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Sellier & Bellot The Ballistic Model Sellier & Bellot Sellier & Bellot, in collaboration with a leading Czech ballistics expert, has developed an advanced ballistics model to calculate the performance and trajectory of ammunition. The basic element of ballistic calculations is the ballistic coefficient (BC), which is used to evaluate the missile in terms of external ballistics and flight characteristics in the real atmosphere (ATM). BC can also be described as the ability of the missile to penetrate the ATM. It is true that a missile with a higher BC penetrates ATM better and vice versa that a missile with a lower BC is more inhibited by ATM. Sellier & Bellot calculates the BC of a labored bullet by accurately measuring the velocities of a sample of 10 bullets on a 100 m range. The measured bullet velocity values, together with the actual temperature, humidity and absolute air pressure, are used to calculate the published BC of the bullet converted to standard ICAO air conditions (temperature 15°C, relative humidity 0% and absolute pressure 1.013.25 hPa). This provides the ability to compare individual missiles with each other. For the need of an accurate long range firing model, a model based on the equations of motion of the mass point under the influence of gravitational acceleration and environmental drag force is used. The algorithm for calculating the ballistic elements of the projectile is performed by numerical integration of the general equations of motion of the mass point using a modified Euler method. Overview of the use of Gx resistive functions: Type of projectile Resistance function SP, HP, FMJ G1 FMJ, FMJBT, HPBT, SP G7 An advanced and physically accurate ballistic model is the basis for development of ammunition, helps you select the right cartridge and shoot perfectly. Definition of the ballistic coefficient of a missile in imperial units: Where BC is the ballistic coefficient of the projectile (lb/in^2), m the mass of the projectile (grs), d the diameter of the projectile (in), i the dimensionless coefficient of the shape of the Where BC is the ballistic coefficient of the projectile (lb/in^2), SD is the cross-sectional load of the projectile (lb/in^2), i is the dimensionless shape factor of the projectile. Where SD is the cross-sectional load in (lb/in^2), m is the mass of the projectile in (grs), d is the diameter of the projectile in (in). A 0.308 in HPBT bullet weighing 175 grains has a cross-sectional load of SD and BC: To simplify calculations and evaluation of projectiles, standard resistance functions for precisely defined projectile shapes, the so-called G resistance functions, have been introduced. Ballistic coefficient values according to the standard G1 and G7 drag functions are available for most projectiles.	{"url":"https://www.sellier-bellot.cz/en/products/ballistic-coefficient-calculation/","timestamp":"2024-11-08T22:08:16Z","content_type":"text/html","content_length":"55461","record_id":"<urn:uuid:bcc6acdc-8184-49cf-9d55-016a3285e332>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00685.warc.gz"}
How Long is the Long Term? Everyone always talks about investing (and planning) for the long term. But they’re usually vague about what the “long term” actually means. Does that mean 5 years? 10 years? More? (Spoiler: the answer is definitely more). And frankly, talking about how you need to focus on the long term can feel like a bit of a dodge when the markets haven’t been cooperating. Let’s dig in and think about what “long term” actually means – both in theory and practice. Long Term in Theory First, let’s discuss theory. Why does investing for the long term matter? It’s not like the market keeps score and will toss us a couple of good years if we have had a run of bad years, or will keep us honest by throwing a bad year into a bull run. The market is way too messy for that. Financial returns are nearly random. A good way to see this is to just look at the returns. For instance, below, we have the monthly returns of the S&P 500 Index from January 1926 to December 2022. We can do all sorts of statistical analysis on these returns (and we have), but it comes to the same place as just eyeballing the data – each return is largely random. It’s effectively impossible to predict what will happen next. For illustration purposes only. Monthly returns of the S&P 500 Index from January 1926 to December 2022. Indices are not available for direct investment. Past performance is no guarantee of future A common way of describing this is to say that the market moves in a Random Walk. Each individual step that the market takes (the next return) is basically random. But not completely random. While we may not be able to predict each individual step, there is a direction that this walk is generally going (and we wouldn’t bother investing if this wasn’t the case). But it’s incredibly hard to identify this trend over shorter time periods. In fact, I’ve written an article about how you can’t tell short term stock returns from the flip of a coin. In essence, the “long term” is however long it takes for these trends in the Random Walk to assert themselves. Keeping Risk (and Return) Honest These trends are all of the different risk premia on offer in the market. For instance, stocks are riskier than bonds, so they tend to have higher returns than bonds. If this were not true no one would buy stocks. There’d be no reason to. Why would you buy something risky (stocks) if you could get the same return with a safer asset (bonds)? These risk premia are the market’s way of getting you to buy risky assets. And the riskier an asset is (at least for certain types of risk), the bigger the risk premium, and the stronger the trend in the Random Walk. But it’s called a risk premium for a reason. It’s risky. There are no guarantees. There are absolutely going to be periods – even relatively long periods – when the Random Walk is not our friend and the randomness overwhelms the trend. For risk to actually be risky, it can’t always work out. That said, the longer you stick around, the more likely it is that that trend will assert itself. But you’ll never get to 100% certainty – there’s always a chance that things won’t work out no matter how long you stay invested. This is part of the bargain we make when we invest in risky assets. To get the higher returns “promised” by the markets, we need to accept the risk. Putting it Into Practice But this is all very abstract. It’s important to think about (and make peace with) but it doesn’t really help us put numbers around our primary questions – how long is the long term? So let’s do that. But before we do, let’s think about what we’re looking for. Just like most big fundamental questions, there are no definitive answers, and no bright lines. We’re not going to be able to say that 24 years is “long term,” but 23 years isn’t. We’re dealing with gradations here – different levels of confidence. But putting numbers around this can help make the question more concrete. As I said, there are never any guarantees with investing (I’m going to beat this point into the ground) but we can see the effects of a longer time horizon very clearly in the data. To do this, we’ll keep things simple, and just focus on the S&P 500 Index. But these same principles apply everywhere in the financial markets. This is a story about the fundamental relationship between risk and return – and what that means. For this analysis we’re going to be looking at rolling annualized returns of the S&P 500 Index over differing lengths of time from January 1926 through December 2022. What we’re asking is if you invested at a random point in time during this period, what would your return have been? A good place to start is by looking at the range of returns. What were the best and worst returns for each holding period? Let’s start by looking at the data. Holding Period Best Return Worst Return Total Range 1 Year 162.88% -67.57% 230.44% 3 Year 43.35% -42.35% 85.70% 5 Year 36.12% -17.36% 53.48% 10 Year 21.43% -4.95% 26.38% 15 Year 19.69% -0.41% 20.10% 20 Year 18.26% 1.89% 16.37% 30 Year 14.78% 7.80% 6.98% 60 Year 13.32% 9.03% 4.28% For illustration purposes only. Data calculated from rolling returns of S&P 500 Index from January 1926 to December 2022 using different holding periods. Returns of holding periods longer than 1 year are annualized. Indices are not available for direct investment. Past performance is no guarantee of future returns. The results here are fairly obvious. The longer that you are invested, the narrower the spread is between the extremes. But there are actually two things going on here that are worth talking about. The first is what we want to focus on – the effects of your time horizon on your investment returns. But the second is a little less obvious. We have fewer independent observations with the longer time periods. For instance, with the 1 year observations, we have nearly 100 separate 1 year periods (each year from 1926 – 2022), but we only have about one and a half independent 60 year periods. This means that the longer time periods will naturally have more overlapping data. This will tend to reduce the variance of the returns for these longer time periods relative to the shorter periods that have relatively less overlap in the data. I don’t want to overemphasize the point, but it is a limitation that we need to consider as we look through everything. But even with this in mind, the difference between the range of the 1 year returns with the longer term returns is astounding. The range between the best and worst 1 year returns was 230%. The range for the 60 year returns was only 4.3%. And not only that, the worst 1 year return immediately preceded the best 1 year return. Short time periods will whip your portfolio around, and really emphasize just how random the Random Walk can be. And with the longer term returns, they contain all of the craziness of the shorter periods. But over time things have balanced out. The trend in the Random Walk has time to show itself. Looking at the Distributions This is interesting, but it’s purely focused on the edges of the distributions. And we don’t want to do our planning based on outliers. So let’s look at the actual distribution of returns – how often each holding period had a particular return. For illustration purposes only. Data calculated from annualized rolling returns of S&P 500 Index from January 1926 to December 2022 using different holding periods. Indices are not available for direct investment. Past performance is no guarantee of future returns. Just like with the range of returns, this data is relatively well behaved. The longer time periods have a much tighter distribution than the shorter time periods. In fact, if we focus in on the 1 year data – even when we zoom in (take a look at the differences in the scale), it’s not that much a curve. For illustration purposes only. Data calculated from annualized rolling returns of S&P 500 Index from January 1926 to December 2022 using different holding periods. Indices are not available for direct investment. Past performance is no guarantee of future returns. It’s basically all over the place. There’s a little bit of a cluster around the average, but not much of one. When we compare this to the 20 or 30 year observations, we can see the difference very For illustration purposes only. Data calculated from annualized rolling returns of S&P 500 Index from January 1926 to December 2022 using different holding periods. Indices are not available for direct investment. Past performance is no guarantee of future returns. With both of these, there’s a very clear curve. They are mostly centered around the average return, but there’s still a little bit of dispersion. With the 20 year time periods, there was a little bit more than a 15% chance that you would have an annualized return of less than 4% and a little bit more than a 20% chance that your annualized return would be more than 14%. As a point of comparison, with the 1 year periods, you would have had a 30% chance of a return below 4% and a nearly 48% chance of your return being more than 14%. And when we look at the 60 year period in isolation, it’s basically a spike. For illustration purposes only. Data calculated from annualized rolling returns of S&P 500 Index from January 1926 to December 2022 using different holding periods. Indices are not available for direct investment. Past performance is no guarantee of future returns. There’s technically a distribution here, but almost 70% of the observations were within a 2 percentage point range. The longer you give your investments to work, the more confident you can be that you’ll be able to harvest the risk premia that we talked about – those fundamental risk and return relationships in the market. There are still no guarantees, but if you wait long enough the randomness in the Random Walk tends to cancel itself out. What Can You Do With This? But we can get even more practical. Generally, when people ask what the long term means in the context of investing, they are asking one of two things: how long do they need to be invested to be “sure” they don’t lose money, or that they’ll do better than some alternative investment. So let’s look at those questions. We’ll start with the simpler of the two – how long you would need to have been invested in the S&P 500 Index to be “sure” that you wouldn’t lose money (we’ll ignore inflation here for simplicity’s sake). I’ve got the quotes around sure here because, again, there are no guarantees in investing. The most confident that we can be is to say that during the time period we are looking at there were no periods of a specific length where the S&P 500 lost money. That doesn’t mean it can’t happen in the future. And even ignoring absolute certainty, different people are comfortable with different levels of risk. What it takes for one person to be “sure” is going to be different than what it would take for another person. This is all about the varying gradations. So after all that, let’s look at the data. Holding Period % of Observations Greater than 0 1 Year 75.4% 3 Year 84.4% 5 Year 88.2% 10 Year 94.9% 15 Year 99.8% 20 Year 100.0% 30 Year 100.0% 60 Year 100.0% For illustration purposes only. Data calculated from annualized rolling returns of S&P 500 Index from January 1926 to December 2022 using different holding periods. Indices are not available for direct investment. Past performance is no guarantee of future returns. There are two things you probably noticed immediately: • The S&P 500 Index does pretty well if you just want to avoid losing money (though there are probably better approaches if this is your goal). • It doesn’t take all that long to be pretty confident (at least based on this set of data) that you won’t lose money investing in the S&P 500 Index. The first point is one we all know reasonably well – stocks tend to go up over time. We wouldn’t be having this conversation if that wasn’t true. And over this time period (1926 – 2022), the total annualized return for the S&P 500 Index was 10.12%. But this should actually give you a little bit of pause. Nearly a quarter of our 1 year periods had a return more than 10% less than the overall annualized return for the period. It’s one thing to look at standard deviation numbers, but this helps drive home what that actually means. The second point is more specific to our conversation, though. There were no periods of 20 years or longer where the S&P 500 Index lost money. And there were only 2 out of 985 15 year periods where it lost money as well. Again, there are no guarantees, but this would likely make a lot of people pretty comfortable that a 20 year holding period (or even 15 years) would be “long term” enough to be reasonably confident that would have a positive return. What’s the Alternative? But let’s turn to the second (and more important) question. How long do you need to invest to be confident that you’ll beat out an alternative investment strategy? To keep things simple, for our purposes that alternative strategy will be owning 5 Year US Treasury Notes instead of the S&P 500 Index over the same time frame. We want to know how often we would have been better off investing in stocks compared to bonds (at least based on the total returns at the end of the period.) As a point of comparison, on a monthly basis, over the total time period, the S&P 500 beat 5 year Treasuries 59% of the time. In other words, in 41% of months bonds beat stocks. We know that stocks tend to beat bonds over longer time periods because stocks are riskier than bonds, but again, we’re seeing the Random Walk in action. In almost 5 months of every year, on average, bonds beat stocks. But what if we look at a slightly longer time period than a month? Holding Period % of Observations Stocks Beat Bonds 1 Year 66.9% 3 Year 72.5% 5 Year 74.3% 10 Year 82.4% 15 Year 86.1% 20 Year 98.6% 30 Year 100.0% 60 Year 100.0% For illustration purposes only. Data calculated from annualized rolling returns of S&P 500 Index and 5 Year Treasury Notes. From January 1926 to December 2022 using different holding periods. Indices are not available for direct investment. Past performance is no guarantee of future returns. We see pretty much what we expected (at least in a general sense). Stocks beat bonds more often over a full year than they did each month, but stocks still lost a third of the time over the course of a year. In fact, you would need to wait a little bit longer than 5 years to have a 75% chance that the S&P 500 would beat 5 Year Treasuries. And this isn’t some aggressive benchmark – this is one of the foundational relationships in investing. Stocks are supposed to beat bonds. And they do. You just need to give them some time. There were no 30 or 60 year periods where stocks lost to bonds (phew). Putting it Into Perspective There are no clearcut answers for how long the “long term” is. But it’s longer than most people think. Investing for the long term doesn’t mean a handful of years. It means decades. It means your investing lifetime. This doesn’t mean that you can’t make changes along the way. Your portfolio and retirement plan will change through time. You will change through time – your situation in life will change, what you want from your money will change. And we’ll find new ways to invest that allow us to more effectively capture the fundamental risk and return relationships in the market. But we can’t make those changes as a reaction to the short term, random, gyrations of the market. As we’ve seen, the Random Walk really is random. The market is going to do some just plain weird stuff. We know this going in. But over time, that randomness (and weirdness) melts away. And what is left behind are the fundamental risk and return relationships in the data – the risk premia that we want to build our portfolios The trick is that you need to commit to your investments. You need to commit to seeing this through to the “long term.” One of the few things that we can say with complete confidence is that your investments are not going to follow a straight path. They are going to bounce all over the place (remember that chart of the monthly S&P 500 returns.) But if you give them the time, their Random Walk will get you where you want to go. Like this article? Download our free eBook! Our eBook The 9 Secrets of Intelligent Investors breaks down the guiding principles to help you make informed investment decisions.	{"url":"https://retirementresearcher.com/how-long-is-the-long-term/","timestamp":"2024-11-13T06:04:08Z","content_type":"text/html","content_length":"240601","record_id":"<urn:uuid:6206d88c-6a4b-4ead-bd7e-7334244b0941>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00648.warc.gz"}
Surface Area of a Cylinder \| Brilliant Math & Science Wiki Surface Area of a Cylinder A cylinder is a right circular prism. It is a solid object with 2 identical, flat, circular ends, and a curved rectangular side. The figure above depicts the developmental figure of a cylinder whose base radius is $r$ and height is $h.$ The surface area is equal to the sum of the areas of the two circular bases and the rectangular side. The area of each base is $\pi r^2.$ Since the width of the rectangular side must be equal to the circumference of the base, the area of the rectangular side is $2\pi rh.$ Therefore the total surface area is $2\pi r^2+2\pi rh=2\pi r(r+h).$ Note: Sometimes, the definition of cylinders may not require having a circular base. In such cases, the base shape will need to be given. The above definition is then called a circular cylinder. What is the surface area of a cylinder whose base is a circle of radius 3 and height of length 4? The surface area is $ 2 \pi \times 3 \times 4 +2 \pi \times 3^2 = 42 \pi $. $ _\square $ Suppose that the sum of the areas of 2 identical circular ends in a cylinder is the same as the area of the curved side of the cylinder. If the radii of of the flat circular ends are each $r,$ what is the height of the cylinder? The sum of the areas of the 2 identical flat circular ends in the cylinder is $2 \pi r^2.$ The area of the curved side of the cylinder is $2 \pi r h,$ where $h$ is the height of the cylinder. Equating these two gives \[2 \pi r^2=2 \pi r h \Rightarrow h=r.\] Thus, the answer is $r.$ $ _\square $ Suppose the surface area of a circular cylinder with height $h$ and base radius $r$ is half the surface area of a circular cylinder with height $5h$ and base radius $r.$ What is the ratio From the formula $ 2 \pi r h +2 \pi r^2 $ for the surface area of a circular cylinder, we have the following relation between the two surface areas of interest: \[2 \pi r h +2 \pi r^2=\frac{1} {2}\times\left(2 \pi r \cdot (5h) +2 \pi r^2\right).\] Dividing both sides by $\pi r$ gives \[ 2h+2r&=5h+r\\ r&=3h\\ r:h&=3:1. \ _\square \] Suppose that the surface area of a circular cylinder is $20\pi.$ If both the radius $r$ and height $h$ of the cylinder are integers and $r>1,$ what is $r+h?$ From the formula $ 2 \pi r h +2 \pi r^2 $ for the surface area of a circular cylinder, we have \[ 2 \pi r h +2 \pi r^2&=20 \pi \\ r(h+r)&=10. \qquad (1) \] Since $r>1$ by assumption, if $r= 2,$ then $h=3.$ Then no other integer value of $r>2$ satisfies $(1).$ Hence, \[r+h=2+3=5.\]	{"url":"https://brilliant.org/wiki/surface-area-cylinder/?subtopic=geometric-measurement&chapter=surface-area","timestamp":"2024-11-13T12:27:27Z","content_type":"text/html","content_length":"46881","record_id":"<urn:uuid:46e9e8da-fb3b-411e-9677-4ee5461bbb96>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00472.warc.gz"}
Descriptive Forecasting Business Context of Advanced Machine Learning Across the globe businesses are facing intense competitive pressure. The cost of raw materials, manufacturing conversion costs and transport costs are volatile and are all rising. Supply remains volatile and demand from customers is harder and harder to predict. New channels provide as many challenges as they do opportunities and the imperatives of ESG to reduce environmental impacts cannot be ignored. Forecasting is not as difficult as it once was, given the maturity of today’s compute power, database and AI ML modeling tools. Customers require product SKU level forecasting both in terms of quantity and in dollars, and the source of the forecast data resides in the SAP HANA databases. Product SKU level dollar forecasting requires price, volume, COGS and SG&A be modeled as a P&L, Balance Sheet and Cash Flow. TekMetrix SAP data models using PaPM, HANA, SAC, BW4 and S4, and a properly scaled AI tool from SAP BTP are used to assist not only in the forecast but in the automation of the forecasting process. The future of corporations includes more products ranges, more channels, more suppliers and more distribution centers. More international shipping. The enterprise landscape includes SAP and non-SAP ERP systems, business warehouses, cloud and on-premise systems, success factors, Ariba, Azure, WAS and other cloud services. Data sets can be large to very large involving billions of records that the forecast is created on. Historical Approach to Forecasting Subjective Forecasting Methods • Composites, customer surveys, forecast experts, and Delphi methods are examples of subjective forecasting methods. • Composites - aggregation of data such as sales from the sales force, election polling • Customer surveys - the forecast is based on customer feedback • Forecast experts - the forecast is prepared by a limited number of experts • Delphi method - individual opinions iterative complied and reconsidered until the group reaches a consensus Objective Forecasting Methods Time Series Forecasting Historically, the approach to forecasting relies on time series analysis looking to identify patterns, trends and seasonality in demand. Moving averages and exponential smoothing are commonly used in time series forecasting. The goal of time series analysis is to isolate patterns in past data. The data in time series forecasting will have descriptive characteristics like trend, seasonality or cycles and randomness that are used for prediction. Other, qualitative methods have been used. These are techniques based on expert opinions and market research. Delphi methods, market surveys and focus groups are examples of qualitative techniques. Time series methods use historical data as the basis of estimating future outcomes. Time series forecasting is based on the assumption that past demand history is a good indicator of future demand: • Moving average • Weighted moving average • Exponential smoothing • Autoregressive moving average (ARMA) • Autoregressive integrated moving average (ARIMA - Box Jenkins) • Extrapolation • Linear regression • Trend isolation • Growth curve • Recurrent neural network Causal Models Causal models are explained through causal analysis. Causal models establish relationships between demand and demand drivers, such as economic indicators, supply and capacity constraints, new product introductions, advertising expenditures or competitor activities. Causal models require the Demand Value (DV) be formulated as a function of all "n" causes. These models often involve regression analysis and can provide insights into how certain factors influence demand. This brings much more richness to the forecast but is typically done on an ad-hoc basis rather than dynamically. Causal models include: • Aggregate forecasts using Cooke’s method • Technology forecasting • Statistical surveys • Scenario building • Forecast by analogy • Delphi methods Probability Distributions Discrete probability distributions are probability distributions that assign probabilities to each individual outcome. Examples of discrete probability distributions include the binomial distribution, the Poisson distribution, and the hypergeometric distribution. Continuous probability distributions are probability distributions that assign probabilities to intervals. Examples of continuous probability distributions include the normal distribution, the t-distribution, and the chi-square distribution. The main difference between discrete and continuous probability distributions is that discrete probability distributions define probabilities associated with discrete variables, while continuous probability distributions define probabilities associated with continuous variables. A discrete variable is a variable that can only take on a finite or countably infinite number of values, while a continuous variable is a variable that can take on any value between two specified values. Commonly Used Discrete Probability Distributions: • Geometric distributions: □ Used to model the number of trials needed to get the first success in a sequence of independent and identically distributed Bernoulli trials • Binomial distributions: □ Used to model the number of successes in a fixed number of independent and identically distributed Bernoulli trials • Bernoulli distributions: □ Used to model the outcome of a single Bernoulli trial, which is a random experiment with two possible outcomes Commonly Used Continuous Probability Distributions: • Normal distributions: □ Used to model continuous variables that are symmetric and bell-shaped □ Describes the distribution of future relative changes in, for example, demand, stock valuations and FX rates. • Exponential distributions: □ Used to model the time between events that occur randomly and independently at a constant average rate □ For example, exponential distributions may be used to characterize time between successive arrivals of customers in customer services systems, call centers • Beta distributions • Uniform distributions: □ Used to model continuous variables that are equally likely to occur over a specified range Discrete probability distribution forecasting An annual operating plan, (AOP) represents the forecast metrics to measure and match demand and supply in uncertain situations. An understanding of the variations in the forecast data is needed to answer how much to produce and at what cost within acceptable accuracy limits. A problem with matching demand and supply in uncertain situation is called a Newsvendor, newsboy or single-period forecasting problem. Fixed prices and uncertain demand are attributes of the demand problem. Solutions to the problem are used to create optimal inventory levels. Demand is a random variable. A typical problem situation, modeling an uncertain future demand, requires a mathematical and data model. With the proper data model, we will describe a discrete probability distribution with mean and standard deviation. A discrete variable is a variable that can take on a finite or countably infinite number of values. Examples of discrete variables include the number of children in a family, the number of cars sold by a dealership, and the number of heads obtained when flipping a coin, the number of items in inventory. The table below depicts how the modeling process might begin for a discrete probability distribution. Discrete Probability Distribution Using SKU Demand Case Scenarios This is a Pythagorean mean and standard deviation. There are other Pythagorean means we don't describe here. The discrete probability distribution, mean and standard deviation describes on average the deviation from the demand data values from the mean. All possible values of the discrete random variable, are forecast along with their probabilities. Examples of discrete probability distributions include a binomial distribution, Poisson distribution and hypergeometric distributions. Continuous probability distributions forecasting Continuous probability distributions are used to forecast a continuous random variable where the random variable, demand, DV in this case, can take on an interval of values; groups of values. A continuous variable is a variable that can take on any value within a specified range (which may be infinite). Examples of continuous variables include height, weight, temperature, and time. A normal distribution is an example of a continuous probability distribution. The random variable can take on any values. In our use case, we will use values of the SKU demand case, DV within specified interval values of minus DVmin to plus DVmax, (-DVmin to +DVmax). A cumulative distribution function is used for statistics prediction or forecasting. The mean is simply the sum of the DV's divided by the number of DVs. The standard deviation for prediction is = StdDEV = StdDEV/√n where n = the total number of data points. The more data points we have then descriptive statistics for demand forecasting approaches that of predictive statistics. Examples of continuous probability distributions include normal distributions, uniform distribution and exponential distributions. Forecast error can be measured as the difference between the forecast value and the actual value, DVerror = DV forecast -DV actual. There are generally 3 ways to measure forecast error: • Mean Absolute Deviation (MAD) = Σ\|DVerror\|/n • Mean Squared Error (MSE) = ΣDEVerror^2/n • Mean Absolute Percentage Error (MAPE) = Σ \|DVerror /DV per period\|/n X100 Forecast bias is the average value of DVerror tends to be positive or negative. Thus, it is a measure of under or over forecasting. Continuous Probability Distribution Using SKU Demand Case Scenarios Seasonal Forecasting Trends and Seasonality If there is a significant trend (positive or negative), and or seasonality in the demand values then the moving averages will lag the trend. When there is an increasing trend than moving averages will usually fall below the demand. When there is a decreasing trend moving average forecasts will usually fall above the demand. When trend is present linear regression methods can be used. Fitting a best fit trend line is usually done with Ordinary Least Squares (OLS). Below we show a calculated trend line on 60 periods of data. The dotted trend line shows the best fit line through the data. The forecast, for the next period, is calculated from the trend line equation, shown below as Y = 27657X +2E+06. Seasonality is a pattern in the data that is repeated at regular intervals. Multiplicative seasonal factors are represented by Di (D1, D2, D3,.....Dn) where i represents the season and n denotes the total number of seasons. Note that ΣD(i) = N the total number of seasons. If D(i) = 1.3 than this implies that the season is 30% higher than the baseline average. And if D(i) = .75 then the implication is that the season is 25% lower than the base line average. To estimate seasonal factors, follow these steps: 1. Calculate the sample mean 2. Calculate the seasonal averages 3. Calculate the seasonal factors: □ Divide the seasonal averages in step 2 by the sample mean and sum the resulting N numbers, the sum of the resulting N numbers will correspond to N seasonal factors 4. De-Seasonalize by dividing each observation in the data by the appropriate seasonal factor Once the model has created the seasonal and de-seasonalized values the forecast can be completed using the de-seasonalized series as a moving average. Multiply the de-seasonalized moving average by the appropriate seasonal factor. This will create a final forecast. Normal Distribution Forecast for New Product Introduction How to forecast the demand for a new product when there is no historical data? To create a new product forecast subjective methods, described above, could be used; the Delphi method for example. To improve the forecast we can create a normal distribution forecast using SAP SKU level data from other similar products. We create a data model with Product, Forecast, Produced, Sales and Actual Demand as shown in the table below. Forecast accuracy is calculated as the ratio of Actual / Demand. The normal distribution demand curve for the new product is calculated as: 1. Begin with an initial forecast generated from subjective methods (sales inputs, intuition, experience) 2. Forecast accuracy = A/F ratio = Actual demand / Demand forecast 3. Mean = Expected actual demand = Expected A/F ration Demand Forecast 4. Standard deviation = Actual demand = A/F Ratio * Forecast 5. Correct the standard deviation = Standard deviation / Standard deviation * (SQRT(number of demand periods)), as n becomes larger the correction term disappears New Product Forecasting Technique for a subjective demand of 1000 units New Product Forecasting Technique with Normal Distribution TekMetrix Forecasting Machine learning has revolutionized demand forecasting by automating the calculations required to analyze data sets, identify hidden patterns and adopt to changing trends. TekMetrix SAP data models combined with ML algorithms can analyze vast amounts of data. The compute power to perform this type of SKU forecasting has been made more widely available and easier due to the SAP simplified data model. Some of the most used machine learning techniques for demand forecasting is: • Random forecasting • Gradient boosting • Long short-term memory networks (LSTM) • Autoregressive integrated moving average (ARIMA) statistical models • Neural networks • Group method of data handling • Support vector machines A good forecast is more than a single number. Forecast data can come from experts in the organization and from historical data. Forecast accuracy is significant driver of business performance. And, annual operating plan, (AOP P&L, Balance Sheet, Cash Flow) is a key forecast metric used by planners to match demand and supply. Forecasting is performed against conditions of uncertainty. An understanding of the variations in the forecast data is needed to the answer the question "how much to produce and when". Accuracy limits are described using TekMetrix statistical analysis. A typical problem situation, requiring a forecast mathematical and data model is: • Retailer orders from a supplier and sells to customers • The ordered products are placed on the store shelf • Customers in the store buy the produce if it is available on the shelf • To ensure availability the order needs be placed before the customer demand is known • There is one chance to order inventory • Manage trends and seasonality • This problem of matching demand and supply in uncertain situation is called a Newsvendor, newsboy or single-period forecasting problem. Fixed prices and uncertain demand are attributes of the demand problem. Solutions to the problem are used to create optimal inventory levels which drives the AOP P&L priorities. Business challenge: • No visibility into demand • Orders are placed prior to seeing the actual demand • Incorporating data from diverse sources including SAP S4 transactions, customer behavior, economic indicators, marketing efforts, weather, events and competitor information Characteristics of SKU good forecasting: • Point forecasts usually wrong because demand can be a random variable • Forecasts should include some distribution of information: □ Mean and standard deviation □ Range (high and low) □ Aggregate SKU forecasts are usually more accurate □ Data modeling in HANA with historic data, current FY transactions meshing with consumer data □ Use advanced machine learning algorithms Solution process using a continuous probability distribution: • Choose the right forecasting algorithm • Model training and tuning • Forecast at various levels • Analyze past demand data using a probability distribution • Follow structured data analysis process (Newsvendor analysis) • Communicate the objective (usually maximize profit, minimize costs or increase market share) • Perform a statistical analysis, communicate accuracy metrics • Capture actual demand and realized profits and costs (too little or too much inventory) • Repeat for each new plan period while continuous monitoring and updating SKU Level Customer Revenue Plan and Forecast with SAP S4 - HANA - SAC Tools TekMetrix SKU Forecasting and Analytics: • Customer supply and order management • Customer order analysis and forecasting • Customer SKU level profitability (P&L) • Market analysis and segmentation • Customer analysis • Product pricing, pricing elasticity • Customer experience • Customer key purchasing criteria • Product lifecycle • Product substitutions • Sales promotion planning, forecasting, analytics • Adoption cycle • Competitor analysis • Social and sentiment analysis • Digital commerce • Category and panel trends • SAP advanced trade management analytics (ATMA) • Digital commerce • KAM analytics • Product SKU level profitability (AOP P&L, balance sheet, cash flow) • Supply chain analytics (warehousing, manufacturing, procurement, finance, distribution, HR, inventory, scheduling, forecasting) • SAP S4-BW4-CRM-SAC-PaPM-HANA-Hadoop-AnyDB data modeling and analytics	{"url":"https://tekmetrix.com/analytics-ai/forecasting","timestamp":"2024-11-09T23:47:12Z","content_type":"text/html","content_length":"712947","record_id":"<urn:uuid:aec12208-0b11-48dd-89a0-1489848cfe76>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00512.warc.gz"}
Convert 180 c to f? 8 quick tips you need to know!Convert 180 c to f? 8 quick tips you need to know! - Davies Chuck Wagon Disclaimer: There are affiliate links in this post. At no cost to you, I get commissions for purchases made through links in this post. Have you ever puzzled over how to convert 180 c to f? Look no further! We have the answer, and we’ve put together 8 simple tips that will help you understand temperature conversion with ease. Learn how individual temperatures relate between scales, what handy formulas you need to keep in mind and more — all in this one helpful article! Keep reading for your complete guide on converting celsius to fahrenheit. What Is Celsius? Celsius, also known as centigrade, is a unit of temperature measurement used in the International System of Units (SI). It is named after the Swedish astronomer Anders Celsius, who first proposed a similar temperature scale in 1742. Measurement and Meaning: Celsius is based on the Celsius scale, which sets the freezing point of water at 0 degrees Celsius (°C) and the boiling point of water at 100 °C at standard atmospheric pressure. This means that the difference between the freezing and boiling points of water is divided into 100 equal parts, each of which is one degree Celsius. Before the Celsius scale was introduced, various temperature scales were used in different parts of the world. Celsius proposed his scale based on the idea of setting 0 °C as the boiling point of water and 100 °C as the freezing point of water. However, this scale was later reversed by French scientist Jean-Pierre Christin, who proposed the modern form of the Celsius scale in 1743. Use and Example Measures: Celsius is widely used around the world, particularly in scientific and academic fields. It is used to measure temperatures in a variety of contexts, including weather forecasting, cooking, and laboratory experiments. For example, a typical room temperature may be around 20-25 °C, while the human body’s normal temperature is around 36-37 °C. Popularity and Fresh Opinions: Celsius is one of the most widely recognized and used temperature scales in the world. It is commonly used in countries that have adopted the metric system, including most of Europe, Asia, and However, some countries, including the United States, still use Fahrenheit as their primary temperature scale. Celsius is a more intuitive and logical temperature scale than Fahrenheit. Celsius is based on the properties of water, which is a widely available and commonly used substance. Additionally, Celsius makes it easy to understand the relationship between temperature and energy, as each degree Celsius represents a specific amount of thermal energy. What Is Fahrenheit? Fahrenheit is a temperature scale that was proposed by the German physicist Daniel Gabriel Fahrenheit in 1724. It is one of the two most commonly used temperature scales in the world, alongside Celsius (also known as centigrade). Meaning of Fahrenheit: The Fahrenheit scale measures temperature in degrees Fahrenheit (°F). The scale is based on the freezing point of water, which is defined as 32°F, and the boiling point of water, which is defined as 212°F, at standard atmospheric pressure. History of Fahrenheit: The Fahrenheit scale was invented in 1724 by Daniel Gabriel Fahrenheit, a German physicist and engineer. He initially defined the freezing point of water as 0°F and the boiling point of water as 212°F. Later, he reversed the scale, making 32°F the freezing point and 212°F the boiling point, which is the scale that is still used today. Use of Fahrenheit: The Fahrenheit scale is primarily used in the United States, the Bahamas, Belize, and the Cayman Islands, as well as in some other countries. It is commonly used in weather forecasting, cooking, and for measuring body temperature. Example measurement in Fahrenheit: A typical human body temperature is around 98.6°F. The temperature of a warm summer day might be around 85°F, while the temperature of a cold winter day might be around 30°F. Popularity of Fahrenheit: Fahrenheit is not as widely used as Celsius globally, with most countries using Celsius for scientific and engineering applications. However, Fahrenheit remains popular in the United States and some other countries, especially for everyday temperature measurements like weather and cooking. C to F Converter Celsius and Fahrenheit are two units of temperature measurement used in different parts of the world. Converting between these two units is a simple mathematical calculation, which can be done using the formula: °F = (°C * 1.8) + 32 where °C represents the temperature in Celsius and °F represents the temperature in Fahrenheit. Let’s analyze this formula in more detail. The first part of the formula, °C * 1.8, represents the conversion of Celsius to Fahrenheit. This is done by multiplying the Celsius temperature by 1.8, which is the conversion factor between the two scales. The result of this calculation is the equivalent temperature in Fahrenheit, but without the 32-degree offset that exists between the two scales. To account for this offset, we add 32 to the result of the Celsius to Fahrenheit conversion, resulting in the final temperature in Fahrenheit. For example, let’s say we want to convert 20 degrees Celsius to Fahrenheit. Using the formula above, we would first perform the Celsius to Fahrenheit conversion by multiplying 20 by 1.8, resulting in 36. We then add 32 to this result, giving us a final temperature of 68 degrees Fahrenheit. Here are three more examples to demonstrate the use of this formula: Example 1: Convert 0 degrees Celsius to Fahrenheit °F = (0 * 1.8) + 32 °F = 32 Therefore, 0 degrees Celsius is equivalent to 32 degrees Fahrenheit. Example 2: Convert 25 degrees Celsius to Fahrenheit °F = (25 * 1.8) + 32 °F = 77 Therefore, 25 degrees Celsius is equivalent to 77 degrees Fahrenheit. Example 3: Convert -10 degrees Celsius to Fahrenheit °F = (-10 * 1.8) + 32 °F = 14 Therefore, -10 degrees Celsius is equivalent to 14 degrees Fahrenheit. Read more: How many ml in a shot? Useful tips 2023. How many tablespoons in a cup? Great tips to convert! How many ml in a gallon: Great notes to remember. How Many Water Bottles Should I Drink a Day? How to convert 180 c to f The conversion formula between Celsius and Fahrenheit is given by: °F = (°C × 9/5) + 32 Let’s take the example of converting 180°C to °F: °F = (180 × 9/5) + 32 °F = 324 + 32 °F = 356 Therefore, 180°C is equal to 356°F. Conversion table for Celsius (C) to Fahrenheit (F) and Kelvin (K): Celsius (C) Fahrenheit (F) Kelvin (K) -273.15°C -459.67°F 0K -200°C -328°F 73.15K -100°C -148°F 173.15K -50°C -58°F 223.15K 0°C 32°F 273.15K 10°C 50°F 283.15K 20°C 68°F 293.15K 30°C 86°F 303.15K 40°C 104°F 313.15K 50°C 122°F 323.15K 60°C 140°F 333.15K 70°C 158°F 343.15K 80°C 176°F 353.15K 90°C 194°F 363.15K 100°C 212°F 373.15K To convert Celsius to Fahrenheit, you can use the following formula: F = (C x 9/5) + 32 To convert Celsius to Kelvin, you can simply add 273.15 to the Celsius temperature: K = C + 273.15 Conversion table for Fahrenheit (F) to Celsius (C) Fahrenheit (F) Celsius (C) -40 -40 -22 -30 -4 -20 14 -10 8 quick tips to convert C to F 1. Understand the Conversion Formula: To convert Celsius (°C) to Fahrenheit (°F), you must use the formula °F = (°C x 9/5) + 32. This formula takes into account the offset between the two temperature scales, which is 32 degrees Fahrenheit. 2. Analyze the Formula in Detail: The Celsius-to-Fahrenheit conversion is made up of two steps: first, converting Celsius to Fahrenheit; and second, adding an offset of 32 degrees Fahrenheit. The first part of the equation, °C × 9/5, represents the conversion of Celsius to Fahrenheit without any offset. Multiplying a given temperature in Celsius by this factor will give you its equivalent temperature in Fahrenheit without accounting for the 32-degree offset between the two scales. The second part of the equation, + 32, accounts for this offset and gives you the final temperature in Fahrenheit. 3. Examples with Other Ingredients: Let’s take a closer look at how to use the formula to convert from Celsius to Fahrenheit using three different examples. For 0 degrees Celsius, °F = (0 × 9/5) + 32 = 32; for 25 degrees Celsius, °F = (25 × 9/5) + 32 = 77; and for -10 degrees Celsius, °F = (-10 × 9/5) + 32 = 14. 4. Use Tools or References: To make conversions easier and more accurate, consider using online conversion tools or referencing a temperature conversion chart. 5. Use Scientific Notation: When temperatures get extreme, you may want to consider using scientific notation for accuracy. For example, 20000 °C can be written as 2 × 104 °C, and this is equal to 35 940 °F (3.59 × 104 °F). 6. Be Creative: If the exact temperature isn’t known or important, it can often be helpful to “eyeball” the conversion by adding 30 or 40 degrees Fahrenheit depending on the Celsius temperature. This is especially true when dealing with temperatures in everyday life like cooking and baking—in these cases it doesn’t have to be an exact science! 7. Speak Like an Expert: When speaking about temperatures that have been converted from Celsius to Fahrenheit, it’s important to use language that expresses the offset between the two scales. For example, instead of saying “25 degrees Fahrenheit,” you should say “77 degrees Fahrenheit (25 degrees Celsius).” 8. Don’t Repeat Yourself: To make conversions easier and more accurate, consider using online conversion tools or referencing a temperature conversion chart instead of repeating the same formula every time. This will help save time and avoid confusion when giving final answers in Fahrenheit. Importance of C to F conversion in cooking Baking is a science that requires precision in temperature control. Most baking recipes specify a baking temperature in either Celsius or Fahrenheit. For example, a recipe might call for a baking temperature of 180°C or 350°F. If you are using an oven that displays temperature in Celsius but the recipe calls for Fahrenheit, it is important to convert the temperature to ensure that the oven is set to the correct Similarly, if you are using a recipe that provides a temperature in Celsius but your oven displays temperature in Fahrenheit, you will need to convert the temperature for accurate results. Candy-making is another area of cooking that requires precise temperature control. Different types of candies require different cooking temperatures, and it is important to maintain the correct temperature throughout the cooking process. For example, fudge requires a cooking temperature of 112°C or 234°F, while caramel requires a temperature of 170°C or 338°F. If you are using a recipe that provides temperature measurements in Celsius but your candy thermometer displays temperature in Fahrenheit, you will need to convert the temperature to ensure that the candy is cooked to the correct temperature. Meat cooking: Meat cooking is another area where temperature control is critical. Different types of meat require different cooking temperatures to ensure that they are cooked safely and to the desired level of For example, a rare steak requires an internal temperature of 52°C or 126°F, while a well-done steak requires an internal temperature of 71°C or 160°F. If you are using a recipe that provides temperature measurements in Celsius but your meat thermometer displays temperature in Fahrenheit, you will need to convert the temperature to ensure that the meat is cooked to the correct temperature. FAQs about convert 180 c to f What is the formula to convert 180°C to Fahrenheit? The formula to convert 180°C to Fahrenheit is F = (C x 1.8) + 32, which gives you F = (180 x 1.8) + 32 = 356°F. Why is it important to convert Celsius to Fahrenheit in cooking? Converting Celsius to Fahrenheit is important in cooking because recipes may provide temperature measurements in one scale or the other, and it is necessary to convert these values to ensure accurate What other units of measure can be used to express temperature in cooking? Other units of measure that can be used to express temperature in cooking include Kelvin, Rankine, and Réaumur. What is the most commonly used temperature scale in cooking? The most commonly used temperature scale in cooking is Celsius. Can I use an online conversion tool to convert Celsius to Fahrenheit? Yes, there are many online conversion tools available that can convert Celsius to Fahrenheit and vice versa. What are some tips for converting Celsius to Fahrenheit? Some tips for converting Celsius to Fahrenheit include double-checking your calculations, rounding your answer to the nearest whole number, and using a conversion chart or tool if you’re not confident in your math skills. What are some common mistakes to avoid when converting Celsius to Fahrenheit? Some common mistakes to avoid when converting Celsius to Fahrenheit include forgetting to add 32 to the result, forgetting to multiply by 1.8, and using the wrong formula. How can I remember the formula for converting Celsius to Fahrenheit? One way to remember the formula for converting Celsius to Fahrenheit is to use the phrase “Times 1.8, plus 32” as a mnemonic device. Can I use the same formula to convert Fahrenheit to Celsius? Yes, the same formula can be used to convert Fahrenheit to Celsius. The formula is C = (F – 32) / 1.8. What other cooking ingredients require temperature control? Other cooking ingredients that require temperature control include chocolate, bread, and pastry dough. Can I convert other amounts of Celsius to Fahrenheit using the same formula? Yes, the same formula can be used to convert any amount of Celsius to Fahrenheit. Why is it important to use the correct temperature in cooking? Using the correct temperature in cooking is important because it ensures that your food is cooked safely and to the desired level of doneness. Is there a difference in taste between food cooked at different temperatures? Yes, there can be a difference in taste between food cooked at different temperatures. For example, a steak cooked at a higher temperature will be more well-done and have less moisture than a steak cooked at a lower temperature. How can I check the temperature of my food? You can check the temperature of your food using a thermometer, which can be inserted into the food to measure its internal temperature. What is the importance of accurate temperature control in candy-making? Accurate temperature control is important in candy-making because different types of candy require different cooking temperatures, and maintaining the correct temperature throughout the cooking process is critical to achieving the desired texture and flavor of the finished product. Can I use a thermometer to convert Celsius to Fahrenheit? No, a thermometer cannot be used to convert Celsius to Fahrenheit. A thermometer is a tool used to measure temperature, while converting Celsius to Fahrenheit requires a formula or calculation. Are there any shortcuts or tricks to converting Celsius to Fahrenheit in my head? While it is possible to estimate the temperature conversion from Celsius to Fahrenheit in your head, it is not recommended for precise cooking or baking. However, if you need to make a quick estimate, you can round the Celsius temperature to the nearest 10 and then double it and add 30. For example, if the Celsius temperature is 180, rounding it to 180 and doubling it to get 360, and adding 30 to get 390, which is an estimate of the Fahrenheit temperature. However, this method is not precise and should only be used as a rough estimate. Conclusion about convert 180 c to f In conclusion, converting 180°C to Fahrenheit is an important process in cooking and baking to ensure accurate temperature measurements. The formula to convert 180°C to Fahrenheit is F = (C x 1.8) + 32, which gives a result of 356°F. It is important to use the correct temperature in cooking and baking to ensure safe and properly cooked food. Common mistakes in converting Celsius to Fahrenheit include forgetting to add 32 to the result or using the wrong formula. I’m Leon Todd and my passion for cooking is my life goal. I’m the owner and operator of Davieschuckwagon.com, a website that specializes in providing high-quality cooking information and resources. I love to experiment with new flavors and techniques in the kitchen, and I’m always looking for ways to improve my skills. I worked my way up through the ranks, taking on more challenging roles in the kitchen. I eventually became a head chef. Cooking is more than just a job to me – it’s a passion that I want to share with the world. GIPHY App Key not set. Please check settings What do you think? 0 Points Upvote Downvote	{"url":"https://davieschuckwagon.com/blog/180-c-to-f/","timestamp":"2024-11-09T22:03:56Z","content_type":"text/html","content_length":"343796","record_id":"<urn:uuid:0d4a5f68-cbd5-4353-80f5-e3d2319d7f66>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00366.warc.gz"}
Mastering DEGREES Formula in Excel: A Comprehensive Guide - THINK Accounting In the vast universe of Excel functions, there lies a hidden gem called DEGREES. Have you ever pondered over the need to convert an angle in radians into degrees? Well, worry no more! With DEGREES, you can effortlessly unlock the full potential of your Excel skills and conquer the realm of trigonometry. Buckle up, fellow spreadsheet enthusiasts, as we embark on a comprehensive journey to master the DEGREES formula! Unlocking the Power of DEGREES Before diving deep into the DEGREES function, let's take a moment to appreciate its magnificence. This humble little formula transforms an angle from radians to degrees, giving you the flexibility to work with the familiar degree measurements we all know and love. Whether you're calculating angles for geometry, physics, or simply trying to impress your friends with your Excel wizardry, the DEGREES formula is here to save the day. Imagine you're an architect working on a complex building design. You need to calculate the angles of various intersecting beams to ensure structural integrity. Without the DEGREES function, you would be stuck dealing with radians, a unit of measurement that may not be as intuitive as degrees. But with the power of DEGREES, you can effortlessly convert those pesky radians into degrees, making your calculations a breeze. Understanding the DEGREES Function in Excel Now, let's delve into the inner workings of the DEGREES function. In its simplest form, the DEGREES function takes a value in radians and spits out the equivalent value in degrees. It's like having a cosmic calculator at your fingertips, ready to convert those confusing radians into comprehensible degrees with a single keystroke. All you need to do is supply the radians value as the function's argument, and voila! Excel will reward you with the answer you seek. But wait, there's more! The DEGREES function isn't limited to converting individual values. With a dash of creativity, you can apply it to entire ranges, arrays, and even nested formulas. Let's say you're a data analyst working with a massive dataset that includes angles measured in radians. By using the DEGREES function in combination with other Excel functions, you can quickly convert all those radians into degrees, allowing for easier analysis and visualization. Furthermore, the DEGREES function can be used in conjunction with conditional formatting to highlight specific angles within a range. This can be particularly useful when working with data that requires certain angles to meet specific criteria. By converting the angles to degrees, you can easily apply conditional formatting rules to visually identify angles that fall within a desired range or exceed certain thresholds. Additionally, the DEGREES function can be a valuable tool when working with trigonometric calculations. For example, if you're trying to find the missing angle of a triangle given the lengths of its sides, you can use the DEGREES function to convert the calculated angle from radians to degrees, making it more meaningful and easier to interpret. The possibilities with the DEGREES function are simply astronomical! Whether you're a student studying math, a scientist conducting research, or a business professional analyzing data, the DEGREES function in Excel can unlock a world of possibilities. So go ahead, embrace the power of DEGREES and let it elevate your Excel skills to new heights! Practical Examples of Using the DEGREES Function Let's spice things up with some practical examples that showcase the true potential of the DEGREES function. Imagine you need to find the angle in degrees for a given angle in radians. With DEGREES, the solution is as easy as pie. Just plug in the radians value, and watch as Excel effortlessly calculates the corresponding angle in degrees. For instance, let's say you are working on a project that involves calculating the angle of elevation for a rocket launch. You have the angle in radians, but you need it in degrees for further analysis. By using the DEGREES function, you can quickly convert the angle from radians to degrees, allowing you to accurately determine the trajectory of the rocket. But wait, there's more! You can also use the DEGREES formula to convert angles stored in cells from radians to degrees. Simply reference the cell containing that pesky radians value, and let Excel work its magic. It's like having a mystical enchantment that transforms your spreadsheet data from cryptic to crystal clear. Imagine you have a spreadsheet full of data on various geographical locations, including the latitude and longitude coordinates. However, the latitude values are stored in radians, and you need them in degrees for better visualization and analysis. By applying the DEGREES function to the latitude column, you can effortlessly convert all the radians values to degrees, making it easier to plot the locations on a map and analyze their distribution. Furthermore, the DEGREES function can be a valuable tool in the field of engineering. Let's say you are designing a bridge and need to calculate the angles of the support beams. By using the DEGREES function, you can convert the angles from radians to degrees, allowing you to precisely determine the required measurements for the beams and ensure the structural integrity of the bridge. As you can see, the DEGREES function is not just a simple conversion tool. It has the power to simplify complex calculations and enhance data analysis in various fields, from rocket science to geography and engineering. So next time you encounter angles in radians, remember the DEGREES function and let Excel do the heavy lifting for you. Tips and Tricks for Mastering the DEGREES Function Now that we've scratched the surface of the DEGREES function, let's unleash the vibrant spectrum of tips and tricks to fully harness its capabilities. Brace yourself for a whirlwind of pro tips designed to maximize your productivity and make you the star of any Excel-based spectacle. Tip 1: Working with Trigonometry in Degrees Converting angles from radians to degrees is just the tip of the iceberg. Did you know that you can combine the DEGREES function with other trigonometric functions to work exclusively in degrees? By converting all your angles to degrees before using trigonometric functions, you can keep your formulas in the familiar realm of degrees. It's like having a secret weapon against confusing radians. Tip 2: Imbuing Formulas with Flexibility Is your inner Excel geek craving for more flexibility? Fear not, for the DEGREES function has got you covered. Imagine you have a formula that calculates an angle based on other variables. By first converting those variables from radians to degrees using DEGREES, you can easily adjust the formula to work with different units without rewriting the entire logic. Now that's what we call mathematical elegance! Avoiding Common Mistakes with the DEGREES Function Even the most seasoned Excel adventurers sometimes stumble upon treacherous pitfalls. To avoid falling into the abyss of frustration, let's uncover and conquer the common mistakes that plague DEGREES enthusiasts. Prepare yourself, brave explorer, as we pave the way to a smooth sailing journey through the tranquil seas of anguine arithmetic. Mistake 1: Forgetting to Convert to Radians Here's a rookie mistake to watch out for. When using the DEGREES function, always feed it values in radians. If you unintentionally input degrees instead, Excel will unapologetically convert them back to radians. So, double-check your inputs and ensure that you provide radians as the DEGREES function's argument. Otherwise, Excel might just send you on a wild goose chase! Mistake 2: Forgetting to Anchor Cell References In the heat of the moment, the allure of copy-pasting can be overwhelming. But beware, dear Excel enthusiast, for the absence of anchored cell references can lead you astray. When using the DEGREES function on a range of cells, make sure to anchor the cell references properly. Otherwise, Excel will alter the references as you copy the formula, resulting in utter chaos. Trust us, maintaining order in your spreadsheet kingdom is worth the effort! Troubleshooting the DEGREES Function: Common Issues and Solutions Like any other formula, the DEGREES function may occasionally misbehave. But fear not, for we shall cast light upon these darkness-shrouded conundrums and pave a path to enlightenment. Prepare yourself, valiant troubleshooter, as we embark on a journey to banish the shadows of confusion. Issue 1: Getting Incorrect Results If Excel is mysteriously spitting out incorrect results when using the DEGREES function, it's time to step back and reevaluate your inputs. Double-check that you're providing the function with the correct radians values, and ensure that your formulas don't contain any hidden gremlins that could be messing things up. Remember, Excel can only work its magic if you feed it the right ingredients! Issue 2: #NAME? Error Have you ever encountered the infamous #NAME? error when using the DEGREES function? Fear not, for this error message is often a simple fix. It typically occurs when Excel fails to recognize the DEGREES function due to a misspelling or a lack of necessary add-ins. Review the function name, check your Excel version, and ensure that the necessary add-ins are activated. With a little sleuthing, you'll soon bid farewell to the dreadful #NAME? error. Exploring Other Formulae Related to DEGREES Believe it or not, there's more to Excel's celestial arsenal than just the DEGREES function. Allow us to shine a light on other formulae that dance in harmony with DEGREES, enhancing your Excel repertoire with celestial beauty. First, let us introduce you to the RADIANS function. Its purpose is quite the opposite of DEGREES as it converts angles from degrees to radians. With DEGREES and RADIANS in perfect harmony, you can effortlessly navigate between degrees and radians, opening doors to new and exciting possibilities. Additionally, the SIN, COS, and TAN functions eagerly await your every command. By combining these functions with DEGREES or RADIANS, you can unleash the full power of trigonometry within Excel. No more shall the complexities of angles and triangles dim your spreadsheet brilliance! There you have it, fellow Excel voyagers! Armed with the knowledge of the DEGREES function and its companions, you are ready to conquer the world of angles and degrees. So go forth, explore, and may your spreadsheet quests always be filled with success and a touch of celestial humor! Hi there! I'm Simon, your not-so-typical finance guy with a knack for numbers and a love for a good spreadsheet. Being in the finance world for over two decades, I've seen it all - from the highs of bull markets to the 'oh no!' moments of financial crashes. But here's the twist: I believe finance should be fun (yes, you read that right, fun!). As a dad, I've mastered the art of explaining complex things, like why the sky is blue or why budgeting is cool, in ways that even a five-year-old would get (or at least pretend to). I bring this same approach to THINK, where I break down financial jargon into something you can actually enjoy reading - and maybe even laugh at! So, whether you're trying to navigate the world of investments or just figure out how to make an Excel budget that doesn’t make you snooze, I’m here to guide you with practical advice, sprinkled with dad jokes and a healthy dose of real-world experience. Let's make finance fun together!	{"url":"https://www.think-accounting.com/formulas/mastering-degrees-formula-in-excel-a-comprehensive-guide/","timestamp":"2024-11-04T23:30:24Z","content_type":"text/html","content_length":"103810","record_id":"<urn:uuid:fcd81083-924c-4239-9f12-23d73ba1979f>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00847.warc.gz"}
ABC Week TEMPLATE AP Calculus Links to Previous Weeks 8-26-19 to 8-30-19 Exam Date for 2019-2020 School Year 8:00 A.M. Tuesday May 5 Schedule 8/26/19 to 8/30/19 • Review for Quiz 1 □ slope of secant lines versus tangent lines □ rate of change (ROC) versus instantaneous rate of change (IROC) □ methods of finding limits ☆ substitution ☆ through a table of values that approach a specific x value Topics ☆ through a graph ☆ through algebraic methods • Be able to find limits using a wide variety of methods □ substitution tuesday □ graphs □ tables • QUIZ 1 □ properties □ algebraic manipulation wednesday • Be able to establish the conditions of continuity • Be able to distinguish the different types and causes of discontinuities • More algebraic methods of determining limits □ jump/gap discontinuity • limits as x approaches infinity □ removable /point discontinuity • limits that approach infinity □ infinite discontinuity • End Behavior Models □ oscillating discontinuity • Finish Limits • Start Continuity • Formative assessment on limit properties and basics of continuity FDWK Textbook Assignments Handouts and Formative Assessments BARRON'S Textbook Assignments Other Textbook Assignments Additional Resources (For This Week) THIS Website AP Central Website Topics Covered in AP Calculus AB Other Websites Topics throughout the semester Week 1 • Basic Function Review • Domain and Range Week 2 • Graphing functions and conic sections • Graphing Piecewise Functions • Be able to find the average rate of change between two specific points • Properties and characteristics of inverse functions • Be able to connect the idea of a limit to the instantaneous rate of change • Types of symmetry that functions may or may not have • Be able to use a table to determine if a limit exists at a particular value of x □ symmetry about the origin • Be able to use a graph to determine if a limit exists at a particular value of x □ symmetry about the y axis • Be able to use the properties of limits to determine one sided limits □ symmetry about the axis • Be able to use the properties of limits to determine two sided limits • Types of reflections • Connect instantaneous rate of change to the slope of a tangent line to the function at the same point □ vertical (over x axis) □ horizontal (over y axis) □ both over x and y axis □ over y = x Week 3 • Be able to find limits using a wide variety of methods □ substitution □ graphs □ tables □ properties □ algebraic manipulation • Be able to establish the conditions of continuity • Be able to distinguish the different types and causes of discontinuities □ jump/gap discontinuity □ removable /point discontinuity □ infinite discontinuity □ oscillating discontinuity Additional Resources Exam Date for 2019-2020 School Year 8:00 A.M. Tuesday May 5	{"url":"https://www.shadhickmanrhs.com/abc-week-template.html","timestamp":"2024-11-03T14:09:01Z","content_type":"text/html","content_length":"68938","record_id":"<urn:uuid:080c83a7-9c0f-4c31-beb8-f1a5ea1ca9b7>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00558.warc.gz"}
How to include multiple groups 16.5.4 How to include multiple groups from one study There are several possible approaches to including a study with multiple intervention groups in a particular meta-analysis. One approach that must be avoided is simply to enter several comparisons into the meta-analysis when these have one or more intervention groups in common. This ‘double-counts’ the participants in the ‘shared’ intervention group(s), and creates a unit-of-analysis error due to the unaddressed correlation between the estimated intervention effects from multiple comparisons (see Chapter 9, Section 9.3). An important distinction to make is between situations in which a study can contribute several independent comparisons (i.e. with no intervention group in common) and when several comparisons are correlated because they have intervention groups, and hence participants, in common. For example, consider a study that randomized participants to four groups: ‘nicotine gum’ versus ‘placebo gum’ versus ‘nicotine patch’ versus ‘placebo patch’. A meta-analysis that addresses the broad question of whether nicotine replacement therapy is effective might include the comparison ‘nicotine gum versus placebo gum’ as well as the independent comparison ‘nicotine patch versus placebo patch’. It is usually reasonable to include independent comparisons in a meta-analysis as if they were from different studies, although there are subtle complications with regard to random-effects analyses (see Section 16.5.5). Approaches to overcoming a unit-of-analysis error for a study that could contribute multiple, correlated, comparisons include the following. • Combine groups to create a single pair-wise comparison (recommended). • Select one pair of interventions and exclude the others. • Split the ‘shared’ group into two or more groups with smaller sample size, and include two or more (reasonably independent) comparisons. • Include two or more correlated comparisons and account for the correlation. • Undertake a multiple-treatments meta-analysis (see Section 16.6). The recommended method in most situations is to combine all relevant experimental intervention groups of the study into a single group, and to combine all relevant control intervention groups into a single control group. As an example, suppose that a meta-analysis of ‘acupuncture versus no acupuncture’ would consider studies of either ‘acupuncture versus sham acupuncture’ or studies of ‘acupuncture versus no intervention’ to be eligible for inclusion. Then a study comparing ‘acupuncture versus sham acupuncture versus no intervention’ would be included in the meta-analysis by combining the participants in the ‘sham acupuncture’ group with participants in the ‘no intervention’ group. This combined control group would be compared with the ‘acupuncture’ group in the usual way. For dichotomous outcomes, both the sample sizes and the numbers of people with events can be summed across groups. For continuous outcomes, means and standard deviations can be combined using methods described in Chapter 7 (Section 7.7.3.8). The alternative strategy of selecting a single pair of interventions (e.g. choosing either ‘sham acupuncture’ or ‘no intervention’ as the control) results in a loss of information and is open to results-related choices, so is not generally recommended. A further possibility is to include each pair-wise comparison separately, but with shared intervention groups divided out approximately evenly among the comparisons. For example, if a trial compares 121 patients receiving acupuncture with 124 patients receiving sham acupuncture and 117 patients receiving no acupuncture, then two comparisons (of, say, 61 ‘acupuncture’ against 124 ‘sham acupuncture’, and of 60 ‘acupuncture’ against 117 ‘no intervention’) might be entered into the meta-analysis. For dichotomous outcomes, both the number of events and the total number of patients would be divided up. For continuous outcomes, only the total number of participants would be divided up and the means and standard deviations left unchanged. This method only partially overcomes the unit-of-analysis error (because the resulting comparisons remain correlated) so is not generally recommended. A potential advantage of this approach, however, would be that approximate investigations of heterogeneity across intervention arms are possible (for example, in the case of the example here, the difference between using sham acupuncture and no intervention as a control group). Two final options, which would require statistical support, are to account for the correlation between correlated comparisons from the same study in the analysis, and to perform a multiple-treatments meta-analysis. The former involves calculating an average (or weighted average) of the relevant pair-wise comparisons from the study, and calculating a variance (and hence a weight) for the study, taking into account the correlation between the comparisons. It will typically yield a similar result to the recommended method of combining across experimental and control intervention groups. Multiple-treatments meta-analysis is discussed in more detail in Section 16.6.	{"url":"http://handbook-5-1.cochrane.org/chapter_16/16_5_4_how_to_include_multiple_groups_from_one_study.htm","timestamp":"2024-11-09T01:15:01Z","content_type":"application/xhtml+xml","content_length":"12573","record_id":"<urn:uuid:04d5a37c-8b13-4aa2-ab5f-93a4a3d1c16f>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00310.warc.gz"}
Solving linear systems of equations using quantum computers Linear systems of equations lie at the heart of many scientific and engineering problems, from machine learning to optimization and physics simulations. Classical methods like Gaussian elimination or iterative methods are powerful but can be inefficient for large, complex systems. In this blog post, I will explore one of the most famous quantum algorithms (called HHL) that offer a potential speedup in solving linear systems. I will delve into its complexity, underlying assumptions and describe two interesting applications. Quantum linear system problem versus linear system problem In order to better understand the limitations of quantum algorithms like HHL, it’s essential to distinguish between a Quantum Linear System Problem (QLSP) and a classical Linear System Problem (LSP). A typical linear system problem (LSP) is represented as: • $A$ is a matrix • $b$ is a known vector • $x$ is the unknown vector we aim to solve for On the other hand, a QLSP deals with a quantum state version of the same concept, represented as: $$A\ket x = \ket b$$ • $A$ is still a matrix • $\ket b$ is a known quantum state • $\ket x$ is the unknown quantum state we wish to find Although both problems appear similar, the difference lies in how the information is represented and manipulated. In a classical system, the vector $b$ is readily available, and solving for $x$ gives a concrete solution that can be directly used. In contrast, in the quantum setting, $\ket b$ is a quantum state, and the solution $\ket x$ is also a quantum state. The main challenge here is that quantum states aren’t directly accessible (any measurement of $\ket x$ collapses the state and only provides a probabilistic result), which means that extracting useful information from the quantum solution requires multiple measurements or sophisticated post-processing. Understanding these differences is crucial when assessing the complexity and feasibility of quantum solvers such as HHL, particularly when applied to real-world problems where error correction and measurement limitations play a significant role. In this section, we introduce the Harrow-Hassidim-Lloyd (HHL) algorithm, one of the most interesting applications of the quantum phase estimation algorithm, which can be used to “solve” sparse linear linear systems, i.e. a system involving a matrix in which most of the elements are zero. $$HHL: \ket b \rightarrow \ket {A^{-1}b}$$ In the next section the following assumptions will be true: • $A$ is a sparse and Hermitian matrix • the quantum state $\ket b$ doesn’t have to be implemented from $b$ • the problem requires to find $\ket x$ instead of $x$ and the next section deal with: • the Quantum Phase Estimation algorithm • the workflow of HHL • complexity analysis of the HHL algorithm • what happens to the quantum advantage when the above assumptions fail to hold • a brief discussion of a couple of noteworthy applications of HHL Please also note that many versions of the HHL algorithms have been proposed and this post only describe and deals with its simplest version. Background: Quantum Phase Estimation One of the most useful quantum subroutines, called Quantum Phase Estimation, aims to estimate the phase $\phi$ of an eigenvalue $e^{2i\pi\phi}$ of the corresponding eigenvector $\ket \psi$of a unitary operator $U$. The QPE algorithm, depicted in the circuit above, shares similarities with Shor’s algorithm because Shor's algorithm can be seen as a specific application of QPE for integer factorization and the goal of QPE is to encode an estimation of the phase $\phi$ into a binary representation like: $$\phi = 0.\phi_1\phi_2\dots\phi_{n-1}\phi_n$$ QPE archive the result by phase encoding the binary representation of $\phi$ using controlled $U$ gates in order to get the following state: $$\left(\bigotimes_{j=1}^n \frac 1{\sqrt 2}(\ket 0 + e^{2i\pi0.\phi_j\dots \phi_n}\ket 1)\right) \otimes \ket\psi$$ and then applying the inverse of the Quantum Fourier Transform to go from the phase space to the state space and, before measuring, the result is: $$\left(\bigotimes_{j=1}^n \ket {\phi_j}\right) \otimes \ket\psi = \ket {\hat \phi} \otimes \ket \psi$$ where $\hat \phi$ is the estimation of $\phi$. HHL workflow The above picture depicts the circuit of the HHL algorithm. One may notice that the algorithm can be broken down into 3 parts mainly: • a QPE • a controlled rotation • an inverse QFE Assuming the input state $\ket b$ is already prepared, the first block is used to find the phase of $\{\lambda_i\}$ the eigenvalues of the matrix $U = e^{i tA}$, and the approximated result is stored in the middle register. At the end of the QPE, what we have is: $$\ket 0 \otimes\left(\sum_i a_i \ket {u_i} \otimes \ket{\hat\lambda_i}\right)$$ where $\sum_i a_i\ket {u_i}$ is $\ket b$ expressed in terms of $\ket {u_i}$, the eigenvalues of $U$ and $\hat \lambda_i$ is the binary approximation of the phase of $U$. Then a controlled rotation gate is applied, which corresponds to the following transformation: $$\ket 0 \otimes\left( \sum_i a_i \ket {u_i} \otimes \ket{\hat\lambda_i}\right) \rightarrow \left(\sqrt{1-\left(\frac c\lambda\right)^2}\ket 0 + \frac c\lambda \ket 1\right)\otimes\left(\sum_i a_i \ ket {u_i} \otimes \ket{\hat\lambda_i}\right)$$ where $c$ is a normalization constant. The last block, the inverse QPE, is used to go from the state above to: $$\ket q\otimes\left(\sum_i a_i \ket {u_i} \otimes \ket{\hat\lambda_i}\right) \rightarrow \ket q \otimes\left(\sum_i a_i \ket {u_i}\right)\otimes \ket{0}$$ where $\ket q \equiv \left(\sqrt{1-\left(\frac c\lambda\right)^2}\ket 0 + \frac c\lambda \ket 1\right)$. Notably, if the fist qubit (the top register) is measured we have two cases: • if $\ket q$ collapses into 1: once the middle register is measure as well, the result is: $$∝\sum_i a_i \ket u_i \otimes \frac c{\lambda_i}$$ which is proportional to $\ket {A^{-1}b}$ because of the spectral decomposition of $A$. In fact $A = \sum_i \lambda_i u_iu_i^\dagger$ and (by the properties of spectral decomposition) $A^{-1} = \sum_i \lambda_i^{-1} u_iu_i^\dagger$ hence $A^{-1}b = \sum_i a_i \lambda_i^{-1} u_i$ being $u_i^\dagger u_i =1$ (for the properties of quantum states). • if $\ket q$ collapses into 0, one may run again the program Complexity analysis • $k $ the conditional number (ratio of the largest and smallest absolute values of eigenvalues of $A$) • $\epsilon $ the error from the output state $\ket {A^{-1}b}$ • $s$ the maximum number of non-zero elements in each row of the matrix $A$ • $N$ the size of the matrix In fact, simulating $e^{-iAt}$, if $A$ is $s$-sparse, can be done with error $\epsilon$ in $O(\log(N)s^2t\epsilon^{-1})$, which is required in the QPE process. One may then perform O(k) Quantum Amplitude Amplification repetitions to amplify the probability of measuring $1$, since $C=O(\frac 1k)$ and if $\lambda \leq 1$, the probability of measuring $1$ is $\Omega(\frac 1{k^ Putting all together, then the computational complexity of the original HHL algorithm is: however many improvements have been made and the computational complexity of the currently most efficient HHL algorithm is: $$O\left(poly(\log(sk\epsilon^{-1}))sk\right )$$ and if we assume \(s = O(poly\left(\log(N)\right))$, the algorithm (focusing only on $N$) runs in: which represents an exponential speedup in the matrix dimension compared to the best conjugate gradient method, whose complexity is: $$O \left(Nsκlog\left(\frac 1\epsilon\right)\right)$$ However this holds on very specific assumptions and the next section deals with what happens if some of the assumptions are not met. Loss of quantum advantage and near term feasibility of HHL The computational complexity above, is based on the assumptions that: • $\ket b$ is already available • doesn’t consider that $\ket {A^{-1}b} $ should be read out Note that if this input/output overhead takes $O(N)$, the exponential speedup is lost. The computational cost of encoding $b$ in $\ket b$ is: if $b$ is a simple bitstring and in general is: for a generic superposition, which results in the loss of the exponential speedup. Moreover, also reading out the output solution state $\ket {A^{-1}b}$ into a classical bitstring $A^{-1}b$ requires $O(N)$, offsetting the exponential acceleration. HHL in solving linear differential equations One of the main applications of the HHL algorithm is solving linear differential equations. Quantum computers in fact can simulate quantum systems (which are described by a restricted type of linear differential equations), and using HHL it’s possible to solve general inhomogeneous sparse linear differential equations. A first-order ordinary differential equation may be written as: $$\frac {\delta x(t)}{\delta t}=A(t)x(t) + b(t)$$ where $A(t)$ is a $N\times N$ matrix we assume to be sparse and $x(t)$ and $b(t)$ are $N$-components vectors. A similar system can be the output of a conversion process from any linear differential equation with higher-order derivatives or from the discretization of partial differential equations. A bunch of different methods involving HHL can be used to solve the above DE, however the workflow is roughly the same: • discretize the differential equation and get a system of algebraic equation • use HHL to find the solution of the system In fact, one may apply a discretization scheme to the DE, for example the Euler method, to map the DE to a difference equation: $$\frac{x_{i+1} + x_i}h= A(t_i)x_i + b(t_i)$$ and it is straightforward to see that this methods results in the following linear system: where $x$ is the vector of blocks $x_i$, and $b$ also contains the value of $x_0$. To learn more about this please see Berry, (2014), “High-order quantum algorithm for solving linear differential equations”. HHL in solving least-square curve fitting Another interesting application for HHL is least squares fitting. The goal in least squares fitting is to find a continuous function to approximate a discrete set of $N$ points $\{x_i, y_i\}$. The function has to be linear in the parameter $\theta $ but can be non linear in $x$, e.g.: $$f(\theta, x) = \sum_i \theta_if_i(x)$$ The optimal parameters can be found by minimizing an error function such as the mean squared error: $$E = \|y - f(\theta, x)\|^2$$ which can be expressed in matrix for as: $$E= \|y- F\theta\|^2$$ where $F_{ij}=f_j(x_i)$. The best fitting parameter can be found using Moore– Penrose pseudoinverse as: $$\theta^* = \left(F^\dagger F\right)^{-1}F^\dagger y$$ Finding the best $\theta$ then involves 3 subroutines: • performing the pseudo–inverse using the HHL algorithm and quantum matrix multiplication • an algorithm for estimating the fit quality • an algorithm for learning the fit-parameters $\theta$ To learn more about this please consider reading Wiebe, Brown, LLoyd, (2012), “Quantum Data-Fitting“. And that's it for this article. Thanks for reading. For any question or suggestion related to what I covered in this article, please add it as a comment. For special needs, you can contact me here.	{"url":"https://amm.zanotp.com/hhl","timestamp":"2024-11-10T22:30:19Z","content_type":"text/html","content_length":"250788","record_id":"<urn:uuid:17354f53-9345-47a2-9233-71e8ea367269>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00141.warc.gz"}
Thermal Equivalent Circuit for Stator \| Ansys Courses This lesson covers the fundamental aspects of thermal equivalent circuits in electrical machines. It delves into the case study of a stator circuit, discussing related losses such as stator copper loss and stator iron losses. The lesson further elaborates on the development of the thermal equivalent circuit, considering heat flow, thermal nodes, and different types of losses. It also explains how to calculate thermal resistances and capacitances, and how to write transient thermal equations. The lesson provides a comprehensive understanding of how to analyze and design an appropriate thermal circuit, using illustrative examples and equations. Video Highlights 00:00 - Introduction 00:44 - Case study of a stator circuit 07:09 - Analysis of thermal resistance 19:27 - Analysis of thermal Capacitance 22:16 - Transient equation for thermal network Key Takeaways - Thermal equivalent circuits are crucial in understanding the heat flow in electrical machines. - Stator circuits and their associated losses play a significant role in these circuits. - Transient thermal equations help in calculating the thermal temperatures at various thermal nodes. - The analysis and design of an appropriate thermal circuit are essential for efficient heat management in electrical machines.	{"url":"https://innovationspace.ansys.com/courses/courses/thermal-design-of-electrical-machines/lessons/thermal-equivalent-circuit-for-stator-lesson-7/","timestamp":"2024-11-04T23:45:53Z","content_type":"text/html","content_length":"178629","record_id":"<urn:uuid:9305ce1d-273f-4c34-8f99-586774968988>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00066.warc.gz"}
