Light Code appears in
FROM ATLANTIS TO THE PROMISED LAND. | 1,145 | 4,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.895455 |
www.vindays.com | 1,501,029,330,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00613.warc.gz | 567,338,205 | 12,802 | Rubik’s Cubes
Rubik’s cubes are a fascinating company. It is exciting to shuffle and interesting to solve them. It has nothing to do with mathematics, but surely it increases your problem solving skills.
Below are some relevant details that every beginner or even a Rubik’s fan must know :
• The Rubik’s cube was invented in 1974 by Erno Rubik, a Hungarian architect, who wanted a working model to help explain three-dimensional geometry. After designing the “magic cube” as he called it (twice the weight of the current toy), he realized he could not actually solve the puzzle.
• It is believed that Rubik’s cube can be a relaxing and productive activity than gaming and social networking.
• It has emerged as a profession for a few. Puzzle creators and solvers have their own respects. Puzzle creators as the name says; create unique Rubik’s cubes which are sometimes sold for lakhs too.
• Puzzle solvers try to crack the cubes in minimum time and develop number of methods to solve the same cube.
• If you have ever tried solving a Rubik’s cube before, you need to try this time. It needs a lot of concentration and thinking to solve cubes.
• You can even test your IQ by solving them. Prefer to solve them on your own rather than taking the help of some YouTube video or the short notation steps.
• For a first few days, it irritates a lot to start arranging the pieces together. But gradually, you will start loving its company.
• Many of them learn finger tricks and start solving cubes faster. But others tend to solve different types of cubes.
• Various competitions are conducted to find the fastest cube solver. According to Guinness book of records, the 3*3 rubix cube was solved in 6 seconds!
• A number of channels are available on YouTube to help you solve cubes. If you are struggling too much then don’t look back to take help of these videos.
Comment on the box below your first rubiks cube feeling if you have tried it once. #My_first_rubiks.
Never miss the joy of solving a Rubik’s cube!
Stay tuned for my next article on types on Rubik’s cube! | 455 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.957225 |
https://mathoverflow.net/questions/215240/is-there-a-non-smooth-algebraic-group-scheme-in-char-p-all-of-whose-defining | 1,695,578,945,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00870.warc.gz | 436,656,634 | 27,886 | # Is there a non-smooth algebraic group scheme in char $p$, all of whose defining relations have degree less than $p$?
Let $k$ be an algebraically closed field of characteristic $p>0$. All the examples of non-smooth algebraic group schemes over $k$ that I have seen (apart from "artificial" examples; see below) have been given by presentations with at least one defining relation of degree a positive power of $p$. Here are the examples I have looked at:
• Frobenius kernels.
• Some automorphism schemes of algebras. The paper "Non-reduced automorphism schemes" by Geiss and Voigt has an interesting example in Section 2, which is given by a presentation including some relations of degree 2. The authors state that this group scheme is not reduced if and only if $p=2$. They also mention that if $G$ is a finite $p$-group, then the group scheme $\mathrm{Aut}(k[G])$ is not reduced.
• Results by Sopkina, "Classification of all connected subgroup schemes of a reductive group containing a split maximal torus", imply, if I have understood things correctly, that every non-smooth subgroup scheme of a reductive group over $k$ (or at least of $\mathrm{GL}_{n}$) containing a maximal torus, has a presentation including a $p$-power relation.
• Non-smooth centralisers in classical groups in not very good characteristic.
On the other hand, it is easy to produce "artificial" examples of a non-smooth group scheme in, for example, char 3 with a presentation involving only quadratic relations. Namely, take $k$ of char 3 and $\alpha_{3}=\mathrm{Spec}\, k[x]/(x^{3})$. Since $k[x]/(x^{3})$ is isomorphic to $k[x,y]/(xy,x-y^{2})$ as $k$-algebras, we can transport the Hopf algebra structure from the former to the latter.
Question. Is there a non-smooth algebraic group scheme $G$ over $k$, and an embedding $G\rightarrow\mathbb{A}^{n}$ of $G$ as a closed subscheme of affine $n$-space, with $n$ minimal, such that every defining relation of $G$ in this embedding is of degree strictly less than $p$? (Even the case $p=3$ would be interesting.)
Using explicit equations and a minimal embedding into affine space may seem a bit unnatural, but results of Kollár and Jelonek (see this previous MO question) - which boil down to estimating degrees and Bézout's theorem - imply that if we take $p>d^{n}$, where $d$ is the maximal degree of a relation, then $p$ does not divide the nilpotency index of any element in the coordinate algebra of $G$, so a proof of Cartier's theorem can be carried through in this case. Hence an example as in the above question must have $d<p<d^n$.
No. Let $f$ be a relation of minimal degree $2 \leq d <p$. Apply the comultiplication. This must be zero in $R \otimes R$, where $R$ is the ring of functions. So if $x_1, \dots, x_n$ are the variables, then it is zero in $k[x_1, \dots, x_n] \otimes k[x_1,\dots x_n]$ modulo the various relations.
Write $f$ as a sum of monomials in the $x_i$. When we apply the comultiplication to a monomial, the leading term does not depend on which monomial we pick. It's just the sum over the ways of splitting that monomial into a product of two monomials. So each pair of monomials in $k[x_1, \dots, x_n] \otimes k[x_1,\dots x_n]$ comes from a unique monomial of $f$, so we may ignore cancellation among different monomials. Moreover, because $d \geq 2$, we may split it into two monomials, each of degree $<d$. Because $d$ was the minimal degree of relations, we may ignore the relations. So we end up with something nonzero unless the number of ways of splitting is nonzero. But the number of ways of splitting is a product of binomial coefficients. All numbers involved in the binomial coefficients are less than $p$, so the binomial coefficients are prime to $p$ and hence nonzer.
• In general, before you get to the $p$th powers, group schemes in characteristic $p$ almost always behave exactly like group schemes in characteristic $0$ - and, in particular, all relations show up in degree $1$. Aug 20, 2015 at 20:43
• Could you clarify what you mean by: 1) "the leading term" of the comultiplication applied to a monomial (my first guess was $x\otimes1+1\otimes x$, but this doesn't seem to be what you mean); 2) "the leading term does not depend on which monomial we pick"; 3) "each pair of monomials in ... comes from a unique monomial of $f$". Also, where does the minimality of $n$ enter in your argument? It seems that the argument aims to derive a contradiction, but if so, where does the non-smoothness enter (supposedly we want to pick a non-zero nilpotent element at some point)? Aug 21, 2015 at 9:32
• @AStasinski 1) I mean you write it as a sum of tensors of monomials and take the tensors corresponding to pairs of monomials of minimal degree. Because the comultiplication applied to $x_i$ gives $x_i \otimes 1 + 1 \otimes x_i$ plus higher-order terms, we can compute the comultiplication of any monomial up to higher-order terms. 2) I meant rather that the leading term doesn't depend on the group law. Aug 25, 2015 at 2:33
• @AStasinski 3) I mean that comultiplicatoin applied to a monomial like $x_1 x_2^2 x_3^3$ will be a sum of terms like $x_1 x_2 x_3 \otimes x_2 x_3^2$ where the product of the two sides of the $\otimes$ is the original monomial. The minimality of $n$ enters because I am taking a relation of minimal degree $d$ and using $d \geq 2$. If $n$ is not minimal, then there is a relation of degree $1$. Yes, the argument is attempting to derive a contradiction. Aug 25, 2015 at 2:34 | 1,488 | 5,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.907708 |
https://geolines.ru/eng/publications/HISTORIC-GEODESY/nonagon-pyramid-system.html | 1,713,824,059,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00820.warc.gz | 234,562,465 | 13,719 | 1
Publications HISTORICAL GEODESY 21.02.2016. BUILDING OF SAMC PART 4. NONAGON.
# HISTORICAL GEODESY
BUILDING OF SAMC PART 4. NONAGON. In this part we will consider directions to key objects of ancient monumental constructions system (SAMC), which can be constructed from the anchor points of main frame (see part 1) with the help of nonagon. The image below shows the anchor points to the east from zero meridian of Great Pyramid (GP). From a point with coordinates 30 °, 67.5 °, with an accuracy of 0.07 you can get the direction to Lalibela. With a little more accuracy - 0.25 ° the direction to Lalibela you can get also from the point of intersection equator with meridian of 67.5 °. From the intersection of equator with 45th meridian you can get two more directions to the St. Sophia in Istanbul and pyramids on an island of Mauritius. In the first case, the error is 0.13 °, in the second - 0.38 °. Let us go further to the east along equator. From a point with coordinates 0 °, 67.5 °, with an error of 0.09 you can get the direction to a mysterious pyramidal structure Gunung Padang on a island of Java. From the point of intersection of equator with perpendicular to meridian of GP using nonagon you can get two more directions to Uluru and megalithic complex in Japan - Park of Asuka. Now let us move to the anchor points (AT) of framework to the west from meridian of Great Pyramid. Let us start with Teotihuacan, which forms a unique combination with three anchor points. The first point is intersection of meridian 112.5 ° with equator with an accuracy of 0.03 a degree gives direction to Teotihuacan. The second point is Teotihuacan itself where this line is the reverse direction when you turn 9-gon on 3 ° counter-clockwise interacts with two other lines in the AT. Besides, in Teotihuacan, a line with an azimuth of 120 °, corresponding to hexagram, which we already considered in the 2nd part forms with the direction to the point of 0 °, 67.5 °, an angle equal to 20 °. I.e. corresponds to nonagon oriented to the cardinal. With the help of nonagon from the anchor point with coordinates 30 °, -135 °, with an accuracy of 0.17 ° you can get direction to Sacsayhuaman. From a point with coordinates 30 °, 157.5 °, with an accuracy of 0.05 nonagon gives direction to geoglyphs in the Nazca desert. Specifically on Cahuachi pyramid located on the plateau. In the point of -30 °, -90 °, with an error of 0.09 of a degree, with the help of a 9-gon it is possible to determine the direction to Machu Picchu. If the intersection of equator with meridian -115 °, which is part of 5 ° of a network to lay angle 220 ° then with an accuracy of 0.05 it will be a direction to Easter Island. If equator divide into 12 equal parts, we will get the meridians corresponding to hexagram. One of them will have a value of -150 °. If from intersection of this meridian with 30th latitude to lay the direction to Easter Island, it will differ from the direction to the south, even on 10 °. Reverse azimuth from Easter Island to this point is also equal to 10 °, with an error of 0.09 of a degree. If from the point with coordinates -30 °, 90 ° to lay direction 50 ° then with accuracy of 0.19 °, it will be the direction to Nan Madol. In the point of Nan Madol there are also directions to the anchor points, the angles between which corresponds to the angles of nonagon. One of the main points of SAMC is fortress Por Bazhyn also is tied to the main frame. If from Por Bazhyn build direction to the anchor point with coordinates 30 °, 90 °, it will differ from the direction to the east exactly on 40 °. The angle between the direction from Por Bazhyn to the points with coordinates of 0 °, 0 ° and 0 °, 22.5 ° is equal to 20 ° with an accuracy of 0.13 °. Let us consider another important point in the system - Rome, which has a mass of megalithic structures. As a point of measurement was selected Egyptian obelisk mounted on the square of St. Peter's in Vatican. This point has a unique location and is connected with the main frame with several correct figures. If on the plane depict equator as straight line and in scale to lay directions to historical objects, the key point of the system, it will look as shown in the image below. Download file .kmz for Google Earth program.