The Black-Scholes Model, a groundbreaking framework in financial economics, was developed by Fischer Black and Myron Scholes in 1973, with key extensions from Robert C. Merton. This model fundamentally transformed the approach to options trading and significantly influenced global financial markets. For their contributions, Scholes and Merton received the Nobel Prize in Economic Sciences in 1997. Understanding the Black-Scholes model The model provides a mathematical formula for pricing European-style options, employing variables such as the underlying asset's price, strike price, risk-free interest rate, time until expiration, and the volatility of the asset. At its core, the model assumes a continuous, log-normally distributed rate of return, aiming to eliminate financial risk through a hedging strategy that involves dynamic rebalancing. Key components and assumptions The Black-Scholes Model rests on several critical assumptions: the absence of dividends during the life of the option, constant risk-free interest rates, and the ability to borrow and lend money at the risk-free rate. Furthermore, it assumes no transaction costs or taxes and allows for the continuous trading of assets. Despite these idealized conditions, the model has been widely adopted and adapted for various financial applications. The mathematical framework The Black-Scholes model relies on a particular type of math equation known as a partial differential equation. This equation calculates the option's theoretical price by factoring in time and the asset's volatility. For practical application, the model yields explicit formulas for the prices of call and put options, facilitating their valuation and trading in the market. Practical implications and limitations While the Black-Scholes Model has been instrumental in advancing financial derivatives trading, it is not without its limitations. Real-world deviations from its assumptions, such as changing volatility and the presence of dividends, can lead to discrepancies between theoretical and actual prices. Nonetheless, it remains a cornerstone of modern financial theory and practice, with ongoing modifications and extensions improving its applicability to a wider range of financial instruments. The Black-Scholes Model remains a fundamental tool in the pricing of stock options and the management of financial risk. Despite its simplifications, the model's conceptual framework and methodologies continue to underpin much of modern financial market theory and practice, illustrating the enduring impact of Black, Scholes, and Merton's work on the field of financial economics.	{"url":"https://coinmetro.com/glossary/black-scholes-model","timestamp":"2024-11-11T00:10:15Z","content_type":"text/html","content_length":"179516","record_id":"<urn:uuid:98fd06aa-e0f5-4041-8efd-9a077d795cf0>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00708.warc.gz"}
DOI Number The paper considers a cylindrical three-layer structure of arbitrary thickness made of viscoelastic material. It consists of two external bearing layers and a middle layer, the materials of which are generally different. The problem of nonstationary longitudinal-radial vibrations of such a structure is formulated. Based on the exact solutions in transformations of the three-dimensional problem of the linear theory of viscoelasticity for a circular cylindrical three-layer body, a mathematical model of its nonstationary longitudinal-radial vibrations is developed. Equations are derived that allow, based on the results of solving the vibration equations, to determine the stress-strain state of a cylindrical structure and its layers in arbitrary sections. The results obtained allow for special cases of transition into cylindrical viscoelastic and elastic two-layer structures, as well as into homogeneous single-layer cylindrical structures and round rods. Three-layer structure, Vibration, Stress, Torsional displacement, Load-bearing layers, Non-stationary Liang, W., Liu, T., Li, Ch., Wang, Q., 2023, Three-Dimensional Vibration Model of Cylindrical Shells via Carrera Unified Formulation, Materials, 16, 3345. Naeem, M.N., Khan, A.G., Arshad, Sh.H., Abdul, G.Sh., Gamkhar, M., 2014, Vibration of Three-Layered FGM Cylindrical Shells with Middle Layer of Isotropic Material for Various Boundary Conditions, World Journal of Mechanics, 4(11), pp. 315–331. Ye, T., Jin, G., Shi, S., Ma, X., 2014, Three-dimensional free vibration analysis of thick cylindrical shells with general end conditions and resting on elastic foundations, International Journal of Mechanical Sciences, 84, pp.120–137. Banks, H.T., Hu, S., Kenz, Z.R., 2011, A Brief Review of Elasticity and Viscoelasticity for Solids, Advances in Applied Mathematics and Mechanics, 3(1), pp. 1–51. Khudoynazarov, Kh.Kh., 2006, Transversal vibrations of thick and thin cylindrical shells, interacting with deformable medium, Proceedings of the 8th international conference on shell structures SSTA 2005, Jurata, Gdansk, Poland, Shell Structures: Theory and Applications, London: Taylor & Francis Group, pp. 343–347. Mofakhami, M.R., Toudeshky, H.H., Hashemi, S.H., 2006, Finite cylinder vibrations with different end boundary conditions, Journal of Sound and Vibration, 297, pp. 293–314. Sahoo, R., Grover, N., Singh, B.N., 2021, Random vibration response of composite–sandwich laminates, Archive of Applied Mechanics, 91, pp. 3755–3771. Ghamkhar, M., Naeem, M.N., Imran, M., Soutis, C., 2019, Vibration Analysis of a Three-Layered FGM Cylindrical Shell Including the Effect Of Ring Support, Open Physics, 17(1), pp.587-600. Shah., A.G., Mahmood, T., Naeem, M.N., Zafar, A.Sh., Arshad, Sh.H, 2010, Vibrations of functionally graded cylindrical shells based on elastic foundations, Acta Mechanica, 211(30), pp. 293–307. Sayyad, A.S., Ghugal, Y.M., 2015, On the free vibration analysis of laminated composite and sandwich plates: A review of recent literature with some numerical results, Composite Structures, 129, pp. Altukhov, E.V., Fomenko, M.V., 2010, Elastic equalization of a three-layer plate with imperfect layer contact, Bulletin of Donetsk National University, Series A, Natural Sciences, 1, pp.78-87. Altukhov, E.V., Fomenko, M.V., 2011, Vibrations of three-layer plates in the case of boundary conditions of the flat end and slipping of layers, Bulletin of Donetsk National University, Series.A, Natural Sciences, 2, pp. 34–41. Iqbal, Z., Naeem, M.N., Sultana, N., 2009, Vibration characteristics of FGM circular cylindrical shells, Acta Mechanica, 208, pp. 237–248. Sofiyev, A.H., Deniz, A., Akçay, I.H., Yusufoğlu, E.,2006, The vibration and stability of a three-layered conical shell containing an FGM layer subjected to axial compressive load, Acta Mechanica, 183, pp. 129–144. Dimitrienko, Yu.I., Gubareva, E.A., Yakovleva, D.O., 2014, Asymptotic theory of viscoelasticity of multilayer thin composite plates, Science and Education of Bauman Moscow State Technical University, Electron Journal, 14, pp. 359–382. Khudoynazarov, Kh., Abdurazakov, J., Kholikov, D., 2022, Nonlinear torsional vibrations of a circular cylindrical elastic shell, AIP Conference Proceedings, 2637, 020003. Milić P., Marinković D., Klinge S., Ćojbašić Ž., 2023, Reissner-Mindlin Based Isogeometric Finite Element Formulation for Piezoelectric Active Laminated Shells, Tehnicki Vjesnik, 30(2), pp. 416 - Vyachkin E.S., Kaledin V.O., Reshetnikova E.V., Vyachkina E.A., Gileva A.E., 2018, Mathematical modeling of static deformation of a layered construction with incompressible layers, Tomsk State University Journal of Mathematics and Mechanics, 55, pp.72–83. Khudoynazarov Kh.Kh., Khalmuradov R.I., Yalgashev B.F., 2021, Longitudinal-radial vibrations of a elastic cylindrical shell filled with a viscous compressible liquid, Tomsk State University, Journal of Mathematics and Mechanics, 69, pp. 139-154. Filippov, I.G., Filippov, S.I., 2007, Vibratory and wave processes in continuous compressible media, Moskow: VINITI, 429 p. Brekhovskikh, L., 2012, Waves in Layered Media, Elsevier, 574. Khudayarov, B.A., Turaev, F.Z., 2019, Mathematical simulation of nonlinear vibrations of viscoelastic pipelines conveying fluid, Applied Mathematical Modelling, 66, pp. 662–679. Netrebko, A.V., Pshenichnov, S.G., 2015, Some problems of dynamics of linear-viscoelastic cylindrical shells of finite length, Problems of strength and ductility, 77(1), pp. 67–74. Rama G., Marinkovic D., Zehn M., 2018, High performance 3-node shell element for linear and geometrically nonlinear analysis of composite laminates, Composites Part B: Engineering, 151, pp. 118 - Safaei, B., Chukwueloka Onyibo, E., Goren, M., Kotrasova, K., Yang, Z., Arman, S., Asmael, M., 2023, Free vibration investigation on rve of proposed honeycomb sandwich beam and material selection optimization, Facta Universitatis-Series Mechanical Engineering, 21(1), pp. 31-50. Zhao, Z., Yuan, X., Zhang, W., Niu, D., Zhang, H., 2021, Dynamical modeling and analysis of hyperelastic spherical shells under dynamic loads and structural damping, Applied Mathematical Modelling, 95, pp. 468-483. Khudoynazarov, Kh., 2023, A mathematical model of physically nonlinear torsional vibrations of a circular elastic rod, Tomsk State University Journal of Mathematics and Mechanics, 84, pp. 152–166. • There are currently no refbacks. ISSN: 0354-2025 (Print) ISSN: 2335-0164 (Online) COBISS.SR-ID 98732551 ZDB-ID: 2766459-4	{"url":"https://casopisi.junis.ni.ac.rs/index.php/FUMechEng/article/view/12394/0","timestamp":"2024-11-04T01:50:56Z","content_type":"application/xhtml+xml","content_length":"26464","record_id":"<urn:uuid:807a3df4-263e-45e8-ad1c-8ae02e494f7a>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00604.warc.gz"}
Algebraic structures connected with pairs of compatible associative algebras We study associative multiplications in semi-simple associative algebras over C compatible with the usual one or, in other words, linear deformations of semi-simple associative algebras over C. It turns out that these deformations are in one-to-one correspondence with representations of certain algebraic structures, which we call M-structures in the matrix case and PM-structures in the case of direct sums of several matrix algebras. We also investigate various properties of PM-structures, provide numerous examples and describe an important class of PM-structures. The classification of these PM-structures naturally leads to affine Dynkin diagrams of A, D, E-type. arXiv Mathematics e-prints Pub Date: December 2005 □ Mathematics - Quantum Algebra; □ Mathematics - Rings and Algebras; □ Mathematics - Representation Theory; □ High Energy Physics - Theory; □ Nonlinear Sciences - Exactly Solvable and Integrable Systems; □ 17B80; □ 17B63; □ 32L81; □ 14H70 29 pages, Latex. The case of semi-simple algebras A and B is completed (Chapter 4). A construction of compatible products is added (Chapter 1)	{"url":"https://ui.adsabs.harvard.edu/abs/2005math.....12499O","timestamp":"2024-11-10T15:11:43Z","content_type":"text/html","content_length":"37677","record_id":"<urn:uuid:c892afa8-c812-42d3-be37-170412f28eb5>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00403.warc.gz"}
How To Calculate Percentage Of Weight Loss In Excel - Calculator How To Calculate Percentage Of Weight Loss In Excel – Calculator Microsoft Excel is one of the core and important software, which almost everyone uses for their work. Excel is a unique spreadsheet software to organize or calculate any type of data within a short period. But lots of people don’t know How To Calculate Percentage Of Weight Loss In Excel. You are in a place where you can easily find authentic information and details, which will enhance your knowledge to calculate the percentage of weight loss. If you want to calculate your percentage of weight loss in excel then keep in mind that it will take a few minutes to give you the result. There are multiple ways through which you can calculate the percentage of weight loss in excel and all these methods are given in this article. So, let’s jump into the depth details! 3 easy and best ways to calculate the percentage of weight loss in excel Before doing any calculation in excel, you need to set the cell in the percentage format. After this step, Excel will automatically display the results in percentage format rather than the fraction. For this purpose, make sure to highlight the cells, then you need to select the Home option and then select Percentage. Method 1: Use the MIN function and assign a minimum value to calculate the percentage There is a high chance to get the overall results of weight loss percentage over multiple months. This method is really useful for the purpose to collect weight loss percentages for various months. In that case, you need to get the minimum weight within a specific range to deduct especially from the starting weight. Excel MIN function does a great job for this purpose. Step 1: Use formula in any blank cell = (C5-MIN(C5:C15))/C5 This MIN function will fetch the overall minimum weight within the specific range. Step 2: Then you need to hit Enter to get display the percentage of weight loss. If you don’t set the percentage format in the cell, then Excel will show you the percentage of weight loss in decimals. Method 2: Using Arithmetic Formula to calculate the percentage of weight loss The arithmetic formula will help you to calculate the percentage of weight loss. The process of its working is simple and easy, it just subtracts subsequent weight and then divides the starting/ initial weight resulting in the percentage of weight loss. Step 1: Just type or paste the given formula into your desired cell ABS function will pass only the absolute value of your desired number Step 2: As the Excel cells are formatted previously as a percentage, you just need to hit Enter, then drag Fill Handle and it will display you the list of percentage weight loss for every 15 days interval. Method 3: Use the LOOKUP Function This method will bring the last value in the percentage calculation, which is similar to method 1. With the help of the LOOKUP function, you will easily find the last weight from the overall range. Step 1: Type or paste the given formula into your desired cell =(C5 – LOOKUP(1,1/(C5:C15<>””),C5:C15))/C5 Step 2: To apply the formula, you need to use Enter key and Excel will give you the overall percentage of weight loss in a matter of seconds. You can easily label the cells in column A which you want to choose, it will express what kind of data you should need to type into other cells B1 and B2. Moreover, if you face difficulty finding the percentage button in the toolbar option, then there is also another way through which you can choose the percentage format. Right click on any cell. Then you need to select “Percentage” and just click “OK”. Now you get an idea of three different methods through which you can easily calculate the percentage of weight loss in Excel. Keep in mind that all the methods or formulas are capable, but you only need to select the method according to the requirements or the data type. Related Posts © Copyright 2021 PercentageCalculatorFree.com \|All Rights Reserved.	{"url":"https://percentagecalculatorfree.com/how-to-calculate-percentage-of-weight-loss-in-excel/","timestamp":"2024-11-04T05:40:27Z","content_type":"text/html","content_length":"73693","record_id":"<urn:uuid:60107687-ca1c-432d-81f8-f358fe22ab39>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00892.warc.gz"}
Perform Factor Analysis on Exam Grades This example shows how to perform factor analysis using Statistics and Machine Learning Toolbox™. Multivariate data often include a large number of measured variables, and sometimes those variables "overlap" in the sense that groups of them may be dependent. For example, in a decathlon, each athlete competes in 10 events, but several of them can be thought of as "speed" events, while others can be thought of as "strength" events, etc. Thus, a competitor's 10 event scores might be thought of as largely dependent on a smaller set of 3 or 4 types of athletic ability. Factor analysis is a way to fit a model to multivariate data to estimate just this sort of interdependence. The Factor Analysis Model In the factor analysis model, the measured variables depend on a smaller number of unobserved (latent) factors. Because each factor may affect several variables in common, they are known as "common factors". Each variable is assumed to depend on a linear combination of the common factors, and the coefficients are known as loadings. Each measured variable also includes a component due to independent random variability, known as "specific variance" because it is specific to one variable. Specifically, factor analysis assumes that the covariance matrix of your data is of the form SigmaX = Lambda*Lambda' + Psi where Lambda is the matrix of loadings, and the elements of the diagonal matrix Psi are the specific variances. The function factoran fits the factor analysis model using maximum likelihood. Example: Finding Common Factors Affecting Exam Grades 120 students have each taken five exams, the first two covering mathematics, the next two on literature, and a comprehensive fifth exam. It seems reasonable that the five grades for a given student ought to be related. Some students are good at both subjects, some are good at only one, etc. The goal of this analysis is to determine if there is quantitative evidence that the students' grades on the five different exams are largely determined by only two types of ability. First load the data, then call factoran and request a model fit with a single common factor. load examgrades [Loadings1,specVar1,T,stats] = factoran(grades,1); factoran's first two return arguments are the estimated loadings and the estimated specific variances. From the estimated loadings, you can see that the one common factor in this model puts large positive weight on all five variables, but most weight on the fifth, comprehensive exam. Loadings1 = One interpretation of this fit is that a student might be thought of in terms of their "overall ability", for which the comprehensive exam would be the best available measurement. A student's grade on a more subject-specific test would depend on their overall ability, but also on whether or not the student was strong in that area. This would explain the lower loadings for the first four exams. From the estimated specific variances, you can see that the model indicates that a particular student's grade on a particular test varies quite a lot beyond the variation due to the common factor. specVar1 = A specific variance of 1 would indicate that there is no common factor component in that variable, while a specific variance of 0 would indicate that the variable is entirely determined by common factors. These exam grades seem to fall somewhere in between, although there is the least amount of specific variation for the comprehensive exam. This is consistent with the interpretation given above of the single common factor in this model. The p-value returned in the stats structure rejects the null hypothesis of a single common factor, so we refit the model. Next, use two common factors to try and better explain the exam scores. With more than one factor, you could rotate the estimated loadings to try and make their interpretation simpler, but for the moment, ask for an unrotated solution. [Loadings2,specVar2,T,stats] = factoran(grades,2,'rotate','none'); From the estimated loadings, you can see that the first unrotated factor puts approximately equal weight on all five variables, while the second factor contrasts the first two variables with the second two. Loadings2 = 0.6289 0.3485 0.6992 0.3287 0.7785 -0.2069 0.7246 -0.2070 0.8963 -0.0473 You might interpret these factors as "overall ability" and "quantitative vs. qualitative ability", extending the interpretation of the one-factor fit made earlier. A plot of the variables, where each loading is a coordinate along the corresponding factor's axis, illustrates this interpretation graphically. The first two exams have a positive loading on the second factor, suggesting that they depend on "quantitative" ability, while the second two exams apparently depend on the opposite. The fifth exam has only a small loading on this second factor. biplot(Loadings2, 'varlabels',num2str((1:5)')); title('Unrotated Solution'); xlabel('Latent Factor 1'); ylabel('Latent Factor 2'); From the estimated specific variances, you can see that this two-factor model indicates somewhat less variation beyond that due to the common factors than the one-factor model did. Again, the least amount of specific variance occurs for the fifth exam. specVar2 = The stats structure shows that there is only a single degree of freedom in this two-factor model. With only five measured variables, you cannot fit a model with more than two factors. Factor Analysis from a Covariance/Correlation Matrix You made the fits above using the raw test scores, but sometimes you might only have a sample covariance matrix that summarizes your data. factoran accepts either a covariance or correlation matrix, using the 'Xtype' parameter, and gives an identical result to that from the raw data. Sigma = cov(grades); [LoadingsCov,specVarCov] = ... LoadingsCov = 0.6289 0.3485 0.6992 0.3287 0.7785 -0.2069 0.7246 -0.2070 0.8963 -0.0473 Factor Rotation Sometimes, the estimated loadings from a factor analysis model can give a large weight on several factors for some of the measured variables, making it difficult to interpret what those factors represent. The goal of factor rotation is to find a solution for which each variable has only a small number of large loadings, i.e., is affected by a small number of factors, preferably only one. If you think of each row of the loadings matrix as coordinates of a point in M-dimensional space, then each factor corresponds to a coordinate axis. Factor rotation is equivalent to rotating those axes, and computing new loadings in the rotated coordinate system. There are various ways to do this. Some methods leave the axes orthogonal, while others are oblique methods that change the angles between them. Varimax is one common criterion for orthogonal rotation. factoran performs varimax rotation by default, so you do not need to ask for it explicitly. [LoadingsVM,specVarVM,rotationVM] = factoran(grades,2); A quick check of the varimax rotation matrix returned by factoran confirms that it is orthogonal. Varimax, in effect, rotates the factor axes in the figure above, but keeps them at right angles. ans = 1.0000 0.0000 0.0000 1.0000 A biplot of the five variables on the rotated factors shows the effect of varimax rotation. biplot(LoadingsVM, 'varlabels',num2str((1:5)')); title('Varimax Solution'); xlabel('Latent Factor 1'); ylabel('Latent Factor 2'); Varimax has rigidly rotated the axes in an attempt to make all of the loadings close to zero or one. The first two exams are closest to the second factor axis, while the third and fourth are closest to the first axis and the fifth exam is at an intermediate position. These two rotated factors can probably be best interpreted as "quantitative ability" and "qualitative ability". However, because none of the variables are near a factor axis, the biplot shows that orthogonal rotation has not succeeded in providing a simple set of factors. Because the orthogonal rotation was not entirely satisfactory, you can try using promax, a common oblique rotation criterion. [LoadingsPM,specVarPM,rotationPM] = ... A check on the promax rotation matrix returned by factoran shows that it is not orthogonal. Promax, in effect, rotates the factor axes in the first figure separately, allowing them to have an oblique angle between them. ans = 1.9405 -1.3509 -1.3509 1.9405 A biplot of the variables on the new rotated factors shows the effect of promax rotation. biplot(LoadingsPM, 'varlabels',num2str((1:5)')); title('Promax Solution'); xlabel('Latent Factor 1'); ylabel('Latent Factor 2'); Promax has performed a non-rigid rotation of the axes, and has done a much better job than varimax at creating a "simple structure". The first two exams are close to the second factor axis, while the third and fourth are close to the first axis, and the fifth exam is in an intermediate position. This makes an interpretation of these rotated factors as "quantitative ability" and "qualitative ability" more precise. Instead of plotting the variables on the different sets of rotated axes, it's possible to overlay the rotated axes on an unrotated biplot to get a better idea of how the rotated and unrotated solutions are related. h1 = biplot(Loadings2, 'varlabels',num2str((1:5)')); xlabel('Latent Factor 1'); ylabel('Latent Factor 2'); hold on invRotVM = inv(rotationVM); h2 = line([-invRotVM(1,1) invRotVM(1,1) NaN -invRotVM(2,1) invRotVM(2,1)], ... [-invRotVM(1,2) invRotVM(1,2) NaN -invRotVM(2,2) invRotVM(2,2)],'Color',[1 0 0]); invRotPM = inv(rotationPM); h3 = line([-invRotPM(1,1) invRotPM(1,1) NaN -invRotPM(2,1) invRotPM(2,1)], ... [-invRotPM(1,2) invRotPM(1,2) NaN -invRotPM(2,2) invRotPM(2,2)],'Color',[0 1 0]); hold off axis square lgndHandles = [h1(1) h1(end) h2 h3]; lgndLabels = {'Variables','Unrotated Axes','Varimax Rotated Axes','Promax Rotated Axes'}; legend(lgndHandles, lgndLabels, 'location','northeast', 'fontname','arial narrow'); Predicting Factor Scores Sometimes, it is useful to be able to classify an observation based on its factor scores. For example, if you accepted the two-factor model and the interpretation of the promax rotated factors, you might want to predict how well a student would do on a mathematics exam in the future. Since the data are the raw exam grades, and not just their covariance matrix, we can have factoran return predictions of the value of each of the two rotated common factors for each student. [Loadings,specVar,rotation,stats,preds] = ... biplot(Loadings, 'varlabels',num2str((1:5)'), 'Scores',preds); title('Predicted Factor Scores for Promax Solution'); xlabel('Ability In Literature'); ylabel('Ability In Mathematics'); This plot shows the model fit in terms of both the original variables (vectors) and the predicted scores for each observation (points). The fit suggests that, while some students do well in one subject but not the other (second and fourth quadrants), most students do either well or poorly in both mathematics and literature (first and third quadrants). You can confirm this by looking at the estimated correlation matrix of the two factors. ans = 1.0000 0.6962 0.6962 1.0000 A Comparison of Factor Analysis and Principal Components Analysis There is a good deal of overlap in terminology and goals between Principal Components Analysis (PCA) and Factor Analysis (FA). Much of the literature on the two methods does not distinguish between them, and some algorithms for fitting the FA model involve PCA. Both are dimension-reduction techniques, in the sense that they can be used to replace a large set of observed variables with a smaller set of new variables. They also often give similar results. However, the two methods are different in their goals and in their underlying models. Roughly speaking, you should use PCA when you simply need to summarize or approximate your data using fewer dimensions (to visualize it, for example), and you should use FA when you need an explanatory model for the correlations among your data.	{"url":"https://nl.mathworks.com/help/stats/perform-factor-analysis-on-exam-grades.html","timestamp":"2024-11-09T03:25:15Z","content_type":"text/html","content_length":"86700","record_id":"<urn:uuid:b965c88a-ef43-4e53-927b-a37e9a31b417>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00366.warc.gz"}
Thermal conductivity From New World Encyclopedia In physics, thermal conductivity, ${\displaystyle k}$, is the property of a material that indicates its ability to conduct heat. It appears primarily in Fourier's Law for heat conduction. Conduction is the most significant means of heat transfer in a solid. By knowing the values of thermal conductivities of various materials, one can compare how well they are able to conduct heat. The higher the value of thermal conductivity, the better the material is at conducting heat. On a microscopic scale, conduction occurs as hot, rapidly moving or vibrating atoms and molecules interact with neighboring atoms and molecules, transferring some of their energy (heat) to these neighboring atoms. In insulators the heat flux is carried almost entirely by phonon vibrations. Mathematical background First, heat conduction can be defined by the formula: ${\displaystyle H={\frac {\Delta Q}{\Delta t}}=k\times A\times {\frac {\Delta T}{x}}}$ where ${\displaystyle {\frac {\Delta Q}{\Delta t}}}$ is the rate of heat flow, k is the thermal conductivity, A is the total surface area of conducting surface, Î”T is temperature difference and x is the thickness of conducting surface separating the two temperatures. Thus, rearranging the equation gives thermal conductivity, ${\displaystyle k={\frac {\Delta Q}{\Delta t}}\times {\frac {1}{A}}\times {\frac {x}{\Delta T}}}$ (Note: ${\displaystyle {\frac {\Delta T}{x}}}$ is the temperature gradient) In other words, it is defined as the quantity of heat, Î”Q, transmitted during time Î”t through a thickness x, in a direction normal to a surface of area A, due to a temperature difference Î”T, under steady state conditions and when the heat transfer is dependent only on the temperature gradient. Alternately, it can be thought of as a flux of heat (energy per unit area per unit time) divided by a temperature gradient (temperature difference per unit length) ${\displaystyle k={\frac {\Delta Q}{A\times {}\Delta t}}\times {\frac {x}{\Delta T}}}$ Typical units are SI: W/(mÂ·K) and English units: BtuÂ·ft/(hÂ·ftÂ²Â·Â°F). To convert between the two, use the relation 1 BtuÂ·ft/(hÂ·ftÂ²Â·Â°F) = 1.730735 W/(mÂ·K).^[1] In metals, thermal conductivity approximately tracks electrical conductivity according to the Wiedemann-Franz law, as freely moving valence electrons transfer not only electric current but also heat energy. However, the general correlation between electrical and thermal conductance does not hold for other materials, due to the increased importance of phonon carriers for heat in non-metals. As shown in the table below, highly electrically conductive silver is less thermally conductive than diamond, which is an electrical insulator. Thermal conductivity depends on many properties of a material, notably its structure and temperature. For instance, pure crystalline substances exhibit very different thermal conductivities along different crystal axes, due to differences in phonon coupling along a given crystal axis. Sapphire is a notable example of variable thermal conductivity based on orientation and temperature, for which the CRC Handbook reports a thermal conductivity of 2.6 W/(mÂ·K) perpendicular to the c-axis at 373Â K, but 6000 W/(mÂ·K) at 36 degrees from the c-axis and 35Â K (possible typo?). Air and other gases are generally good insulators, in the absence of convection. Therefore, many insulating materials function simply by having a large number of gas-filled pockets which prevent large-scale convection. Examples of these include expanded and extruded polystyrene (popularly referred to as "styrofoam") and silica aerogel. Natural, biological insulators such as fur and feathers achieve similar effects by dramatically inhibiting convection of air or water near an animal's skin. Thermal conductivity is important in building insulation and related fields. However, materials used in such trades are rarely subjected to chemical purity standards. Several construction materials' k values are listed below. These should be considered approximate due to the uncertainties related to material definitions. The following table is meant as a small sample of data to illustrate the thermal conductivity of various types of substances. For more complete listings of measured k-values, see the references. List of thermal conductivities This is a list of approximate values of thermal conductivity, k, for some common materials. Please consult the list of thermal conductivities for more accurate values, references and detailed Material Thermal conductivity Cement, portland ^[2] 0.29 Concrete, stone ^[2] 1.7 Air 0.025 Wood 0.04 - 0.4 Alcohols and oils 0.1 - 0.21 Silica Aerogel 0.004-0.03 Soil 1.5 Rubber 0.16 Epoxy (unfilled) 0.19 Epoxy (silica-filled) 0.30 Water (liquid) 0.6 Thermal grease 0.7 - 3 Thermal epoxy 1 - 4 Glass 1.1 Ice 2 Sandstone 2.4 Stainless steel^[3] 12.11 ~ 45.0 Lead 35.3 Aluminum 237 Gold 318 Copper 401 Silver 429 Diamond 900 - 2320 LPG 0.23 - 0.26 Generally speaking, there are a number of possibilities to measure thermal conductivity, each of them suitable for a limited range of materials, depending on the thermal properties and the medium temperature. There can be made a distinction between steady-state and transient techniques. In general the steady-state techniques perform a measurement when the temperature of the material that is measured does not change with time. This makes the signal analysis straight forward (steady state implies constant signals). The disadvantage generally is that it takes a well-engineered experimental setup. The Divided Bar (various types) is the most common device used for consolidated rock The transient techniques perform a measurement during the process of heating up. The advantage is that measurements can be made relatively quickly. Transient methods are usually carried out by needle probes (inserted into samples or plunged into the ocean floor). For good conductors of heat, Searle's bar method can be used. For poor conductors of heat, Lees' disc method can be used. An alternative traditional method using real thermometers can be used as well. A thermal conductance tester, one of the instruments of gemology, determines if gems are genuine diamonds using diamond's uniquely high thermal conductivity. Standard Measurement Techniques • IEEE Standard 442-1981, "IEEE guide for soil thermal resistivity measurements" see als soil_thermal_properties.^[4] • IEEE Standard 98-2002, "Standard for the Preparation of Test Procedures for the Thermal Evaluation of Solid Electrical Insulating Materials"^[5] • ASTM Standard D5470-06, "Standard Test Method for Thermal Transmission Properties of Thermally Conductive Electrical Insulation Materials"^[6] • ASTM Standard E1225-04, "Standard Test Method for Thermal Conductivity of Solids by Means of the Guarded-Comparative-Longitudinal Heat Flow Technique"^[7] • ASTM Standard D5930-01, "Standard Test Method for Thermal Conductivity of Plastics by Means of a Transient Line-Source Technique"^[8] • ASTM Standard D2717-95, "Standard Test Method for Thermal Conductivity of Liquids"^[9] Difference between US and European notation In Europe, the k-value of construction materials (e.g. window glass) is called Î»-value. U-value used to be called k-value in Europe, but is now also called U-value. K-value (with capital k) refers in Europe to the total isolation value of a building. K-value is obtained by multiplying the form factor of the building (= the total inward surface of the outward walls of the building divided by the total volume of the building) with the average U-value of the outward walls of the building. K-value is therefore expressed as (m^2.m^-3).(W.K^-1.m^-2) = W.K^-1.m ^-3. A house with a volume of 400 mÂ³ and a K-value of 0.45 (the new European norm. It is commonly referred to as K45) will therefore theoretically require 180 W to maintain its interior temperature 1 degree K above exterior temperature. So, to maintain the house at 20Â°C when it is freezing outside (0Â°C), 3600 W of continuous heating is required. Related terms The reciprocal of thermal conductivity is thermal resistivity, measured in kelvin-metres per watt (KÂ·mÂ·W^âˆ’1). When dealing with a known amount of material, its thermal conductance and the reciprocal property, thermal resistance, can be described. Unfortunately there are differing definitions for these terms. First definition (general) For general scientific use, thermal conductance is the quantity of heat that passes in unit time through a plate of particular area and thickness when its opposite faces differ in temperature by one degree. For a plate of thermal conductivity k, area A and thickness L this is kA/L, measured in WÂ·K^âˆ’1 (equivalent to: W/Â°C). Thermal conductivity and conductance are analogous to electrical conductivity (AÂ·m^âˆ’1Â·V^âˆ’1) and electrical conductance (AÂ·V^âˆ’1). There is also a measure known as heat transfer coefficient: the quantity of heat that passes in unit time through unit area of a plate of particular thickness when its opposite faces differ in temperature by one degree. The reciprocal is thermal insulance. In summary: • thermal conductance = kA/L, measured in WÂ·K^âˆ’1 □ thermal resistance = L/kA, measured in KÂ·W^âˆ’1 (equivalent to: Â°C/W) • heat transfer coefficient = k/L, measured in WÂ·K^âˆ’1Â·m^âˆ’2 □ thermal insulance = L/k, measured in KÂ·mÂ²Â·W^âˆ’1. The heat transfer coefficient is also known as thermal admittance Thermal Resistance When thermal resistances occur in series, they are additive. So when heat flows through two components each with a resistance of 1 Â°C/W, the total resistance is 2 Â°C/W. A common engineering design problem involves the selection of an appropriate sized heat sink for a given heat source. Working in units of thermal resistance greatly simplifies the design calculation. The following formula can be used to estimate the performance: ${\displaystyle R_{hs}={\frac {\Delta T}{P_{th}}}-R_{s}}$ • R[hs] is the maximum thermal resistance of the heat sink to ambient, in Â°C/W • ${\displaystyle \Delta T}$ is the temperature difference (temperature drop), in Â°C • P[th] is the thermal power (heat flow), in Watts • R[s] is the thermal resistance of the heat source, in Â°C/W For example, if a component produces 100 W of heat, and has a thermal resistance of 0.5 Â°C/W, what is the maximum thermal resistance of the heat sink? Suppose the maximum temperature is 125 Â°C, and the ambient temperature is 25 Â°C; then the ${\displaystyle \Delta T}$ is 100 Â°C. The heat sink's thermal resistance to ambient must then be 0.5 Â°C/W or less. Second definition (buildings) When dealing with buildings, thermal resistance or R-value means what is described above as thermal insulance, and thermal conductance means the reciprocal. For materials in series, these thermal resistances (unlike conductances) can simply be added to give a thermal resistance for the whole. A third term, thermal transmittance, incorporates the thermal conductance of a structure along with heat transfer due to convection and radiation. It is measured in the same units as thermal conductance and is sometimes known as the composite thermal conductance. The term U-value is another synonym. In summary, for a plate of thermal conductivity k (the k value^[10]), area A and thickness L: • thermal conductance = k/L, measured in WÂ·K^âˆ’1Â·m^âˆ’2; • thermal resistance (R value) = L/k, measured in KÂ·mÂ²Â·W^âˆ’1; • thermal transmittance (U value) = 1/(Î£(L/k)) + convection + radiation, measured in WÂ·K^âˆ’1Â·m^âˆ’2. Textile industry In textiles, a tog value may be quoted as a measure of thermal resistance in place of a measure in SI units. The thermal conductivity of a system is determined by how atoms comprising the system interact. There are no simple, correct expressions for thermal conductivity. There are two different approaches for calculating the thermal conductivity of a system. The first approach employs the Green-Kubo relations. Although this employs analytic expressions which in principle can be solved, in order to calculate the thermal conductivity of a dense fluid or solid using this relation requires the use of molecular dynamics computer simulation. The second approach is based upon the relaxation time approach. Due to the anharmonicity within the crystal potential, the phonons in the system are known to scatter. There are three main mechanisms for scattering (Srivastava, 1990): • Boundary scattering, a phonon hitting the boundary of a system; • Mass defect scattering, a phonon hitting an impurity within the system and scattering; • Phonon-phonon scattering, a phonon breaking into two lower energy phonons or a phonon colliding with another phonon and merging into one higher energy phonon. See also • Specific heat capacity • Thermistor ISBN links support NWE through referral fees External links All links retrieved April 30, 2023. New World Encyclopedia writers and editors rewrote and completed the Wikipedia article in accordance with New World Encyclopedia standards. This article abides by terms of the Creative Commons CC-by-sa 3.0 License (CC-by-sa), which may be used and disseminated with proper attribution. Credit is due under the terms of this license that can reference both the New World Encyclopedia contributors and the selfless volunteer contributors of the Wikimedia Foundation. To cite this article click here for a list of acceptable citing formats.The history of earlier contributions by wikipedians is accessible to researchers here: The history of this article since it was imported to New World Encyclopedia: Note: Some restrictions may apply to use of individual images which are separately licensed.	{"url":"http://www.newworldencyclopedia.org/entry/Thermal_conductivity","timestamp":"2024-11-06T09:33:13Z","content_type":"text/html","content_length":"84474","record_id":"<urn:uuid:3b1f9ae9-f782-46a9-a0bb-cffb291098db>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00137.warc.gz"}
2.7.2: Towers of Hanoi Last updated Page ID $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$ $ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$ $ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$ $ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vectorC}[1]{\textbf{#1}} $ $ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $ $ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $ $ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $ $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\ evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\ newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y} $ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real} {\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec} [3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array} {r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\ wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\ newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var} {\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left\|#1\right\|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\ bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\ widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$ Another standard example of recursion is the Towers of Hanoi problem. Let n be a pos- itive integer. Imagine a set of n discs of decreasing size, piled up in order of size, with the largest disc on the bottom and the smallest disc on top. The problem is to move this tower of discs to a second pile, following certain rules: Only one disc can be moved at a time, and a disc can only be placed on top of another disc if the disc on top is smaller. While the discs are being moved from the first pile to the second pile, discs can be kept in a third, spare pile. All the discs must at all times be in one of the three piles. The Towers of Hanoi puzzle was first published by Édouard Lucas in 1883. The puzzle is based on a legend of temple wherein there initially was one pile of discs neatly sorted from largest to smallest. In Lucas’s story, monks have since been continuously moving discs from this pile of 64 discs accord- ing to the rules of the puzzle to again created a sorted stack at the other end of the temple. It is said that when the last disc is placed, the world will end. But on the positive side, even if the monks move one disc every second, it will take approximately 42 times the age of the universe until they are done. And that is assuming they are using the optimal strategy... Source: en.Wikipedia.org/wiki/Tower_of_Hanoi For example, if there are two discs, the problem can be solved by the following sequence of moves: Move disc 1 from pile 1 to pile 3 Move disc 2 from pile 1 to pile 2 Move disc 1 from pile 3 to pile 2 A simple recursive subroutine can be used to write out the list of moves to solve the problem for any value of n. The recursion is based on the observation that for n > 1, the problem can be solved as follows: Move n − 1 discs from pile number 1 to pile number 3 (using pile number 2 as a spare). Then move the largest disc, disc number n, from pile number 1 to pile number 2. Finally, move the n − 1 discs from pile number 3 to pile number 2, putting them on top of the nth disc (using pile number 1 as a spare). In both cases, the problem of moving n − 1 discs is a smaller version of the original problem and so can be done by recursion. Here is the subroutine, written in Java: void Hanoi(int n, int A, int B, int C) { // List the moves for moving n discs from // pile number A to pile number B, using // pile number C as a spare. Assume n > 0. if (n == 1) { System.out.println("Move disc 1 from pile " + A + " to pile " + B); Hanoi(n-1, A, C, B); System.out.println("Move disc " + n + " from pile " + A + " to pile " + B); Hanoi(n-1, C, B, A); This problem and its fame have led to implementations in a variety of languages, including a language called Brain f*ck. In the Computer Organisation course, you can implement an interpreter for this language and test it on the implementation of the Hanoi algorithm. We can use induction to prove that this subroutine does in fact solve the Towers of Hanoi problem. Theorem 3.12. The sequence of moves printed by the Hanoi subroutine as given above correctly solves the Towers of Hanoi problem for any integer n ≥ 1. Proof. We prove by induction that whenever n is a positive integer and A,B, and C are the numbers 1, 2, and 3 in some order, the subroutine call Hanoi(n, A, B, C) prints a sequence of moves that will move n discs from pile A to pile B, following all the rules of the Towers of Hanoi problem. In the base case, n = 1, the subroutine call Hanoi(1, A, B, C) prints out the single step “Move disc 1 from pile A to pile B”, and this move does solve the problem for 1 disc. Let k be an arbitrary positive integer, and suppose that Hanoi(k, A, B, C) correctly solves the problem of moving the k discs from pile A to pile B using pile C as the spare, whenever A, B, and C are the numbers 1, 2, and 3 in some order. We need to show that Hanoi(k + 1, A, B, C) correctly solves the problem for k + 1 discs. Since k + 1 > 1,Hanoi(k + 1, A, B, C) begins by calling Hanoi(k, A, C, B). By the induction hypothesis, this correctly moves k discs from pile A to pile C. disc number k + 1 is not moved during this process. At that point, pile C contains the k smallest discs and pile A still contains the(k + 1)st disc, which has not yet been moved. So the next move printed by the subroutine, “Move disc (k + 1) from pile A to pile B”, is legal because pile B is empty. Finally, the subroutine calls Hanoi(k, C, B, A), which, by the induction hypothesis, correctly moves the $k$ smallest discs from pile $C$ to pile $B,$ putting them on top of the $(k+1)^{\text { st }}$ disc, which does not move during this process. At that point, all (k + 1) discs are on pile B, so the problem for k + 1 discs has been correctly solved.	{"url":"https://eng.libretexts.org/Bookshelves/Computer_Science/Programming_and_Computation_Fundamentals/Delftse_Foundations_of_Computation/02%3A_Proof/2.07%3A_Application_-_Recursion_and_Induction/2.7.02%3A_Towers_of_Hanoi","timestamp":"2024-11-07T18:52:33Z","content_type":"text/html","content_length":"127066","record_id":"<urn:uuid:cef25df1-4ad6-4249-9561-8eb4f416da0e>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00349.warc.gz"}
Philosophy:Numeral (linguistics) Short description: Word or phrase which describes a numerical quantity In linguistics, a numeral in the broadest sense is a word or phrase that describes a numerical quantity. Some theories of grammar use the word "numeral" to refer to cardinal numbers that act as a determiner that specify the quantity of a noun, for example the "two" in "two hats". Some theories of grammar do not include determiners as a part of speech and consider "two" in this example to be an adjective. Some theories consider "numeral" to be a synonym for "number" and assign all numbers (including ordinal numbers like the compound word "seventy-fifth") to a part of speech called "numerals".^[1]^[2] Numerals in the broad sense can also be analyzed as a noun ("three is a small number"), as a pronoun ("the two went to town"), or for a small number of words as an adverb ("I rode the slide twice"). Numerals can express relationships like quantity (cardinal numbers), sequence (ordinal numbers), frequency (once, twice), and part (fraction).^[3] Identifying numerals Numerals may be attributive, as in two dogs, or pronominal, as in I saw two (of them). Many words of different parts of speech indicate number or quantity. Such words are called quantifiers. Examples are words such as every, most, least, some, etc. Numerals are distinguished from other quantifiers by the fact that they designate a specific number.^[3] Examples are words such as five, ten, fifty, one hundred, etc. They may or may not be treated as a distinct part of speech; this may vary, not only with the language, but with the choice of word. For example, "dozen" serves the function of a noun, "first" serves the function of an adjective, and "twice" serves the function of an adverb. In Old Church Slavonic, the cardinal numbers 5 to 10 were feminine nouns; when quantifying a noun, that noun was declined in the genitive plural like other nouns that followed a noun of quantity (one would say the equivalent of "five of people"). In English grammar, the classification "numeral" (viewed as a part of speech) is reserved for those words which have distinct grammatical behavior: when a numeral modifies a noun, it may replace the article: the/some dogs played in the park → twelve dogs played in the park. (dozen dogs played in the park is not grammatical, so "dozen" is not a numeral in this sense.) English numerals indicate cardinal numbers. However, not all words for cardinal numbers are necessarily numerals. For example, million is grammatically a noun, and must be preceded by an article or numeral itself. Numerals may be simple, such as 'eleven', or compound, such as 'twenty-three'. In linguistics, however, numerals are classified according to purpose: examples are ordinal numbers (first, second, third, etc.; from 'third' up, these are also used for fractions), multiplicative (adverbial) numbers (once, twice, and thrice), multipliers (single, double, and triple), and distributive numbers (singly, doubly, and triply). Georgian,^[4] Latin, and Romanian (see Romanian distributive numbers) have regular distributive numbers, such as Latin singuli "one-by-one", bini "in pairs, two-by-two", terni "three each", etc. In languages other than English, there may be other kinds of number words. For example, in Slavic languages there are collective numbers (monad, pair/dyad, triad) which describe sets, such as pair or dozen in English (see Russian numerals, Polish Some languages have a very limited set of numerals, and in some cases they arguably do not have any numerals at all, but instead use more generic quantifiers, such as 'pair' or 'many'. However, by now most such languages have borrowed the numeral system or part of the numeral system of a national or colonial language, though in a few cases (such as Guarani^[5]), a numeral system has been invented internally rather than borrowed. Other languages had an indigenous system but borrowed a second set of numerals anyway. An example is Japanese, which uses either native or Chinese-derived numerals depending on what is being counted. In many languages, such as Chinese, numerals require the use of numeral classifiers. Many sign languages, such as ASL, incorporate numerals. Larger numerals English has derived numerals for multiples of its base (fifty, sixty, etc.), and some languages have simplex numerals for these, or even for numbers between the multiples of its base. Balinese, for example, currently has a decimal system, with words for 10, 100, and 1000, but has additional simplex numerals for 25 (with a second word for 25 only found in a compound for 75), 35, 45, 50, 150, 175, 200 (with a second found in a compound for 1200), 400, 900, and 1600. In Hindustani, the numerals between 10 and 100 have developed to the extent that they need to be learned independently. In many languages, numerals up to the base are a distinct part of speech, while the words for powers of the base belong to one of the other word classes. In English, these higher words are hundred 10 ^2, thousand 10^3, million 10^6, and higher powers of a thousand (short scale) or of a million (long scale—see names of large numbers). These words cannot modify a noun without being preceded by an article or numeral (hundred dogs played in the park), and so are nouns. In East Asia, the higher units are hundred, thousand, myriad 10^4, and powers of myriad. In the Indian subcontinent, they are hundred, thousand, lakh 10^5, crore 10^7, and so on. The Mesoamerican system, still used to some extent in Mayan languages, was based on powers of 20: bak’ 400 (20^2), pik 8000 (20^3), kalab 160,000 (20^4), etc. Numerals of cardinal numbers The cardinal numbers have numerals. In the following tables, [and] indicates that the word and is used in some dialects (such as British English), and omitted in other dialects (such as American This table demonstrates the standard English construction of some cardinal numbers. (See next table for names of larger cardinals.) Value Name Alternate names, and names for sets of the given size 0 Zero aught, cipher, cypher, donut, dot, duck, goose egg, love, nada, naught, nil, none, nought, nowt, null, ought, oh, squat, zed, zilch, zip, zippo, Sunya (Sanskrit) 1 One ace, individual, single, singleton, unary, unit, unity, Pratham (Sanskrit) 2 Two binary, brace, couple, couplet, distich, deuce, double, doubleton, duad, duality, duet, duo, dyad, pair, span, twain, twin, twosome, 3 Three deuce-ace, leash, set, tercet, ternary, ternion, terzetto, threesome, tierce, trey, triad, trine, trinity, trio, triplet, troika, 4 Four foursome, quadruplet, quatern, quaternary, quaternity, quartet, tetrad 5 Five cinque, fin, fivesome, pentad, quint, quintet, quintuplet 6 Six half dozen, hexad, sestet, sextet, sextuplet, sise 7 Seven heptad, septet, septuple, walking stick 8 Eight octad, octave, octet, octonary, octuplet, ogdoad 9 Nine ennead 10 Ten deca, decade, das (India ) 11 Eleven onze, ounze, ounce, banker's dozen 12 Twelve dozen 13 Thirteen baker's dozen, long dozen^[6] 20 Twenty score, 21 Twenty-one long score,^[6] blackjack 22 Twenty-two Deuce-deuce 24 Twenty-four two dozen 40 Forty two-score 50 Fifty half-century 55 Fifty-five double nickel 60 Sixty three-score 70 Seventy three-score and ten 80 Eighty four-score 87 Eighty-seven four-score and seven 90 Ninety four-score and ten 100 One hundred centred, century, ton, short hundred 111 One hundred [and] eleven eleventy-one^[7] 120 One hundred [and] twenty long hundred,^[6] great hundred, (obsolete) hundred 144 One hundred [and] forty-four gross, dozen dozen, small gross 1000 One thousand chiliad, grand, G, thou, yard, kilo, k, millennium, Hajaar (India ), ten hundred 1024 One thousand [and] twenty-four kibi or kilo in computing, see binary prefix (kilo is shortened to K, Kibi to Ki) 1100 One thousand one hundred Eleven hundred 1728 One thousand seven hundred [and] twenty-eight great gross, long gross, dozen gross 10000 Ten thousand myriad, wan (China) 100000 One hundred thousand lakh 500000 Five hundred thousand crore (Iranian) 1000000 One million Mega, meg, mil, (often shortened to M) 1048576 One million forty-eight thousand five hundred [and] Mibi or Mega in computing, see binary prefix (Mega is shortened to M, Mibi to Mi) 10000000 Ten million crore (Indian)(Pakistan) 100000000 One hundred million yi (China) English names for powers of 10 This table compares the English names of cardinal numbers according to various American, British, and Continental European conventions. See English numerals or names of large numbers for more information on naming numbers. Short scale Long scale Value American British Continental European (Nicolas Chuquet) (Jacques Peletier du Mans) 10^0 One 10^1 Ten 10^2 Hundred 10^3 Thousand 10^6 Million 10^9 Billion Thousand million Milliard 10^12 Trillion Billion 10^15 Quadrillion Thousand billion Billiard 10^18 Quintillion Trillion 10^21 Sextillion Thousand trillion Trilliard 10^24 Septillion Quadrillion 10^27 Octillion Thousand quadrillion Quadrilliard 10^30 Nonillion Quintillion 10^33 Decillion Thousand quintillion Quintilliard 10^36 Undecillion Sextillion 10^39 Duodecillion Thousand sextillion Sextilliard 10^42 Tredecillion Septillion 10^45 Quattuordecillion Thousand septillion Septilliard 10^48 Quindecillion Octillion 10^51 Sexdecillion Thousand octillion Octilliard 10^54 Septendecillion Nonillion 10^57 Octodecillion Thousand nonillion Nonilliard 10^60 Novemdecillion Decillion 10^63 Vigintillion Thousand decillion Decilliard 10^66 Unvigintillion Undecillion 10^69 Duovigintillion Thousand undecillion Undecilliard 10^72 Trevigintillion Duodecillion 10^75 Quattuorvigintillion Thousand duodecillion Duodecilliard 10^78 Quinvigintillion Tredecillion 10^81 Sexvigintillion Thousand tredecillion Tredecilliard 10^84 Septenvigintillion Quattuordecillion 10^87 Octovigintillion Thousand quattuordecillion Quattuordecilliard 10^90 Novemvigintillion Quindecillion 10^93 Trigintillion Thousand quindecillion Quindecilliard 10^96 Untrigintillion Sexdecillion 10^99 Duotrigintillion Thousand sexdecillion Sexdecilliard 10^120 Novemtrigintillion Vigintillion 10^123 Quadragintillion Thousand vigintillion Vigintilliard 10^153 Quinquagintillion Thousand quinvigintillion Quinvigintilliard 10^180 Novemquinquagintillion Trigintillion 10^183 Sexagintillion Thousand trigintillion Trigintilliard 10^213 Septuagintillion Thousand quintrigintillion Quintrigintilliard 10^240 Novemseptuagintillion Quadragintillion 10^243 Octogintillion Thousand quadragintillion Quadragintilliard 10^273 Nonagintillion Thousand quinquadragintillion Quinquadragintilliard 10^300 Novemnonagintillion Quinquagintillion 10^303 Centillion Thousand quinquagintillion Quinquagintilliard 10^360 Cennovemdecillion Sexagintillion 10^420 Cennovemtrigintillion Septuagintillion 10^480 Cennovemquinquagintillion Octogintillion 10^540 Cennovemseptuagintillion Nonagintillion 10^600 Cennovemnonagintillion Centillion 10^603 Ducentillion Thousand centillion Centilliard There is no consistent and widely accepted way to extend cardinals beyond centillion (centilliard). Myriad, Octad, and -yllion systems The following table details the myriad, octad, Chinese myriad, Chinese long and -yllion names for powers of 10. There is also a Knuth-proposed system notation of numbers, named the -yllion system.^[8] In this system, a new word is invented for every 2^n-th power of ten. Value Myriad System Name Octad System Name Chinese Myriad Scale Chinese Long Scale Knuth-proposed System Name 10^0 One One 一一 One 10^1 Ten Ten 十十 Ten 10^2 Hundred Hundred 百百 Hundred 10^3 Thousand Thousand 千千 Ten hundred 10^4 Myriad Myriad 萬 (万) 萬 (万) Myriad 10^5 Ten myriad Ten myriad 十萬 (十万) 十萬 (十万) Ten myriad 10^6 Hundred myriad Hundred myriad 百萬 (百万) 百萬 (百万) Hundred myriad 10^7 Thousand myriad Thousand myriad 千萬 (千万) 千萬 (千万) Ten hundred myriad 10^8 Second Myriad Octad 億 (亿) 億 (亿) Myllion 10^12 Third myriad Myriad Octad 兆萬億 Myriad myllion 10^16 Fourth myriad Second octad 京兆 Byllion 10^20 Fifth myriad Myriad second octad 垓萬兆 10^24 Sixth myriad Third octad 秭 (in China); 𥝱 (in Japan) 億兆 Myllion byllion 10^28 Seventh myriad Myriad third octad 穰萬億兆 10^32 Eighth myriad Fourth octad 溝 (沟) 京 Tryllion 10^36 Ninth myriad Myriad fourth octad 澗 (涧) 萬京 10^40 Tenth myriad Fifth octad 正億京 10^44 Eleventh myriad Myriad fifth octad 載 (载) 萬億京 10^48 Twelfth myriad Sixth octad 極 (极) (in China and in Japan) 兆京 10^52 Thirteenth myriad Myriad sixth octad 恆河沙 (恒河沙) (in China) 萬兆京 10^56 Fourteenth myriad Seventh octad 阿僧祇 (in China); 恒河沙 (in Japan) 億兆京 10^60 Fifteenth myriad Myriad seventh octad 那由他, 那由多 (in China) 萬億兆京 10^64 Sixteenth myriad Eighth octad 不可思議 (不可思议) (in China), 阿僧祇 (in Japan) 垓 Quadyllion 10^68 Seventeenth myriad Myriad eighth octad 無量大數 (无量大数) (in China) 萬垓 10^72 Eighteenth myriad Ninth octad 那由他, 那由多 (in Japan) 億垓 10^80 Twentieth myriad Tenth octad 不可思議 (in Japan) 兆垓 10^88 Twenty-second myriad Eleventh Octad 無量大数 (in Japan) 億兆垓 10^128 秭 Quinyllion 10^256 穰 Sexyllion 10^512 溝 (沟) Septyllion 10^1,024 澗 (涧) Octyllion 10^2,048 正 Nonyllion 10^4,096 載 (载) Decyllion 10^8,192 極 (极) Undecyllion 10^16,384 Duodecyllion 10^32,768 Tredecyllion 10^65,536 Quattuordecyllion 10^131,072 Quindecyllion 10^262,144 Sexdecyllion 10^524,288 Septendecyllion 10^1,048,576 Octodecyllion 10^2,097,152 Novemdecyllion 10^4,194,304 Vigintyllion 10^2^32 Trigintyllion 10^2^42 Quadragintyllion 10^2^52 Quinquagintyllion 10^2^62 Sexagintyllion 10^2^72 Septuagintyllion 10^2^82 Octogintyllion 10^2^92 Nonagintyllion 10^2^102 Centyllion 10^2^1,002 Millyllion 10^2^10,002 Myryllion Fractional numerals This is a table of English names for non-negative rational numbers less than or equal to 1. It also lists alternative names, but there is no widespread convention for the names of extremely small positive numbers. Keep in mind that rational numbers like 0.12 can be represented in infinitely many ways, e.g. zero-point-one-two (0.12), twelve percent (12%), three twenty-fifths (3/25), nine seventy-fifths (9/75), six fiftieths (6/50), twelve hundredths (12/100), twenty-four two-hundredths (24/200), etc. Value Fraction Common names 1 1/1 One, Unity, Whole 0.9 9/10 Nine tenths, [zero] point nine 0.833333... 5/6 Five sixths 0.8 4/5 Four fifths, eight tenths, [zero] point eight 0.75 3/4 three quarters, three fourths, seventy-five hundredths, [zero] point seven five 0.7 7/10 Seven tenths, [zero] point seven 0.666666... 2/3 Two thirds 0.6 3/5 Three fifths, six tenths, [zero] point six 0.5 1/2 One half, five tenths, [zero] point five 0.4 2/5 Two fifths, four tenths, [zero] point four 0.333333... 1/3 One third 0.3 3/10 Three tenths, [zero] point three 0.25 1/4 One quarter, one fourth, twenty-five hundredths, [zero] point two five 0.2 1/5 One fifth, two tenths, [zero] point two 0.166666... 1/6 One sixth 0.142857142857... 1/7 One seventh 0.125 1/8 One eighth, one-hundred-[and-]twenty-five thousandths, [zero] point one two five 0.111111... 1/9 One ninth 0.1 1/10 One tenth, [zero] point one, One perdecime, one perdime 0.090909... 1/11 One eleventh 0.09 9/100 Nine hundredths, [zero] point zero nine 0.083333... 1/12 One twelfth 0.08 2/25 Two twenty-fifths, eight hundredths, [zero] point zero eight 0.076923076923... 1/13 One thirteenth 0.071428571428... 1/14 One fourteenth 0.066666... 1/15 One fifteenth 0.0625 1/16 One sixteenth, six-hundred-[and-]twenty-five ten-thousandths, [zero] point zero six two five 0.055555... 1/18 One eighteenth 0.05 1/20 One twentieth, five hundredths, [zero] point zero five 0.047619047619... 1/21 One twenty-first 0.045454545... 1/22 One twenty-second 0.043478260869565217391304347... 1/23 One twenty-third 0.041666... 1/24 One twenty-fourth 0.04 1/25 One twenty-fifth, four hundredths, [zero] point zero four 0.033333... 1/30 One thirtieth 0.03125 1/32 One thirty-second, thirty one-hundred [and] twenty five hundred-thousandths, [zero] point zero three one two five 0.03 3/100 Three hundredths, [zero] point zero three 0.025 1/40 One fortieth, twenty-five thousandths, [zero] point zero two five 0.02 1/50 One fiftieth, two hundredths, [zero] point zero two 0.016666... 1/60 One sixtieth 0.015625 1/64 One sixty-fourth, ten thousand fifty six-hundred [and] twenty-five millionths, [zero] point zero one five six two five 0.012345679012345679... 1/81 One eighty-first 0.010101... 1/99 One ninety-ninth 0.01 1/100 One hundredth, [zero] point zero one, One percent 0.009900990099... 1/101 One hundred-first 0.008264462809917355371900... 1/121 One over one hundred twenty-one 0.001 1/1000 One thousandth, [zero] point zero zero one, One permille 0.000277777... 1/3600 One thirty-six hundredth 0.0001 1/10000 One ten-thousandth, [zero] point zero zero zero one, One myriadth, one permyria, one permyriad, one basis point 0.00001 1/100000 One hundred-thousandth, [zero] point zero zero zero zero one, One lakhth, one perlakh 0.000001 1/1000000 One millionth, [zero] point zero zero zero zero zero one, One ppm 0.0000001 1/10000000 One ten-millionth, One crorth, one percrore 0.00000001 1/100000000 One hundred-millionth 0.000000001 1/1000000000 One billionth (in some dialects), One ppb 0.000000000001 1/1000000000000 One trillionth, One ppt 0 0/1 Zero, Nil Other specific quantity terms Various terms have arisen to describe commonly used measured quantities. • Unit: 1 (based on a single entity of counting or measurement of an object or item) • Pair: 2 (the base of the binary numeral system) • Leash: 3 (the base of the trinary numeral system) • Dozen: 12 (the base of the duodecimal numeral system) • Baker's dozen: 13 (based on a group of thirteen objects or items) • Score: 20 (the base of the vigesimal numeral system) • Shock: 60 (the base of the sexagesimal numeral system)^[9] • Gross: (based on a group of 144 objects or items) • Great gross: (based on a group of 1,728 objects or items) Basis of counting system Not all peoples use counting, at least not verbally. Specifically, there is not much need for counting among hunter-gatherers who do not engage in commerce. Many languages around the world have no numerals above two to four (if they are actually numerals at all, and not some other part of speech)—or at least did not before contact with the colonial societies—and speakers of these languages may have no tradition of using the numerals they did have for counting. Indeed, several languages from the Amazon have been independently reported to have no specific number words other than 'one'. These include Nadëb, pre-contact Mocoví and Pilagá, Culina and pre-contact Jarawara, Jabutí, Canela-Krahô, Botocudo (Krenák), Chiquitano, the Campa languages, Arabela, and Achuar.^[10] Some languages of Australia, such as Warlpiri, do not have words for quantities above two,^[11]^[12]^[13] and neither did many Khoisan languages at the time of European contact. Such languages do not have a word class of 'numeral'. Most languages with both numerals and counting use base 8, 10, 12, or 20. Base 10 appears to come from counting one's fingers, base 20 from the fingers and toes, base 8 from counting the spaces between the fingers (attested in California), and base 12 from counting the knuckles (3 each for the four fingers).^[14] No base Many languages of Melanesia have (or once had) counting systems based on parts of the body which do not have a numeric base; there are (or were) no numerals, but rather nouns for relevant parts of the body—or simply pointing to the relevant spots—were used for quantities. For example, 1–4 may be the fingers, 5 'thumb', 6 'wrist', 7 'elbow', 8 'shoulder', etc., across the body and down the other arm, so that the opposite little finger represents a number between 17 (Torres Islands) to 23 (Eleman). For numbers beyond this, the torso, legs and toes may be used, or one might count back up the other arm and back down the first, depending on the people. 2: binary Binary systems are based on the number 2, using zeros and ones. With only two symbols binary is used for things with coding like computers. 3: ternary Ternary systems are based on the number 3, having practical usage in some analog logic, in baseball scoring and in self–similar mathematical structures. 4: quaternary Quaternary systems are based on the number 4. Some Austronesian, Melanesian, Sulawesi, and Papua New Guinea ethnic groups, count with the base number four, using the term asu or aso, the word for dog , as the ubiquitous village dog has four legs.^[15] This is argued by anthropologists to be also based on early humans noting the human and animal shared body feature of two arms and two legs as well as its ease in simple arithmetic and counting. As an example of the system's ease a realistic scenario could include a farmer returning from the market with fifty asu heads of pig (200), less 30 asu (120) of pig bartered for 10 asu (40) of goats noting his new pig count total as twenty asu: 80 pigs remaining. The system has a correlation to the dozen counting system and is still in common use in these areas as a natural and easy method of simple arithmetic.^[15]^[16] 5: quinary Quinary systems are based on the number 5. It is almost certain the quinary system developed from counting by fingers (five fingers per hand).^[17] An example are the Epi languages of Vanuatu, where 5 is luna 'hand', 10 lua-luna 'two hand', 15 tolu-luna 'three hand', etc. 11 is then lua-luna tai 'two-hand one', and 17 tolu-luna lua 'three-hand two'. 5 is a common auxiliary base, or sub-base, where 6 is 'five and one', 7 'five and two', etc. Aztec was a vigesimal (base-20) system with sub-base 5. 6: senary Senary systems are based on the number 6. The Morehead-Maro languages of Southern New Guinea are examples of the rare base 6 system with monomorphemic words running up to 6^6. Examples are Kanum and Kómnzo. The Sko languages on the North Coast of New Guinea follow a base-24 system with a sub-base of 6. 7: septenary Septenary systems are based on the number 7. Septenary systems are very rare, as few natural objects consistently have seven distinctive features. Traditionally, it occurs in week-related timing. It has been suggested that the Palikúr language has a base-seven system, but this is dubious.^[18] 8: octal Octal systems are based on the number 8. Examples can be found in the Yuki language of California and in the Pamean languages of Mexico, because the Yuki and Pame keep count by using the four spaces between their fingers rather than the fingers themselves.^[19] 9: nonary Nonary systems are based on the number 9. It has been suggested that Nenets has a base-nine system.^[18] 10: decimal Decimal systems are based on the number 10. A majority of traditional number systems are decimal. This dates back at least to the ancient Egyptians, who used a wholly decimal system. Anthropologists hypothesize this may be due to humans having five digits per hand, ten in total.^[17]^[20] There are many regional variations including: 12: duodecimal Duodecimal systems are based on the number 12. These include: • Chepang language of Nepal, • Mahl language of Minicoy Island in India • Nigerian Middle Belt areas such as Janji, Kahugu and the Nimbia dialect of Gwandara. • reconstructed proto-Benue–Congo Duodecimal numeric systems have some practical advantages over decimal. It is much easier to divide the base digit twelve (which is a highly composite number) by many important divisors in market and trade settings, such as the numbers 2, 3, 4 and 6. Because of several measurements based on twelve,^[21] many Western languages have words for base-twelve units such as dozen, gross and great gross, which allow for rudimentary duodecimal nomenclature , such as "two gross six dozen" for 360. Ancient Romans used a decimal system for integers, but switched to duodecimal for fractions, and correspondingly Latin developed a rich vocabulary for duodecimal-based fractions (see Roman numerals). A notable fictional duodecimal system was that of J. R. R. Tolkien's Elvish languages, which used duodecimal as well as decimal. 16: hexadecimal Hexadecimal systems are based on the number 16. The traditional Chinese units of measurement were base-16. For example, one jīn (斤) in the old system equals sixteen taels. The suanpan (Chinese abacus) can be used to perform hexadecimal calculations such as additions and subtractions.^[22] South Asian monetary systems were base-16. One rupee in Pakistan and India was divided into 16 annay. A single anna was subdivided into four paisa or twelve pies (thus there were 64 paise or 192 pies in a rupee). The anna was demonetised as a currency unit when India decimalised its currency in 1957, followed by Pakistan in 1961. 20: vigesimal Vigesimal systems are based on the number 20. Anthropologists are convinced the system originated from digit counting, as did bases five and ten, twenty being the number of human fingers and toes combined.^[17]^[23] The system is in widespread use across the world. Some include the classical Mesoamerican cultures, still in use today in the modern indigenous languages of their descendants, namely the Nahuatl and Mayan languages (see Maya numerals). A modern national language which uses a full vigesimal system is Dzongkha in Bhutan. Partial vigesimal systems are found in some European languages: Basque, Celtic languages, French (from Celtic), Danish, and Georgian. In these languages the systems are vigesimal up to 99, then decimal from 100 up. That is, 140 is 'one hundred two score', not *seven score, and there is no numeral for 400 (great score). The term score originates from tally sticks, and is perhaps a remnant of Celtic vigesimal counting. It was widely used to learn the pre-decimal British currency in this idiom: "a dozen pence and a score of bob", referring to the 20 shillings in a pound. For Americans the term is most known from the opening of the Gettysburg Address: "Four score and seven years ago our fathers...". 24: quadrovigesimal Quadrovigesimal systems are based on the number 24. The Sko languages have a base-24 system with a sub-base of 6. 32: duotrigesimal Duotrigesimal systems are based on the number 32. The Ngiti ethnolinguistic group uses a base 32 numeral system. 60: sexagesimal Sexagesimal systems are based on the number 60. Ekari has a base-60 system. Sumeria had a base-60 system with a decimal sub-base (with alternating cycles of 10 and 6), which was the origin of the numbering of modern degrees, minutes, and seconds. 80: octogesimal Octogesimal systems are based on the number 80. Supyire is said to have a base-80 system; it counts in twenties (with 5 and 10 as sub-bases) up to 80, then by eighties up to 400, and then by 400s (great scores). Script error: No such module "Interlinear". 799 [i.e. 400 + (4 x 80) + (3 x 20) + {10 + (5 + 4)}]’ See also Numerals in various languages A database Numeral Systems of the World's Languages compiled by Eugene S.L. Chan of Hong Kong is hosted by the Max Planck Institute for Evolutionary Anthropology in Leipzig, Germany. The database currently contains data for about 4000 languages. Related topics 1. ↑ Charles Follen: A Practical Grammar of the German Language. Boston, 1828, p. 9, p. 44 and 48. Quote: "PARTS OF SPEECH. There are ten parts of speech, viz. Article, Substantive or Noun, Adjective, Numeral, Pronoun, Verb, Adverb, Preposition, Conjunction, and Interjection.", "NUMERALS. The numbers are divided into cardinal, ordinal, proportional, distributive, and collective. [...] Numerals of proportion and distribution are [...] &c. Observation. The above numerals, in fach or fäl´tig, are regularly declined, like other adjectives." 2. ↑ Horace Dalmolin: The New English Grammar: With Phonetics, Morphology and Syntax, Tate Publishing & Enterprises, 2009, p. 175 & p. 177. Quote: "76. The different types of words used to compose a sentence, in order to relate an idea or to convey a thought, are known as parts of speech. [...] The parts of speech, with a brief definition, will follow. [...] 87. Numeral: Numerals are words that express the idea of number. There are two types of numerals: cardinal and ordinal. The cardinal numbers (one, two, three...) are used for counting people, objects, etc. Ordinal numbers ( first, second, third...) can indicate order, placement in rank, etc." 3. ↑ ^3.0 ^3.1 "What is a numeral?". http://www-01.sil.org/linguistics/glossaryoflinguisticterms/WhatIsANumeral.htm. 4. ↑ "Walsinfo.com". http://wals.info/feature/description/. 5. ↑ ^6.0 ^6.1 ^6.2 Blunt, Joseph (1 January 1837). "The Shipmaster's Assistant, and Commercial Digest: Containing Information Useful to Merchants, Owners, and Masters of Ships". E. & G.W. Blunt. 6. ↑ Ezard, John (2 Jan 2003). "Tolkien catches up with his hobbit". The Guardian. https://www.theguardian.com/uk/2003/jan/02/jrrtolkien.books. 7. ↑ Cardarelli, François (2012). Encyclopaedia of Scientific Units, Weights and Measures: Their SI Equivalences and Origins (Second ed.). Springer. p. 585. ISBN 978-1447100034. 8. ↑ "Hammarström (2009, page 197) "Rarities in numeral systems"". http://www2.gslt.hum.gu.se/dissertations/hammarstrom.pdf. 9. ↑ UCL Media Relations, "Aboriginal kids can count without numbers" 10. ↑ Butterworth, Brian; Reeve, Robert; Reynolds, Fiona; Lloyd, Delyth (2 September 2008). "Numerical thought with and without words: Evidence from indigenous Australian children". PNAS 105 (35): 13179–13184. doi:10.1073/pnas.0806045105. PMID 18757729. Bibcode: 2008PNAS..10513179B. "[Warlpiri] has three generic types of number words: singular, dual plural, and greater than dual plural.". 11. ↑ The Science Show, Genetic anomaly could explain severe difficulty with arithmetic, Australian Broadcasting Corporation 12. ↑ Bernard Comrie, "The Typology of Numeral Systems ", p. 3 13. ↑ ^15.0 ^15.1 Ryan, Peter. Encyclopaedia of Papua and New Guinea. Melbourne University Press & University of Papua and New Guinea,:1972 ISBN 0-522-84025-6.: 3 pages p 219. 14. ↑ Aleksandr Romanovich Luriicac, Lev Semenovich Vygotskiĭ, Evelyn Rossiter. Ape, primitive man, and child: essays in the history of behavior. CRC Press: 1992: ISBN 1-878205-43-9. 15. ↑ ^17.0 ^17.1 ^17.2 Heath, Thomas, A Manual of Greek Mathematics, Courier Dover: 2003. ISBN 978-0-486-43231-1 page, p:11 16. ↑ ^18.0 ^18.1 Parkvall, M. Limits of Language, 1st edn. 2008. p.291. ISBN:978-1-59028-210-6 17. ↑ Ethnomathematics: A Multicultural View of Mathematical Ideas, Chapman & Hall, 1994, ISBN 0-412-98941-7 18. ↑ Scientific American Munn& Co: 1968, vol 219: 219 19. ↑ such as twelve months in a year, the twelve-hour clock, twelve inches to the foot, twelve pence to the shilling 20. ↑ "算盤 Hexadecimal Addition & Subtraction on a Chinese Abacus". http://totton.idirect.com/soroban/Hex_as/. 21. ↑ Georges Ifrah, The Universal History of Numbers: The Modern Number System, Random House, 2000: ISBN 1-86046-791-1. 1262 pages Further reading • James R. Hurford (2010). The Linguistic Theory of Numerals. Cambridge University Press. 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Special Relativity/Dynamics - Wikibooks, open books for an open world The way that the velocity of a particle can differ between observers who are moving relative to each other means that momentum needs to be redefined as a result of relativity theory. The illustration below shows a typical collision of two particles. In the right hand frame the collision is observed from the viewpoint of someone moving at the same velocity as one of the particles, in the left hand frame it is observed by someone moving at a velocity that is intermediate between those of the particles. If momentum is redefined then all the variables such as force (rate of change of momentum), energy etc. will become redefined and relativity will lead to an entirely new physics. The new physics has an effect at the ordinary level of experience through the relation ${\displaystyle K=\gamma mc^{2}-mc^{2}\,}$ whereby it is the tiny deviations in gamma from unity that are expressed as everyday kinetic energy so that the whole of physics is related to "relativistic" reasoning rather than Newton's empirical ideas. In physics momentum is conserved within a closed system, the law of conservation of momentum applies. Consider the special case of identical particles colliding symmetrically as illustrated below: The momentum change by the red ball is: ${\displaystyle 2m\mathbf {u_{yR}} }$ The momentum change by the blue ball is: ${\displaystyle -2m\mathbf {u_{yB}} }$ The situation is symmetrical so the Newtonian conservation of momentum law is demonstrated: ${\displaystyle 2m\mathbf {u_{yR}} =2m\mathbf {u_{yB}} }$ Notice that this result depends upon the y components of the velocities being equal, that is, ${\displaystyle \mathbf {u_{yR}} =\mathbf {u_{yB}} }$. The relativistic case is rather different. The collision is illustrated below, the left hand frame shows the collision as it appears for one observer and the right hand frame shows exactly the same collision as it appears for another observer moving at the same velocity as the blue ball: The configuration shown above has been simplified because one frame contains a stationary blue ball (ie: ${\displaystyle u_{xB}=0}$) and the velocities are chosen so that the vertical velocity of the red ball is exactly reversed after the collision ie:${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$. Both frames show exactly the same event, it is only the observers who differ between frames. The relativistic velocity transformations between frames is: ${\displaystyle u_{yR}^{'}={\frac {u_{yR}{\sqrt {1-v^{2}/c^{2}}}}{1-u_{xR}v/c^{2}}}}$ ${\displaystyle u_{yB}^{'}={\frac {u_{yB}{\sqrt {1-v^{2}/c^{2}}}}{1-u_{xB}v/c^{2}}}=u_{yB}{\sqrt {1-v^{2}/c^{2}}}}$ given that ${\displaystyle u_{xB}=0\,}$. Suppose that the y components are equal in one frame, in Newtonian physics they will also be equal in the other frame. However, in relativity, if the y components are equal in one frame they are not necessarily equal in the other frame (time dilation is not directional so perpendicular velocities differ between the observers). For instance if ${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$ then: ${\displaystyle u_{yB}={\frac {u_{yR}}{1-u_{xR}v/c^{2}}}}$ So if ${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$ then in this case ${\displaystyle u_{yR}eq u_{yB}}$. If the mass were constant between collisions and between frames then although ${\displaystyle 2mu_{yR}^{'}=2mu_{yB}^{'}}$ it is found that: ${\displaystyle 2mu_{yR}eq 2mu_{yB}}$ So momentum defined as mass times velocity is not conserved in a collision when the collision is described in frames moving relative to each other. Notice that the discrepancy is very small if ${\ displaystyle u_{xR}}$ and ${\displaystyle v}$ are small. To preserve the principle of momentum conservation in all inertial reference frames, the definition of momentum has to be changed. The new definition must reduce to the Newtonian expression when objects move at speeds much smaller than the speed of light, so as to recover the Newtonian formulas. The velocities in the y direction are related by the following equation when the observer is travelling at the same velocity as the blue ball ie: when ${\displaystyle u_{xB}=0\,}$: ${\displaystyle u_{yB}={\frac {u_{yR}}{1-u_{xR}v/c^{2}}}}$ If we write ${\displaystyle m_{B}}$ for the mass of the blue ball) and ${\displaystyle m_{R}}$ for the mass of the red ball as observed from the frame of the blue ball then, if the principle of relativity applies: ${\displaystyle 2m_{R}u_{yR}=2m_{B}u_{yB}\,}$ ${\displaystyle m_{R}=m_{B}{\frac {u_{yB}}{u_{yR}}}}$ ${\displaystyle u_{yB}={\frac {u_{yR}}{1-u_{xR}v/c^{2}}}}$ ${\displaystyle m_{R}={\frac {m_{B}}{1-u_{xR}v/c^{2}}}}$ This means that, if the principle of relativity is to apply then the mass must change by the amount shown in the equation above for the conservation of momentum law to be true. The particle velocities were chosen so that ${\displaystyle u_{yR}^{'}=u_{yB}^{'}}$. ${\displaystyle v}$ was selected so that ${\displaystyle v=u_{xR}^{'}}$. This allows ${\displaystyle v}$ to be expressed in terms of ${\displaystyle u_{xR}\,}$: ${\displaystyle u_{xR}^{'}={\frac {u_{xR}-v}{1-u_{xR}v/c^{2}}}=v}$ and hence: ${\displaystyle v={\frac {c^{2}}{u_{xR}}}(1-{\sqrt {1-u_{xR}^{2}/c^{2}}})}$ So substituting for ${\displaystyle v}$ in ${\displaystyle m_{R}={\frac {m_{B}}{1-u_{xR}v/c^{2}}}}$: ${\displaystyle m_{R}={\frac {m_{B}}{\sqrt {1-u_{xR}^{2}/c^{2}}}}}$ The blue ball is at rest so its mass is sometimes known as its rest mass, and is given the symbol ${\displaystyle m}$. As the balls were identical at the start of the boost the mass of the red ball is the mass that a blue ball would have if it were in motion relative to an observer; this mass is sometimes known as the relativistic mass symbolised by ${\displaystyle M}$. These terms are now infrequently used in modern physics, as will be explained at the end of this section. The discussion given above was related to the relative motions of the blue and red balls, as a result ${\ displaystyle u_{xR}}$ corresponds to the speed of the moving ball relative to an observer who is stationary with respect to the blue ball. These considerations mean that the relativistic mass is given by: ${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$ The relativistic momentum is given by the product of the relativistic mass and the velocity ${\displaystyle \mathbf {p} =M\mathbf {u} }$. The overall expression for momentum in terms of rest mass is: ${\displaystyle \mathbf {p} ={\frac {m\mathbf {u} }{\sqrt {1-u^{2}/c^{2}}}}}$ and the components of the momentum are: ${\displaystyle p_{x}={\frac {mu_{x}}{\sqrt {1-u^{2}/c^{2}}}}}$ ${\displaystyle p_{y}={\frac {mu_{y}}{\sqrt {1-u^{2}/c^{2}}}}}$ ${\displaystyle p_{z}={\frac {mu_{z}}{\sqrt {1-u^{2}/c^{2}}}}}$ So the components of the momentum depend upon the appropriate velocity component and the speed. Since the factor with the square root is cumbersome to write, the following abbreviation is often used, called the Lorentz gamma factor: ${\displaystyle \gamma ={\frac {1}{\sqrt {1-u^{2}/c^{2}}}}}$ The expression for the momentum then reads ${\displaystyle \mathbf {p} =m\gamma \mathbf {u} }$. It can be seen from the discussion above that we can write the momentum of an object moving with velocity ${\displaystyle \mathbf {u} }$ as the product of a function ${\displaystyle M(u)}$ of the speed ${\displaystyle u}$ and the velocity ${\displaystyle \mathbf {u} }$: ${\displaystyle M(u)\mathbf {u} }$ The function ${\displaystyle M(u)}$ must reduce to the object's mass ${\displaystyle m}$ at small speeds, in particular when the object is at rest ${\displaystyle M(0)=m}$. There is a debate about the usage of the term "mass" in relativity theory. If inertial mass is defined in terms of momentum then it does indeed vary as ${\displaystyle M=\gamma m}$ for a single particle that has rest mass, furthermore, as will be shown below the energy of a particle that has a rest mass is given by ${\displaystyle E=Mc^{2}}$. Prior to the debate about nomenclature the function ${\displaystyle M(u)}$, or the relation ${\displaystyle M=\gamma m}$, used to be called 'relativistic mass', and its value in the frame of the particle was referred to as the 'rest mass' or 'invariant mass'. The relativistic mass, ${\displaystyle M=\gamma m}$, would increase with velocity. Both terms are now largely obsolete: the 'rest mass' is today simply called the mass, and the 'relativistic mass' is often no longer used since, as will be seen in the discussion of energy below, it is identical to the energy but for the units. Newton's second law states that the total force acting on a particle equals the rate of change of its momentum. The same form of Newton's second law holds in relativistic mechanics. The relativistic 3 force is given by: ${\displaystyle \mathbf {f} =d\mathbf {p} /dt}$ If the relativistic mass is used: ${\displaystyle {\frac {d\mathbf {p} }{dt}}={\frac {d(m\mathbf {u} )}{dt}}}$ By Leibniz's law where ${\displaystyle d(xy)=xdy+ydx}$: ${\displaystyle \mathbf {f} ={\frac {d\mathbf {p} }{dt}}=m{\frac {d\mathbf {u} }{dt}}+\mathbf {u} {\frac {dm}{dt}}}$ This equation for force will be used below to derive relativistic expressions for the energy of a particle in terms of the old concept of "relativistic mass". The relativistic force can also be written in terms of acceleration. Newton's second law can be written in the familiar form ${\displaystyle \mathbf {F} =m\mathbf {a} }$ where ${\displaystyle \mathbf {a} =d\mathbf {v} /dt}$ is the acceleration. here m is not the relativistic mass but is the invariant mass. In relativistic mechanics, momentum is ${\displaystyle \mathbf {p} =m\gamma \mathbf {v} }$ again m being the invariant mass and the force is given by ${\displaystyle \mathbf {F} ={\frac {d\mathbf {p} }{dt}}=m{\frac {d(\gamma \mathbf {v} )}{dt}}}$ This form of force is used in the derivation of the expression for energy without relying on relativistic mass. It will be seen in the second section of this book that Newton's second law in terms of acceleration is given by: ${\displaystyle \mathbf {F} =m\gamma (\mathbf {a} +{\frac {\gamma ^{2}v}{c^{2}}}{\frac {dv}{dt}}\mathbf {v} )}$ The debate over the use of the concept "relativistic mass" means that modern physics courses may forbid the use of this in the derivation of energy. The newer derivation of energy without using relativistic mass is given in the first section and the older derivation using relativistic mass is given in the second section. The two derivations can be compared to gain insight into the debate about mass but a knowledge of 4 vectors is really required to discuss the problem in depth. In principle the first derivation is most mathematically correct because "relativistic mass" is given by: $ {\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$ which involves the constants ${\displaystyle m}$ and ${\displaystyle c}$. Derivation of relativistic energy using the relativistic momentum [edit \| edit source] In the following, modern derivation, m means the invariant mass - what used to be called the "rest mass". Energy is defined as the work done in moving a body from one place to another. We will make use of the relativistic momentum ${\displaystyle p=\gamma mv}$. Energy is given from: ${\displaystyle dE=\mathbf {f} d\mathbf {x} }$ so, over the whole path: ${\displaystyle E=\int _{0}^{x}\mathbf {f} d\mathbf {x} }$ Kinetic energy (K) is the energy used to move a body from a velocity of 0 to a velocity ${\displaystyle \mathbf {u} }$. Restricting the motion to one dimension: ${\displaystyle K=\int _{u=0}^{u=u}\mathbf {f} dx}$ Using the relativistic 3 force: ${\displaystyle K=\int _{u=0}^{u=u}{\frac {d(m\gamma u)}{dt}}dx=\int _{u=0}^{u=u}m{\frac {d(\gamma u)}{dt}}dx=\int _{u=0}^{u=u}md(\gamma u){\frac {dx}{dt}}}$ substituting for ${\displaystyle d(\gamma u)}$ and using ${\displaystyle dx/dt=u}$: ${\displaystyle K=\int _{u=0}^{u=u}m(\gamma du+ud\gamma )u}$ Which gives: ${\displaystyle K=\int _{u=0}^{u=u}m(u\gamma du+u^{2}d\gamma )}$ The Lorentz factor ${\displaystyle \gamma }$ is given by: ${\displaystyle \gamma ={\frac {1}{\sqrt {1-u^{2}/c^{2}}}}}$ meaning that : ${\displaystyle d\gamma ={\frac {u}{c^{2}}}\gamma ^{3}du}$ ${\displaystyle du={\frac {c^{2}}{u\gamma ^{3}}}d\gamma }$ So that ${\displaystyle K=\int _{\gamma =1}^{\gamma =\gamma }m(u\gamma {\frac {c^{2}}{u\gamma ^{3}}}d\gamma +u^{2}d\gamma )=\int _{\gamma =1}^{\gamma =\gamma }m({\frac {c^{2}}{\gamma ^{2}}}+u^{2})d\gamma =\ int _{\gamma =1}^{\gamma =\gamma }mc^{2}d\gamma }$ Alternatively, we can use the fact that: ${\displaystyle \gamma ^{2}c^{2}-\gamma ^{2}u^{2}=c^{2}\,}$ ${\displaystyle 2\gamma c^{2}d\gamma -\gamma ^{2}2udu-u^{2}2\gamma d\gamma =0\,}$ So, rearranging: ${\displaystyle \gamma udu+u^{2}d\gamma =c^{2}d\gamma \,}$ In which case: ${\displaystyle K=\int _{u=0}^{u=u}m(u\gamma du+u^{2}d\gamma )=\int _{u=0}^{u=u}mc^{2}d\gamma \,}$ As ${\displaystyle u}$ goes from 0 to ${\displaystyle u}$, the Lorentz factor ${\displaystyle \gamma }$ goes from 1 to ${\displaystyle \gamma }$, so: ${\displaystyle K=mc^{2}\int _{\gamma =1}^{\gamma =\gamma }d\gamma \,}$ and hence: ${\displaystyle K=\gamma mc^{2}-mc^{2}\,}$ The amount ${\displaystyle \gamma mc^{2}}$ is known as the total energy of the particle. The amount ${\displaystyle mc^{2}}$ is known as the rest energy of the particle. If the total energy of the particle is given the symbol ${\displaystyle E}$: ${\displaystyle E=\gamma mc^{2}=mc^{2}+K\,}$ So it can be seen that ${\displaystyle mc^{2}}$ is the energy of a mass that is stationary. This energy is known as mass energy. The Newtonian approximation for kinetic energy can be derived by using the binomial theorem to expand ${\displaystyle \gamma =(1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$. The binomial expansion is: ${\displaystyle (a+x)^{n}=a^{n}+na^{n-1}x+{\frac {n(n-1)}{2!}}a^{n-2}x^{2}....}$ So expanding ${\displaystyle (1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$: ${\displaystyle K={\frac {1}{2}}mu^{2}+{\frac {3mu^{4}}{8c^{2}}}+{\frac {5mu^{6}}{16c^{4}}}+...}$ So if ${\displaystyle u}$ is much less than ${\displaystyle c}$: ${\displaystyle K={\frac {1}{2}}mu^{2}}$ which is the Newtonian approximation for low velocities. Derivation of relativistic energy using the concept of relativistic mass [edit \| edit source] Energy is defined as the work done in moving a body from one place to another. Energy is given from: ${\displaystyle dE=\mathbf {F} d\mathbf {x} }$ so, over the whole path: ${\displaystyle E=\int _{0}^{x}\mathbf {F} d\mathbf {x} }$ Kinetic energy (K) is the energy used to move a body from a velocity of 0 to a velocity ${\displaystyle u}$. So: ${\displaystyle K=\int _{u=0}^{u=u}Fdx}$ Using the relativistic force: ${\displaystyle K=\int _{u=0}^{u=u}{\frac {d(Mu)}{dt}}dx}$ ${\displaystyle K=\int _{u=0}^{u=u}d(Mu){\frac {dx}{dt}}}$ substituting for ${\displaystyle d(Mu)}$ and using ${\displaystyle dx/dt=u}$: ${\displaystyle K=\int _{u=0}^{u=u}(Mdu+udM)u}$ Which gives: ${\displaystyle K=\int _{u=0}^{u=u}(Mudu+u^{2}dM)}$ The relativistic mass is given by: ${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$ Which can be expanded as: ${\displaystyle M^{2}c^{2}-M^{2}u^{2}=m^{2}c^{2}}$ ${\displaystyle 2Mc^{2}dM-M^{2}2udu-u^{2}2MdM=0}$ So, rearranging: ${\displaystyle Mudu+u^{2}dM=c^{2}dM}$ In which case: ${\displaystyle K=\int _{u=0}^{u=u}(Mudu+u^{2}dM)}$ is simplified to: ${\displaystyle K=\int _{u=0}^{u=u}c^{2}dM}$ But the mass goes from ${\displaystyle m}$ to ${\displaystyle M}$ so: ${\displaystyle K=c^{2}\int _{M=m}^{M=M}dM)}$ and hence: ${\displaystyle K=Mc^{2}-mc^{2}}$ The amount ${\displaystyle Mc^{2}}$ is known as the total energy of the particle. The amount ${\displaystyle mc^{2}}$ is known as the rest energy of the particle. If the total energy of the particle is given the symbol ${\displaystyle E}$: ${\displaystyle E=mc^{2}+K}$ So it can be seen that ${\displaystyle mc^{2}}$ is the energy of a mass that is stationary. This energy is known as mass energy and is the origin of the famous formula ${\displaystyle E=mc^{2}}$ that is iconic of the nuclear age. The Newtonian approximation for kinetic energy can be derived by substituting the rest mass for the relativistic mass ie: ${\displaystyle M={\frac {m}{\sqrt {1-u^{2}/c^{2}}}}}$ ${\displaystyle K=Mc^{2}-mc^{2}}$ ${\displaystyle K={\frac {mc^{2}}{\sqrt {1-u^{2}/c^{2}}}}-mc^{2}}$ ${\displaystyle K=mc^{2}((1-u^{2}/c^{2})^{-{\frac {1}{2}}}-1)}$ The binomial theorem can be used to expand ${\displaystyle (1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$: The binomial theorem is: ${\displaystyle (a+x)^{n}=a^{n}+na^{n-1}x+{\frac {n(n-1)}{2!}}a^{n-2}x^{2}....}$ So expanding ${\displaystyle (1-u^{2}/c^{2})^{-{\frac {1}{2}}}}$: ${\displaystyle K={\frac {1}{2}}mu^{2}+{\frac {3mu^{4}}{8c^{2}}}+{\frac {5mu^{6}}{16c^{4}}}+...}$ So if ${\displaystyle u}$ is much less than ${\displaystyle c}$: ${\displaystyle K={\frac {1}{2}}mu^{2}}$ Which is the Newtonian approximation for low velocities. When protons and neutrons (nucleons) combine to form elements the combination of particles tends to be in a lower energy state than the free neutrons and protons. Iron has the lowest energy and elements above and below iron in the scale of atomic masses tend to have higher energies. This decrease in energy as neutrons and protons bind together is known as the binding energy. The atomic masses of elements are slightly different from that calculated from their constituent particles and this difference in mass energy, calculated from ${\displaystyle E=mc^{2}}$, is almost exactly equal to the binding energy. The binding energy can be released by converting elements with higher masses per nucleon to those with lower masses per nucleon. This can be done by either splitting heavy elements such as uranium into lighter elements such as barium and krypton or by joining together light elements such as hydrogen into heavier elements such as deuterium. If atoms are split the process is known as nuclear fission and if atoms are joined the process is known as nuclear fusion. Atoms that are lighter than iron can be fused to release energy and those heavier than iron can be split to release energy. When hydrogen and a neutron are combined to make deuterium the energy released can be calculated as follows: The mass of a proton is 1.00731 amu, the mass of a neutron is 1.00867 amu and the mass of a deuterium nucleus is 2.0136 amu. The difference in mass between a deuterium nucleus and its components is 0.00238 amu. The energy of this mass difference is: ${\displaystyle E=mc^{2}=1.66\times 10^{-27}\times 0.00238\times (3\times 10^{8})^{2}}$ So the energy released is ${\displaystyle 3.57\times 10^{-13}}$ joules or about ${\displaystyle 2\times 10^{11}}$ joules per gram of protons (ionised hydrogen). (Assuming 1 amu = ${\displaystyle 1.66\times 10^{-27}}$ Kg, Avogadro's number = ${\displaystyle 6\times 10^{23}}$ and the speed of light is ${\displaystyle 3\times 10^{8}}$ metres per second) Present day nuclear reactors use a process called nuclear fission in which rods of uranium emit neutrons which combine with the uranium in the rod to produce uranium isotopes such as ^236U which rapidly decay into smaller nuclei such as Barium and Krypton plus three neutrons which can cause further generation of ^236U and further decay. The fact that each neutron can cause the generation of three more neutrons means that a self sustaining or chain reaction can occur. The generation of energy results from the equivalence of mass and energy; the decay products, barium and krypton have a lower mass than the original ^236U, the missing mass being released as 177 MeV of radiation. The nuclear equation for the decay of ^236U is written as follows: ${\displaystyle _{92}^{236}U\rightarrow _{56}^{144}Ba+_{36}^{89}Kr+3n+177MeV}$ Nuclear explosion If a large amount of the uranium isotope ^235U (the critical mass) is confined the chain reaction can get out of control and almost instantly release a large amount of energy. A device that confines a critical mass of uranium is known as an atomic bomb or A-bomb. A bomb based on the fusion of deuterium atoms is known as a thermonuclear bomb, hydrogen bomb or H-bomb.	{"url":"https://en.wikibooks.org/wiki/Special_Relativity/Dynamics","timestamp":"2024-11-03T13:23:15Z","content_type":"text/html","content_length":"270952","record_id":"<urn:uuid:52d0b728-f308-470e-bb91-cb00a50ac091>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00697.warc.gz"}
[EM] Thoughts on Burial Juho juho4880 at yahoo.co.uk Sun Jul 25 08:26:14 PDT 2010 On Jul 24, 2010, at 3:46 AM, fsimmons at pcc.edu wrote: > You guys have come up with some interesting ideas about the > likelihood of sincere cycles, but my idea > is not that complicated: > Usually in the high stakes elections that I have witnessed there are > just a few issues that most voters > feel strongly about, and opinions on these issues are highly > correlated (or anti-correlated) so that the > voter distribution in issue space is basically cigar shaped. > Perpendicular to the long axis of that cigar find a plane that > divides the voters into two equal subsets > (plus or minus one). The candidate closest to that plane is very > likely a Condorcet candidate. > But this Condorcet candidate can be buried as easily as a Condorcet > candidate can be buried in a > precisely one dimensional issue space. > I like Condorcet methods that discourage burial in one dimensional > cases. I don't care so much about > the case where the candidates are distributed on the vertices of an > acute triangle, i.e. the triangle is > close to equilateral. In that case burial may serve a useful > purpose of decreasing the probability of > winning for a low utility Condorcet candidate. > In particular, the sincere profile > 40 A>C>>B > 30 B>C>A > 30 C>A>>B > could easily come from a one dim or cigar shaped issue space. Any > condorcet method that doesn't > make burial of C risky for the A faction in this context is going to > end up with more artificial cycles than > real ones. Thanks, this is at least a well defined case where strategic cycles might occur. (I guess "B>C>A" should be read "either B>C>A" or "B>>C>A", and in addition to "C>A>>B" votes there could be also some "C>A>B" and even few "C>B>A" votes.) I'm not sure if this case would lead to artificial cycles very easily. 75% of the A supporters should vote strategically to make the strategy work. A smaller number of strategic voters (50%) is sufficient to create an artificial cycle. There are many possible ways this strategy can fail. For example the B supporters prefer C to A. If they know that A supporters will try a strategy and win, then the B supporters might vote directly for C and thereby guarantee that the strategy of the A supporters will not work. The preferences may also change before the election, and part of the C supporters may sincerely rank A lower because of the attempted strategy that tries to steal the victory from their favourite. In short, if some society is so strategic that they would try this strategy then there could be also other strategic moves, and the whole election (and future elections) might become a chaos. It is possible that in some "very strategically oriented" societies with very stable opinions (e.g. assuming that B can not win even if A supporters would rank B higher, and C supporters will not stop liking A because of the strategy) we would get strategic cycles this way, but it seems probable to me that in most societies this kind of chaos would not emerge (assuming that some percentage of voters want to vote sincerely rather than steal the victory etc.) (sorry, no clear proof available). > Note that random ballot on Smith is adequate for preventing the > burial without any defensive strategy on > the part of the C supporters. > On the other hand the profile > 40 A>>C>B > 30 B>>C>A > 30 C>>A>B > could not arise from a one dimensional or cigar shaped issue space. > And candidate C has such low > uility, it wouldn't be bad if A got a share of the probability > through a burial of C. One could also consider low utility Condorcet winners to be worth being elected with 100% probability. The difference in philosophy is if one tries to find a winner that would offer best sum of utility to the voters (=> sum of ratings like philosophy) or if one wants to find a winner that can rule the society thanks to having majority support. > Random Ballot Smith doesn't discourage burial in this case, in which > C retains only 30% of the > probability. Without more detailed information it would be > impossible to prove that C deserved more than > that amount. > ---- > Election-Methods mailing list - see http://electorama.com/em for > list info More information about the Election-Methods mailing list	{"url":"http://lists.electorama.com/pipermail/election-methods-electorama.com/2010-July/124731.html","timestamp":"2024-11-04T05:06:32Z","content_type":"text/html","content_length":"8029","record_id":"<urn:uuid:5588cf13-6d87-4a52-a740-4185eefb2513>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00807.warc.gz"}
Tidy t-test — t_test Tidy t-test A tidier version of t.test() for two sample tests. response = NULL, explanatory = NULL, order = NULL, alternative = "two-sided", mu = 0, conf_int = TRUE, conf_level = 0.95, A data frame that can be coerced into a tibble. A formula with the response variable on the left and the explanatory on the right. Alternatively, a response and explanatory argument can be supplied. The variable name in x that will serve as the response. This is an alternative to using the formula argument. The variable name in x that will serve as the explanatory variable. This is an alternative to using the formula argument. A string vector of specifying the order in which the levels of the explanatory variable should be ordered for subtraction, where order = c("first", "second") means ("first" - "second"). Character string giving the direction of the alternative hypothesis. Options are "two-sided" (default), "greater", or "less". A numeric value giving the hypothesized null mean value for a one sample test and the hypothesized difference for a two sample test. A logical value for whether to include the confidence interval or not. TRUE by default. A numeric value between 0 and 1. Default value is 0.95. For passing in other arguments to t.test(). # t test for number of hours worked per week # by college degree status gss %>% tidyr::drop_na(college) %>% t_test(formula = hours ~ college, order = c("degree", "no degree"), alternative = "two-sided") #> # A tibble: 1 × 7 #> statistic t_df p_value alternative estimate lower_ci upper_ci #> <dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl> #> 1 1.12 366. 0.264 two.sided 1.54 -1.16 4.24 # see vignette("infer") for more explanation of the # intuition behind the infer package, and vignette("t_test") # for more examples of t-tests using infer	{"url":"https://infer.netlify.app/reference/t_test","timestamp":"2024-11-13T23:01:50Z","content_type":"text/html","content_length":"14951","record_id":"<urn:uuid:92ebd721-1efd-4ed7-91ca-103061b002f4>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00056.warc.gz"}
Rounding Game - Rounding for Seats ⋆ PreAlgebraCoach.com Rounding Game – Rounding for Seats Students sitting in a Pre-Algebra class will have varying degrees of ability when it comes to rounding and estimating. This activity is meant to be used with those who need remedial help and to assess the overall ability of the class. It is a fun way to accomplish two things throughout the year rounding and your seating chart! Rounding for Seats – Rounding Game This Rounding Game can be used when you change the seating arrangement in the class. Four sets of cards have been made. One set for each grading period. We have provided enough cards for 5 rows or groups of 5 students in each row or group (25 total students). Here is the Rounding Game Print Out: Rounding for Seats – Rounding Game (PDF) 1. Cut out the Cards below 2. Give Each Student a Card • Numbers that Rounds to ___ will Sit in the First Row • Numbers that Rounds to ___ will Sit in the Second Row • Numbers that Rounds to ___ will Sit in the Third Row • Numbers that Rounds to ___ will Sit in the Fourth Row • Numbers that Rounds to ___ will Sit in the Fifth Row 3. After they find their Row or Group they will sit in the desks in order by value. Highest Number in the First Seat and the Lowest Number in the Last Seat. Like So… Sample Card: Teacher Notes: Print out to attached PDF and cut out the cards for the first grading period. Make 2 copies because one will serve as your answer document. The cards are organized in columns. So the first column will all round to the same number the second will be the second number and so on… Each of the four grading periods have a different place value to round to. The cards get more difficult by grading period. So you will know if they are getting better throughout the year. The number of cards that round to a particular number is equal to the number of students sitting at the table. Each table / columb has a different answer to the number of cards having a problem that rounds to it. If students are seated in rows of desks, there is a different answer for each row. The number of cards that round to that are each different but round to that answer. You can also add decimals. Put a card like 235.74 on the table or row and and give out cards that say ‘Round 235.735 to the nearest hundredths’, ‘..235.740..’, ‘..235.736..’, and 235.741..’ This activity has been used by handing out cards as students walk into the classroom. They find their new seat by matching the card to a number on a table or row. The class starts off in an unusual manner, the students are engaged from the minute they walk in, and it is an excellent way to introduce the lesson on Rounding and Estimating with a Rounding Game. You check students cards as you collect them when everyone is seated. If two students are vying for the same seat, you have a teachable moment for them to discuss why they think they belong in that Here is your Free Content for this lesson! 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Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. 1-1 Assignment Student Edition – Rounding Worksheet (Members Only) 1-1 Assignment Teacher Edition – Rounding Worksheet (Members Only) 1-1 Bell Work Student Edition- Rounding Activity (Members Only) 1-1 Bell Work Teacher Edition- Rounding Activity (Members Only) 1-1 Exit Quiz Student Edition- Rounding Quiz (Members Only) 1-1 Exit Quiz Teacher Edition- Rounding Quiz (Members Only) 1-1 Guided Notes Student Edition – Rounding Notes (Members Only) 1-1 Guided Notes Teacher Edition – Rounding Notes (Members Only) 1-1 Lesson Plan – Rounding (Members Only) 1-1 Online Activities – Rounding (Members Only) 1-1 Slide Show – Rounding Power Point (Members Only) Want access to everything? Simply click the image below to Get All of Our Lessons! 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Torque and Angular Momentum in context of magnitude of torque 30 Aug 2024 Title: The Magnitude of Torque: A Study on the Relationship between Torque and Angular Momentum This paper delves into the fundamental concepts of torque and angular momentum, with a focus on the magnitude of torque. We will explore the mathematical relationships between these two physical quantities and discuss their implications in various contexts. Torque (τ) is a measure of the rotational force that causes an object to rotate or twist around a pivot point. Angular momentum (L), on the other hand, is a measure of the tendency of an object to keep rotating or revolving around its axis. The magnitude of torque is crucial in understanding the dynamics of rotational motion. Mathematical Formulation: The magnitude of torque (τ) can be calculated using the following formula: τ = r × F where r is the radius from the pivot point to the point where the force (F) is applied. In terms of angular momentum, the relationship between torque and angular momentum can be expressed as: L = τ / ω where ω is the angular velocity. Theoretical Framework: From a theoretical perspective, the magnitude of torque plays a crucial role in determining the rate of change of angular momentum. The equation for the time derivative of angular momentum (dL/dt) can be written as: dL/dt = τ - ω × L This equation highlights the interplay between torque and angular momentum, demonstrating how changes in torque affect the magnitude of angular momentum. Physical Insights: The magnitude of torque has significant implications for various physical systems. For instance, in mechanical systems, a larger torque can lead to increased rotational speed or acceleration. In electrical systems, the magnitude of torque is critical in determining the performance of motors and generators. In conclusion, this paper has explored the fundamental relationship between torque and angular momentum, with a focus on the magnitude of torque. The mathematical formulations and theoretical framework presented provide a comprehensive understanding of the interplay between these two physical quantities. Further research can build upon these findings to investigate the applications of torque and angular momentum in various fields. • [1] Goldstein, H. (1980). Classical Mechanics. Addison-Wesley. • [2] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons. Note: The references provided are for general information purposes only and do not constitute a comprehensive bibliography on the topic. Related articles for ‘magnitude of torque ‘ : Calculators for ‘magnitude of torque ‘	{"url":"https://blog.truegeometry.com/tutorials/education/556fbe8f2f9b2a8d1d75cf79881966aa/JSON_TO_ARTCL_Torque_and_Angular_Momentum_in_context_of_magnitude_of_torque_.html","timestamp":"2024-11-12T05:51:08Z","content_type":"text/html","content_length":"16890","record_id":"<urn:uuid:e380a546-d1f4-4f00-8e4b-35b95afc926e>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00467.warc.gz"}
operating cash flow Financial ratios or accounting ratios are most commonly used by every businesses and companies to determine or evaluate the overall financial health of the business and companies. These ratios are frequently used by financial analyst, managers, shareholders, creditors to find out about the strength and weaknesses of the any organization. The data used in calculating financial ratios comes from either income statement, profit and loss account, cash flow statement or company balance sheet. These financial ratios allow the companies to compare its financial strength between companies, industries, different time period for one company. These rations are measured always against benchmark set by a company. Without benchmark, these ratios are not so useful. Company have to have some kind of industries benchmark set to compare against its financial ratios. Why Ignoring #VarianceAnalysis Will Cost You Sales? http://t.co/Hf0W5WtpGw pic.twitter.com/jfHdkLVmb9 — Tutor Pace (@TutorPace) July 26, 2015 Most publicly traded companies are required by law to use generally accepted accounting principles (GAAP) for their home country. However, the private companies such as LLC, partnership, private companies are not required to use GAAP method. There are primarily four main categories of financial ratios that all business used to analyze its data: 1. Profitability Ratios 2. Liquidity Ratios 3. Debt Ratios 4. Activity Ratios 3 ways companies can lower its break-even point. http://t.co/Oz2V3eSVf2 pic.twitter.com/52Eu5GvTro — Tutor Pace (@TutorPace) July 25, 2015 Let’s elaborate further about all the above financial ratios: 1: Profitability Ratios: These ratios allow companies to measure its ability to make adequate return on sales, total assets and invested capital. In other words, these ratios measure how effectively a company utilizes its resources. Some of the profitability ratios are as follows: Profit margin ratio: profit margin ratio is calculated by dividing the net income by sales over a reporting period. For example: if company earns net income for $25,000 in a reporting period and its sales amounted to $250,000, the profit margin ratio can be calculated as follows: Profit Margin Ratio: Net Income/Sales Profit Margin: $25,000/$250,000 = 10% Return on Investment Assets: Return on investment can be arrived by dividing the Net Income by Total Assets. For example if company Total Assets are $200,000, its return on assets ratio will be as Return on Assets: Net Income/Total Assets ROI: $25,000/$200,000 = 12.5% Return on Equity: Return on equity ratio is calculated as dividing the Net Income by Net equity. In other words if company Net equity is worth at $100,000, its Return on Equity ratios will look like as follows: Return on Equity: Net Income/ Net Equity = $25,000/$100,000 = 25% Gross Margin Ratio: Gross Margin Ratio can be calculated by dividing the Gross profit by Net Sales. For example: If company gross profit is $50,000 and its net Sales are $250,000. The Gross Profit Margin ration will look like as follows: Gross Profit Margin Ratio: Gross Profit/ Net Sales = $50,000/$250,000 = 20% 2.Liquidity Ratios: Liquidity ratios determined the company ability to pay it short term obligations normally due within 12 months. There are mainly four liquidity ratio that business or companies would like to find out if they have enough cash to pay its short term debt: Current Ratio: Current ratio is also known as working capital ratio. The current ratio is calculated by dividing the current assets to current liabilities. For example: If Company net current assets are $150,000 and its net current liabilities are $75,000, The company current ratio will look like as follows: Current Ratio: Current Assets/ Current Liabilities = $150,000/$75,000 = 2 Times Quick Ratio: Quick ratio is calculated by subtracting Inventory from Current Assets divided by Current Liabilities. For Example: Company Inventory in hand at the end of reporting period amounted to Quick Ratio: Current Assets-Inventory/ Current Liabilities = 1:1 Cash Ratio: Cash ratios are calculated adding Cash and Marketable Securities divided by Current Liabilities. For Example: If company balance sheet shows cash in hand equal to $100,000 and Its marketable securities on books amounted to $50,000, the its cash ratio should look like this: Cash Ratio: Cash + Marketable Securities/ Current Liabilities $100,000 + $50,000/$75,000 = 2 times Operating Cash Flow: Operating cash flow ratio is calculated as dividing the Operating Cash Flow by Total Debts. For example: company operating cash flow shows $150,000 and Total debt shows $75,000, its operating cash flow ratio should look like this: Operating Cash Flow: Operating Cash Flow/Total Debts = 2 Times 3. Debt Ratios: Debit ratios are also known as leveraging ratios. The ratio is defined as the ratio of total debt to total assets expressed as percentage. These ratios can be interpreted as the proportion of total company’s assets that are financed by company’s debt. The higher of these ratios represent that the company is more leveraged with its debt associated with more financial risk. There are mainly four debt ratios Companies would like to know: Debit Equity Ratio: Debt equity ratio represent the shareholders equity and the debt used to finance company’s assets. The debt equity ratio can be interpreted as proportion of long term debt plus Value of Leases divided by Total Assets. Debit Equity Ratio: Total Liabilities/ Shareholder Equity For Example: If a company has its total liabilities of $200,000 and total shareholders’ equity of $800,000. Its debt to equity ratio will look this: Debt Equity Ratio: $200,000/$800,000 = .25 Total Debt Ratio: Total debt ratio represent the company total liabilities to total assets. The lower the ratio means the company is less dependent on its leverage. In other words, the higher the ratio, the more risk company is taking. For Example: Let’s assume company total assets at the end of reporting period amounted to $900,000 on the balance sheet. So the total debt ratio will look like this: Debt Ratio: Total Liabilities/Total Assets Debit Ratio: $200,000/$900,000 = 22% Interest Coverage Ratio: Interest coverage ratio is commonly used by companies to determine if it can pay interest expenses on its outstanding debt. The ration is calculated by dividing the company earnings before interest and taxes (EBIT) by the total interest expenses. The lower the ratio, the more the company is burdened by its debt expenses. The ratio of less than 1.5 is considered risky as company ability to pay interest expenses will be questionable. For Example: ABC LTD has its earnings before interest and taxes amounted to $200,000 and Interest expenses amounted to $28,000. Interest Coverage Ratio: Earnings before Interest and Taxes/ Interest Expenses Interest Coverage Ratio: $200,000/$28,000 = 7.14 This represent that the company has good margin of safety to cover its interest expenses. Cash Flow to Debt Ratio: The cash flow to debt ratio is calculated by dividing the company operating cash flow by total debit. The ratio tell the business owner if they have the ability to cover their debit from its operating cash flow earning. The higher the ratio is, the chances that better the business owner to carry its total debt. For Example: The Company ABC Ltd. have total operating cash flow amounted to $100,000 and its total debt at the end of reporting period are $130,000. The cash flow to debt ratio will look this: Cash Flow to Debt Ratio: Operating Cash Flow/Total Debt Cash Flow to Debt Ratio: $100,000/ $130,000 = .77 4. Activity Ratios: Activity ratios are those ratios that all business owners like to know if they have the ability to convert different accounts of balance sheet in to cash or sales. These ratios widely used to measure the relative efficiency of a company assets, leverage or other balance sheet items. There are mainly three activity ratios that businesses would like to know: Stock Turnover Ratio: Stock turnover ratio can be calculated by dividing the cost of goods sold by average inventory. Generally the stock turnover ratio can also be calculated by dividing the sales by Inventory. A low turnover is normally considered a bad sign because products value tend to deteriorate as they sit in the warehouse for longer than average period of time. Stock Turnover Ratio: Sales/ Inventory Stock Turnover Ratio: Cost of Goods Sold/ Average Inventory For Example: If a company shows its sales at the end of reporting period amounted to $500,000 and its inventory shows total to $200,000, then stock turnover ratio should look like as follows: Stock Turnover Ratio $500,000/$200,000 = 2.5 Assets Turnover Ratios: Assets turnover ratio is calculated by dividing the sales or Revenues by Total assets. Generally speaking, the higher the ratio, the better it is as company generating more revenue per dollar of assets. For Example: ABC LTD. has total sales at the end of reporting period amounted to $500,000 and its total assets appears on balance sheet at the end of reporting period amounted to $750,000. Then assets turnover ratio can be calculated as follows: Assets Turnover Ratio: $500,000/$750,000 = 67% Inventory Conversion Ratio: Inventory conversion ratio is calculated by total inventory to cost of sales divided by 365. The inventory conversion is measured as against the time required to acquire raw materials for a product, manufacture and then sell it. Inventory Conversion Ratio Inventory/Cost of Sales/365 In addition to helping management and owners of business in diagnosing the financial health of their company or business, ratios can also helpful for managers to make decisions about investments or projects that the company is considering to take, such as acquisitions, or expansion. Still confused or need to brush up your knowledge on financial ratios? Connect with our Online Accounting Tutor and get the help right away.	{"url":"https://freeonlinetutoring.edublogs.org/tag/operating-cash-flow/","timestamp":"2024-11-12T02:23:11Z","content_type":"text/html","content_length":"56362","record_id":"<urn:uuid:d30350dc-a51a-4428-bce8-40f0d4d14a37>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00375.warc.gz"}
Lesson 5 Measure with Connecting Cubes Warm-up: Notice and Wonder: Measure a Pencil (10 minutes) The purpose of this warm-up is for students to compare lengths of objects and notice when they are longer, shorter, or equal to each other in length. While students may notice and wonder many things about these images, comparing the length is an important discussion point. • Groups of 2 • Display the image. • “What do you notice? What do you wonder?” • 1 minute: quiet think time • “Discuss your thinking with your partner.” • 1 minute: partner discussion • Share and record responses. Student Facing What do you notice? What do you wonder? Activity Synthesis • “How can you describe the length of the pencil?” (The pencil is longer than the yellow cubes. The pencil is the same length as the purple cubes.) Activity 1: Lengths of Creepy, Crawly Things (15 minutes) The purpose of this activity is for students to create a connecting cube tower that is the same length as a given image. In the activity synthesis, students transition from describing the length of an object by comparing it to another object ("The earthworm is the same length as a tower of 8 cubes.") to describing the length of an object as a measurement ("The earthworm is 8 cubes long."). Required Materials Materials to Gather Materials to Copy • Lengths of Creepy, Crawly Things • Groups of 2 • Give each student connecting cubes and a copy of the blackline master. • “We just saw a picture that showed a pencil that was the same length as the purple tower. Use connecting cubes to build a tower that is the same length as each creepy, crawly thing.” • 10 minutes: partner work time • Monitor for students who carefully line up the cubes with the endpoints of the images. Activity Synthesis • Invite previously identified students to share. • Display the image of the caterpillar. • “How did you measure the caterpillar?” (I lined the first cube up with the end of the caterpillar. Then I added cubes until I got all the way to the other end of the caterpillar.) • “Since the tower of 4 cubes is the same length as the caterpillar and each cube has the same length, we can say the caterpillar is 4 cubes long.” • For each animal, invite students to say, “The ______ is ___ cubes long.” Activity 2: Measure More Creepy, Crawly Things (10 minutes) The purpose of this activity is for students to measure the length of images using connecting cubes. Although students have compared the lengths of objects in previous activities, length is defined in this activity since it is the first time students measure and describe the length of objects as a number of same-size length units. Students determine how many connecting cube sides long each image is. They make statements such as “The grasshopper is five cubes long.” Some students may disagree on how to measure with their partner based on where they start and end the measurement, which is the focus of the activity synthesis. When students disagree with each other and explain how they decided to measure each image, they critique the reasoning of others (MP3). MLR8 Discussion Supports. During partner work, invite students to take turns sharing their responses. Ask students to restate what they heard using precise mathematical language and their own words. Display the sentence frame: “I heard you say . . .” Original speakers can agree or clarify for their partner. Advances: Listening, Speaking Engagement: Provide Access by Recruiting Interest. Synthesis: Invite students to generate a list of additional examples of objects that can be measured with connecting cubes. Supports accessibility for: Conceptual Processing, Visual-Spatial Processing Required Materials Materials to Gather Materials to Copy • More Creepy, Crawly Things • Groups of 2 • Give each student connecting cubes and a copy of the blackline master. • “In the previous activity, we found the length of animals using connecting cubes. Length is the measure of how long an object is in same-size units without gaps or overlaps.” • “Use connecting cubes to find the length of more creepy, crawly things. First, measure on your own. Then compare your thinking with your partner. If you and your partner don’t agree on the length, work together to come to an agreement. Complete each statement with the number that makes it true.” • 6–8 minutes: partner work time • As students work, consider asking: □ “How long is the _____? How do you know?” Activity Synthesis • Display answers. • “Check your measurements with these. Did you find the same measurements of length?” • Consider asking: □ “If you found the same measurements, what did you and your partner do to make sure you found the right measurement of the length?” □ “If you found a different measurement, what do you think happened when you measured?” Activity 3: Centers: Choice Time (15 minutes) The purpose of this activity is for students to choose from activities that offer practice adding two-digit numbers within 100. Students choose from any stage of previously introduced centers. • How Close? • Target Numbers • Five in a Row Required Preparation • Gather materials from: □ How Close? Stages 1–3 □ Target Numbers, Stages 1–3 □ Five in a Row, Stages 1–6 • Groups of 2 • “Now you are going to choose from centers we have already learned.” • Display the center choices in the student book. • “Think about what you would like to do.” • 30 seconds: quiet think time • Invite students to work at the center of their choice. • 10 minutes: center work time Student Facing Choose a center. How Close? Target Numbers Five in a Row Activity Synthesis • “Diego and Elena are playing How Close. Diego has a sum of 91. Elena has a sum of 89. Who gets a point for being closer to 100? How do you know?” Lesson Synthesis Display an item from the classroom with connecting cubes lined up from endpoint to endpoint or use the image from the warm-up. “Today we measured length with connecting cubes. What is the length of the pencil? How do you know?” (It is 6 cubes long. I know because the cubes are lined up with the top of the pencil and go to the end of the pencil and I counted 6 cubes.) Display the same item from the classroom, using a connecting cube tower with extra cubes on each end of the item. For example: “What is the length of the rectangle? How do you know?” (It is 6 cubes long. There are extra cubes before it and after it, but those aren’t counted because they are not starting or ending on the As needed, “Even though there are some extra cubes before and after the rectangle, we can still measure the length by counting the cube that begins where the rectangle begins. We can stop counting when we get to the end of the rectangle.” Cool-down: Unit 6, Section B Checkpoint (0 minutes)	{"url":"https://curriculum.illustrativemathematics.org/k5/teachers/grade-1/unit-6/lesson-5/lesson.html","timestamp":"2024-11-07T23:12:14Z","content_type":"text/html","content_length":"97202","record_id":"<urn:uuid:a69cb066-a93c-41ac-bed1-565c61cbdd74>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00893.warc.gz"}
Math Mama Writes... My publisher, Natural Math, uses crowdfunding to support each book they publish. I'm excited about this book. Check it out, and contribute (which is really the same as making an advance order). I think you'll enjoy it. Farzanah and the 17 Camels celebrates the excitement and the rewards of solving a challenging and intriguing math problem. Set against the backdrop of the ancient Silk Road, with bustling markets, stunning carpets, fun characters, and camels, the story draws readers into the magic of Farzanah's surroundings. As Farzanah searches for an unusual approach, a way of solving the problem that no one else could think of, she follows the wise advice of her mother: "My dear Farzanah, don't be discouraged,” said Mama. “Sometimes, being stuck is exactly where you need to be. I find the best thing I can do is to step away. I free my mind to think about other things. It is in that space that the magic happens. I am able to look at things from a different perspective. With wait time and wishful thinking comes the solution.” Join the crowdfunding campaign here. The Playful Math Carnival is a collection of blog posts and articles from around the internet, putting lots of goodies in one place for your enjoyment. The theme for this issue is fractions and division. Why are division and fractions so much harder than what came before? And how can we explore them in playful, delightful, engaging ways? This carnival includes lots of perspectives, and approaches the topic from many levels, elementary to college. A puzzle for 174: What are all the factors of 174? Learning how to find factors goes hand in hand with division and fractions. It's easy to see that 2 is a factor of 174. Can you see any others before you divide by 2? There's a "trick" for 3 (and 9), but everything in math has a reason. Do you know why that "trick" works? I see that 1 plus 7 plus 4 is 12, and I know that 3 goes into 12. Why would that tell me something about whether or not 3 is a factor of 174? [Solutions at end. Hint: It's got to do with 10 being 9 plus 1.] Before I started teaching, I had no idea that fractions might be hard. Part of what makes fractions difficult for some students is how many meanings fractions can have: a fraction of one whole, a fraction of some collection, a fraction of a measurement, etc. My own troubles with division came from a slight case of (undiagnosed) dyslexia. Why is it that we write a / b, but then we have b going into a, with the numbers in the opposite order? The way we write it made no sense to me. And I got confused if the numbers were big ones. Because of this challenge for me, I learned one of my first problem-solving lessons: Make a simpler problem with the same structure. If I saw 158 ÷ 79, I could think to myself, "That's like 6 ÷ 3." And then I knew what to do - find out how many 79s in 158. Aha, it's 2, just like 6 ÷ 3! I used to hate the words divisor and dividend. I could not keep them straight. And I still don't know which is which (but if I care, the internet is my friend). And I, my friends, am a math professor. I tell my students often that my bad memory has helped me learn math, because I always tried to make sense out of it, instead of memorizing. My personal favorite division issue now is why division by 0 is undefined. I wrote about it in my forthcoming book, Althea and the Mysteries of Triangles, Circles, and Pi. I'll share that passage at the end of this post. It's written at about high school level. We can go even higher level with the math and explore 0 / 0, an important concept for calculus that took mathematicians 150 years to come to terms with, which I did in a post a few years back. John Golden is the Math Hombre. Denise Gaskins writes at Let's Play Math! Who the heck is Professor Smudge? Shayla Heavner (aka SJ Bennett) created MathBait, and is the author of Marcos the Great and the History of Numberville. She brings us two factoring games. Maria Droujkova, founder of Natural Math (my publisher), is conducting a crowdfunding campaign for a lovely book, Farzanah and the 17 Camels, by Dr. Sue Looney, which tells the story of an ancient math puzzle. One part of that puzzle asks: How can we possibly give one heir half of the 17 camels? Join that campaign here (your donation is basically an advance order of the book). Do you want more?! The Ontario Math Links blog is updated weekly. Browse to your heart's content. (That's where I found Professor Smudge.) The Math Teachers at Play Blog Carnival was created in 2009. Its name changed to Playful Math Carnival along the way, and it's been going strong for 15 years! (15 years online feels like a century anywhere else.) Links to all past posts available here. I used to include dozens of bloggers in my posts. This one only includes 5 people. (When Google evilly got rid of Google Reader, it really devastated the "math blogosphere".) If you have written something you think we'd like to see, please add a comment. Puzzle Solutions: The factors of 174 are 1, 2, 3, 6, 29, 58, 87, and 174. (There are 8 of them. Do all numbers have an even number of factors, or do some have an odd number of factors? Which are which?) Understanding that factoring "trick" for 3 and 9: Add the digits of your number. If 3 or 9 goes into the sum, then it goes into the original. Why? Let's consider 174. The sum of the digits is 12, and 3 goes into 12. Hmm. 174 means 1100 + 710 + 4, and that can be written 1(99+1) + 7 (9+1) + 4. If I distribute, I get 199 + 1 + 79 +7 + 4. 99 and 9 are multiples of 3. So we have 199 + 79 + (1+7+4). Each term is a multiple of 3. The last term is that sum of the digits we looked at. After reading this, could you explain to someone else why the 3 and 9 factoring "tricks" work? p.s Here's that ... Sneak Preview from Althea and the Mysteries of Triangles, Circles, and Pi: Sofia nods. “I messed up. I had 1 over 0, so I wrote 0 for my final answer. I’m not really sure why it’s supposed to be undefined instead. Can you explain that? It did feel kind of tricky to me.” Mom says, “That’s a great question. And to answer it, we actually need to go back to some basics. The problem is that division doesn’t always work. It turns out that dividing by 0 doesn’t make sense. But to see why, we have to go back and look at how we define division." She starts writing on the whiteboard and explaining at the same time. “We know that 6 over 2 is 3, because 2 times 3 is 6. I want to take that relationship and write it in a more generic way. I’m going to use T for top, B for bottom, and A for answer. When I was younger, I think I had trouble remembering numerator and denominator. That might be why I like saying top and bottom. Or maybe I just like shorter words. Anyway, now I can look at multiplication to help me think about weird division problems. I look at Sofia. It seems like she’s deep in thought. Kiara’s taking notes, even though she seemed to know this. I think Mom has shown me this before. “So 0 over 5 equals A becomes 5 times A equals 0. So what’s A?” Sofia says, “It can only be 0.” Mom nods. “And there are other good ways to think about this one. But for the one that tripped you up, this is the only way I know of to make it really make sense. So now 5 over 0 equals A. What does that become?” Aiden starts to talk, but Sofia gives him a look. She says, “I’m the one who doesn’t get this, so let me try. It turns into 0 times A equals 5. But Miss Annie, you can’t get 5. If you have 0 times anything, you’ll get 0.” Mom nods and waits. Sofia continues, “So there is no A that works in this one, and that means there’s no A for the first one. So it has no answer, and that’s why they say undefined?” Mom nods again. Kiara says, “Whoa! I just knew it was supposed to be undefined, but I definitely did not know why. And until this moment, I would not have known I was missing something.” Aiden is nodding too. “I always knew one of those was undefined, but sometimes I mix up which one is which. I don’t think I’ll have that problem anymore.” Sofia looks at him. “So you didn’t get it either?” Aiden says, “I had number 5 right, but I think it was a lucky guess.” We picked a time. We're meeting for nine weeks, each Saturday from March 2 to April 27, for an hour, at 3pm PT / 6pm ET. We still have a few spots open. We'll be playing with Triangles, Circles, and Pi, along with the fictional Althea and her friends. Participants will get an introduction to geometry, proof, and trigonometry. I'm writing a new book series, Althea's Math Mysteries. In four young adult novels, Althea and her friends explore some of the mysteries of mathematics. The first two books are nearing publication, and the second book needs folks to test it out. In Althea and the Mysteries of Triangles, Circles, and Pi, Althea and friends, with the help of Althea’s mom, explore geometry and proof in order to then learn the basics of trigonometry. We'd like to find some eager math students to join us in an online math circle, led by me, to explore along with Althea and her friends. Students will participate in 9 weeks of lively small-group sessions: in part a deep and friendly math course, and also a unique book club, with the author refining the story based on student reactions. Do you know any students who enjoy math, know a bit of algebra, and would enjoy "user testing” Althea and the Mysteries of Triangles, Circles, and Pi? We're looking for a few more young people to try out the activities in this book together. • Attend 9 weeks of 60-minute live online sessions from March 2 to April 27, each Saturday at 3pm PT / 6pm ET. • Read and comment on 1 to 3 chapters of the book each week. • Keep an informal math journal during this time. For all who stay the course: • You'll learn the foundations of geometry and trigonometry (and will get a certificate for completing the course). • You'll get a signed copy of the published book. • Your name or alias will appear in the book's acknowledgements, and you will receive a letter of appreciation for your help with this STEM project. (If you’d like a letter of recommendation later, we will be happy to write one for you.) • You’ll get to build community with math friends and mentors. Interested? Please email me at mathanthologyeditor@gmail.com for more information, or to sign up. Join an online math circle for students ages 12 to 15 in March and April, exploring geometry, proof, and the basics of trigonometry. As most of you know, I'm writing a new book series. In four young adult novels, Althea and her friends will be exploring some of the mysteries of mathematics. The first two books are nearing publication at Natural Math. In Althea and the Mysteries of Triangles, Circles, and Pi, Althea and friends, with the help of Althea’s mom, explore geometry and proof in order to then learn the basics of trigonometry. My publisher and I would like to find some eager math students to join me in an online math circle, exploring some math mysteries along with Althea and her friends. Participants will join lively small-group sessions: in part a deep and friendly math course, and also a unique book club, allowing me to refine the story based on student reactions. Althea and the Mysteries of Triangles, Circles, and Pi is a fictional story set in the present, in which the characters discuss math, with Mom throwing in a few true stories from the past. Like The Number Devil and Math Girls, this book gives you more the more you put into it by doing the math yourself. Do you know any students who enjoy math, know a bit of algebra, and would enjoy user testing Althea and the Mysteries of Triangles, Circles, and Pi? We’re looking for 5 to 8 young people to try out the activities in this book together. If interested, please add your name and information here. • Attend 9 weeks of 60-minute live online sessions in March and April. Times to be determined, most likely 4 p.m. EST / 1 p.m. PST, on Saturday or a weekday (whichever works for more students). • Read and comment on 1 to 3 chapters of the book each week. • Keep an informal math journal during this time. For all who stay the course: • You’ll learn the foundations of geometry and trigonometry (and will get a certificate for completing the course). • You’ll get a signed copy of the published book. • Your name or alias will appear in the book’s acknowledgements, and you will receive a letter of appreciation for your help with this STEM project. (If you’d like a letter of recommendation later, we will be happy to write one for you.) • You’ll get to build community with math friends and mentors. Anyone here reading knows that I'm working on my series of young adult novels with math at the center - Althea's Math Mysteries. But did you know that this drive to tell math stories is growing among budding storytellers across the lands? Sue in California (me!) is writing Althea's Math Mysteries. Four of them! Shayla (aka SK Bennett) in New Mexico is writing the next book after Marco the Great and the History of Numberville. (I'm loving this one. I'm so glad there will be another.) Sarah in Washington has written some wonderful fairy tales about physics and math. I'm reading Newton's Laws: A Fairy Tale right now. (Currently free.) And of course there are about a dozen lovely stories from the authors who work with Natural Math. Who else is out there, writing tales of mathjoy that I haven't discovered yet?!	{"url":"https://mathmamawrites.blogspot.com/2024/","timestamp":"2024-11-10T18:55:12Z","content_type":"application/xhtml+xml","content_length":"154027","record_id":"<urn:uuid:81724f7b-f4bd-48a6-a4f9-cffe8c734dd9>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00769.warc.gz"}
American Mathematical Society Recurrence relations for multivariate $B$-splines HTML articles powered by AMS MathViewer by Carl de Boor and Klaus Höllig Proc. Amer. Math. Soc. 85 (1982), 397-400 DOI: https://doi.org/10.1090/S0002-9939-1982-0656111-8 PDF \| Request permission We prove recurrence relations for a general class of multivariate ${\text {B}}$-splines, obtained as ’projections’ of convex polyhedra. Our results are simple consequences of Stokes’ theorem and include, as special cases, the recurrence relations for the standard multivariate simplicial ${\text {B}}$-spline. References C. de Boor, Splines as linear combinations of $B$-splines, Approximation Theory II, G. G. Lorentz, C. K. Chui and L. L. Schumaker (eds.), Academic Press, New York, 1976, pp. 1-47. C. de Boor and R. DeVore, Approximation by smooth multivariate splines, Math. Res. Center Tech. Summary Rep. 2319, Univ. of Wisconsin-Madison, 1981. C. de Boor and K. Höllig, $B$-splines from parallelepipeds, Math. Res. Center Tech. Summary Rep. 2320, Univ. of Wisconsin-Madison, 1982. H. B. Curry and I. J. Schoenberg, On spline distributions and their limits: the Pólya distribution functions, Bull. Amer. Math. Soc. 53 (1947), 1114, Abstract 380t. —, Multivariate $B$-splines—recurrence relations and linear combinations of truncated powers, Multivariate Approximation Theory, W. Schempp and K. Zeller (eds.), Birkhäuser, Basel, 1979, 64-82. —, Konstruktion mehrdimensionaler $B$-splines und ihre Anwendungen auf Approximationsprobleme, Numerische Methoden der Approximationstheorie, Bd. 5, L. Collatz, G. Meinardus and H. Werner (eds.), Birkhäuser, Basel, 1980, pp. 84-110. —, Approximation by smooth multivariate splines on non-uniform grids, Quantitative Approximation, R. DeVore and K. Scherer (eds.), Academic Press, New York, 1980, pp. 99-114. —, On the linear independence of multivariate $B$-splines. I. Triangulations of simploids, SIAM J. Numer. Anal. (to appear). T. N. T. Goodman and S. L. Lee, Spline approximation operators of Bernstein-Schoenberg type in one and two variables, J. Approximation Theory (to appear). H. Hakopian, On multivariate $B$-splines, SIAM J. Numer. Anal. (to appear). K. Höllig, A remark on multivariate $B$-splines, J. Approximation Theory (to appear). —, Multivariate splines, Math. Res. Center Tech. Summary Rep. 2188, Univ. of Wisconsin-Madison, 1981; SIAM J. Numer. Anal. (to appear). P. Kergin, Interpolation of ${C^K}$ functions, Thesis, University of Toronto, 1978. —, On a numerically efficient method for computing multivariate $B$-splines, Multivariate Approximation Theory, W. Schempp and K. Zeller (eds.), Birkhäuser, Basel, 1979, pp. 211-248. I. J. Schoenberg, letter to Philip J. Davis dated May 31, 1965. Similar Articles • Retrieve articles in Proceedings of the American Mathematical Society with MSC: 41A15 • Retrieve articles in all journals with MSC: 41A15 Bibliographic Information • © Copyright 1982 American Mathematical Society • Journal: Proc. Amer. Math. Soc. 85 (1982), 397-400 • MSC: Primary 41A15 • DOI: https://doi.org/10.1090/S0002-9939-1982-0656111-8 • MathSciNet review: 656111	{"url":"https://www.ams.org/journals/proc/1982-085-03/S0002-9939-1982-0656111-8/home.html","timestamp":"2024-11-09T00:38:16Z","content_type":"text/html","content_length":"61495","record_id":"<urn:uuid:56ecc8c5-aba2-4e5b-9b60-a82d863b4d5d>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00432.warc.gz"}
McGraw Hill My Math Grade 4 Chapter 10 Answer Key Fractions and Decimals All the solutions provided in McGraw Hill My Math Grade 4 Answer Key PDF Chapter 10 Fractions and Decimals will give you a clear idea of the concepts. McGraw-Hill My Math Grade 4 Answer Key Chapter 10 Fractions and Decimals Essential Question How are fractions and decimals related? Answer: Noninteger, or partial, numbers can be expressed using both fractions and decimals. The ratio between two numbers is a fraction. Often, these numbers are each whole numbers, represented in \ (\frac{p}{q}\) form such as $\frac{1}{2}$ or $\frac{7}{10}$. Decimals are defined as the digits that come after the decimal point and are numbers that fall between integers. The spaces past the decimal point are known as tenths, hundredths, thousandths and so on because decimals use a system of numbers based on units of tens. Additionally, fractions can be expressed as decimals by performing the division of the ratio. (For example, $\frac{1}{4}$ is equivalent to 1 divided by 4, or 0.25.) Decimals can also be expressed as fractions in terms of tenths, hundredths, thousandths and so on. (For example, 0.25 is equivalent to 25 hundrendths, which is equivalent to $\frac{25}{100}$.) Am I Ready Write a fraction to describe the part that is green. Question 1. Answer: $\frac{7}{10}$ There are 10 parts in total, 7 of which are green. So, Numerator indicates green color parts and Denominator indicates Total parts. Thus, Green color parts (Numerator) / Total parts (Denominator) = $\frac{7}{10}$. Question 2. Answer: $\frac{2}{10}$ There are 10 parts in total, 2 of which are green. So, Numerator indicates green color parts and Denominator indicates Total parts. Thus, Green color parts (Numerator) / Total parts (Denominator) = $\frac{2}{10}$. Question 3. Answer: $\frac{68}{100}$ There are 100 parts in total, 68 of which are green. So, Numerator indicates green color parts and Denominator indicates Total parts. Thus, Green color parts (Numerator) / Total parts (Denominator) = $\frac{68}{100}$. Write each as a fraction. Question 4. four tenths Answer: $\frac{4}{10}$ 4 tenths as a fraction is $\frac{4}{10}$since it is 4 over ten. Question 5. eight tenths Answer: $\frac{8}{10}$ 8 tenths as a fraction is $\frac{8}{10}$ since it is 8 over ten. Question 6. twenty hundredths Answer: $\frac{20}{100}$ 20 Hundredths as a fraction is $\frac{20}{100}$ since it is 20 over hundred. Question 7. On Tuesday, seven tenths of an inch of rain fell. Write the amount of rain as a fraction. Answer: $\frac{7}{10}$ Here, the total number of parts is 10. We have to take only 7 equal parts out of 10 equal part Thus, the required fraction =$\frac{7}{10}$ Algebra Find each unknown. Question 8. Answer: $\frac{2}{10}$ $\frac{1}{5}$ is indicated as blue colored parts in total parts. Similarly, there are 10 parts in total, 2 of which are in blue. This shows that the parts in left hand side are doubled in right hand side. Hence the fraction got doubled. Question 9. Answer: $\frac{8}{10}$ $\frac{4}{5}$ is indicated as blue colored parts in total parts. Similarly, there are 10 parts in total, 8 of which are in blue. This shows that the parts in left hand side are doubled in right hand side. Hence the fraction got doubled. My Math Words Review Vocabulary place value Making Connections Use the review vocabulary to make equivalent fractions. Find the unknown in the fraction or shade to complete the area model. Answer: i) $\frac{3}{10}$ equivalent fraction is $\frac{30}{100}$ There are 100 parts in total, 30 of which are in blue color. ii) $\frac{80}{100}$ equivalent fraction is $\frac{8}{10}$ There are 10 parts in total, 8 of which are in green color. To determine equivalent fractions, divide the fraction’s numerator and denominator by the same number (common factor). Although $\frac{3}{10}$ may look like a different fraction, it is actually equivalent to {$\frac{30}{100}$} [$\frac{3}{10}$divided by $\frac{10}{10}$ = $\frac{30}{100}$] Similarly, $\frac{80}{100}$may look like a different fraction, it is actually equivalent to {$\frac{8}{10}$} [($\frac{80}{100}$ divided by $\frac{10}{10}$ = $\frac{8}{10}$] My Vocabulary Cards Ideas for Use • Draw or write examples for each card. Be sure your examples are different from what is shown on each card. • Use a blank card to write a word from a previous chapter that you would like to review. • Use a blank card to write this chapter’s essential question. Use the back of the card to write or draw examples that help you answer the question. One of one hundred equal parts. Write thirty-one hundredths as a decimal. Answer: 0.31 If you divide 31 by one hundred you get 31 hundredths as a decimal which is 0.31 A number that uses place value and a decimal point to show part of a whole. Explain how decimals and fractions are alike. Answer: Noninteger, or partial, numbers can be expressed using both fractions and decimals. The ratio between two numbers is a fraction. Often, these numbers are each whole numbers, represented in \ (\frac{p}{q}\) form such as $\frac{1}{2}$ or $\frac{7}{10}$. Decimals are defined as the digits that come after the decimal point and are numbers that fall between integers. The spaces past the decimal point are known as tenths, hundredths, thousandths and so on because decimals use a system of numbers based on units of tens. Additionally, fractions can be expressed as decimals by performing the division of the ratio. (For example, $\frac{1}{4}$ is equivalent to 1 divided by 4, or 0.25.) Decimals can also be expressed as fractions in terms of tenths, hundredths, thousandths and so on. (For example, 0.25 is equivalent to 25 hundrendths, which is equivalent to $\frac{25}{100}$.) One of ten equal parts. Write six tenths as a decimal. Answer: 0.6 six tenths is 6 divided by 10 and is written as 0.6. Thus, the decimal part here indicated six tenths. My Foldable Follow the steps on the back to make your Foldable. Answer: 3 ($\frac{1}{10}$) = $\frac{3}{10}$ In green color circle, there are total 10 parts where each part thus becomes $\frac{1}{10}$th part of total. In step 3, blue color circle folds green color circle in such a way that only 3 parts are visible out of 10 parts. Therefore, $\frac{3}{10}$ is the required fraction from the folding. Leave a Comment You must be logged in to post a comment.	{"url":"https://ccssmathanswers.com/mcgraw-hill-my-math-grade-4-chapter-10-answer-key/","timestamp":"2024-11-07T02:52:04Z","content_type":"text/html","content_length":"259902","record_id":"<urn:uuid:f59103bb-6a6b-4e9f-afb7-ce18503ef081>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00425.warc.gz"}
How hexadecimal notation works for CSS colors Hexadecimal notation are colors that start with a "#". For example, #ff0000 is red and #ff00ff is pink. But how do I know which colors they are? Read on to find out. How hecadecimal works Each color starts with a #. Then there are three pairs of numbers, where each pair is the red, green and blue component of a color. Visually, that looks like this: red green blue # 00 00 00 These numbers are in hexadecimal (16 steps), so they don't count from 0 to 9 like we do, but from 0 to F. To make up for the missing numbers after 9, they go to "A", "B" all the way to "F". It doesn't matter if you use lower case or uppercase. Because there's a pair of numbers it means there are 255 steps, from 00, to 02, 03, all the way to FF. How does that make colors? The color #ff0000 has as much red as possible (ff), no green (00) and no blue (also 00). In other words, it's fully red. The color #ffff00 is likewise as much red as possible, as much green as possible and no blue. Red and green together make yellow. Lastly, #ffffff is all red, all green and all blue or in other words, full white (and #000000 is full black). When all the colors are the same it means not one color is more visible than the other, making the result grey. #111111, #666666 and #9a9a9a are all shades of grey. Likewise, when the numbers are close together, they are desaturated (closer to gray) In hexadecimal notation, 88 is the middle point . Anything above that is light, anything below it is dark. Color notation variations In CSS there are three variations on the hexadecimal notation. You can add a fourth pair of numbers, which correspond to the Alpha of a color, the transparency. So #ff000088 would be fully red at half transparency. There is also the short notation, which has just three numbers. In it #f00 is the same as #ff0000. The single numbers are automatically expanded by browsers. Likewise this three number notation can also get a fourth number that encodes the transparency. #f008 is fully red at half transparency. "Reading" a color When I read a color, I find it most useful to ignore each second number in a pair since it doesn't have a drastic effect. So for example the color#e5e7b1 would be: • E for red, which is not fully red (that would be F) but very close to it. • Same for green which also has an E • The blue component is a B, so it has a bit less blue. The result of this is then a light yellow. And fo another color, #123456: • 12 for red, so basically no red • 34 for green, so a little bit of green • 45 for blue, so a bit more of blue All are way below 88 though, so this would be a dark, somewhat desaturated (since the colors are close to each other) blue. This was adapted from an explanation I gave to someone not able to see colors but that still wanted to understand how they worked. I hope this is useful to other people as well! Top comments (0) For further actions, you may consider blocking this person and/or reporting abuse	{"url":"https://dev.to/kilianvalkhof/how-hexadecimal-notation-works-for-css-colors-2693","timestamp":"2024-11-12T22:53:50Z","content_type":"text/html","content_length":"66226","record_id":"<urn:uuid:2c30023d-97e3-4e2b-8afe-20179bf0aa11>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00239.warc.gz"}
Cart Pole In this fourth tutorial we’ll be creating a simulation of a cart pole using machine learning. The pole is attached by an un-actuated joint to the cart. The cart is controlled by a linear actuator, that drives the cart left or right. The goal is to swing the pole up and balance it above the cart using motor control. Start by downloading the CAD geometry. The completed model can be downloaded. RL Problem The reinforcement learning problem describes how an intelligent agent can take actions in an environment (simulated world) to maximize the cumulative reward. The problem is described in the paper Neuronlike adaptive elements that can solve difficult learning control problems by Andrew Barto, Richard Sutton and Charles Anderson. Markov Decision Process The RL problem can be formalised as a Markov Decision Process (MDP). A MDP is a mathematical framework used to describe discrete-time stochastic systems where we would like to model decision making. At each time step, the MDP is in some state s and the agent can select an action a. At the next time step, the MDP responds by moving into a new state and giving the agent a reward R(s,a). The MDP can be formally defined as a 5-tuple: $$\langle \mathcal{A}, \mathcal{S}, \mathcal{P}, \mathcal{R}, \gamma \rangle$$ where: $$ \begin{align} & \boldsymbol{\cdot} \mathcal{A} \text{ is a set of actions (action space) } \newline & \ boldsymbol{\cdot} \mathcal{S} \text{ is a set of states (state space) } \newline & \boldsymbol{\cdot} \mathcal{P} \text{ is a probabilistic state transition function } \newline & \boldsymbol{\cdot} \ mathcal{R} \text{ is a probabilistic reward function } \newline & \boldsymbol{\cdot} \gamma \text{ is a discount factor in the interval } [0, 1] \newline \end{align} $$ Markov Decision Process refers to the fact that the system obeys the Markov property: transitions only depend on the most recent state and action, and no prior history of states and actions. The discount factor γ is used to trade-off short-term rewards against long-term future rewards. Optimization Objective The goal is to find the optimal policy 𝜋* for the agent which maximizes the expected discounted cumulative reward. It has the following objective function: $$\max_{\pi} \ \mathbb{E} \left[ \sum_{t=0}^{\infty} \gamma^t \ \mathcal{R}(s_t, a_t) \right]$$ where: $$ \begin{align} & \boldsymbol{\cdot} s_t \text{ is the state at time } t \newline & \ boldsymbol{\cdot} a_t \text{ is the action taken at time } t \newline & \boldsymbol{\cdot} \gamma \text{ is the discount factor } \newline & \boldsymbol{\cdot} \pi \text{ is the policy function s.t. $a_t \sim \pi(s_t)$ } \newline & \boldsymbol{\cdot} \mathcal{R}(s_t, a_t) \text{ is the reward received at time } t \newline \end{align} $$ ProtoTwin Model In this tutorial, we train the cart pole inside ProtoTwin Connect. The state transition function P is deterministic, and is modeled by stepping the simulation. The reward function R is deterministic. The policy function 𝜋 is deterministic (a = 𝜋(s)). The discount factor γ = .99 and the episode is finite-horizon with total number of time steps T corresponding to a time limit of 10 seconds. At each time step, the agent sees a partial observation of the state of the virtual world. The goal is to maximize the cumulative reward: $$\max_{\pi} \ \sum_{t=0}^{T} \gamma^t \ \mathcal{R}(o_t)$$ Action Space The action space is an ndarray with shape (1,) containing: NUM ACTION MIN MAX 0 Cart Target Velocity -1 1 Observation Space The observation space is an ndarray with shape (4,) containing: NUM OBSERVATION MIN MAX 0 Cart Position -1 1 1 Pole Angle -1 1 2 Cart Velocity -inf inf 3 Pole Angular Velocity -inf inf The observation space is a subset of the state space. • Cart Position is a measure of the cart’s distance from the center, where 0 is at the center and +/-1 is at the limit. • Pole Angle is a measure of the pole’s angular distance from the upright position, where 0 is at the upright position and +/-1 is at the down position. Reward Function Since we want to balance the pole upright for as long as possible, our reward function is defined as: def reward(self, obs): distance = 1 - math.fabs(obs[0]) # How close the cart is to the center angle = 1 - math.fabs(obs[1]) # How close the pole is to the upright position force = math.fabs(self.get(address_cart_force)) # How much force is being applied to drive the cart's motor reward = angle * 0.8 + distance * 0.2 - force * 0.004 return max(reward * self.dt, 0) Episode End The episode ends if any one of the following conditions are met: • Termination: Cart goes beyond the limits [-1, 1]. These limits correspond to a distance of ±0.65m from the center. • Truncation: Episode time is greater than 10 seconds. To solve the cart pole problem, we will use an algorithm called Proximal Policy Optimization (PPO). PPO is a policy gradient method for training the agent’s policy neural network. The learned policy 𝜋 is a Multi-Layer Perceptron (MLP) which takes as input an observation and outputs a probability distribution over the actions. PPO is an actor-critic algorithm meaning it uses MLP to learn both the optimal policy function (actor network) and value function (critic network). The action to take is the one with the highest probability: $$a_t = \arg\max_{a} \ \pi(a\|s_t)$$ Signals represent I/O for components defined in ProtoTwin. Signals are either readable or writable. You can find the signals provided by each component inside ProtoTwin under the I/O dropdown menu. The I/O window lists the name, address and type of each signal along with its access (readable/writable). The signals used in this tutorial are: • The target velocity of the cart motor. • The current position of the cart motor. • The current velocity of the cart motor. • The current force applied by the cart motor. • The current position of the pole motor. • The current velocity of the pole motor. Make sure to install the following packages: pip install prototwin pip install prototwin-gymnasium pip install stable-baselines3 pip install torch pip install numpy pip install asyncio The prototwin package provides a client for starting and connecting to an instance of ProtoTwin Connect. Using this client you can issue commands to load a model, step the simulation forwards in time, read signal values and write signal values. The prototwin gymnasium package provides a base environment for Gymnasium for being used in RL workflows. The stable baselines3 package provides a reliable set of RL algorithm implementations in PyTorch. We also use NumPy when working with arrays and asyncio for writing concurrent code using the async/await syntax. Python Script The complete python script is provided below: # STEP 1: Import dependencies import asyncio import os import torch import numpy as np import math import gymnasium import prototwin import stable_baselines3.ppo from stable_baselines3 import PPO from stable_baselines3.common.vec_env import VecMonitor from stable_baselines3.common.callbacks import CheckpointCallback from prototwin_gymnasium import VecEnvInstance, VecEnv # STEP 2: Define signal addresses (obtain these values from ProtoTwin) address_cart_target_velocity = 3 address_cart_position = 5 address_cart_velocity = 6 address_cart_force = 7 address_pole_angle = 12 address_pole_angular_velocity = 13 # STEP 3: Create your vectorized instance environment by extending the base environment class CartPoleEnv(VecEnvInstance): def __init__(self, client: prototwin.Client, instance: int) -> None: super().__init__(client, instance) self.dt = 0.01 # Time step self.x_threshold = 0.65 # Maximum cart distance def reward(self, obs): distance = 1 - math.fabs(obs[0]) # How close the cart is to the center angle = 1 - math.fabs(obs[1]) # How close the pole is to the upright position force = math.fabs(self.get(address_cart_force)) # How much force is being applied to drive the cart's motor reward = angle * 0.8 + distance * 0.2 - force * 0.004 return max(reward * self.dt, 0) def observations(self): cart_position = self.get(address_cart_position) # Read the current cart position cart_velocity = self.get(address_cart_velocity) # Read the current cart velocity pole_angle = self.get(address_pole_angle) # Read the current pole angle pole_angular_velocity = self.get(address_pole_angular_velocity) # Read the current pole angular velocity pole_angular_distance = math.atan2(math.sin(pole_angle), math.cos(math.pi - pole_angle)) # Calculate the pole's angular distance from upright position return np.array([cart_position / self.x_threshold, pole_angular_distance / math.pi, cart_velocity, pole_angular_velocity]) def reset(self, seed = None): return self.observations(), {} def apply(self, action): self.set(address_cart_target_velocity, action[0]) # Apply action by setting the cart's target velocity def step(self): obs = self.observations() reward = self.reward(obs) # Calculate reward done = abs(obs[0]) > 1 # Terminate if cart goes beyond limits truncated = self.time > 10 # Truncate after 10 seconds return obs, reward, done, truncated, {} # STEP 4: Setup the training session async def main(): # Start ProtoTwin Connect client = await prototwin.start() # Load the ProtoTwin model filepath = os.path.join(os.path.dirname(__file__), "CartPole.ptm") await client.load(filepath) # Create the vectorized environment entity_name = "Main" num_envs = 64 # The observation space contains: # 0. A measure of the cart's distance from the center, where 0 is at the center and +/-1 is at the limit # 1. A measure of the pole's angular distance from the upright position, where 0 is at the upright position and +/-1 is at the down position # 2. The cart's current velocity (m/s) # 3. The pole's angular velocity (rad/s) observation_high = np.array([1, 1, np.finfo(np.float32).max, np.finfo(np.float32).max], dtype=np.float32) observation_space = gymnasium.spaces.Box(-observation_high, observation_high, dtype=np.float32) # The action space contains only the cart's target velocity action_high = np.array([1.0], dtype=np.float32) action_space = gymnasium.spaces.Box(-action_high, action_high, dtype=np.float32) env = VecEnv(CartPoleEnv, client, entity_name, num_envs, observation_space, action_space) monitored = VecMonitor(env) # Monitor the training progress # Create callback to regularly save the model save_freq = 10000 # Number of timesteps per instance checkpoint_callback = CheckpointCallback(save_freq=save_freq, save_path="./logs/checkpoints/", name_prefix="checkpoint", save_replay_buffer=True, save_vecnormalize=True) # Define learning rate schedule def lr_schedule(progress_remaining): initial_lr = 0.003 return initial_lr * (progress_remaining ** 2) # Define the ML model model = PPO(stable_baselines3.ppo.MlpPolicy, monitored, device=torch.cuda.current_device(), verbose=1, batch_size=4096, n_steps=1000, learning_rate=lr_schedule, tensorboard_log="./tensorboard/") # Start training! model.learn(total_timesteps=10_000_000, callback=checkpoint_callback) Exporting to ONNX It is possible to export trained models to the ONNX format. This can be used to embed trained agents into ProtoTwin models for inferencing. Please refer to the Stable Baselines exporting documentation for further details. The complete python script is provided below: import torch as th from typing import Tuple from stable_baselines3 import PPO from stable_baselines3.common.policies import BasePolicy # Export to ONNX for embedding into ProtoTwin models using ONNX Runtime Web def export(): class OnnxableSB3Policy(th.nn.Module): def __init__(self, policy: BasePolicy): self.policy = policy def forward(self, observation: th.Tensor) -> Tuple[th.Tensor, th.Tensor, th.Tensor]: return self.policy(observation, deterministic=True) # Load the trained ML model model = PPO.load("model", device="cpu") # Create the Onnx policy onnx_policy = OnnxableSB3Policy(model.policy) observation_size = model.observation_space.shape dummy_input = th.randn(1, *observation_size) th.onnx.export(onnx_policy, dummy_input, "CartPole.onnx", opset_version=17, input_names=["input"], output_names=["output"]) Inference in ProtoTwin It is possible to embed trained agents into ProtoTwin models. To do this, you must create a scripted component that loads the ONNX model, feeds observations into the model, and applies the output actions. This example assumes the ONNX file has been included into the model by dragging the file into the script editor’s file explorer. Alternatively, the ONNX file can be loaded from a URL. The complete source code for the inference component is provided below: import { type Entity, type Handle, InferenceComponent, MotorComponent, Util } from "prototwin"; export class CartPole extends InferenceComponent { public cartMotor: Handle<MotorComponent>; public poleMotor: Handle<MotorComponent>; constructor(entity: Entity) { this.cartMotor = this.handle(MotorComponent); this.poleMotor = this.handle(MotorComponent); public override async initializeAsync() { // Load the ONNX model from the local filesystem. this.loadModelFromFile("CartPole.onnx", 4, new Float32Array([-1]), new Float32Array([1])); public override async updateAsync() { const cartMotor = this.cartMotor.value; const poleMotor = this.poleMotor.value; const observations = this.observations; if (cartMotor === null \|\| poleMotor === null \|\| observations === null) { return; } // Populate observation array const cartPosition = cartMotor.currentPosition; const cartVelocity = cartMotor.currentVelocity; const poleAngularDistance = Util.signedAngularDifference(poleMotor.currentPosition, Math.PI); const poleAngularVelocity = poleMotor.currentVelocity; observations[0] = cartPosition / 0.65; observations[1] = poleAngularDistance / Math.PI; observations[2] = cartVelocity; observations[3] = poleAngularVelocity; // Apply the actions const actions = await this.run(); if (actions !== null) { cartMotor.targetVelocity = actions[0];	{"url":"https://prototwin.com/docs/tutorials/cart-pole","timestamp":"2024-11-11T17:12:51Z","content_type":"text/html","content_length":"79845","record_id":"<urn:uuid:0ee9a113-5f7a-49d1-86e1-219616bf751b>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00255.warc.gz"}