Author: GeoLines.ru
MORE ON THIS TOPIC:
BUILDING OF SAMC. PART 5: OCTAGON.
BUILDING OF SYSTEM OF ANCIENT MONUMENTAL CONSTRUCTIONS. (SAMC). PART 1. THE MAIN FRAIM.
BUILDING OF SAMC. PART 2. HEXAGRAM. | 1,183 | 4,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.838574 |
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# How do I obtain the standard deviation from the standard error of the mean (s.e.m.)?
$$\frac{\mbox{The standard deviation}}{\sqrt{\mbox{sample size}}}$$= standard error of the mean
i.e.
standard deviation = $$\sqrt{\mbox{sample size}} \mbox{multiplied by the standard error of the mean}$$
It follows from this (e.g. p.218 of Babbie) that the standard error fo the mean decreases with sample size, N.
This follows since: The variance of the mean = 1/N x variance of the response = 1/N x The variance of N randomly sampled responses from a parent population = 1/ N2 x (Nsigma2)
= sigma2/N where sigma2 is the (unobserved true) variance of the (parent population of the) response. Since this involves a term 1/N the variance (or its square root = standard error) of the mean decreases with sample size.
The mean (= to the midpoint or median in a Normal distribution) on the other hand is not proportional to sample size so is uneffected by N. One can see this easily by considering an example: suppose we have a sample of size 3 of a responses = 1 2 3 then the mean is 2. Suppose I take a sample of size 7 say of the same response and get values of 1 1 2 2 2 3 3 then the mean = 2 there since it is symmetric about the (hypothesised true) mean of 2 (which follows from sampling from a response following a normal distribution).
So it follows that the one sample z-test statistic = mean / s.e.(mean) will increase with increasing sample size because the mean stays the same and the s.e.(mean) decreases.
Reference
Babbie, E. (2008). The Basics of Social Research. Fourth Edition. Thomson Wadsworth: Belmont.CA. | 522 | 2,000 | {"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} | 3.953125 | 4 | CC-MAIN-2022-33 | latest | en | 0.866317 |
http://www.thirteen.org/edonline/ntti/resources/lessons/tipperary/orgb.html | 1,521,427,486,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646213.26/warc/CC-MAIN-20180319023123-20180319043123-00035.warc.gz | 481,954,588 | 2,575 | 'Tis a Long, Long Way to Tipperary! Data Sheet #2 Calculating Longitude and Time
(a) From the prime meridian to City A, the distance in degrees is ___________. (b) From the prime meridian to City B, the distance in degrees is ___________. (c) Therefore, the total distance from City A to City B in degrees is _________. (d) Convert degrees apart into time apart. ____________________________ (e) If every 15° is equivalent to 1 hour, how many hours are in 90°? _________ (f) City B is west of City A. Therefore it must be earlier in City B than City A. If the time in City A is midnight, what is the time in City B? _________________ | 168 | 637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-13 | latest | en | 0.790422 |
https://books.google.com.jm/books?qtid=77e7848&lr=&id=-YUdTJ55ezkC&sa=N&start=30 | 1,719,164,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00570.warc.gz | 115,686,571 | 6,528 | Books Books
The angle at the centre of a circle is double the angle at the circumference on the same arc.
The Elements of geometry [Euclid book 1-3] in general terms, with notes &c ... - Page 128
by Euclides - 1821
## A Treatise on Surveying: In which the Theory and Practice are Fully ...
Samuel Alsop - Surveying - 1865 - 440 pages
...angled triangle is equal to the sum of the similar figures similarly described on the sides. (31.6.) 83. The angle at the centre of a circle is double the angle at the circumference on the same base. Thus, the angle at 0 (Fig. 11) is double e'ther D or E. (20.3.) 84. Angles in the...
## Report on Education in the Parochial Schools of the Counties of Aberdeen ...
...produced, what change ought, from the nature of the case, to be made in the enunciation? 4. The angle ut the centre of a circle is double the angle at the circumference. Assuming that there is no limitation as to the magnitude of an angle, deduce from this proposition,...
## Elements of Plane and Solid Geometry: And of Plane and Spherical ...
Gerardus Beekman Docharty - Geometry - 1867 - 474 pages
...the arc AEB ; and thus, having equal measures, they are equal to each other (ax. 11). THEOREM XI. An angle at the centre of a circle is double the angle at the circumference, when both stand on the same arc. Let C be an angle at the centre C, and D an angle at the circumference,...
## The elements of mechanism
Thomas Minchin Goodeve - 1870 - 294 pages
...equal to CP or c R ; draw any fixed diameter A c B, and join A R. Then it is proved in Euclid that the angle at the centre of a circle is double the angle at the circumference when Fig. 248. both angles stand upon the same arc, that is<lPCA = 2Z.PRA or the angular velocity of c P...
## Four Phases of Morals: Socrates, Aristotle, Christianity, Utilitarianism
John Stuart Blackie - Christian ethics - 1871 - 432 pages
...present, or to come. When I say that all the angles of a triangle are equal to two right angles, or that the angle at the centre of a circle is double the angle at the circumference, I prove this from certain necessary relations of lines to lines drawn under conditions of which my...
## Euclid's Elements of Geometry
Euclid - Geometry - 1872 - 284 pages
...AB, is the centre. PROPOSITION XX. THEOREM. The angle (ACD) at the centre of a circle, is double of the angle at the circumference, when they have the same part of the circumference for their base. Firstly — Let either leg of the angle at the periphery pass through the centre ; and because in the...
## An Introduction to the Practical and Theoretical Study of Nautical Surveying
John Knox Laughton - Hydrographic surveying - 1872 - 222 pages
...DA, will pass through B. The segment A GB is the segment required. It is proved (Euclid III. 20) that the angle at the centre of a circle is double the angle at the circumference, standing on the same base, or part of the circumference ; that is, the angle ADB is double of the angle...
## Annual Register of the United States Naval Academy, Annapolis, Md, Volumes 25-32
United States Naval Academy - 1874 - 888 pages
...iu E and D, prove that e straight lines BD, DE, and E С will be equal to each other. 2. Prove that the angle. -at the centre of a circle is double the angle at the circum•ence on the same arc. AR and CD are two chords of a circle which intersect, зеп produced,... | 855 | 3,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.826473 |
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USE CODE: chait6
``` If x3/4(log3X)^2 + log3X - 5/4 = root 3. then x has
a) one negative integral value
b) two irrational value
c) two positive rational value
d) none of these
```
7 years ago
Share
``` Hi,
just take the log base 3 of both side and put log(x)to the base 3 = some t
now the eqn will
=> {3/4(t)^2 +t - 5/4}t =1/2
by putng 0 & 1 u'll cm to now that one root is one
then rearrange the eqn like 3(t)^3 +4(t)^2 -5t -2 =0
now 1 is a root then divide it by (t-1) u'll get:
3(t)^2+7t+2=0
=> (3t+1)(t+2)=0
=> t = -2 , -1/3
=> x = (3)^1, (3)^-2 ,(3)^-1/3
=> x= 3, 1/9, 1/cube root of(3)
3 & 1/9 r rational & +ve but -2 is _ve integral value
Hence option a) & c) r correct
hav gd lk
```
7 years ago
# Other Related Questions on Algebra
If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...
Ajay 6 months ago
Small Mistake in last para posting again..............................................................................................................
Ajay 6 months ago
We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.
mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??
The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below
Ajay 5 months ago
Thankyou so much............................. …......................................................................!
Anshuman Mohanty 5 months ago
Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7
Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above
If |z-i| = ?? PLs complete the question
Nishant Vora one month ago
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
Divya one month ago
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
Divya one month ago
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent
Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...
Kumar 3 months ago
Not any people get my answer why. You can no give answer my question I am join this site
Vivek kumar 5 months ago
Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?
Ajay 5 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2
What needs to be solved here ? The question is incomplete....................................................................
Ajay 6 months ago
i don’t know how to do this...............................................................................................
Saravanan 2 months ago
this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)
Naveen Shankar 6 months ago
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H. The following can be proved easily: 1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends...
mycroft holmes one month ago
Draw which is Isoceles as OB = OC. Now which means . Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required...
mycroft holmes one month ago
a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)
mycroft holmes one month ago
View all Questions »
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USE CODE: chait6
More Questions On Algebra | 1,508 | 4,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-04 | longest | en | 0.760138 |
https://www.socketloop.com/tutorials/golang-converting-a-negative-number-to-positive-number | 1,590,815,699,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00258.warc.gz | 940,676,836 | 98,723 | # Golang : Converting a negative number to positive number
Tags :
Helping out a friend here to figure out why her totalizer formula is spitting out incorrect result. Apparently, what she wanted to achieve is to convert a negative number to positive number so that her totalizer formula will calculate properly.
According to her co-workers, there are times when the field instrument will pump out negative number (flow rate), but it should be assumed to be a positive number and the numbers from a few measurement instruments have to be added up.
Sometimes, the simplest thing can be hardest to spot and we just assume the final calculation is correct. Writing this down as a reminder on how to convert a negative number to positive number properly with `math.Abs()` function.
Not going to show you the actual formula in totaling up the flow rates, but a simple example will do.
Here you go!
`````` package main
import (
"fmt"
"math"
)
func main() {
total := 1 + 1 + (-1)
fmt.Println("Total : ", total)
correctTotal := 1 + 1 + math.Abs(-1) // convert negative 1 to positive
fmt.Println("Correct total with absolute : ", correctTotal)
}
``````
Output:
Total : 1
Correct total with absolute : 3
Tags :
IF you gain some knowledge or the information here solved your programming problem. Please consider donating to the less fortunate or some charities that you like. Apart from donation, planting trees, volunteering or reducing your carbon footprint will be great too. | 317 | 1,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-24 | latest | en | 0.866189 |
https://www.anonymouschristian.org/blog/a-tough-geometry-question/ | 1,716,771,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00811.warc.gz | 544,805,949 | 32,227 | # A tough Geometry Question
Question
Two circles (⊙O and ⊙O’, the radius of ⊙O > the radius of ⊙O’) intersect at Point A and Point B. Line CD, which is their common tangent line at the bottom, is drawn tangent to ⊙O and ⊙O’ at Point C and Point D respectively. Extend CB to intersect AD at E, extend DB to intersect AC at F.
PROVE: DA * DE = DB * DF
Aha, geometry!
Draw a diagram and follow the steps below.
The trick is to show that the triangles DBE and
triangle DAF are similar.
Proof That triangles DBE and triangle DAF are similar.
Let <ABC> donote angle ABC.
We need to show any two corresponding angles of these traingles are equal.
First angle,
<FDA> = <BDE> because D,B,F collinear and D,E,A collinear.
Second angle,
<DAF> = <DAC> because A,F,C collinear
= (pi) – <ACD> – <ADC> angles in triangle ACD
= (pi) – <ACD> – <ADC> …(1)
<ACD> = <ACB> + <BCD> splitting <ACD> with line CB
= <ACB> + <CAB>, tangent property <BCD> = <CAB>
= (pi) – <CBA> angles in triangle ABC
= <ABE> because C,B,E are all colinear
Now, similarly
= (pi) – <ABD> angles in triangle ABD
Substituiting these two info into (1), we get,
<DAF> = (pi) – <ACD> – <ADC>
= (pi) – <ABE> – [ (pi) – <ABD> ]
= <ABD> – <ABE>
= <BDE> because BE splits <ABD>
Done. In showing the triangle DAF and triangle DBE are similar.
Hence, by ratio for similar triangles,
DA/DF = DB/DE
Or Rewriting we get
DA * DE = DB * DF
Proved. | 460 | 1,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.568881 |
http://www.circuitdiagramworld.com/unclassified_circuit/E_Series__Why_Resistors_come_only_in_certain_values__5427.html | 1,516,673,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891705.93/warc/CC-MAIN-20180123012644-20180123032644-00338.warc.gz | 396,137,584 | 4,763 | E Series: Why Resistors come only in certain values?_Circuit Diagram World
Position: Index > Unclassified >
# E Series: Why Resistors come only in certain values?
2017-08-24 19:50
Declaration:We aim to transmit more information by carrying articles . We will delete it soon, if we are involved in the problems of article content ,copyright or other problems.
Why is it easy to find 4.7k resistors, but not 4.8k resistors? Where do common values like 1.2k, 2.7k, 560, and 820 come from and who decides them? As you may know, resistors come in different tolerances, as indicated by the 4th band (gold = 5%, silver = 10%). A 100 ohm resistor with a 10% tolerance is expected to have a value somewhere between 90 and 110 ohms, so it wouldn't make much sense to buy a 101 Ohm resistor when it's actual value could be less than a 95 Ohm, 10% resistor. The Electronic Industries Association (EIA) is the primary body that standardizes the values for resistors, and they publish value lists called "E" series. In the 10% series, known as E12, each value is spaced so that there won't be overlap. The min and max values are listed:
(min)value(max)
(90)100(110)
(108)120(132)
(135)150(165)
(162)180(198)
(198)220(242)
(243)270(297)
The number following the "E" stands for the number of logarithmic steps per decade. Logwell has a table that lists common values from 10% through 1% Here is a neat resistor selection tool from uCHobby that allows you to select only legal values | 401 | 1,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | latest | en | 0.908837 |
https://studiegids.universiteitleiden.nl/en/courses/114185/advanced-statistical-computing | 1,720,874,359,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00489.warc.gz | 464,797,962 | 5,603 | # Advanced Statistical Computing
Course
2022-2023
This course on computational statistics requires good knowledge of probability theory, inferential statistics, and programming with R. The courses “Statistical Computing with R” and “Statistics and Probability” from the MSc in Statistics and Data Science are prerequisites for this course.
## Description
Advanced Statistical Computing will introduce you to the field of computational statistics, which can be loosely defined as a set of numerical methods and algorithms that can be employed to solve a well-defined statistical problem. During the course, we will focus on the following topics:
1. Methods for the generation of random variables
2. The Monte Carlo method
3. Applications of the Monte Carlo method
4. Variance reduction methods
5. Numerical integration and root finding algorithms
6. The bootstrap
7. Permutation tests
## Course objectives
By the end of the course, students should:
1. understand the mathematical and statistical foundations of the methods covered in the course;
2. be able to explain how each of the aforementioned methods work;
3. know how to solve relevant statistical problems using such methods;
4. be able to implement the methods and algorithms covered in this course using R.
## Timetable
You will find the timetables for all courses and degree programmes of Leiden University in the tool MyTimetable (login). Any teaching activities that you have sucessfully registered for in MyStudyMap will automatically be displayed in MyTimeTable. Any timetables that you add manually, will be saved and automatically displayed the next time you sign in.
MyTimetable allows you to integrate your timetable with your calendar apps such as Outlook, Google Calendar, Apple Calendar and other calendar apps on your smartphone. Any timetable changes will be automatically synced with your calendar. If you wish, you can also receive an email notification of the change. You can turn notifications on in ‘Settings’ (after login).
For more information, watch the video or go the the 'help-page' in MyTimetable. Please note: Joint Degree students Leiden/Delft have to merge their two different timetables into one. This video explains how to do this.
## Mode of instruction
A combination of lectures and computer practicals. To fully benefit from the practicals, it is recommended that you bring your own laptop with R and RStudio installed.
## Assessment method
A written exam at the end of the course. The grade of the written exam should be at least 5.5 to get a pass.
• Rizzo, M. L. (2019). Statistical Computing with R. CRC Press.
## Registration
From the academic year 2022-2023 on every student has to register for courses with the new enrollment tool MyStudyMap. There are two registration periods per year: registration for the fall semester opens in July and registration for the spring semester opens in December. Please see this page for more information.
Please note that it is compulsory to both preregister and confirm your participation for every exam and retake. Not being registered for a course means that you are not allowed to participate in the final exam of the course. Confirming your exam participation is possible until ten days before the exam.
Extensive FAQ's on MyStudymap can be found here. | 669 | 3,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.851066 |
https://www.doubtnut.com/question-answer/find-the-particular-solution-of-the-differential-equation-exsqrt1-y2dx-y-x-dy0-given-that-y1-when-x0-13255 | 1,632,847,031,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00358.warc.gz | 746,226,372 | 78,196 | Home
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# Find the particular solution of the differential equation e^xsqrt(1-y^2)dx+y/x dy=0, given that y=1 when x=0
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 10-3-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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13.8 K+
Text Solution
Solution :
e^xdx(sqrt(1-y^2))+ydy(1/x)=0<br> xe^xdxsqrt(1-y^2)+ydy=0<br> xe^xdx=((-y)dy)/sqrt(1-y^2)<br> integrating both side<br> intxe^xdx=int(-y)dy/sqrt(1-y^2)<br> intxe^xdx=x inte^xdx-intdx/dx*inte^xdxdx<br> xe^x-inte^xdx=xe^x-e^x<br> 1/2t^(1/2)/(1/2)=sqrtt<br> sqrt(1-y^2)<br> xe^x-e^x=sqrt(1-y^2)+c<br> c=-1<br> xe^x-e^x=sqert(1-y^2)-1<br> (x-1)e^x+1=sqrt(1-y^2).
Image Solution
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5:05 | 607 | 1,352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-39 | latest | en | 0.352598 |
https://www.wordunscrambler.net/unscramble-qtruoae | 1,708,682,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00307.warc.gz | 1,093,268,055 | 13,948 | # Unscramble QTRUOAE
QTRUOAE unscrambles into 97 different words! We have all of them and the meanings below! Enter any word and we will UNSCRAMBLE IT!
### 2 letter words made by unscrambling QTRUOAE
#### Unscrambled 16 2 Letter Words
Above are the words made by unscrambling QTRUOAE (AEOQRTU). To further help you, here are a few lists related to/with the letters QTRUOAE
### The Value of QTRUOAE In Word Scramble Games
The letters QTRUOAE are worth 16 points in Scrabble
The letters QTRUOAE are worth 17 in points Words With Friends
• Q = 10 points in WWF & 10 points in Scrabble
• T = 1 points in WWF & 1 points in Scrabble
• R = 1 points in WWF & 1 points in Scrabble
• U = 2 points in WWF & 1 points in Scrabble
• O = 1 points in WWF & 1 points in Scrabble
• A = 1 points in WWF & 1 points in Scrabble
• E = 1 points in WWF & 1 points in Scrabble
## What Does QTRUOAE Mean... If you Unscramble it?
### Possible Definitions of QTRUOAE
If we unscramble these letters, QTRUOAE, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters:
### Equator
• The great circle of the celestial sphere, coincident with the plane of the earth's equator; -- so called because when the sun is in it, the days and nights are of equal length; hence called also the equinoctial, and on maps, globes, etc., the equinoctial line.