Generalised Linear Models with brms - Rens van de Schoot Intro to Bayesian (Multilevel) Generalised Linear Models (GLM) in R with brms Last modified: date: 14 October 2019 This tutorial provides an introduction to Bayesian GLM (genearlised linear models) with non-informative priors using the brms package in R. If you have not followed the Intro to Frequentist (Multilevel) Generalised Linear Models (GLM) in R with glm and lme4 tutorial, we highly recommend that you do so, because it offers more extensive information about GLM. If you are not familar with Bayesian inference, we also recommend that you read this tutorial Building a Multilevel Model in BRMS Tutorial: Popularity Data prior to using this tutorial. The current tutorial specifically focuses on the use of Bayesian logistic regression in both binary-outcome and count/porportion-outcome scenarios, and the respective approaches to model evaluation. The tutorial uses the Thai Educational Data example in Chapter 6 of the book Multilevel analysis: Techniques and applications. Furthermore, the tutorial briefly demonstrates the multilevel extension of Bayesian GLM models. This tutorial follows this structure: 1. Preparation; 2. Introduction to GLM; 3. Thai Educational Data; 4. Data Preparation; 5. Bayesian Binary (Bernoulli) Logistic Regression; 6. Bayesian Binomial Logistic Regression; 7. Bayesian Multilevel Logistic Regression. Note that this tutorial is meant for beginners and therefore does not delve into technical details and complex models. For a detailed introduction into frequentist multilevel models, see this LME4 Tutorial. For an extensive overview of GLM models, see here. If you want to use the Bayesian approach for your own research, we recommend that you follow the WAMBS-checklist. 1. Preparation This tutorial expects: – Installation of R packages brms for Bayesian (multilevel) generalised linear models (this tutorial uses version 2.9.0). Because of some special dependencies, for brms to work, you still need to install a couple of other things. See this tutorial on how to install brms. Note that currently brms only works with R 3.5.3 or an earlier version; – Installation of R package tidyverse for data manipulation and plotting with ggplot2; – Installation of R package haven for reading sav format data; – Installation of R package ROCR for calculating area under the curve (AUC); – Installation of R package sjstats for calculating intra-class correlation (ICC). Remember to install version 0.17.5 (using the command install_version("sjstats", version = "0.17.5") after loading the package devtools, because the latest version of sjstats does not support the ICC function anymore); – Installation of R package modelr for data manipulation; – Installation of R package tidybayes for extraction, manipulation, and visualisation of posterior draws from Bayesian models; – Basic knowledge of hypothesis testing and statistical inference; – Basic knowledge of Bayesian statistical inference; – Basic knowledge of coding in R; – Basic knowledge of plotting and data manipulation with tidyverse. 2. Introduction to Genearlised Linear Models (GLM) If you are already familar with generalised linear models (GLM), you can proceed to the next section. Otherwise, click “Read More” to learn about GLM. Recall that in a linear regression model, the object is to model the expected value of a continuous variable, $Y$, as a linear function of the predictor, $\eta = X\beta$. The model structure is thus: $E(Y) = X\beta + e$, where $e$ refers to the residual error term. The linear regression model assumes that $Y$ is continous and comes from a normal distribution, that $e$ is normally distributed and that the relationship between the linear predictor $\eta$ and the expected outcome $E(Y)$ is strictly linear. However, these assumptions are easily violated in many real world data examples, such as those with binary or proportional outcome variables and those with non-linear relationships between the predictors and the outcome variable. In these scenarios where linear regression models are clearly inappropriate, generalised linear models (GLM) are needed. The GLM is the genearlised version of linear regression that allows for deviations from the assumptions underlying linear regression. The GLM generalises linear regression by assuming the dependent variable $Y$ to be generated from any particular distribution in an exponential family (a large class of probability distributions that includes the normal, binomial, Poisson and gamma distributions, among others). In this way, the distribution of $Y$ does not necessarily have to be normal. In addition, the GLM allows the linear predictor $\eta$ to be connected to the expected value of the outcome variable, $E(Y)$, via a link function $g(.)$. The outcome variable, $Y$, therefore, depends on $\eta$ through $E(Y) = g^{-1}(\eta) = g^{-1}(X\beta)$. In this way, the model does not assume a linear relationship between $E(Y)$ and $\eta$; instead, the model assumes a linear relationship between $E(Y)$ and the transformed $g^{-1}(\eta)$. This tutorial focuses on the Bayesian version of the probably most popular example of GLM: logistic regression. Logistic regression has two variants, the well-known binary logistic regression that is used to model binary outcomes (1 or 0; “yes” or “no”), and the less-known binomial logistic regression suited to model count/proportion data. Binary logistic regression assumes that $Y$ comes from a Bernoulli distribution, where $Y$ only takes a value of 1 (target event) or 0 (non-target event). Binary logistic regression connects $E (Y)$ and $\eta$ via the logit link $\eta = logit(\pi) = log(\pi/(1-\pi))$, where $\pi$ refers to the probability of the target event ($Y = 1$). Binomial logistic regression, in contrast, assumes a binomial distribution underlying $Y$, where $Y$ is interpreted as the number of target events, can take on any non-negative integer value and is binomially distributed with regards to $n$ number of trials and $\pi$ probability of the target event. The link function is the same as that of binary logistic regression. The next section details the exampler data (Thai Educational Data) in this tutorial, followed by the demonstration of the use of Bayesian binary, Bayesian binomial logistic regression and Bayesian multilevel binary logistic regression. For the frequentist versions of these models, see the Intro to Frequentist (Multilevel) Generalised Linear Models (GLM) in R with glm and lme4 tutorial. 3. Thai Educational Data The data used in this tutorial is the Thai Eduational Data that is also used as an example in Chapter 6 of Multilevel analysis: Techniques and applications. The data can be downloaded from here. The data stems from a national survey of primary education in Thailand (Raudenbush & Bhumirat, 1992). Each row in the data refers to a pupil. The outcome variable REPEAT is a dichotomous variable indicating whether a pupil has repeated a grade during primary education. The SCHOOLID variable indicates the school of a pupil. The person-level predictors include: SEX (0 = female, 1 = male) and PPED (having had preschool education, 0 = no, 1 = yes). The school-level is MSESC, representing school mean SES (socio-economic status) scores. The main research questions that this tutorial seeks to answer using the Thai Educational Data are: 1. Ignoring the clustering structure of the data, what are the effects of gender and preschool education on whether a pupil repeats a grade? 2. Ignoring the clustering structure of the data, what is the effect of school mean SES on the proportion of pupil repeating a grade? 3. Considering the clustering structure of the data, what are the effects of gender, preschool education and school mean SES on whether a pupil repeats a grade? These three questions are answered by using these following models, respectively: Bayesian binary logistic regressioin; Bayesian binomial logistic regression; Bayesian multilevel binary logistic 4. Data Preparation 4.1. Load necessary packages # if you dont have these packages installed yet, please use the install.packages("package_name") command. library(tidyverse) # for data manipulation and plots library(haven) #for reading sav data library(sjstats) #for calculating intra-class correlation (ICC) library(ROCR) #for calculating area under the curve (AUC) statistics library(brms) #for Bayesian (multilevel) generalised linear modelling library(modelr) #for data manipulation library(tidybayes) #for analysis of posterior draws of a Bayesian model 4.2. Import Data ThaiEdu_Raw <- read_sav("https://github.com/MultiLevelAnalysis/Datasets-third-edition-Multilevel-book/blob/master/chapter%206/Thaieduc/thaieduc.sav?raw=true") ## # A tibble: 6 x 5 ## SCHOOLID SEX PPED REPEAT MSESC ## <dbl> <dbl+lbl> <dbl+lbl> <dbl+lbl> <dbl> ## 1 10101 0 [girl] 1 [yes] 0 [no] NA ## 2 10101 0 [girl] 1 [yes] 0 [no] NA ## 3 10101 0 [girl] 1 [yes] 0 [no] NA ## 4 10101 0 [girl] 1 [yes] 0 [no] NA ## 5 10101 0 [girl] 1 [yes] 0 [no] NA ## 6 10101 0 [girl] 1 [yes] 0 [no] NA Alternatively, you can download the data directly from here and import it locally. 4.3. Data Processing ThaiEdu_New <- ThaiEdu_Raw %>% mutate(SCHOOLID = factor(SCHOOLID), SEX = if_else(SEX == 0, "girl", "boy"), SEX = factor(SEX, levels = c("girl", "boy")), PPED = if_else(PPED == 0, "no", "yes"), PPED = factor(PPED, levels = c("no", "yes"))) ## # A tibble: 6 x 5 ## SCHOOLID SEX PPED REPEAT MSESC ## <fct> <fct> <fct> <dbl+lbl> <dbl> ## 1 10101 girl yes 0 [no] NA ## 2 10101 girl yes 0 [no] NA ## 3 10101 girl yes 0 [no] NA ## 4 10101 girl yes 0 [no] NA ## 5 10101 girl yes 0 [no] NA ## 6 10101 girl yes 0 [no] NA 4.4. Inspect Missing Data ThaiEdu_New %>% summarise_each(list(~sum(is.na(.)))) %>% ## # A tibble: 5 x 2 ## key value ## <chr> <int> ## 1 SCHOOLID 0 ## 2 SEX 0 ## 3 PPED 0 ## 4 REPEAT 0 ## 5 MSESC 1066 The data has 1066 observations missing for the MSESC variable. The treatment of missing data is a complicated topic on its own. For the sake of convenience, we simply list-wise delete the cases with missing data in this tutorial. ThaiEdu_New <- ThaiEdu_New %>% 5. Bayesian Binary Logistic Regression (with Non-Informative Priors) 5.1. Explore Data: number of REPEAT by SEX and PPED ThaiEdu_New %>% group_by(SEX) %>% summarise(REPEAT = sum(REPEAT)) ## # A tibble: 2 x 2 ## SEX REPEAT ## <fct> <dbl> ## 1 girl 428 ## 2 boy 639 ThaiEdu_New %>% group_by(PPED) %>% summarise(REPEAT = sum(REPEAT)) ## # A tibble: 2 x 2 ## PPED REPEAT ## <fct> <dbl> ## 1 no 673 ## 2 yes 394 It seems that the number of pupils who repeated a grade differs quite a bit between the two genders, with more male pupils having to repeat a grade. More pupils who did not have preschool education repeated a grade. This observation suggests that SEX and PPED might be predictive of REPEAT. 5.2. Fit a Bayesian Binary Logistic Regression Model The brm function from the brms package performs Bayesian GLM. The brm has three basic arguments that are identical to those of the glm function: formula, family and data. However, note that in the family argument, we need to specify bernoulli (rather than binomial) for a binary logistic regression. The brm function has a few more additional (and necessary) arguments that glm does not offer: warmup specifies the burn-in period (i.e. number of iterations that should be discarded); iter specifies the total number of iterations (including the burn-in iterations); chains specifies the number of chains; inits specifies the starting values of the iterations (normally you can either use the maximum likelihood esimates of the parameters as starting values, or simply ask the algorithm to start with zeros); cores specifies the number of cores used for the algorithm; seed specifies the random seed, allowing for replication of results. See below the specification of the binary logistic regression model with two predictors, without using informative priors. Bayes_Model_Binary <- brm(formula = REPEAT ~ SEX + PPED, family = bernoulli(link = "logit"), warmup = 500, iter = 2000, chains = 2, inits= "0", seed = 123) ## Compiling the C++ model ## Start sampling 5.3. Model Convergence Before looking at the model summary, we should check whether there is evidence of non-convergence for the two chains. To do so, we can use the stanplot function from the brms package. First, we plot the caterpillar plot for each parameter of interest. type = "trace") ## No divergences to plot. The plot only shows the iterations after the burn-in period. The two chains mix well for all of the parameters and therefore, we can conclude no evidence of non-convergence. We can also check autocorrelation, considering that the presence of strong autocorrelation would bias variance estimates. type = "acf_bar") The plot shows no evidence of autocorrelation for all model variables in both chains, as the autocorrelation parameters all quickly diminish to around zero. 5.4. Interpretation Now, we can safely proceed to the interpretation of the model. Below is the model summary of the Bayesian binary logistic regression model. ## Family: bernoulli ## Links: mu = logit ## Formula: REPEAT ~ SEX + PPED ## Data: ThaiEdu_New (Number of observations: 7516) ## Samples: 2 chains, each with iter = 2000; warmup = 500; thin = 1; ## total post-warmup samples = 3000 ## Population-Level Effects: ## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat ## Intercept -1.77 0.06 -1.88 -1.65 2621 1.00 ## SEXboy 0.43 0.07 0.30 0.57 2470 1.00 ## PPEDyes -0.61 0.07 -0.74 -0.48 2451 1.00 ## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample ## is a crude measure of effective sample size, and Rhat is the potential ## scale reduction factor on split chains (at convergence, Rhat = 1). For comparison, below is the model summary of the frequentist binary logistic regression model. Model_Binary <- glm(formula = REPEAT ~ SEX + PPED, family = binomial(link = "logit"), data = ThaiEdu_New) ## Call: ## glm(formula = REPEAT ~ SEX + PPED, family = binomial(link = "logit"), ## data = ThaiEdu_New) ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -0.6844 -0.5630 -0.5170 -0.4218 2.2199 ## Coefficients: ## Estimate Std. Error z value Pr(>\|z\|) ## (Intercept) -1.76195 0.05798 -30.387 < 2e-16 * ## SEXboy 0.42983 0.06760 6.358 2.04e-10 * ## PPEDyes -0.61298 0.06833 -8.971 < 2e-16 * ## --- ## Signif. codes: 0 '' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 ## (Dispersion parameter for binomial family taken to be 1) ## Null deviance: 6140.8 on 7515 degrees of freedom ## Residual deviance: 6016.2 on 7513 degrees of freedom ## AIC: 6022.2 ## Number of Fisher Scoring iterations: 4 From the model summary above, we can see that the Bayesian model estimates are almost identical to those of the frequentist model. The interpretation of these estimates are the same in both frequentist and Bayesian models. Nevertheless, note that the interpretation of the uncertainty intervals is not the same between the two models. In the frequentist model, the idea behind using a 95% uncertainty interval (confidence interval) is that, under repeated sampling, 95% of the resulting uncertainy intervals would cover the true population value. That allows us to say that, for a given 95% confidence interval, we are 95% confident that this confidence interval contains the true population value. However, it does not allow us to say that there is a 95% chance that the confidence interval contains the true population value (i.e. frequentist uncertainty intervals are not probability statements). In contrast, in the Bayesian model, the 95% uncertainty interval (called credibility interval), which is more interpretable, states that there is 95% chance that the true population value falls within this interval. When the 95% credibility intervals do not contain zero, we conclude that the respective model parameters are likely meaningful. Let’s visualise the point estimates and their associated uncertainty intervals, using the stanplot function. type = "areas", prob = 0.95) The plot above shows the densities of the parameter estimates. The dark blue line in each density represents the point estimate, while the light-blue area indicates the 95% credibility intervals. We can easily see that both SEX and PPED are meaningful predictors, as their credibility intervals do not contain zero and their densities have a very narrow shape. SEX positively predicts a pupil’s probability of repeating a grade, while PPED negatively so. Specifically, in comparison to being a girl, being a boy is more likely to repeat a grade, assuming everything else stays constant. Having previous schooling is less likely to result in repeating a grade, assuming everything else stays constant. To interpret the value of the parameter estimates, we need to exponentiate the estimates. See below. ## Estimate Q2.5 Q97.5 ## Intercept 0.1711235 0.1529269 0.1917295 ## SEXboy 1.5411834 1.3490342 1.7598020 ## PPEDyes 0.5418504 0.4755864 0.6217185 We can also plot densities of these parameter estimates. For this, we again use the stanplot function from brms. type = "areas", prob = 0.95, transformations = "exp") + geom_vline(xintercept = 1, color = "grey") Note that the interpretation of the parameter estimates is linked to the odds rather than probabilities. The definition of odds is: P(event occurring)/P(event not occurring). In this analysis, assuming everything else stays the same, being a boy increases the odds of repeating a grade by 54%, in comparison to being a girl; having preschool education lowers the odds of repeating a grade by (1 – 0.54)% = 46%, in comparison to not having preschool education, assuming everything else stays constant. The baseline odds (indicated by the intercept term) of repeating a grade, namely if you’re a girl with no previous schooling, is about 17%. 5.4. Visualisation of Parameter Effects We can plot the marginal effects (i.e. estimated probabilities of repeating a grade) of the variables in the model. Below, we show how different combinations of SEX and PPED result in different probability estimates. The advantage of this approach is that probabilities are more interpretable than odds. ThaiEdu_New %>% data_grid(SEX, PPED) %>% add_fitted_draws(Bayes_Model_Binary) %>% ggplot(aes(x = .value, y = interaction(SEX, PPED))) + stat_pointintervalh(.width = c(.68, .95)) + coord_flip() + xlab("predicted probability") + scale_x_continuous(breaks = seq(0, 0.24, 0.02)) As we can see, being a male pupil with no preschool education has the highest probability (~0.21), followed by being a girl with no preschool education (~0.15), being a boy with preschool education (~0.13), and lastly, being a girl with preschool education (~0.09). Note that both 68% (thicker inner lines) and 95% (thinner outer lines) credibility intervals for the estimates are included to give us some idea of the uncertainties of the estimates. 5.5. Model Evaluation In the Intro to Frequentist (Multilevel) Generalised Linear Models (GLM) in R with glm and lme4 tutorial, we learn that we can use the likelihood ratio test and AIC to assess the goodness of fit of the model(s). However, these two approaches do not apply to Bayesian models. Instead, Bayesian models make use of so-called Posterior Predictive P-values (PPPs) to assess the fit of the model. In addition, many also use Bayes factors to quantify support from the data for the model. This tutorial does not delve into PPPs or Bayes factors because of the complexity of the topics. The other two measures mentioned in Intro to Frequentist (Multilevel) Generalised Linear Models (GLM) in R with glm and lme4 are correct classification rate and area under the curve (AUC). They are model-agnostic, meaning they can be applied to both frequentist and Bayesian models. 5.5.1. Correct Classification Rate The percentage of correct classification is a useful measure to see how well the model fits the data. #use the `predict()` function to calculate the predicted probabilities of pupils in the original data from the fitted model Pred <- predict(Bayes_Model_Binary, type = "response") Pred <- if_else(Pred[,1] > 0.5, 1, 0) ConfusionMatrix <- table(Pred, pull(ThaiEdu_New, REPEAT)) #`pull` results in a vector #correct classification rate ## [1] 0.8580362 ## Pred 0 1 ## 0 6449 1067 We can see that the model correctly classifies 85.8% of all the observations. However, a closer look at the confusion matrix reveals that the model predicts all of the observations to belong to class “0”, meaning that all pupils are predicted not to repeat a grade. Given that the majority category of the REPEAT variable is 0 (No), the model does not perform better in classification than simply assigning all observations to the majority class 0 (No). 5.5.2. AUC (area under the curve). An alternative to using correct classification rate is the Area under the Curve (AUC) measure. The AUC measures discrimination, that is, the ability of the test to correctly classify those with and without the target response. In the current data, the target response is repeating a grade. We randomly pick one pupil from the “repeating a grade” group and one from the “not repeating a grade” group. The pupil with the higher predicted probability should be the one from the “repeating a grade” group. The AUC is the percentage of randomly drawn pairs for which this is true. This procedure sets AUC apart from the correct classification rate because the AUC is not dependent on the imblance of the proportions of classes in the outcome variable. A value of 0.50 means that the model does not classify better than chance. A good model should have an AUC score much higher than 0.50 (preferably higher than 0.80). # Compute AUC for predicting Class with the model Prob <- predict(Bayes_Model_Binary, type="response") Prob <- Prob[,1] Pred <- prediction(Prob, as.vector(pull(ThaiEdu_New, REPEAT))) AUC <- performance(Pred, measure = "auc") AUC <- AUC@y.values[[1]] ## [1] 0.6014733 With an AUC score of close to 0.60, the model does not discriminate well. 6. Bayesian Binomial Logistic Regression (with Non-Informative Priors) As explained in the Intro to Frequentist (Multilevel) Generalised Linear Models (GLM) in R with glm and lme4 tutorial, logistic regression can also be used to model count or proportion data. Binary logistic regression assumes that the outcome variable comes from a bernoulli distribution (which is a special case of binomial distributions) where the number of trial $n$ is 1 and thus the outcome variable can only be 1 or 0. In contrast, binomial logistic regression assumes that the number of the target events follows a binomial distribution with $n$ trials and probability $q$. In this way, binomial logistic regression allows the outcome variable to take any non-negative integer value and thus is capable of handling count data. The Thai Educational Data records information about individual pupils that are clustered within schools. By aggregating the number of pupils who repeated a grade by school, we obtain a new data set where each row represents a school, with information about the proportion of pupils repeating a grade in that school. The MSESC (mean SES score) is also on the school level; therefore, it can be used to predict proportion or count of pupils who repeat a grade in a particular school. See below. 6.1. Tranform Data ThaiEdu_Prop <- ThaiEdu_New %>% group_by(SCHOOLID, MSESC) %>% summarise(REPEAT = sum(REPEAT), TOTAL = n()) %>% ## # A tibble: 6 x 4 ## SCHOOLID MSESC REPEAT TOTAL ## <fct> <dbl> <dbl> <int> ## 1 10103 0.88 1 17 ## 2 10104 0.2 0 29 ## 3 10105 -0.07 5 18 ## 4 10106 0.47 0 5 ## 5 10108 0.76 3 19 ## 6 10109 1.06 9 21 In this new data set, REPEAT refers to the number of pupils who repeated a grade; TOTAL refers to the total number of students in a particular school. 6.2. Explore data ThaiEdu_Prop %>% ggplot(aes(x = exp(MSESC)/(1+exp(MSESC)), y = REPEAT/TOTAL)) + geom_point() + geom_smooth(method = "lm") We can see that the proportion of students who repeated a grade is (moderately) negatively related to the inverse-logit of MSESC. Note that we model the variable MSESC as its inverse-logit because in a binomial regression model, we assume a linear relationship between the inverse-logit of the linear predictor and the outcome (i.e. proportion of events), not linearity between the predictor itself and the outcome. 6.3. Fit a Binomial Logistic Regression Model To fit a Bayesian binomial logistic regression model, we also use the brm function like we did with the previous Bayesian binary logistic regression model. There are, however, two differences: First, to specify the outcome variable in the formula, we need to specify both the number of target events (REPEAT) and the total number of trials (TOTAL) wrapped in trials(), which are separated by \|. In addition, the family should be “binomial” instead of “bernoulli”. Bayes_Model_Prop <- brm(REPEAT \| trials(TOTAL) ~ MSESC, data = ThaiEdu_Prop, family = binomial(link = "logit"), warmup = 500, iter = 2000, chains = 2, inits = "0", cores = 2, seed = 123) ## Compiling the C++ model ## Start sampling ## Family: binomial ## Links: mu = logit ## Formula: REPEAT \| trials(TOTAL) ~ MSESC ## Data: ThaiEdu_Prop (Number of observations: 356) ## Samples: 2 chains, each with iter = 2000; warmup = 500; thin = 1; ## total post-warmup samples = 3000 ## Population-Level Effects: ## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat ## Intercept -1.81 0.03 -1.87 -1.74 2743 1.00 ## MSESC -0.44 0.09 -0.62 -0.26 2478 1.00 ## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample ## is a crude measure of effective sample size, and Rhat is the potential ## scale reduction factor on split chains (at convergence, Rhat = 1). The frequentist model (for comparison): Model_Prop <- glm(formula = cbind(REPEAT, TOTAL-REPEAT) ~ MSESC, family = binomial(logit), data = ThaiEdu_Prop) ## Call: ## glm(formula = cbind(REPEAT, TOTAL - REPEAT) ~ MSESC, family = binomial(logit), ## data = ThaiEdu_Prop) ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -3.3629 -1.8935 -0.5083 1.1674 6.9494 ## Coefficients: ## Estimate Std. Error z value Pr(>\|z\|) ## (Intercept) -1.80434 0.03324 -54.280 < 2e-16 * ## MSESC -0.43644 0.09164 -4.763 1.91e-06 * ## --- ## Signif. codes: 0 '*' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 ## (Dispersion parameter for binomial family taken to be 1) ## Null deviance: 1480.7 on 355 degrees of freedom ## Residual deviance: 1457.3 on 354 degrees of freedom ## AIC: 2192 ## Number of Fisher Scoring iterations: 5 We can see that the model estimates between the Bayesian and the frequentist binomial logistic regression models are very similar. Note that we skipped the step of checking model convergence, for the sake of keeping this tutorial shorter. You can use the same codes we showed before (with the binary logistic regression model) to check the convergence of this model. 6.4. Interpretation The parameter interpretation in a binomial regression model is the same as that in a binary logistic regression model. We know from the model summary above that the mean SES score of a school is negatively related to the odds of students repeating a grade in that school. To enhance interpretability, we again calculate the exponentiated coefficient estimate of MSESC. Since MSESC is a continous variable, we can standardise the exponentiated MSESC estimate (by multiplying the original estimate with the SD of the variable, and then then exponentiating the resulting number). exp(fixef(Bayes_Model_Prop)[2,-2]sd(pull(ThaiEdu_Prop, MSESC), na.rm = T)) ## Estimate Q2.5 Q97.5 ## 0.8465065 0.7897812 0.9056835 We can see that with a SD increase in MSESC, the odds of students repeating a grade is lowered by about (1 – 85%) = 15%. “Q2.5” and “Q97.5” refer to the lower bound and the upper bound of the uncertainty interval, respectively. This credibility interval does not contain zero, suggesting that the variable is likely meaningful. We can visualise the effect of MSESC. Bayes_Model_Prop %>% spread_draws(b_Intercept, b_MSESC) %>% mutate(MSESC = list(seq(-0.77, 1.49, 0.01))) %>% #the observed value range of MSESC unnest(MSESC) %>% mutate(pred = exp(b_Intercept + b_MSESCMSESC)/(1+exp(b_Intercept + b_MSESCMSESC))) %>% group_by(MSESC) %>% summarise(pred_m = mean(pred, na.rm = TRUE), pred_low = quantile(pred, prob = 0.025), pred_high = quantile(pred, prob = 0.975)) %>% ggplot(aes(x = MSESC, y = pred_m)) + geom_line() + geom_ribbon(aes(ymin = pred_low, ymax = pred_high), alpha=0.2) + ylab("Predicted Probability of Repeating a Grade") + scale_y_continuous(breaks = seq(0, 0.22, 0.01)) The plot above shows the expected influence of MSESC on the probability of a pupil repeating a grade. Holding everything else constant, as MSESC increases, the probability of a pupil repeating a grade lowers (from 0.19 to 0.08). The grey shaded areas indicate the 95% credibility intervals of the predicted values at each value of MSESC. 6.5. Model Evaluation Similar to the Bayesian binary logistic regression model, we can use the PPPS and Bayes factor (which are not discussed in this tutorial) to evaluate the fit of a Bayesian binomial logistic regression model. Correct classification rate and AUC are not suited here, as the model is not concerned with classification. 7. Bayesian Multilevel Binary Logistic Regression (with Non-Informative Priors) The Bayesian binary logistic regression model introduced earlier is limited to modelling the effects of pupil-level predictors; the Bayesian binomial logistic regression is limited to modelling the effects of school-level predictors. To incorporate both pupil-level and school-level predictors, we can use multilevel models, specifically, Bayesian multilevel binary logistic regression. If you are unfamiliar with multilevel models, you can use Multilevel analysis: Techniques and applications for reference and this tutorial for a good introduction to multilevel models with the lme4 package in In addition to the motivation above, there are more reasons to use multilevel models. For instance, as the data are clustered within schools, it is likely that pupils from the same school are more similar to each other than those from other schools. Because of this, in one school, the probability of a pupil repeating a grade may be high, while in another school, low. Furthermore, even the relationship between the outcome (i.e. repeating a grade) and the predictor variabales (e.g. gender, preschool education, SES) may be different across schools. Also note that there are missing values in the MSESC variable. Using multilevel models can appropriately address these issues. See the following plot as an example. The plot shows the proportions of students repeating a grade across schools. We can see vast differences across schools. Therefore, we need multilevel models. ThaiEdu_New %>% group_by(SCHOOLID) %>% summarise(PROP = sum(REPEAT)/n()) %>% We can also plot the relationship between SEX and REPEAT by SCHOOLID, to see whether the relationship between gender and repeating a grade differs by school. ThaiEdu_New %>% mutate(SEX = if_else(SEX == "boy", 1, 0)) %>% ggplot(aes(x = SEX, y = REPEAT, color = as.factor(SCHOOLID))) + geom_point(alpha = .1, position = "jitter")+ geom_smooth(method = "glm", se = F, method.args = list(family = "binomial")) + theme(legend.position = "none") + scale_x_continuous(breaks = c(0, 1)) + scale_y_continuous(breaks = c(0, 1)) In the plot above, different colors represent different schools. We can see that the relationship between SEX and REPEAT appears to be quite different across schools. We can make the same plot for PPED and REPEAT. ThaiEdu_New %>% mutate(PPED = if_else(PPED == "yes", 1, 0)) %>% ggplot(aes(x = PPED, y = REPEAT, color = as.factor(SCHOOLID))) + geom_point(alpha = .1, position = "jitter")+ geom_smooth(method = "glm", se = F, method.args = list(family = "binomial")) + theme(legend.position = "none") + scale_x_continuous(breaks = c(0, 1)) + scale_y_continuous(breaks = c(0, 1)) The relationship between PPED and REPEAT also appears to be quite different across schools. However, we can also see that most of the relationships follow a downward trend, going from 0 (no previous schooling) to 1 (with previous schooling), indicating a negative relationship between PPED and REPEAT. Because of the observations above, we can conclude that there is a need for multilevel modelling in the current data, with not only a random intercept (SCHOOLID) but potentially also random slopes of the SEX and PPED. 7.1. Center Variables Prior to fitting a multilevel model, it is necessary to center the predictors by using an appropriately chosen centering method (i.e. grand-mean centering or within-cluster centering), because the centering approach matters for the interpretation of the model estimates. Following the advice of Enders and Tofighi (2007), we should use within-cluster centering for the first-level predictors SEX and PPED, and grand-mean centering for the second-level predictor MSESC. ThaiEdu_Center <- ThaiEdu_New %>% mutate(SEX = if_else(SEX == "girl", 0, 1), PPED = if_else(PPED == "yes", 1, 0)) %>% group_by(SCHOOLID) %>% mutate(SEX = SEX - mean(SEX), PPED = PPED - mean(PPED)) %>% ungroup() %>% mutate(MSESC = MSESC - mean(MSESC, na.rm = T)) ## # A tibble: 6 x 5 ## SCHOOLID SEX PPED REPEAT MSESC ## <fct> <dbl> <dbl> <dbl+lbl> <dbl> ## 1 10103 -0.647 -0.882 0 [no] 0.870 ## 2 10103 -0.647 -0.882 0 [no] 0.870 ## 3 10103 -0.647 0.118 0 [no] 0.870 ## 4 10103 -0.647 0.118 0 [no] 0.870 ## 5 10103 -0.647 0.118 0 [no] 0.870 ## 6 10103 -0.647 0.118 0 [no] 0.870 7.2. Intercept Only Model To specify a multilevel model, we again use the brm function from the brms package. Note that the random effect term should be included in parentheses. In addition, within the parentheses, the random slope term(s) and the cluster terms should be separated by \|. We start by specifying an intercept-only model, in order to assess the impact of the clustering structure of the data. Note that we will skip the step of model convergence diagnostics. Bayes_Model_Multi_Intercept <- brm(REPEAT ~ 1 + (1\|SCHOOLID), data = ThaiEdu_Center, family = bernoulli(link = "logit"), warmup = 500, iter = 2000, chains = 2, inits = "0", cores = 2, seed = 123) ## Compiling the C++ model ## Start sampling Below we calculate the ICC (intra-class correlation) of the intercept-only model. Note that for non-Gaussian Bayesian models (e.g. logistic regression), we need to set “ppd = T” such that the variance calculation is based on the posterior predictive distribution. icc(Bayes_Model_Multi_Intercept, ppd = T) ## # Random Effect Variances and ICC ## Family: bernoulli (logit) ## Conditioned on: all random effects ## ## Variance Ratio (comparable to ICC) ## Ratio: 0.29 HDI 89%: [0.20 0.37] ## ## Variances of Posterior Predicted Distribution ## Conditioned on fixed effects: 0.09 HDI 89%: [0.08 0.10] ## Conditioned on rand. effects: 0.12 HDI 89%: [0.12 0.13] ## ## Difference in Variances ## Difference: 0.03 HDI 89%: [0.02 0.05] A variance ratio (comparable to ICC) of 0.29 means that 29% of the variation in the outcome variable can be accounted for by the clustering stucture of the data. This provides evidence that a multilevel model may make a difference to the model estimates, in comparison with a non-multilevel model. Therefore, the use of multilevel models is necessary and warrantied. 7.3. Full Model It is good practice to build a multilevel model step by step. However, as this tutorial’s focus is not on muitilevel modelling, we go directly from the intercept-only model to the full-model that we are ultimately interested in. In the full model, we include not only fixed effect terms of SEX, PPED and MSESC and a random intercept term, but also random slope terms for SEX and PPED. Note that we specify family = bernoulli(link = "logit"), as this model is essentially a binary logistic regression model. Bayes_Model_Multi_Full <- brm(REPEAT ~ SEX + PPED + MSESC + (1 + SEX + PPED\|SCHOOLID), data = ThaiEdu_Center, family = bernoulli(link = "logit"), warmup = 500, iter = 2000, chains = 2, inits = "0", cores = 2, seed = 123) ## Compiling the C++ model ## Start sampling ## Family: bernoulli ## Links: mu = logit ## Formula: REPEAT ~ SEX + PPED + MSESC + (1 + SEX + PPED \| SCHOOLID) ## Data: ThaiEdu_Center (Number of observations: 7516) ## Samples: 2 chains, each with iter = 2000; warmup = 500; thin = 1; ## total post-warmup samples = 3000 ## Group-Level Effects: ## ~SCHOOLID (Number of levels: 356) ## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat ## sd(Intercept) 1.34 0.08 1.19 1.50 851 1.00 ## sd(SEX) 0.38 0.18 0.05 0.73 492 1.00 ## sd(PPED) 0.26 0.18 0.01 0.69 800 1.00 ## cor(Intercept,SEX) 0.42 0.29 -0.22 0.93 1298 1.00 ## cor(Intercept,PPED) -0.20 0.43 -0.90 0.75 2392 1.00 ## cor(SEX,PPED) 0.01 0.49 -0.86 0.87 1502 1.00 ## Population-Level Effects: ## Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat ## Intercept -2.29 0.09 -2.47 -2.12 801 1.00 ## SEX 0.45 0.11 0.22 0.66 1615 1.00 ## PPED -0.60 0.13 -0.86 -0.34 2826 1.00 ## MSESC -0.51 0.22 -0.94 -0.07 743 1.00 ## Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample ## is a crude measure of effective sample size, and Rhat is the potential ## scale reduction factor on split chains (at convergence, Rhat = 1). We can plot the densities of the relevant model parameter estimates. type = "areas", prob = 0.95) The results (pertaining to the fixed effects) are similar to the results of the previous Bayesian binary logistic regression and binomial logistic regression models. On the pupil-level, SEX has a positive influence on the odds of a pupil repeating a grade, while PPED has a negative influence. On the school-level, MSESC has a negative effect on the outcome variable. Among three predictors, SEX and PPED have credibility intervals (indicated by the shaded light blue regions in the densities) that clearly do not contain zero. Therefore, they should be treated as meaningful predictors. In contrast, MSESC, despite having a 95% credibility interval without zero, the upper bound of the credibility interval is very close to zero, and its density only contains zero. Because of this, MSESC is likely a less relevant predictor than SEX and PPED. Now let’s look at the random effect terms (sd(Intercept), sd(SEX) and sd(PPED)). The density of sd(Intercept) in the plot is clearly away from zero, indicating the relevance of including this random intercept term in the model. The variance of the random slope of SEX is $0.38^2 = 0.14$, and that of PPED is $0.26^2 = 0.07$. Both variances are not negligible. However, if we look at the density plot, the lower bounds of the credibility intervals of both sd(SEX) and sd(PPED) are very close to zero, and their densities also not clearly separate from zero. This suggests that including these two random slope terms may not be necessary. We can also plot the random effect terms across schools. #extract posterior distributions of all the random effect terms data_RandomEffect <- ranef(Bayes_Model_Multi_Full) #extract posterior distributions of `sd(Intercept)` r_Intercept <- data_RandomEffect$SCHOOLID[, , 1] %>% as_tibble() %>% rownames_to_column(var = "SCHOOLID") %>% mutate(Variable = "sd(Intercept)") #extract posterior distributions of `sd(SEX)` r_SEX <- data_RandomEffect$SCHOOLID[, , 2] %>% as_tibble() %>% rownames_to_column(var = "SCHOOLID") %>% mutate(Variable = "sd(SEX)") #extract posterior distributions of `sd(PPED)` r_PPED <- data_RandomEffect$SCHOOLID[, , 3] %>% as_tibble() %>% rownames_to_column(var = "SCHOOLID") %>% mutate(Variable = "sd(PPED)") r_Intercept %>% bind_rows(r_SEX) %>% bind_rows(r_PPED) %>% mutate(Contain_Zero = if_else(Q2.5Q97.5 > 0, "no", "yes")) %>% ggplot(aes(x = SCHOOLID, y = Estimate, col = Contain_Zero)) + geom_point() + geom_errorbar(aes(ymin=Q2.5, ymax=Q97.5)) + facet_grid(. ~ Variable, scale = "free") + coord_flip() + theme(legend.position = "top") Again, we can see that the posterior distributions of the random intercept term (sd(Intercept)) have a large variance across schools. Quite a number of them are also away from zero. Therefore, we can conclude that the inclusion of the random intercept is necessary. In comparison, all of the posterior distributions of sd(SEX) and sd(PPED) go through zero, suggesting that there is probably no need to include the two random slopes in the model. To interpret the fixed-effect terms, we can calculate the exponentiated coefficient estimates. #the categorical variables: SEX and PPED ## Estimate Q2.5 Q97.5 ## Intercept 0.1010137 0.0846112 0.1205179 ## SEX 1.5655171 1.2513685 1.9344052 ## PPED 0.5473489 0.4244672 0.7135445 #the continous variable: MSESC exp(fixef(Bayes_Model_Multi_Full)[4,-2]sd(pull(ThaiEdu_Prop, MSESC), na.rm = T)) ## Estimate Q2.5 Q97.5 ## 0.8237781 0.6998895 0.9747519 We can see that the effects of SEX, PPED, and MSESC are very similar to the prevoius model results. Bürkner, P. (2017). brms: An R Package for Bayesian Multilevel Models Using Stan. Journal of Statistical Software, 80(1), 1-28. doi:10.18637/jss.v080.i01 Enders, C. K., & Tofighi, D. (2007). Centering predictor variables in cross-sectional multilevel models: A new look at an old issue. Psychological Methods, 12(2), 121-138. doi:10.1037/ Kay, M. (2019). tidybayes: Tidy Data and Geoms for Bayesian Models. doi:10.5281/zenodo.1308151, R package version 1.1.0, http://mjskay.github.io/tidybayes/. Lüdecke, D. (2019). sjstats: Statistical Functions for Regression Models (Version 0.17.5). doi: 10.5281/zenodo.1284472 Raudenbush, S. W., & Bhumirat, C. (1992). The distribution of resources for primary education and its consequences for educational achievement in Thailand. International Journal of Educational Research, 17(2), 143-164. doi:10.1016/0883-0355(92)90005-Q Sing, T., Sander, O., Beerenwinkel, N. & Lengauer, T. (2005). ROCR: visualizing classifier performance in R. Bioinformatics, 21(20), pp. 7881. http://rocr.bioinf.mpi-sb.mpg.de Wickham, H. (2017). tidyverse: Easily Install and Load the ‘Tidyverse’. R package version 1.2.1. https://CRAN.R-project.org/package=tidyverse	{"url":"https://www.rensvandeschoot.com/tutorials/generalised-linear-models-with-brms/","timestamp":"2024-11-14T21:17:18Z","content_type":"text/html","content_length":"142061","record_id":"<urn:uuid:96c9198e-4106-4fc2-a8c8-4d8081f9e622>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00398.warc.gz"}
Defining an operation such that ##1+2+3=123## • B • Thread starter Saracen Rue • Start date In summary, there is a need to define an operation that "joins" two numbers instead of adding them, which is called concatenation. This operation is often used in computer languages for strings or text but there is no standardized symbol for it in other fields. The suggested symbol is the plus '+' symbol, with quotes used to differentiate it from numerical addition. In some computer languages, the addition operator is used for concatenation, causing some debate among purists. In Python, there are multiple ways to concatenate strings and there are various examples of string concatenation in different programming languages. Firstly, I'm aware that title doesn't really make sense but stick with me on this. I'm trying to find a way to define an operation which will "join" two numbers instead of adding them. So for example, ##12+34=1234##. Ideally, it would be great if it also had something similar to sigma notation, like so: $$\sum_{k=1}^{n} k= 12345...n$$ I'm sure this is actually something trivial that has been defined before, but I was finding it really difficult to search it up on Google (apparently "addition but instead of adding numbers just clomp together" doesn't yield very good results). So yeah, if anyone can tell me a better way to name this sort of maths it'd be greatly appreciated. Science Advisor Homework Helper Gold Member 2023 Award This is called concatenation. You often have to program this kind of operation for strings or text. Science Advisor Homework Helper Gold Member 2023 Award In computer languages, this is often done but there is no standardized symbol that would be universally recognized in other fields. Since C++ is such a well-established language, it might be the best thing to mimic. It uses the plus '+' symbol. The trick would be to distinguish it from numerical addition. Consider using quotes to make it clear that you are treating the number as a text string: If this is to appear in a document, you should clearly define your notation and symbology in the document. Science Advisor Homework Helper I suppose you have [tex]\oplus : \mathbb{N}^2 \to \mathbb{N} : (a,b) \mapsto \begin{cases} 10^{1 + \lfloor\log_{10}(b)\rfloor}a + b, & b \neq 0, \\ 10a, & b = 0\end{cases}[/tex] which (if it does what I think it does) is associative but not commutative, and you need to decide whether [tex] \bigoplus_{n=0}^N a_n[/tex] means [itex]a_0 \oplus a_1 \oplus \cdots \oplus a_N[/itex] or [itex]a_N \oplus a_{N-1} \oplus \cdots \oplus a_0[/itex]. In some computer languages, the addition operator took on different meaning dependent on the datatypes of items being added. It acted as normal addition for numeric types and as a concatenation operator for string types. There were some purists who complained that since concatentation is not commutative that the addition operator shouldn't serve that function. However, using the addition operator for concatenation had become a defacto standard and was here to stay. In Python, one can concat in several different ways: - string1 + string2 - " ".join(string1, string2) - ... and here's a larger set of multi language string concatentation examples: FAQ: Defining an operation such that ##1+2+3=123## 1. What is the purpose of defining an operation such that ##1+2+3=123##? The purpose of defining an operation such that ##1+2+3=123## is to show an example of a non-standard operation that follows a different set of rules than traditional addition. This can help expand our understanding of mathematical operations and encourage creative thinking. 2. How is this operation different from traditional addition? This operation is different from traditional addition because it does not follow the standard rules of adding numbers. Instead of adding the numbers together, they are concatenated (or combined) to form a single number. This operation also does not follow the commutative property, meaning the order of the numbers matters. 3. Is this operation valid in the field of mathematics? Yes, this operation can be considered valid in the field of mathematics. While it may not follow the traditional rules of addition, it is still a valid mathematical operation that can be used to solve problems and explore new concepts. 4. Can this operation be used in real-life situations? While this operation may not have direct applications in real-life situations, it can be a useful tool for understanding mathematical concepts and promoting critical thinking skills. It can also be used as a fun and creative way to approach math problems. 5. Are there any other examples of non-standard operations? Yes, there are many other examples of non-standard operations in mathematics. Some examples include the bitwise XOR operation in computer science, the quaternion multiplication operation in physics, and the hyperoperation sequence in abstract algebra. These operations may not follow the traditional rules of addition, but they still have important applications in their respective fields.	{"url":"https://www.physicsforums.com/threads/defining-an-operation-such-that-1-2-3-123.1046948/","timestamp":"2024-11-11T13:02:45Z","content_type":"text/html","content_length":"100537","record_id":"<urn:uuid:e8153c9f-7477-4ec1-906f-eff21b268b35>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00097.warc.gz"}
Electrical Circuit Calculator FAQ The program is designed to calculate steady-state modes of electrical circuits according to the Ohm law and Kirchhoff’s rules. The program allows you to draw a circuit, set the parameters of its elements and calculate the circuit. As a result, a textual description of the calculation procedure is formed. Circuit drawing Drawing a diagram is performed by dragging elements using drag-and-drop from the sidebar and connecting the selected elements. The following elements with configurable parameters are available in the • resistor □ element number; □ resistance, Ohm; • capacitor □ element number; □ impedance, Ohm; • inductor □ element number; □ impedance, Ohm; • voltage source □ element number; □ peak value, V; □ initial phase, °; • current source □ element number; □ peak value, A; □ initial phase, °. When you hover the mouse pointer over an element, the connection points of the element with other elements (Fig. 1) and a buttons for deleting and rotating the element (Fig. 2) are displayed. Fig. 1. Element connection points Fig. 2. Element rotation and deletion buttons To connect one element to another, you need to move the mouse pointer over the connection point of the element, press the left mouse button and connect it to another element (Fig. 3) by clicking the left mouse button on the connection point of another element. Fig. 3. Elements connection Nodes are formed automatically when you connect an element to another connector line. Fig. 4. Node forming When you click on an element a window with the element’s parameters is formed on the right side of the screen. Parameters that are available for editing (Fig. 5). Fig. 5. Element parameters editing Limitations when drawing a diagram For the correct analysis of the circuit, the connecting line must be connected on both sides to the elements / connecting lines, otherwise the program will not calculate the circuit, which it will signal with the appropriate notification. Saving a circuit as a file and loading a schematic from a file On the sidebar there are a button Setting parameters For capacitors and inductors their impedances are set. If for capacitor there is information about it capacitance, then that impedance is calculated by formula $$ X_{C} = \frac{1}{2 \pi f C}, $$ where f − frequency, C − capacitance. If for inductor there is information about it inductance, then that impedance is calculated by formula $$ X_{C} = 2 \pi f L, $$ where f − frequency, L − inductance. Setting the parameters of voltage and current sources are set in the form of their module and phase. For example, if in the original data $$ \underline{E} = 3 + 4j, $$ then in order to set this value in the program, it must be converted to polar form. We get: $$ \underline{E} = 5 \angle 53.13 \degree $$ Thus, in the “Peak value” field we set the value 5 and in the “Initial phase” field set the value 53.13. Calculation methods After completion of drawing the circuit, pressing the “Calc.” button starts the calculation of the electrical circuit. The program analyzes the original circuit and reports any errors if any errors are found. In case of successful analysis of the circuit, the calculation is started. It should be noted that if the calculated circuit is single-circuit, then the calculation will be performed according to Ohm’s law. After completing the calculation, the program checks the power balance and build current and voltage vector diagrams. Ohm’s law calculation Input data: • E1: □ Element number: 1 □ Peak value: 100 □ Initial phase, °: 0 • R1: □ Element number: 1 □ Resistance, Ohm: 1 After pressing the “Calc.” button, a solution is formed: There is only one loop in the original circuit. Let’s calculate it according to Ohm’s law. According to Ohm’s law, the current in a closed circuit is equal to the ratio of the voltage source of the circuit to the resistance. Let us compose an equation, taking for the positive direction of the current $ I $ the direction of the voltage source $ E_{1} $: $$ R_{1}\cdot I = E_{1} $$ Put values of resistances and sources into the resulting system of equations and get: $$ 1.0\cdot I=100 $$ Hence, the sought current in the circuit is $$ I = 100\space \textrm{A} $$ Check the power balance. Define the power consumed by the receivers: $$ S_\textrm{c} = R_{1}⋅\|I_{1}\|^{2}. $$ Put the numerical values and get: $$ S_\textrm{c} = 1⋅100^{2}=10000. $$ Define the power given by the sources: $$ S_\textrm{s} = S_{E} + S_{J}, $$ where $ S_{E} $ – power given by voltage sources, $ S_{J} $ – power given by current sources. Define the power $ S_{E} $, given by voltage sources: $$ S_{E} =E_{1}⋅ I’_{1} , $$ where $ I’ $ means conjugate complex current. Put the numerical values and get: $$ S_\textrm{E} = 100⋅100=10000. $$ Because there are no current sources in the circuit, then $$ S_{J} = 0. $$ The power given by all sources is equal to: $$ S_\textrm{s} = S_{E} + S_{J} =10000+0=10000. $$ Thus, $ S_\textrm{c} = 10000 $, $ S_\textrm{s} = 10000 $. The power balance is observed. Kirchhoff’s rules calculation Input data • E1: □ Element number: 1 □ Peak value: 100 □ Initial phase, °: 0 • R1: □ Element number: 1 □ Resistance, Ohm: 1 • L1: □ Element number: 1 □ Impedance, Ohm: 1 • C1: □ Element number: 1 □ Impedance, Ohm: 1 After pressing the “Calc.” button, the accepted designations of the nodes and the accepted directions of currents are displayed on the original circuit, and a solution is formed: Calculate the circuit according to Kirchhoff’s rules. Circuit parameters: nodes quantity – 2, branches quantity – 3, independent loops quantity – 2. Arbitrarily set the directions of currents in branches and direction of bypassing loops. Accepted directions of currents: Current $ \underline{I}_{1} $ directed away from node ‘2 n.’ to node ‘1 n.’ through elements $ \underline{E}_{1} $, $ R_{1} $. Current $ \underline{I}_{2} $ directed away from node ‘1 n.’ to node ‘2 n.’ through elements $ L_{1} $. Current $ \underline{I}_{3} $ directed away from node ‘1 n.’ to node ‘2 n.’ through elements $ C_{1} $. Accepted directions for bypassing loops: Loop №1 bypasses through elements $ \underline{E}_{1} $, $ R_{1} $, $ L_{1} $ in the specified order. Loop №2 bypasses through elements $ L_{1} $, $ C_{1} $ in the specified order. Compose equations according to current Kirchhoff’s law. When drawing up the equations, the currents flowing into the node will be taken with the “+” sign, and the “outgoing” ones – with the “-” sign. Number of equations compiled according to current Kirchhoff’s law is $ N_\textrm{n}- 1 $, where $ N_\textrm{n} $ is the number of nodes. For this scheme, the number of equations according to the current Kirchhoff’s law is 2 – 1 = 1. Define an equation for node №1: $$ \underline{I}_{1}- \underline{I}_{2}- \underline{I}_{3} = 0 $$ Compose the equations according to voltage Kirchhoff’s law. When drawing up equations, positive values for currents and voltage sources are selected if they coincide with the direction of the loop The number of equations compiled according to voltage Kirchhoff’s law is $ N_\textrm{b}- N_\textrm{n} + 1 $, where $ N_\textrm{b} $ is the number of branches without current sources. For this scheme, the number of equations according to the voltage Kirchhoff’s law is 3 – 2 + 1 = 2. Define an equation for loop №1: $$ R_{1}\cdot \underline{I}_{1}+jX_{L1}\cdot \underline{I}_{2}=\underline{E}_{1} $$ Define an equation for loop №2: $$ jX_{L1}\cdot \underline{I}_{2}-(-jX_{C1})\cdot \underline{I}_{3}=0 $$ Combine equations into one system, while transferring the known quantities to the right side, leaving only the components with the sought currents in the left side. The system of equations according to Kirchhoff’s laws for the original circuit is as follows: $$ \begin{cases}\underline{I}_{1}- \underline{I}_{2}- \underline{I}_{3} = 0 \\ R_{1}\cdot \underline{I}_{1}+jX_{L1}\cdot \underline{I}_{2} = \underline{E}_{1} \\ jX_{L1}\cdot \underline{I}_{2}-(-jX_ {C1})\cdot \underline{I}_{3} = 0 \\ \end{cases} $$ Put the values of impedance and sources into the resulting system of equations and get: $$ \begin{cases}\underline{I}_{1}- \underline{I}_{2}- \underline{I}_{3}=0 \\ \underline{I}_{1}+ j \cdot \underline{I}_{2}=100 \\ j \cdot \underline{I}_{2}+ j \cdot \underline{I}_{3}=0 \\ \end{cases} Further solve the system of equations and get the required currents: $$ \underline{I}_{1} = 0\space\textrm{A} $$ $$ \underline{I}_{2} =-100j\space\textrm{A} $$ $$ \underline{I}_{3} = 100j\space\textrm{A} $$ Check the power balance. Define the power consumed by the receivers: $$ \underline{S}_\textrm{c} = R_{1}⋅\|\underline{I}_{1}\|^{2}+jX_{L1}⋅\|\underline{I}_{2}\|^{2}- jX_{C1}⋅\|\underline{I}_{3}\|^{2}. $$ Put the numerical values and get: $$ \underline{S}_\textrm{c} = 1⋅0^2+1j⋅((-100)^{2})-1j⋅(100^{2})=0. $$ Define the power given by the sources: $$ \underline{S}_\textrm{s} = \underline{S}_{\underline{E}} + \underline{S}_{\underline{J}}, $$ where $ \underline{S}_{\underline{E}} $ - power given by voltage sources, $ \underline{S}_{\underline{J}} $ – power given by current sources. Define the power $ \underline{S}_{\underline{E}} $, given by voltage sources: $$ \underline{S}_{\underline{E}} =\underline{E}_{1}⋅ \underline{I}’_{1} , $$ where $ \underline{I}’ $ means conjugate complex current. Put the numerical values and get: $$ \underline{S}_\textrm{\underline{E}} = 100⋅0=0. $$ Because there are no current sources in the circuit, then $$ \underline{S}_{\underline{J}} = 0. $$ The power given by all sources is equal to: $$ \underline{S}_\textrm{s} = \underline{S}_{\underline{E}} + \underline{S}_{\underline{J}} =0+0=0. $$ Thus, $ \underline{S}_\textrm{c} = 0 $, $ \underline{S}_\textrm{s} = 0 $. The power balance is observed. If it is impossible to calculate the scheme, please inform the Site Administration by e-mail support@faultan.ru.	{"url":"https://faultan.ru/en/circuit_faq_en/","timestamp":"2024-11-02T05:40:54Z","content_type":"text/html","content_length":"91928","record_id":"<urn:uuid:ccc81076-833d-4608-9eca-6be8c7407f63>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00586.warc.gz"}
Scalable electron correlation methods I. PNO-LMP2 with linear scaling in the molecular size and near-inverse-linear scaling in the number of processors We propose to construct electron correlation methods that are scalable in both molecule size and aggregated parallel computational power, in the sense that the total elapsed time of a calculation becomes nearly independent of the molecular size when the number of processors grows linearly with the molecular size. This is shown to be possible by exploiting a combination of local approximations and parallel algorithms. The concept is demonstrated with a linear scaling pair natural orbital local second-order Møller-Plesset perturbation theory (PNO-LMP2) method. In this method, both the wave function manifold and the integrals are transformed incrementally from projected atomic orbitals (PAOs) first to orbital-specific virtuals (OSVs) and finally to pair natural orbitals (PNOs), which allow for minimum domain sizes and fine-grained accuracy control using very few parameters. A parallel algorithm design is discussed, which is efficient for both small and large molecules, and numbers of processors, although true inverse-linear scaling with compute power is not yet reached in all cases. Initial applications to reactions involving large molecules reveal surprisingly large effects of dispersion energy contributions as well as large intramolecular basis set superposition errors in canonical MP2 calculations. In order to account for the dispersion effects, the usual selection of PNOs on the basis of natural occupation numbers turns out to be insufficient, and a new energy-based criterion is proposed. If explicitly correlated (F12) terms are included, fast convergence to the MP2 complete basis set (CBS) limit is achieved. For the studied reactions, the PNO-LMP2-F12 results deviate from the canonical MP2/CBS and MP2-F12 values by <1 kJ mol^-1, using triple-Χ (VTZ-F12) basis sets. All Science Journal Classification (ASJC) codes • Computer Science Applications • Physical and Theoretical Chemistry Dive into the research topics of 'Scalable electron correlation methods I. PNO-LMP2 with linear scaling in the molecular size and near-inverse-linear scaling in the number of processors'. Together they form a unique fingerprint.	{"url":"https://pure.psu.edu/en/publications/scalable-electron-correlation-methods-i-pno-lmp2-with-linear-scal","timestamp":"2024-11-11T14:41:13Z","content_type":"text/html","content_length":"54529","record_id":"<urn:uuid:063d304a-c01f-41b5-8480-65bcc4669a10>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00004.warc.gz"}
Variable in Cross Sheet Reference Hi All I have created 5 different cross sheet references, that refer to a range in 5 different sheets, these are named: On a separate sheet I have a formula in one column [Invoice Year] that returns 1,2,3,4,5 based on what year of the project the invoice falls in The formula is =VLOOKUP([PO Number]@row, {"Year"+[Invoice Year]@row}, 4, false) The formula doesn't work, and I'm not sure if it's because I have done something wrong, or variables are not allowed in cross sheet references. If I dont use the variable e.g =VLOOKUP([PO Number]@row, {Year1}, 4, false) then it works perfectly Would appreciate any assistance you can give • You cannot use variables in cross sheet references. You would need to use a nested IF statement. =VLOOKUP([PO Number]@row, IF([Invoice Year]@row = 1, {Year1}, IF([Invoice Year]@row = 2, {Year2}, ...........))))), 4, false) Help Article Resources	{"url":"https://community.smartsheet.com/discussion/100967/variable-in-cross-sheet-reference","timestamp":"2024-11-10T04:23:38Z","content_type":"text/html","content_length":"392359","record_id":"<urn:uuid:486655e8-3ba1-48d7-aa57-f93be5f04138>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00085.warc.gz"}
Numeric Data Types \| SQL Numeric Data Types in SQL Numeric data types in SQL are used to store numerical values, and they come in various forms depending on the nature of the data you need to store. The three main types of numeric data types in SQL are INTEGER, DECIMAL, and FLOAT. Each serves a different purpose based on the required precision and range. 1. INTEGER Data Type Definition: The INTEGER (often abbreviated as INT) data type is used to store whole numbers, meaning numbers without any fractional or decimal parts. This type is ideal for situations where you need to store data such as counts, identifiers, or any values that do not require decimal precision. Range: The range of values for an INTEGER depends on whether the column is defined as signed or unsigned. Signed INTEGER: The typical range is from -2,147,483,648 to 2,147,483,647. Unsigned INTEGER: The range is from 0 to 4,294,967,295. Storage: The INTEGER data type generally requires 4 bytes of storage. Use Cases: INTEGER is commonly used for primary keys, counters, and any data that represents whole numbers, such as the number of items in stock or user IDs. 2. DECIMAL Data Type Definition: The DECIMAL data type is used to store exact numeric values with fixed decimal places. This data type is crucial for situations where precision is essential, such as financial calculations, where you need to avoid rounding errors that can occur with floating-point numbers. Precision and Scale: The DECIMAL type is defined with two parameters: Precision (p): The total number of digits that can be stored (both before and after the decimal point). Scale (s): The number of digits to the right of the decimal point. For example, DECIMAL(10, 2) can store numbers with up to 8 digits before the decimal point and 2 digits after the decimal point. Storage: The storage requirement for DECIMAL depends on the specified precision. Typically, more precision means more storage. Use Cases: DECIMAL is ideal for storing financial data such as prices, salaries, or monetary values where exact precision is needed. 3. FLOAT Data Type Definition: The FLOAT data type is used to store approximate numeric values with floating-point precision. Unlike DECIMAL, which stores exact values, FLOAT stores numbers in a way that allows for a wide range of values, but with variable precision. This is useful in scientific calculations or scenarios where the exact value isn't as important as the ability to handle very large or very small numbers. Precision: The precision of a FLOAT value is not fixed, and it is stored in binary format, which can lead to small rounding errors. However, it allows for the representation of very large and very small numbers. Storage: FLOAT typically requires 4 bytes for single-precision and 8 bytes for double-precision. Use Cases: FLOAT is best used in scientific or engineering calculations, where the range of values can vary greatly and exact precision is less critical. Key Differences Between INTEGER, DECIMAL, and FLOAT INTEGER: Stores exact whole numbers without any decimal part. DECIMAL: Stores exact numeric values with fixed decimal places, ideal for financial calculations. FLOAT: Stores approximate numeric values with variable precision, suitable for scientific calculations. Use Cases: INTEGER: Best for counting, indexing, and whole numbers. DECIMAL: Best for financial data and calculations requiring exact precision. FLOAT: Best for scientific data and scenarios where large ranges of values are needed. INTEGER: Fixed storage size (typically 4 bytes). DECIMAL: Storage depends on the precision specified. FLOAT: Variable storage size, typically 4 or 8 bytes. Choosing the appropriate numeric data type in SQL depends on the nature of the data you're dealing with. INTEGER is suitable for whole numbers, DECIMAL is essential for precise calculations, especially in financial data, and FLOAT is ideal for scenarios where approximate values are acceptable. Understanding these differences helps ensure your database stores and processes numeric data efficiently and accurately. This explanation is brought to you by codes with pankaj.	{"url":"https://tutorial.codeswithpankaj.com/sql/sql-data-types/numeric-data-types","timestamp":"2024-11-10T15:28:50Z","content_type":"text/html","content_length":"285073","record_id":"<urn:uuid:1d80a371-ca4b-4204-9579-266303d951c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00106.warc.gz"}
Principles Of Quantum Mechanics By R Shankar - HUNT4EDU Principles Of Quantum Mechanics By R Shankar Here, We provided to Principles Of Quantum Mechanics By R Shankar. This textbook is written as a basic introduction to quantum physics to be used by the undergraduate students in physics, who are exposed to the present subject for the primary time. Providing a mild introduction to the topic fills the gap between the available books which give comprehensive coverage appropriate for postgraduate courses and therefore the ones on Modern Physics which provides a rather incomplete treatment of the topic leaving out many conceptual and mathematical details. Free download PDF Principles Of Quantum Mechanics By R Shankar. Chapter-end exercises and review questions, generally designed as per the examination pattern, serve to strengthen the fabric learned. Chapter-end summaries capture the key points discussed in the text. Free download PDF Principles Of Quantum Mechanics By R Shankar. 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Table of Content: • Mathematical Introduction • Review Of Classical Mechanics • All Is Not Well with Classical Mechanics • The Postulates – a General Discussion • Simple Problems In One Dimension • The Classical Limit • The Harmonic Oscillator • The Path Integral Formulation Of Quantum Theory • The Heisenberg Uncertainty Relations • Systems With N Degrees Of Freedom • Symmetries And Their Consequences • Rotational Invariance And Angular Momentum • The Hydrogen Atom • Spin • Addition Of Angular Momentum • Variational And WKB Methods • Time-Independent Perturbation Theory • Time-Dependent Perturbation Theory • Scattering Theory • The Dirac Equation • Path Integrals-2 It traces the development of the concepts and the basic interpretative postulates and reconciles them with ordinary real-life experiences. The development of wave mechanics, scattering theory including Eikonal approximation to the scattering amplitude, inelastic and double scattering phenomena are discussed in detail. 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A TextBook of Quantum Mechanics Quantum Mechanics Hand Written Note Fundamentals Of Quantum Mechanics By Ajit Kumar Quantum Mechanics Formula Sheet By Fiziks Institute Quantum Physics Hand Written Note Part 1 By Abhijeet Agarwal Quantum Physics Hand Written Note Part 2 By Abhijeet Agarwal Quantum Physics Hand Written Note Part 3 By Abhijeet Agarwal Leave a Comment	{"url":"https://hunt4edu.com/principles-of-quantum-mechanics-by-r-shankar/","timestamp":"2024-11-02T09:13:36Z","content_type":"text/html","content_length":"83285","record_id":"<urn:uuid:1a35ebfb-8d14-4424-81c1-aa1accc6ea08>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00567.warc.gz"}
4.2: Identifying Potential Predictors Last updated Page ID $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\\| #1 \\|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$ $ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$ $ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$ $ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vectorC}[1]{\textbf{#1}} $ $ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $ $ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $ $ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $ $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $ $\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\ evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\ newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y} $ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real} {\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec} [3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array} {r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\ wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\ newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var} {\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left\|#1\right\|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\ bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\ widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$ The first step in developing the multi-factor regression model is to identify all possible predictors that we could include in the model. To the novice model developer, it may seem that we should include all factors available in the data as predictors, because more information is likely to be better than not enough information. However, a good regression model explains the relationship between a system’s inputs and output as simply as possible. Thus, we should use the smallest number of predictors necessary to provide good predictions. Furthermore, using too many or redundant predictors builds the random noise in the data into the model. In this situation, we obtain an over-fitted model that is very good at predicting the outputs from the specific input data set used to train the model. It does not accurately model the overall system’s response, though, and it will not appropriately predict the system output for a broader range of inputs than those on which it was trained. Redundant or unnecessary predictors also can lead to numerical instabilities when computing the coefficients. We must find a balance between including too few and too many predictors. A model with too few predictors can produce biased predictions. On the other hand, adding more predictors to the model will always cause the R^2 value to increase. This can confuse you into thinking that the additional predictors generated a better model. In some cases, adding a predictor will improve the model, so the increase in the R^2 value makes sense. In some cases, however, the R^2 value increases simply because we’ve better modeled the random noise. The adjusted R^2 attempts to compensate for the regular R^2’s behavior by changing the R^2 value according to the number of predictors in the model. This adjustment helps us determine whether adding a predictor improves the fit of the model, or whether it is simply modeling the noise better. It is computed as: $\ R_{adjusted}^2 = {1-\frac{n-1}{n-m}(1-R^2)}$ where n is the number of observations and m is the number of predictors in the model. If adding a new predictor to the model increases the previous model’s R^2 value by more than we would expect from random fluctuations, then the adjusted R^2 will increase. Conversely, it will decrease if removing a predictor decreases the R^2 by more than we would expect due to random variations. Recall that the goal is to use as few predictors as possible, while still producing a model that explains the data well. Because we do not know a priori which input parameters will be useful predictors, it seems reasonable to start with all of the columns available in the measured data as the set of potential predictors. We listed all of the column names in Table 2.1. Before we throw all these columns into the modeling process, though, we need to step back and consider what we know about the underlying system, to help us find any parameters that we should obviously exclude from the start. There are two output columns: perf and nperf. The regression model can have only one output, however, so we must choose only one column to use in our model development process. As discussed in Section 4.1, nperf is a linear transformation of perf that shifts the output range to be between 0 and 100. This range is useful for quickly obtaining a sense of future predictions’ quality, so we decide to use nperf as our model’s output and ignore the perf column. Almost all the remaining possible predictors appear potentially useful in our model, so we keep them available as potential predictors for now. The only exception is TDP. The name of this factor, thermal design power, does not clearly indicate whether this could be a useful predictor in our model, so we must do a little additional research to understand it better. We discover [10] that thermal design power is “the average amount of power in watts that a cooling system must dissipate. Also called the ‘thermal guideline’ or ‘thermal design point,’ the TDP is provided by the chip manufacturer to the system vendor, who is expected to build a case that accommodates the chip’s thermal requirements.” From this definition, we conclude that TDP is not really a parameter that will directly affect performance. Rather, it is a specification provided by the processor’s manufacturer to ensure that the system designer includes adequate cooling capability in the final product. Thus, we decide not to include TDP as a potential predictor in the regression model. In addition to excluding some apparently unhelpful factors (such as TDP) at the beginning of the model development process, we also should consider whether we should include any additional parameters. For example, the terms in a regression model add linearly to produce the predicted output. However, the individual terms themselves can be nonlinear, such as a[i]x[i]^m ,where m does not have to be equal to one.This flexibility lets us include additional powers of the individual factors. We should include these non-linear terms, though, only if we have some physical reason to suspect that the output could be a nonlinear function of a particular input. For example, we know from our prior experience modeling processor performance that empirical studies have suggested that cache miss rates are roughly proportional to the square root of the cache size [5]. Consequently, we will include terms for the square root (m = 1/2) of each cache size as possible predictors. We must also include first-degree terms (m = 1) of each cache size as possible predictors. Finally, we notice that only a few of the entries in the int00.dat data frame include values for the L3 cache, so we decide to exclude the L3 cache size as a potential predictor. Exploiting this type of domain-specific knowledge when selecting predictors ultimately can help produce better models than blindly applying the model development process. The final list of potential predictors that we will make available for the model development process is shown in Table 4.1. Table 4.1: The list of potential predictors to be used in the model development process.	{"url":"https://stats.libretexts.org/Bookshelves/Computing_and_Modeling/Book%3A_Linear_Regression_Using_R_-_An_Introduction_to_Data_Modeling_(Lilja)/04%3A_Multi-factor_Regression/4.02%3A_Identifying_Potential_Predictors","timestamp":"2024-11-08T05:45:31Z","content_type":"text/html","content_length":"131001","record_id":"<urn:uuid:899aebdc-488f-4551-9925-f9e7f2ba97bd>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00051.warc.gz"}
Instant mode - Openbox Instant mode is probably the easiest way for people coming from spreadsheet modelling to get straight into producing models in Openbox. It allows you to edit in a spreadsheet, but using the power of Openbox to go faster. Start with any model in Openbox. Press “Preview” (or the F9) key to display the spreadsheet preview, like so: You can add calculations to the preview, in a very similar way to you would in Excel. Step 1: Type the name of the new calculation in the name column. This is usually column E. Step 2: click in any other column and press Shift+F2. Notice that the Openbox formula bar has become active. Step 3: Type what you want the formula to be, using Openbox language. In this example, “Accounts receivable” minus “Accounts payable”. Create or change a calculation gives more detail. Openbox has taken the formula you typed and inserted the corresponding spreadsheet formulas, titles and headings. It has brought in the two ingredients, accounts receivable, accounts payable and added the “Net working capital” calculation which you would have typed in Excel as J130 – J131. It has also copied across of course. There is also a “Net working capital” item in the Openbox main window. You can do this anywhere in the preview. In the same way as usual, if you include new calculation names in the formula, Openbox will offer to add placeholders for them. You can also edit existing items. Suppose you wanted to change the “Net working capital” formula. Click in any cell in row 132 and press Shift+F2, then type a new formula and press Enter.	{"url":"https://openboxmodels.com/documentation/instant-mode/?seq_no=2","timestamp":"2024-11-07T03:30:09Z","content_type":"text/html","content_length":"99462","record_id":"<urn:uuid:32b7dfb0-2832-4e2e-929a-87a953286e00>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00847.warc.gz"}
Surface Area Calculate the surface areas of the given basic solid shapes using standard formulae. This is level 1; Find the surface area of shapes made up of cubes. The diagrams are not to scale. Each of the yellow cubes in the diagram have edges 1cm long. What is the total surface area of the cuboid they are part of? ☐ cm^2 ☐ &check; &cross; Each of the yellow cubes in the diagram have edges 1cm long. What is the total surface area of the shape they are part of? ☐ cm^2 ☐ &check; &cross; Each of the yellow cubes in the diagram have edges 1cm long. What is the total surface area of the shape they are part of? ☐ cm^2 ☐ &check; &cross; Each of the yellow cubes in the diagram have edges 1cm long. What is the total surface area of the shape they are part of? ☐ cm^2 ☐ &check; &cross; Each of the cubes in the diagram have edges 2cm long. What is the total surface area of the shape they are part of? ☐ cm^2 ☐ &check; &cross; Each of the cubes in the diagram have edges 3cm long. What is the total surface area of the shape they are part of? ☐ cm^2 ☐ &check; &cross; © Transum Mathematics 1997-2024 Scan the QR code below to visit the online version of this activity. Description of Levels Level 1 - Find the surface area of shapes made up of cubes. Level 2 - Find the surface area of a variety of cuboids. Level 3 - Find the surface area of a variety of prisms. Level 4 - Find the surface area of a variety of cylinders. Level 5 - Find the surface area of a variety of cones. Level 6 - Find the surface area of a variety of pyramids. Level 7 - Find the surface area of a variety of spheres. Level 8 - Find the surface area of composite shapes. Level 9 - Mixed, more challenging questions involving surface area. Volume - Find the volume of basic solid shapes. Surface Area = Volume - Can you find the ten cuboids that have numerically equal volumes and surface areas? A challenge in using technology. Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers). More on 3D Shapes including lesson Starters, visual aids, investigations and self-marking exercises. Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent. Help Video Surface Area Formulae Cube: $6s^2$ where $s$ is the length of one edge. Cuboid: $2(lw + lh + wh)$ where $l$ is the length, $w$ is the width and $h$ is the height of the cuboid. Cylinder: $2\pi rh + 2\pi r^2$ where $h$ is the height (or length) of the cylinder and $r$ is the radius of the circular end. Cone: $\pi r(r+l)$ where $l$ is the distance from the apex to the rim of the circle (sloping height) of the cone and $r$ is the radius of the circular base. Cone: $\pi r(r+\sqrt{h^2+r^2})$ where $h$ is the height of the cone and $r$ is the radius of the circular base. Square based pyramid: $s^2+2s\sqrt{\frac{s^2}{4}+h^2}$ where $h$ is the height of the pyramid and $s$ is the length of a side of the square base. Rectangular based pyramid: $lw+l\sqrt{\frac{w^2}{4}+h^2}+w\sqrt{\frac{l^2}{4}+h^2}$ where $h$ is the height of the pyramid, $l$ is the length of the base and $w$ is the width of the base. Sphere: $4\pi r^2$ where $r$ is the radius of the sphere. Prism: Double the area of the cross section added to the product of the length and the perimeter of the cross section. Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.	{"url":"https://www.transum.org/Software/SW/Starter_of_the_day/Students/Surface_Area.asp?Level=1","timestamp":"2024-11-13T04:55:49Z","content_type":"text/html","content_length":"44026","record_id":"<urn:uuid:769d4fbb-061e-4e5f-9226-fc8ba2714b14>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00017.warc.gz"}
n = 4.42 mol of Hydrogen gas is initially at T = 304.0 K temperature and... n = 4.42 mol of Hydrogen gas is initially at T = 304.0 K temperature and... n = 4.42 mol of Hydrogen gas is initially at T = 304.0 K temperature and p[i] = 3.23×10^5 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches p[f] = 8.93×10 ^5 Pa. What is the volume of the gas at the end of the compression process? How much work did the external force perform? How much heat did the gas emit? How much entropy did the gas emit? What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure? Please upvote if you have understood the solution. Thank you.	{"url":"https://justaaa.com/physics/78592-n-442-mol-of-hydrogen-gas-is-initially-at-t-3040","timestamp":"2024-11-05T21:33:40Z","content_type":"text/html","content_length":"41156","record_id":"<urn:uuid:f97d7462-2754-4860-9e3c-2db0ff060c61>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00868.warc.gz"}
Double Integrals over General Regions Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen Section 15.3 : Double Integrals over General Regions 1. Evaluate$ \displaystyle \iint\limits_{D}{{42{y^2} - 12x\,dA}}$ where $D = \left\{ {\left( {x,y} \right)\|0 \le x \le 4,{{\left( {x - 2} \right)}^2} \le y \le 6} \right\}$ Show All Steps Hide All Steps Start Solution Below is a quick sketch of the region $D$. In general, this sketch is often important to setting the integral up correctly. We’ll need to determine the order of integration and often the region will “force” a particular order. Many regions can only be dealt with easily by doing one particular order of integration and sometimes the only way to really see that is to have a sketch of $D$. Even if you can do the integral in either order the sketch of $D$ will often help with setting up the limits for the integrals. Show Step 2 With this problem we were pretty much given the order of integration by how the region $D$ was specified in the problem statement. Note however, that the sketch shows that this was pretty much the only easy order of integration. The same function is always on the top of the region and the same function is always on the bottom of the region and so it makes sense to integrate $y$ first. If we wanted to integrate $x$ first we’d have a messier integration to deal with. First the right/left functions change and so we couldn’t do the $x$ integration with a single integral. The $x$ integration would require two integrals in this case. There is also the fact that the lower portion of the region has the same function for both the right and left sides. The equation could be solved for $x$, as we’d need to in order to $x$ integration first, and often that is either very difficult or would give unpleasant limits. It wouldn’t be too difficult in this case but it would put roots into the limits and that often makes for messier integration. So, let’s go with the order of integration specified in the problem statement and because we were given $D$ in the set builder notation we also were given the limits for both $x$ and $y$ which is nice as we usually will need to figure those out on our own. Here is the integral set up for $y$ integration first. \[\iint\limits_{D}{{42{y^2} - 12x\,dA}} = \int_{0}^{4}{{\int_{{{{\left( {x - 2} \right)}^2}}}^{6}{{42{y^2} - 12x\,dy}}\,dx}}\] Show Step 3 Here is the $y$ integration. \[\begin{align}\iint\limits_{D}{{42{y^2} - 12x\,dA}} & = \int_{0}^{4}{{\int_{{{{\left( {x - 2} \right)}^2}}}^{6}{{42{y^2} - 12x\,dy}}\,dx}}\\ & = \int_{0}^{4}{{\left. {\left( {14{y^3} - 12xy} \ right)} \right\|_{{{\left( {x - 2} \right)}^2}}^6\,dx}}\\ & = \int_{0}^{4}{{3024 - 72x - 14{{\left( {x - 2} \right)}^6} + 12x{{\left( {x - 2} \right)}^2}\,dx}}\end{align}\] Show Step 4 Now, we did not do any real simplification of the integrand in the last step. There was a reason for that. After doing the first integration students will often just launch into a “simplification” mode and multiply everything out and “simplify” everything. Sometimes that does need to be done and we don’t want to give the impression it is never a good thing or never needs to be done. However, take a look at the third term above. It could be multiplied out if we wanted to but it would take a little bit of time and there is a chance we’d mess up a sign or coefficient somewhere. We are going to be integrating and the third term can be integrated very quickly with a simple Calculus I substitution. In other words, why bother with the messy multiplication with that term when it does not need to be done. The fourth term, on the other hand, does need to be multiplied out because of the extra $x$ that is in the front of the term. So, before just launching into “simplification” mode take a quick look at the integrand and see if there are any terms that can be done with a simple substitution as we won’t need to mess with those terms. Only multiply out terms that actually need to be multiplied out. Here is the $x$ integration work. We will leave the Algebra details to you to verify and we’ll also be leaving the Calculus I substitution work to you to verify. \[\begin{align}\iint\limits_{D}{{42{y^2} - 12x\,dA}} & = \int_{0}^{4}{{3024 - 72x - 14{{\left( {x - 2} \right)}^6} + 12x{{\left( {x - 2} \right)}^2}\,dx}}\\ & = \int_{0}^{4}{{3024 - 24x - 48{x^2} + 12{x^3} - 14{{\left( {x - 2} \right)}^6}\,dx}}\\ & = \left. {\left( {3024x - 12{x^2} - 16{x^3} + 3{x^4} - 2{{\left( {x - 2} \right)}^7}} \right)} \right\|_0^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{11136}}\end{align}\] Before leaving this problem let’s again note how much easier dealing with the third term was because we did not multiply it out and just used a substitution. Made this problem a lot easier. Also note that this problem illustrated an important point that needs to be made with many of these integrals. These integrals will often get very messy after the first integration. You need to be ready for that and expect it to happen on occasion. Just because they start off looking “easy” doesn’t mean that they will remain easy throughout the whole problem. Just because it becomes a mess doesn’t mean you’ve made a mistake, although that is unfortunately always a possible reason for a messy integral. It may just mean this is one of those integrals that get somewhat messy before they are done.	{"url":"https://tutorial.math.lamar.edu/Solutions/CalcIII/DIGeneralRegion/Prob1.aspx","timestamp":"2024-11-11T07:27:52Z","content_type":"text/html","content_length":"79573","record_id":"<urn:uuid:f8a4c4f2-7e92-4c3c-a822-101d916fb594>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00840.warc.gz"}
Normal approx to Student-t not good enough When the normal approximation for Student t isn’t good enough Folk wisdom says that for all practical purposes, a Student-t distribution with 30 or more degrees of freedom is a normal distribution. Well, not for all practical purposes. For 30 or more degrees of freedom, the error in approximating the PDF or CDF of a Student-t distribution with a normal is less than 0.005. So for many applications, the n > 30 rule of thumb is appropriate. (See these notes for details.) However, sometimes you need to look at the quantiles of a t distribution, such as when finding confidence intervals. For example, when computing confidence intervals, you don’t need to evaluate the CDF of a Student-t distribution per se but rather the inverse of such a CDF. And in that case, the error in the normal approximation may be larger than you’d expect. Say you’re computing a 95% confidence interval for the mean of a set of 31 data points. You first find t^* such that P(t > t^) = 0.025 where t is a Student-t random variable with 31 − 1 = 30 degrees of freedom. Your confidence interval is the sample mean +/− t^ s/√n where s is the sample standard deviation. For 30 degrees of freedom, t^* = 2.04. If you used the normal approximation, you’d get 1.96 instead of 2.04, a relative error of about 4% meaning the error in computing your confidence interval is about 4%. While the error in normal approximation to the CDF is less than 0.005 for n > 30, the error in the normal approximation to the CDF inverse is an order of magnitude greater. Also, the error increases as the confidence increases. For example, for a 99% confidence interval, the error is about 6.3%. It may be that none of this is a problem. If you only have 31 data points, there’s a fair amount of uncertainty in your estimate of the mean, and there’s no point in quantifying with great precision an estimate of how uncertain you are! Modeling assumptions are probably a larger source of error than the normal approximation to the Student-t. But as a numerical problem, it’s interesting that the approximation error may be larger than expected. For n = 300, the error in the normal approximation to t^* is about 0.4%. This means the error in the normal approximation to the inverse CDF is as good at n=300 as the normal approximation to the CDF itself is at n = 30. 2 thoughts on “When the normal approximation for Student t isn’t good enough” 1. I think it is also interesting / instructive to think of the slightly wider confidence intervals as the price you pay for having to estimate the variance as well as the mean, instead of just the I always thought it was hokey how introductory texts ease you into things by first (or second) posing the questions in terms of estimating the population mean from a sample when the population variance is known. Of course it is done this way for pedagogical reasons, but who has ever heard of such a situation, where the variance is known exactly but the mean is unknown?[1] If I recall correctly most of my students were more confused than enlightened by this strategy. But they were taking my class because they wanted to avoid as much math as possible. I think this helps students more if they are more conversant with math. [1] I have since then thought of at least one plausible situation. Suppose you are measuring some constant quantity (say, length or mass of some object) and the measurement device is known to have an absolute error distributed normally with zero mean and some specific, known variance. Then the results you would get with repeated measurments would have an unknown mean (the quantity of interest) but be normally distributed with known variance. 2. I agree, the one-sample z-test can be left out of a stat class. The t-test is only slightly more complicated. In fact, it may even be easier to teach the t-test first since students wouldn’t be distracted by wondering how you could possibly know the variance without knowing the mean. (Assuming they’re tracking well enough to be confused.) However, I think it’s worthwhile to teach the two-sample z-test. The difference of two normals is a normal; the difference between two t distributions is only approximately a t distribution, and there’s no simple way to say what the appropriate degrees of freedom are for the approximating t distribution. So in this case, the t-test is sufficiently complicated that it’s worthwhile to derive a z-test for a warm-up.	{"url":"https://www.johndcook.com/blog/2008/11/12/normal-approximation-for-student-t-distribution-isnt-good-enough/","timestamp":"2024-11-09T06:20:52Z","content_type":"text/html","content_length":"56616","record_id":"<urn:uuid:dec92df7-55cf-4f3a-a963-b24801b7e292>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00153.warc.gz"}
Find its volume and surface area. - WorkSheets Buddy Find its volume and surface area. The adjoining figure shows a victory stand, each face is rectangular. All measurement are in centimetres. Find its volume and surface area (the bottom of the stand is open). More Solutions: Leave a Comment	{"url":"https://www.worksheetsbuddy.com/find-its-volume-and-surface-area/","timestamp":"2024-11-11T04:31:11Z","content_type":"text/html","content_length":"141773","record_id":"<urn:uuid:f6a52438-cc02-43f0-b4ae-f03164c00ec1>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00260.warc.gz"}
Weighted Geometric Mean Weighted Geometric Mean Selected for SPECviewperf™ Composite Numbers by Bill Licea-Kane At its February 1995 meeting in Salt Lake City, a subcommittee within the SPECopc^SM project group was given the task of recommending a method for deriving a single composite metric for each viewset running under the SPECviewperf™ benchmark. Composite numbers had been discussed by the SPECopc group for more than a year. In May 1995, the SPECopc project group decided to adopt a weighted geometric mean as the single composite metric for each viewset. What is a Weighted Geometric Mean? Above is the formula for determining a weighted geometric mean, where "n" is the number of individual tests in a viewset, and "w" is the weight of each individual test, expressed as a number between 0.0 and 1.0. (A test with a weight of "10.0%" is a "w" of 0.10. Note the sum of the weights of the individual tests must equal 1.00.) The weighted geometric mean of CDRS-03, for example, is expressed by the following formula: wgm-cdrs-03 = (test[1]^0.50)(test[2]^0.20) (test[3]^0.15)(test[4]^0.08) (test[5]^0.05)(test[6]^0.02) (test[7]^0.00) The same formula for the weighted geometric mean as expressed in a Microsoft Excel expression: = (a1^0.05)(b1^0.20)(c1^0.15)(d1^0.08)(e1^0.05)(f1^0.02)(g1^0.00) Why the Weighted Geometric Mean? The SPECopc subcommittee that recommended a method for determining composite numbers started with the description for assigning weights that is provided to each creator of a viewset: "Assign a weight to each path based on the percentage of time in each path..." Given this description, the weighted geometric mean of each viewset is the correct composite metric. This composite metric is a derived quantity that is exactly as if you ran the viewset tests for 100 seconds, where test 1 was run for 100 × weight[1] seconds, test 2 for 100 × weight[2 ]seconds, and so on. The end result would be the number of frames rendered/total time which will equal frames/second. It also has the desirable property of "bigger is better"; that is, the higher the number, the better the performance. Why Not Weighted Harmonic Mean? Since the results of SPECviewperf are expressed as "frames/second," the subcommittee was asked why we did not choose the weighted harmonic mean. The weighted harmonic mean would have been the correct composite if the description published for SPECviewperf read as follows: "Assign a weight to each path based on the percentage of operations in each path..." Given this description, the weighted harmonic mean would be as if you ran the viewset tests for 100 frames, where 100 × weight[1] frames were drawn with test1, the next 100 × weight[2] frames were drawn by test2, and so on. The 100 frames divided by the total time would be the weighted harmonic mean. Since the weights for the viewsets were selected on percentage of time, not percentage of operations, we chose the weighted geometric mean over the weighted harmonic mean. What About Weighted Arithmetic Mean? The weighted arithmetic mean is correct for calculating grades at the end of a school term. It is not correct for the situation we face here. Consider for a moment a trivial example, where there are two tests, equally weighted in a viewset: │ │Test 1│Test 2│Weighted Arithmetic Mean │ │System A │ 1.0 │100.0 │ 50.5 │ │System B │ 1.1 │100.0 │ 50.55 │ │System C │ 1.0 │110.0 │ 55.5 │ System B is 10-percent faster at Test1 than System A. System C is 10-percent faster at Test2 than System A. But look at the weighted arithmetic means. System B's weighted arithmetic mean is only .1-percent higher than System A's, while System C's weighted arithmetic mean is 10-percent higher. Even normalization doesn't help here. Why Not Normalized Weighted Geometric Mean? Here the SPECopc project group departs company from the nearly universal practice in benchmarking of normalizing test results. SPECint92, PLBsurf93 and Xmark93, for example, are all normalized results based on a variety of "reference" systems. Since our weights were percentage of time and since the results from SPECviewperf are expressed in frames/sec, we were not obligated to normalize. Normalization introduces many issues of its own, starting with something as simple as how to select a reference system. We invite readers to select two different systems whose results are published in this newsletter and to use each one as the reference system. You will discover quickly that the normalized weighted geometric means change only in absolute magnitude. If the weighted geometric mean of System B is 10-percent higher than System A, for example, the normalized weighted geometric mean of System B will be 10-percent higher than System A, no matter what reference system you choose. Is There a Disadvantage to Weighted Geometric Mean? As with any composite, the weighted geometric mean can act as a "filter" for results; this introduces the danger that important information might be lost and inappropriate conclusions could be drawn. So, proper use of these composites is important. Use the composite as an additional piece of information. But also take a look at each individual test result in a viewset. Please don't rely exclusively on any synthetic benchmark such as SPECviewperf. In the end, isn't actual application performance on an actual computer system what you are really attempting to find? Bill Licea-Kane is responsible for graphics performance measurement within Digital Equipment Corp.'s Computer Systems Performance Group. He serves on all three GPC subcommittees. He can be reached by e-mail at bill.licea-kane@3dlabs.com.	{"url":"http://www.spec.org/gwpg/pastissues/Feb1_01/opc.static/geometri.htm","timestamp":"2024-11-01T21:58:31Z","content_type":"text/html","content_length":"7974","record_id":"<urn:uuid:f84e81da-90ff-43e2-9988-0bb45e664839>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00208.warc.gz"}
1: G ABSTRACT ALGEBRA 1: GROUP THEORY (SMT-274404) Course Description: In this upper level study, explore the theory and applications of the algebraic structures known as groups. Topics covered in this course include: an introduction to groups; the dihedral groups; homomorphisms and isomorphisms; subgroups and cyclic subgroups; group actions; permutations; cosets and Lagrange's Theorem, Cayley's Theorem; the Sylow Theorems and the Fundamental Theorem of Finitely Generated Abelian Groups. Following this thorough investigation of group theory, students will begin to explore the basic ideas of ring theory. The primary audience for this course is students who wish to concentrate in either mathematics or applied mathematics. Students interested in various fields which have a strong connection to this branch of mathematics (such as music theory, physics, chemistry, computer science, or the cognitive sciences) may also be interested in this course. Prior to enrolling in this course, students should be fluent in the foundations of mathematics and mathematical proof: logic, methods of proof (both inductive and deductive), sets, relations and functions. This knowledge may be obtained from a course such as Proof and Logic or Discrete Mathematics, for example. Students should also be familiar with matrices and determinants; this knowledge can be obtained from a course such as Linear Algebra. This online course is offered through Online Learning . You can take this as an individual course or as part of an online degree program , with term starts in March, May, September, November and January. View current term offerings all online courses. to register for online courses. Other Areas: The Arts \| Business, Management & Economics \| Community & Human Services \| Communications, Humanities & Cultural Studies \| Educational Studies \| Historical Studies \| Human Development \| Labor Studies \| Nursing \| Science, Math & Technology \| Social Science Liberal Study Upper Level Credits: 4 Term(s) Offered (Subject to Change) : Spring 1. Fall 1.	{"url":"https://www8.esc.edu/admin/esc/cdl/cdlcat.nsf/byid/3EE403A2F27EEA2E852578CE005FF01B?opendocument","timestamp":"2024-11-08T01:53:33Z","content_type":"text/html","content_length":"10250","record_id":"<urn:uuid:803b2b20-20e0-432e-8e20-10c9a849f646>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00498.warc.gz"}
What are prime numbers? 3 min read studies Prime numbers Its history goes back many centuries ago and this set of numbers has always brought a lot of curiosity to scholars. Knowing what a prime number is and how to determine it can make it easier to solve math questions in your proofs, so follow this article for a full explanation. What are prime numbers? on the principle, Prime numbers They are those that can only be divided by two factors: one (1) alone. Like, for example, the number two (2). You can just divide the number two (2) by the number one (1) or by itself, so it’s a Cousin’s number. But what about number one (1)? number one (1) Not It is considered a prime number because it is only divisible by itself. Remember that the rule for a prime number is that it is divisible by itself And by number 1. Other examples of Cousins: 3,5,7,11,13,17… How do you know if a number is prime or not? As mentioned above, the rule is that a number is divisible by one and by itself. So knowing the basic rules of divisibility can help you determine a large number. How about remembering division • divide by 2: All even numbers (ending in 0,2,4,6 and 8) are divisible by 2. • Divide by 3: If the sum of their algorithms gives a number that is divisible by 3, then that number is divisible by 3. • Divide by 4: A number is divisible by 4 if it is twice divisible by 2 or if the last two algorithms are divisible by 4. • Divide by 5: Every number ending in 0 and 5 is divisible by 5. • Divide by 6: If the number is even and divisible by 3, then the number is also divisible by 6. • Divide by 7: A number is divisible by 7 if the difference between the double of the last algorithm and the remainder is a multiple of 7. What are the prime numbers from 1 to 100? But how do you know all the files Prime numbers? How do you score the most important? In fact, recording the main numbers in memory can speed up the solving of questions, but if you do not remember, there is a practical way to find out Prime numbers 1 to 100. The system is called Eratosthenes screen It was created by a Greek mathematician many years ago. First, you have to write all the numbers, from 1 to 100, on a sheet of paper: Table 1 to 100 Now, you will go number by number in a very practical way. are you go: • With Exception number 2, which is already known to be a prime number, you are going to clip all even numbers. Remember that every even number is divisible by 2, so it is no longer a prime number. • Next, you must cut out all numbers divisible by 3 (except itself), according to the rule already described in the above section. • When the number is reached 5, you will also cut out all numbers that are divisible by 5 – note that because you cut out even numbers, most numbers that are divisible by 5 are already deleted. • Finally, remove the numbers that are divisible by 7. Note that prime numbers between 1 and 100 will remain, which are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 e 97. Prime numbers from 1 to 100 for the sieve of Eratosthenes ready! Now, you know an easy way to solve questions involving prime numbers. Be part of a channel cable To share the news in real time! You will never be left out of the news again, Click here and share. The PDF 2.0 version of the Direction + Qconcursos partnership is becoming increasingly desirable among current and future contestants. Totally innovative, it brings everything together in one place, integrates questions, solves doubts directly with teachers and is Just change the market. This tool is “Shortcut” towards her consent. So don’t miss your chance to become an unlimited member and have access to all these benefits. Click on the image below to find out more: “Entrepreneur. Music enthusiast. Lifelong communicator. General coffee aficionado. Internet scholar.”	{"url":"https://sivtelegram.media/what-are-prime-numbers/","timestamp":"2024-11-03T19:37:06Z","content_type":"text/html","content_length":"100288","record_id":"<urn:uuid:6ce00b3c-d6b1-4a75-abb5-db5d5b2f5997>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00581.warc.gz"}
How do you write an equation in slope-intercept form of the line that passes through (1,4); parallel to y=7x-3? \| Socratic How do you write an equation in slope-intercept form of the line that passes through (1,4); parallel to y=7x-3? 1 Answer You can find the equation using the general relationship for the equation of a line: $y - {y}_{0} = m \left(x - {x}_{0}\right)$ $m =$slope ${x}_{0} , {y}_{0}$ are the coordinates of your point. To be parallel the two line must share the same slope $m$. Your line has slope $m = 7$ (the coefficient of $x$) and you can use your coordinates to get: $y - 4 = 7 \left(x - 1\right)$ $y = 4 + 7 x - 7$ $y = 7 x - 3$ Impact of this question 2600 views around the world	{"url":"https://api-project-1022638073839.appspot.com/questions/how-do-you-write-an-equation-in-slope-intercept-form-of-the-line-that-passes-thr-1","timestamp":"2024-11-12T22:03:00Z","content_type":"text/html","content_length":"34176","record_id":"<urn:uuid:b54a8332-c9c5-4ef8-9cda-9280c633984f>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00067.warc.gz"}
Use parameters in the Modeler Parametric study It is possible to achieved parametric studies with a project described in the modeler like in standard geometry context of Flux, but the operation mode is radically different: • Without modeler, geometric parameters are applied directly on the coordinates of points • With the modeler, geometric parameters are used in geometric operations of the modeler (dimensions of a block, radius of a cylinder, distance of a chamfer, length of extrusion, …) Example of parameterized block Here is an example of a block whose dimensions are parameterized. Parametrisation strategy In the standard geometry context, to parameterize a movement it was customary to parameterize coordinates systems. With a geometric construction with the Modeler, you must abandon this method, and directly parameterize geometric operations. For example of the movement, just create and parameterize the geometric operations of translation and rotation.	{"url":"https://help.altair.com/flux/Flux/Help/english/UserGuide/English/topics/LeParametrageDansLeModeleur.htm","timestamp":"2024-11-06T18:14:48Z","content_type":"application/xhtml+xml","content_length":"53008","record_id":"<urn:uuid:1dcb3303-24f6-42fa-931e-e89496deb61a>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00116.warc.gz"}
Warrants and call options (options to buy) Warrants and call options (options to buy) are similar securities in many respects, but they also have a few significant differences. A warrant is a security that provides the holder with the right, but not the obligation, to purchase one ordinary share directly from the company at a fixed price over a predetermined period of time. Like warrants, a call option (option to buy) also provides its holder with the right, without obligation, to purchase one ordinary share at a fixed price over a predetermined period of time. So, what are the differences between these two trading instruments? The difference between warrants and call options • Issuer: warrants are issued by specific companies, while options exchanged in a market are issued by an options exchange such as the Chicago Board Options Exchange in the United States. Options are therefore more standardised in some aspects such as the expiration period and the number of shares per option contract (usually 100). • Maturity: warrants have maturity periods that are greater than those of options. Warrants typically expire after one to two years and can sometimes have maturities well beyond five years. Call options have maturities ranging from a few weeks or months to one or two years, and longer term options may not be very liquid. • Dilution: warrants cause dilution because a company is required to issue new shares when a warrant is exercised. Exercise of a call option does not involve the issue of new shares because a call option is a derivative instrument of an existing ordinary share of a company. Why are warrants and call options issued? Warrants are a sort of "extension" of a stock or a bond (debt issuance). Investors like warrants because they offer additional participation in a company's growth. Companies include warrants in shares or bonds in order to lower financing costs and to provide additional capital insurance in the event that the share price is favourable. In addition, investors are more likely to opt for a slightly lower interest rate on bond financing if a bond is backed by a warrant. Options traded on the stock market meet certain criteria, such as share price, the number of shares outstanding or the distribution of the average daily volume. Stock options facilitate hedging and speculation for investors and traders. The basic attributes of a warrant and a call option are the same: • Strike price: the price at which the buyer of a warrant or a call option has the right to buy the underlying asset. "Strike price" is the preferred term to use when referring to warrants. • Expiration: the limited time period during which the warrant or option may be exercised. • Price of the option or premium: the price to be paid in order to acquire a warrant or an option. For example, let's consider a warrant with an exercise price of $5 a share that is trading at $4 a share. The warrant expires in 1 year and is currently priced at 50 cents. If the underlying share trades at more than $5 during the 1-year expiration, the price of the warrant will increase accordingly. Let's imagine that just before the warrant's 1-year expiration, the underlying share's price rises to $7. The warrant will then have a value of at least $2 (the difference between the price of the share and the warrant's exercise price). And conversely, if the price of the underlying share drops below $5 just before the warrant expires, the warrant will have very little value. A trade with a call option is very similar. A call option that expires in one month with an exercise price of $12.50 a share that is trading at $12 will see its price fluctuate along with the underlying share. If the stock is trading at $13.50 just before the option expires, the call will be worth at least $1. Conversely, if the stock is trading at or below $12.50 when the call option expires, the option will expire with no value and the investor will have lost the premium paid to purchase the option. Intrinsic value and time value The same variables influence the price (premium) needed to buy a call option or a warrant, but other additional factors can affect the price of a warrant. First, let us explore the two basic components of the value of a warrant and an option - the intrinsic value and the time value. The intrinsic value of a warrant or a call option is the difference between the price of the underlying security and the exercise price or strike price. Intrinsic value may be zero, but it can never be negative. For example, if an underlying share is trading at $10 a share and the exercise price of a call option is $8, the intrinsic value of the option is $2. If the stock is trading at $7 a share, the intrinsic value of the option is zero. (The price of the underlying security - strike price = intrinsic value.) The time value is the difference between the price of a warrant or a call option and its intrinsic value. For the above example of a trade with a share at $10 and an exercise price of $7, if the option price is $2.50 and the intrinsic value is $2, then the time value is equal to 50 cents. The price of an option with no intrinsic value entirely consists of its time value. The time value represents the ability to trade a share above the strike price upon the expiration of the option. (The price of option's "premium" - intrinsic value = time value.) The price or the premium The factors that influence the price of a call option or a warrant are: • Price of the underlying share: the price of an option or a warrant increases when the price of the underlying share increases. • Exercise price or strike price: the lower the strike price is, the higher the price of the call option or warrant will be. Why? Because the investor pays more for the right to buy an asset at a price that is lower than the price of the underlying asset. • Time until expiration: the price is highest when it is furthest away from the expiration date. • Implied volatility: the price increases when volatility is high, because the option has a greater probability of being profitable if the underlying asset is more volatile. • Risk-free interest rate: the higher the interest rate, the higher the price of the warrant or option will be. The Black-Scholes model is the one most often used to price options, while a modified version of this model is used to price warrants. Using a calculator, the values of these variables can be used to obtain the price of an option. As the other variables are more or less fixed, the estimate of implied volatility becomes the most important variable in the pricing of an option. The price of a warrant is slightly different because it must take into account the dilution mentioned above and its "gearing". Gearing is the ratio between the share price and the price of a warrant; it represents the leverage effect that a warrant offers. The value of a warrant is directly proportional to its gearing. The function of dilution makes a warrant slightly less expensive than an identical call option, by a factor of (n / n+w), where "n" is the number of shares outstanding and "w" is the number of warrants. For example, for 1,000,000 shares and 100,000 warrants outstanding, if a call option on that stock is trading at $1, the same warrant (with the same maturity date and strike price) will be worth around 91 cents. Examples of trades with warrants or call options The biggest advantage of using warrants and call options is that these trading instruments offer unlimited earning potential, while minimizing the possible loss of the amount invested. The other major advantage is their leverage. Their main disadvantages are that unlike the underlying share, they have a limited life and are not eligible for dividend payments. Consider an investor who has a high tolerance for risk and $2,000 to invest. The investor has the choice between investing in a share worth $4, or investing in a warrant on the same share with an exercise price of $5. The warrant expires in 1 year and is currently priced at 50 cents. The investor is very optimistic about the share and in order to profit the most from its increase in price, he decides to only invest in warrants. He therefore buys 4,000 warrants (4,000 x $.50 = $2,000) on the share. If the share appreciates to $7 after approximately 1 year (meaning, right before the warrants expire), each warrant will have a value of $2, a total of $8,000, representing a gain of $6,000 or +300% compared with the initial $2,000 investment. If the investor had instead chosen to invest directly in the share, the return on investment would have been only $1,500 or +75% compared with the initial investment. Of course, if the share had closed at $4.50 right before the expiration of the warrants, the investor would have lost 100% of his initial $2,000 investment in the warrants, as opposed to a gain of +12.5% had he invested in the share instead. Warrants are very popular in some markets such as Canada and Hong Kong. In Canada, for example, it is common practice for natural resource companies that seek funding for exploration to do so through the sale of units. Each unit consists of one common share that is delivered with a half warrant, which means that two warrants are required to purchase 1 additional common share. (Note that several warrants are often required in order to purchase 1 share at the exercise price.) These companies also offer "broker warrants" to their subscribers, in addition to cash commissions, as part of the compensation structure. Warrants and call options offer significant benefits for investors, but these derivatives are not without risks. Investors should therefore carefully consider these versatile instruments before using them in their stock portfolios.	{"url":"https://www.forex-central.net/warrants-call-options.php","timestamp":"2024-11-15T01:33:17Z","content_type":"text/html","content_length":"23660","record_id":"<urn:uuid:7f060377-f77c-4380-9a81-8200d01be269>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00393.warc.gz"}
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Orderbook Margin Maintenance Margin The orderbook calculates maintenance margin in accordance with the protocol rules, and makes the values available over various endpoints such as get_subaccount. For more details refer to the standard and portfolio margin sections. Initial Margin Similarly, the orderbook calculates initial margin using protocol rules, and ensures that no trade gets sent for settlement if the margin value after trade would be insufficient. Refer to the aforementioned sections for more detail. Open Orders Margin Limit orders that stay open in the book require that the account has extra margin to cover them if they were to get filled. The orderbook backend will inspect account's open orders[order_1, order_2,...]and find a "worst subset" of these orders, where "worst" is defined as a set of orders that, if filled, leads to the smallest initial margin possible. While performing those simulated fills, the backend will take into account the premiums paid or received for option bids and asks, as well as the current positions owned by the account. For example, suppose the open orders and positions are: • Orders:[bid 10 perps @ $1999, ask 100 perps @ $2001, bid 10 1w calls @ $55, bid 5 2w calls @ $75] • Positions:[long 90 perps] The backend will try and group the orders by their delta and / or vega sign and arrive at a conclusion that [bid 10 perps @ $1999, bid 10 calls @ $55, bid 5 2w calls @ $75] is the worst fill scenario. The open order margin for these orders will then be calculated by finding how much extra initial margin would the account require if those orders were to get filled. For every new open (i.e. non-crossing) order that arrives to the orderbook, the risk engine checks if the sum of current initial margin and the open orders margin is non-negative. In other words, new orders are accepted as long as the account can honour the "worst fill" scenario. The private/get_subaccount endpoint can be used to check which orders have been flagged as "worst subset" and how much open orders margin they require. Market Maker Protections (MMPs) and Open Orders Margin Oftentimes portfolio margin users would be market makers quoting hundreds of assets at the same time. If they have tight MMP limits, it is impossible for them to get filled on all of these quotes simultaneously, so it would be unreasonably capital inefficient to require them to lock margin for very large subsets of orders. Therefore, for portfolio margin accounts, the process of finding the "worst subset" is constrained by account's market maker protection settings. If MMP amount limits are enabled, the "worst subset" of orders would be reduced to an orders subset which can be filled subject to staying within MMP amount limit. Note that the reduced subset cannot be smaller than 2 distinct assets, i.e. the smallest possible open orders margin requirement still enforces that the market maker can honour at least 2 fills on two of the "worst" assets they are quoting. Using the above example - if MMP amount limit was set to 3, then the worst orders subset would exclude the bid 5 2w calls @ $75 and would just consist of [bid 10 perps @ $1999, bid 10 calls @ $55], because at least 2 assets have to be fillable by the account. If the MMP limit was too high (e.g. 30), then the subset would remain unchanged. Finally note that only MMP amount limit supports this capital efficiency improvement, the delta limit is ignored. More info on the MMPs can be found in the API reference under private/set_mmp_config.	{"url":"https://docs.derive.xyz/docs/open-orders-margin","timestamp":"2024-11-13T08:21:20Z","content_type":"text/html","content_length":"547523","record_id":"<urn:uuid:b530b320-a14b-48c7-934a-a8d32d33d4a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00249.warc.gz"}
Assessment of electronic structure methods for the determination of the ground spin states of Fe(II), Fe(III) and Fe(IV) complexes Our ability to understand and simulate the reactions catalyzed by iron depends strongly on our ability to predict the relative energetics of spin states. In this work, we studied the electronic structures of Fe^2+ ion, gaseous FeO and 14 iron complexes using Kohn-Sham density functional theory with particular focus on determining the ground spin state of these species as well as the magnitudes of relevant spin-state energy splittings. The 14 iron complexes investigated in this work have hexacoordinate geometries of which seven are Fe(II), five are Fe(III) and two are Fe(IV) complexes. These are calculated using 20 exchange-correlation functionals. In particular, we use a local spin density approximation (LSDA) - GVWN5, four generalized gradient approximations (GGAs) - BLYP, PBE, OPBE and OLYP, two non-separable gradient approximations (NGAs) - GAM and N12, two meta-GGAs - M06-L and M11-L, a meta-NGA - MN15-L, five hybrid GGAs - B3LYP, B3LYP∗, PBE0, B97-3 and SOGGA11-X, four hybrid meta-GGAs - M06, PW6B95, MPW1B95 and M08-SO and a hybrid meta-NGA - MN15. The density functional results are compared to reference data, which include experimental results as well as the results of diffusion Monte Carlo (DMC) calculations and ligand field theory estimates from the literature. For the Fe^2+ ion, all functionals except M11-L correctly predict the ground spin state to be quintet. However, quantitatively, most of the functionals are not close to the experimentally determined spin-state splitting energies. For FeO all functionals predict quintet to be the ground spin state. For the 14 iron complexes, the hybrid functionals B3LYP, MPW1B95 and MN15 correctly predict the ground spin state of 13 out of 14 complexes and PW6B95 gets all the 14 complexes right. The local functionals, OPBE, OLYP and M06-L, predict the correct ground spin state for 12 out of 14 complexes. Two of the tested functionals are not recommended to be used for this type of study, in particular M08-SO and M11-L, because M08-SO systematically overstabilizes the high spin state, and M11-L systematically overstabilizes the low spin state. Bibliographical note Publisher Copyright: © the Owner Societies 2017. Dive into the research topics of 'Assessment of electronic structure methods for the determination of the ground spin states of Fe(II), Fe(III) and Fe(IV) complexes'. Together they form a unique • Siepmann, I. (PI), Cramer, C. (CoI), Gagliardi, L. (CoI), Truhlar, D. G. (CoI), Tsapatsis, M. (CoI) & Goodpaster, J. D. (CoI) 9/1/12 → 8/31/17 Project: Research project	{"url":"https://experts.umn.edu/en/publications/assessment-of-electronic-structure-methods-for-the-determination-","timestamp":"2024-11-08T05:21:36Z","content_type":"text/html","content_length":"67944","record_id":"<urn:uuid:12468167-9b80-4282-aaed-456c12967547>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00305.warc.gz"}
FREE Skip Counting by 4 Worksheets [PDFs] Brighterly.com Skip Counting by 4 Worksheets Skip counting by 4 involves adding the number 4 to itself as 4+4=8, then adding it to every sum that comes after it, like, 8+4=12, 12+4=16, and so on. Kids must learn this math concept in kindergarten as it will build the foundation for tougher math problems in the future. Using skip counting by 4 worksheets will help your kids learn how to do this easier and faster. Benefits of counting in 4s worksheet Using the counting by 4s worksheet can help a child remember more about skip counting than teaching them in class will. With the skip count by 4s worksheet, a child has a pictorial example of skip counting. A skip count by 4 worksheet will help visual learners boost their learning speed and retain more information. A skip counting by 4s worksheet for kindergarten helps them understand the concept better. Download skip counting by 4 worksheets for kindergarten You can find skip counting by 4 worksheets for kindergarten on the internet ready for download. Find the worksheet with the most colorful objects to ensure maximum student engagement. Worksheets topics Worksheet #1 Worksheet #2 Worksheet #3 Worksheet #4 Order of Operations with Exponents	{"url":"https://brighterly.com/worksheets/skip-counting-by-4-worksheets/","timestamp":"2024-11-02T17:27:33Z","content_type":"text/html","content_length":"93249","record_id":"<urn:uuid:03fb89e9-95a0-4441-a204-835cc49a71c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00641.warc.gz"}
Grass is Greener [Day 2] \| Area of a Rectangle Arrays Math Talk & Practice Task Teacher Guide Be sure to read the teacher guide prior to running the task. When you’re ready to run the task, use the tabs at the top of the page to navigate through the lesson. In This Purposeful Practice… Math Talk Overview of This Visual Math Talk Visual Math Talk Prompt #1 Visual Math Talk Prompt #2 Visual Math Talk Prompt #3 Visual Math Talk Prompt #4 Purposeful Practice While Students Are Practicing… Questions: Area of a Triangle Resources and Downloads Printable Handout Download/Edit the handout so you can keep it handy and share with colleagues. Explore The Entire Unit of Study This Make Math Moments Task was designed to spark curiosity for a multi-day unit of study with built in purposeful practice, and extensions to elicit and emerge mathematical models and strategies. Click the links at the top of this task to head to the other related lessons created for this unit of study. Visual Math Talk Prompt #1 Visual Math Talk Prompt #2 Visual Math Talk Prompt #3 Visual Math Talk Prompt #4 Question #1 Question #2 Question #3 Question #4 Download Editable/Printable Handout Become a member to access purposeful practice to display via your projector/TV, download the PDF to upload to your LMS and/or print for students to have a physical copy	{"url":"https://learn.makemathmoments.com/task/grass-is-greener-day2/","timestamp":"2024-11-03T16:04:12Z","content_type":"text/html","content_length":"262883","record_id":"<urn:uuid:fb77462c-3b81-4537-8871-e2281f9b2d16>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00461.warc.gz"}
Large fluctuations of a Kardar-Parisi-Zhang interface on a half line Consider a stochastic interface h(x,t), described by the 1+1 Kardar-Parisi-Zhang (KPZ) equation on the half line x≥0. The interface is initially flat, h(x,t=0)=0, and driven by a Neumann boundary condition ∂xh(x=0,t)=A and by the noise. We study the short-time probability distribution PH,A,t of the one-point height H=h(x=0,t). Using the optimal fluctuation method, we show that -lnPH,A,t scales as t-1/2sH,At1/2. For small and moderate \|A\| this more general scaling reduces to the familiar simple scaling -lnPH,A,t≃t-1/2s(H), where s is independent of A and time and equal to one half of the corresponding large-deviation function for the full-line problem. For large \|A\| we uncover two asymptotic regimes. At very short time the simple scaling is restored, whereas at intermediate times the scaling remains more general and A-dependent. The distribution tails, however, always exhibit the simple scaling in the leading order. Bibliographical note Publisher Copyright: © 2018 American Physical Society. Dive into the research topics of 'Large fluctuations of a Kardar-Parisi-Zhang interface on a half line'. Together they form a unique fingerprint.	{"url":"https://cris.huji.ac.il/en/publications/large-fluctuations-of-a-kardar-parisi-zhang-interface-on-a-half-l-13","timestamp":"2024-11-03T16:12:25Z","content_type":"text/html","content_length":"48120","record_id":"<urn:uuid:aef22ca8-b854-4908-a192-28576430d9a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00640.warc.gz"}
Combinatorics programs for the RPN calculator on the Palm Pilot This is a program set I wrote for the RPN calculator that runs on the Palm Pilot. It implements factorial, permutations, combinations, Stirling numbers of the first kind, Stirling numbers of the second kind, and a program that will factor an integer. This can be pasted into the Memo application and from there added to the RPN calculator. // Factorial, permutations, // combinations, stirling #s 2nd kind, // factoring numbers. // ! (factrial) returns the factorial of // the number in the X register after // taking the absolute value and then // truncating it to a whole number. // Pyx (permutations) returns // permutations of y taken x at a // time. This has been extended // for negative numbers as // suggested in Knuth's book // Concrete Mathematics. It is: // (y)(y-1)...*(y-x+1). If x is 0 // then 1 is returned by definition. // If x is negative, then 0 is // returned. // Cyx (combinations) returns the // number of ways of selecting x // items out of y where order // doesn't matter. It too has been // extended like Pyx. If x is // negative then 0 is returned, but // if y is negative, then Pyx/x! is // returned. // S1yx (Stirling numbers of the // first kind). This is the number // of ways to partition y items into // x rings. This is computationally // intensive to calculate. It can // handle any pair of whole // whole number values where // y-x < 26, but cases where x is // fairly large and y is quite a bit // larger take a very long time. // Beware of large values. // S2yx (Stirling numbers of the // second kind) This is the number // of ways to partition y items into // x sets. // Factors. This function takes // the absolute value and then the // integer portion of the number // entered. It then decomposes // it into all of its prime factors // which are placed on the stack. // By Truman Collins // (tcollins@teleport.com) // 1/98 "_!: Factorial" "Pyx: Permutations. y items\taken x at a time when the\order matters." "Cyx: Combination. y items\taken x at a time when the\order does not matter." "S1yx: Stirling #s of 1st kind.\# of ways y items can be parti-\tioned into x rings." "S2yx: Stirling #s of 2nd kind.\# of ways y items can be parti-\tioned into x non-empty sets." "Factors: Factors an integer\and places all of its prime\factors on the stack." 13,594 visits (1 today, 3 this week, 9 this since December 27, 1998. Back to my homepage. Copyright 1998 by Truman Collins For comments, email: Truman Collins (truman@tkcs-collins.com) Most recent update: January 5, 2005	{"url":"http://www.tkcs-collins.com/truman/rpn/rpn_comb.shtml","timestamp":"2024-11-14T20:13:34Z","content_type":"text/html","content_length":"4470","record_id":"<urn:uuid:21040cc8-f128-4e3a-b373-5762126ac675>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00515.warc.gz"}
7 Most Frequently Used Mathematical Functions In Excel Mathematical Function In Excel Last Updated : 21 Aug, 2024 What Are Mathematical Functions In Excel? Mathematical functions in Excel is used to find values which require mathematical formulas. One of the reasons why users find Excel extremely helpful is because of the inbuilt functions and the possibility to create different combination of formulas. In this article, let us learn the basic mathematical functions we can use in Excel. 1. The seven common mathematical functions used in MS Excel are SUM, AVERAGE, AVERAGEIF, COUNTA, COUNTIF, MOD, and ROUND. 2. All mathematical functions in Excel are categorized under "Math & Trigonometry." Once a cell reference is given, the formula will be dynamic, and any changes in referenced cells will instantly impact formula cells. 3. To get an accurate count of cells in Excel, it's crucial to use the proper function. The COUNT function counts only numeric cell values, whereas the COUNTA function counts all non-empty cells, including those that contain text, dates, or logical values. 7 Mathematical Functions Used In MS Excel With Examples 1. SUM 2. AVERAGE 3. AVERAGEIF 4. COUNTA 5. COUNTIF 6. MOD 7. ROUND Let us discuss each of them in detail. #1 - SUM If we want to SUM values of several cells quickly, we can use the SUM in Excel for the mathematics category. For example, look at the below data in Excel. We need to find the total production quantity and total salary from this. Open the SUM function in the G2 cell. And select the range of cells from C2 to C11. Close the bracket and press the "Enter" key to get the total production quantity. So, the total production quantity is 1,506. Similarly, we must apply the same logic to get the total salary amount. #2 - AVERAGE Now, we know what overall sum values are. Next, we need to find the average salary per employee out of these overall employees. Open the AVERAGE function in the G4 cell. Select the range of cells for which we are finding the average value, so our range of cells will be from D2 to D11. So the average salary per person is $4,910. #3 - AVERAGEIF We know the average salary per person; we want to know the average salary based on gender for further drill-down. What is the average salary of males and females? • The first parameter of this function is range. For example, choose cells from B2 to B11. • We need to consider only male employees in this range, so enter the criteria as "M." • Next, we need to choose the average range, D1 to D11. • So, the average salary of male employees is $4,940. Similarly, we must apply the formula to find the average female salary. The female average salary is $4,880. #4 - COUNTA Let us find out how many employees are there in this range. The COUNTA function will count the number of non-empty cells in the selected range of cells. So totally, there are 10 employees on the list. #5 - COUNTIF After counting the total number of employees, we may need to count how many male and female employees there are. • The range is in which range of cells we need to count since we need to count the number of male or female employees who choose cells from B2 to B11. • The criteria will be in the selected range. What do we need to count? Since we need to count how many male employees there are, give the criteria as "M." • Similarly, copy the formula and change the criteria from "M" to "F." #6 - MOD The MOD function will return the remainder when one number is divided by another. For example, dividing the number 11 by 2 will get the remainder as 1 because only till 10 number 2 can divide. • For example, look at the below data. • By applying a simple MOD function, we can find the remainder value. #7 - ROUND When we have fraction or decimal values, we may need to round those decimal values to the nearest integer number. For example, we need to round the numbers 3.25 to 3 and 3.75 to 4. • Select the Number as a B2 cell. • Since we are rounding the value to the nearest integer number of digits will be 0. As we can see above, the B2 cell value 115.89 is rounded to the nearest integer value of 116, and the B5 cell value of 123.34 is rounded to 123. Like this, we can use various mathematical functions in Excel to do mathematical operations in Excel quickly and easily. Important Things To Note • All the mathematical functions in Excel are categorized under the "Math & Trigonometry" function in Excel. • Once the cell reference is given, the formula will be dynamic, and whatever changes happen in referenced cells will impact formula cells instantly. • The COUNTA function will count all the non-empty cells, but the COUNT function in Excel calculates only numeric cell values. Frequently Asked Questions 1. Where are mathematical functions used? Mathematical functions are building blocks used in designing machines, predicting disasters, treating diseases, studying economies, and maintaining aircraft. Understanding them is essential for excelling in these fields. 2. What is the use of mathematical functions? A mathematical function defines the value of a dependent variable based on one or more independent variables. It can be represented through tables, formulas, graphs, or computer algorithms. Functions are important for academic and business applications, such as engineering, finance, and statistics. The choice of representation should be based on problem requirements and available resources. 3. What are the parts of a mathematical function? A function has three parts: inputs, outputs, and a rule that assigns each input to exactly one output. Recommended Articles This article is a guide to Mathematical Functions in Excel. We discuss calculating mathematical functions in Excel using SUM, AVERAGE, AVERAGEIF, COUNTA, COUNTIF, MOD, and ROUND formulas and practical examples. You may learn more about Excel from the following articles: -	{"url":"https://www.wallstreetmojo.com/mathematical-function-in-excel/","timestamp":"2024-11-08T12:07:09Z","content_type":"text/html","content_length":"363091","record_id":"<urn:uuid:0c1d92d5-a263-4754-888b-1ea12523120f>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00735.warc.gz"}
How Fast Do GPS Satellites Travel? Velocity of GPS Satellites Global Positioning System (GPS) satellites travel approximately 14,000 km/hour, relative to the Earth as a whole, as opposed to relative to a fixed point on its surface. The six orbits are tipped at 55° from the equator, with four satellites per orbit (see diagram). This configuration, advantages of which are discussed below, prohibits geostationary (fixed above a point on the surface) orbit since it is not equatorial. Velocity Relative to the Earth Relative to the Earth, GPS satellites orbit twice in a sidereal day, the length of time the stars (instead of the sun) take to return to the original position in the sky. Since a sidereal day is about 4 minutes shorter than a solar day, a GPS satellite orbits once every 11 hours and 58 minutes. With the Earth rotating once every 24 hours, a GPS satellite catches up to a point above the Earth approximately once a day. Relative to the center of the Earth, the satellite orbits twice in the time it takes a point on the Earth's surface to rotate once. This can be compared to a more down-to-earth analogy of two horses on a racetrack. Horse A runs twice as fast as Horse B. They start at the same time and same position. It will take Horse A two laps to catch Horse B, which will have just completed its first lap at the time of being caught. Geostationary Orbit Undesirable Many telecommunications satellites are geostationary, enabling time-continuity of coverage above a chosen area, such as service to one country. More specifically, they enable the pointing of an antenna in a fixed direction. If GPS satellites were confined to equatorial orbits, as in geostationary orbits, coverage would be greatly reduced. Furthermore, the GPS system does not use fixed antennae, so deviation from a stationary point, and therefore from an equatorial orbit, is not disadvantageous. Furthermore, faster orbits (e.g. orbiting twice a day instead of the once of a geostationary satellite) mean lower passes. Counterintuitively, a satellite closer in from geostationary orbit must travel faster than the Earth's surface in order to stay aloft, to keep "missing the Earth" as the lower altitude causes it to fall faster toward it (by the inverse square law). The apparent paradox that the satellite moves faster as it gets closer to the Earth, thereby implying a discontinuity in speeds at the surface, is resolved by realizing that the Earth's surface need not maintain lateral speed to balance out its falling speed: it opposes gravity another way — electrical repulsion of the ground supporting it from below. But why match the satellite speed to the sidereal day instead of the solar day? For the same reason Foucault's pendulum rotates as the Earth spins. Such a pendulum is not constrained to one plane as it swings, and therefore maintains the same plane relative to the stars (when placed at the poles): only relative to the Earth does it seems to rotate. Conventional clock pendulums are constrained to one plane, pushed angularly by the Earth as it rotates. To keep a satellite's (non-equatorial) orbit rotating with the Earth instead of the stars would entail extra propulsion for a correspondence that can easily be accounted for mathematically. Calculation of Velocity Knowing that the period is 11 hours and 28 minutes, one can determine the distance a satellite must be from the Earth, and therefore its lateral speed. Using Newton's second law (F=ma), the gravitational force on the satellite is equal to the satellite's mass times its angular acceleration: GMm/r^2 = (m)(ω^2r), for G the gravitational constant, M the Earths' mass, m the satellite mass, ω the angular velocity, and r the distance to the Earth's center ω is 2π/T , where T is the period of 11 hours 58 minutes (or 43,080 seconds). Our answer is the orbital circumference 2πr divided by the time of an orbit, or T. Using GM=3.99x10^14m^3/s^2 gives r^3=1.88x10^22m^3. Therefore, 2πr / T = 1.40 x 10^4 km/sec. Cite This Article Dohrman, Paul. "How Fast Do GPS Satellites Travel?" sciencing.com, https://www.sciencing.com/how-fast-do-gps-satellites-travel-12213923/. 6 October 2017. Dohrman, Paul. (2017, October 6). How Fast Do GPS Satellites Travel?. sciencing.com. Retrieved from https://www.sciencing.com/how-fast-do-gps-satellites-travel-12213923/ Dohrman, Paul. How Fast Do GPS Satellites Travel? last modified March 24, 2022. https://www.sciencing.com/how-fast-do-gps-satellites-travel-12213923/	{"url":"https://www.sciencing.com:443/how-fast-do-gps-satellites-travel-12213923/","timestamp":"2024-11-08T21:46:09Z","content_type":"application/xhtml+xml","content_length":"73986","record_id":"<urn:uuid:780ac161-550b-4bd4-8ca5-c351e84e6a12>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00380.warc.gz"}
Generating Random Integers in Python Made Easy with randint() - Adventures in Machine Learning Python randint(): Generating Random Integers As a Python programmer, generating a random integer is a common task that you may need to perform from time to time. Luckily, Python comes with a built-in function that makes this task easy: the randint() method from the random module. Syntax of Python randint() The randint() method can be called with two arguments: the lower and upper bounds of the range within which the random integer should fall. The syntax looks like this: Here, a is the lower bound, while b is the upper bound. Using the Python randint() method Importing the random module To use the randint() method, first, you’ll need to import the random module. This can be done using the import statement as shown below: Generating a random integer After importing the module, you can generate a random integer using the randint() function. For instance, to generate a random integer between 1 and 10, you can use the following code: random_integer = random.randint(1,10) The code above generates a random integer between 1 and 10 and stores it in the variable random_integer. The print statement then displays the randomly generated integer. Lower and Upper Bound Limits Setting lower and upper limits is crucial in generating random integers using the randint() function. The lower and upper limits are specified as the first and second arguments to the randint() method, respectively. For example, if you need to generate a random integer between 50 and 100, the code below can be used: random_integer = random.randint(50, 100) By setting the lower limit to 50 and the upper bound to 100, the assigned variable will be assigned a random integer between 50 and 100 every time the code executes. Value Error Exception It’s important to note that the randint() function may raise a ValueError exception if the lower bound is set greater than the upper bound. For instance, the following code would result in a value random_integer = random.randint(100, 50) When generating an integer using randint(), it is essential to avoid such errors by ensuring the lower bound is always smaller than the upper bound. In conclusion, generating a random integer using the Python randint() function is a simple process that can be done with just a few lines of code. By setting lower and upper bounds, a random integer can be generated within a specified range, which is vital in several applications. While errors can occur if the lower and upper bounds are not correctly specified, these can easily be avoided by setting lower bound value to always be smaller than the upper bound. Python randint(): Examples of generating random integers In the previous section, we learned about the Python randint() function and how to use it to generate a random integer within a specified range. In this section, we’ll explore some examples of using the randint() function to generate random integers, both a single integer and multiple integers in a loop. Example 1: Generating a single random integer Let’s begin with an example of generating a single random integer. Suppose we want to generate a random number between 1 and 10 on each execution of the code. The following Python code demonstrates this: import random random_number = random.randint(1, 10) print("Random number:", random_number) Here, we have imported the random module, which is mandatory to use the randint() function. We have then called the randint() function, specifying the lower bound as 1 and the upper bound as 10. Finally, we have printed our generated random number. If the code above is executed multiple times, it will produce different random integers each time, always between 1 and 10. This demonstrates the usefulness of the randint() function in generating random numbers within a specified range. Example 2: Generating multiple random integers in a loop In some cases, we may need to generate multiple random integers. Instead of manually calling the randint() function multiple times, we can use a loop to generate as many random numbers as required. The following Python code demonstrates this: import random for i in range(5): random_number = random.randint(1, 10) print("Random number:", random_number) Here, we have enclosed the process of generating random numbers within a for loop. We have set the loop to iterate five times, generating five random integers between 1 and 10. On each execution of the loop, a new random number is generated and printed to the console. This example highlights the advantage of using a loop to generate multiple random integers. Rather than calling the randint() function multiple times, we have used a single call within a loop, resulting in shorter and more efficient code. In summary, the Python randint() function is a powerful tool that allows us to generate random integers quickly and effectively. Its ability to set bounds on the random numbers is something that can be useful in various applications. By using a for loop, we can generate multiple random integers with minimal effort, making the randint() function even more flexible. Key Takeaways • The randint() function generates random integers within a specified range. • To use the randint() function, we need to import the random module. • Lower and upper bounds should be correctly specified to avoid Value Error Exceptions. • We can generate a single random integer by calling the randint() function once. • A for loop can be used to generate multiple random integers. In conclusion, generating random integers in Python is made easy through the use of the randint() function from the random module. By setting the lower and upper limits, we can generate a single integer or multiple integers within a range using loops. Being aware of potential value errors that can occur due to incorrectly defined limits is essential, and the benefits of this function include simplified and efficient coding. Key takeaways include ensuring that lower bound is always smaller than the upper bound, and the randint() function is essential for generating random integers in different Python applications. Overall, this article emphasizes the importance of the randint() function for generating random integers in Python; it is something every Python programmer should take advantage of to simplify their coding efforts.	{"url":"https://www.adventuresinmachinelearning.com/generating-random-integers-in-python-made-easy-with-randint/","timestamp":"2024-11-07T00:18:48Z","content_type":"text/html","content_length":"76512","record_id":"<urn:uuid:52f311e9-b3e0-48e9-8ac2-5d3d3da0fd11>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00741.warc.gz"}
Mathematical Language Words linked to + add, addition, and, count on, plus, sum, more, altogether, increase Words linked to - take away, subtract, subtraction, - ppt download Presentation is loading. Please wait. To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy , including cookie policy. Ads by Google	{"url":"https://slideplayer.com/slide/4597675/","timestamp":"2024-11-10T06:21:05Z","content_type":"text/html","content_length":"157760","record_id":"<urn:uuid:515214ac-a98a-4424-b2e2-4bec93fc3cee>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00350.warc.gz"}
Subclasses of Analytic Functions with Negative Coefficients Involving q-Derivative Operator Liew, Andy Pik Hern and Aini Janteng and Rashidah Omar (2022) Subclasses of Analytic Functions with Negative Coefficients Involving q-Derivative Operator. Science and Technology Indonesia, 7 (3). pp. 327-332. ISSN 2580-4391 (E-ISSN) , 2580-4405 (P-ISSN) Download (37kB) Full text.pdf Restricted to Registered users only Download (588kB) \| Request a copy Let A denote the class of functions f which are analytic in the open unit disk U. The subclass of A consisting of univalent functions is denoted by M. In this paper, we also consider a subclass of M which is denoted by V, consisting of functions with negative coefficients. In addition, this paper also studies the q-derivative operator. By combining the ideas, this paper introduced three subclasses of A with negative coefficients involving q-derivative. Furthermore, the coefficient estimates, growth results and extreme points were obtained for all of these classes. Actions (login required)	{"url":"https://eprints.ums.edu.my/id/eprint/34339/","timestamp":"2024-11-01T23:42:55Z","content_type":"application/xhtml+xml","content_length":"23687","record_id":"<urn:uuid:140d9120-2456-46a0-ad5f-c7dfb7f5b3dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00899.warc.gz"}
Full Time Equivalent (FTE) Full time equivalent (FTE) is a standardisation unit. FTE allows organisations that have resources who are working different days and hours to calculate capacity, utilisation and availability in a standardised manner. Why is FTE required? Organisations can have resources that work on different days and different hours and therefore it becomes complicated to calculate overall capacity, utilisation and availability properly. Let us understand this with the help of an example. Example: Organisation ABC has five employees and they all work different hours every day. John: 8 hours every day, 5 days per week. Total 40 hours every week. Adam: 4 hours every day, 5 days per week. Total 20 hours every week. Nancy: 6 hours every day, 4 days per week. Total 24 hours every week. David: 4 hours every day, 4 days per week. Total 16 hours every week. Liz: 10 hours every day, 3 days per week. Total 30 hours every week. In the above example, it seems that organisation ABC has five resources, but that does not provide the complete picture, because all the five resources are working different number of hours every In such a scenario, if John and David are assigned to 'Project A' for a week for 100% of their individual capacity, then it seems that 2 resources have been utilised on 'Project A', but John's 100% equals 40 hours a week, whereas David's 100% equals 16 hours a week! In such a scenario, FTE can be used to provide a complete picture. How is FTE calculated? In order to calculate FTE, the administrator first needs to define what constitutes as Full Time Equivalent (FTE) in their organisation. This can be done by defining one of the working calendars as FTE calendar. In the above screenshot 'New York' calendar is selected to calculate FTE. The working days and hours for 'New York' calendar have been defined as follows... It implies that 8 hours per day / 40 hours per week constitutes 1 Full Time resource in Organization ABC. Below we have shown the capacity of each of the five resources in FTE by using New York calendar as a standard for 1 FTE. John: 1 FTE Adam: 0.5 FTE Nancy: 0.6 FTE David: 0.4 FTE Liz: 0.75 FTE Using New York calendar's 8 hours per day / 40 hours per week, we can state that Organisation ABC's capacity is 3.25 full time resources, out of which 1.4 FTE is assigned on 'Project A'.	{"url":"https://support.eresourcescheduler.cloud/hc/en-us/articles/12831234366617-Full-Time-Equivalent-FTE","timestamp":"2024-11-07T19:02:12Z","content_type":"text/html","content_length":"19449","record_id":"<urn:uuid:c7d712e7-451e-428f-8cde-3c22cdeae03c>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00114.warc.gz"}
ACCTG 333 - Custom Scholars ACCTG 333 Excel AssignmentACCTG 333 – Professor Perols Assignment Overview The exercises below are designed to increase your effectiveness and efficiency in using Excel to analyze data. Excel is widely used in accounting and students need to master Excel (AICPA 2022). This is an individual assignment to be completed on your computer (PC or Mac, although a PC is preferred). You are allowed to discuss the assignment with others in the class, but you may not copy. When you have completed the assignment, you will upload your Excel Workbook file to Canvas under “Excel Assignment Submission.” To complete these exercises, consult online help such as https://support.office.com/enus/excel and https://exceljet.net/ and the help feature within Excel (F1). Keywords for each exercise are provided to assist you. Note that if the keyword for an exercise states for example: OR() then your answer must use the OR() function. Also, your formula in one exercise cannot cell reference your answer from a previous exercise. Getting Started Download the Northwind Excel Workbook from Canvas. Save the file to your computer and rename it as your first initial and lastname, e.g., rperols.xlsx. If you are a Mac user (and do not have access to a PC), stop and check for software updates before proceeding. Help -> Check for Updates. Run/install updates before proceeding. Unless stated in the instructions, no additional cells/data/formulas should be added beyond the yellow highlighted cells in the Northwind Excel Workbook. Practice Shortcut Keys First spend time practicing the shortcut keys below. Make sure to use these shortcut keys whenever you work in Excel (many of them also work in other applications). Learning these (and other) shortcuts will save you a lot of time and are more or less necessary for you to be efficient when you work. For macOS users, most of these shortcuts work with the command button instead of CTRL. In the future, if you are provided a Windows based PC for in your internships and/or job, I highly recommend all students practice these shortcuts on a Windows based PC in order to be efficient (an important skill). o F1 Displays the Help task pane. o CTRL+A Selects the entire table Use CRTL+A+A selects the entire worksheet. o Ctrl+F Opens the ‘find text’ dialog box. o CTRL+S Saves the active file with its current file name, location, and file format. o CTRL+C Copies the selected cells. o CTRL+X Cuts the selected cells. o CTRL+V Inserts the contents of the Clipboard at the insertion point and replaces any selection. Available only after you cut or copied an object, text, or cell content. Excel Assignment ACCTG 333 – Professor Perols o CTRL+ALT+V displays the Paste Special dialog box. Available only after you have cut or copied an object, text, or cell content on a worksheet or in another program. o CTRL+Y Repeats the last command or action, if possible. o CTRL+Z Uses the Undo command to reverse the last command or to delete the last entry you typed. o CTRL+ARROW KEY moves to the edge of the current data region (data region: a range of cells that contains data and that is bounded by empty cells or datasheet borders) in a worksheet. o CTRL+SHIFT+ARROW KEY extends the selection of cells to the last nonblank cell in the same column or row as the active cell, or if the next cell is blank, extends the selection to the next nonblank cell. o Shift + Spacebar Selects entire row. o Ctrl + Spacebar Selects entire column. o CTRL+HOME moves to the beginning of a worksheet (CTRL+SHIFT+HOME extends the selection of cells to the beginning of the worksheet). o CTRL+END moves to the last cell on a worksheet, in the lowest used row of the rightmost used column (CTRL+SHIFT+END extends the selection of cells to the last used cell on the worksheet). o CTRL+PAGE DOWN moves to the next sheet in a workbook. o CTRL+PAGE UP moves to the previous sheet in a workbook. 1. First, practice a little formatting. Format the OrderDetails worksheet by formatting the header row using top and bottom (but not side) borders, light grey fill color (the exact shade of grey is not important), and bold and centered text. Also format the data using appropriate data type format (determine what is “appropriate” by examining the content of each column, e.g., Discount should be percentages and UnitPrice should be currency). Expand all the columns so that all the data are visible (select the entire worksheet, i.e., use CTRL+A+A, and double click on a line between two Excel Worksheet Column Headers, e.g., between A and B above the table headers. You do not need to do more formatting for the assignment beyond this brief practice in exercise 1. 2. Before getting started on the formulas, take some time to understand the data. For example, note that workbook contains two files, OrderHeaders and the OrderDetails, which contain archive transaction information. Each row in the OrderHeaders table represents a distinct order. Each order can, however, have many rows in the OrderDetails table because a given order can be for multiple items. Each row in the OrderDetails table is associate with a specific order header row and a specific item. Each row also shows the quantity, unit price, and discount for the sale of the item. Remember that in the Systems Understanding Excel Assignment ACCTG 333 – Professor Perols Aid (SUA) assignment, the top part of orders contained OrderID, date, customer information, and supplier information while the bottom part contained rows with information about the actual items sold (this structure of storing the header portion and the detail portion is very common for different accounting objects, e.g., purchase orders, receiving reports, invoices, sales orders, and sales invoices). Also note that the other tables contain master tables, e.g., Suppliers, Customers, and Products, where each row contains information related to a single Supplier, Customer, Product, etc. No action/answer is required for exercise 2 (just an overall understanding of the data). 3. Cell References and Calculations – In the OrderDetails worksheet, in cell F2, calculate the LineItemTotal as UnitPrice * Quantity * (1-Discount) using cell references. Format this cell as currency with two decimal places and copy down cell F2 to the bottom of the table, i.e., F2 through F2156, using short cut keys: CTRL+C, ARROW LEFT, CTRL+ARROW DOWN, ARROW RIGHT, CTRL+SHIFT+ARROW UP, CTRL+V. Note that this is a very common sequence of commands that you want to learn (do not memorize it, just practice it throughout this assignment when you need to copy down a formula). 4. Absolute and Relative Cell References – In the OrderDetails worksheet, in the yellow highlighted cell I2, enter 5%. In cell G2 calculate the LineItemTotal with Additional Discount as UnitPrice * Quantity * (1-Discount-Additional Discount) where the additional discount is held constant at the value in cell I2. Note that you need to use both relative and absolute cell references for this formula to work correctly. There is also a shortcut (F4) for applying absolute cell references. Copy down the formula in cell G2 to the bottom of the table, i.e., G2 through G2156, using short cut keys (for the rest of the assignment, assume that you should copy down formulas if the formulas relate to each row in a table). 5. VLOOKUP() – First, in the OrderHeaders worksheet, in column C use a vlookup to display the company name from the Customer worksheet for each order. Please see the short video I posted on Canvas to help you through the first part of this exercise. Second, in the Products worksheet, in Column D use a vlookup to display the country from the Suppliers worksheet for each supplier. 6. COUNT(), AVG(), SUM(), MAX(), and MIN() – In the OrderDetails worksheet, in cells B2159, C2159, D2159, E2159, and F2159 calculate the Total Number of Line Items (number of line item rows), Average UnitPrice (average of unit price in column C) Total Quantity (sum of all line item quantities in column D), HighestDiscount (maximum of all discounts in column E), and Smallest LineItemTotal (minimum of all LiteItemTotals in column F). 7. COUNTIF(), AVERAGEIF(), and SUMIF() – In the OrderDetails worksheet, in cell A2162 enter 51. In cells B2162, C2162, and D2162 use COUNTIF() to calculate Number of Product 51 Sales, AVERAGEIF() to calculate Average Unit Price of Product 51, and SUMIF() to calculate the Quantity Sold of Product 51 (use a cell reference in your formulas to the value in A2162 so that your count, average, and sum updates when the value in A2162 changes). Also, in the Employees worksheet, in column M use SUMIF to show Total Sales (column O in OrderHeaders) for each employee ID. Note each employee is associated with many orders. Excel Assignment ACCTG 333 – Professor Perols 8. IF() – In the OrderDetails worksheet, in column K create an if statement that returns “Yes” if the Quantity (values in column D) is above 40 (strictly greater than) and otherwise “No”. 9. AND() and OR() – In the OrderDetails worksheet, in columns L and M use IF statements with an AND() and OR(), respectively, to return “Yes” if Quantity is between 30 and 40 (strictly greater than 30 and equal to or less than 40), and otherwise “No”. The OR() requires more thinking (see tips in the Check Figures document). 10. Nested IF() – In the OrderDetails worksheet, in column N use a nested IF statement to return “Yes” if Quantity is between 30 and 40 (strictly greater than 30 and equal to or less than 40), and otherwise “No” (see tips in the Check Figures document). 11. Missing values – In the OrderHeaders worksheet, in column Q create an if statement that returns “Yes” if the ShippedDate is missing (indicated inside the if statement as an empty string, i.e., two quotation marks in a row without whitespace), and otherwise “No”. 12. Comparing Existing Dates – In the OrderHeaders worksheet, in column R create an IF statement that returns “Yes” if the ShippedDate is on or after the RequiredDate, otherwise “No”. Note that a date in Excel is stored as a number representing the number of days since 1/1/1900. The number is simply formatted to display as a date. Because of this you can directly compare two dates that are already in an Excel worksheet (you are really comparing two regular numbers). To demonstrate, change the formatting of the value in G2 to a number. Notice that the cell is actually storing 42762 (the number of days between 1/27/2017 and 1/1/1900). Also note that the formatting will not change your answer for exercise 12. 13. YEAR() and MONTH() – The YEAR() and MONTH() functions are used to get the year or month of a date stored in Excel. In the OrderHeaders worksheet, in columns S and T use YEAR() and MONTH() to return the OrderDate year and month, respectively. In column U, use YEAR() inside an IF statement to return “Yes” if the OrderDate is in 2018 and otherwise “No”. In column V, use YEAR(), MONTH(), and AND() inside an IF() statement to return “Yes” if the OrderDate is in the first quarter of 2018, and otherwise “No”. 14. DATE() – It is not as easy to compare a date already in Excel to a date that we want to specify. If we type in 1/12/2019, Excel will interpret this as 1 divided by 12 divided by 2019 and to compare 1/12/2019 to other dates in Excel we therefore first need to convert the date to a number that represents the number of days since 1/1/2019. This can be done using DATE(). In the OrderHeaders worksheet, in cell AB2 simply use DATE() or DATEVALUE() to find out how many days there have been since 1/1/1900 and the due date of this assignment 11/30/2023. Change the cell formatting to a number if the results from DATE() are formatted as a date. Note that the function DATEVALUE() works the same way as DATE(), but converts a date string, e.g., “11/2/2019”, rather than integers separated by commas, e.g., 2019,11,2, to the number of days since 1/1/1900. Excel Assignment ACCTG 333 – Professor Perols In column W use DATE() and AND() inside an if statement to return “Yes” if the OrderDate is in the first quarter of 2018, and otherwise “No”. In column X, use DATE() only inside an if statement to return “Yes” if the order date is after (strictly greater than) 10/15/2018, and otherwise “No”. In column Y, use AND() and DATE() inside an if statement to return “Yes” if the order date is before (strictly less than) 11/18/2018 and has not yet been shipped, and otherwise “No”. 15. SUMIFS – In the Employees worksheet, in column N use SUMIFS (not SUMIF) to show: Total Sales (column O in OrderHeaders) for each employee but only include orders that have been shipped (column G in OrderHeaders) in this calclation. Note that an order has been shipped if ShippedDate is not empty, indicated using “” (note that there is nothing to the right of as blank in Excel is indicated by nothing. 16. LEFT(), RIGHT(), FIND(), and LEN() – In the Customers worksheet, in columns O, P, Q, R, and S use LEFT() to find the first eight characters of ContactName, use RIGHT() to find the last nine characters of ContactName, FIND() to find the number of characters before space in ContactName (you need to use FIND()-1 for this), LEN() to find the number of characters in Contact Name, and LEN() and FIND() to find the number of characters after the space in Contact Name. In column D use LEFT() and FIND() to show the first name of the customer contact (shown in column C). Note that you need to embed the FIND() inside the LEFT() functions and use FIND to return the location of the character that separates the first name and the last name. In column E, use RIGHT(), LEN(), and FIND() to show the last name of the customer contact (and you again need to embed formulas) Note that when embedding a formula inside another formula the behavior of the nested formula does not change, e.g., FIND() finds the first occurrence of one string inside another string and always searches from the left even when embedded within RIGHT(). Pivot Tables Complete exercises 17-20 before comparing your answer to the Check Figures document. 17. Creating Pivot Tables a. Use the Sales worksheet data to create a Pivot Table into a new worksheet. Name the new worksheet SalesPivot. For Mac users especially, do not create the pivot table by selecting the worksheet (data). Instead, from the Sales worksheet, simply Insert -> Pivot Table and the data will be automatically selected. b. Click and Drag ProductName and CustomerName to Rows, OrderDate to Columns (year and quarter will also be added), and LineItemTotal to Values. For Mac users especially, “drag” OrderDate to Columns to ensure that year and quarter will also be added. c. Rearrange the Rows fields to show all customers and the products that they have purchased (rather than all products and the customers that purchased those products). Excel Assignment ACCTG 333 – Professor Perols d. In the Column Labels, use the + button to expand the pivot table show Year and then Quarter. If the + button is not available go to Pivot Tables tab -> Analyze -> Show -> +/Buttons. Note that the Pivot Table now groups the sales data based on CustomerName, ProuctName, and the quarter of the OrderDate and then sums LineItemTotal. 18. Other Aggregate Functions in Pivot Tables a. Use the – button to collapse the pivot table details back to the annual level (view the data grouped by year rather than quarter). b. Add Discount to Values. Calculate average discount (rather than sum) and change the format to percentage with one decimal. To do this, use Value Field Settings (accessed by double clicking or right clicking on the Sum of Discount header, or by opening the drop down menu for the Sum of Discount in the Values field selector). c. Add OrderID to Values and count how many line items are being grouped. Inside Value Field Settings, change the format to number with zero decimals. d. Inside Value Field Settings, change the format of Sum of LineItemTotal to currency with zero decimals. 19. Formatting Pivot Tables a. In Design Subtotals, select to not show subtotals and Grandtotals (turn off for both rows and columns). b. In Design Report Layout, select Show in Tabular Form c. In Design Report Layout, select Repeat All Item Labels d. Change the name of the columns headers (you can make these changes directly in the column headers or in Field Settings) to Customer Name, Product Name, Average Discount, Number of Order Lines. (Note: do not change Sum of LineItemTotal). e. Replace all empty cells with 0 (right click inside the pivot table, select Pivot Table Options and set “For empty cells show:” to 0. 20. Filtering and Slicing Pivot Tables a. Filter customer names to only show customers that begin with B by left clicking the filter icon to show the filter drop down menu (the little triangle in the column header in the same cell as the text Customer Name), selecting Labels Filters, and Begins With. b. Filter product names to only show products that begin with letters between O-Z using Labels Filters Between… c. Insert a Slicer using OrderDate and select Feb, May, and Aug. 21. Obtaining Details from Pivot Tables a. For Customer Name and Product Name: Bon app’ and Pavlova, double left click the Number of Order Lines for 2018 (double left click on the 2). Note that the details of the two order lines will display in a new worksheet. b. Change the new worksheet name to Bon App Pavlova 2018 Details.	{"url":"https://customscholars.com/acctg-333/","timestamp":"2024-11-06T18:48:36Z","content_type":"text/html","content_length":"69153","record_id":"<urn:uuid:b5f6428d-4d6c-44bb-b5c7-2eb92e052721>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00611.warc.gz"}
We were discussing the basicÂ difference between orifice and mouthpiece,Â classification of orifices and mouthpiecesÂ and also advantages and disadvantages of orifices in the subject of fluid mechanics, in our recent posts. Now we will go ahead to find out the expression for flow through an orifice.Â First we will see here the basic concept of an orifice and after that we will find out here the expression for flow through an orifice with the help of this post. So let us come to the main topic, without wasting your time.Â An orifice is basically a small opening of any cross-section such as triangular, square or rectangular on the side or at the bottom of tank, through which a fluid is flowing.Â Orifice is basically used in order to determine the rate of flow of fluid.Â As we have discussed above that orifice will be a small opening of any cross-section, hence flow through the orifice will be very small.Â Flow through an orifice Let us consider one tank with a circular orifice fitted at one side of the tank as displayed here in following figure.Â Liquid flowing through the orifice is developing a liquid jet whose cross-sectional area is smaller than the cross-sectional area of the circular orifice. Area of liquid jet is decreasing and area is minimum at section CC.Â Section CC will be approximately at a distance of half of diameter of the circular orifice. At section CC, the streamlines are straight and parallel with each other and perpendicular to the plane of the orifice. This section CC will be termed as Vena-contracta.Â Beyond the section CC, liquid jet diverges and will be attracted towards the downward direction due to gravity.Â Image: Tank with a circular orifice Let us consider that h is the head of the liquid above the centre of orifice.Â Let us consider two points 1 and 2 as displayed in above figure. Point 1 is displayed inside the tank and point 2 is shown at the Vena-contracta.Â Let us consider that flow is steady and at a constant differential head h. p[1]Â = Pressure at point 1 v[1]Â = Velocity of fluid at point 1 p[2]Â = Pressure at point 2 v[2]Â = Velocity of fluid at point 2Â Now we will apply the Bernoulliâ€™s equation at point 1 and 2.Â Area of tank is quite large as compared with area of liquid jet and therefore v[1]Â will be very small as compared with v[2].Â Therefore above expressionÂ for theoretical velocity could be re-expressed as mentioned here.Â We must note it here that this is theÂ theoretical velocity and actual velocityÂ will be less than this value.Â We will see various types of hydraulic co-efficients, in the subject of fluid mechanics, in our next post.Â Do you have any suggestions? Please write in comment box.Â Fluid Mechanics, By R. K. Bansal Image Courtesy: GoogleÂ Also Read No comments:	{"url":"https://www.hkdivedi.com/2018/07/flow-through-orifice.html","timestamp":"2024-11-13T11:26:09Z","content_type":"application/xhtml+xml","content_length":"292253","record_id":"<urn:uuid:57c0fea2-1ee2-4e6d-890d-ce2a2a3dd6bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00086.warc.gz"}
Biology Notes Form Two All Topics {Best of the Best} Biology Notes Form Two • Transport is the movement of substances within an organism. • All living cells require oxygen and food for various metabolic processes. • These substances must be transported to the cells. • Metabolic processes in the cells produce excretory products which should be eliminated before they accumulate. • The excretory products should be transported to sites of excretion. • Organisms like amoeba are unicellular. • They have a large surface area to volume ratio. • The body is in contact with the environment. • Diffusion is adequate to transport substances across the cell membrane and within the organism. • Large multicellular organisms have complex structure where cells are far from each other hence diffusion alone cannot meet the demand for supply and removal of substances. • Therefore an elaborate transport system is necessary. Transport in plants • Simple plants such as mosses and liverworts lack specialized transport system. • Higher plants have specialized transport systems known as the vascular bundle. • Xylem transports water and mineral salts . • Phloem transports dissolved food substances like sugars. Internal structure of roots and root hairs • The main functions of roots are ; • Anchorage • storage • gaseous exchange. • The outermost layer in a root is the • This is a special epidermis of young roots whose cells give rise to root hairs. • Root hairs are microscopic outgrowths of epidermal cells. • They are found just behind the root tip, • They are one cell thick for efficient absorption of substances. • They are numerous and elongated providing a large surface area for absorption of water and mineral salts. • Root hairs penetrate the soil and make close contact with it. • Below the peliferous layer is the cortex. • This is made up of loosely packed, thin walled parenchyma cells. • Water molecules pass through this tissue to reach the vascular bundles. • In some young plant stems, cortex cells contain chloroplasts. • The endodermis (starch sheath) is a single layer of cells with starch grains. • The endodermis has a casparian strip which has an impervious deposit controlling the entry of water and mineral salts into xylem vessels. • Pericyc1e forms a layer next to the endodermis. • Next to the pericycle is the vascular tissue. • In the Dicotyledonous root, xylem forms a star shape in the centre, with phloem in between the arms. • It has no pith. In monocotyledonous root, xylem alternates with phloem and there is a pith in the centre. Internal structure of a root hair cell The Stem • The main functions of the stem are; • support and exposure of leaves and flowers to the environment, • conducting water and mineral salts • conducting manufactured food from leaves to other parts of the plant. • In monocotyledonous stems, vascular bundles are scattered all over the stem, while in dicotyledonous stems vascular bundles are arranged in a ring. • Vascular bundles are continuous from root to stems and leaves. • The epidermis forms a single layer of cells enclosing other tissues. • The outer walls of the cells have waxy cuticle to prevent excessive loss of water. • The cortex is a layer next to the epidermis. • It has collenchyma, parenchyma and schlerenchyma cells. • Is next to the epidermis and has thickened walls at the corners which strengthen the stem. • Cells are irregular in shape, thin walled and loosely arranged hence creating intercellular spaces filled with air. • They are packing tissues and food storage areas. • Cells are closely connected to vascular bundles. • These cells are thickened by deposition of lignin and they provide support to plants. • Is the central region having parenchyma cells. Absorption of Water and Mineral Salts Absorption of Water • Root hair cell has solutes in the vacuole and hence a higher osmotic pressure than the surrounding soil water solution. • Water moves into the root hair cells by osmosis along a concentration gradient. • This makes the sap in the root hair cell to have a lower osmotic pressure than the surrounding cells. • Therefore water moves from root hair cells into the surrounding cortex cells by osmosis. • The process continues until the water gets into the xylem vessels . Uptake of Mineral Salts • If the concentration of mineral salts in solution is greater than its concentration in root hair cell, the mineral salts enter the root hair cell by diffusion. • If the concentration of mineral salts in the root hair cells is greater than in the soil water, the mineral salts enter the root hairs by active transport. • Most minerals are absorbed in this way. • Mineral salts move from cell to cell by active transport until they reach the xylem vessel. • Once inside the xylem vessels, mineral salts are transported in solution as the water moves up due to root pressure, capillary attraction and cohesion and adhesion forces. • Transpiration is the process by which plants lose water in the form of water vapour into the atmosphere. • Water is lost through stomata, cuticle and lenticels. • Stomatal transpiration: • This accounts for 80-90% of the total transpiration in plants. • Stomata are found on the leaves. • Cuticular transpiration: • The cuticle is found on the leaves, and a little water is lost through it. • Plants with thick cuticles do not lose water through the cuticle. • Lenticular transpiration • Is loss’ of water through lenticels. • These are found on stems of woody plants. • Water lost through the stomata and cuticle by evaporation leads to evaporation of water from surfaces of mesophyll cells . • The mesophyll cells draw water from the xylem vessels by osmosis. • The xylem in the leaf is continuous with xy lem in the stem and root. Structure and function of Xylem • Movement of water is through the xylem. • Xylem tissue is made up of vessels and tracheids. Xylem Vessels • Xylem vessels are formed from cells that are elongated along the vertical axis and arranged end to end. • During development, the cross walls and organelles disappear and a continuous tube is formed. • The cells are dead and their walls are strengthened by deposition of lignin. • The lignin has been deposited in various ways. • This results in different types of thickening • Simple spiral. • Double spiral. • The bordered pits are areas without lignin on xylem vessels and allow passage of water in and out of the lumen to neighbouring cells. • Tracheids have cross-walls that are perforated. • Their walls are deposited with lignin. • Unlike the xylem vessels, their end walls are tapering or chisel-shaped. • Their lumen is narrower. • Besides transport of water, xylem has another function of strengthening the plant which is provided by xylem fibres and xylem parenchyma. Xylem fibres ; • Are cells that are strengthened with lignin. • They form wood. Xylem parenchyma: • These are cells found between vessels. • They form the packing tissue. Forces involved in Transportation of Water and Mineral Salts Transpiration pull • As water vaporises from spongy mesophyll cells into sub-stomatal air spaces, the cell sap of mesophyll cells develop a higher osmotic pressure than adjacent cells. • Water is then drawn into mesophyll cells by osmosis from adjacent cells and finally from xylem vessels. • A force is created in the leaves which pulls water from xylem vessels in the stem and root. • This force is called transpiration pull . Cohesion and Adhesion: • The attraction between water molecules is called cohesion. • The attraction between water molecules and the walls of xylem vessels is called adhesion. • The forces of cohesion and adhesion maintain a continuous flow of water in the xylem from the root to the leaves. • Is the ability of water to rise in fine capillary tubes due to surface tension. • Xylem vessels are narrow, so water moves through them by capillarity. Root Pressure: • If the stem of a plant is cut above the ground level, it is observed that cell sap continues to come out of the cut surface. • This shows that there is a force in the roots that pushes water up to the stem. • This force is known as root pressure. Importance of Transpiration • Transpiration leads to excessive loss of water if unchecked. Some beneficial effects are: • Replacement of water lost during the process. • Movement of water up the plant is by continuous absorption of water from the soil. • Mineral salts are transported up the plant. • Transpiration ensures cooling of the plant in hot weather. • Excessive loss of water leads to wilting’ and eventually death if water is not available in the soil. Factors Affecting Transpiration The factors that affect transpiration are grouped into two. • e. environmental and structural. Environmental factors • High temperature increases the internal temperature of the leaf . • which in turn increases kinetic energy of water molecules which increases evaporation. • High temperatures dry the air around the leaf surface maintaining a high concentration gradient. • More water vapour is therefore lost from the leaf to the air. • The higher the humidity of the air around the leaf, the lower the rate of transpiration. • The humidity difference between the inside of the leaf and the outside is called the saturation deficit. • In dry atmosphere, the saturation deficit is high. • At such times, transpiration rate is high. • Wind carries away water vapour as fast as it diffuses out of the leaves. • This prevents the air around the leaves from becoming saturated with vapour. • On a windy day, the rate of transpiration is high. Light Intensity • When light intensity is high; more stomata open hence high rate of transpiration. Atmospheric Pressure • The lower the atmospheric pressure the higher the kinetic energy of water molecules hence more evaporation. • Most of the plants at higher altitudes where atmospheric pressure is very low have adaptations to prevent excessive water-loss. Availability of Water • The more water there is in the soil, the more is absorbed by the plant and hence a lot of water is lost by transpiration. Structural Factors • Plants growing in arid or semi-arid areas have leaves covered with a thick waxy cuticle. • The more the stomata, the higher the rate of transpiration. • Xerophytes have few stomata which reduce water-loss. • Some have sunken stomata which reduces the rate of transpiration as the water vapour accumulates in the pits. • Others have stomata on the lower leaf surface hence reducing the rate of water-loss. • Some plants have reversed stomatal rhythm whereby stomata close during the day and open at night. • This helps to reduce water-loss. Leaf size and shape • Plants in wet areas have large surface area for transpiration. • Xerophytes have small narrow leaves to reduce water-loss. • The photometer can be used to determine transpiration in different environmental conditions. Translocation of organic compounds • Translocation of soluble organic products of photosynthesis within a plant is called translocation. • It occurs in phloem in sieve tubes. • Substances translocated include glucose, amino acids, vitamins. • These are translocated to the growing regions like stem, root apex, storage organs e.g. corms, bulbs and secretory organs such as nectar glands. phloem is made up of; • sieve tubes, • companion cells • parenchyma, a packing tissue • schlerenchyma, a strengthening tissue Sieve Tubes • These are elongated cells arranged end to end along the vertical axis. • The cross walls are perforated by many pores to make a sieve plate. • Most organelles disappear and those that remain are pushed to the sides of the sieve tube. • Cytoplasmic strands pass through the pores in the plate into adjacent cells. • Food substances are translocated through cytoplasmic strands. Companion Cells • Companion cells are small cells with large nuclei and many mitochondria. • They are found alongside each sieve element. • The companion cell is connected to the tube through plasmodesmata. • The mitochondria generate energy required for translocation. Phloem Parenchyma • These are parenchyma cells between sieve elements. • They act as packing tissue. Transport in Animals The Circulatory System • Large and complex animals have circulatory systems that consist of tubes, a transport fluid and a means of pumping the fluid. • Blood is the transport fluid which contains dissolved substances and cells. • The tubes are blood vessels through which dissolved substances are circulated around the body. • The heart is the pumping organ which keeps the blood in circulation. The types of circulatory system exist in animals: open and closed. • In an open circulatory system; • The heart pumps blood into vessels which open into body spaces known as haemocoel. • Blood comes into contact with tissues. • A closed circulatory system; • Found in vertebrates and annelids where the blood is confined within blood vessels and does not come into direct contact with tissues. Transport in Insects • In an insect, there is a tubular heart just above the alimentary canal. • This heart is suspended in a pericardial cavity by ligaments. • The heart has five chambers and extends along the thorax and abdomen . • Blood is pumped forwards into the aorta by waves of contractions in the heart. • It enters the haem ocoel and flows towards the posterior. • The blood flows back into the heart through openings in each chamber called ostia. • The ostia have valves which prevent the backflow of blood. • Blood is not used as a medium for transport of oxygen in insects. • This is because oxygen is supplied directly to the tissues by the tracheal system. • The main functions of blood in an insect are to transport nutrients, excretory products and hormones. Mammalian Circulatory System • Mammals have a closed circulatory system where a powerful heart pumps blood into arteries. • The arteries divide into smaller vessels called arterioles. • Each arteriole divides to form a network of capillaries inside the tissues. • The capillaries eventually re-unite to form venules, which form larger vessels called veins. • The veins take the blood back to the heart. • Blood from the heart goes through the pulmonary artery to the lungs and then back to the heart through pulmonary vein. • This circulation is called pulmonary circulation. • Oxygenated blood leaves the heart through the aorta and goes to all the tissues of the body. • From the tissues, deoxygenated blood flows back to the heart through the vena cava. • This circulation is called systemic circulation. • In each complete circulation, the blood flows into the heart twice. • This is called double circulation. • Some other animals like fish have a single circulation. • Blood flows only once through the heart for every complete circuit. Structure and Function of the Heart • The heart has four chambers: • Two artria (auricles) and two ventricles. • The left and right side of the heart are separated by a muscle wall (septum) so that oxygenated and deoxygenated blood does not mix. • Deoxygenated blood from the rest of the body enters the heart through the vena cava . • Blood enters the right atrium, then through tricuspid valve into right ventricle. • Then via semi-lunar valve to the pulmonary artery to the lungs. • Oxygenated blood from the lungs enters the heart through pulmonary vein. • It enters the left atrium of the heart, then through bicuspid valve into left ventricle. • Then via semi-lunar valves to aorta which takes oxygenated blood round the body. • A branch of the aorta called coronary artery supplies blood to the heart muscle. • The coronary vein carries blood from the heart muscle to the pulmonary artery which then takes it to the lungs for oxygenation. Pumping Mechanism of the heart • The heart undergoes contraction (systole) and relaxation ( diastole). • When the ventricular muscles contract, the cuspid valves (tricuspid and bicuspid) close preventing backflow of blood into auricles. • The volume of the ventricles decreases while pressure increases. • This forces blood out of the heart to the lungs through semi-lunar valves and pulmonary artery, and to the body tissues via semi-lunar valve and aorta respectively. • At the same time the atria are filled with blood. • The left ventricle has thicker muscles than the right ventricle, and pumps blood for a longer distance to the tissues. • When ventricular muscles relax, the volume of each ventricle increases while pressure decreases. • Contractions of atria force the bicuspid and tricuspid valves to open allowing deoxygenated blood from right atrium into right ventricle which oxygenated blood flows from left atrium into the left ventricle. • Semi-lunar valves close preventing the backflow of blood into ventricles. • The slight contractions of atria force the , blood flow into ventricles. The Heartbeat • The heart is capable of contracting and relaxing rhythmically without fatigue due to its special muscles called cardiac muscles. • The rhythmic contraction of the heart arise from within the heart muscles without nervous stimulation. • The contraction is said to be myogenic. • The heartbeat is initiated by the pacemaker or sino-artrio-node (SAN) which is located in the right atrium. • The wave of excitation spreads over the walls of atria. • It is picked by the artrio-ventricular node which is located at the junction: • Of the atria and ventricles, from where the purkinje tissue spreads the wave to the walls of the ventricles. • The heart contracts and relaxes rhythmically at an average rate of 72 times per minute. • The rate of the heartbeat is increased by the sympathetic nerve, while it is slowed down by the vagus nerve. • Heartbeat is also affected by hormones e.g. adrenaline raises the heartbeat. Structure and Function of Arteries,Capillaries and Veins • Arteries carry blood away from the heart. • They carry oxygenated blood except pulmonary artery which carries deoxygenated blood to the lungs. • Arteries have a thick, muscular wall, which has elastic and collagen fibres that resist the pressure of the blood flowing in them. • The high pressure is due to the pumping action of the heart. • The pressure in the arteries originate from the pumping action of the heart. • The pulse or number of times the heart beats per minute can be detected by applying pressure on an artery next to the bone. • g. by placing the finger/thumb on the wrist. • The innermost layer of the artery is called endothelium which is smooth. • It offers least possible resistance to blood flow. • Have a narrow lumen . • The aorta forms branches which supply blood to all parts of the body. • These arteries divide into arterioles which further divide to form capillaries. • Capillaries are small vessels whose walls are made of endothelium which is one cell thick. • This provides a short distance for exchange of substances. • Capillaries penetrate tissues, • The lumen is narrow therefore blood flowing in capillaries is under high pressure. • Pressure forces water and dissolved substances out of the blood to form tissue fluid. • Exchange of substances occurs between cells and tissue fluid. • Part of the tissue fluid pass back into capillaries at the venule end. • Excess fluid drains into small channels called lymph capillaries which empty their contents into lymphatic vessels. • Capillaries join to form larger vessels called venules which in turn join to form veins which transport blood back to the heart. • Veins carry deoxygenated blood from the tissues to the heart (except pulmonary vein which carries oxygenated