• The imaginary great circle on the earth's surface, everywhere equally distant from the two poles, and dividing the earth's surface into two hemispheres.
## Permutations of QTRUOAE
According to our other word scramble maker, QTRUOAE can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word.
#### Definition of Permutation
a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.
How is this helpful? Well, it shows you the letters qtruoae scrambled in different ways That way you will recognize the set of letters more easily. It will help you the next time QTRUOAE comes up in a word scramble game.
We stopped it at 50, but there are so many ways to scramble QTRUOAE!
### Scramble Words
Unscramble these letters to make words...
scrambled using word scrambler... | 634 | 2,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.877705 |
https://mathmunch.org/tag/dimensions/page/2/ | 1,590,875,004,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00262.warc.gz | 446,543,409 | 26,228 | # Sandpiles, Prime Pages, and Six Dimensions of Color
Welcome to this week’s Math Munch!
Four million grains of sand dropped onto an infinite grid. The colors represent how many grains are at each vertex. From this gallery.
We got our first snowfall of the year this past week, but my most recent mathematical find makes me think of summertime instead. The picture to the right is of a sandpile—or, more formally, an Abelian sandpile model.
If you pour a bucket of sand into a pile a little at a time, it’ll build up for a while. But if it gets too tall, an avalanche will happen and some of the sand will tumble away from the peak. You can check out an applet that models this kind of sand action here.
A mathematical sandpile formalizes this idea. First, take any graph—a small one, a medium sided one, or an infinite grid. Grains of sand will go at each vertex, but we’ll set a maximum amount that each one can contain—the number of edges that connect to the vertex. (Notice that this is four for every vertex of an infinite square grid). If too many grains end up on a given vertex, then one grain avalanches down each edge to a neighboring vertex. This might be the end of the story, but it’s possible that a chain reaction will occur—that the extra grain at a neighboring vertex might cause it to spill over, and so on. For many more technical details, you might check out this article from the AMS Notices.
This video walks through the steps of a sandpile slowly, and it shows with numbers how many grains are in each spot.
A sandpile I made with Sergei’s applet
You can make some really cool images—both still and animated—by tinkering around with sandpiles. Sergei Maslov, who works at Brookhaven National Laboratory in New York, has a great applet on his website where you can make sandpiles of your own.
David Perkinson, a professor at Reed College, maintains a whole website about sandpiles. It contains a gallery of sandpile images and a more advanced sandpile applet.
Hexplode is a game based on sandpiles.
I have a feeling that you might also enjoy playing the sandpile-inspired game Hexplode!
Next up: we’ve shared links about Fibonnaci numbers and prime numbers before—they’re some of our favorite numbers! Here’s an amazing fact that I just found out this week. Some Fibonnaci numbers are prime—like 3, 5, and 13—but no one knows if there are infinitely many Fibonnaci primes, or only finitely many.
A great place to find out more amazing and fun facts like this one is at The Prime Pages. It has a list of the largest known prime numbers, as well as information about the continuing search for bigger ones—and how you can help out! It also has a short list of open questions about prime numbers, including Goldbach’s conjecture.
Be sure to peek at the “Prime Curios” page. It contains intriguing facts about prime numbers both large and small. For instance, did you know that 773 is both the only three-digit iccanobiF prime and the largest three-digit unholey prime? I sure didn’t.
Last but not least, I ran across this article about how a software company has come up with a new solution for mixing colors on a computer screen by using six dimensions rather than the usual three.
The arithmetic of colors!
Well, there are actually several ways that computers store colors. Each of them encodes colors using three numbers. For instance, one method builds colors by giving one number each to the primary colors yellow, red, and blue. Another systems assigns a number to each of hue, saturation, and brightness. More on these systems here. In any of these systems, you can picture a given color as sitting within a three-dimensional color cube, based on its three numbers.
A color cube, based on the RGB (red, green, blue) system.
If you numerically average two colors in these systems, you don’t actually end up with the color that you’d get by mixing paint of those two colors. Now, both scientists and artists think about combining colors in two ways—combining colored lights and combining colored pigments, or paints. These are called additive and subtractive color models—more on that here. The breakthrough that the folks at the software company FiftyThree made was to assign six numbers to each color—that is, to use both additive and subtractive ideas at the same time. The six numbers assigned to a given number can be thought of as plotting a point in a six-dimensional space—or inside of a hyper-hyper-hypercube.
I think it’s amazing that using math in this creative way helps to solve a nagging artistic problem. To get a feel for why mixing colors using the usual three-coordinate system is such a problem, you might try your hand at this color matching game. For even more info about the math of color, there’s some interesting stuff on this webpage.
Bon appetit!
# 3D Printer MArTH, Polyhedra, and Hart Videos
Welcome to this week’s Math Munch!
It’s my turn now to post about how much fun we had at Bridges! One of the best parts of Bridges was seeing the art on display, both in the galleries and in the lobby where people were displaying and selling their works of art. We spent a lot of time oogling over the 3D printed sculptures of Henry Segerman. Henry is a research fellow at the University of Melbourne, in Australia, studying 3-dimensional geometry and topology. The sculptures that he makes show how beautiful geometry and topology can be.
These are the sculptures that Henry had on display in the gallery at Bridges. They won Best Use of Mathematics! These are models of something called 4-dimensional regular polytopes. A polytope is a geometric object with flat sides – like a polygon in two dimensions or a polyhedron in three dimensions. 4-dimensional polyhedra? How can we see these in three dimensions? The process Henry used to make something 4-dimensional at least somewhat see-able in three dimensions is called a stereographic projection. Mapmakers use stereographic projections to show the surface of the Earth – which is a 3-dimensional object – on a flat sheet of paper – which is a 2-dimensional object.
A stereographic projection of the Earth.
To do a stereographic projection, you first set the sphere on the piece of paper, or plane. It’ll touch the plane in exactly 1 point (and will probably roll around, but let’s pretend it doesn’t). Next, you draw a straight line starting at the point at the top of the sphere, directly opposite the point set on the plane, going through another point on the sphere, and mark where that line hits the plane. If you do that for every point on the sphere, you get a flat picture of the surface of the sphere. The point where the sphere was set on the plane is drawn exactly where it was set – or is fixed, as mathematicians say. The point at the top of the sphere… well, it doesn’t really have a spot on the map. Mathematicians say that this point went to infinity. Exciting!
A stereographic projection like this draws a 3-dimensional object in 2-dimensions. The stereographic projection that Henry did shows a 4-dimensional object in 3-dimensions. Henry first drew, or projected, the vertices of his 4-dimensional polytope onto a 4-dimensional sphere – or hypersphere. Then he used a stereographic projection to make a 3D model of the polytope – and printed it out! How beautiful!
Here are some more images of Henry’s 3D printed sculptures. We particularly love the juggling one.
Henry will be dropping by to answer your questions! So if you have a question for him about his sculptures, the math he does, or something else, then leave it for him in the comments.
Speaking of polyhedra, check out this site of applets for visualizing polyhedra. You can look at, spin, and get stats on all kinds of polyhedra – from the regular old cube to the majestic great stellated dodecahedron to the mindbogglingly complex uniform great rhombicosidodecahedron. You can also practice your skills with Greek prefixes and suffixes.
Finally, two Math Munches ago, we told you about some videos made by the mathematical artist George Hart. He’s the man who brought us the Yoshimoto cube. And now he’s brought us… Pentadigitation. In this video, George connects stars, knots, and rubber bands. Enjoy watching – and trying the tricks!
Bon appetit!
# Numberphile, Cube Snakes, and the Hypercube.
Welcome to this week’s Math Munch!
Each one of those pictures takes you to a math video. Numberphile is a YouTube channel full of fantastic math videos by Brady Haran, each one about a different number. Is one Googolplex bigger than the universe? Why does Pac Man end after level 255? Is 1 a prime number? Click the numbers to watch the related video. They also feature James Grime, one of my favorite math people on the internet.
Next up, let’s work on the Saint Ann’s School Problem of the Week. You can read the fully worded question by following the link, but here it is in short: If we start in the center, we can snake our way through the 9 small squares of a 3×3 square. Can we snake our way through the 27 small cubes a 3x3x3 cube? Can we do it if we start in the middle?
Can we snake our way through the 3x3x3 cube starting in the center?
There’s a new question posted every week (obviously), and if you check the Problem of the Week Archives, you can find more than 4 years of previous questions! How many do you think we could solve if we did a 24 hour math marathon?
Finally, let’s have a mind-blowing look at higher dimensions. The problem above is about whether a property of the square (a 2-dimensional object) can be carried over to the cube (its 3D counterpart). So what is the 4-dimensional version of a cube? The Hypercube!
The "cube" idea, from 1D to 4D
I’ve heard a lot of people say the 4th dimension is “time” or “duration,” but what would the 5th dimension be? Well, here’s a video called “Imagining the Tenth Dimension.” And if you’re hungry for more, there’s a series of 9 math videos called “Dimesions.” All together it’s 2 amazing hours of math. You can watch the first chapter online by clicking here.
Bon appetit! | 2,377 | 10,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-24 | longest | en | 0.919882 |
http://www.gurufocus.com/term/ChangeInInventory/WRLD/Change%2BIn%2BInventory/World%2BAcceptance%2BCorporation | 1,480,735,437,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540804.14/warc/CC-MAIN-20161202170900-00153-ip-10-31-129-80.ec2.internal.warc.gz | 510,385,767 | 30,825 | Switch to:
World Acceptance Corp (NAS:WRLD)
Change In Inventory
\$0.0 Mil (As of Sep. 2016)
World Acceptance Corp's change in inventory for the quarter that ended in Sep. 2016 was \$0.0 Mil. It means World Acceptance Corp's inventory stayed the same from Jun. 2016 to Sep. 2016 .
World Acceptance Corp's change in inventory for the fiscal year that ended in Mar. 2016 was \$0.0 Mil. It means World Acceptance Corp's inventory stayed the same from Mar. 2015 to Mar. 2016 .