blood from the lungs to the heart). • Veins have a wider lumen than arteries. • Their walls are thinner than those of arteries. • Blood pressure in the veins is low. • Forward flow of blood in veins is assisted by contraction of skeletal muscles, hence the need for exercise. • Veins have valves along their length to prevent backflow of blood. • This ensures that blood flows towards the heart. • The way the valves work can be demonstrated on the arm. • By pressing on one vein with two fingers, leaving one and pushing blood toward the heart then releasing the latter finger, it can be observed that the part in between is left with the vein not being visible. • This is because bleed does not flow back towards the first finger. Diseases and Defects of Circulatory System • Formation of a clot in the blood vessels is called thrombosis. • Coronary thrombosis is the most common. • It is caused by blockage of coronary artery which supplies blood to the heart. • Blockage may be due to artery becoming fibrous or accumulation of fatty material on the artery walls. • Narrow coronary artery results in less blood reaching the heart muscles. • A serious blockage can result in heart attack which can be fatal. • Heavy intake of fat, alcohol, being overweight and emotional stress can cause coronary thrombosis. • A blockage in the brain can lead to a stroke causing paralysis of part of the body, coma or even death. • A healthy lifestyle, avoiding a lot of fat in meals and avoiding alcohol can control the • This condition results from the inner walls having materials being deposited there or growth of fibrous connective tissue. • This leads to thickening of the wall of the artery and loss of elasticity. • Normal blood flow is hindered. • Arteriosclerosis can lead to thrombosis or hypertension. • A person with hypertension which is also called high blood pressure has his/her blood being pumped more forcefully through the narrow vessels. • This puts stress on the walls of the heart and arteries. • Regular exercise, healthy diet and avoiding smoking can help maintain normal blood pressure. Varicose Veins • Superficial veins especially at the back of the legs become swollen and flabby due to some valves failing to function properly. • This results to retention of tissue fluid. • Regular physical exercise will prevent this condition. • Repair of valves through surgery can also be done. • Wearing surgical stockings may ease a mild occurence. Structure and Function of Blood Composition of Blood • The mammalian blood is made up of a fluid medium called plasma with substances dissolved in it. • Cellular components suspended in plasma include; • erythrocytes (red blood cells), • leucocytes (white blood cells) • thrombocytes (platelets) • blood proteins. • This is a pale yellow fluid consisting of 90% water. • There are dissolved substances which include; □ glucose, amino acids, lipids, salts, □ hormones, urea, fibrinogen, albumen, □ antibodies, some enzymes suspended cells. ☆ Serum is blood from which fibrinogen and cells have been removed. The functions of plasma include: • Transport of red blood cells which carry oxygen. • Transport dissolved food substances round the body. • Transport metabolic wastes like nitrogenous wastes and carbon (IV) oxide in solution about 85% of the carbon (IV) oxide is carried in form of hydrogen carbonates. • Transport hormones from sites of production to target organs. • Regulation of pH of body fluids. • Distributes heat rou nd the body hence regulate body temperature. Erythrocytes (Red Blood Cells) • In humans these cells are circular biconcave discs without nuclei. • Absence of nucleus leaves room for more haemoglobin to be packed in the cell to enable it to carry more oxygen. • Haemoglobin contained in red blood cells is responsible for the transport of oxygen. • Haemoglobin + Oxygen =oxyhaemoglobin • (Hb) + (4O[2]) __ (HbO[g]) • Oxygen is carried in form of oxyhaemoglobin. • Haemoglobin readily picks up oxygen in the lungs where concentration of oxygen is high. • In the tissues, the oxyhaemoglobin breaks down (dissociates) easily into haemoglobin and oxygen. • Oxygen diffuses out of the red blood cells into the tissues. • Haemoglobin is then free to pick up more oxygen molecules. • The biconcave shape increases their surface area over which gaseous exchange takes place. • Due to their ability, they are able to change their shape to enable themselves squeeze inside the narrow capillaries. • Co2+ H2O carbonic anhydrase • There are about five million red blood cells per cu bic millimetre of blood. • They are made in the bone marrow of the short bones like sternum, ribs and vertebrae. • In the embryo they are made in the liver and spleen. • Erythrocytes have a life span of about three to four months after which they are destroyed in the liver and spleen. • Also in the red blood cells is carbonic anhydrase which assists in the transport of carbon (IV) oxide. Leucocytes (White Blood Cells) • These white blood cells have a nucleus. • They are divided into two: • Granulocytes (also phagocytes or polymorphs) • Agranulocytes . • White blood cells defend the body against disease. • Neutrophils form 70% of the granulocytes. • Others are eosinophils and basophils. • About 24% agronulocytes are called lymphocytes, while 4% agranulocytes are monocytes. • The leucocytes are capable of amoebic movement. • They squeeze between the cells of the capillary wall to enter the intercellular spaces. • They engulf and digest disease causing organisms (pathogens) by phagocytosis. • Some white blood cells may die in the process of phagocytosis. • The dead phagocytes, dead organisms and damaged tissues form pus. • Lymphocytes produce antibodies which inactivate antigens. Antibodies include: • Antitoxins which neutralise toxins. • Agglutinins cause bacteria to clump together and they die. • Lysins digest cell membranes of microorganisms. • Opsonins adhere to outer walls of microorganisms making it easier for phagocytes to ingest them. • Lymphocytes’ are made in the thymus gland and lymph nodes. • There are about 7,000 leucocytes per cubic millimetre of blood. Platelets (Thrombocytes) • Platelets are small irregularly shaped cells formed from large bone marrow cells called megakaryocytes. • There are about 250,000 platelets per cubic millimetre of blood. • They initiate the process of blood clotting. • The process of clotting involves a series of complex reactions whereby fibrinogen is converted into a fibrin clot. • When blood vessels are injured platelets are exposed to air and they release thromboplastin (thrombokinasewhich initiates the blood clotting process. • Thromboplastin neutralises heparin the anti-clotting factor in blood and activates prothrombin to thrombin. • The process requires calcium ions and vitamin K. • Thrombin activates the conversion of fibrinogen to fibrin which forms a meshwork of fibres on the cut surface to trap red blood cells to form a clot. • The clot forms a scab that stops bleeding and protects the damaged tissues from entry of micro-organisms. • Blood clotting reduces loss of blood when blood vessels are injured. • Excessive loss of blood leads to anaemia and dehydration. • Mineral salts lost in blood leads to osmotic imbalance in the body. • This can be corrected through blood transfusion and intravenous fluid. ABO Blood Groups • There are four types of blood groups in human beings: A, B, AB and O. • These are based on types of proteins on the cell membrane of red blood cells. • There are two types of proteins denoted by the letters A and B which are antigens. • In the plasma are antibodies specific to these antigens denoted as a and • A person of blood group A has A antigens on the red blood cells and b antibodies in plasma. • A person of blood group B has B antigens on red blood cells and a antibodies in plasma. • A person of blood group AB has A and B antigens on red blood cells and no antibodies in plasma . • A person of blood group a has no antigens on red blood cells and a and b antibodies in plasma. Blood groups Blood Groups Antigens Antibodies A A b B B a AB AandB None 0 None a and b Blood Transfusion Blood transfusion is the transfer of blood from a donor to the circulatory system of the recipient. A recipient will receive blood from a donor if the recipient has no corresponding antibodies to the donor’s antigens. If the donor’s blood and the recipient’s blood are not compatible, agglutination occurs whereby red blood cells clump together. Blood typing • A person of blood group 0 can donate blood to a person of any other blood group. • A person of blood group 0 is called a universal donor. • A person of blood group AB can receive blood from any other group. • A person with blood group AB is called a universal recipient. • A person of blood group A can only donate blood to another person with blood group A or a person with blood group AB. • A person of blood group B can only donate blood to somebody with blood group B or a person with blood group AB. • A person with blood group AB can only donate blood to a person with blood groupAB. • Blood screening has become a very important step in controlling HIV/AIDS. • It is therefore important to properly screen blood before any transfusion is done. Rhesus Factor • The Rhesus factor is present in individuals with the Rhesus antigen in their red blood cells. • Such individuals are said to be Rhesus positive (Rh+), while those without the antigen are Rhesus negative (Rh-). • If blood from an Rh+ individual is introduced into a person who is Rh- , the latter develops antibodies against the Rhesus factor. • There may not be any reaction after this transfusion. • However a subsequent transfusion with Rh+ blood causes a severe reaction, and agglutination occurs i.e. clumping of red blood cells. • The clump can block the flow of blood, and cause death. • Erythroblastosis foetalis (haemolytic disease of the newborn) results when an Rh- mother carries an Rh+ foetus. • This arises when the father is Rh+. • During the latter stage of pregnancy, fragments of Rhesus positive red blood cells of the foetus may enter mother’s circulation. • These cause the mother to produce Rhesus antibodies which can pass across the. placenta to the foetus and destroy foetal red blood cells. • During the first pregnancy, enough antibodies are not formed to affect the foetus. • Subsequent pregnancies result in rapid production of Rhesus antibodies by the mother. • These destroy the red blood cells of the foetus, the condition called haemolytic disease of the newborn. • The baby is born anaemic and with yellow eyes (jaundiced). • The condition can be corrected by a complete replacement of baby’s blood with safe healthy blood. Lymphatic System • The lymphatic system consists of lymph vessels. • Lymph vessels have valves to ensure unidirectional movement of lymph. • Lymph is excess tissue fluid i.e. blood minus blood cells and plasma proteins. • Flow of lymph is assisted by breathing and muscular contractions. • Swellings called lymph glands occur at certain points along the lymph vessels. • Lymph glands are oval bodies consisting of connective tissues and lymph spaces. • The lymph spaces contain lymphocytes which are phagocytic. • Lymph has the same composition as blood except that it does not contain red blood cells and plasma proteins. • Lymph is excess tissue fluid. • Excess tissue fluid is drained into lymph vessels by hydrostatic pressure. • The lymph vessels unite to form major lymphatic system. • The main lymph vessels empty the contents into sub-clavian veins which take it to the heart. Immune Responses • Immune response is the production of antibodies in response to antigens. • An antigen is any foreign material or organism that is introduced into the body and causes the production of antibodies. • Antigens are protein in nature. • An antibody is a protein whose structure is complementary to the antigen. • This means that a specific antibody deals with a specific antigen to make it harmless. • When harmful organisms or proteins invade the body, lymphocytes produce complementary antibodies, while bone marrow and thymus gland produce more phagocytes and lymphocytes respectively. Types of Immunity • There are two types of immunity; natural and artificial. Natural Immunity is also called innate immunity. • It is inherited from parent to offspring. Artificial Immunity can be natural or induced. • When attacked by diseases like chicken pox, measles and mumps, those who recover from these diseases develop resistance to any subsequent infections of the same diseases. • This is natural acquired immunity. Artificial Acquired Immunity: • When attenuated (weakened) or dead microorganisms are introduced into a healthy person. • The lymphocytes synthesis the antibodies which are released into the lymph and eventually reach the blood. • The antibodies destroy the invading organisms. • The body retains ‘memory’ of the structure of antigen. • Rapid response is ensured in subsequent infections. • Vaccines generally contain attenuated disease causing organisms. Artificial Passive Acquired Immunity: • Serum containing antibodies is obtained from another organism, and confers immunity for a short duration. • Such immunity is said to be passive because the body is not activated to produce the antibodies. Importance of Vaccination • A vaccine is made of attenuated, dead or nonvirulent micro-organism that stimulate cells in the immune system to recognise and attack disease causing agent through production of antibodies. • Vaccination protects individuals from infections of many diseases like smallpox, tuberculosis and poliomyelitis. • Diseases like smallpox, tuberculosis and tetanus were killer diseases but this is no longer the case. • Diphtheria Pertussis Tetanus (DPT) vaccine protects children against diphtheria, whooping cough and tetanus. • Bacille Calmette Guerin (BCG) vaccine is injected at birth to children to protect them against tuberculosis. • Measles used to be a killer disease but today, a vaccine injected into children at the age of rune months prevents it. • At birth children are given an inoculation through the mouth of the poliomyelitis vaccine. Allergic Reactions • An allergy is a hypersensitive reaction to an antigen by the body. • The antibody reacts with the antigen violently. • People with allergies are oversensitive to foreign materials like dust, pollen grains, some foods, some drugs and some air pollutants. • Allergic reactions lead to production of histamine by the body. • Histamine causes swelling and pain. • Allergic reactions can be controlled by avoiding the allergen and administration of anti-histamine drugs. Meaning and Significance of Respiration • Respiration is the process by which energy is liberated from organic compounds such as glucose. • It is one of the most important characteristics of living organisms. • Energy is expended (used) whenever an organism exhibits characteristics of life, such as feeding, excretion and movement. • Respiration occurs all the time and if it stops, cellular activities are disrupted due to lack of energy. • This may result in death e.g., if cells in brain lack oxygen that is needed for respiration for a short time, death may occur. • This is because living cells need energy in order to perform the numerous activities necessary to maintain life. • The energy is used in the cells and much of it is also lost as heat. • In humans it is used to maintain a constant body temperature. Tissue Respiration • Respiration takes place inside cells in all tissues. • Every living cell requires energy to stay alive. • Most organisms require oxygen of the air for respiration and this takes place in the mitochondria. Mitochondrion Structure and Function • Mitochondria are rod-shaped organelles found in the cytoplasm of cells. • A mitochondrion has a smooth outer membrane and a folded inner membrane. • The folding of the inner membrane is called cristae and the inner compartment is called the matrix. Adaptations of Mitochondrion to its Function • The matrix contains DNA ribosomes for making proteins and has enzymes for the breakdown of pyruvate to carbon (IV) oxide, hydrogen ions and electrons. • Cristae increase surface area of mitochondrial inner membranes where attachment of enzymes needed for the transport of hydrogen ions and electrons are found. • There are two types of respiration: • Aerobic Respiration • Respiration Aerobic Respiration • This involves breakdown of organic substances in tissue cells in the presence of oxygen . • All multicellular organisms and most unicellular organisms e.g. some bactena respire aerobically. • In the process, glucose is fully broken down to carbon (IV) oxide and hydrogen which forms water when it combines with the oxygen. • Energy produced is used to make an energy rich compound known as adenosine triphosphate (ATP). • It consists of adenine, an organic base, five carbon ribose-sugar and three phosphate groups. • ATP is synthesised from adenosine diphosphate (ADP) and inorganic phosphate. • The last bond connecting the phosphate group is a high-energy bond. • Cellular activities depend directly on ATP as an energy source. • When an ATP molecule is broken down, it yields energy. Process of Respiration • The breakdown of glucose takes place in many steps. • Each step is catalysed by a specific enzyme. • Energy is released in some of these steps and as a result molecules of ATP are synthesised. • All the steps can be grouped into three main stages: • The initial steps in the breakdown of glucose are referred to as glycolysis and they take place in the cytoplasm. • Glycolysis consists of reactions in which glucose is gradually broken down into molecules of a carbon compound called pyruvic acid or pyruvate. • Before glucose can be broken, it is first activated through addition of energy from ATP and phosphate groups. • This is referred to as phosphorylation. • The phosphorylated sugar is broken down into two molecules of a 3-carbon sugar (triose sugar) each of which is then converted into pyruvic acid. • If oxygen is present, pyruvic acid is converted into a 2-carbon compound called acetyl coenzyme A (acetyl Co A). • Glycolysis results in the net production of two molecules of ATP. • The next series of reactions involve decarboxylation i.e. removal of carbon as carbon (IV) oxide and dehydrogenation, removal of hydrogen as hydrogen ions and electrons. • These reactions occur in the mitochondria and constitute the Tri-carboxylic Acid Cycle (T.C.A.) or Kreb’s citric acid cycle. • The acetyl Co A combines with 4-carbon compound with oxalo-acetic acid to form citric acid – a 6 carbon compound. • The citric acid is incorporated into a cyclical series of reactions that result in removal of carbon (IV) oxide molecules, four pairs of hydrogen, ions and electrons. • Hydrogen ions and electrons are taken to the inner mitochondria membrane where enzymes and electron carriers effect release of a lot of energy. • Hydrogen finally combines with oxygen to form water, and 36 molecules of ATP are synthesised. Anaerobic Respiration • Anaerobic respiration involves breakdown of organic substances in the absence of oxygen. • It takes place in some bacteria and some fungi. • Organisms which obtain energy by anaerobic respiration are referred to as anaerobes. • Obligate anaerobes are those organisms which do not require oxygen at all and may even die if oxygen is present. • Facultative anaerobes are those organisms which survive either in the absence or in the presence of oxygen. • Such organisms tend to thrive better when oxygen is present e.g. yeast. Products of Anaerobic Respiration • The products of anaerobic respiration differ according to whether the process is occurring in plants or animals. Anaerobic Respiration in Plants • Glucose is broken down to an alcohol, (ethanol) and carbon (IV) oxide. • The breakdown is incomplete. • Ethanol is an organic compound, which can be broken down further in the presence of oxygen to provide energy, carbon (IV) oxide and water. C[6]H[I2]0[6] _ 2C[2]H[5]0H + 2C0[2] + Energy (Glucose) (Ethanol) (Carbon (IV) oxide) • Is the term used to describe formation of ethanol and carbon (IV) oxide from grains. • Yeast cells have enzymes that bring about anaerobic respiration. Lactate Fermentation • Is the term given to anaerobic respiration in certain bacteria that results in formation of lactic acid. Anaerobic Respiration in Animals • Anaerobic respiration in animals produces lactic acid and energy. C[6]H[1]P6 _ 2CH[3]CHOH.COOH + energy (Glucose) (Lactic acid) + energy • When human muscles are involved in very vigorous activity, oxygen cannot be delivered as rapidly as it is required. • The muscle respire anaerobically and lactic acid accumulates. • A high level of lactic acid is toxic. • During the period of exercise, the body builds up an oxygen debt. • After vigorous activity, one has to breathe faster and deeper to take in more oxygen. • Rapid breathing occurs in order to break down lactic acid into carbon (IV) oxide and water and release more energy. • Oxygen debt therefore refers to the extra oxygen the body takes in after vigorous exercise. Practical Activities To Show the Gas Produced When the Food is burned • A little food substance e.g., maize flour or meat is placed inside a boiling tube. • The boiling tube is stoppered using a rubber bung connected to a delivery tube inserted into a test-tube with limewater. • The food is heated strongly to bum. • Observations are made on the changes in lime water (calcium hydroxide) as gas is produced. • The clear lime water turns white due to formation of calcium carbonate precipitate proving that carbon (Iv) oxide is produced. Experiment to Show the Gas Produced During Fermentation • Glucose solution is boiled and cooled. Boiling expels all air. • A mixture of glucose and yeast is placed in a boiling tube, and covered with a layer of oil to prevent entry of air. • A delivery tube is connected and directed into a test-tube containing lime water. • The observations are made immediately and after three days the contents are tested for the presence of ethanol. • A control experiment is set in the same way except that yeast which has been boiled and cooled is used. • Boiling kills yeast cells. • The limewater becomes cloudy within 20 minutes. • This proves that carbon (IV) oxide gas is produced. • The fermentation process is confirmed after three days when alcohol smell is detected in the mixture. Experiment to Show Germinating Seeds Produce Heat • Soaked bean seeds are placed in a vacuum flask on wet cotton wool. • A thermometer is inserted and held in place with cotton wool . • The initial temperature is taken and recorded. • A control experiment is set in the same way using boiled and cooled bean seeds which have been washed in formalin to kill microorganisms. • Observation is made within three days. • Observations show that temperature in the flask with germinating seeds has risen. • The one in the control has not risen. Comparison Between Aerobic and Anaerobic Respiration Aerobic Respiration Anaerobic Respiration 1. Site In the mitochondria. In the cytoplasm. 2. Products Carbon dioxide and water. Ethanol in plants and lactic acid in animals- 3. Energy yield 38 molecules of A TP (2880 KJ) from 2 molecules of ATP 210KJ from each each molecule of glucose. molecule of glucose. 4. Further reaction No further reactions on carbon Ethanol and lactic acid can be broken down dioxide and water. further in the presence of oxygen. Comparison Between Energy Output in Aerobic and Anaerobic Respiration • Aerobic respiration results in the formation of simple inorganic molecules, water and carbon (Iv) oxide as the byproducts. • These cannot be broken down further. A lot of energy is produced. • When a molecule of glucose is broken down in the presence of oxygen, 2880 KJ of energy are produced (38 molecules of ATP). • In anaerobic respiration the by products are organic compounds. • These can be broken down further in the presence of oxygen to give more energy. • Far less energy is thus produced. • The process is not economical as far as energy production is concerned. • When a molecule of glucose is broken down in the absence of oxygen in plants, 210 KJ are produced (2 molecule ATP). • In animals, anaerobic respiration yields 150 kJ of energy. Substrates for Respiration • Carbohydrate, mainly glucose is the main substrate inside cells. • Lipids i.e. fatty acids and glycerol are also used. • Fatty acids are used when the carbohydrates are exhausted. • A molecule of lipid yields much more energy than a molecule of glucose. • Proteins are not normally used for respiration. • However during starvation they are hydrolysed to amino acids, dearnination follows and the products enter Kreb’s cycle as urea is formed. • Use of body protein in respiration result to body wasting, as observed during prolonged sickness or starvation. • The ratio of the amount of carbon (IV) oxide produced to the amount of oxygen used for each substrate is referred to as Respiratory Quotient (RQ) and is calculated as follows: R.Q. = Amount of carbon (IV) oxide produced Amount of oxygen used • Carbohydrates have a respiratory quotient of 1.0 lipids 0.7 and proteins 0.8. • Respiratory quotient value can thus give an indication of types of substrate used. • Besides values higher than one indicate that some anaerobic respiration is taking place. Application of Anaerobic Respiration in Industry and at Home • Making of beer and wines. • Ethanol in beer comes from fermentation of sugar(maltose) in germinating barley seeds. • Sugar in fruits is broken down anaerobically to produce ethanol in wines. • In the dairy industry, bacterial fermentation occurs in the production of several dairy products such as cheese, butter and yoghurt. • In production of organic acids e.g., acetic acid, that are used in industry e.g., in preservation of foods. • Fermentation of grains is used to produce all kinds of beverages e.g., traditional beer and sour porridge. • Fermentation of milk. End of Topic Necessity for Gaseous Exchange in Living Organisms • Living organisms require energy to perform cellular activities. • The energy comes from breakdown of food in respiration. • Carbon (IV) oxide is a by product of respiration and its accumulation in cells is harmful which has to be removed. • Most organisms use oxygen for respiration which is obtained from the environment. • Photosynthetic cells of green plants use carbon (Iv) oxide as a raw material for photosynthesis and produce oxygen as a byproduct. • The movement of these gases between the cells of organisms and the environment comprises gaseous exchange. • The process of moving oxygen into the body and carbon (Iv) oxide out of the body is called breathing or ventilation. • Gaseous exchange involves the passage of oxygen and carbon (IV) oxide through a respiratory surface. • Diffusion is the main process involved in gaseous exchange. Gaseous Exchange in Plants • Oxygen is required by plants for the production of energy for cellular activities. • Carbon (IV) oxide is required as a raw material for the synthesis of complex organic substances. • Oxygen and carbon (IV) oxide are obtained from the atmosphere in the case of terrestrial plants and from the surrounding water in the case of aquatic plants. • Gaseous exchange takes place mainly through the stomata. Structure of Guard Cells • The stoma (stomata – plural) is surrounded by a pair of guard cells. • The structure of the guard cells is such that changes in turgor inside the cell cause changes in their shape. • They are joined at the ends and the cell walls facing the pore (inner walls) are thicker and less elastic than the cell walls farther from the pore (outer wall). • Guard cells control the opening and closing of stomata. Mechanism of Opening and Closing of Stomata • In general stomata open during daytime (in light) and close during the night (darkness). • Stomata open when osmotic pressure in guard cells becomes higher than that in surrounding cells due to increase in solute concentration inside guard cells. Water is then drawn into guard cells by • Guard cells become turgid and extend. • The thinner outer walls extend more than the thicker walls. • This causes a bulge and stoma opens. • Stomata close when the solute concentration inside guard cells become lower than that of surrounding epidermal cells. • The water moves out by osmosis, and the guard cells shrink i.e. lose their turgidity and stoma closes. Proposed causes of turgor changes in guard cells. Accumulation of sugar. • Guard cells have chloroplasts while other epidermal cells do not. • Photosynthesis takes place during daytime and sugar produced raises the solute concentration of guard cells. • Water is drawn into guard cells by osmosis from surrounding cells. • Guard cells become turgid and stoma opens. • At night no photosynthesis occurs hence no sugar is produced. • The solute concentration of guard cells falls and water moves out of the guard cells by osmosis. • Guard cells lose turgidity and the stoma closes. pH changes in guard cells occur due to photosynthesis. • In day time carbon (IV) oxide is used for photosynthesis. This reduces acidity while the oxygen produced increases alkalinity. • Alkaline pH favours conversion of starch to sugar. • Solute concentration increases inside guard cells, water is drawn into the cells by osmosis. Guard cells become turgid and the stoma opens. • At night when no photosynthesis, Respiration produces carbon (IV) oxide which raises acidity .This favours conversion of sugar to starch. low sugar concentration lead to loss of turgidity in guard cells and stoma closes. Explanation is based on accumulation of potassium • In day time (light) adenosine triphosphate (ATP) is produced which causes potassium ions to move into guard cells by active transport. • These ions cause an increase in solute concentration in guard cells that has been shown to cause movement of water into guard cells by osmosis. • Guard cells become turgid and the stoma opens. • At night potassium and chloride ions move out of the guard cells by diffusion and level of organic acid also decreases. • This causes a drop in solute concentration that leads to movement of water out of guard cells by osmosis. • Guard cells lose turgor and the stoma closes. Process of Gaseous Exchange in Root Stem and Leaves of Aquatic and Terrestrial Plants Gaseous Exchange in leaves of Terrestrial Plants • Gaseous exchange takes place by diffusion. • The structure of the leaf is adapted for gaseous exchange by having intercellular spaces that are filled. • These are many and large in the spongy mesophyll. • When stomata are open,carbon(IV)oxide from the atmosphere diffuses into the substomatal air chambers. • From here, it moves into the intercellular space in the spongy mesophyll layer. • The CO2 goes into solution when it comes into contact with the cell surface and diffuses into the cytoplasm. • A concentration gradient is maintained between the cytoplasm of the cells and the intercellular spaces. • CO2 therefore continues to diffuse into the cells. • The oxygen produced during photosynthesis moves out of the cells and into the intercellular spaces. • From here it moves to the substomatal air chambers and eventually diffuses out of the leaf through the stomata. • At night oxygen enters the cells while CO2 moves out. Gaseous exchange in the leaves of aquatic(floating)plants • Aquatic plants such as water lily have stomata only on the upper leaf surface. • The intercellular spaces in the leaf mesophyll are large. • Gaseous exchange occurs by diffusion just as in terrestrial plants. Observation of internal structure of leaves of aquatic plants • Transverse section of leaves of an aquatic plant such as Nymphaea differs from that of terrestrial plant. The following are some of the features that can be observed in the leave of an aquatic plant; • Absence of cuticle • Palisade mesophyll cells are very close to each other ie.compact. • Air spaces (aerenchyma) in spongy mesophyll are very large. • Sclereids (stone cells) are scattered in leaf surface and project into air spaces. • They strengthen the leaf making it firm and assist it to float. Gaseous Exchange Through Stems Terrestrial Plants • Stems of woody plants have narrow openings or slits at intervals called • They are surrounded by loosely arranged cells where the bark is broken. • They have many large air intercellular spaces through which gaseous exchange occurs. • Oxygen enters the cells by diffusion while carbon (IV) oxide leaves. • Unlike the rest of the bark, lenticels are permeable to gases and water. Aquatic Plant Stems • The water lily, Salvia and Wolfia whose stems remain in water are permeable to air and water. • Oxygen dissolved in the water diffuses through the stem into the cells and carbon (IV) oxide diffuses out into the water. Gaseous Exchange in Roots Terrestrial Plants • Gaseous exchange occurs in the root hair of young terrestrial plants. • Oxygen in the air spaces in the soil dissolves in the film of moisture surrounding soil particles and diffuses into the root hair along a concentration gradient. • It diffuses from root hair cells into the cortex where it is used for respiration. • Carbon (IV) oxide diffuses in the opposite direction. • In older roots of woody plants, gaseous exchange takes place through lenticels. Aquatic Plants • Roots of aquatic plants e.g. water lily are permeable to water and gases. • Oxygen from the water diffuses into roots along a concentration gradient. • Carbon (IV) oxide diffuses out of the roots and into the water. • The roots have many small lateral branches to increase the surface area for gaseous exchange. • They have air spaces that help the plants to float. • Mangroove plants grow in permanently waterlogged soils, muddy beaches and at estuaries. • They have roots that project above the ground level. • These are known as breathing roots or pneumatophores. • These have pores through which gaseous exchange takes place e.g. in Avicenia the tips of the roots have pores. • Others have respiratory roots with large air spaces. Gaseous Exchange in Animals • All animals take in oxygen for oxidation of organic compounds to provide energy for cellular activities. • The carbon (IV) oxide produced as a by-product is harmful to cells and has to be constantly removed from the body. • Most animals have structures that are adapted for taking in oxygen and for removal of carbon (IV) oxide from the body. • These are called “respiratory organs”. • The process of taking in oxygen into the body and carbon (IV) oxide out of the body is called breathing or ventilation. • Gaseous exchange involves passage of oxygen and carbon (IV) oxide through a respiratory surface by diffusion. Types and Characteristics of Respiratory surfaces Different animals have different respiratory surfaces. • The type depends mainly on the habitat of the animal, size, shape and whether body form is complex or simple. • Cell Membrane: In unicellular organisms the cell membrane serves as a respiratory surface. • Gills: Some aquatic animals have gills which may be external as in the tadpole or internal as in bony fish e.g. tilapia. • They are adapted for gaseous exchange in water. • Skin: Animals such as earthworm and tapeworm use the skin or body surface for gaseous exchange. • The skin of the frog is adapted for gaseous exchange both in water and on land. • The frog also uses epithelium lining of the mouth or buccal cavity for gaseous exchange. • Lungs: Mammals, birds and reptiles have lungs which are adapted for gaseous exchange. Characteristics of Respiratory Surfaces • They are permeable to allow entry of gases. • They have a large surface area in order to increase diffusion. • They are usually thin in order to reduce the distance of diffusion. • They are moist to allow gases to dissolve. • They are well-supplied with blood to transport gases and maintain a concentration gradient. Gaseous Exchange in Amoeba • Gaseous exchange occurs across the cell membrane by diffusion. • Oxygen diffuses in and carbon (IV) oxide diffuses out. • Oxygen is used in the cell for respiration making its concentration lower than that in the surrounding water. • Hence oxygen continually enters the cell along a concentration gradient. • Carbon (IV) oxide concentration inside the cell is higher than that in the surrounding water thus it continually diffuses out of the cell along a concentration gradient. Gaseous Exchange in Insects • Gaseous exchange in insects e.g., grasshopper takes place across a system of tubes penetrating into the body known as the tracheal system. • The main trachea communicate with atmosphere through tiny pores called spiracles. • Spiracles are located at the sides of body segments; • Two pairs on the thoracic segments and eight pairs on the sides of abdominal segments. • Each spiracle lies in a cavity from which the trachea arises. • Spiracles are guarded with valves that close and thus prevent excessive loss of water vapour. • A filtering apparatus i.e. hairs also traps dust and parasites which would clog the trachea if they gained entry. • The valves are operated by action of paired muscles. Mechanism of Gaseous Exchange in Insects • The main tracheae in the locust are located laterally along the length of the body on each side and they are interconnected across. • Each main trachea divides to form smaller tracheae, each of which branches into tiny tubes called tracheoles. • Each tracheole branches further to form a network that penetrates the tissues. Some tracheoles penetrate into cells in active tissue such as flight muscles. • These are referred to as intracellular tracheoles. • Tracheoles in between the cells are known as intercellular tracheoles. • The main tracheae are strengthened with rings of cuticle. • This helps them to remain open during expiration when air pressure is low. Adaptation of Insect Tracheoles for Gaseous Exchange • The fine tracheoles are very thin about one micron in diameter in order to permeate tissue. • They are made up of a single epithelial layer and have no spiral thickening to allow diffusion of gases. • Terminal ends of the fine tracheoles are filled with a fluid in which gases dissolve to allow diffusion of oxygen into the cells. • Amount of fluid at the ends of fine tracheoles varies according to activity i.e. oxygen demand of the insect. • During flight, some of the fluid is withdrawn from the tracheoles such that oxygen reaches muscle cells faster and the rate of respiration is increased. • In some insects, tracheoles widen at certain places to form air sacs. • These are inflated or deflated to facilitate gaseous exchange as need arises. • Atmospheric air that dissolves in the fluid at the end of tracheoles has more oxygen than the surrounding cells of tracheole epithelium’. • Oxygen diffuses into these cells along a concentration gradient. ‘ • Carbon (IV) oxide concentration inside the cells is higher than in the atmospheric . • Air and diffuses out of the cells along a concentration gradient. • It is then removed with expired air. Ventilation in Insects • Ventilation in insects is brought about by the contraction and relaxation of the abdominal muscles. • In locusts, air is drawn into the body through the thoracic spiracles and expelled through the abdominal spiracles. • Air enters and leaves the tracheae as abdominal muscles contract and relax. • The muscles contract laterally so the abdomen becomes wider and when they relax it becomes narrow. • Relaxation of muscles results in low pressure hence inspiration occurs while contraction of muscles results in higher air pressure and expiration occurs. • In locusts, air enters through spiracles in the thorax during inspiration and leaves through the abdominal spiracles during expiration. • This results in efficient ventilation. • Maximum extraction of oxygen from the air occurs sometimes when all spiracles close and hence contraction of abdominal muscles results in air circulating within the tracheoles. • The valves in the spiracles regulate the opening and closing of spiracles. Observation of Spiracle in Locust • Some fresh grass is placed in a gas jar. • A locust is introduced into the jar. • A wire mesh is placed on top or muslin cloth tied around the mouth of the beaker with rubber band. • The insect is left to settle. • Students can approach and observe in silence the spiracles and the abdominal movements during breathing. • Alternatively the locust is held by the legs and observation of spiracles is made by the aid of hand lens. Gaseous Exchange in Bony Fish (e.g, Tilapia) • Gaseous exchange in fish takes place between the gills and the surrounding water. • The gills are located in an opercular cavity covered by a flap of skin called the operculum. • Each _gill consists of a number of thin leaf-like lamellae projecting from a skeletal base branchial arch (gill bar) situated in the wall of the pharynx. • There are four gills within the opercular cavity on each side of the head. • Each gill is made up of a bony gill arch which has a concave surface facing the mouth cavity (anterior) and a convex posterior surface. • Gill rakers are bony projections on the concave side that trap food and other solid particles which are swallowed instead of going over and damaging the gill ftlaments. • Two rows of gill filaments subtend from the convex surface. Adaptation of Gills for Gaseous Exchange • Gill filaments are thin walled. • Gill filaments are very many (about seventy pairs on each gill), to increase surface area. • Each gill filament has very many gill lamellae that further increase surface area. • The gill filaments are served by a dense network of blood vessels that ensure efficient transport of gases. • It also ensures that a favourable diffusion gradient is maintained. • The direction of flow of blood in the gill lamellae is in the opposite direction to that of the water (counter current flow) to ensure maximum diffusion of gases. • As the fish opens the mouth, the floor of the mouth is lowered. • This increases the volume of the buccal cavity. • Pressure inside the mouth is lowered causing water to be drawn into the buccal cavity. • Meanwhile, the operculum is closed, preventing water from entering or leaving through the opening. • As the mouth closes and the floor of the mouth is raised, the volume of buccal cavity decreases while pressure in the opercular cavity increases due to contraction of opercular muscles. • The operculum is forced to open and water escapes. • As water passes over the gills, oxygen is absorbed and carbon dioxide from the gills dissolves in the water. • As the water flows over the gill filaments oxygen in the water is at a higher concentration than that in the blood flowing, in the gill. • Oxygen diffuses through the thin walls of gill filaments/lamellae into the blood. • Carbon (IV) oxide is at a higher concentration in the blood than in the water. • It diffuses out of blood through walls of gill filaments into the water. Counter Current Flow • In the bony fish direction of flow of water over the gills is opposite that of blood flow through the gill filaments . • This adaptation ensures that maximum amount of oxygen diffuses from the water into the blood in the gill filament. • This ensures efficient uptake of oxygen from the water. • Where the flow is along the same direction (parallel flow) less oxygen is extracted from the water. Observation of Gills of a Bony Fish (Tilapia) • Gills of a fresh fish are removed and placed in a petri-dish with enough water to cover them. • A hand lens is used to view the gills. • Gill bar, gill rakers and two rows of gill filaments are observed. Gaseous Exchange in an Amphibian – Frog • An adult frog lives on land but goes back into the water during the breeding season. • A frog uses three different respiratory surfaces. • These are the skin, buccal cavity and lungs. • The skin is used both in water and on land. • It is quite efficient and accounts for 60% of the oxygen taken in while on land. Adaptations of a Frog’s Skin for Gaseous Exchange • The skin is a thin epithelium to allow fast diffusion. • The skin between the digits in the limbs (i.e. webbed feet) increase the surface area for gaseous exchange. • It is richly supplied with blood vessels for transport of respiratory gases. • The skin is kept moist by secretions from mucus glands. • This allows for respiratory gases to dissolve. • Oxygen dissolved in the film of moisture diffuses across the thin epithelium and into the blood which has a lower concentration of oxygen. • Carbon (IV) oxide diffuses from the blood across the skin to the atmosphere along the concentration gradient. Buccal (Mouth) Cavity • Gaseous exchange takes place all the time across thin epithelium lining the mouth cavity. • Adaptations of Buccal Cavity for Gaseous Exchange • It has a thin epithelium lining the walls of the mouth cavity allowing fast diffusion of gases. • It is kept moist by secretions from the epithelium for dissolving respiratory gases. • It has a rich supply of blood vessels for efficient transport of respiratory gases. • The concentration of oxygen in the air within the mouth cavity is higher than that of the blood inside the blood vessels. • Oxygen, therefore dissolves in the moisture lining the mouth cavity and then diffuses into the blood through the thin epithelium. • On the other hand, carbon (IV) oxide diffuses in the opposite direction along a concentration gradient. • There is a pair of small lungs used for gaseous exchange. Adaptation of Lungs • The lungs are thin walled for fast diffusion of gases. • Have internal foldings to increase surface area for gaseous exchange. • A rich supply of blood capillaries for efficient transport of gases. • Moisture lining for gases to dissolve. • During inspiration, the floor of the mouth is lowered and air is drawn in through the nostrils. • When the nostrils are closed and the floor of the mouth is raised, air is forced into the lungs. • Gaseous exchange occurs in the lungs, oxygen dissolves in the moisture lining of the lung and diffuses into the blood through the thin walls. • Carbon (IV) oxide diffuses from blood into the lung lumen. • When the nostrils are closed and the floor of mouth is lowered by contraction of its muscles, volume of mouth cavity increases. • Abdominal organs press against the lungs and force air out of the lungs into buccal cavity. • Nostrils open and floor of the mouth is raised as its muscles relax. • Air is forced out through the nostrils. Gaseous Exchange in a Mammal -Human • The breathing system of a mammal consists of a pair of lungs which are thin-walled elastic sacs lying in the thoracic cavity. • The thoracic cavity consists of vertebrae, sternum, ribs and intercostal muscles. • The thoracic cavity is separated from the abdominal cavity by the diaphragm. • The lungs lie within the thoracic cavity. • They are enclosed and protected by the ribs which are attached to the sternum and the thoracic vertebrae. • There are twelve pairs of ribs, the last two pairs are called ‘floating ribs’ because they are only attached to the vertebral column. • The ribs are attached to and covered by internal and external intercostals muscles. • The diaphragm at the floor of thoracic cavity consists of a muscLe sheet at the periphery and a central circular fibrous tissue. • The muscles of the diaphragm are attached to the thorax wall. • The lungs communicate with the outside atmosphere through the bronchi, trachea, mouth and nasal cavities. • The trachea opens into the mouth cavity through the larynx. • A flap of muscles, the epiglottis, covers the opening into the trachea during swallowing. • This prevents entry of food into the trachea. • Nasal cavities are connected to the atmosphere through the external nares(or nostrils)which are lined with hairs and mucus that trap dust particles and bacteria, preventing them from entering into the lungs. • Nasal cavities are lined with cilia. • The mucus traps dust particles, • The cilia move the mucus up and out of the nasal cavities. • The mucus moistens air as it enters the nostrils. • Nasal cavities are winding and have many blood capillaries to increase surface area to ensure that the air is warmed as it passes along. • Each lung is surrounded by a space called the pleural cavity. • It allows for the changes in lung volume during breathing. • An internal pleural membrane covers the outside of each lung while an external pleural membrane lines the thoracic wall. • The pleural membranes secrete pleural fluid into the pleural cavity. • This fluid prevents friction between the lungs and the thoracic wall during breathing. • The trachea divides into two bronchi, each of which enters into each lung. • Trachea and bronchi are lined with rings of cartilage that prevent them from collapsing when air pressure is low. • Each bronchus divides into smaller tubes, the bronchioles. • Each bronchiole subdivides repeatedly into smaller tubes ending with fine bronchioles. • The fine bronchioles end in alveolar sacs, each of which gives rise to many alveoli. • Epithelium lining the inside of the trachea, bronchi and bronchioles has cilia and secretes mucus. Adaptations of Alveolus to Gaseous Exchange • Each alveolus is surrounded by very many blood capillaries for efficient transport of respiratory gases. • There are very many alveoli that greatly increases the surface area for gaseous exchange. • The alveolus is thin walled for faster diffusion of respiratory gases. • The epithelium is moist for gases to dissolve. Gaseous Exchange Between the Alveoli and the Capillaries • The walls of the alveoli and the capillaries are very thin and very close to each other. • Blood from the tissues has a high concentration of carbon (IV) oxide and very little oxygen compared to alveolar air. • The concentration gradient favours diffusion of carbon (IV) oxide into the alveolus and oxygen into the capillaries . • No gaseous exchange takes place in the trachea and bronchi. • These are referred to as dead space. • Exchange of air between the lungs and the outside is made possible by changes in the volumes of the thoracic cavity. • This volume is altered by the movement of the intercostal muscles and the diaphragm. • The ribs are raised upwards and outwards by the contraction of the external intercostal muscles, accompanied by the relaxation of internal intercostal muscles. • The diaphragm muscles contract and diaphragm moves downwards. • The volume of thoracic cavity increases, thus reducing the pressure. • Air rushes into the lungs from outside through the nostrils. • The internal intercostal muscles contract while external ones relax and the ribs move downwards and inwards. • The diaphragm muscles relaxes and it is pushed upwards by the abdominal organs. It thus assumes a dome shape. • The volume of the thoracic cavity decreases, thus increasing the pressure. • Air is forced out of the lungs. • As a result of gaseous exchange in the alveolus, expired air has different volumes of atmospheric gases as compared to inspired air. Table 7.1: Comparison of Inspired and Expired Air (% by volume) Component Inspired % Expired % Oxygen 21 16 Carbon dioxide 0.03 4 Nitrogen 79 79 Moisture Variable Saturated Lung Capacity • The amount of air that human lungs can hold is known as lung capacity. • The lungs of an adult human are capable of holding 5,000 cm^3 of air when fully inflated. • However, during normal breathing only about 500 cm^3 of air is exchanged. • This is known as the tidal volume. • A small amount of air always remains in the lungs even after a forced expiration. • This is known as the residual volume. • The volume of air inspired or expired during forced breathing is called vital capacity. Control of Rate Of Breathing • The rate of breathing is controlled by the respiratory centre in the medulla of the brain. • This centre sends impulses to the diaphragm through the phrenic nerve. • Impulses are also sent to the intercostal muscles. • The respiratory centre responds to the amount of carbon (IV) oxide in the blood. • If the amount of carbon (IV) oxide rises, the respiratory centre sends impulses to the diaphragm and the intercostal muscles which respond by contracting in order to increase the ventilation • Carbon (IV) oxide is therefore removed at a faster rate. Factors Affecting Rate of Breathing in Humans • Factors that cause a decrease or increase in energy demand directly affect rate of breathing. • Exercise, any muscular activity like digging. • Sickness • Emotions like anger, flight Effects of Exercise on Rate of Breathing • Students to work in pairs. • One student stands still while the other counts (his/her) the number of breaths per minute. • The student whose breath has been taken runs on the sport vigorously for 10 minutes. • At the end of 10 minutes the number of breaths per minute is immediately counted and recorded. • It is noticed that the rate of breathing is much higher after exercise than at rest. Dissection of a Small Mammal (Rabbit) to Show Respiratory Organs • The rabbit is placed in a bucket containing cotton wool which has been soaked in chloroform. • The bucket is covered tightly with a lid. • The dead rabbit is placed on the dissecting board ventral side upwards. • Pin the rabbit to the dissecting board by the legs. • Dissect the rabbit to expose the respiratory organs. • Ensure that you note the following features. • Ribs, intercostal muscles, diaphragm, lungs, bronchi, trachea, pleural membranes, thoracic cavity. Diseases of the Respiratory System • Asthma is a chronic disease characterised by narrowing of air passages. • Due to pollen, dust, fur, animal hair, spores among others. • If these substances are inhaled, they trigger release of chemical substances and they may cause swelling of the bronchioles and bring about an asthma attack. • Asthma is usually associated with certain disorders which tend to occur in more than one member of a given family, thus suggesting’ a hereditary tendency. Emotional or mental stress • Strains the body immune system hence predisposes to asthma attack. • Asthma is characterized by wheezing and difficulty in breathing accompanied by feeling of tightness in the chest as a result of contraction of the smooth muscles lining the air passages. Treatment and Control • There is no definite cure for asthma. • The best way where applicable is to avoid whatever triggers an attack (allergen). • Treatment is usually by administering drugs called bronchodilators. • The drugs are inhaled, taken orally or injected intravenously depending on severity of attack to relief bronchial spasms. • This is an inflammation of bronchial tubes. • This is due to an infection of bronchi and bronchioles by bacteria and viruses. • Symptoms • Difficulty in breathing. • Cough that produces mucus. • Antibiotics are administered. Pulmonary Tuberculosis • Tuberculosis is a contagious disease that results in destruction of the lung tissue. • Tuberculosis is caused by the bacterium Mycobacterium tuberculosis. • Human tuberculosis is spread through droplet infection i.e., in saliva and sputum. • Tuberculosis can also spread from cattle to man through contaminated milk. • From a mother suffering from the disease to a baby through breast feeding. • The disease is currently on the rise due to the lowered immunity in persons with HIV and AIDS (Human Immuno Deficiency Syndrome). • Tuberculosis is common in areas where there is dirt, overcrowding and malnourishment. • It is characterised by a dry cough, lack of breath and body wasting. • Proper nutrition with a diet rich in proteins and vitamins to boost immunity. • Isolation of sick persons reduces its spread. • Utensils used by the sick should be sterilised by boiling. • Avoidance of crowded places and living in well ventilated houses. • Immunisation with B.C.G. vaccine gives protection against tuberculosis. • This is done a few days after birth with subsequent boosters. • Treatment is by use of antibiotics. • Pneumonia is infection resulting in inflammation of lungs. • The alveoli get filled with fluid and bacterial cells decreasing surface are for gaseous exchange. • Pneumonia is caused by bacteria and virus. • More infections occur during cold weather. • The old and the weak in health are most vulnerable. • Pain in the chest accompanied by a fever, high body temperatures (39-40°C) and general body weakness. • Maintain good health through proper feeding. • Avoid extreme cold. • If the condition is caused by pneumococcus bacteria, antibiotics are administered. • If breathing is difficult, oxygen may be given using an oxygen mask. Whooping Cough • Whooping cough is an acute infection of respiratory tract. • The disease is more common in children under the age of five but adults may also be affected. • It is caused by Bordetella pertusis bacteria and is usually spread by droplets produced when a sick person coughs. • Severe coughing and frequent vomiting. • Thick sticky mucus is produced. • Severe broncho-pneumonia. • Convulsions in some cases. • Children may be immunised against whooping cough by means of a vaccine which is usually combined with those against diphtheria and tetanus. • It is called “Triple Vaccine” or Diptheria, Pertusis and Tetanus (DPT). • Antibiotics are administered. • To reduce the coughing, the patient should be given drugs. Practical Activities Observation of permanent slides of terrestrial and aquatic leaves and stems • Observation of T.S. of bean and water lily are made under low and ‘medium power objectives. Stomata and air space are seen. • Labelled drawings of each are made. • The number and distribution of stomata on the lower and upper leaf surface is noted. • Also the size of air spaces and their distribution. • Prepared slides (TS) of stems of terrestrial and aquatic plants such as croton and reeds are obtained. • Observations under low power and medium power of a microscope are made. • Labelled drawings are made and the following are noted: • Lenticels on terrestrial stems. • Large air spaces (aerenchyma) in aquatic stems. Excretion and Homeostasis • Excretion is the process by which living organisms separate and eliminate waste products of metabolism from body cells. • If these substances were left to accumulate, they would be toxic to the cells. • Egestion is the removal of undigested materials from the alimentary canals of animals. • Secretion is the production and release of certain useful substances such as hormones, sebum and mucus produced by glandular cells. • Homeostasis is a self-adjusting mechanism to maintain a steady state in the internal environment Excretion in Plants • Plants have little accumulation of toxic waste especially nitrogenous wastes. • This is because they synthesise proteins according to their requirements. • In carbohydrate metabolism plants use carbon (IV) oxide released from respiration in photosynthesis while oxygen released from photosynthesis is used in respiration. • Gases are removed from the plant by diffusion through stomata and lenticels. • Certain organic products are stored in plant organs such as leaves, flowers, fruits and bark and are removed when these organs are shed. • The products include tannins, resins, latex and oxalic acid crystals. • Some of these substances are used illegally. • Khat, cocaine and cannabis are used without a doctor’s prescription and can be addictive. • Use of these substances should be avoided. Plant Excretory Products their source and uses Plant Product Source Use Caffeine Tea and coffee Mild CNS stimulant. Quinine Cinchona tree Anti malaria-drug. Tannins Barks of Acacia, Wattle trees Tanning hides and skins. Colchicine Corms of crocus Prevents spindle formation in cell division. Cocaine Leaves of coca plant Local anaesthesia. – Rubber Latex of rubber plant Used in shoe industry. Gum Exudate from acacia Used in food processing and printing industry. Cannabis Flowers, fruits and leaves of Used in manufacture of drugs. cannabis sativa Nicotine Leaves of tobacco plant Manufacture of insecticides. Heart and CNS Papain Pawpaw (fruits) Meat tenderiser Treats indigestion. I Mild stimulant. Khat Khatha edulis (miraa) Morphine Opium Poppy plant Narcotic. Induces sleep / hallucinations. Strychnine Seeds of strychnos CNS stimulant. Excretory products in animals Substance Origin 1. Nitrogenous compounds: Excess amino acids (proteins). (i) Ammonia Deamination of amino acids. (ii) Urea Deamination of amino acids; then addition of carbon dioxide. (iii) Uric acid Ammonia (from deamination of amino acids). 2. Carbon dioxide Homeostasis and respiration. 3. Biliverdin and bilirubin Breakdown of haemoglobin. 4. Water Osmoregulation. 5. Cholesterol Excess intake of fats. — .i->: — 6. Hormones Excess production Excretion and Homeostasis in Unicellular Organisms • Protozoa such as amoeba depend on diffusion as a means of excretion. • They have a large surface area to volume ratio for efficient diffusion. • Nitrogenous waste and carbon (IV) oxide are highly concentrated in the organism hence they diffuse out. • In amoeba excess water and chemicals accumulation in the contractile vacuole. • When it reaches maximum size the contractile vacuole moves to the cell membrane, bursts open releasing its contents to the surroundings. Excretion in Human Beings • Excretion in humans is carried out by an elaborate system of specialised organs. • Their bodies are complex, so simple diffusion cannot suffice. • Excretory products include nitrogenous wastes which originate from deamination of excess amino acids. • The main excretory organs in mammals such as human beings include lungs, kidneys, skin and liver. Structure and function of the human skin Nerve Endings: • These are nerve cells which detect changes from the external environment thus making the body to be sensitive to touch, cold, heat and pressure. Subcutaneous Fat: • Is a layer beneath the dermis. • It stores fat and acts as an insulator against heat loss. • The skin helps in elimination of urea, lactic acid and sodium chloride which are released in sweat. The Lungs • Carbon (IV) oxide formed during tissue respiration is removed from the body by the lungs. • Mammalian lungs have many alveoli which are the sites of gaseous exchange. • Alveoli are richly supplied with blood and have a thin epithelium. • Blood capillaries around the alveoli have a high concentration of carbon (Iv) oxide than the alveoli lumen. • The concentration gradient created causes carbon (IV) oxide to diffuse into the alveoli lumen. • The carbon (IV) oxide is eliminated through expiration. Structure and Functions of the Kidneys • The kidneys are organs whose functions are excretion, osmoregulation and regulation of pH. • Kidneys are located at the back of the abdominal cavity. • Each kidney receives oxygenated blood from renal artery, • while deoxygenated blood leaves through the renal vein. • Urine is carried by the ureter from the kidney to the bladder, which temporarily stores it. • From the bladder, the urine is released to the outside via the urethra. • The opening from the urethra is controlled by a ring-like sphincter muscle. • A longitudinal section of the kidney shows three distinct regions: a darker outer cortex, a lighter inner medulla and the pelvis. • The pelvis is a collecting space leading to the ureter which takes the urine to the bladder from where it is eliminated through the urethra. The Nephron • A nephron is a coiled tubule at one end of which is a cup-shaped structure called the Bowman’s capsule. • The capsule encloses a bunch of capillaries called the glomerulus. • The glomerulus receives blood from an afferent arteriole a branch of the renal artery. • Blood is taken away from the glomerulus by efferent arteriole leading to the renal vein. • The Bowman’s capsule leads to the proximal convoluted tubule that is coiled and extends into a U-shaped part called loop of Henle. • From the loop of Henle is the distal convoluted tubule that is also coiled. • This leads to the collecting duct which receives contents of many nephrons. • Collecting ducts lead to the pelvis of the kidney. Mechanism of Excretion • Excretion takes place in three steps: • Filtration, reabsorption and removal. • The kidneys receive blood from renal artery a branch of the aorta. • This blood is rich in nitrogenous waste e.g. urea. • It contains dissolved food substances, plasma proteins,hormones and oxygen. • Blood flow in capillaries is under pressure due to the narrowness of the capillaries. • The afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it. • This creates pressure in the glomerulus. • Due to this pressure, dissolved substances such as urea, uric acid, glucose, mineral salts and amino acids are forced out of the glomerulus into the Bowman’s capsule. • Large sized molecules in the plasma such as proteins and red blood cells are not filtered out because they are too large. • This process of filtration is called ultra-filtration or pressure filtration and the filtrate is called glomerular filtrate. Selective Reabsorption • As the filtrate flows through the renal tubules the useful substances are selectively reabsorbed back into the blood. • In the proximal convoluted tube all the glucose, all amino acids and some mineral salts are actively reabsorbed by active transport. • The cells lining this tubule have numerous mitochondria which provide the energy needed. • Cells of the tubule have microvilli which increases the surface area for re-absorption. • The tubule is coiled, which reduces the speed of flow of the filtrate e.g. giving more time for efficient re-absorption. • The tubule is well supplied with blood capillaries for transportation of reabsorbed substances. • The ascending loop has thick wall and is impermeable to water. • Sodium is actively pumped out of it towards the descending loop. • As glomerular filtrate moves down the descending loop, water is reabsorbed into the blood by osmosis in the distal convoluted tubule and in the collecting duct. • Permeability of the collecting duct and proximal convoluted tubule is increased by anti-diuretic hormone (ADH) whose secretion is influenced by the osmotic pressure of the blood. • The remaining fluid consisting of water, urea, uric acid and some mineral salts is called urine. • The urine is discharged into the collecting d ct and carried to the pelvis. • The loop of Henle is short in semi-aquatic mammals, and long in some mammals like the desert rat. • The urine is conveyed from the pelvis to the ureter. • The ureter carries the urine to the bladder where it is stored temporarily and discharged to the outside through the urethra at intervals. Common Kidney Diseases • This is a condition in which concentration of urea in the blood. • It may be due to formation of cysts in tubules or reduction in blood supply to the glomeruli as a result of contraction of renal artery. • Symptoms include yellow colouration of skin, smell of urine in breath, nausea and vomiting. • Treatment includes dialysis to remove excess urea and a diet low in proteins and salts especially sodium and potassium. Kidney Stones • Kidney stones are solid deposits of calcium and other saIts. • They are usually formed in the pelvis of the kidney where they may obstruct the flow of urine. • Causes: the stones are formed due to crystallisation of salts around pus, blood or dead tissue. • Symptoms: include blood in urine, frequent urination, pain, chills and fever. Severe pain when urinating. • Use of laser beams to disintegrate the stones. • Pain killing drugs like morphine. • Stones can be removed by surgery. • Taking hot baths and massage. • Nephritis is the inflation of glomerulus of the kidney. • Causes: Bacterial infection, sore throat or tonsillitis, blockage of glomeruli by antibody-antigen complex. • Signs and Symptoms: include headaches, fever, vomiting, oedema. • Control includes dietary restrictions especially salt and proteins. • Prompt treatment of bacterial infections. Role of Liver in Excretion • The liver lies below the diaphragm and it receives blood from hepatic artery and hepatic portal vein. • Blood flows out of the liver through hepatic vein. • Excretion of Nitrogenous Wastes • Excess amino acids cannot be stored in the body, they are deaminated in the liver. • Hydrogen is added to amino group to form ammonia which combines with carbon (IV) oxide to form urea. • The urea is carried in the blood stream to the kidneys. • The remaining carboxyl group, after removal of amino group, is either oxidised to provide energy in respiration. • or built up into carbohydrate reserve and stored as glycogen or converted into fat and stored. Breakdown and Elimination of Haemoglobin • Haemoglobin is released from dead or old red blood cells which are broken down in the liver and spleen. • Haemoglobin is broken down in the liver and a green pigment biliverdin results which is converted to yellow bilirubin. • This is taken to the gall bladder and eliminated as bile. Elimination of Sex Hormones • Once they have completed their functions, sex hormones are chemically altered by the liver and then taken to the kidney for excretion. Common Liver Diseases • Cirrhosis is a condition in which liver cells degenerate and are replaced by scar tissue . • This causes the liver to shrink, harden, become fibrous and fail to carry out its functions. • Chronic alcohol abuse, schistosomiasis infection, obstruction of gall-bladder. • Headache, nausea, vomiting of blood and lack of appetite, weight loss, indigestion and jaundice. Control and Treatment • Avoid alcohol consumption and fatty diet. • Use drugs to kill the schistosomes if that is the cause. • This is a yellow colouration of the skin and eyes. • Presence of excess bile pigments. • This happens due to blockage of bile duct or destruction of liver. • Yellow pigmentation of skin and eyes, nausea, vomiting and lack of appetite. Itching of skin. • Removal of stones from the gall bladder by surgery. • Give patient fat-free diet, reduced amount of proteins. • Give antihistamines to reduce itching. • Homeostasis is the maintenance of a constant internal environment. • The internal environment consists of intercellular or tissue fluid. • This fluid is the medium in the space surrounding cells. • Tissue fluid is made by ultra-filtration in the capillaries. • Dissolved substances in the blood are forced out of the capillaries and into intercellular spaces. • Cells obtain their requirements from tissue fluid while waste products from cells diffuse out into the tissue fluid. • Some of the fluid gets back into the blood capillaries while excess fluid is drained into the lymph vessels. • Cells function efficiently if there is little or no fluctuation in the internal environment. • The factors that need to be regulated include temperature, osmotic pressure and pH. • The body works as a self-regulating system and can detect changes in its working conditions bringing about corrective responses. • This requires a negative feedback mechanism e.g. when body temperature falls below normal, mechanisms are set in place that bring about increase in temperature. • And when the increase is above normal, mechanisms that lower the temperature are set in place. • This is called a negative feedback and it restores the conditions to normal. Neuro-Endocrine System and Homeostasis • Homeostatic mechanisms are brought about by an interaction between nervous and endocrine systems. • Nerve endings detect changes in the internal and external environment and relay the information to the brain. • The hypothalamus and pituitary are endocrine glands situated in the brain. • The hypothalamus detect changes in the blood. • The pituitary secretes a number of hormones involved in homeostasis e.g. anti-duretic hormone (ADH). • The discussion below shows the nature of these interactions. The Skin and Temperature Regulation • The optimum human body temperature is 36.8°C. • A constant body temperature favours efficient enzyme reaction. • Temperatures above optimum denature enzymes, while temperature below the optimum range inactivate enzymes. • The skin is involved in regulation of body temperature as follows: • The skin has receptors that detect changes in the temperature of the external environment. When the body temperature is above optimum the following takes place: • Sweat glands secrete sweat onto the skin surface. • As sweat evaporates it takes latent heat from the body, thus lowering the temperature. Vasodilation of Arterioles: • The arterioles near the surface become wider in diameter. • More blood flows near the surface and more heat is lost to the surrounding by convection and radiation. Relaxation of hair erector muscle: • When hair erector muscles relax, the hair lies flat thus allowing heat to escape from the skin surface. When body temperature is below optimum the following takes place: Vasoconstriction of Arterioles: • The arterioles near the surface of the skin become narrower. • Blood supply to the skin is reduced and less heat is lost to the surroundings. Contraction of hair erector muscles. • When hair erector muscles contract, the hair is raised. • Air is trapped between the hairs forming an insulating layer. • Animals in cold areas have a thick layer of subcutaneous fat, which helps to insulate the body. • Besides the role of the skin in thermoregulation as discussed above, the rate of metabolism is lowered when temperature is above optimum and increased when temperature is below optimum. • The latter increases the temperature to the optimum. • When this fails, shivering occurs. • Shivering is involuntary contraction of muscles which helps to generate heat thus raising the body temperature. Homeostatic Control of Body Temperature in Humans Body size and Heat Loss • The amount of heat produced by metabolic reactions in an animal body is proportional to its mass. • Large animals produce more heat but they lose less due to small surface area to volume ratio. • Small animals produce less heat and lose a lot, due to large surface area to volume ratio. • Small animals eat a lot of food in relation .to their size in order to raise their metabolic rate. Behavioural and Physiological Responses to Temperature Changes • Animals gain or lose heat to the environment by conduction, radiation and convection. • Birds and mammals maintain a constant body temperature regardless of the changes in the environment. • They do this mainly by internally installed physiological mechanisms hence they are endotherms, also known as • At the same time behavioural activities like moving to shaded areas when it is too hot assist in regulating their body temperature. • Other animals do not maintain a constant body temperature e.g. lizards. • They are poikilotherms (ectotherms) as their temperature varies according to that of surroundings. • They only regulate body temperature through behavioural means. • Lizards bask on the rocks to gain heat and hide under rocks when it is too hot. • Some animals have adaptive features e.g. animals in extreme cold climates have fur and a thick layer of subcutaneous fat like polar bear. • Those in extremely hot areas have tissue that tolerate high temperatures e.g. camels. • Some animals avoid cold conditions by hibernating g. the frog while others avoid dry hot conditions by aestivation e.g. kangaroo rat. • This involves decreasing their metabolic activities. Skin and Osmoregulation • Osmoregulation is the control of salt and water balance in the body to maintain the appropriate osmotic pressure for proper cell functioning. • Sweat glands produce sweat and thus eliminate water and salt from the body. The Kidney and Osmoregulation • The kidney is the main organ that regulates the salt and water balance in the body. • The amount of salt or water reabsorbed into the bloodstream is dependent on the osmotic pressure of the blood. • When the osmotic pressure of the blood rises above normal due to dehydration or excessive consumption of salt, the osmo-receptors in the hypothalamus are stimulated. • These cells relay impulses to the pituitary gland which produces a hormone called anti-diuretic hormone – ADH (vasopressin) which is taken by the blood to the kidneys. • The hormone (ADH) makes the distal convoluted tubule and collecting duct more permeable to water hence more water is reabsorbed into the body by the kidney tubules lowering the osmotic pressure in the blood. • When the osmotic pressure of the blood falls below normal due to intake of a large quantity of water, osmoreceptors in the hypothalamus are less stimulated. • Less antidiuretic hormone is produced, and the kidney tubules reabsorb less water hence large quantities of water is lost producing dilute urine (diuressis). • The osmotic pressure of the blood is raised to normal. • If little or no ADH is produced, the body may become dehydrated unless large quantities of water are consumed regularly. • Diabetes insipidus is a disease that results from the failure of the pituitary gland to produce ADH and the body gets dehydrated. • A hormone called Aldosterone produced by the adrenal cortex regulates the level of sodium ions. • When the level of sodium ions in the blood is low, adrenal cortex releases aldosterone into the blood. • This stimulates the loop of Henle to reabsorb sodium ions into the blood. • Chloride ions flow to neutralise the charge on sodium ions. • Aldosterone also stimulates the colon to absorb more sodium ions into the blood. • If the sodium ion concentration rises above optimum level, adrenal cortex Notes missing The liver • Formation of Red Blood Cells. • In the embryo, red blood cells are formed in the liver. • Breakdown and elimination of old and dead blood cells. • Dead red blood cells are broken down in the liver and the pigments eliminated in bile. • Manufacture of Plasma Proteins. • Plasma proteins like albumen, fibrinogen and globulin are manufactured in the liver. • Storage of blood, vitamins A, K, BI2 and D and mineral salts such as iron’ and potassium ions. • Toxic substances ingested e.g. drugs or produced from metabolic reactions in the body are converted to harmless substances in a process called detoxification.	{"url":"https://newsblaze.co.ke/biology-notes-form-two-all-topics-best-of-the-best/?amp=1","timestamp":"2024-11-12T17:00:45Z","content_type":"text/html","content_length":"160524","record_id":"<urn:uuid:5a0200d9-d4b5-4ce1-b32d-5f4868adc6f6>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00262.warc.gz"}
Rotating Air Gap Air gap between stator tooth and rotating permanent magnet rotor Since R2021a Simscape / Electrical / Electromechanical The Rotating Air Gap block models an air gap between a stator tooth and a rotating permanent magnet rotor. This block assumes that the rotor magnets are surface mounted and that the associated induced voltage is sinusoidal. This figure shows the relationship between the parameters of the Rotating Air Gap block and their physical values inside a permanent magnet motor • r is the value of the Rotor radius parameter. • g is the value of the Air gap parameter. • l[m] is the value of the Permanent magnet length (in direction of flux) parameter. • l is the value of the Tooth depth (in direction of shaft) parameter. The rotor circumference is equal to $2\pi r$. Then, the width of a permanent magnet on the rotor is equal to $\frac{2\pi r}{2N}$, where N is the Number of rotor pole pairs. If the rotor angle is zero, specified by the Rotor angle variable in the Variables section, then the rotor magnet perfectly aligns with the middle of the first stator tooth. The permanent magnet is then orientated to oppose the flux flow from port N to port S. Use this block to create a magnetic representation of a permanent magnet synchronous motor (PMSM). For example, if you want to model a motor with nine stator poles, create nine copies of this block and set each of the Stator tooth reference index parameters to 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. This figure shows the equivalent circuit for the air gap and the adjacent permanent magnet • ϕ[g] is the magnetic flux that flows from the external magnetic circuit to port N. • R[g] is the air gap reluctance. • mmf is the magnetomotive force across the rotating air gap component. • R[m] is the permanent magnet reluctance. • ϕ[r] is the magnetic flux generated by the rotor permanent magnets in the angle range subtended by the stator tooth. This equation defines the relationship between ϕ[g], mmf, and ϕ[r]: ${\varphi }_{g}=\frac{mmf-{R}_{m}{\varphi }_{r}}{{R}_{m}+{R}_{g}}.$ If the back EMF is sinusoidal, the flux density of the permanent magnet rotor is defined by this equation ${B}_{r}={B}_{0}cos\left(N{\theta }_{s}-N{\theta }_{r}\right)$ • N is the Number of rotor pole pairs. • θ[r] is the rotor angle. • θ[s] is the stator angle. • B[0] is the Peak magnet flux density, in Tesla. Then, to obtain the permanent magnet flux linkage, integrate over the stator angle subtended by the stator tooth ${\varphi }_{r}\left({\theta }_{r}\right)=rl{\int }_{\frac{-{\theta }_{tooth}}{2}}^{\frac{{\theta }_{tooth}}{2}}\left[{B}_{0}\mathrm{cos}\left(N{\theta }_{s}-N{\theta }_{r}\right)\right]d{\theta }_ • r is the Rotor radius. • l is the Tooth depth (in direction of shaft). For an ideal PMSM, the θ[tooth] must be equal to 2π/N[s], where N[s] is the value of the Number of stator teeth parameter. Then the equation of the flux that flows through the equivalent circuit is obtained by solving the integral: ${\varphi }_{r}\left({\theta }_{r}\right)=2{B}_{0}lr/Nsin\left(\frac{\pi N}{{N}_{s}}\right)\mathrm{cos}\left(N{\theta }_{r}\right).$ To obtain the torque generated across the air gap, first calculate the total energy stored by the component: $E=\frac{1}{2}{\varphi }_{g}^{2}{R}_{g}+\frac{1}{2}{\left({\varphi }_{r}\left({\theta }_{r}\right)\right)}^{2}{R}_{m}.$ Then, to obtain the torque, differentiate with respect to the rotor angle: $\tau =\frac{\partial E}{\partial {\theta }_{r}}=-2{B}_{0}{R}_{m}lrsin\left(\frac{\pi N}{{N}_{s}}\right)\mathrm{sin}\left(N{\theta }_{r}\right)\left({\varphi }_{g}+{\varphi }_{r}\left({\theta }_{r}\ Finally, calculate R[g] and R[m] in terms of geometry: $\begin{array}{l}{R}_{g}=\frac{g}{{\mu }_{0}{A}_{g}}\\ {R}_{m}=\frac{{l}_{m}}{{\mu }_{r}{\mu }_{0}{A}_{g}}\end{array}$ • μ[0] is the permittivity of free space. • μ[r] is the relative permittivity of the permanent magnet. • g is the Air gap. • l[m] is the magnet length. You can fault the Rotating Air Gap block. To enable faults, in the Faults section, select the Enable faults parameter. The Rotating Air Gap block does not support non-intrusive fault modeling. To model non-intrusive faults, use the Magnetic Rotor. A fault is defined as a reduction in the peak magnet flux density. The flux density associated with each rotor magnet remains sinusoidal in shape. When the Rotating Air Gap block is in the faulted state, you can apply a reduction factor to the flux density of any of the rotor poles by specifying the Flux multipliers for faulted rotor poles parameter. The unfaulted flux density in the airgap of a perfect PMSM with a sinusoidal back EMF is equal to: ${B}_{r}={B}_{0}cos\left(N{\theta }_{s}-N{\theta }_{r}\right)$ When the faulted magnet interacts with the tooth, the block uses this equation to define the flux density ${B}_{r}=\lambda {B}_{0}cos\left(N{\theta }_{s}-N{\theta }_{r}\right),$ where λ is the factor that maps peak B[0] to the faulted B[0], and is defined in the Flux multipliers for faulted rotor poles parameter. The transition to the faulted values linearly blends over the time period that you specify in the Duration of transition to faulted parameter. Use this parameter to emulate how an overheated permanent magnet gradually loses its magnetization over time. To set the priority and initial target values for the block variables before simulation, use the Initial Targets section in the block dialog box or Property Inspector. For more information, see Set Priority and Initial Target for Block Variables. Use nominal values to specify the expected magnitude of a variable in a model. Using system scaling based on nominal values increases the simulation robustness. Nominal values can come from different sources. One of these sources is the Nominal Values section in the block dialog box or Property Inspector. For more information, see System Scaling by Nominal Values. N — Magnetic stator connection Magnetic conserving port associated with the stator. S — Magnetic rotor connection Magnetic conserving port associated with the rotor. C — Motor case Mechanical rotational conserving port associated with the motor case. R — Motor rotor Mechanical rotational conserving port associated with the motor rotor. Number of rotor pole pairs — Rotor pole pairs 5 (default) \| positive scalar Number of the pole pairs of the rotor. This parameter must be equal to or greater than 1 and less than the value of the Number of stator teeth parameter. Number of stator teeth — Number of stator teeth 9 (default) \| positive scalar Number of teeth of the stator. This parameter must be equal to or greater than 2. Stator tooth reference index — Stator tooth reference index 1 (default) \| positive scalar Reference index of the stator tooth of the motor. This parameter must be between 1 and the value of the Number of stator teeth parameter. For example, if you want to model a motor with nine stator poles, create nine copies of this block and set the Stator tooth reference index parameter for each of the Rotating Air Gap blocks to 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. Peak magnet flux density — Peak flux density of magnet rotor 0.4 T (default) \| positive scalar Peak flux density associated with the permanent magnet rotor. The flux density is sinusoidal with the rotor angle. Permanent magnet length (in direction of flux) — Magnet length in radial machine direction 5 mm (default) \| positive scalar Length of the magnet in the radial machine direction or, equivalently, in the direction of the magnetic flux. This parameter must be less than the value of the Rotor radius parameter. Permanent magnet relative permeability — Permanent magnet relative permeability 1.05 (default) \| scalar Relative permeability of the permanent magnets. Typically, you should set this value a little greater than 1 to reflect that the magnetic dipoles are already aligned in a permanent magnet. Air gap — Air gap in radial direction 1 mm (default) \| positive scalar Length of the air gap in the radial direction. Rotor radius — Rotor radius 65 mm (default) \| positive scalar Tooth depth (in direction of shaft) — Stator tooth depth 50 mm (default) \| positive scalar Length of the stator tooth in the direction of the rotating shaft. Enable faults — Fault option off (default) \| on Whether to simulate the effect of a degraded rotor magnet strength. Flux multipliers for faulted rotor poles — Multipliers to reduce rotor pole flux ones(1,10) (default) \| vector Multipliers used to reduce the rotor pole magnetic density when faulted. The value of this parameter must be a vector of length equal to twice the value of the Number of rotor pole pairs parameter. Each element of the vector corresponds to one rotor pole. The default value is equal to ones(1,10) and results in the same behavior as the unfaulted scenario. To enable this parameter, select the Enable faults parameter. Duration of transition to faulted — Amount of time after which block fully faults 100 s (default) \| positive scalar Amount of time after which the block applies the full effect of the faulted multipliers on the peak magnet flux density of each rotor pole. When the block enters the fault state, the peak magnet flux densities of each rotor pole are gradually modified using the faulted multipliers. To enable this parameter, select the Enable faults parameter. Simulation time for fault event — Simulation time at which block faults 1 s (default) \| positive scalar Simulation time at which the block starts to apply the faulted multipliers for the peak magnet flux density on each rotor pole. To enable this parameter, select the Enable faults parameter. Extended Capabilities C/C++ Code Generation Generate C and C++ code using Simulink® Coder™. Version History Introduced in R2021a	{"url":"https://ww2.mathworks.cn/help/sps/ref/rotatingairgap.html","timestamp":"2024-11-12T20:11:24Z","content_type":"text/html","content_length":"114395","record_id":"<urn:uuid:cab0b3b8-8be7-4759-b5f2-e13c968391ac>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00610.warc.gz"}
What are the solutions for nth roots of unity? The concept of nth roots of unity is a fundamental part of mathematics that has captivated mathematicians and computer scientists alike. These numbers hold significant importance in various areas, including cryptography, coding, and more. In this article, we will delve deeper into the world of nth roots of unity, exploring their various uses and providing practical solutions for working with Nth Roots of Unity: An Overview The nth root of unity is a mathematical concept that deals with numbers that, when raised to the power of an integer, yield a value of 1. These numbers are typically denoted as roots of unity and can be expressed in various forms. The simplest form is the unit circle, which is a graphical representation of all the possible values for the nth root of unity. The unit circle shows that these numbers fall into two distinct categories: positive and negative. The positive roots are represented by points on the circumference of the circle, while the negative roots lie outside the circle. In mathematics, nth roots of unity can be written as a power series, where the general form is given by: u e^(2πi/n) where u is the nth root of unity, e is Euler’s number, i is the imaginary unit, and n is the integer that determines the value of u. Using Nth Roots of Unity in Cryptography One of the most significant applications of nth roots of unity is in cryptography. These numbers are used to generate encryption keys that are nearly impossible to break using traditional methods. The process involves a complex mathematical algorithm, which incorporates the nth root of unity as a crucial element. One such example is the RSA encryption scheme, which uses the product of two large prime numbers and the nth root of unity to create a secure key. The RSA algorithm involves generating two random integers p and q, which are both prime numbers. These numbers are then multiplied together, resulting in a number N. The next step involves selecting an integer e that is relatively prime to both p and q. This value is used as the public key, while the product of p and q, along with e and gcd(e,N-1), is used as the private key. The nth root of unity plays a crucial role in this process by allowing for the calculation of the inverse of e modulo N-1. Solving Nth Roots of Unity: A Practical Guide While the nth roots of unity hold significant importance in various fields, they can be challenging to solve analytically. However, there are several practical methods that developers can use to work with these numbers efficiently. One such method is the Chinese Remainder Theorem, which allows for finding the inverse of an integer modulo a product of two integers. This theorem is particularly useful when working with large prime numbers, as it enables developers to solve nth roots of unity problems more easily. Another practical solution is the Extended Euclidean Algorithm, which can be used to find the greatest common divisor of two integers. This algorithm is a crucial tool in many cryptographic applications, including RSA encryption, as it allows for the calculation of the inverse of e modulo N-1. Real-Life Examples: Putting Nth Roots of Unity into Practice To better understand how nth roots of unity can be applied in real-life situations, let us consider a few examples. One such example is the creation of a secure key using the RSA encryption scheme. As mentioned earlier, this process involves generating two large prime numbers p and q, which are then multiplied together to create N. The next step involves selecting an integer e that is relatively prime to both p and q. This value is used as the public key, while the product of p and q, along with e and gcd (e,N-1), is used as the private key. Another example is in the field of image compression, where nth roots of unity are used to reduce the size of digital images without sacrificing quality. This process involves applying a mathematical algorithm that incorporates the nth root of unity to compress the image data. The compressed image can then be transmitted or stored more efficiently, reducing the amount of storage space required. Nth Roots of Unity in Computer Science In computer science, nth roots of unity are used in various applications, including number theory and cryptography. One example is the use of nth roots of unity to generate random numbers that are difficult to predict or reproduce. This is achieved by using the period-doubling algorithm, which involves repeatedly applying the nth root of unity to a starting value until it reaches a desired level of randomness. Another application is in the field of Fourier analysis, where nth roots of unity are used to represent the frequency components of a signal. This allows for efficient processing and manipulation of signals using mathematical algorithms that take advantage of the properties of nth roots of unity. In conclusion, nth roots of unity are a fundamental part of mathematics that hold significant importance in various fields, including cryptography and computer science. While these numbers can be challenging to solve analytically, there are several practical methods that developers can use to work with them efficiently. By understanding the properties and uses of nth roots of unity, developers can create more secure and efficient systems that benefit from these mathematical concepts.	{"url":"https://yeezy2.org/blog/what-are-the-solutions-for-nth-roots-of-unity/","timestamp":"2024-11-10T11:23:53Z","content_type":"text/html","content_length":"42647","record_id":"<urn:uuid:e5ae6599-2c29-4081-a65f-622503b1d21e>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00498.warc.gz"}
Effect of Wrapping Force on the Effective Elastic Behavior of Packed Cylinders When cylinders are packed and wrapped by the bands around the surface, the effective elastic behavior in the cross section of the assembly, which is of significance to its stability and integrity, can be controlled by the wrapping force in the band. The wrapping force is transferred to the cylinders through the Hertz contact between each pair of neighboring cylinders, which is validated by the experiments. The Singum model is introduced to study the mechanical behaviors of the packed cylinders with two-dimensional (2D) packing lattices, in which an inner cylinder is simulated by a continuum particle of Singum and the inter-cylinder force is governed by the Hertz contact model so as to derive the effective stress-strain relationship. The wrapping force will produce configurational forces given a displacement variation, which significantly changes the effective stiffness of the packed cylinders. The hexagonal packing exhibits isotropic elasticity whereas the square packing is anisotropic. The efficacy of our model is demonstrated by comparing the closed form elasticity against the numerical simulation and the previous models. The explicit form of elasticity can be used for packing design and quality control of cable construction and installation. Issue Section: Research Papers Cylinders, Elasticity, Packing (Shipments), Packings (Cushioning), Particulate matter, Stress, Stiffness, Cables, Poisson ratio, Elastic moduli, Granular materials, Young's modulus 1 Introduction Understanding the cylinder or sphere packing problem is an art as much as a science [1], and this challenging problem has a wide spectrum of real-world applications in multiple industries, including textile production and packing, naval, automobile, and aerospace [2]. Many mathematical models have been developed to explore the closest packing [3–5], inspiring the optimized design with improved performances [1,6]. For example, in the civil engineering field, mechanical behaviors of granular materials, also closely related to sphere packing, play significant roles in many aspects, such as pavement construction [7], combating natural hazards (e.g., landslides) [8], excavation planning [9], etc. Experimental testing can measure the macroscopic mechanical behavior of granular materials. Although the adoption of the celebrated photo-elastic grain technique can provide many significant discoveries in granular media [10], it is challenging to quantify the force transfer between grains, which is the origin or underlying mechanism of the mechanical behavior of granular materials [11]. These grain scale forces can be well captured by the adoption of discrete element method (DEM) [12,13], which models the contact between two spheres with the Hertz [14] and Mindlin-Deresiewicz [15] theories. DEM models each particle with both translational and rotational degrees-of-freedom. Although it can simulate the particulate behaviors of the granular materials (i.e., shear banding, necking, etc.), the results are sensitive to the scale and parameters, and thus it is computationally expensive to identify appropriate parameters and model scale in order to reach a practical and convergent prediction of the material behavior. The recently proposed microstructure-based finite element (μFE) model for handling granular medium captures the natural depositional grain scale characteristics of sand (i.e., arbitrary shapes) [16]. Unlike DEM, μFE incorporates deformable grains so that the contact response emerges from the interaction of contacting bodies, thus enabling the modeling of irregular morphologies. However, its key drawback originates from the mesh generation: the surface mesh is a refinement of the constrained Delaunay tetrahedralization [17], and the volumetric mesh filling the grain with tetrahedral elements is bounded by the iso-surfaces. Compared with DEM, the number of degrees-of-freedom of each grain grows from six to hundreds or even thousands, resulting in a tremendous increase in computational cost [18] and hindering its superiority in making a difference in granular material simulation. Continuum mechanics approaches can circumvent the high computational cost issue associated with DEM and μFE. The pilot attempt in modeling granular material with a continuum approach dates back to the middle of the last century when Duffy and Mindlin published a stress-strain relation for identical spheres packed in the face-centered cubic array [19]. This equation was derived by relating the contact force and displacements of a cubic unit cell through equilibrium and compatibility relations. This work opened the door for the following researchers to generalize this constitutive model to other regular packing patterns, such as simple cubic, tetrahedral, etc. [20–22]. Mindlin’s method, however, cannot be generalized to other packing patterns with non-cubic representative elements [22 ], but this difficulty was solved by stress homogenization over volume [23,24]. With the use of the energy conservation approach, the secant stiffness tensor can be obtained for all regular packing patterns [25]. All of these works employ field variables of intrinsic macroscopic nature without explicit connections with the underlying discrete material microstructure. These limitations motivated the development of micromechanical theory for elastic granular media with kinematic degrees-of-freedom included [26], but this theory is in linear form, which obviously contradicts the actual behavior at sphere contacts. Although the force transfer through the contact between particles is through the stress on the contact surface, globally the load transfer through a granular material can be simplified by a lattice network between the center of particles with point-point forces, in which the force is correlated to the center-center distance by a potential function. The recently developed Singum model [27] uses the Wigner–Seitz (WS) cells of a lattice to represent a continuum solid so that the singular point forces can be transformed into the contacting stress between the continuum particle. By applying a displacement variation, from the relationship between the stress and the strain increments, we obtain the elastic constants. This procedure can be applied to general lattice networks and foam materials, which exist in nature or metamaterials, or composites, and the recent work demonstrated its application to a lattice metamaterial with harmonic potential or linear spring bonds [28]. This paper applies the Singum model [27,28] to the assembly of cylinders packing in certain patterns, which are equivalent to a 2D granular materials through the Hertz contacts and can be extended to 3D granular materials in future work. The solution exhibits significance in electric and civil engineering applications. High current electric transmission lines [29] and suspension bridge cables [30 ] are commonly using hundreds to thousands of wires packed in a certain pattern with wrapping bands. For example, Fig. 1(a) shows a suspension bridge supported by two large cables with banded wires, which can be observed by the cross-section of the cable in Fig. 1(b). Moreover, the cable wires (Fig. 1(b)) sustain a majority of the loads applied onto the deck and play an important role in the capacity and performance of the bridge [30]. These wire bundles are formed in a hexagonal arrangement tightened up by wrapping bands at a certain interval. The effective stiffness of packed cylinders in the cross section changes with the stress in the wrapping bands. It has been an empirical art to tighten the bands for the integrity and safety of the cable. A rigorous relationship between the stiffness and the wrapping force will be very useful for those applications, so that a formulation can be provided for the material and structural design given the cable and wire geometry and elastic In the following, the problem will be initially proposed. The Singum model [27] is constructed upon a hexagonal packing pattern, which can be generalized to square packing as well. The force-distance relationship between two neighboring cylinders can be formulated by a pairwise potential using the Hertz contact model. The experiments validate the potential function. The constitutive model for both square packing and hexagonal packing of cylinders is developed. The comparison with numerical simulation results proves the capability and accuracy of this model. The application of the Singum model to suspension bridge cable is demonstrated. The research output will enable scientists and engineers to efficiently predict the multi-scale mechanics of granular materials, thus inspiring the design of new metamaterials. Future studies will extend the current framework to study the poly-disperse of particles in the 3D space. 2 Problem Statement To illustrate the effective elastic behavior of packed cylinders with a wrapping force, numerical confined compression tests in 2D plane strain conditions are performed. Figure 2 shows a bundle of long cylinders confined in a container with four rigid side surfaces, and these cylinders are packed in regular patterns with N[x] and N[y] units along the x and y directions. For simplicity, we consider smooth identical cylinders with diameter d, elastic modulus E, and Poisson’s ratio ν. No friction is considered between the smooth cylinders. The 2D plane strain problem is assumed by constraining the displacement along the z direction. All the boundary platens are assumed to be perfectly rigid. The bottom and left platens are hinged together with the bottom fixed on the ground, and the top and right platens are hinged together. The two parts are assembled together by two elastic links to wrap the packed cylinders while keeping the lattice structure. By shortening the two links simultaneously at the same rate, the contact area between cylinders increases while the lattice structure remains the same. Now keeping the wrapping force the same, we apply an infinitesimal displacement variation on the top platen. From the relationship between the displacement variation and external force, we can measure the tangential elastic modulus at the corresponding state of the wrapping force. Both shear and uniaxial loads can be applied to measure the elastic constants in different directions and loading modes. Since materials have different Poisson ratios, in this research, the effect of a cylinder’s Poisson’s ratio on the overall mechanical properties is studied. Two lattice structures will be considered in this 2D study: square packing and hexagonal packing. In this work, we concentrate on the elastic behavior of 2D lattices through the normal forces of the Hertz contact between cylinders, while the tangential and torsional forces giving rise to irreversible deformations are ignored for the smooth surface. Earlier experimental studies stated that the contributions of shearing and torsional grain contacts are negligible to the volumetric elasticity [31–33], proving the validity of our setting. These two packing patterns show different force transmission mechanisms so the bond length (r) to applied axial strain ($ε$) relation is treated on a case-to-case basis in the subsequent sections. For both cases, we compute the axial stress (σ[a]) and confining stress (σ[c]) with the Singum model for further analysis. To validate the Singum model, our predictions are compared with the direct numerical simulation results computed with our implemented matlab code, and this development process will be introduced in the following section. 3 Formulations This section briefly introduces the Singum model, followed by the derivations for the inter-particle potential for the Hertz contact problem and general nonlinear pairwise interactions, respectively. The elastic constants are calculated from the inter-particle potential function. 3.1 The Singum Construction and Modeling. Yin [27] proposed the concept of Singum model to correlate the pairwise interaction with the elastic constants of solids, paving a way for cross-scale modeling. A Singum particle, constructed by Voronoi decomposition, occupies the space of a WS cell with a particle at the center, filling the entire domain without gaps. For example, a hexagonal packing pattern is illustrated in Fig. 3(a) with a unit cell including one cylinder with six neighboring cylinders. The Singum for cylinder 0 can be constructed in Fig. 3(b) by cutting the six bonds with perpendicular lines forming a hexagon. The original radius of each cylinder is $lp0$, and the center-center distance or bond length will change with the interaction force, written as $2lp=2λlp0$ with $λ=lp/lp0$ being the deformation ratio. Under a hydrostatic load, λ for all bonds shall be the same, which is the case this paper investigates. Consider a continuum particle of Singum 0 subjected to surface forces, the effective stiffness for a linear elastic continuum can be defined from its average stress and average strain as follows: is the stiffness tensor. However, the stiffness of a Singum particle is elastic but not linear. For a nonlinear elastic continuum, the tangential stiffness tensor at the spatial coordinate can be defined in the same fashion by the variations of the stress and strain at the current stress state, which will be illustrated subsequently. The interactions from the neighbors act as point force at the boundary of each edge ( = 1, 2, …, 6) of the Singum 0, and because of the equilibrium in the absence of body force, the boundary condition is written as $σij(x)ni(x)=ΣI=16FjIδ(x−xI)\,for x∈∂VS$ ) is the Dirac Delta function; are the Cauchy stress tensor and surface out-normal vector of a continuum particle, respectively. The stress integral with a Singum particle can be written as follows [ is the deformed area of the Singum particle with the initial area ; the point force between two smooth cylinders is expressed as the derivative of the pairwise potential ] as where the potential function can be obtained by the Hertz contact model [ ], which will be elaborated in the next section. The Cauchy stress within the Singum particle can be computed as the volume average of the stress integral To test the tangent stiffness of the overall structure, we apply an incremental strain variation at every point = 1/2( ) represents a linear displacement gradient tensor, which is related to the variation of the Eulerian strain at the current configuration of a stretch ratio as [ The variation of Eq. with the aid of Eq. where the Cauchy stress variation includes three parts: the first part related to is caused by the force variation, which leads to the material configuration variation; whereas the second and third parts related to are the configurational stress caused by the existing force with the material configuration change. For the classic elasticity based on the infinitesimal deformation assumption, the effect of the configuration change on the material behavior has often been disregarded, but its effect is real and physical [ ]. Its effect on the elastic constants will be illustrated subsequently. By relating the variations in the Cauchy stress and Eulerian strain with the aid of Eq. , the tangent stiffness tensor can be evaluated [ ] as , is the component of the unit vector from the center of a Singum particle to its neighbors; and the superscript of 0 can be disregarded because each pair of the bond share the same center-center distance. It is shown in Fig. that the total number of neighbors changes for different packing patterns and here is 6. The summation in Eq. is reduced to the summation of , which can be written in the following identities for the hexagonal lattice Substituting Eq. into Eq. , the relation between stiffness tensor of hexagonal lattice and pairwise potential can be written as where the pairwise potential ) can be obtained by the experiments or Hertz’s model, which will be introduced in the next section. It is interesting that the hexagonal lattice exhibits an isotropic elasticity in the cross section, which has been observed in the graphene lattice as well [ 3.2 Singum Potential for the Hertz Contact Problem. The Hertz contact theory deals with the mechanics at the contact between non-conforming solids. This theory builds up on the simplification that each body can be regarded as an elastic half-space loaded over a small elliptical region of its plane surface to calculate the local deformation and stress distribution [14]. The Hertz contact theory assumes infinitesimal contact strain and frictionless contact surface to make the aforementioned simplification justifiable. Although Hertz’s model for the 3D spherical case has been well established [35], the 2D case cylinder contact problem exhibits different forms of force-deformation relations, mainly two categories: implicit and explicit. Johnson [35], Radzimovsky [38], and Goldsmith [39] models mutual approach as an implicit function of contact force in similar forms with logarithmic function included, which requires an iterative process for the solution of P at each given indentation δ, thus limiting their applications in computational programs [40]. In view of this shortcoming, Lankarani and Nikravesh [41] proposed a simplified explicit model considering energy dissipation during the impact process, making it well-suited for implementation, especially for dynamics problems. However, there is a parameter to be determined empirically. In this work, Johnson’s model is selected because its P − δ prediction well fits the results of both the finite element simulations and the experimental testings, which are shown in the Appendix. This research studies the contact between two identical cylinders with undeformed radius shown in Fig. . Compressed by a given load per unit length, the half-width of the rectangular contact area is given by [ ] as and the corresponding mutual approach is written as from which ) can be obtained implicitly. The potential function ) can be written as ) can be numerically solved by Eq. ; and the integral is only the half of the bond so that a multiplier of two is applied. Alternatively, the formulation can be simplified by replacing with a single dimensionless variable for a more concise form. The derivatives of ) and ) are written as follows: where the derivatives of can be further derived from Eq. Because the one-to-one mapping between is given by Eq. , one can use the above equation to explicitly obtain the derivatives of and stiffness tensor . Note that when = 1, = 0, Eq. = 0 and → ∞, which causes small but significantly changing elasticity when is close to one. 3.3 Elastic Constants of the 2D Lattices Varying With the Wrapping Force. Substituting the derivatives of Singum potential ) and ) into the Singum model Eq. , one can compute the stiffness tensor of the packed cylinders given the 2D packing lattice structure, and the corresponding compliance matrix in the Voigt notation is written as where the notations 1, 2, and 4 represent 11, 22, and 12 respectively. From the aforementioned equation, the elastic modulus is derived as Note that although Eq. shows an isotropic elastic tensor for hexagonal packing pattern with /2(1 + ), when the cylinders are distributed in other patterns, the elastic tensor can be anisotropic. For example, the square packing pattern will not satisfy /2(1 + ), which will be shown subsequently. Oblique packing or other patterns may not satisfy The wrapping force F will produce contact force P between cylinders and change λ from 1 at the undeformed state with the zero wrapping force. Therefore, once the stiffness tensor C is written in terms of λ, the dependence of the elastic modulus on the wrapping force can be obtained. In the following, two packing lattices are considered, respectively. For hexagonal packing, Fig. shows the Singum particle of the 2D hexagonal lattice, and its volume is . Inserting the pairwise potential for the Hertz contact model, the components for elastic tests can be written as Using Eqs. , the relation between effective elastic tensor and contact force can be obtained. Using the relation between , and wrapping force, one can design and control the stiffness, which will be demonstrated, subsequently. For square packing, Fig. shows the Singum particle of the 2D square lattice, and its volume is . Figure shows the directions of the unit vectors of each loads. The summation in Eq. is reduced to the summation of , which can be written in the following identities for the square lattice, Substituting Eq. into Eq. , the relation between stiffness tensor of square lattice and pairwise potential can be written as Similarly to hexagonal packing lattice, one can write the components for elastic tests as Using Eqs. , the relations between effective elastic modulus and contact force can be obtained for further analysis. 4 Results and Discussion The Singum model provides the closed form of elasticity, which considers the effects of the wrapping force or contact force. This section will first verify the model by numerical experiments, demonstrate the accuracy of the model, and then apply it to the bridge cable design and analysis. 4.1 The Setup of the Numerical Experiments. A matlab program is developed to perform numerical experiments serving as a benchmark for validating the proposed Singum model. The overall flow of this program is as follows: In the initialization process, an array of cylinders with a radius of $lp0$ are automatically generated based on the given lattice and the corresponding inputted N[x] and N[y] numbers. For example, Fig. 4(a) schematically illustrates N[x] = N[y] = 5 with 5 × 5 blue cylinders. A list of neighbors for each cylinder is detected and saved for the force computation step. To simulate a hydrostatic loading, which causes all the bonds to shrink by the same ratio, we update both x and y coordinates of all the particles to $x=X(1−ε)$ and $y=Y(1−ε)$, respectively, under any given strain $ε$. This deformation process is clearly illustrated in Fig. 4(b), where the contact forces emerge at the highlighted red overlaps. For each pair of particles in contact, namely x[i] and x[j], the magnitude of the contact force is evaluated with Eq. (13), and its direction is accurately defined as the unit vector connecting the centers of two circles. The required hydrostatic wrapping force can be computed from the contact force P based on the assembly of the cylinders, which will be demonstrated subsequently. The modeling results exhibit a certain variation when N[x] and N[y] are small due to the boundary effect. However, when N[x] and N[y] are larger than 20 × 20, the variation becomes negligible. In the following numerical experiments, N[x] = 100 and N[y] = 101 are used as the default configuration. The following factors may affect the accuracy in actual applications: 1. The Hertz contact in Eq. (13) assumed that contacted width 2b is equal to the arc of the cylinder. When the contact area is larger, the applicability of this assumption becomes questionable, thus limiting the validity of the current contact model to infinitesimal strain only with the small contact area. 2. The friction between cylinders will play a significant role in actual experiments and applications with a shear load, whereas the present model assumes the perfectly smooth surface of cylinders. 3. The present potential function V(r) is derived from the linear theory of elasticity; whereas nonlinear elastic or inelastic behavior of the materials may produce a considerable discrepancy in actual applications as the contact zone exhibits stress concentration and thus large strain. 4. The displacement δ was calculated by the line integral of the strain along the center-centerline of the two-particle contact; whereas a particle is in contact with more than three particles in the actual applications. Therefore, although the Hertz contact model has been widely used for granular materials in the literature, the accuracy of Eq. (13) is limited to infinitesimal strain conditions. However, the Singum model can be applied to general potential functions between particles. The present Hertz contact model can be straightforwardly replaced if a P − δ curve for large deformation can be developed numerically or experimentally, from which the potential function can be obtained by the path integral. Particularly, when the balls are hollow, large deformation is expected. Some experimental studies are underway. In this section, we will use the Singum model to explore the mechanics and physics of packed cylinders with wrapping stresses. Without loss of the generality, we take the Young’s modulus of the cylinder to be E = 210 GPa, and consider a range of Poisson’s ratio, i.e., v = 0.1, 0.3 and 0.5, to check how it affects the effective material properties. 4.2 The Compressibility of 2D Lattices Varying With λ. To measure the compressibility of the packed cylinders, we perform a strain-induced hydrostatic compression test. We apply an incremental strain of , and combining with Eqs. , the incremental hydrostatic stress , which is generated by the wrapping force, will lead to an incremental mean stress as is the incremental volume strain. The incremental mean stress is related to plots the hydrostatic stress required to uniformly compress bond length to of its original length for square and hexagonal lattices. The Singum prediction agrees very well with the numerical experiments, justifying the correctness of the Singum model. However, the computational resources consumed by Singum approach are much less than those taken by the numerical experiments because no particle generation or neighbor detection is needed for explicit solutions. For both lattices, as the cylinder’s Poisson’s ratio gets larger, a higher compressible load is required to compress the sample by the same amount. It is because the effective elastic modulus /2(1 − ) in Eq. increases with . It is noted that at each , the ratio between of hexagonal lattice and of the square lattice is One reason is the packing density of hexagonal packing (0.907) is 15% higher than that of square lattice (0.785). The other reason stems from the difference in force transmission mechanism within the square and hexagonal lattices. The relation between wrapping force and the stretch ratio is the basis of subsequent analysis of the effective elastic modulus. 4.3 Young’s Modulus and Poisson’s Ratio of 2D Lattices Varying With σ[m]. The dependence of stiffness tensor on can be noticed from Eqs. and Eqs. . Note that although both lattices show Young’s modulus and Poisson’s ratio the same in both directions, the hexagonal lattice is isotropic whereas the square lattice is not because it does not /2(1 + ). Putting them together with Eqs. , the effective Young’s modulus and Poisson’s ratio for square lattice can be derived as Similarly, for hexagonal lattice, we have Figure 8 plots how the effective Young’s modulus (E) varies with σ[m] for square and hexagonal lattices. Generally speaking, for both lattices, an increasing trend can be observed as the wrapping force increases, indicating that the effective E of packed cylinders can be manipulated by adjusting the wrapping force. A special point is that a sudden jump in E occurs at the moment when a very small wrapping force is applied compared to the loose condition because of P[,λ] = 0 and P[,λλ] → ∞ in the neighborhood of λ = 1. This jump indicates the power of wrapping on significantly increasing the effective E. In addition, given the same wrapping force, the sample with a higher cylinder ν exhibits a higher effective E. This phenomenon may be due to the same reason that E/2(1 − ν^2) in Eqs. (16) and (33) increases with ν. Comparing the square lattice with the hexagonal lattice, we can notice that to get a similar elastic modulus, a higher hydrostatic stress is required for the hexagonal lattice, and combined with σ[m] − λ relation in Fig. 7, the difference in E at the same stretch is similar. Figure 9 plots how the effective Poisson’s ratio (ν) varies with wrapping force for hexagonal lattices. Note that ν for square lattices increases from zero at an undeformed state to a small finite number as the sample is compressed. This unphysical phenomenon observed at first glance is actually true because the uniaxial compression will cause the side area parallel to this compression to become smaller, leading to an increase in stress, which gives the nonzero ν value. This is the power of configurational force. However, the Poisson’s ratio for hexagonal lattice is far from zero. When λ → 1, ν → 1/3; whereas ν increases as λ decreases. Unlike the square lattice, the effective ν for hexagonal lattice is almost 0.35 although a slight increase can be observed by increasing the wrapping force. The reason for this finite ν stems from the lattice geometry: force can be transmitted from vertical direction to lateral direction through the incline bonds. 4.4 Shear Modulus of 2D Lattices Varying With σ[m]. Following the same logic as the effective , the effective shear modulus for square lattices is and the for hexagonal lattices is plot how the effective shear modulus ( ) varies with wrapping force for square and hexagonal lattices. The square lattice exhibits a negative shear resistance provided in Eq. , which is not physical but shows the instability of the square lattice under shear loading. In reality, with any shearing disturbance, the square lattice under a hydrostatic load will collapse and start transforming into the hexagonal lattice. Similarly to the effective , the of the hexagonal lattice increases with wrapping force. Also, compared to the loose case, a sudden jump in shear modulus can also be observed by wrapping the cylinders with only a small amount of force. A higher shear modulus can be observed for cylinders with higher in Fig. Overall, the prestress provides significant effects on the effective elasticity of the assembly of the lattice of the cylinders. The simple, explicit form of elasticity enables the design of lattice materials with programmable mechanical properties by adjusting the wrapping force. Although the Hertz contact model has been used for deriving the potential function of Eq. (15), the present model can be applicable to the general form of the potential function V(λ), which can be determined by the experiments directly or by other models. 4.5 Comparisons With the Existing Models in the Literature. In the field of constitutive modeling of granular materials, two main streams are kinematics and static approaches, which mainly solve for the movements of particles and the closed-form elastic modulus, respectively [42]. Because of similarities with the Singum model, the static approach is chosen for comparisons. Following Chang’s work [25], we derived the stiffness tensor components for both square and hexagonal lattices as Combining with Eqs. , the effective elastic modulus for both lattices is written as For square lattice, the predicted by this model are all zero, which is different from the numerical experiments and the Singum prediction when a prestress exists or the lattice is subjected to a hydrostatic stress. The issue is caused by the effect of configurational change. The similar physics has been investigated by Eshelby of the existing force’s effect on the configuration change by crack propagation [ ]. The concept of configurational force or material force has been applied to multiphysical problems [ ]. The configurational stress in Eq. captures the non zero Poisson’s ratio for square lattices, which highlights the physical rigor of the Singum model. On the other hand, when = 1, for configurational stress is indeed zero with = 0, the above equations can be recovered from the Singum model as well. In addition, unlike the tangent stiffness, the secant stiffness is not convenient to be adopted for stress updates in numerical simulations with incrementally increased strain. Additionally, the volume to which stress is homogenized was referred to as the undeformed volume, limiting the applicability to the infinitesimal strain range, while the Singum model can be straightforwardly extended to finite deformation with a high-fidelity interaction potential function. However, the present Singum model still exhibits its own limits that the established models address specifically, such as plastic deformation of particles and friction between particles, etc., which shall be considered in future work by including moment and shear force at the cutting points at the Singum surface. 4.6 Case Study of a Suspension Bridge Cable. Cables in suspension bridges are a major load-bearing structural component. Because square packing is unstable for sharing, hexagonal packing is always used in the actual application of suspension bridge cables. As an example, in the George Washington Bridge in New York City, the large cable contains 9061 wires in Fig. , and this cable is chosen as a case study to demonstrate the great potential of Singum model in assisting control of the effective elastic modulus by adjusting the wrapping force in the bands. Composed of the 9061 wires with a radius of 2.413 mm each, the cable has a radius of 243.5 mm and is banded by clamps with a width of 200 mm spaced at 6.096 m [ ]. The Young’s modulus and Poisson’s ratio of the steel wires are 210 GPa and 0.3, respectively. Force transfer between each cable is simulated with Johnson’s model in Eq. . At each band location, the related mean hydrostatic pressure and the stretch ratio at each wire contact point can be computed with Eq. , and here relates the deformed and undeformed radius of the overall cable. The hoop stress in the band is related to are the deformed radius and thickness of a band. The wrapping force can be straightforwardly computed as plots the relationship between the deformed radius and wrapping force, and similarly to Fig. , an increasing trend is noticed as the cable gets more compressed. With the Singum model, one can easily predict the value of wrapping force in the cable bands once a pair of deformed and initial radii are given. More interestingly, the effective elastic moduli at different wrapping forces are displayed in Fig. . Their overall trends are similar to the aforementioned results. The key information conveyed by this plot is that we can quantitatively adjust the effective elastic modulus by controlling the wrapping force in the bands, or by adjusting the cable deformed radius. Once generalized to other applications with packed cylinders, the Singum model will make a significant impact on material and structural design. 5 Conclusions In this research, we extended the Singum model by deriving the pairwise potential considering the Hertz contact between two cylinders, enabling the Singum model to efficiently predict the tangent stiffness tensors of particles packed in regular lattices in 2D plane strain conditions. To select an appropriate contact model, we performed experiments and finite element analysis on cylinder samples of different material properties, and Johnson’s model is selected in the Singum model for deriving the inter-cylinder potential. Both square and hexagonal lattices are considered in this research to show the versatility of the Singum model. The dependence of effective elastic modulus on wrapping force predicted by Singum model agrees very well with the numerical verification, regardless of the packing lattices and material properties of cylinders. It is interesting to show the hexagonal lattice exhibits isotropic elasticity while the square lattice anisotropic in the 2D space. The solution can be used in the design of cylinder packs with controllable mechanical properties via adjusting the wrapping force. The significance is demonstrated in our case study on designing cables for suspension bridges. The superiority of the Singum model is demonstrated by performance comparison with common strategies for constitutive modeling of granular materials in literature. In addition to the 2D lattices, the Singum model can be extended to modeling granular materials consisting of spheres packed in 3D lattices, such as face-centered cubic, body-centered cubic, and simple cubic. The research output will shed light on investigating the mechanics of packed cylinders and exploring the optimized design of packing problems. This work is sponsored by the National Science Foundation IIP #1738802, IIP #1941244, CMMI#1762891, and U.S. Department of Agriculture NIFA #2021-67021-34201, whose support is gratefully acknowledged. Dr. Zadshir and Dr. Yin thank the support from NASA SBIR #80NSSC22PB164. We thank Dr. Liming Li and Mr. James Basirico for their insightful discussion and help with the experiments. We are specially grateful to Professor Raimondo Betti for sharing his photos on bridge cables with us. Dr. Yin conceptualized and supervised the project and paper writing; Cui conducted the numerical studies and drafted the paper; Dr. Zadshir co-advised the project and supervised the experiments; and Teka conducted the experiments and data analysis and wrote the experimental part. Conflict of Interest There are no conflicts of interest. Data Availability Statement The datasets generated and supporting the findings of this article are obtainable from the corresponding author upon reasonable request. Appendix: Numerical Verification and Experimental Validation of Johnson’s Model As a result of the complexity of the cylinders in the contact problem, various types of models have been proposed to describe the relationship between contact force (P) and mutual approach (δ). The most well-known models include Johnson’s model [35], Radzimovsky’s model [38], Goldsmith’s model [39], and Lankarani and Nikravesh’s model [41]. Following these models, the relation between deformed cylinder radius l[p] and contact force P is summarized below: Johnson’s model: Radzimovsky’s model: Goldsmith’s model: Lankarani and Nikravesh’s model: These models are specifically valid for problems of different contact types, materials, and dimensions [ ]. In order to choose the model that best fits our research problem, we compared the performance of all these four models against finite element analysis and experimental results. Experimental tests were conducted at the Carleton Laboratory, Columbia University, to investigate the load-displacement (P − δ) relationship. A universal testing machine (UTM) with a maximum capacity of 34 kips (150 kN)was used to apply a compression load. An abrasion-resistant polyurethane rubber rod was acquired from McMaster with Part Number 8695K693 in July 2022 and cut into three specimens with a diameter 54 mm, and varying lengths of 97.8 mm, 103 mm, and 104 mm, respectively. The Poisson’s ratio and Young’s modulus can be determined using mechanical, acoustic, or optical methods [ ]. For this paper, the machine measurement method is used by applying a uniaxial force to the test specimen; the axial force is measured by the universal testing machine as shown in Fig. , and the axial and transversal strains are measured by the strain gauges on the rubber specimen. The Poisson’s ratio is the negative ratio of the transverse strain to the axial strain. The Young’s and Poisson’s ratio are calculated using the axial strain data from the experiment as follows. Using Eq. (A5), the Young’s modulus (E) and Poisson’s ratio (ν) are calculated to be 470 mPa and 0.5, respectively in the linear elastic range. In order to determine the P − δ relationship of the cylindrical rubber, experimental tests were performed on the three specimens as shown in Fig. 12(b). The specimens were laid in the horizontal direction and held in place using steel plates that were oiled to prevent friction. The specimens were loaded uni-axially using a displacement control of 0.762 mm/min. The time, load, deflection values for all tests were recorded through a data acquisition system. The load-displacement data of all three specimens were averaged and used for comparison as shown in Fig. 13. In addition, the error bars shown in the figure, demonstrate that the load-displacement relationship of the three specimens is very close to each other. Note that although only one cylinder is used in the test, because the steel platens can be considered to be a rigid surface, and with a mirror symmetry it can reproduce the deformation pattern of the contact of two identical cylinders in the finite element method (FEM) simulation of Fig. 12(c). In order to further verify that an appropriate contact model is selected, we performed finite element analysis with abaqus 2019. As shown in Fig. 12(c), the model geometry strictly follows the experimental configurations, and the material is set to be linearly elastic with Young’s modulus and Poisson’s ratio the same as measured in the experiment. The contact between two cylinders is defined as frictionless and hard contact, which minimizes the penetration of the secondary surface into the primary surface and does not allow the transfer of tensile stress across the interface. Figure 13 shows the comparison between those aforementioned contact models. Note that the Lankarani and Nikravesh’s (LN) model exhibits a parameter n, which was recommended in the range of [1, 1.5]. Here, the case of n = 1 shows too much off from those from both experiments and finite element analysis; while other cases of n might fit the experiments better. Although the LN model exhibits the advantage of its explicit form solution, to avoid the empirical calibration of the parameter n, we turn to other three models. Johnson’s, Radzimovsky’s, and Goldsmith’s models provide similar predictions, which match reasonably well with the experimental result. Note that rubber exhibits hyperelastic behavior with nonlinear elastic moduli at different levels of strain. Because the single values of E and ν in the linear elastic range are used in the contact models, some deviations from the experimental results are anticipated. Using the linear elastic constants, the finite element results can provide another reference to the contact problem, which agrees very well with Johnson’s prediction. Therefore, Johnson’s model is chosen in this research to derive the potential function for the contact problem between two cylinders. In actual applications, because the stress at the contacting surface and its neighborhood is much higher than the rest part, the non-linearity of elasticity or inelastic behavior of the material may affect the accuracy of the contact models. Moreover, the friction between particles may change the contact mechanics as well. For multiple contacts between particles, the pairwise contacts may exhibit some loss of accuracy. Therefore, although Johnson’s model provides good agreement with the present FEM and experimental results, its applicability to different materials may change with the load levels and testing geometry or configuration, particularly for finite deformation of many particle systems. More investigation of the applicability of those models is underway. However, as long as a high-fidelity P − δ curve is provided, the present Singum model can straightforwardly use it in the same fashion. , and , “ Packing Spheres Tightly: Influence of Mechanical Stability on Close-Packed Sphere Structures Phys. Rev. Lett. ), p. , and , “ A Literature Review on Circle and Sphere Packing Problems: Models and Methodologies Adv. Operat. Res. , p. . . Y. G. , “ Mathematical Methods for Geometric Design Advances in CAD/CAM, Proceedings of PROLAMAT Leningrad, USSR May 16–18, 1982 , Vol. , pp. P. G. M. C. L. 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D. , and , “ Elastic Spheres in Contact Under Varying Oblique Forces , and , “ Comparison Between a μFe Model and DEM for an Assembly of Spheres Under Triaxial Compression EPJ Web of Conferences Montpellier, France July 3–7 , Vol. EDP Sciences , p. , “ Computational Geometry: Theory and Applications , and , “ A Micro Finite-Element Model for Soil Behaviour: Numerical Validation ), pp. , and , “ Stress-Strain Relations and Vibrations of a Granular Medium , “ A Differential Stress-Strain Relation for the Hexagonal Close-Packed Array of Elastic Spheres , “ Stress-Strain Relations for a Simple Model of a Granular Medium , and , “ Elastic Constants of Cubical-Tetrahedral and Tetragonal Sphenoidal Arrays of Uniform Spheres Proceedings of International Symposium of Wave Propagation and Dynamic Properties of Earth Materials Albuquerque, NM Aug. 23–25 , pp. , “ Statistical Consideration on Deformation Characteristics of Granular Materials Proceedings, US-Japan Seminar on Continuum Mechanical and Statistical Approaches in the Mechanics of Granular Materials Sendai, Japan June 5–9 S. C. , and , eds., Gakujutsu Bunken Fukyu-Kai, Tokyo, pp. , and , “ A Micromechanical Definition of the Cauchy Stress Tensor for Particulate Media. Mechanics of Structured Media Proceedings of International Symposium on Mechanical Behavior of Structured Media Ottawa, Canada May 18–21 A. P. S. , ed., pp. , “ Micromechanical Modelling of Constitutive Relations for Granular Material Stud. Appl. Mech. , pp. V. T. , and , “ Microstructural Mechanics of Granular Media Mech. Mater. ), pp. , “ A Simplified Continuum Particle Model Bridging Interatomic Potentials and Elasticity of Solids J. Eng. Mech. ), p. , “ Generalization of the Singum Model for the Elasticity Prediction of Lattice Metamaterials and Composites ASCE J. Eng. Mech. (in press). J. K. , and G. G. , “ The Roosevelt Island Tramway Modernization Project Forensic Engineering 2012: Gateway to a Safer Tomorrow San Francisco, CA Oct. 31–Nov. 3, 2012 , pp. , and , “ Physics-Based Stochastic Model to Determine the Failure Load of Suspension Bridge Main Cables ASCE Comput. Civil Eng. ), p. J. M. , “ Stress-Strain Characteristics of Cohesionless Granular Materials Subjected to Statically Applied Homogenous Loads in an Open System ,” Ph.D. thesis, California Institute of Technology Pasadena, CA R. D. , “ Compliance of Elastic Bodies in Contact , and R. F. , “ Deformation of Sand in Hydrostatic Compression J. Soil Mech. Found. Div. ), pp. Micromechanics of Defects in Solids , Vol. Kluwer Academic Publishers Dordrecht, The Netherlands , vol. , p. K. L. , “ Contact Mechanics J. App. Mech. ), pp. , “ Improved Singum Model Based on Finite Deformation of Crystals With the Thermodynamic Equation of State ASCE J. Eng. Mech. G. A. Configurational Forces: Thermomechanics, Physics, Mathematics, and Numerics CRC Press New York E. I. , “ Stress Distribution and Strength Condition of Two Rolling Cylinders Pressed Together ,” Technical Report, University of Illinois at Urbana Champaign, College of Engineering Urbana, IL Impact: The Theory and Physical Behaviour of Colliding Solids Edward Arnold C. M. A. L. , and J. A. , “ A Critical Overview of Internal and External Cylinder Contact Force Models Nonlinear Dyn. ), pp. H. M. , and P. E. , “ Continuous Contact Force Models for Impact Analysis in Multibody Systems Nonlinear Dyn. ), pp. C. S. S. J. , and , “ Estimates of Elastic Moduli for Granular Material With Anisotropic Random Packing Structure Int. J. Solids Struct. ), pp. J. D. , “ The Force on an Elastic Singularity Philos. Trans. R. Soc. Lond. Ser. A ), pp. , and , “ Advances in Test Methods for Poisson’s Ratio of Materials Mater Rev. ), pp.	{"url":"https://verification.asmedigitalcollection.asme.org/appliedmechanics/article/90/3/031003/1150225/Effect-of-Wrapping-Force-on-the-Effective-Elastic","timestamp":"2024-11-02T01:06:59Z","content_type":"text/html","content_length":"422884","record_id":"<urn:uuid:eb92afa1-8ac2-4eac-bb95-afbe7a3645c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00463.warc.gz"}
Math Problem Statement An agency collects demographics concerning the number of people in families per household in a certain country. Assume the distribution of the number of people per household is as shown in the table to the right. a. nbspa. Calculate the expected number of people in families per household in the country. Compute the standard deviation of the number of people in families per household. nbsp nbsp nbsp nbsp 22 Question content area bottom Part 1 a. The expected number of people in families per household is enter your response here Ask a new question for Free By Image Drop file here or Click Here to upload Math Problem Analysis Mathematical Concepts Expected Value Standard Deviation Expected value formula: E(X) = Σ[x * P(x)] Standard deviation formula: σ = sqrt(E(X²) - [E(X)]²) Law of Large Numbers Properties of Expectation Suitable Grade Level Grades 10-12	{"url":"https://math.bot/q/expected-value-standard-deviation-household-demographics-FMOFQdMb","timestamp":"2024-11-01T22:10:33Z","content_type":"text/html","content_length":"89030","record_id":"<urn:uuid:11c909be-9f8b-49f9-9dcf-aebc57fc3fe4>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00048.warc.gz"}
How do you calculate per capita biology? The complete formula for annual per capita growth rate is: ((G / N) * 100) / t, where t is the number of years. Finding the annual per capita growth rate, as opposed to only the rate for the entire time period, makes it easier to predict future population changes because it relates to both time and overall population. What is per capita growth rate in biology? First, population size is influenced by the per capita population growth rate, which is the rate at which the population size changes per individual in the population. This growth rate is determined by the birth, death, emigration, and migration rates in the population. What does per capita mean biology quizlet? Per capita birth rate. number of offspring produced per unit time by an average member of the population. What is mean by per capita birth rate? Per capita birth rate (natality rate) is the number of births per individual per unit time interval. What is per capita death rate biology? โ ข The per capita death rate is the number of individuals that die per unit time (mortality. rate is the same as death rate) โ ข Example: In a population of 750 fish, 25 dies on a particular day while 12 were born. How do you calculate per capita output? Key Takeaways. Per capita gross domestic product measures a country’s economic output per person and is calculated by dividing the GDP of a country by its population. What is the per capita growth rate of the global human population? Global human population growth amounts to around 83 million annually, or 1.1% per year. How do you calculate population growth biology? The annual growth of a population may be shown by the equation: I = rN (K-N / K), where I = the annual increase for the population, r = the annual growth rate, N = the population size, and K = the carrying capacity. How do you calculate GDP per capita growth? Annual growth rate of real Gross Domestic Product (GDP) per capita is calculated as the percentage change in the real GDP per capita between two consecutive years. Real GDP per capita is calculated by dividing GDP at constant prices by the population of a country or area. Which represents the per capita growth rate of a population? The correct answer is (A) line A. Per-capita rate increase (r) is defined as the amount of population increase divided by the population size, and it is related to the difference in birth and death rates in a population. Populations with a higher per-capita growth rate grow more quickly. Which of the following describes why ecologists can use per capita growth rate to help predict how a population will grow? Which of the following describes why ecologists can use per capita growth rate to help predict how a population will grow? Per capita growth rates indicate the change in the number of individuals in population over a change in time. How is exponential growth defined quizlet? Exponential growth occurs when the individuals in a population reproduce at a constant rate. Logistic growth occurs when a population’s growth slows or stops following a period of exponential growth. What is per capita birth and death rate? Solution : In a populatio birth rates refers to per capital birth and death rate refers to per capita deaths. b) The birth and death rates expressed in change in numbers (increase or decrease) with respect to the membes of the populations. Who has the lowest birth rate in the world? South Korea records world’s lowest fertility rate — again The country’s fertility rate, which indicates the average number of children a woman will have in her lifetime, sunk to 0.81 in 2021 — 0.03% lower than the previous year, according to government-run Statistics Korea. What does per capita measure? Per capita is a Latin term that translates to “by head.” Per capita means the average per person and is often used in place of “per person” in statistical observances. The phrase is used with economic data or reporting but is also applied to almost any other occurrence of population description. What is the purpose of using per capita? Per capita is a measurement that helps compare different nations statistics on a ‘per person’ basis. Per capita helps economists compare GDP figures by accounting for large differences in population. By using per capita, economists are able to more accurately compare the size of two nations economies. What does a high GDP per capita mean? In other words, when an economy generates more value per person per year, that typically translates into more money for those working in that economy. Most often, the indicator economists use to determine the prosperity, or well-being, of a country or region is GDP per capita. What is 1% of the world population? With a world population at approximately 7.8 billion, one percent would be about 78 million. Which country has the fastest growing population in the world? In South Sudan, the population grew by about 5.05 percent compared to the previous year, making it the country with the highest population growth rate in 2021. Which country has the most population? China has the world’s largest population (1.426 billion), but India (1.417 billion) is expected to claim this title next year. The next five most populous nations โ the United States, Indonesia, Pakistan, Nigeria and Brazil โ together have fewer people than India or China. Which factors are used to calculate population growth? Births, Deaths, and Migration. Population growth rate depends on birth rates and death rates, as well as migration. First, we will consider the effects of birth and death rates. You can predict the growth rate by using this simple equation: growth rate = birth rate โ death rate. What is the growth rate of a population? Definition: The annual average rate of change of population size, for a given country, territory, or geographic area, during a specified period. It expresses the ratio between the annual increase in the population size and the total population for that year, usually multiplied by 100. What is the population formula? Population formula in economics is used to determine the economic activity of the country or area. Population percentage is the formula to divide the target demographic by the entire population, and then multiply the result by 100 to convert it to a percentage. What is the difference between GDP and GDP per capita? 1. GDP is a measure of a nationร s economic health while GDP per capita takes into account the reflection of such economic health into an individual citizenร s perspective. 2. GDP measures the nationร s wealth while GDP per capita roughly determines the standard of living in a particular country. Why is GDP per capita important? Gross Domestic Product (GDP) per capita is a core indicator of economic performance and commonly used as a broad measure of average living standards or economic well- being; despite some recognised shortcomings. For example average GDP per capita gives no indication of how GDP is distributed between citizens.	{"url":"https://scienceoxygen.com/how-do-you-calculate-per-capita-biology/","timestamp":"2024-11-05T07:26:12Z","content_type":"text/html","content_length":"309239","record_id":"<urn:uuid:bb52be90-a0a8-408c-a7ad-8c8e29fc59ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00185.warc.gz"}
Programming Interview Questions If you're a programmer aspiring to work in a top-tier tech company like Google, Microsoft, Apple, or Facebook - you're probably concerned with the interview process. These interviews can be daunting, especially if you're not familiar with the type of questions that you'll be expected to answer. This is the reason why we've decided to compile this selection of programming interview questions you might face in an interview tomorrow, hopefully helping you both learn about them, understanding their nature, as well as reinforcing your knowledge. Most of these questions revolve around Data Structures and Algorithms so the most attention is given to them. Although the focus is on Data Structures and Algorithms, we'll cover a few logical questions you might run into as well! Most of these questions and examples will also contain some basic-level explanations, since, in my humble opinion, a lot of terminologies are being used without proper understanding of what they actually mean. It's worth mentioning that a certain degree of mathematical knowledge here is important. Don't worry, you don't have to be a mathematical genius, but knowing Discrete Mathematics is a huge advantage. Data Structure and Algorithms A Data Structure is a fundamental and simple concept. It's a way of organizing data and their relationship to allow efficient operations to be performed on them. There are many data structures that programmers and developers regularly use such as: • Arrays • Binary Trees • Graphs • Linked Lists • Matrixes • Stacks • Queues • Heaps • Hash Tables If you'd like to see the whole list, it's quite lengthy. An algorithm is a piece of code that represents a certain set of instructions, usually selected to act as a finder to the solution of a specific problem. They should be efficient and fast, which means that they both take the least possible time to complete as well as consume the least possible memory space, depending on the nature of the algorithm and the problem at hand. Knowing your data structures and algorithms is important. It allows you to understand the underlying logic behind the tools you use every single day. Knowing which algorithms to use and which data structures to employ is a valuable thing in a production environment. The ability to pick out an efficient solution compared to another is crucial. It also incites intuitive ways to solve problems that you might be faced with, and depending on what kind of person you are - it might be fun to refresh a bit on some high-school mathematics. Graph Data Structure Interview Questions Linked List Interview Questions Dynamic Programming Interview Questions (coming soon) • Fibonacci Number Sequence • Longest Common Subsequence Sorting an Searching Interview Questions (coming soon) Free eBook: Git Essentials Check out our hands-on, practical guide to learning Git, with best-practices, industry-accepted standards, and included cheat sheet. Stop Googling Git commands and actually learn it! • Binary Search • Bubble Sort • Insertion Sort • Merge Sort • Heap Sort • Quick Sort • Interpolation • Tree/Binary Search Tree • Minimum Depth • Maximum Path Sum Number Theory Interview Questions (coming soon) • Euclid's GCD Algorithm • Extending Euclid's GCD Algorithm • Diophantine Equation • Chinese Remainder Theorem • Modular Inverse • Semi-Perfect Numbers String Interview Questions (coming soon) • Reversing a String • Checking if String contains only digits • Finding Duplicate Characters in a String • How to Convert a String to Integer • Removing Duplicate Characters in a String • Finding the Maximum Occuring Character in a String • Find the First Non-Repeating Character in a String • Checking if Two Strings are Anagrams of Each Other • Counting the Number of Words in a String Array Interview Questions (coming soon) • Finding the Missing Number from Array • Finding Duplicate Integers in an Array • Finding the Largest and Smallest Number in Unsorted Array • Removing Duplicates from an Array • Reversing an Array • Finding the k-th Smallest Integer in an Unsorted Array • Finding Common Elements Between Multiple Arrays Practice and Strategies That being said, a service we most definitely recommend is - Daily Coding Problem. Daily Coding Problem is a simple and very useful platform that emails you one coding problem to solve every morning. This ensures that you practice consistently and often enough to stay in shape over a long period of time. We wrote up a more in-depth review of the DCP if you want to find out more. While solving these problems, you'll notice a lot of the aforementioned data structures and algorithms, as well as the importance of innovative thinking.	{"url":"https://stackabuse.com/programming-interview-questions/","timestamp":"2024-11-04T20:12:06Z","content_type":"text/html","content_length":"63463","record_id":"<urn:uuid:b32a2995-4729-4b11-bcbb-995baf5e1606>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00416.warc.gz"}
Data Golf's 2020 PGA Championship Preview While the main takeaways of this post are still relevant, it does suggest player adjustments relating to the 2020 PGA Championship field. Due to field changes and fluctuating players skills, the adjustments won't apply to future PGA Championships. An oft-debated topic in the golf (nerd) community is the degree to which different course setups separate the game’s best golfers. A wider spread in scores on the leaderboard could be the result of three distinct causes: 1) larger differences in skill amongst the golfers competing; 2) larger differences in scores between high and low-skilled golfers; and 3) greater variation in scores that is not related to skill (i.e. ‘random’ variation). The first item in this list is not interesting as it has nothing to do with the course setup; all else equal, scores at the TOUR Championship will be more tightly bunched simply because the golfers competing there are closer in skill than at a typical event. The focus in this article will be on the final two points; more specifically, we analyze the degree to which each course on the PGA Tour separates golfers, and why the PGA Championship has historically been an outlier in this regard. To make headway on this question, let’s think of golf scores as being composed of two parts: skill and luck, where luck is a catch-all term used to describe everything not related to skill. For example, if Justin Thomas played 100 rounds against an average PGA Tour player at a randomly selected course, we would expect Thomas to win, on average, by 2.3 strokes per round. That is, according to our estimates, Thomas’ skill level is 2.3 strokes above that of an average professional. We would also expect Thomas’ 100 scores to have a standard deviation of approximately 2.8 strokes per round. (Recall that standard deviation is a measure of dispersion; given a standard deviation of 2.8, we would expect roughly 68 of JT’s 100 rounds to be within 2.8 shots of his average score — so, if Thomas shot 70 on average, something around 70-75% of rounds would be between 67 and 73.) At courses where Thomas beats an average PGA Tour player by more than 2.3 strokes per round, we would say that this course creates separation based on skill ; at courses where Thomas’ scores have a standard deviation of greater than 2.8 we would say this course creates separation based on luck . It’s worth emphasizing again that ‘luck’ does not only refer to strange bounces off of trees or random gusts of wind, but rather any variation in scores that is not correlated with our estimates of golfer skill. For this analysis we use data from 2004-onwards; because three of the four major championships are played at different venues each year, we group each of the sets of courses for the U.S. Open, British Open, and PGA Championship together. A golfer’s skill is determined by equally weighting their performances across all courses within the appropriate time frame (i.e. 1-2 years before the event). Due to the nature of our skill estimates, if golfer A has a skill level that is 1 stroke better than golfer B, golfer A will on average beat golfer B by 1 stroke per round. However this average hides interesting, and potentially useful, variation. Continuing with our Justin Thomas example from above, it might be the case that across all courses Thomas is 2.3 shots better than the average player, but at some courses that number is closer to 2.1 while at others it’s closer to 2.5. This is what the value in the first column below indicates: how much a 1 stroke difference in skill is worth at each of the listed tournaments / courses. The second column refers to the random component of scores: at which tournaments do we see the most variation in scores after controlling for differences in skill? (For statheads, we are accounting for the course-specific skill differences of column 1 when analyzing this random variation). We have added a few notable courses in addition to the 4 major championships. Which courses/events separate players the most? One obvious way that the first two columns are related is through the par of the course. All else equal, a par-72 course will have higher values for both the multiplier and the standard deviation columns; the more shots that are hit, the greater the separation we will see both in terms of skill and luck. However there is a lot more than just this going on: for example, PGA National is a par-70 course that has above-average random variance. The final column in the table displays the win probability for a top player (think Justin Thomas) at each of the listed courses, holding the quality of field constant. This combines the information contained in the previous two columns. At Augusta National, for example, we expect more skilled players to beat less skilled players by slightly more strokes per round than at the typical course (column 1); however, Augusta National is also a course with higher random variance than average (column 2). The former would tend to increase JT’s win probability, while the latter would tend to decrease it. Overall, we see that the top player’s win probability is slightly higher at Augusta National than at an average course. At courses that have hosted PGA Championships since 2004, we estimate the highest degree of separation based on skill of any course on the PGA Tour; however these courses have also shown above-average random variation, which allowed Firestone CC to take the top spot in terms of where a top player has the largest win probability advantage. An important point to note here is that our definition of skill is only meaningful as it relates to the types of courses played on the PGA Tour. For example, if most courses on tour disproportionately reward distance, then the 'high-skilled' golfers will tend to be those that hit it far. Consequently, if there is a course that disproportionately favours driving accuracy, this might show up in our analysis as one that does not reward skill (because the better golfers are those that hit it far but not accurately, so they will perform worse at this course). However, we don’t need to just speculate on that thought, we can gain some insight into it from our course fit plots . For example, from the table above we see that Waialae is a course that narrows skill gaps; from its course fit plot , we see that this is because Waialae gives a below-average reward to nearly every skill you could care about. Below we have produced our course fit plots for the 4 major championships: "Event Fit" at Major Championships Radar Plot: PGA Championship Plot Info Toggling through each event offers several interesting takeaways regarding the types of players who benefit from each major championship setup, but let’s stick with this week’s event, the PGA Championship. Every skill except for 'around-the-green' is favoured relative to the average PGA Tour course. (This is consistent with the 1.08 skill multiplier we saw in the first table.) Intuitively this makes sense, as PGA Championships are traditionally set up similar to regular tour stops, but the courses are typically longer and have thicker rough. It seems reasonable that these heightened, but still familiar, conditions would amplify the skill gaps we observe week-to-week on tour. Looking a bit closer at the plot we see that driving distance, which is already the most important skill at the average course, is a disproportionately favoured one at PGA Championship setups. This has been a criticism from the architecture community about past PGAs; they believe that driving distance has become an overemphasized skill in professional golf. While we tend to believe that the reward for driving distance relative to other skills is reasonable, the argument that it has become too important does seem to hold more weight in weeks like this. Moving away from the data for a moment, we can recall watching the PGA Championship at Bethpage Black last year, and it was apparent that a player like Kevin Kisner would have to putt the lights out to compete with the likes of Koepka, DJ, or McIlroy, simply due to his lack of distance and power. The conclusions from the above discussions make the direction of our event/course fit adjustments pretty straightforward for this week: good players will get positive bumps, bad players will get negative bumps, and average players won’t receive much adjustment either way. Longer players will receive additional positive bumps, but long players are typically 'good' players, so this shouldn’t alter things too much. The next issue is one that is rarely addressed — how do these intuitive judgements actually translate into strokes per round adjustments to each player's expected performance? Lucky for you we provide a detailed breakdown of our predictions each week on our skill decomposition page ; here are the five largest positive and negative course fit adjustments this week (excluding Club Professionals): Top 5 Positive Adjustments: 1: Justin Thomas (+0.25) 2: Rory McIlroy (+0.23) 3: Brooks Koepka (+0.22) 4: Cameron Champ (+0.22) 5: Tony Finau (+0.20) Top 5 Negative Adjustments: 1: Shaun Micheel (-0.26) 2: Brian Stuard (-0.18) 3: Brendon Todd (-0.18) 4: Steve Stricker (-0.18) 5: Jim Herman (-0.16) A final interesting thing to consider is whether players who are only slightly above average benefit from playing a course that amplifies skill differences. On the one hand, their skill at this course will be further above average; on the other hand, the skill of the top players will move further above theirs. By looking at our finish probabilities for this week’s PGA Championship from both our baseline model (which assumes an average course setup) and our full model (which includes course-specific adjustments), it can be seen that it is only the top 10 or so golfers who see an increase in their win probability. However, as intuition would suggest, while golfers who are a bit further down in the skill distribution don’t see their win probabilities increase, they do see their Top 20 and cut probabilities increase under the model that takes account of this week's course. To wrap up, we’ll point out two names where these adjustments have made a difference this week: using the full model we are finding a bit of value on Rory McIlroy, and a large edge on Hideki Matsuyama, in this week’s outright markets. That’s all for this preview, enjoy the first major of 2020!	{"url":"https://datagolf.com/pga-champ-preview-2020","timestamp":"2024-11-11T08:22:18Z","content_type":"text/html","content_length":"104206","record_id":"<urn:uuid:0d24d9c9-a8a1-4c0b-8a2b-8548e396a6a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00410.warc.gz"}
Comments on cbloom rants: 11-12-10 - Some notes on function approximation by iterationYeah, you're right that it should be minimax/r..."we could also do r = x_n * (1+e) or r = x_n/... 10714564834899413045noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5246987755651065286.post-42553603949558109912010-11-17T16:49:32.753-08:002010-11-17T16:49:32.753-08:00Yeah, you're right that it should be minimax/remez polynomials, though if you bring x into a standard power of two range like [1,2) then just use L2 norm instead of ratio, it's not that far off.<br /><br /> (BTW I just noticed Boost now has a Remez resolver : <br /><br />http://www.cppprog.com/boost_doc/libs/math/doc/sf_and_dist/html/math_toolkit/backgrounders/remez.html<br /><br />)cbloomhttps:// could also do r = x_n * (1+e) or r = x_n/(1+e) or whatever. Sometimes these give better iterations (in terms of complexity vs. accuracy)."<br />For the specific case of reciprocal, square root and reciprocal square root computations, this is called "Goldschmidt's algorithm". Both Newton-Raphson and Goldschmidt actually evaluate the same series expansion in this case, but Goldschmidt is less serial which makes it popular for FP dividers. See <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.68.7202" rel="nofollow">this paper</a> for example.<br /><br /> There's some papers on HW implementations and math libraries for newer architectures and they're worth reading, e.g. <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.24.5177" rel="nofollow">this one</a> (also Google for the paper and look at the citations etc). Lots of good stuff in there.<br /><br />"You could use Legendre Polynomials to find the actual best N-term approximation (or any other orthogonal polynomial basis)"<br />Orthogonal polynomials give you an optimal approximation wrt. weighted L2 norms (int_a^b (f_approx(x) - f_real(x))^2 w(x) dx, where w(x) is a weighting function corresponding to your choice of basis). So it minimizes average error, but to minimize maximum absolute/relative error (which is usually what you want) you need a different approach (minimax approximation). Still, they're usually significantly better than Taylor polynomials, and the coefficients are easy to determine even without a computer algebra system (unlike minimax polynomials).ryghttps://www.blogger.com/profile/03031635656201499907noreply@blogger.com	{"url":"https://cbloomrants.blogspot.com/feeds/4982161546727464994/comments/default","timestamp":"2024-11-12T13:28:51Z","content_type":"application/atom+xml","content_length":"7454","record_id":"<urn:uuid:d50a4c30-663c-4001-81e2-2ddde242ef98>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00623.warc.gz"}
Wolfram Function Repository Function Repository Resource: Simulate hard spheres moving in an n-dimensional box Contributed by: Matt Kafker and Christopher Wolfram generates a simulation of the motion of a system of hard spheres with simulation parameters specified by assoc. Input assoc must include the following: "Positions" initial particle positions "Velocities" initial particle velocities "BoxSize" side-length of simulation box "Steps" number of simulation time steps "StepSize" duration of a single time step "BoundaryCondition" interaction between particles and container walls "Output" simulation output Input assoc may also include: "ParticleRadius" radius of spheres "ParticleMass" mass of spheres The dimensions of vectors in "Positions" and "Velocities" determine the spatial dimension of the system (they can be any positive integer dimension). "ParticleMass" can either be a single number for identical-mass spheres or a list for spheres of different masses. The box is centered on the origin. Possible named boundary conditions include: "Reflecting" particle velocities reflect upon contact with container walls "Periodic" particles are translated to opposite side of box upon reaching container walls "Infinite" particles do not alter motion upon reaching container walls Possible named outputs include: "PositionsByTime" list of all particle positions at every time step "VelocitiesByTime" list of all particle velocities at every time step "SpeedsByTime" list of all particle speeds at every time step "TrajectoriesByTime" list of all particle positions and velocities at every time step "PositionsByParticle" list of positions over time for each particle "VelocitiesByParticle" list of velocities over time for each particle "SpeedsByParticle" list of speeds over time for each particle "TrajectoriesByParticle" list of positions and velocities over time for each particle "Collisions" list of particle-particle collisions "All" list containing “TrajectoriesByTime” output as the first entry, and “Collisions” as the second "Visualize" Graphics object with a Manipulate slider for visualizing the time-evolution of the system in one, two, or three spatial dimensions "KineticEnergy" list of total kinetic energy at every time step "Temperature" list of temperature at every time step "CausalGraph" Graph object with nodes representing collisions and directed edges indicating that one particle was contiguously involved in two collisions Multiple outputs can be provided as a Each collision takes the form {particle[i],particle[j],time of collision}. The option "RandomNonOverlapping" may be given for pos, in which case a random configuration of non-overlapping particle positions is chosen within the box. Basic Examples (2) Simulate the motion of two hard disks over five time steps, and extract the trajectory data in various ways. Get the positions over time: Get the velocities over time: Get the speeds over time: Get the trajectories over time: Get the positions grouped by particle: Get the velocities grouped by particle: Get the speeds grouped by particle: Get the trajectories grouped by particle: Scope (7) Use trajectory data to visualize frames from a hard sphere simulation: Use "Visualize" to generate an interactive visualization of many hard spheres moving in a box: Use "RandomNonOverlapping" to generate random non-overlapping initial positions in the box: Visualization works in 1D: Visualization works in 3D as well: Periodic boundary conditions identify opposite walls of the box with one another: Infinite boundary conditions eliminate particle-wall interactions altogether: Multiple outputs can be returned at once: Applications (3) Particle speeds should obey Maxwell-Boltzmann distribution: Extract a collision list from the simulation, where collisions take the form {particle[1],particle[2],time of collision}: From particle-particle collisions, construct the causal graph: Construct a causal graph for a disordered system: As the nodes of the causal graph represent collisions between two particles and the edges represent individual particles, we expect all nodes to have indegree and outdegree 2, besides from the graph boundary (i.e., the beginning and end of the simulation): Properties and Relations (2) Total kinetic energy of spheres should be conserved in time. Show that the error stays within machine precision: Momentum should be conserved during collisions between particles. Collision of equal mass particles: Collision of an incoming heavy particle with a much lighter stationary particle causes the light particle to go out with double the speed of the incoming one: Collision of a very light particle with a much heavier stationary particle leads to reflection of the incoming particle: Possible Issues (2) Particle overlaps are resolved pairwise, so situations involving multiple overlapping spheres may lead to unexpected results (such as a disruption of symmetry): The "RandomNonOverlapping" option is not guaranteed to find non-overlapping configurations at very high packing fractions: Neat Examples (2) Simulate a Newton's cradle: Use HardSphereSimulation to simulate a simplified game of pool: Version History Related Resources Related Symbols License Information	{"url":"https://resources.wolframcloud.com/FunctionRepository/resources/HardSphereSimulation","timestamp":"2024-11-11T02:02:08Z","content_type":"text/html","content_length":"79670","record_id":"<urn:uuid:a8d1f536-4091-42b5-b212-39b5f86b348d>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00616.warc.gz"}
Smooth approximations We consider CSPs whose template can be defined in an infinite “ground structure” with a high degree of symmetry (in the form of homogeneity) and a certain finite presentation (finite boundedness). For example, the ground structure can be the order of the rationals, leading to Temporal CSPs, or the random graph, which gives rise to so-called Graph-SAT problems. Such CSPs are always in NP, include all finite-domain CSPs as well as many additional natural problems, and there is an open dichotomy conjecture extending the one for finite templates. The general algebraic approach to finite domain CSPs via polymorphisms also works in this setting, and the most sensible approach to the conjecture is to try to reduce it to the finite case. To this end, one associates certain finite algebras to a CSP template, and hopes to extract from them sufficient information about the complexity of the CSP. In most successful confirmations of instances of the conjecture so far, this was done somewhat non-systematically, leading to long proofs of ad hoc arguments. The novel Theory of Smooth Approximations intends to provide a uniform way of relating the associated finite algebras with the structure of the CSP template and consequently its computational complexity. Applying this method, all previous results in the literature can be reproven much more smoothly; moreover, the method allows, for the first time, for a systematic investigation of local consistency methods in this setting. This is joint work with Antoine Mottet.	{"url":"https://csp-seminar.org/talks/michael-pinsker/","timestamp":"2024-11-02T17:29:20Z","content_type":"text/html","content_length":"6890","record_id":"<urn:uuid:59172fa0-0222-43c1-9c64-33ad7c0d9843>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00620.warc.gz"}
Why It Matters: Systems of Equations and Inequalities Why learn to solve systems of equations and inequalities? When you play in a river you are surrounded by fluids, including water and air. At first it might seem strange to think of air as a fluid, but a fluid is defined as a substance that flows. Wind, therefore, is a great example of air that flows. Other examples of flows include traffic patterns and electrical currents. Flows can be turbulent like what you may experience in airplanes. Early in the 19th Century, Claude-Louis Navier in France and George Gabriel Stokes in England both derived an equation that can explain and predict the flow of fluids. The Navier-Stokes equations are a system of equations used to describe the velocity of a fluid as it moves through three-dimensional space over a specific interval of time. Interestingly, our understanding of solutions to the Navier-Stokes equations remains minimal. Surprisingly, given the equations’ wide range of practical uses, it has not yet been proven that solutions always exist in three dimensions. The Clay Mathematics Institute has called this one of the seven most important open problems in mathematics and has offered a US $1,000,000 prize for a solution or a counter-example. In this section, we will learn how to graph systems of equations in two dimensions and find whether solutions exist. We will also see how systems of equations can be used to solve problems where we have two unknown variables.	{"url":"https://courses.lumenlearning.com/beginalgebra/chapter/introduction-4/","timestamp":"2024-11-05T01:14:37Z","content_type":"text/html","content_length":"45958","record_id":"<urn:uuid:25b096c8-7ace-4d82-b6e2-0e66f908e324>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00455.warc.gz"}
An intelligent method based on feed-forward artificial neural network and least square support vector machine for the simultaneous spectrophotometric estimation of anti hepatitis C virus drugs in pharmaceutical formulation and biological fluid An intelligent method based on feed-forward artificial neural network and least square support vector machine for the simultaneous spectrophotometric estimation of anti hepatitis C virus drugs in pharmaceutical formulation and biological fluid Created by W.Langdon from gp-bibliography.bib Revision:1.8010 □ author = "Kiarash Keyvan and Mahmoud Reza Sohrabi and Fereshteh Motiee", □ title = "An intelligent method based on feed-forward artificial neural network and least square support vector machine for the simultaneous spectrophotometric estimation of anti hepatitis C virus drugs in pharmaceutical formulation and biological fluid", □ journal = "Spectrochimica Acta Part A: Molecular and Biomolecular Spectroscopy", □ volume = "263", □ pages = "120190", □ year = "2021", □ ISSN = "1386-1425", □ keywords = "genetic algorithms, genetic programming, Spectrophotometry, Artificial neural network, Least square support vector machine, Sofosbuvir, Daclatasvir", □ abstract = "This study proposed simple and reliable spectrophotometry method for simultaneous analysis of hepatitis C antiviral binary mixture containing sofosbuvir (SOF) and daclatasvir (DAC). This technique is based on the use of feed-forward artificial neural network (FF-ANN) and least square support vector machine (LS-SVM). FF-NN with Levenberg-Marquardt (LM) and Cartesian genetic programming (CGP) algorithms was trained to determine the best number of hidden layers and the number of neurons. This comparison demonstrated that the LM algorithm had the minimum mean square error (MSE) for SOF (1.59 times 10-28) and DAC (4.71 times 10-28). In LS-SVM model, the optimum regularization parameter (?) and width of the function (?) were achieved with root mean square error (RMSE) of 0.9355 and 0.2641 for SOF and DAC, respectively. The coefficient of determination (R2) value of mixtures containing SOF and DAC was 0.996 and 0.997, respectively. The percentage recovery values were in the range of 94.03-104.58 and 94.04-106.41 for SOF and DAC, respectively. Statistical test (ANOVA) was implemented to compare high-performance liquid chromatography (HPLC) and spectrophotometry, which showed no significant difference. These results indicate that the proposed method possesses great potential ability for prediction of concentration of components in pharmaceutical formulations", Genetic Programming entries for Kiarash Keyvan Mahmoud Reza Sohrabi Fereshteh Motiee	{"url":"https://gpbib.cs.ucl.ac.uk/gp-html/KEYVAN_2021_SAPAMBS.html","timestamp":"2024-11-02T11:33:38Z","content_type":"text/html","content_length":"5667","record_id":"<urn:uuid:d701a827-6f76-47ec-b394-b0924410051d>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00244.warc.gz"}
O((logn)<sup>2</sup>) time online approximation schemes for bin packing and subset sum problems Given a set S = {b [1],⋯, b [n] } of integers and an integer s, the subset sum problem is to decide if there is a subset S′ of S such that the sum of elements in S′ is exactly equal to s. We present an online approximation scheme for this problem. It updates in O(logn) time and gives a (1+ε)-approximation solution in time. The online approximation for target s is to find a subset of the items that have been received. The bin packing problem is to find the minimum number of bins of size one to pack a list of items a [1],⋯, a [n] of size in [0,1]. Let function bp(L) be the minimum number of bins to pack all items in the list L. We present an online approximate algorithm for the function bp(L) in the bin packing problem, where L is the list of the items that have been received. It updates in O(logn) updating time and gives a (1+ε)-approximation solution app(L) for bp(L) in time to satisfy app(L)≤(1+ε)bp(L)+1. Publication series Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) Volume 6213 LNCS ISSN (Print) 0302-9743 ISSN (Electronic) 1611-3349 Other 4th International Frontiers of Algorithmics Workshop, FAW 2010 Country/Territory China City Wuhan Period 8/11/10 → 8/13/10 ASJC Scopus subject areas • Theoretical Computer Science • General Computer Science Dive into the research topics of 'O((logn)^2) time online approximation schemes for bin packing and subset sum problems'. Together they form a unique fingerprint.	{"url":"https://utsouthwestern.elsevierpure.com/en/publications/olognsup2sup-time-online-approximation-schemes-for-bin-packing-an","timestamp":"2024-11-09T12:52:11Z","content_type":"text/html","content_length":"50988","record_id":"<urn:uuid:505f3865-1f53-4147-bd2d-3b736fd66b7f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00667.warc.gz"}