World Acceptance Corp's inventory for the quarter that ended in Sep. 2016 was \$0.0 Mil.
Days inventory indicates the number of days of goods in sales that a company has in the inventory.
Inventory can be measured by Days Sales of Inventory (DSI). World Acceptance Corp's days sales of inventory (DSI) for the quarter that ended in Sep. 2016 was 0.00.
Inventory turnover measures how fast the company turns over its inventory within a year.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. World Acceptance Corp's inventory to revenue ratio for the quarter that ended in Sep. 2016 was 0.00.
Definition
Change In Inventory is the difference between last periodÂ’s ending inventory and the current periodÂ’s ending inventory.
Explanation
1. Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
World Acceptance Corp's Days Inventory for the quarter that ended in Sep. 2016 is calculated as:
Days Inventory = Average Inventory / Cost of Goods Sold * Days in Period = 0 / 0 * 365 / 4 = N/A
2. Inventory can be measured by Days Sales of Inventory (DSI).
World Acceptance Corp's Days Sales of Inventory for the quarter that ended in Sep. 2016 is calculated as
Days Sales of Inventory (DSI) = Average Inventory / Revenue * Days in Period = 0 / 123.75 * 365 / 4 = 0.00
3. Inventory Turnover measures how fast the company turns over its inventory within a year.
World Acceptance Corp's Inventory Turnover for the quarter that ended in Sep. 2016 is calculated as
Inventory Turnover = Cost of Goods Sold / Average Inventory = 0 / 0 = N/A
4. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
World Acceptance Corp's Inventory to Revenue for the quarter that ended in Sep. 2016 is calculated as
Inventory to Revenue = Average Inventory / Revenue = 0 / 123.75 = 0.00
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
World Acceptance Corp Annual Data
Mar07 Mar08 Mar09 Mar10 Mar11 Mar12 Mar13 Mar14 Mar15 Mar16 ChangeInInventory 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
World Acceptance Corp Quarterly Data
Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 ChangeInInventory 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 857 | 3,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-50 | longest | en | 0.917587 |
https://www.primidi.com/monty_hall_problem | 1,638,547,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00194.warc.gz | 1,020,366,438 | 5,028 | # Monty Hall Problem
The Monty Hall problem is a probability puzzle loosely based on the American television game show Let's Make a Deal and named after the show's original host, Monty Hall. The problem, also called the Monty Hall paradox, is a veridical paradox because the result appears impossible but is demonstrably true. The Monty Hall problem, in its usual interpretation, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older Bertrand's box paradox.
The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a) (Selvin 1975b). One well known statement of the problem was published in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990a):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Vos Savant's response was that the contestant should switch to the other door. If the car is initially equally likely to be behind each door, a player who picks door 1 and doesn't switch has a 1 in 3 chance of winning the car while a player who picks door 1 and does switch has a 2 in 3 chance, because the host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car.
Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Decision scientist Andrew Vazsonyi described how Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until Vazsonyi showed him a computer simulation confirming the predicted result (Vazsonyi 1999).
The Monty Hall problem has attracted academic interest because the result is surprising and the problem is simple to formulate. Furthermore, variations of the Monty Hall problem are made by changing the implied assumptions, and the variations can have drastically different consequences. For example, if Monty only offered the contestant a chance to switch when the contestant had initially chosen the car, then the contestant should never switch. Variations of the Monty Hall problem are given below.
Read more about Monty Hall Problem: The Problem, Vos Savant and The Media Furor, Sources of Confusion, Criticism of The Simple Solutions, Strategic Solution By Dominance, Simulation, Variants, History
### Other articles related to "monty hall problem, problem, monty, monty hall":
Monty Hall Problem - History
... The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889 in his Calcul des probabilités (Barbeau ... As in the Monty Hall problem the intuitive answer is 1/2, but the probability is actually 2/3 ... The Three Prisoners problem, published in Martin Gardner's Mathematical Games column in Scientific American in 1959 (1959a, 1959b), is equivalent to the Monty ...
Let's Make A Deal - The Monty Hall Problem
... The Monty Hall Problem, also called the Monty Hall paradox, is a veridical paradox because the result appears impossible but is demonstrably true ... The Monty Hall problem, in its usual interpretation, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older ... The problem examines the counter-intuitive effect of switching one's choice of doors, one of which hides a "prize." The problem has been analyzed many times, in books, articles and online ...
### Famous quotes containing the words problem, monty and/or hall:
Will women find themselves in the same position they have always been? Or do we see liberation as solving the conditions of women in our society?... If we continue to shy away from this problem we will not be able to solve it after independence. But if we can say that our first priority is the emancipation of women, we will become free as members of an oppressed community.
Ruth Mompati (b. 1925)
Nobody expects the Spanish Inquisition.
Monty Python’s Flying Circus. first broadcast Sept. 22, 1970. Michael Palin, in Monty Python’s Flying Circus (BBC TV comedy series)
When Western people train the mind, the focus is generally on the left hemisphere of the cortex, which is the portion of the brain that is concerned with words and numbers. We enhance the logical, bounded, linear functions of the mind. In the East, exercises of this sort are for the purpose of getting in tune with the unconscious—to get rid of boundaries, not to create them.
—Edward T. Hall (b. 1914) | 1,138 | 5,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-49 | latest | en | 0.968376 |
https://siilats.com/docs/statistics/ex95q4.htm | 1,723,117,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00730.warc.gz | 418,482,447 | 10,135 | Custom Search
Q4 Take home exam a. a) Compute the OLS regression of the number of crimes on population and population density for all 51 observations. Test (using both the Goldfeld-Quandt test and the test used by Micro-Fit) whether the null hypothesis that the residuals of the estimated equation are homoscedastic can be accepted. Why might the two tests give different results? Dependent variable is CRIM93 51 observations used for estimation from 1 to 51 ******************************************************************************* Regressor Coefficient Standard Error T-Ratio[Prob] CONSTANT -37411.8 15369.2 -2.4342[.019] POP93 62.3154 2.0370 30.5915[.000] ******************************************************************************* R-Squared .95025 R-Bar-Squared .94923 S.E. of Regression 81486.7 F-stat. F( 1, 49) 935.8409[.000] Mean of Dependent Variable 277567.9 S.D. of Dependent Variable 361646.9 Residual Sum of Squares 3.25E+11 Equation Log-likelihood -648.0637 Akaike Info. Criterion -650.0637 Schwarz Bayesian Criterion -651.9955 * A:Serial Correlation*CHSQ( 1)= 3.2907[.070]*F( 1, 48)= 3.3108[.075]* * B:Functional Form *CHSQ( 1)= 5.6735[.017]*F( 1, 48)= 6.0081[.018]* * C:Normality *CHSQ( 2)= 139.6596[.000]* Not applicable * * D:Heteroscedasticity*CHSQ( 1)= 2.7690[.096]*F( 1, 49)= 2.8131[.100]* Microfit test: H0:errors have an increasing variance H1:errors have the same variance For X2 p value is 0.096 For F test it is 0.1 At 5% level accept H0, there is no heteroscedacity Goldfeld-Quant H0:errors have an increasing variance H1:errors have the same variance I let c=11 and order the population for the first 20 Regressor Coefficient Standard Error T-Ratio[Prob] CONSTANT 970.1958 9217.7 .10525[.917] POP93 46.2374 6.8773 6.7232[.000] R-Squared 0.71519 for the last 20 Regressor Coefficient Standard Error T-Ratio[Prob] CONSTANT -103894.2 49308.7 -2.1070[.049] POP93 66.8360 4.2202 15.8370[.000] ******************************************************************************* R-Squared 0.93304 lamda=RSS2/RSS1=(1-r2(2))/(1-r2(1))= 0.23510 df=(51-11-4)/2= 18 Fcrit= 2.2 Thus there is likely homoscedacity The second model is so much more restrictive. It depends on the c value used etc The first model uses much more complicated techniques to spot heterosced. Not just assuming that the error term variance depends on the square of ii b) Plot scatter graphs of the squared residuals from the estimated equation against population and population squared, Do these plots provide additional help to enable you to decide whether heteroscedasticity is present in your estimated equation? RES2 POP93 POP2 766719095.1 470 220900 595208004.6 576 331776 4820165130 579 335241 1118492260 598 357604 245872698.4 637 405769 779652136.5 698 487204 195252979.3 716 512656 639508012.1 841 707281 403465632.3 1000 1000000 124543776.3 1100 1210000 464.3909849 1124 1263376 1439589654 1166 1359556 561947.5822 1240 1537600 1346859740 1382 1909924 10918731.32 1613 2601769 1441629340 1616 2611456 889683797.1 1818 3305124 357718488.3 1860 3459600 8700467.328 2426 5885476 30898941.04 2535 6426225 109527336 2640 6969600 893060053.9 2821 7958041 542088929.5 3035 9211225 50427931.87 3233 10452289 302103622.8 3278 10745284 151343624.5 3564 12702096 649534469.6 3630 13176900 5674345602 3794 14394436 7185974738 3945 15563025 366251383.2 4181 17480761 4072383680 4290 18404100 2123390606 4524 20466576 975773446.1 4958 24581764 2472580407 5044 25441936 171531825.7 5094 25948836 487781730.7 5235 27405225 515791803.6 5259 27657081 4017875529 5706 32558436 1855981642 6018 36216324 9905582453 6473 41899729 1207709802 6902 47637604 8280285.049 6952 48330304 5627096725 7859 61763881 1313168202 9460 89491600 2426505575 11061 122345721 1175526129 11686 136562596 1015955886 12030 144720900 1078000806 13726 188403076 5595257545 18022 324792484 7417663339 18153 329531409 1161983188 31217 974501089 There are some values that are way out. Heteroscedacity is not present, just There are some very weird constituents with high (or low) crime rate. iii One should exclude the outliers. Ie countys with unusually high or low pop or crime Result is an OLS model that only applies to "normal" areas. Alternatively, there are some more complicated estimation techniques that take heteroscedacity into account. (GARCH). However, the resulting equation wont be BLUE. Finaly one should use population density instead of population to predict crime. Population density might be a better explanatory variable for crime iv Correlating crime rate and population density (pop/area) Dependent variable is CRIM93 51 observations used for estimation from 1 to 51 ******************************************************************************* Regressor Coefficient Standard Error T-Ratio[Prob] CONSTANT 299992.9 53053.6 5.6545[.000] POPDEN -189.9849 143.7568 -1.3216[.192] ******************************************************************************* R-Squared .034417 R-Bar-Squared .014711 S.E. of Regression 358976.9 F-stat. F( 1, 49) 1.7465[.192] Mean of Dependent Variable 277567.9 S.D. of Dependent Variable 361646.9 Residual Sum of Squares 6.31E+12 Equation Log-likelihood -723.6874 Akaike Info. Criterion -725.6874 Schwarz Bayesian Criterion -727.6192 * D:Heteroscedasticity*CHSQ( 1)= .59156[.442]*F( 1, 49)= .57503[.452]* doing the density gets rid of heteroscedacity, however, popdensity is not significant. But this model is restricted. Instead we can use the 2 variables, pop and area, separately: Dependent variable is CRIM93 51 observations used for estimation from 1 to 51 ******************************************************************************* Regressor Coefficient Standard Error T-Ratio[Prob] CONSTANT -48292.5 17359.5 -2.7819[.008] POP93 62.0446 2.0326 30.5253[.000] AREA .068207 .051916 1.3138[.195] ******************************************************************************* R-Squared .95197 R-Bar-Squared .94997 * D:Heteroscedasticity*CHSQ( 1)= 2.2226[.136]*F( 1, 49)= 2.2327[.142]* As seen, area is not signifficant, and heteroscedacity has increased, although it is not critical. This is the best i can do, i am afraid | 2,133 | 7,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.302404 |
https://web2.0calc.com/questions/35-5-5-70-11-6-3636363636 | 1,558,881,666,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259177.93/warc/CC-MAIN-20190526125236-20190526151236-00177.warc.gz | 689,879,904 | 5,966 | +0
# 35/5.5 = 70/11 = 6.3636363636
0
451
3
35/5.5 = 70/11 = 6.3636363636
how did 35/5.5 turn into 70/11?
Sep 13, 2017
#1
+865
0
It was multiplied by 2.
Sep 13, 2017
#2
+7612
+2
It was multiplied by 2 2
Sep 13, 2017
#3
0
## 2^12
2^12
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$
.
Sep 13, 2017 | 174 | 305 | {"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} | 3.5625 | 4 | CC-MAIN-2019-22 | latest | en | 0.879765 |
https://sachinchitta.org/ai/frequent-question-why-are-the-artificial-neural-networks-called-the-universal-approximators.html | 1,642,524,377,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00096.warc.gz | 577,104,050 | 19,748 | # Frequent question: Why are the artificial neural networks called the Universal Approximators?
Contents
The Universal Approximation Theorem tells us that Neural Networks has a kind of universality i.e. no matter what f(x) is, there is a network that can approximately approach the result and do the job! This result holds for any number of inputs and outputs.
## What is universal Approximators in machine learning?
The universal approximation theorem states that a feed-forward neural network with a single hidden layer containing a finite number of neurons can approximate any continuous function (provided some assumptions on the activation function are met).
## Which are universal Approximators?
In the mathematical theory of artificial neural networks, universal approximation theorems are results that establish the density of an algorithmically generated class of functions within a given function space of interest. … Most universal approximation theorems can be parsed into two classes.
## Why artificial neural network is called an intelligent network?
The term “Artificial neural network” refers to a biologically inspired sub-field of artificial intelligence modeled after the brain. An Artificial neural network is usually a computational network based on biological neural networks that construct the structure of the human brain.
## What is universality theorem?
Summing up, a more precise statement of the universality theorem is that neural networks with a single hidden layer can be used to approximate any continuous function to any desired precision.
## What is a universal function?
A universal function is a function that can, in some defined way, imitate all other functions. This occurs in several contexts: In computer science, a universal function is a computable function capable of calculating any other computable function. It is shown to exist by the utm theorem.
## Are polynomials universal approximators?
We actually do use other function approximators: in fact polynomials were the first provable universal approximators, this having been shown in 1885 via the so-called Stone–Weierstrass approximation theorem.
## Why is Universal Approximation Theorem used?
The Universal Approximation Theorem states that a neural network with 1 hidden layer can approximate any continuous function for inputs within a specific range. If the function jumps around or has large gaps, we won’t be able to approximate it.
## What is meant by artificial neural network?
An artificial neural network is an attempt to simulate the network of neurons that make up a human brain so that the computer will be able to learn things and make decisions in a humanlike manner. ANNs are created by programming regular computers to behave as though they are interconnected brain cells.
## Is Perceptron a universal approximator?
Although the multilayer perceptron (MLP) can approximate any functions [1, 2], traditional SNP is not universal approximator. MLP can learn through the error backpropagation algorithm (EBP), whereby the error of output units is propagated back to adjust the connecting weights within the network.
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## Why artificial neural network is used?
Artificial Neural Network(ANN) uses the processing of the brain as a basis to develop algorithms that can be used to model complex patterns and prediction problems. … In our brain, there are billions of cells called neurons, which processes information in the form of electric signals.
## What is artificial neural network in machine learning?
Artificial Neural networks (ANN) or neural networks are computational algorithms. It intended to simulate the behavior of biological systems composed of “neurons”. ANNs are computational models inspired by an animal’s central nervous systems. It is capable of machine learning as well as pattern recognition.
## How is the human brain different from the artificial neuron network models?
Answer: Unlike humans, artificial neural networks are fed with massive amount of data to learn. While artificial neural nets were initially designed to function like biological neural networks, the neural activity in our brains is far more complex than might be suggested by simply studying artificial neurons.
## What is the output of neural network?
Computing neural network output occurs in three phases. The first phase is to deal with the raw input values. The second phase is to compute the values for the hidden-layer nodes. The third phase is to compute the values for the output-layer nodes. … Each hidden-layer node is computed independently.
## What are the caveats of neural networks as universal Approximator?
No, there are no specific functions that a neural network cannot approximate. However, there are some important caveats: Neural networks do not encode the actual functions, only numeric approximations. This means there are practical limits on the ranges of inputs for which you can achieve a good approximation.
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## Why neural network is known as best function Approximator?
Artificial neural networks learn to approximate a function. … We say “approximate” because although we suspect such a mapping function exists, we don’t know anything about it. The true function that maps inputs to outputs is unknown and is often referred to as the target function.
Categories AI | 1,027 | 5,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-05 | latest | en | 0.898437 |
https://ncna19.kattis.com/problems/weirdflecksbutok | 1,656,577,024,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00433.warc.gz | 478,393,765 | 6,868 | Hide
# Weird Flecks, But OK
An artist who wanted to create an installation where his works appeared to be floating in midair has cast a large cube of clear acrylic to serve as a base. Unfortunately, during the casting, some small flecks of dirt got into the mix, and now appear as a cluster of pinpoint flaws in the otherwise clear cube.
He wants to drill out the portion of the cube containing the flaws so that he can plug the removed volume with new, clear acrylic. He would prefer to do this in one drilling step. For stability’s sake, the drill must enter the cube only once, perpendicular to one of its faces. The cube’s faces are parallel to the coordinate axes.
Given the $(x,y,z)$ positions of the flaws, and treating the size of the flaws as negligible, what is the smallest diameter drill bit that can be used to remove the flaws in one operation??
## Input
The first line of input contains an integer $N$ denoting the number of flaws. $3 \leq N \leq 5\, 000$
This is followed by $N$ lines of input, each containing three real numbers in the range $-1\, 000.0\ldots 1\, 000.0$, denoting the $(x,y,z)$ coordinates of a single flaw. Each number contains at most $6$ digits following a decimal point. The decimal point may be omitted if all succeeding digits are zero.
## Output
Print the diameter of the smallest drill bit that would remove all the flaws.
The answer is considered correct if the absolute or relative error is less than $10^{-4}$
Sample Input 1 Sample Output 1
3
1.0 0.0 1.4
-1.0 0.0 -1.4
0.0 1.0 -0.2
2.0000000000
Sample Input 2 Sample Output 2
5
1.4 1.0 0.0
-0.4 -1.0 0.0
-0.1 -0.25 -0.5
-1.2 0.0 0.9
0.2 0.5 0.5
2.0000000000
Sample Input 3 Sample Output 3
8
435.249 -494.71 -539.356
455.823 -507.454 -539.257
423.394 -520.682 -538.858
446.507 -501.953 -539.37
434.266 -503.664 -560.631
445.059 -549.71 -537.501
449.65 -506.637 -513.778
456.05 -499.715 -561.329
49.9998293198
CPU Time limit 3 seconds
Memory limit 1024 MB | 605 | 1,965 | {"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} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.841247 |
https://www.coursehero.com/file/6060439/Lect6/ | 1,490,828,632,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191405.12/warc/CC-MAIN-20170322212951-00177-ip-10-233-31-227.ec2.internal.warc.gz | 875,344,389 | 52,775 | # Lect6 - Conditional Probability Multiplication Rule Independence of Events Independent versus Mutually Exclusive The Birthday Prob Outline
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Conditional Probability Multiplication Rule l e m Outline Conditional Probability Multiplication Rule Independence of Events “Independent” versus “Mutually Exclusive” The Birthday Problem Simpson’s Paradox 1 / 21 Xinghua Zheng Probability II
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Conditional Probability Multiplication Rule l e m Example, Stein’s class ± What is the probability for a randomly selected student from Professor Stein’s class this year to receive an A? In the last 5 years Professor Stein has awarded 190 A’s out of 1000 students. Can use long run relative frequency : the probability can be estimated as 190 / 1000 = 0 . 19 ± Amy studies 10 hours or more every week for professor Stein’s course. Amy is really interested in knowing “What’s the probability for a student who studies 10 hours or more to be awarded an A?" Conditional probability 2 / 21 Xinghua Zheng Probability II
Conditional Probability Multiplication Rule l e m Conditional Probability ± The probability of an event A, given that the event B has occurred, is called the conditional probability of A given B Denoted by P ( A | B ) ± Mathematically, P ( A | B ) := P ( A B ) P ( B ) , Assuming P ( B ) > 0 3 / 21 Xinghua Zheng Probability II
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Conditional Probability Multiplication Rule l e m Interpretation P ( A | B ) = P ( A B ) / P ( B ) Geometrically, = the ratio of the area of the shaded part to that of the right disk P ( B | A ) = P ( A B ) / P ( A ) 4 / 21 Xinghua Zheng Probability II
Conditional Probability Multiplication Rule l e m Interpretation, ctd The conditional probability P ( A | B ) : Restrict sample space to just event B P ( A | B ) measures the chance of event A occurring in this new sample space In other words, if B occurs, then what is the chance of A occurring 5 / 21 Xinghua Zheng Probability II
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Conditional Probability Multiplication Rule l e m Example, Stein’s class, ctd ± What’s the probability for a randomly selected student from Prof. Stein’s class who studies 10 hours or more to be awarded an A? Past data shows that 200 students worked 10 hours or more for Stein’s course, and 120 of them were awarded A. Define events A={to be awarded an A}, B = {studies 10 hours or more} To find the answer to our question, we need to find out 6 / 21 Xinghua Zheng Probability II
Conditional Probability Multiplication Rule l e m Example, Stein’s class, ctd P ( A | B ) = P ( A B ) / P ( B ) A B ={studies 10 hours or more and is to be awarded A} By the long-run relative frequency, P ( A B ) = 120 / 1000 = 0 . 12, P ( B ) = 200 / 1000 = 0 . 2 Hence
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## This note was uploaded on 12/28/2010 for the course ISOM ISOM111 taught by Professor Anthonychan during the Fall '09 term at HKUST.
### Page1 / 21
Lect6 - Conditional Probability Multiplication Rule Independence of Events Independent versus Mutually Exclusive The Birthday Prob Outline
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# How do you graph $y=-\dfrac{3}{5}x+1$ ?
Last updated date: 20th Jun 2024
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Hint: For answering this question we need to draw the graph of $y=-\dfrac{3}{5}x+1$ . For drawing the graph of the straight line we will find the intercepts of the given equation of the straight line and plot them and join them and extend the line formed. The intercepts are the respective points when the other variable is zero.
Complete step-by-step solution:
Now considering from the question we have to draw the graph of $y=-\dfrac{3}{5}x+1$ .
From the basics of the concept we know that for drawing the graph of the straight line we need to find the intercepts of the given equation of the straight line and plot them and join them and extend the line formed.
The intercepts are the respective points when the other variable is zero.
The $\text{x-intercept}$ is given by substituting $y=0$ in the given equation of the straight line.
After substituting we will have
\begin{align} & \Rightarrow 0=-\dfrac{3}{5}x+1 \\ & \Rightarrow -1=-\dfrac{3}{5}x \\ & \Rightarrow x=\dfrac{5}{3} \\ \end{align}
The $\text{y-intercept}$ is given by substituting $x=0$ in the given equation of the straight line.
After substituting we will have
\begin{align} & \Rightarrow y=-\dfrac{3}{5}\left( 0 \right)+1 \\ & \Rightarrow y=1 \\ \end{align}
So now we have two points lying on the straight line $\left( 0,1 \right)$ and $\left( \dfrac{5}{3},0 \right)$ .
Now we will plot these two points and join them and extend them to form a straight line. The graph of the straight line is shown below.
Note: For answering this question we need to draw the graph of the given equation of the straight line so we need to be careful while plotting the points and extending the line. We should be careful while performing the calculations. The slope of this given straight line $y=-\dfrac{3}{5}x+1$ is $\dfrac{-3}{5}$ as we know that the slope intercept form is given as $y=mx+c$ where $m$ is the slope. | 578 | 2,088 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2024-26 | latest | en | 0.861579 |
http://www.typologycentral.com/forums/myers-briggs-and-jungian-cognitive-functions/18306-visualizing-functions-type-2.html | 1,477,110,380,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718423.65/warc/CC-MAIN-20161020183838-00232-ip-10-171-6-4.ec2.internal.warc.gz | 768,020,487 | 14,857 | # Thread: Visualizing the Functions of each Type
1. Now you're just making my head spin.
2. Originally Posted by TheChosenOne
Now you're just making my head spin.
MBTI does that if you get beyond the basics.
Without Ti or Ni, you can't get very far into the guts of MBTI, I'm afraid.
3. It's pretty simple conditional program, really.
Where J is judgment function and P is perception function,
For reverse-engineering type from a dominant function:
Code:
```If J, then:
if Ji, TYPE = IxJP.
else if Je, TYPE = ExJJ
Else if P:
if Pi, TYPE = IPxJ.
else if Pe, TYPE = EPxP.```
Where 1, 2, 3, and 4 refer to dominant, auxiliary, tertiary, and inferior functions, and (a) and (b) refer to opposite functions (t/f, s/n),
For determining remaining functions from a dominant function:
Code:
```If 1 = J(a/b)(i/e), then:
4 = J(b/a)(e/i) and
[2, 3] = [P(a/b)(e/i), P(b/a)(i/e)].
Else if 1 = P(a/b)(i/e), then:
4 = P(b/a)(e/i) and
[2, 3] = [J(a/b)(e/i), J(b/a)(i/e)].```
And so on. It's all very yin-yang.
4. So, what does visualizing the "functions" of a hypothetical "type" get for you?
I mean, other than a pretty diagram.
5. The theory is that extraverts always have an extraverted dominant function (vice-versa for introverts); extraverted judgers have dominant judging, while extraverted perceivers have dominant perception (vice-versa for introverts).
The fourth function is the opposite of the first; the third is the opposite of the second. If the first function is perception, the second will be judging (and vice-versa). If the first function is extraverted, the second will be introverted (and vice-versa).
As for what the visualization gives you, it's a tool to intuitively clarify the abstract patterns behind the logic.
As for what the visualization gives you, it's a tool to intuitively clarify the abstract patterns behind the logic.
Ah yes, this. Ni + Te = Lots of pretty colors to look at so I can understand the rest of the jibberish! No hard logic for me like Aderack posted. I looked at the colors and essentially said 'Ohh hey! I get it! I can tell you the functions of the rest of the types now!' Once I get the pattern memorized to where I can call on it at need that is. I'm still in the rememberization phase... But now it makes SENSE, which is a step farther than I was before.
7. Originally Posted by Misty_Mountain_Rose
But now it makes SENSE, which is a step farther than I was before.
Okaay..well, congrats I guess. It still doesn't make sense to me.
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• | 708 | 2,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2016-44 | longest | en | 0.889831 |
http://spmath816.blogspot.com/2008/03/nathaniels-third-growing-post.html | 1,490,350,187,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187792.74/warc/CC-MAIN-20170322212947-00423-ip-10-233-31-227.ec2.internal.warc.gz | 343,976,403 | 13,430 | ## Friday, March 14, 2008
### Nathaniel's Pythagoras Growing Post
Third Growing Post!
What Is A Pythagorean Triple?
A Pythagorean Triple is three square numbers in which when two square numbers are added they equal another square number. The formula being Asquared + Bsquared = Csquared. Asquared and Bsquared are the legs of the trianged. Csquared is the hypotenuse.
For example 3squared + 4squared = 5squared. Which equals 9 + 16 = 25
Another Pythagorean Triple is 5squared + 12squared = 13squared. Which equals 25 + 144 = 169
How to find a missing leg/hypotenuse
BubbleShare: Share photos - Find great Clip Art Images.
How to solve a word problem!
Here is a word problem that I will use to help me explain how to solve a word problem.
You picked up a ground ball at first base, and you see the other team's player running towards third base. How far do you have to throw the ball to get it from first base to third base, and throw the runner out.
There was also a picture
The key thing you look for is numbers! They usually are very important. The first thing you should underline is First Base because thats where you are. And then third base you underline because thats where the other person is going to. And the question, you have to find out how far you have to throw the ball. The picture shows that from 1st base to 2nd base is 90 feet, and 2nd base to third base is 90 feet. If you look at it, it is a right triangle! The legs being 90 feet and the hypotenuse being first base to third base! All you have to do is solve for the hypotenuse.
a = 90 feet
b = 90 feet
c = ?
a2 + b2 = c2
90squared + 90squared = c2
8100 + 8100 = c2
16200 = c2
square root of 16200 = c
127.28 = c
My Own Word Problem!
Stretch Armstrong sees a little boy crying because his ball got stuck on the roof, so he decides to help get the ball down. He is 10 feet from the wall, and the roof is 40 feet high, how far does Stretch Armstrong have to stretch his arm to reach the ball if his arm is 20 feet high. | 532 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-13 | longest | en | 0.961935 |
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# 35 374 006 380 b calculate the ph if 0050 mol
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Unformatted text preview: [HA]) = -log(1.8×10-4) + log(0.40 / 0.35) = 3.74 + (0.06) = 3.80 (b) Calculate the pH if 0.050 mol of NaOH is added to 1.0 L of of the buffer solution above. HA(aq) H + (aq) + A - (aq) 0.350 0.350 M - 0.050M 0M 0.400M 0.050M 0.400M + 0.050M 0.3000.300- x x 0.450 + x pH = pK a + log[A - ] [HA] -4 = − log(1.8 ×10 ) + log(0.450 0.300) = 3.74 + 0.176 = 3.92 How How many moles of sodium acetate must be added to 1.50 L of 0.100 M acetic acid to form a pH = 5.00 buffer? Ka for acetic acid = 1.8×10-5 pKa = 4.74 pK pH = pKa + log ([A-]/[HA]) 5.00 = 4.74 + log (acetate/0.100) 5.00 - 4.74 = log (acetate/0.100) 0.26 = log (acetate) – (-1) 0.26-1 = -0.74 = log(acetate) Acetate = 10-0.74 = 0.18M Moles = Molarity × Liter = 0.18 M × 1.50 L = 0.27 moles acetate 0.27 Specify Specify the chemicals and proportions you would choose chemicals proportions from among these...
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## This document was uploaded on 02/26/2014.
Ask a homework question - tutors are online | 493 | 1,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-22 | longest | en | 0.784227 |
https://mathhelpforum.com/tags/possibilities/ | 1,575,748,597,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540501887.27/warc/CC-MAIN-20191207183439-20191207211439-00000.warc.gz | 455,814,880 | 17,817 | # possibilities
1. ### math of possibilities, help needed.
Hi I need a math equation to get me all the possible order's I can make for certain numbers combination. For example : If I have these numbers "1234" it's have 24 possible order's. And if I have these numbers "1233" it's have 9 possible order's. And if I have these numbers "1122" it's have...
2. ### How many possibilities are there for dividing a string to substrings, of any length?
Hi. I just can wrap my head around this. Let S be a string of length n: S=\sigma_1\sigma_2... \sigma_n how many possibilities are there for dividing it to substrings? for example: S_1=\sigma_1\sigma_2\hspace{12pt} S_2=\sigma_3...\sigma_i\hspace{12pt} S_3=\sigma_{i+1}...\sigma_n is one...
3. ### Coin Purchase Possibilities
A puzzle problem involving coinageA shopkeeper and her customer each have an unlimited number of coins. However, they are of only two denominations – 3¢ and 5¢. 1. What amount purchases are not possible using only these two denominations of coinage, if the shopkeeper is allowed to give change...
4. ### Calculating the amount of possibilities
Hi there, I need some help to resolve a problem... Let's say that I have 30 options spread in 3 blocks where a person need to choose 10 options according to these rules: The person can only choose 3 options in the last block The person can only choose 3 options in the second block There is...
5. ### possibilities story problems
A campus coffee shop offers 15 types of hot beverages (coffee tea etc), 8 varieties of sandwich 12 types of cold beverage (soda energy drinks juice) and 11 different types of bagel. How many different meals could be purchased if you wanted either a hot beverage and a bagel , or cold a...
6. ### possibilities story problems
n the world series, The American league Champion plays The National League Champion. How many world series match-ups are possible if there are 14 teams in the American league and 16 teams in the national league? A travel app will suggest 3 hotels and 5 restaurants at random for any city it...
7. ### Possibilities in a given set
There are 3 men A,B and C. They have 4 sons in total. What are the total possibilities of the order of no. of sons they have. How do we find it. Plz help I am totally new to this topic.
8. ### Calculating Possibilities and Combinations
Hi everyone, I have a situation which I need help calculating. I have 15 selections to choose from and each selection has a 50/50 ratio of being correct. I want to figure out how many combinations I can choose for each selection in order to cover all possibilities. On each of the 15...
9. ### Number of possibilities?
A class of 20 is selecting a new president, with 5 candidates. How many different ways can the tally come out?
10. ### Maximizing over a continuum of possibilities
Let x be a random variable with density function f and [a,b] be the possible set of values of x. Let w(x) be a value associated with each value of x. I.e. w() is not known, and suppose that U(w(x)) is a C^{\infty} function. Let H(w(x)) = \int_{a}^b f(x)U(w(x))dx Find the derivative of H...
11. ### Permute number of on/off possibilities for a 6x6 buss matrix
I'm an electroacoustic musician using a single microphone input into a laptop Logic Pro audio mixing environment with 6 auxiliary channels. Each auxiliary has unique fx/eq and I can send its audio to any of the other 5 auxiliary channels via their respective input busses: Auxiliary 1 -> Buss 2...
12. ### Microeconomics help: Production Possibilities Frontier
Might seem quite simple, so don't judge lol (a) Draw country A’s Production Possibilities Frontier (PPF) by using Excel (position food on the horizontal axle). (b) What is happening to the opportunity cost of food in terms of clothing when the production of food is increasing? (c) If the...
13. ### How many different possibilities?
A professor writes 40 discreate mathematics true/false questions. of the statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible.
14. ### possibilities
Imagine a city grid that is like a rectangular coordinate system. You are currently at (0,0) and you want to get to corner (3,4). If a path consists of a series of eastward and northward moves, how many shortest paths are possible assuming that all streets and avenues are of the same length...
15. ### theoretical possibilities
I had to use a tree diagram to work out all the possible outcomes (arrangements of boys and girls) in a family of 4 children. I worked out that there are 14 possibilities. From here I am stuck, I have to calculate the theoretical possibility of obtaining 2 girls and 2 boys as a fraction, decimal...
16. ### finding possibilities when guessing answers
Multiple choice questions each have 5 possible answers, one is correct. Assume that you guess the answers to 3 such questions. Using multiplication rule to find the probability that the first 2 guesses are wrong and the 3rd is correct. Find, P(WWC), where C is correct and W is wrong...
17. ### How Many Possibilities?
Suppose five cards are drawn from a standard (52 card) deck of cards. From the 5 cards chosen, 3 are red and 2 are black. How many possibilities are there for this hand?
18. ### Possibilities? (I don't remember the exact word)
Hi! First time poster. So I'm doing some homework, due in six hours. I'm almost done, but there's a couple of questions I'm not too sure of: one has to do with possibilities, the other not so much. 1. The easier one: What is the coefficient of X^11 in (3 + X)^15? I'm guessing it's 15!/11!4! =...
19. ### number of possibilities
How many three-digit numbers can be made if the first digit can be 1,2,3 or 4, the second digit can be, 1,2,3,4,5 or 6 and the third digit can be 0,1 or 2?. Really got me stumped?? Please help
20. ### number of possibilities
how many 4 digit numbers can you form with (1,2,3,4,5,6,7) ? (No repeated numbers) The answer is 840 (7P4) BUT How many of these are odd numbers? I know that there are 4 odd numbers... | 1,498 | 6,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-51 | longest | en | 0.894854 |
http://www.idealessaywriters.com/prepare-a-loan-amortization-schedule-based-on-monthly-payments-for-the-1600000-if-royal-dutch-shipping-can-pay-10-down-on-a-loan-for-1600000-and-can-get-a-loan-for-6-interest-for-10-years-do-no/ | 1,526,809,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00587.warc.gz | 429,551,396 | 14,008 | ## Prepare a loan amortization schedule based on monthly payments for the \$1,600,000 if Royal Dutch Shipping can pay 10% down on a loan for \$1,600,000 and can get a loan for 6% interest for 10 years (do not include this in your Net Present Value computations.
Assignment: Based on the information provided below, compute the Net Present Value of the project (CO 3). A Net Present Value Template is Attached. (Hint: Don’t forget to update the discount rate to the amount required for this project and add your cash flow numbers.)
Royal Dutch Shipping is planning on Investing \$1,600,000 to buy a freighter. Prepare a net present value analysis based on the assumption that the freighter will be sold for 10% of its cost at the end of the year 5. Assume a 10% cost of capital. Annual operating cash flows for the project are:
Year 1: \$380,000
Year 2: \$390,000
Year 3: \$400,000
Year 4: \$410,000
Year 5: \$420,000
Prepare a loan amortization schedule based on monthly payments for the \$1,600,000 if Royal Dutch Shipping can pay 10% down on a loan for \$1,600,000 and can get a loan for 6% interest for 10 years (do not include this in your Net Present Value computations. This is a separate issue. (CO 3). (Hint: WWW.bank rate.com/calculators/mortgages/amortization-calculator.asp )
Instructions: Prepare the homework in Excel template provided. Save your assignment file as ‘LastnameFirstinitial-FINC300-3″, and submit by 11:55 pm ET, Day 7 (Sunday). | 388 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-22 | latest | en | 0.874814 |
https://socratic.org/questions/how-do-you-solve-2x-4-10x-2-0 | 1,714,043,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00887.warc.gz | 478,140,645 | 6,107 | # How do you solve 2x^4 - 10x^2 = 0?
Oct 31, 2015
$x = + \sqrt{5}$
$x = - \sqrt{5}$
#### Explanation:
Here,
It is easy to solve for quadratic function but difficult if the power rises up to more than 3,
in this case you let ${x}^{2} = a$
Now replace the value of ${x}^{2}$ with $a$
The new equation is,
$2 {a}^{2} - 10 a$ =0
Take 2 common for both the terms,
2 ( ${a}^{2} - 5 a$) =0
${a}^{2} - 5 a = 0$
${a}^{2} = 5 a$
Cancel one a ,
SO,
$a = 5$
Now replace this value of a in ${x}^{2}$
SO,
${x}^{2}$ = 5
$x = \pm \sqrt{5}$
So, either, $x = + \sqrt{5}$
or, $x = - \sqrt{5}$ | 246 | 592 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "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} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.680667 |
https://www.oreilly.com/library/view/digital-circuit-boards/9781118278116/c03_level1_13.xhtml | 1,576,035,364,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529745.80/warc/CC-MAIN-20191211021635-20191211045635-00161.warc.gz | 832,260,952 | 14,154 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
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# 3.13 Radiation from Board Edges
The energy stored between conducting planes is in parallel with the energy stored in decoupling capacitors. When a demand for energy is made, waves propagate into both the decoupling capacitors and into the space between the planes. Reflections return energy to the point of demand. This wave action was discussed in Section 2.4. Consider waves that travel between ground and power planes. When a wave reaches the edge of the board, the open circuit reflects the wave and doubles the voltage. Wave fronts reach the board edge at different times. Waves double in amplitude as they reflect. The wave amplitude is inversely proportional to the distance from the point of demand. If the wave amplitude is 1 mV at the board edge and the planes have a 2 mm spacing, then the E field strength is 2 mV/2 mm or 1 V/m. This E field level would be measurable.
N.B.
Vias that are used to take energy from the ground/power planes should be located away from board edges.
In multilayer boards, each ground/power plane stores field energy. Energy is also stored in parallel decoupling capacitors. Every logic transition requires field energy. This energy comes from the nearest sources, which includes decoupling capacitors, ground/power plane capacitances, and connected logic. When logic crosses layers, it carries field energy to a new region. This energy must eventually be radiated or dissipated. Field energy cannot travel through conductors. It can only travel through holes ...
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No credit card required | 392 | 1,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.914734 |
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