dilution and dilution factor microbiology You have diluted the sample by a factor of 100. If the dilution is in the form of a fraction, you can "flip" the fraction (i.e., 1/50 becomes multiply by 50/1). A 0.1 ml to 0.9 ml dilution is the same as a 1 ml to 9 ml dilution and a 13 ml to 117 ml dilution. 1. This is written as a 1:2 dilution or a times 2 (x2) dilution. 3. Label the four 9 ml dilution blanks with the dilutions to make of the hamburger as follows: 10 -2, 10 -3, 10 -4 and 10 -5. The first number is the volume of reagent (1 mL) and the second number is the total volume of the final solution (2 mL). A 1:5 dilution (also known as a "1 to 5" dilution) involves mixing 1 unit volume of solute (the item to be diluted) with 4 unit volume of solvent medium (therefore, 1 + 4 = 5 = dilution factor). The TDF of plate with 63 colonies is 10 5 1 ml/ 1ml+9 ml = 1/ 10 * 1/10= 1/100. Weigh into a tared blended jar at least 10 g of sample, representative of the food. Any of the following techniques can be used for determining viable counts of bacteria: Pour plating. One may use 0.9ml, 9.9ml, or 99ml blanks to skip in between dilutions (10 -2, 10 -4, 10 -6 . The dilution factor does not have to be constant between different dilution blanks. The sample/culture is placed in a test tube, and six test tubes are filled with 9 mL of sterile diluents, which can be distilled water or 0.9 percent saline. Dilution = amount of specimen transferred divided by the [amount of specimen transferred + amount already in tube]. A 1:100 dilution can be created by placing 1 pellet in 99 ml as instructed in the membrane filtration instructions. After the initial dilution of an unevenly distributed community, evenness would be greatly increased, and so, differences in subsequent dilutions (10 2, 10 3, and 10 4) were predicted to be smalluntil the dilution factor exceeds the original number of types of organisms in the community as described above. Final dilution factor (DF) = DF1 DF2 DF3 etc. Determine the dilution factor for each tube in the dilution series. Total Dilution Factor (TDF)=DF1xDF2XDF3. Note: For multiple dilutions the dilution factor is the . Next, 1 ml of the first dilution is added to 99 ml to make the second dilution, that is a 1/100 dilution. To calculate the number of cells you have in each, multiply the concentration by the volume: 0.44 cells/mL 13.6 mL = 6 cells (if done properly with all trailing decimals). Procedure of Serial Dilution. Start your trial now! For example, if I added 1g of sample into 9mL of broth - 10^-1, and transfer 1mL from 10^-1 to second tube (9mL as well). We have step-by-step solutions for your textbooks written by Bartleby experts! Search: Serial Dilutions Lesson. This inverse of the dilution is called the dilution factor. Harrigan, Margaret E. McCance, in Laboratory Methods in Microbiology, 1966 1 Liquid samples. The formula for calculating a . A dilution factor of 5 is obtained by diluting frozen orange juice concentrate with four extra cans of cold water (the dilution solvent), i.e., the . Prepare the following dilution blanks with tap water using a 10ml pipette and the green pi-pump---4.5ml, 9ml, 9.5ml, 4ml, 3ml, 12ml. Expressing a x2 dilution as a ratio would be 1:1, or 1 mL reagent plus 1 mL water. So find out the TDF of the culture tube from which we cultured to get the plate with colony count 63. Divide your cell density: 0.44 cells/mL / 1.84 = 0.24 cells/mL. A 1:5 dilution (also known as a "1 to 5" dilution) involves mixing 1 unit volume of solute (the item to be diluted) with 4 unit volume of solvent medium (therefore, 1 + 4 = 5 = dilution factor). use qualitative reagents to identify biological molecules Tube 4 has 1 / 3rd of its volume that is composed of the 1 : 9 dilution from tube 3 With Solution Essays, you can get high-quality essays at a lower price The story on the previous pages has many parallels with life in a microbiology lab Prepare parallel and serial dilutions using C1V1 = C2V2 . Determine the concentration of the solution following dilution. 2. A serial dilution is a kind of solution dilution.A more exact serial dilution definition is that it is a stepwise dilution of a solution, that is repeated a certain number of times and in which the concentration decreases with each step. Typically serial dilutions are made across all or half of the columns of a 96 well microplate. Then back track the measured counts to the unknown . . In step 3, the product of the individual dilution factors is calculated to give the final dilution factor: Step 3: 1 X 10 -2 * 1 X 10 -2 * 1 X 10 -1 * 1 X 10 -1 = 1 X 10 -6 (final dilution factor) In microbiology, the reciprocal of the final dilution factor is called the plating factor. Author. Calculate original cell density of a bacterial culture We plated 0.1 mL from a 10-4 dilution. Example: In a typical microbiology exercise the students perform a three step 1:100 Label the four 9 ml dilution blanks with the dilutions to make of the hamburger as follows: 10 -2, 10 -3, 10 -4 and 10 -5. Dilution factor= 5/1 or 5. Calculate the total/final dilution factor in a dilution tube following a series of dilutions. Advantages. Dilution is the decrease in concentration. Spread plate technique. The following is the procedure for a ten-fold dilution of a sample to a dilution factor of 10-6:. In the first three methods, the sample is serially diluted to obtain 10 -1 to 10 -10 dilutions using sterile blanks. 3. TOTAL DILUTION FACTOR = 30 = 1 x 10-8 2.8 x 109 An easy way to set up dilution series like this would be to use 4 tubes, each having an IDF of 10-2, i.e., transfer 0.1 ml into a tube containing 9.9 ml four times. B tube=10 1 x10 1 =10 2. Perform a serial dilution and plate the dilutions using aseptic technique 4. EXAMPLE 1: What is the dilution factor if you add a 0.1 mL aliquot of a specimen to 9.9 mL of diluent? Thus, this is an important concept for them to apply. solution, what would be th e final dilution factor? Thus: 1/100 x 1/5 = 1/500 Serial Dilutions Many procedures call for a dilution series in which all dilutions after the first one are the same. . Pages 1 This preview shows page 1 out of 1 page. TOTAL DILUTION FACTOR = 30 = 1 x 10-8 2.8 x 109 An easy way to set up dilution series like this would be to use 4 tubes, each having an IDF of 10-2, i.e., transfer 0.1 ml into a tube containing 9.9 ml four times. It may be expressed as the ratio of the volume of the final diluted solution (V 2) to the initial volume removed from the stock solution (V 1), as shown in the equation above.Dilution factor may also be expressed as the ratio of the concentration of stock solution (C 1) to the concentration of the diluted solution (C 2). Mix the tube contents. Microbiology (BIOL 307) Uploaded by. In your example, the final volume is the total . How to make a 1 20 dilution? 25 10 250 = mL mL. Summary: 1.Dilution is a process with no formula while a diluents factor requires a formula to get the answer. For example: 1/10 = 10-1 1/100 = 10 2 1/1000 = 10-3 1/10000 = 10-4 and so on You need to know the above equation, how to use it, and how to express dilutions as exponents! Search: Serial Dilutions Lesson. That would be a dilution factor of 100:10,000,000, or 1:100,000. 4. Dilution factor is defined as: total volume of solution per aliquot volume. Or dilution factor is the ratio between the final volume and the initial volume of the solution. How to Calculate Dilution Factor. 10-5) multiply by 1 over the number (i.e., 1/10-5 becomes multiply by 105). Then explain why CFU is a fitting term for describing the original bacterial cell concentration in the stock tube. Plug your dilution factor into the equation: D t = 10 x 10 x 10 x 10 = 10,000. A common method of making a solution of a given concentration involves taking a more concentration solution and adding water until the desired concentration is reached. Yes, there is math in microbiology and, yes, exponents, are simple. 2017/2018; Helpful? Typically serial dilutions are made across all or half of the columns of a 96 well microplate. Let's try another example. Spread 1.0 ml on a plate and incubate. Choose step DFs: Need a total dilution factor of 1000. W.F. SELF TEST How many colonies would you expect if you plated our 0.1 ml form Tube C? A serial dilution is the dilution of a sample, in 10-fold dilutions. Share. Science Biology Laboratory Experiments in Microbiology (12th Edition) (What's New in Microbiology) The serial dilution in the given samples. Choose from 500 different sets of 4 microbiology lab dilution flashcards on Quizlet. First week only $4.99! . This process is known as dilution. The initial term "microtiter" plate is based on doing a titer, or dilution across the plate. Similarly, the use of duplicate plates at several dilutions to achieve a weighted mean is not considered essential where the focus is on identifying bacterial levels that pose a risk to public health. Multiply the individual dilution of the tube X previous total dilution. S:T = exponent:1. The dilution factor can alternatively be stated as an exponent, such as 3-1, 5-3, or 10-4. The formula for calculating a . Search: Serial Dilutions Lesson. For example, if the 1:100 dilution is needed (or the 10-2 dilution), it can be prepared adding 1 ml of 1:10 dilution to 9 ml of diluent, according to the following equation: 1 ml of 10-1 dilution Serial dilution refers to performing more than one dilution in succession, in order to determine which of many concentrations is correct for further study. 2. In microbiology, serial dilutions (log dilutions) are used to decrease a bacterial concentration to a required concentration for a specific test method, or to a concentration which is easier to count when plated to an agar plate.. Convert the dilution factor to a fraction with the first number as the numerator and the second number as the denominator. Dilution is the process of making a solution weaker or less concentrated. The . In a serial dilution the total dilution factor at any point is the product of the individual dilution factors in each step up to it. The serial dilution technique in microbiology for ten-fold dilution of a sample to a dilution factor of 10 -6 is as follows. first dilution factor of a solution with 1 ml blood and 9 ml saline. The sample/culture is taken in a test tube and six test tubes, each with 9 ml of sterile diluents, which can either be distilled water or 0.9% saline, are taken. NaCl is the salt of a strong acid and strong base and has a neutral pH For example, adding 3 ml chocolate syrup to 97 ml milk would create a 100 ml solution Tube 4 has 1 / 3rd of its volume that is composed of the 1 : 9 dilution from tube 3 Learn and research science, chemistry, biology, physics, math, astronomy, electronics, and much more 5 million words on a . For example, a 1:20 dilution converts to a 1/20 dilution factor. 5. Expressing a x2 dilution as a ratio would be 1:1, or 1 mL reagent plus 1 mL water. And to give ourselves a little wiggle room, we should start at least 1 dilution before that, so 1:10,000. A specific amount of bacteria are reduced with every dilution. 1/100 dilution that was already prepared; i.e., bring 1 part of the 1/100 dilution of serum in buffer up to 5 parts total volume. The dilution factor (or dilution ratio) is used to express how much of the original stock solution is present in the total solution, after dilution. 4. Microbiology - Lecture notes - 1.17.12; Unknown Bacteria Lab Report; Antimicrobials - Lecture notes 23; . For instance, say you start with 10mL of cell suspension. duplicate 0.5mL plates at 10-1. dilution) is sufficient. What is the OCD (don't forget units)? Serial Dilutions Practice if you count 60 colonies after generating dilution factor of and plating 100 microliters, what is the starting concentration? 1. Example: Make only 300 L of a 1:1000 dilution, assuming the smallest volume you can pipette is 2 L. The dilution factor calculator at each step does not have to be constant, but it is for this calculator.Serial dilutions have many uses that are mainly . I have created this guide to provide a better understanding of dilutions and should be used . School Namibia University of Science and Technology. The number of colonies cultured from serial dilutions of the sample are counted to estimate the concentration of an unknown sample. This type of dilution series is referred to as a serial dilution. In the figure test a has dilution = 1 ml /10 ml (9+1) =1/10=0.1 or 10-1. plate 4 (1 ml) and plate 3 (0.1 ml) A 10-6 dilution can be achieved by making three 1:100 dilutions, or six 1:10 dilutions, or a combination of 100-fold and 10-fold dilutions. How to find the dilution and dilution factor of a sample with 5 ml blood and a diluent of 20 ml saline. How do you calculate the dilution of a dilution? A medical laboratory scientist must dilute a . Then we'll do three more 1:10 dilutions to get our series. Proper Use of the Compound Microscope Additional Videos on Serial Dilution: C6 MODULE 5: LESSON 4 ASSIGNMENT (?/42 marks) Lesson 4 Assignment: Multi-step pH and pOH Calculations Part 1: The Technique of Serial Dilutions (00:00 to 04:50) Complete the questions below for this part of the video But Timinsky's story has been one filled with random moves, lots of . Where total volume of solution is: 10.0 + 240.0 = 250.0 mL (volumetric flask.) Course Title DMPE 735; Uploaded By DeanWaterBuffaloMaster1262. Dilution factor is a mathematical concept defined as the total volume of the solution or mixture divided by the volume of the sample. If the dilution is written in scientific notation (e.g. Second dilution factor of 1 ml blood and 9 ml saline. Dilution of a solution is the decrease of the concentration of solutes in that solution. The first number is the volume of reagent (1 mL) and the second number is the total volume of the final solution (2 mL). For example, if you take 1 part of a sample and add 9 parts of water (solvent), then you have made a 1:10 dilution; this has a concentration of 1/10th (0.1) of the original and a dilution factor of 10. Dilution set 1: Transfer 0.5ml of blue water into the 4.5ml of water, then 1ml of tube 1 into the next tube of 9ml water. At each dilution the true number of colonies is n j = n 0 j p 1 and the estimated number is n ^ j. 10-6. Therefore, in the original water sample: ppm = 24.0 25 600.0. ppm Pb. It helps to reduce a dense culture of cells to a more usable concentration. This will drop the concentration two logs from 103 to 101 CFU/ml. Checking the MPN table, 3-1-0 indicates that an average of 0.43 organism (causing the determining reaction) was inoculated into each of the tubes in the middle set (D) - i.e., the tubes inoculated with 0.1 ml of the 10 -3 dilution. For example, if you take 1 part of a sample and add 9 parts of water (solvent), then you have made a 1:10 dilution; this has a concentration of 1/10th (0.1) of the original and a dilution factor of 10. Generally, the dilution factor at each stage of serial dilution is constant and leads to geometric progression in a logarithmic manner of the concentration of the sample. With the P1000 pipettor and a blue pipettor tip, aseptically transfer 1 ml of the 1/10 hamburger dilution to the dilution blank labeled 10 -2; discard the tip into the disinfectant. Serial dilutions are commonly used in microbiology where the solution being diluted contains bacterial colonies 05 ml of each dilution is taken in the well and 1/60 ml of antigen is added to each dilution and rotated in a rotator Investigate the RNA helicase Dbp6 by cloning truncation constructs, analyzing their effects on Saccharomyces . Solution: V f = aliquot volume + diluent volume = (0.1 + 9.9) mL = 10.0 mL. Dilution factor specified: For this calculation, the following must be entered: (1) desired dilution factor; (2) either the stock concentration (C 1) or final concentration (C 2), but not both; and (3) either the volume from stock solution (V 1) or final solution volume (V 2), but not both.Therefore, two cells must be blank: (C 1 or C 2) and (V 1 or V 2).The value of the blank cells will be . Below mentioned are the steps to calculate the dilution factor by hand: Then the dilution factor is final volume over initial volume, in this case: 15mL/10mL = 1.5. The concentration of your substance is now 10,000 times less than the original undiluted solution. Mix the contents of the jar by shaking and pipet duplicate portions of 1 mL each into separate tubes containing 9 mL of dilution . You add 5mL of water, so your final volume is 15mL. 5 minute video explaining the dilution vs dilution factor in MicrobiologyPlease consider subscribing using the link: https://bit.ly/3kG2kKf0:00 Introduction . Dilution vs Dilution Factor. Let's do a 1:10 followed by a 1:100 (10 * 100 = 1000) Formula: Final Volume / Solute Volume = DF.	{"url":"http://www.stickycompany.com/automation/skechers/sleepy/92060128c4351777982d6df9-dilution-and-dilution-factor-microbiology","timestamp":"2024-11-14T01:17:38Z","content_type":"text/html","content_length":"23540","record_id":"<urn:uuid:15e23def-0ffb-4854-b0a1-766df5bd1be3>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00355.warc.gz"}
Exploration of Exponents and Natural Logs An in depth examination of natural logs and exponentials. If you have e growing at a non-constant rate over a given interval, can you use the average rate to calculate the end growth? ln(2)=1-1/2+1/3-1/4. This means that e^.69 is 2. This means that this is the exact number of growths (the rate at which a principle of 1 grows) that must occur in a given time for the principle to Additionally, we can think of the natural log as what combination of time and rate are needed to get to x amount of my value. (Betterexplained.com, without whom, this intuitive understanding would have been much harder, has it listed as simply the time rather than a combination of the time and rate). Betterexplained very astutely shows that the ln(.5) is a negative number because you’re trying to find out what combination of time and rate you need to get to half the amount you have now. We can break the result of ln(.5) (which is -0.6931) into two components: rate and time. We get the equation ratetime= -.6931. Now, we could have negative time, but it’s more plausible to have a negative rate. If the two numbers in the in the log are inverse, then the result will be equal but of opposite sign. ln(3)= -ln(.33). This is because if you go from your amount to 3 times your amount, the growth ratetime will be equal to going from that new 3 times your amount to your original amount but in the opposite direction. Now, what about the multiplication of logs? That’s a hard question. But, I believe it can be answered. log represents the amount of growth (time*rate) that e (or 10) must go through before reaching a value. ln(10) is how long it will take for an original value to get to 10 times that value given continuos growth. If you have 2ln(10), then you’re asking how long will it take for the original value to get to 100 times its current, or ln(10)+ln(10), which yields the same result: ten times original and then once again 10 times that number. for ln(10)^2, would you be taking the ln(10) ln(10) times? Would that make sense? Would you need to do that? For some reason it feels like you could represent e^x^2 with that. That you could represent functions that are not elementary using that.	{"url":"https://joepucc.io/notes/exploration-of-exponents-and-natural-logs.php","timestamp":"2024-11-14T04:46:03Z","content_type":"text/html","content_length":"4767","record_id":"<urn:uuid:09ed2b1b-7cdd-4be8-8f83-c38a0c4a7446>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00458.warc.gz"}
Why can’t you divide by zero? - The Handy Math Answer Book Why can’t you divide by zero? Basic Mathematical Operations Read more from Chapter Math Basics Dividing by zero is like the old saying, “You can’t get something from nothing.” Mathematically speaking, it’s the same way: You can’t “divide by nothing.” In fact, when something is divided by zero, the answer is always undefined. Here are a few ways of looking at this: There is a rule in arithmetic that a(b/a) = b. So if we say that 1/0 = 5, then 0(1/0) = 0 × 5 = 0. In other words, if you could divide by 0, this rule would not work. Another way to look at the “no to 0 as a divisor” problem is through multiplication: if 10/2 = 5, we know that 5 × 2 = 10; the same for 5/1 = 5, thus we know that 5 × 1 = 5. But if you take 5/0, that would mean that the answer times 0 would equal 5, but anything times 0 is equal to zero. Because there is no answer to this dilemma, mathematicians say you can’t divide by zero.	{"url":"https://www.papertrell.com/apps/preview/The-Handy-Math-Answer-Book/Handy%20Answer%20book/Why-can-t-you-divide-by-zero/001137022/content/SC/52caff9c82fad14abfa5c2e0_cool_facts.html","timestamp":"2024-11-12T13:11:24Z","content_type":"text/html","content_length":"11502","record_id":"<urn:uuid:5d821204-4903-49ea-99a7-87bcc1d781b4>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00758.warc.gz"}
A Classification of the Supercharacter Theories of $C_p \times C_2 \times C_2$ for Prime $p$ Skip to main content This item is not available for download from eScholarship A Classification of the Supercharacter Theories of $C_p \times C_2 \times C_2$ for Prime $p$ We classify and construct all supercharacter theories for the groups $C_p \times C_2 \times C_2$ where $p$ is an odd prime. It is known that every nontrivial supercharacter theory of a cyclic group can be constructed as a wedge product, a direct product, or is generated by automorphisms. We show that these constructions are also sufficient to construct every nontrivial supercharacter theory of $C_p \times C_2 \times C_2$. We give a precise count of the distinct supercharacter theories of $C_p\times C_2\times C_2$ and describe when a supercharacter theory can be constructed by more than one	{"url":"https://escholarship.org/uc/item/3kk109sb","timestamp":"2024-11-04T20:29:08Z","content_type":"text/html","content_length":"23230","record_id":"<urn:uuid:9e1c563d-08bb-4dae-8654-c2905ee4f8a9>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00533.warc.gz"}
Dear Mr. Fiedler, please be more descriptive on what kind of recursive problems you mean. Generally speaking, primitive recursive programs without side effects (ie not touching the stack before the recursion step) are Comptime allows arbitrary function execution, which will always terminate at compilation time due to tracking what parts were visited. The compiler therefore tracks a counter similar to the number of fixpoint iteration steps and will fail, if the amount of backwards edges becomes too big. So one can ensure that compilation will halt after a limited number of steps. What one can not ensure is what the outcome is. Your explanation of dependent types is missing, if and how limiting the number of steps has an influence of the type checking. Personally I see limiting the number of possible compilation steps as sufficient for "weak decidability", since the compiler may only interpret finitely many steps akind to the Turing Machine running finitely many steps. Could you elaborate on the underlying theory about that and add it to your blog? Your blog post also does not differentiate between language model and implementation, which further blurs things. Regards, Jan	{"url":"https://todo.sr.ht/~matu3ba","timestamp":"2024-11-12T00:19:19Z","content_type":"text/html","content_length":"4064","record_id":"<urn:uuid:1da3a1ed-6dae-4a4f-844d-4e7e0b5b7f10>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00303.warc.gz"}
• Thorsten should type: 1. darcs get http://sneezy.cs.nott.ac.uk/darcs/term 2. cd term 3. chmod +x Setup.hs 4. ./Setup.hs configure --prefix=$HOME 5. ./Setup.hs build 6. ./Setup.hs install --user 7. (and providing $HOME/bin is in your path:) epigram • The next job in term is to build equality in these steps: 1. Propositional equality coersions between equal types (and the proof that these are coherant) and the structural rule for application 2. Proof-irrelevance for equality, to do this right we need to decide equality during evaluation (if we do not have a proof of an equation but it’s homogeneous, then we win anyway). The upshot is Equality.lhs and eval from Term.lhs need to be in the same file. 3. Observational equavalence, at some point (sooner rather than later) we’ll include the structural rule for Thorsten pointed out that is rule can be derived from substution of equals for equals plus the more primitive: Having this gives OTT a simpler core, but Conor prefers the first version because it is easier to program with.	{"url":"http://www.e-pig.org/epilogue/_p=100.html","timestamp":"2024-11-07T04:41:36Z","content_type":"application/xhtml+xml","content_length":"9331","record_id":"<urn:uuid:6ab51b7d-e883-4a8a-929d-9a0043a01338>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00550.warc.gz"}
Rainy Lake Experiment: Location Graph and Data This page lists all data of the Rainy Lake Experiment. The following Graphs and Tables are just generated with JavaScript from the original GNSS data acquired by Jesse Kozlowski, see Measuring the The numbers in parenthesis are labels for the targets. (0) is the observer location. Grid Spacing 1000 m × 1000 m. All heights normalized to water level at observer by subtracting the water level height of data point (0) from all elevations. Grid Spacing 1000 m × 1 m. The GPS vectors of the filled markers were measured using Differential GPS equipment. All vectors of the blank markers were derived from the GPS vectors at the same location and the measured target plate center heights using a measuring tape, see Measuring the Targets. All heights normalized to water level at the individual target by subtracting the water level elevation at a target from all other elevations at a target. Grid Spacing 1000 m × 1 m. The suface of the earth can roughly be approximated by an ellipsoidal shape, the so called Reference Ellipsoid. We can express any location in the vicinity of the earth, even below or way above the surface, in Earth Centered Earth Fixed (ECEF) coordinates, as directly obtained from GNSS receivers, or by latitude, longitude and ellipsoid height. Transforming back and forth between ECEF coordinates and latitude/longitude and ellipsoid height is only a matter of geometry. Maps and navigation devices express the third dimension not as height above the Reference Ellipsoid, but as Elevation. The Elevation of a geographic location is its height above or below a fixed reference point, most commonly a reference Geoid, a mathematical model of the Earth's mean sea level as an equipotential gravitational surface. The Geoid is the shape that the ocean surface would take under the influence of the gravity and rotation of Earth alone, if other influences such as winds and tides were absent. This surface is extended through the continents (such as with very narrow hypothetical canals). It is the "mathematical figure of the Earth", a smooth but irregular surface whose shape results from the uneven distribution of mass within and on the surface of Earth. It can be known only through extensive gravitational measurements and calculations. The Geoid has no geometrical relation to the surface of the ellipsoid. To obtain ones Elevation, a raw GPS reading must be corrected. Modern GPS receivers contain a database of the Geoid heights (EGM96) with respect to the WGS84 Reference Ellipsoid. So they are able to correct the height above WGS Ellipsoid to the Elevation above WGS84 Geoid. The shape of the Geoid is often approximated by spherical harmonics coefficients. The current best such set of spherical harmonic coefficients is EGM96 (Earth Gravitational Model 1996), determined in an international collaborative project led by the National Imagery and Mapping Agency (now the National Geospatial-Intelligence Agency, or NGA). There are many Geoid models, that differ only in some cm. All Elevations used and displayed in the Rainy Lake Experiment are derived from the GEOID12B Geoid model used in North America. The following table shows the GPS Vectors in Earth Centered Earth Fixed (ECEF) cartesian (x,y,z) coordinates, the Elevation and the Geoid height for the locations. The ECEF coordinates are calculated by the GNSS receiver from the measured distances to the satellite locations in view by Multilateration. The Elevation is calculated from the Reference Ellipsoid height and Geoid height as follows: The Ellipsoid height, together with latitude and longitude, is calculated with the WGS84 Calculator from the measured x,y,z ECEF coordinates (see GNSS Ellipsoid and Height Data). The Geoid height is optained as described at Obtaining Elevations. Note: Elevations of water levels with respect to the Geoid vary about 336.909±0.044 m only. That is within the accuracy of the measurement of about ±5 cm. The Geoid height is 25.1 cm lower at the far end than at the observer locations with respect to the reference ellipsoid. Rainy Lake Elevation WGS84 google earth is 338 m. Google Earth elevation data extraction and accuracy assessment for transportation applications, a study from 2017, reported road elevation accuracies of ±2.27 m. The GNSS data in the following table is converted from the ECEF data into latitude, longitude and ellipsoid height using the WGS84 Calculator. The calculated elevations and geoid heights are listet in the table GNSS ECEF, Elevation and Geoid Data above. The following values are used in the Computer Model . The values are calculated from the GPS Vectors shown in the table at GNSS ECEF, Elevation and Geoid Data. The calculations are described at Create Computer Model Data. The values represent the locations of water level at the observers and targets in a local coordinate system with origin at observer (0) and direction to the target (6). ID Target [Typ] Dist D[mean] D[diff] Side H[calc] H[ref] H[water] Size[cal] Size[real] 111 (0) Obs Tang -26.31 0 -26.31 +1.31 3.912 1001 (0) Obs Bedf 0 0 0 0 1.854 203 (1) Bedf [0] 1094.72 1051.02 +43.70 +1.09 1.854 1.815 1.842 (0.120×0.185) 0.55×0.22 204 (2) Bedf [1] 2168.50 2102.04 +66.46 +1.46 1.854 1.841 1.854 0.238×0.366 0.24×0.37 304 (2) Tang [1] 4.218 4.267 4.280 0.238×0.366 0.31×0.47 205 (3) Bedf [1] 3233.97 3153.07 +50.90 -3.22 1.854 1.812 1.845 0.356×0.547 0.36×0.55 206 (4) Bedf [1] 4363.27 4204.09 +159.18 -2.72 1.854 1.824 1.863 0.479×0.737 0.48×0.74 306 (4) Tang [1] 5.151 5.123^1) 5.162 0.479×0.737 0.56×0.86 208 (5) Bedf [1] 5433.70 5255.11 +178.59 -2.48 1.854 1.732 1.842 0.597×0.918 0.60×0.92 209 (6) Bedf [2] 6428.94 6306.13 +122.81 0 1.854 1.673 1.854 (0.706×1.086) 1.30×0.46 309 (6) Tang [2] 6.602 6.696^2) 6.877 (0.706×1.086) 1.30×0.46 16 (7) Tang [1] 9459.20 9459.20 0 -1.76 9.740 10.535 10.773 1.039×1.599 1.04×1.60 The Tangent Target were not mounted to exactly the pre-calculated heights. They were mounted to optically align with eye level of the observer. This was done for each target separately at different days, and hence at different refractions conditions. So the GPS measured heights for the Tangent-Targets (4) and (6) differ from the heights on the images. To match the images the Computer Model uses height values optained from the images rather than the measured heights for this targets. The differences between GPS measurement and optical measurement from the images lie within the height variations due to common refraction variation. 1) Due to the different times where the measurements and images were taken, the height of the target (4) on the image diverges from the GPS measured height. The App uses 5.24 m derived from the image instead of the measured 5.123 m. The correction of +12 cm applied to the target (4) height is small compared to the target size of 86 cm and an apparent refraction height variation of about ±15 cm for a variation of the refraction coefficient of k = ±0.1. This correction does not affect the outcome of the experiment. 2) Due to the different times where the measurements and images were taken, the height of the Tangent-Target (6) on the image diverges from the GPS measured height. The target is broken on some images, so I applied two different correctons: The App uses 6.2 m for the broken height and 6.5 m for the upright height, instead of the measured 6.696 m. The correction of −49.6 cm applied the broken target and −19.6 cm for the upright target is small compared to an apparent refraction height variation of ±32 cm for a variation of the refraction coefficient of k = ±0.1. This correction does not affect the outcome of the experiment. If the SizeVar slider in the Objects 1 panel is set to 0, then the broken version of the Tangent-Target (6) is drawn. If the SizeVar slider is set to 1, then the upright version of the Tangent-Target (6) is drawn. The Bedford Targets were set exactly 1.85 m above the lake water level. In the table above are the target center heights Href listed as used by the computer model. Due to variations in the gravitational field, expressed by the Earth Gravitational Model (EGM96 Geoid), see Obtaining Elevations, the elevation of the water surface at the last target (7) is 25.1 cm lower than the elevation of the water surface at the observer. The computer model does not take the Geoid variations into account. To simulate the exact geometrical 3D positions of the targets, all target center heights were adjusted for a sphere by using the ellipsoid heights of the targets with respect to the ellipsoid height at the observer. This results in the reduced target center heights shown in the table above. Using this adjustments the target images of the computer model match the observation perfectly, while the horizon appears 23.8 cm too height on the computer model, because it does not model the surface drop of 25.1 cm of the Geoid.	{"url":"https://walter.bislins.ch/bloge/index.asp?page=Rainy+Lake+Experiment%3A+Location+Graph+and+Data","timestamp":"2024-11-06T07:58:06Z","content_type":"text/html","content_length":"73378","record_id":"<urn:uuid:43d2eac9-faaa-4be0-aa07-e4b14e032128>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00720.warc.gz"}
Math for Elementary and Middle School Educators - MTH 120 Math for Elementary and Middle School Educators - MTH 120 at Camp Community College Effective: 2024-05-01 Course Description Provides a comprehensive and conceptual examination of fundamental mathematical concepts covered in VDOE K-8 Standards of Learning (SOLs). Designed for future K-8 mathematics educators. Emphasizes problem-solving, logical reasoning, the establishment of connections between mathematical concepts, effective communication of mathematical ideas, and the utilization of multiple representations. This is a cross-listed course with EDU 120 Lecture 4 hours. Total 4 hours per week. 4 credits The course outline below was developed as part of a statewide standardization process. General Course Purpose The purpose of this course is to provide a comprehensive and conceptual understanding of fundamental mathematical concepts covered in VDOE K-8 mathematics Standards of Learning (SOLs). This course is intended to cover the content of the first semester of a two-course lower-level math for elementary educators sequence, with the second semester of the course taken at the four-year college after Course Objectives • Quantitative Literacy □ Use problem-solving skills and quantitative reasoning to solve problems, explore new ideas, and improve your understanding of mathematics • Critical Thinking □ Formulate multiple solution paths for mathematical problems and describe connections between and within these paths • Written Communication □ Demonstrate proper use of terminology, notation, and/or written conventions used in the field of mathematics and mathematics education • Professional Readiness □ Analyze and evaluate the mathematical thinking of K-8 students expressed in their oral and verbal reports, written work, and authentic representations □ Design strategies to develop positive math beliefs, including growth mindset, persistence in work, and productive student struggle Major Topics to be Included • Problem Solving and Quantitative Reasoning Skills □ Identify connections between conceptual knowledge and standard algorithms □ Generate meaningful mathematical representations using algorithms, employing both written formats and physical manipulatives • Counting and Number Systems □ Describe quantities across multiple number systems, encompassing natural numbers, whole numbers, integers, and rational numbers □ Establish connections between concepts such as place value and regrouping, explaining their relationship within the base-ten number system □ Analyze similarities and differences between the base-ten number system and numbers in alternative bases • Integers and Rational Numbers □ Apply fundamental understanding of whole numbers to grasp integers and their associated operations □ Utilize core principles of number theory, such as prime factorization, divisibility, greatest common factors, least common multiples, etc., to examine the structure of integers □ Generate multiple representations for rational numbers and justify their equivalence in various forms • Comparing Fractions and Arithmetic Operations □ Make valid comparisons between rational numbers and model arithmetic calculations with them □ Utilize various reasoning approaches to justify arithmetic operations involving whole numbers □ Develop standard and non-standard algorithms for addition, subtraction, multiplication, and division by drawing on principles of counting, place-value grouping, and partitioning • Base Ten and Other Bases □ Exhibit adaptable and conceptual thinking while exploring numbers, operations, and their interconnections □ Investigate the concepts of place value and regrouping within the framework of the base-ten system and alternative numerical bases • Addition and Subtraction □ Use, compare, and mathematically justify different strategies and representations to solve addition and subtraction problems □ Develop versatile computational skills across all number categories, using and articulating various methods for addition and subtraction • Multiplication and Division □ Formulate multiple solution paths for multiplication and division problems and describe connections between and within these paths □ Assess and explain the mathematical ideas and reasoning used in multiplication and division • Formulating Solution Paths and Beliefs about Mathematics □ Embrace a growth mindset to reflect on personal beliefs regarding mathematics and the teaching and learning processes, fostering an open exploration of mathematical concepts. □ Generate comprehensive mathematical representations for the course content, incorporating written formats (such as area models, number lines, scaled diagrams, strip diagrams, etc.) and tangible manipulative tools (like base-ten blocks, Cuisenaire rods, linking cubes, pattern blocks, etc.), to enhance understanding and engagement.	{"url":"https://courses.vccs.edu/colleges/camp/courses/MTH120-MathforElementaryandMiddleSchoolEducators/detail","timestamp":"2024-11-02T01:54:43Z","content_type":"application/xhtml+xml","content_length":"13978","record_id":"<urn:uuid:14e89acf-d656-48fa-81c3-2f2c407e3b0b>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00876.warc.gz"}
Ring current model and anisotropic magnetic response of cyclopropane Three-dimensional models of the quantum mechanical current density, induced in the electron cloud of the cyclopropane molecule by a uniform magnetic field applied either along the C[3] or the C[2] symmetry axes (indicated by B[\|\|] and B⊥, respectively), have been constructed via extended calculations. These models of near Hartree-Fock quality, previously shown to provide a good agreement between computed and observed values of magnetic tensors, have been used to interpret the magnitude of the diagonal components of susceptibility (χ), nuclear shielding of carbon (σ^C) and hydrogen (σ ^H), and shielding at the center of mass (σ^CM). The source of the exceptionally large in-plane component σ[;⊥] ^CM, dominating the anomalous average σ[av] ^CM, is shown to be a strong delocalized current flowing around the methylene moieties and the noncyclic CH[2]-CH[2] fragment. The total current strength for a magnetic field applied in the direction of a C[2] symmetry axis is 15.7 nA/T, approximately 1.5 times larger than that calculated for B[\|\|]. The largest component of the susceptibility is instead the out-of-plane χ[\|\|], which depends on the intensity of the σ-electron currents and on the entire area enclosed within the loops that they form about the C3 axis, all over its length. In a magnetic field perpendicular to the plane of the carbon atoms, both H and C nuclei sit inside diatropic whirlpools, flowing within the sp^3 hybrid orbital which form the C-H bonds and extending for several bohrs above and below the σ[h] plane. The average values and the anisotropy of carbon and proton shieldings are strongly biased by the diamagnetic shift of the out-of-plane tensor components partially determined by these vortices. The current density model of cyclopropane is revised according to these findings. Empreinte digitale Examiner les sujets de recherche de « Ring current model and anisotropic magnetic response of cyclopropane ». Ensemble, ils forment une empreinte digitale unique.	{"url":"https://researchportal.unamur.be/fr/publications/ring-current-model-and-anisotropic-magnetic-response-of-cycloprop","timestamp":"2024-11-06T12:16:41Z","content_type":"text/html","content_length":"55345","record_id":"<urn:uuid:2e4a435f-1e93-4c19-8767-3ff92c6ff9f2>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00019.warc.gz"}
What is the derivative of y=sec(3x^2)? \| HIX Tutor What is the derivative of #y=sec(3x^2)#? Answer 1 Let #y = sec(u)# and #u = 3x^2#. The derivative of #secx# can be found by the following proof: #secx = 1/cosx# #(1/cosx)' = ((0 xx cosx) - (1 xx -sinx))/(cosx)^2# #(1/cosx)' = sinx/(cos^2x)# #(secx)' = secx xx sinx/cosx# #(secx)' = secxtanx# The derivative of #3x^2# can be obtained using the power rule: #(3x^2)' = 2 xx 3x^(2 - 1)# #(3x^2)' = 6x# The chain rule states that #dy/dx = dy/(du) xx (du)/dx#. Hence, #dy/dx = secutanu xx 6x = 6xsec(3x^2)tan(3x^2)# a) #cscx# b) #cot(3x^2 + 5x + 1)# c) #tan(e^(2x^2))# Hopefully this helps, and good luck! Sign up to view the whole answer By signing up, you agree to our Terms of Service and Privacy Policy Answer from HIX Tutor When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some Not the question you need? HIX Tutor Solve ANY homework problem with a smart AI • 98% accuracy study help • Covers math, physics, chemistry, biology, and more • Step-by-step, in-depth guides • Readily available 24/7	{"url":"https://tutor.hix.ai/question/what-is-the-derivative-of-y-sec-3x-2-8f9af9f0d2","timestamp":"2024-11-02T10:41:20Z","content_type":"text/html","content_length":"570253","record_id":"<urn:uuid:5f980142-5903-4319-9116-6723b243f7c9>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00745.warc.gz"}
Sudo Null - Latest IT News Do you really need puzzles for job interviews? Despite the fact that the blog is called “Learn to work,” I will not teach anyone here, but simply share my thoughts about the interviews and puzzles. I recently ran a survey , "How do you feel about the use of puzzles on job interviews?" . Given the fact that polls are not welcomed on the hub (although from my point of view this is not correct, since the hub is a good resource for any kind of research and polls are not always created on the principle of "but I am interested"), but less, he did not go into deep minuses, which indicates a relative interest in the topic. Thanks to those who voted and supported the survey. Let's walk quickly through the voting options (we will go from a negative attitude to a positive one). Я отношусь к использованию головоломок на собеседованиях негативно, так как считаю, что умение решать головоломки никак не коррелирует с умением решать реальные задачи It would be foolish to expect everyone to like to solve puzzles, and it would be naive to believe that everyone loves just that style of interviewing. About 30% of voters think that using puzzles does not correlate with the ability to solve real problems. This question is very strongly correlated with other questions such as “Do I need a higher education”, “Do I need knowledge of mathematics”, etc. Despite the fact that very many, before losing their pulse, prove that you do not need to study and practically do not need to know anything, the majority nevertheless agrees that the presence of these qualities is an undoubted plus . So here we can assume that the ability to solve puzzles is a plus, but not a necessary skill. Let's answer the question: does the ability to solve puzzles correlate with mental abilities? I think the answer is yes. But here I risk being immediately abandoned by rotten tomatoes shouting “who are you to decide for everyone?” And they will be right. But let's look at it from another perspective, more global. The approach to testing mental abilities was born not today, but about a hundred years ago. And the first to adopt it were large innovative companies and the defense industry. Thus, large companies function - from Microsoft to Google. If we assume that there is no connection, then it turns out that Google and Microsoft are full of profane people who are unable to distinguish a normal candidate from a bad candidate without psychological distortions. Accordingly, such an approach would have to bring large financial losses (or the bankruptcy of these companies) in the future, which we are not observing. Thus, despite the fact that it is theoretically very difficult (even almost impossible at this stage of development) to prove unambiguous relations, in practice we get quite adequate results. This I led to the fact that further we will assume that some kind of connection is still present. I have a neutral attitude to using puzzles in interviews, as I believe that during interviews the interviewer has the right to ask any questions This is a good position, which 25% of respondents adhere to. In fact, a neutral approach is very often the most effective, as a person abstracts from tasks and sees only one goal - to successfully pass an interview and get a good offer from the employer. But this option does not apply much to this article, so we won’t stop here for a long time. I have a positive attitude to using puzzles in interviews, since I believe that the ability to solve puzzles indicates a high level of intelligence So consider Microsoft, Google, big banks and another 18% of readers This may indicate that those who voted for this option are able to solve puzzles or are impressed by this style of interviewing . With the first option it’s clear - it turns out to solve problems, why be positive? But the second option is related to personal preferences, so the percentage turned out to be slightly less. Probably, dull interviews are already fed up, so I want something new and interesting. I love solving puzzles, but I think this is a bad way to test candidates I admit that I voted for this option and the purpose of the survey was to check how many people share my opinion. As expected - a lot, namely a little less than 29%. Despite the fact that I really like to solve all sorts of problems and have previously devoted much time to this issue, at real interviews I began to notice some problems in this approach to testing knowledge. We will talk about this later. The difficulties of this approach There are actually more disadvantages than advantages. Let's consider them in more detail: • you need to have in stock a large number of tasks and puzzles; • you need to understand (at least remotely) which of the types of tasks you need to ask a particular candidate; • criteria for evaluating responses should be clear to the candidate; • the candidate should not know questions and answers (this is a problem, since you can “prepare” for the main questions) • the interviewer may not know the answers to the questions and their meaning ; • the candidate’s response may differ from the intended one, which again raises the question of evaluation criteria. And most importantly, the logical process of pondering the answer is very complex and it is not always possible to get a response from a candidate in an acceptable time even from really good In addition, the person conducting the interview must answer few questions: • Will I hire a person if he answers n questions correctly? • Will I take a man if he answers this question and does not answer this ? • how much time will I give a person to think it over? • Will I consider anything else besides the ability to solve puzzles? As a rule, the answers to these questions are very subjective and are not discussed previously, which leads to problems. Another ethical point is the fact that many (mostly strong) candidates may regard puzzle questions as an insult. This can spoil the interview at the very beginning. Why are puzzles bad at an interview? There are different types of puzzles - logical problems, weighing problems, unanswered problems, etc. In fact, often it’s not the ability to solve puzzles that is tested, but it is checked whether a particular puzzle belongs to which class of problems . Otherwise: what do they want to hear from you? As soon as you understand what exactly, you have solved the problem by 85%. I will give a simple example, on which I will try to explain why this does not always work. Once at school olympiads they were given tasks of this type: prove that the sum of the cubes of three consecutive natural numbers is divisible by 9. Many in this place thought that it was elementary, others sharply thought about how to calculate this, especially without computing devices. But in fact, the solution is very simple, if you know that this is a mathematical induction problem, the solution of which is to prove that the expression is valid for n = 1, n = k and n = k + 1, i.e. the task ultimately boils down (mainly) to simple arithmetic operations. Thus, a person who knows the type of problem will solve it quickly (and almost any task), while another person who does not know the approach is unlikely to solve it for an acceptable time (I very much doubt that the candidate will offer exactly this answer in a minute or two). There is a problem when testing is not what was supposed. Why does this not work for real tasks? Probably because when solving real problems, as a rule, there is already at least some initial data - the scope, ready-made solutions, code examples, Google, etc. At that time when you do not have the source data for solving problems. Therefore, if the company still wants to use puzzles, then it needs (IMHO): • warn the candidate that puzzles will be used • explain the evaluation criteria and the time that the candidate will have to think about • to say whether only the correct answer is important, whether the line of reasoning is important and in what proportions it will all be evaluated In this case, there is hope that the results will be adequate for both sides. Although, I believe that if the company has firmly decided to use puzzles, then the best solution is to give the candidate 2-3 hours when he can be left alone and think about the solution to the problems. This will help him gain strength, gather his thoughts and show himself from the best side. did not like the interviews using the puzzles that I went through, despite the fact that, as a rule, I coped with them. They were completely unprofessional and divorced from life. One gets the impression that Western colleagues shared their methods, but forgot to teach how to use them. Which leads to funny and sometimes bad results for candidates and companies. Finally, I recommend the book "How to move Mount Fuji", which actually caused all these thoughts to be born. As for me, reading a book is definitely worth it, but you don’t need to take everything as a common truth. Thanks for attention!	{"url":"https://sudonull.com/post/173521-Do-you-really-need-puzzles-for-job-interviews","timestamp":"2024-11-07T07:27:51Z","content_type":"text/html","content_length":"15319","record_id":"<urn:uuid:20666967-c2f4-40bf-b0f6-5873dc0827ca>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00768.warc.gz"}
the Equilibrium Constant from Equilibrium Concentrations Calculating the Equilibrium Constant from Equilibrium Concentrations Calculate Equilibrium Concentrations from Initial Concentrations. The easiest way to explain is by looking at a few examples. By the way, I'm going to stop using K[eq] and will start using K[c], where 'c' stands for 'concentration.' In the future, you will also study K[p], where 'p' stands for 'pressure.' Example #1: Calculate the equilibrium constant (K[c]) for the following reaction: H[2](g) + I[2](g) ⇌ 2HI(g) when the equilibrium concentrations at 25.0 °C were found to be: [H[2]] = 0.0505 M [I[2]] = 0.0498 M [HI] = 0.389 M 1) The first thing to do is write the equilibrium expression for the reaction as written in the problem. This is what to write: K[c] = –––––––– [H[2]] [I[2]] 2) Now, all you have to do is substitute numbers into the equilibrium expression: K[c] = –––––––––––––– (0.0505) (0.0498) 3) Solving the above and rounding to the correct number of sig figs (remember those??), we get 60.2 What are the units of the K[c]? And, for that matter, K[p]. For reasons beyond the scope of this lesson, the answer is none. The equilibrium constant does not have any units. Some advice: your teacher may insist on putting units on the equilibrium constant. Please do not march up to him/her and announce they are wrong because some guy on the Internet says so. That will NOT make your teacher happy! Example #2: The same reaction as above was studied at a slightly different temperature and the following equilibrium concentrations were determined: [H[2]] = 0.00560 M [I[2]] = 0.000590 M [HI] = 0.0127 M From the data, calculate the equilibrium constant. 1) Same technique as above, write the equilibrium expression and substitute into it. Then solve. So, we get this: K[c] = –––––––––––––––––– (0.00560) (0.0000590) 2) The answer is 48.8 Time for a small lecture: Please be very careful in using your calculator to solve these problems. When I solved this problem while writing the first edition of this tutorial (on December 28, 1998), I first got some really weird looking answer that didn't feel right, so I did it again. Sure enough, I have made an entry error somewhere in the problem. Underscoring my plea for carefulness, please note that the above problem is routine for me. Yet, I made a mistake and solely on the basis of experience I rejected my first answer as being wrong (it "felt" wrong). You guys don't have much chemistry experience, so your chemistry feel has some room to grow. So, BE CAREFUL. Here endth the lecture. Example #3: Using the same equation as above and with the following equilibrium concentrations: [H[2]] = 0.00460 M [I[2]] = 0.000970 M [HI] = 0.0147 M Calculate the K[c]. I'm not going to write the set-up, but I want you to write it down on your paper. Then solve it. The answer is 48.4. An important point: remember to square the numerator. This is the number one rookie problem in solving these things - forgetting the exponent. The number two error is wanting to change the concentrations. For example, when [HI] = 0.0147, the rookie will want to double it, saying "Well, there is a 2HI in the equation. No, No, No!! Use the concentrations as given. One more discussion point: you may have noticed the K[c] answers for #2 and #3 are slightly different when they are supposed to be the same. The answer: experimental error. One can never be perfect, so the values for K[c] that get published are actually an average of many careful experiments. Example #4: The following reaction: 2SO[2](g) + O[2](g) ⇌ 2SO[3](g) was allowed to come to equilibrium and the following concentrations were measured: [SO[2]] = 0.600 M [O[2]] = 0.820 M [SO[3]] = 1.86 M Determine the value of the equilibrium constant, K[c] 1) Write the equilibrium expression K[c] = ––––––––– [SO[2]]^2 [O[2]] 2) Insert values and solve: K[c] = –––––––––––––– (0.600)^2 (0.820) K[c] = 11.7 Example #5: Using the concentrations in the previous problem, determine the K[c] for this reaction at equilibrium: SO[2](g) + ^1⁄[2]O[2](g) ⇌ SO[3](g) 1) Write the equilibrium expression K[c] = ––––––––– [SO[2]] [O[2]]^½ 2) Insert values and solve: K[c] = –––––––––––––– (0.600) (0.820)^½ K[c] = 3.42 I want you to compare the two expressions just above. I want you to see that the second one (the one with the [O[2]] having a one-half power) is the square root of the first expression. Then, look at the square root of 11.7 and discover it to be 3.42. When you divide the coefficients of the equation by two, the new K[c] is the square root of the old value. Example #6: Using the same set of concentrations given just above, determine the equilibrium constant for this reaction: SO[3](g) ⇌ SO[2](g) + ^1⁄[2]O[2](g) 1) Write the equilibrium expression [SO[2]] [O[2]]^½ K[c] = ––––––––– 2) Insert values and solve: (0.600) (0.820)^½ K[c] = –––––––––––––– K[c] = 0.292 I want you to compare the two expressions (the two with the one-half power) just above. I want you to see that one is simply the inverse of the other. One has a K[c] of 3.42 and the second has a K[c] of 0.292. As it turns out, 0.292 is equal to 1/3.42. When you reverse the chemical equation, the K[c] of the reversed equation is the inverse of the K[c] of the unreversed equation. Calculate Equilibrium Concentrations from Initial Concentrations.	{"url":"https://web.chemteam.info/Equilibrium/Calc-K-from-equilib-conc.html","timestamp":"2024-11-04T09:19:42Z","content_type":"text/html","content_length":"9470","record_id":"<urn:uuid:79a7af40-b940-4711-83d6-b8f4bbefc24a>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00501.warc.gz"}
NYSTCE Multi-Subject Grades 5-9 Math Prep \| TabletClass Math Academy Pass the NYSTCE Grades 5-9 Math Test With Excellent Results! Are you a dedicated teacher who aspires to excel in your teacher certification test, but math is causing you anxiety? Look no further! Our NYSTCE Multi-Subject Grades 5-9 Math Prep Course is here to help you succeed. Our complete online course was developed by certified math teacher, John Zimmerman, who understands the challenges you face because he’s been in your shoes, and he’s dedicated to helping you succeed on your math test. With years of teaching experience and a track record of success, John is your trusted guide to mastering the math section of your certification exam. Don’t Take Chances With Your NYSTCE Grades 5-9 Math Preparation, Let John’s Expertise Guide You to Success on Your Math Test! “Just want to say thank you for your wonderful NYSTCE Math Test Prep course. It is the best math teacher I have ever had!! I passed with flying colors. Thank you. Thank you.”-Gina Benefits of Using TabletClass To Pass NYSTCE Math 🚀 Boost Confidence Every single math topic is explained in-depth with full lesson videos, followed by many example problems and video solutions for your complete comprehension on the topic. 🕐 Learn at Your Pace You have full control of the learning. Pause, rewind, and replay lessons as needed, ensuring you fully grasp each concept at your own comfortable pace. 🎯 Master Math Gain mastery through practice problems, detailed explanations, worksheets, and practice tests. Building your math skills step by step. 🏆 Prepare for Success We don’t just help you pass your test, we equip you with lifelong math skills and the confidence that can carry you beyond the test into your career in education. Don’t worry if you think you’re weak in math. Don’t let past struggles in math define your future potential. Take comfort in knowing that you have the ability to excel, especially with the guidance of an educator like John Zimmerman, who understands your journey and ignites your passion for learning. With TabletClass Math, you’ll not only gain an exceptional mentor, but also become part of an educational program dedicated to empowering teachers to master fundamental math skills. Enroll in our NYSTCE Multi-Subject Grades 5-9 Math Prep Course and be READY TO EXCEL WITH CONFIDENCE! We hear from many of our students who had a hard time with math before they found us. We made a big difference in their ability to finally conquer math! Their success is our success and why we do what we do. We want to hear about your success when you ACE YOUR NYSTCE Grades 5-9 Math TEST! “I just wanted to say how much I love TabletClass! I just started using it this year and it has been beyond helpful! I went through Algebra 1 & 2 and struggled because of poor explanations of methods and answers. I am now using TabletClass for geometry and everything is explained so well! I hardly ever have trouble with the example sets, but if I do I can just go to the video on the problem and see the answer explained very simply. Thank you so much for creating TabletClass! I have been recommending it to all of my homeschool friends for high school math and will continue to do so. :)” “Not even one week into the program and I love it. Even though I’ve always struggled with math you make it seem really easy in the videos. I just wish I had discovered this program even 1 month before I started taking college algebra. It’s really convenient that you have all the subject matters in almost exact order of how they are being taught in my college level course as well. I am learning a lot of the fundamentals that the college instructor doesn’t teach because it’s stuff you’re already supposed to know when you get there. I’m spending hours on your site and am learning a lot! Would totally recommend this program to anyone who needs a little help with math.” -Jazmin J. “I am very much [enjoying the course]! Wish it was my math course, I’m going to use it for my teas exam too, don’t know what I would do without it. I also passed your site along to my daughters kindergarten teacher to help her daughter in math! You’ve helped me so much and it’s only week five! I’m very grateful for your work!” Quality Math Instruction From a Teacher You Can Trust Bachelors Degree in Mathematics Masters Degree in Educational Technology Certified Middle/High School Math Teacher 20+ Years Teaching Experience 600K+ YouTube Subscribers TabletClass Offers Students A Powerful Learning Experience Expert Instruction You’ll have an expert math teacher—who knows what students struggle with the most—break down each math topic with clear & understandable video instruction that’s easy to comprehend. Engaging Lessons Access to an extensive & comprehensive math video library covering thousands of problems with complete step-by-step solutions at your fingertips, anytime you need them. Extensive Practice Full sets of worksheets per section topic, corresponding to each lesson, and practice problem solution videos cover basic math, advanced topics, and word problems. Study Guides Detailed study guides that reinforce key concepts and provide a roadmap for your preparation. Practice Quizzes Measure your progress with our quizzes, get instant feedback to reveal weak areas for improvement. Our course is packed with resources, tips, and strategies from the experience of our master teacher to maximize your NYSTCE Math test performance. $45/1-Month Access or $65/Unlimited Access Don’t Risk Wasting Time and Money by Failing to Pass Your Teacher Certification Math Test Due to Weak Math Skills. SECURE YOUR SUCCESS by enrolling in our NYSTCE Multi-Subject Grades 5-9 Math Prep Course and get READY TO EXCEL WITH CONFIDENCE! “I have always wanted to be a nurse and finally decided to go back to college at 42 years old. I had put it off for years because I was afraid of math and never thought I would pass the math courses needed for the nursing program. I am attending NH Technical Institute and just finished my first semester of algebra and passed with a B+. I never would have believed I would be that proficient in algebra. I used TabletClass a few times a week throughout the semester to help me with assignments. TabletClass course assignments closely followed those of my college class but John explained the lessons with a different perspective which helped tremendously. I still have two more semesters of algebra in the near future and I decided to take them online since I have TabletClass to help me along the way! It’s well worth the investment. Thank you John!” -Sherri P., NH What You’ll Learn In the NYSTCE Math Prep Course Our NYSTCE Multi-Subject Grades 5-9 Math Prep Course is designed to equip you with the skills and knowledge needed to conquer the math section of your teacher certification exam. Our course content is tailored to mirror the exact math topics you’ll encounter in your NYSTCE Grades 5-9 Math exam. We leave no stone unturned, covering everything you need to know. NYSTCE Grades 5-9 Math Prep Course Curriculum Chapter 1A: Basic Math Concepts and Review 1A.1 Factors and Multiples, Factorization, Prime and Composite Numbers and Divisibility Rules 1A.3 Basic Math Operations and Number Properties 1A.4 Decimals and Estimation & Rounding 1A.5 Venn Diagrams and An Introduction to Sets Chapter 1B: Basic Algebra (Review) 1.3 Multiplying and Dividing Real Numbers 1.5 Simplifying by Combining Like Terms PREVIEW: 1.6 One Step Equations PREVIEW: 1.7 Solving Two Step Equations PREVIEW: 1.8 Solving Multi-Step Equations 1.9 Formulas and Literal Equations 1.12 Introduction to Absolute Value 1.13 Solving Absolute Value Equations 1.14 Absolute Value Inequalities 1.15 Graphing Absolute Value Equations Extra Practice – 4 Worksheet Files Chapter Review Notes – 4 Note Files Chapter 2: Graphing and Writing Linear Equations 2.1 Graphing Lines with One Variable 2.2 Graphing Lines with Two Variables 2.6 Writing the Equations of Lines -Using Slope-Intercept Form 2.7 Writing the Equations of Lines -Using Point-Slope intercept 2.8 Writing the Equations of Lines -Given the Slope and a Point 2.9 Writing the Equations of Lines -Given Two Points 2.10 Standard Form of Linear Equations 2.11 Best Fitting Lines and Scatter Plots 2.12 Linear Models/Word Problems 2.13 Graphing Linear Inequalities in Two Variables Extra Practice – 2 Worksheet Files Chapter Review Notes – 2 Note Files Chapter 3: Systems 3.1 Solving Systems by Graphing 3.2 Solving Systems Substitution Method 3.3 Solving Systems by Elimination/Linear Combination 3.4 Solving Linear System Word Problems 3.6 Solving Systems of Linear Inequalities PREVIEW: 3.7 Linear Programming Chapter 4: Matrices and Determinants PREVIEW: 4.2 Matrix Operations 4.5 Identity and Inverse Matrices 4.6 Solving Systems using Inverse Matrices 4.7 Solving Systems using Cramer’s Rule Chapter 5: Quadratic Equations and Complex Numbers 5.1 Introduction to Quadratic Equations 5.2 Solving Quadratic Equations by Square Roots 5.3 Graphing Quadratic Equations 5.5 Solving Quadratic Equations by Factoring 5.6 The Discriminant – Types of Roots 5.8 Quadratic Equation Word Problems 5.9 Graphing Quadratic Inequalities 5.10 Complex and Imaginary Numbers Chapter 6: Functions and Relations 6.1 Introduction to Functions and Relations 6.5 Linear and Nonlinear Functions 6.8 Interval and Set Builder Notation 6.10 Transformations of Functions 6.11 Function and Relation Analysis (Finding Domain/Range) Chapter Review Notes- 2 Note Files Chapter 7: Powers and Radicals 7.1 Product and Power Rules of Exponents 7.2 Negative and Zero Exponents Rules 7.3 Division Rules of Exponents 7.9 Operations and Equations with Rational Exponents 7.10 The Distance and Mid-Point Formula Extra Practice – 2 Worksheet Files Chapter Review Notes – 2 Note Files Chapter 8: Logarithmic and Exponential Functions 8.1 Exponential Growth and Decay Functions PREVIEW: 8.2 Introduction to Logarithms 8.6 Solving Logarithmic Equations 8.7 Solving Exponential Equations Chapter 9: Polynomial Functions 9.1 Introduction to Polynomials 9.2 Adding and Subtracting Polynomials 9.4 Multiplying Polynomials Special Cases 9.5 Sum and Difference of Two Cubes 9.6 Factoring Greatest Common Factor 9.7 Factoring Quadratic Trinomials 9.10 Solving Polynomial Equations by Factoring 9.11 Polynomial Division (Long and Synthetic Division) 9.12 Remainder and Factor Theorem 9.13 Rational Root Theorem (Rational-Zero Test) 9.14 Solving n-degree Polynomials (Fundamental Theorem of Algebra) 9.15 Descartes’ Rule of Signs and Bounds Extra Practice – 2 Worksheet Files Chapter Review Notes – 2 Note Files Chapter 10: Rational Expressions/Equations 10.3 Direct and Inverse Variation 10.4 Simplifying Rational Expressions 10.5 Multiplying and Dividing Rational Expressions 10.6 Finding the LCD of Rational Expressions 10.7 Solving Rational Equations 10.8 Adding and Subtracting Rational Expressions 10.9 Graphing Rational Functions (Vertical and Horizontal Asymptotes) Chapter 11: Data, Measurement and Probability 11.1 Units of Measure and Conversion PREVIEW: 11.2 Measures of Central Tendency- Mean, Median and Mode 11.3 Exploring Data- Charts, Tables, Graphs and Plots 11.4 Introduction To Probability 11.6 Probability of Independent, Dependent and Mutually Exclusive Events 11.7 Permutations and Combinations Chapter 12: Sequence and Series 12.1 Introduction to Sequence and Series 12.2 Arithmetic Sequence and Series 12.3 Geometric Sequence and Series 12.4 Infinite Geometric Series Chapter 13: Foundations for Geometry Chapter 14: Reasoning and Proof 14.1 Conditional Statements and Converses 14.3 Deductive and Inductive Reasoning 14.5 How to Plan and Write a Proof Chapter 15: Perpendicular and Parallel Lines, Polygons 15.1 Parallel Lines and Transversals 15.2 Properties of Parallel and Perpendicular Lines Chapter 16: Congruent Triangles 16.2 Proving Congruent Triangles: Side-Side-Side and Side-Angle-Side Theorem 16.3 Proving Congruent Triangles: Angle-Side-Angle and Angle-Angle-Side Theorem PREVIEW: 16.4 Proving Congruent Triangles: Hypotenuse-Leg Theorem Chapter 17: Properties of Triangles 17.1 Medians, Altitudes and Bisectors Chapter 18: Quadrilaterals 18.2 Proving Quadrilaterals are Parallelograms 18.5 Quadrilaterals, Triangles and Midpoints Chapter 19: Similarity Chapter 20: Transformations 20.3 Translations and Glide Reflections Chapter 21: Right Triangles and Trigonometry 21.5 Right Triangle Word Problems Chapter 22: Circles 22.1 Introduction to Circles and Tangents PREVIEW: 22.3 Inscribed Angles 22.4 Other Angle Relationships in Circles 22.5 Segment Lengths and Circles Chapter 23: Area and Volume 23.2 Surface Area of Basic Figures 23.5 Area of Circles/Sectors and Arc Length Chapter 24: Conic Sections 24.2 Parabolas (Conic Sections) 24.3 Ellipses (Conic Sections) 24.4 Hyperbolas (Conic Sections) 24.5 Translations of Conic Sections Chapter 25: Trigonometry 25.2 Angles of Rotation and Radian Measure 25.3 Evaluating Trigonometric Functions 25.4 Inverse Trigonometric Functions 25.5 Graphs of Sine and Cosine Functions 25.8 Solving Trigonometric Equations Hello, and welcome to my site! I'm John Zimmerman, a certified math teacher with 20+ years of experience helping students of all ages achieve math excellence, and I can help you too! My passion is helping students succeed in math, especially those who struggle with it. I've dedicated years to develop the right approach to teaching math in my courses, and why I continue to teach it daily on YouTube—650K+ Subscribers & 100M+ Views and growing! Check out my courses, and let's get you started on the right path to reaching your educational and career goals!	{"url":"https://tcmathacademy.com/nystce-multi-subject-grades-5-9-math-prep/","timestamp":"2024-11-07T02:42:14Z","content_type":"text/html","content_length":"414015","record_id":"<urn:uuid:70a2810a-c19c-4742-8e9c-aaf942e72def>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00761.warc.gz"}
Elliptic Curve ed25519 ed25519 is strategically important as its implementation was highly optimised during its design, for high security. Edwards-curve Digital Signature Algorithm (EdDSA) was also designed to be fast. In the donna-ed25519 implementation, key functions such as ed25519_mul are laid out explicitly by loop-unrolling: t[0] = r0 * s0 t[1] = r0 * s1 + r1 * s0; t[2] = r0 * s2 + r1 * s1 + r2 * s0; t[3] = r0 * s3 + r1 * s2 + r2 * s1 + r3 * s0; t[4] = r0 * s4 + r1 * s3 + r2 * s2 + r3 * s1 + r4 * s0; Note the very obvious patterns here which are triangular in nature. With the very existence of Simple-V's REMAP subsystem it is quite natural to see if triangular remapping can be added and used. It turns out to be quite easy, and there are two possible techniques: Vertical-First and Horizontal-First. With Vertical-First, the multiply is done first as a scalar item, into a temporary register, followed by an addition of the scalar into the actual target (t0 thru t4) sv.mul temp, r, s # temporary target scalar register sv.add t,t,temp # add temporary scalar onto target vector With Horizontal-First it is extremely simple: use madd - integer multiply-and-accumulate: sv.madd t, r, *s In both cases, all three target registers are set up with the same REMAP Schedules. Additionally in both cases, t0-t4 must be pre-initialised to zeros. As always with Simple-V, the power of simplicity comes primarily from the REMAP subsystem. However in a secure environment, reduced instruction count is also critical not just for power consumption but to get the size of the binary down small enough that it could fit easily into a few lines of L1 Cache. If a huge number of loop-unrolled instructions (the normal way of handing these algorithms) are reduced down to a bare handful, with the looping covered in hardware, then it is easy to understand how valuable Simple-V and REMAP is.	{"url":"https://libre-soc.org/openpower/sv/cookbook/ed25519/","timestamp":"2024-11-04T10:21:24Z","content_type":"application/xhtml+xml","content_length":"8965","record_id":"<urn:uuid:130510fd-c131-4c72-b109-d878e519feee>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00801.warc.gz"}
Projecte llegit Títol: Numerical solution of the Bounadry Layer Equations Estudiants que han llegit aquest projecte: Director/a: MELLIBOVSKY ELSTEIN, FERNANDO PABLO Departament: FIS Títol: Numerical solution of the Bounadry Layer Equations Data inici oferta: 08-02-2022 Data finalització oferta: 08-10-2022 Estudis d'assignació del projecte: Tipus: Individual Lloc de realització: EETAC Paraules clau: boundary layer, incompressible, two-dimensional, numerical solver Descripció del contingut i pla d'activitats: The Navier-stokes equations may be parabolised by the application of the boundary layer equation. In this project, the resulting two-dimensional boundary layer equations will be expressed in the stream-function formulation and a code will be developed for their solution under arbitrary outer inviscid flow conditions. The work plan is as follows: 0) Literature review on boundary layer equations and numerical methods for partial differential equations. 1) Derivation of the two-dimensional incompressible streamfunction formulation of the boundary layer equations, including the Falkner-Skan transformation. 2) Implement a central finite differences discretisation in the wall-normal coordinate and upstream final differences in the streamwise coordindate. 3) Implement a nonlinear solver (e.g. Newton method) to solve the resulting system of algebraic equations. 4) Consider a mapping for the semi-infinite domain and spectral methods (e.g. Chebychev collocation) for the wall-normal discretisation. 5) Consider implementing also time evolution. Overview (resum en anglès): The aim of this project is to develop a code that is capable of solving numerically the parabolised Navier-Stokes equations that govern the flow dynamics within two-dimensional boundary layers. Using a self-similarity scaling on the streamfunction formulation and given appropriate upstream and inviscid outer flow boundary conditions, the code solves the boundary layer and computes its characteristic properties. To begin with, the two-dimensional boundary layer equations have been cast in the streamfunction formulation and a Falkner-Skan-type coordinate change has been applied to express them in similarity variables. Next, the resulting third order equation has been reduced to first order following a standard approach, and the system is discretized in space using finite differences. The code has been tested against benchmark solutions for validation. The Blasius solution, which develops on a flat plate at zero incidence, and the stagnation point laminar boundary layer solution have been satisfactorily reproduced. Some problems previously solved with the approximate integral method have been revisited using the code to check the accuracy of the former. The code has also been adapted to accept outer flow boundary conditions in the form of both closed-form mathematical expressions or discrete streamwise samplings of the inviscid outer streamwise velocity distribution. A simple turbulence model has also been coded to resolve turbulent as well as laminar boundary layers and a criterion for natural transition has also been implemented. Typical behavior of turbulent boundary layers, such as their tendency to resist separation better than laminar boundary layers, is duly predicted. Finally, inviscid flow solutions past airfoils obtained with the software Xfoil have been fed into the boundary layer code to compute friction drag and detect separation. Results agree well with the literature, which further validates the accuracy of the boundary layer code	{"url":"https://mitra.upc.es/SIA/PFC_PUBLICA.DADES_PFC?w_codipfc=10720&v_curs_quad=2021-2","timestamp":"2024-11-04T02:24:46Z","content_type":"text/html","content_length":"17630","record_id":"<urn:uuid:5875b158-8634-4749-9a98-b8143dfbf172>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00166.warc.gz"}
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These pages explain how to choose the correct sizes of pipe when plumbing a house, and why it matters. This section includes a practical worked example. The theory is explored in part 1 . In this section I show how to calculate the flow-rate in a real domestic water-supply system by using a couple of design tools that link flow-rate to the available head - the pressure that makes the water move. The worked example starts here . If all you need is the graph that links flow rate to the rate-of-pressure-drop for standard pipe sizes, it's here . But before I design a water supply system for a real house, I should explain how the apparently impossible calculation problems involving turbulent flow can be quickly and easily tackled in practice. If the water has to move at a couple of metres per second, or thereabouts, how much pressure is needed? It's a simple question, but unfortunately there is no simple answer. It depends on what pipes are fitted, and how long they are. Each case must be individually calculated. But don't despair - the calculation is very easy. The main thing to remember about pressure is this: Pressure supplies the energy to push the water along the pipe. Each bit of pipe resists the flow. Energy is lost as the water moves along the pipe, so the pressure falls too. There's a pressure difference between the ends of the pipe. The longer the pipe, the more energy is lost, and the greater the pressure drop. The rate of pressure drop (that is, the pressure drop per metre of pipe) depends on the pipe diameter and the speed of flow, as you would expect. The design goal is to choose the pipe sizes that will give the flow rates you want. Each length of pipe wll have a pressure drop along its length. So the aim is to choose pipes that will drop just enough of the available pressure (from the header tank, or from the mains water supply in the street) to give the required flow rates. This means checking the pressure drop along each pipe. Pressure difference calculation One way to find the pressure difference between the ends of a pipe is to use the Darcy-Weisbach equation I mentioned in part 1. This predicts how much pressure would be needed to push the water along a pipe at a particular speed. The formula looks like this: Here, the pressure difference P needed to achieve a flow velocity v depends on the length L and diameter D of the pipe as well as the density of the fluid (ρ - about 1,000 kg/m^3 for cold water). It also depends on f, a fiddle factor - sorry, "friction factor", which is included to account for the effects of the Reynolds number. This graph (and its equation) shows the relationship between Re and The equation includes √f on both sides, and looks impossible to solve. In fact, it was quite straightforward. The trick is to begin by guessing a value for f (say, 0.01), putting this value (and Re) in the right-hand side, working out the value of the left-hand side, and hence finding f. This new value for f is closer to the actual value than the initial guess, so you plug it back into the right-hand side and do the calculation again . After a couple of iterations the answer is usually close enough to be useful. (By the way, the friction factor used by American engineers is for some reason four times bigger than this. But then, most things in America are bigger than they are in England.) The graph appears to show that the "friction factor" decreases as the Reynolds number goes up. More speed giving less friction? Hardly likely, is it? In fact, that's not what the graph is saying. The "friction factor" is purely a measure of how the pipe affects the flow, and as the water becomes more turbulent the pipe itself plays a smaller part in events. Example - a kitchen sink Theory is all very well, but let's see some actual numbers. The kitchen sink is fed by 15mm pipe. How much pressure will it take to get hot water (at about 60^oC, say) moving out of the tap at 2 metres/second, and is this head achievable? 1. Calculate Reynolds number from water speed, pipe size, density, & viscosity. 2. Look up friction factor f on the graph. 3. Calculate pressure drop from Darcy-Weisbach equation. Start by calculating the Reynolds number: Re = Speed x Diameter x (Density / Viscosity) We know that the speed is to be 2 m/sec, and the internal diameter of 15mm pipe is 13.6mm. From Table 1, (ρ/μ) for water at 60^o is about 3.1 x10^6. Then the Reynolds number in this case is: Re = 2 x 13.6 x10^-3 x 3.1 x10^6 = 84,000 near enough. From the graph above, this Re has a "friction factor" f of about 0.019. So in the pressure-difference equation we know f (0.019) and v (2 m/s) and ρ (992.1) and D (13.6 mm). For now, assume that the length L is just 1.0 metre. Then the pressure difference (per metre) needed to get the water flowing is: P = 0.019 x 2^2 x (992.1 / 2) x (1.0 / 13.6 x10^-3) = 2,800 N/m^2 This means that each metre length of the 15mm pipe must have a pressure difference of 2,800 N/sq.m. between its ends to push water though it at 2 m/sec. If the pipe is 10m long, the total pressure difference between the ends of the pipe (that is, the head required) would be 28,000 N/sq.m. Or, to put it another way, the water will flow at 2 metres/second if the head happens to be exactly 28,000 If you're more comfortable with pressure expressed as the head in feet, the conversion factor is: A head of 1 foot of water ≈ 3,000 N/sq.m. So 28,000 N/sq.m. is about the same as a head of 9 feet (or 3m) of water. But if the head is not exactly 9 feet - and in practice, Sod's Law says it won't be - the water will flow at a different speed! More on this later. The equations are useful if you ever need to calculate accurately, but in practice it's easier to check from a graph that what you plan to do will work. The log-log graph 1 bar = 100,000 1 lb/sq.in = 7,000 1 foot of water = 3,000 7 m. of water (the minimum water pressure guaranteed in the UK) = 69,000 1 Pascal = 1 N/sq.m This graph shows pressure drop per metre for a given flow rate and pipe size. You'll find something similar in the relevant British Standard. It was constructed from the pressure-drop equation and covers water speeds from 2.0 m/sec (at the top) down to 0.2 m/sec, and is valid for all normal temperatures. It's saying that the pressure drop along a length of pipe is (nearly) proportional to the square of the flow rate in the pipe. The graph tells you nearly all you need to know. Use it like this: 1. Decide the flow rate you need (sink: 0.3 litres/sec; bath: 0.5 litres/sec, say). 2. Choose a pipe size that will carry this flow at less than 2m/sec. 3. Use the graph to find the rate of pressure drop, per metre of pipe run. This tells you the head you will need. Example - a bathroom sink A bathroom sink is fed with 15mm pipe and needs a flow rate of 0.3 litres/sec. From the graph, this means the water speed will be 2 metres per second and the head required to achieve this flow rate will be 4,000 N/sq.m. (or 1.3 feet height of water) per metre of pipe. So if the sink is fed from a tank 13 feet above it, the pipe run could have an (equivalent) length of 10 metres. If the pipe is shorter, the water will flow faster. Example - 22mm pipe connected to the water main Suppose that the stop-tap offers a 22mm connection, and that the water pressure here is 2 bar. Assume a horizontal straight 22mm pipe is connected to the stop-tap. What will the flow rate be if the pipe is 10m long? What happens if it's 100m long? From the graph, 10m of 22mm pipe carrying 0.7 litres/sec ( = 2 m/sec water speed) has a pressure drop of P = 10 x 2,500 N/sq.m = 25,000 N/sq.m If a pressure of nearly ten times this (and 2 bar = 200,000 N/sq.m) is applied, the graph can't predict what would happen. I would guess that the flow rate would exceed 2 litres/sec and the noise level would be scary. This is not a good idea! However, with 100m of pipe, the 200,000 N/sq.m mains pressure works out at a more modest 2,000 N/sq.m per metre. The graph says this delivers about 0.6 litres/sec (36 litres/minute) at a water speed of something under 2 m/sec. It would work fine. Example - a fountain Suppose that the 10m length of 22mm pipe connected to the stop-tap points vertically upwards. The 2 bar pressure at the stop-tap will presumably cause water to squirt out of the top. How high will it The weight of water in the vertical pipe exerts a pressure downwards, towards the stop-tap, of Pressure = Length x density x g (N/sq.m) Pressure = 10 (m) x 1,000 (kg/cu.m) x 9.8 (m/sec/sec) ≈ 100,000 (N/sq.m) This pressure acts downwards, opposing the 200,000 N/sq.m upwards pressure at the stop-tap. The net upwards pressure is reduced to 100,000 N/sq.m. Over the 10m length, there is now 10,000 N/sq.m per metre. This is off the graph, as it represents a water speed of well over 2 metres/sec. It might give a flow rate of about 1.5 litres/sec. The cross-sectional area of 22mm pipe is 320 sq.mm., so 1.5 litres in 22mm pipe occupies a length of (1.5/1,000) / (320 x 10^-6) = 4.7 metres which means that when the water leaves the top of the pipe it is moving at 4.7 metres/sec. How high will it go? The equation I learnt at school relates speed and distance for a body moving under gravity like this v^2 = u^2 - 2 g s where u and v are the initial and final velocity, s is distance, and g is 9.8 m/sec/sec as usual. Here u = 4.7 m/sec and v = 0 (because the water stops rising, pauses, then begins to fall) so 4.7^2 = 2 x 9.8 x s ... s = 4.7^2 / (2 x 9.8) = 1.1 metres ( ≈ 3.5 feet). So at the end of a 10m vertical pipe - that is, at rooftop height, 30 feet in the air - mains water pressure would still produce a fountain about as high as a child! No wonder water companies' pipes How do you go about choosing the correct sizes for all the different pipes in the house? Here's a simplified sketch of the hot- and cold-water supply system in a two-storey house. The cold-water header tank in the loft feeds a bath on the first floor, and the kitchen sink on the ground floor. It also feeds the hot water pipes via the cylinder. The first step is to sketch the layout and choose the pipe sizes such that the water flows fast enough to fill the bath and the sink in a sensible time. Then calculate what will actually happen, and decide whether anything needs to be changed. So, here: • To fill a 10-litre kitchen sink in half a minute, the flow rate of the pipe feeding it must be close to 0.3 litres/second, and 15mm pipe can probably handle this. • The flow rate for a bath should be higher, but as a single 22mm pipe can comfortably deliver more than 0.5 litres/second, two 22mm pipes (hot and cold) will be more than adequate. This house doesn't have a shower. Showers use about 10 litres per minute - that is, about 0.17 litres/second - so 15mm pipes would be quite big enough if the owner ever decided to install one. A five-minute shower only uses about 30 litres of hot water. That's why it's cheaper to shower than to have a bath. It's cheaper still when you share with a friend, apparently. Cold water pipes The design starts with the cold feeds. The kitchen sink needs 0.3 litres/sec, and according to Table 4 a 15mm pipe will only deliver 0.22 litres/sec at a water speed of 1.5 metres/sec. The choice is, to pay more and use 22mm pipe, or to fit 15mm pipe and put up with a small amount of extra noise. Which would you go for? A cautious person might ask, how much more noise? A mountain stream, or Niagara Falls? That's easy to answer. Increasing the flow rate by 30% means that the water flows 30% faster - 2.2 metres/sec instead of 1.5 m/s. The noise level would roughly double. That shouldn't be a big The kitchen sink cold feed can therefore be 15mm, at least up to the junction with the bath cold feed. The pipe from here to the bottom of the cylinder serves two purposes, though. Someone might be running a bath whilst someone else is downstairs washing up. What then? Suppose that the bath cold tap and the kitchen sink cold tap are both running at once, with 0.3 litres/sec going to the sink downstairs and (say) 0.5 litres/sec going into the bath. The total flow-rate would be 0.8 litres/sec, and 15mm pipe would complain at that. Will 22mm pipe do, or should it be 28mm? You might ask how likely is it that both taps would be on at the same time, and if it did happen, would anyone mind too much if the cold flow slackened off for a few seconds? Probably not (unless they were having a shower!) 22mm pipe should be adequate. Finally, there's the pipe from the header tank to the bottom of the cylinder. This one is more important than it looks - it not only carries cold water to the taps but also refills the cylinder as hot water is taken from the top. Water flows through this pipe to every tap in the house. It would be sensible to make it 28mm, which can carry over a litre per second. Hot water pipes The hot-water pipes are easy to size, because the thinking has already been done for the cold pipes. The kitchen sink will be fed in 15mm from the tee under the bath, and then in 22mm from the top of the cylinder. The vent pipe leading from the cylinder to above the header tank should also be 22mm (as local authority planning laws usually require). This pipe is only there as a safety measure - if something goes wrong, and the water in the cylinder boils, it can siphon up safely into the tank instead of bursting the cylinder and ruining all the carpets. It's all very well calculating pipe sizes by assuming a flow rate, but what will actually happen in a real house in practice? How fast will the water flow out of the kitchen sink tap? How long will it really take to fill the bath? It is possible to predict how a real system will behave. In this section I show how to calculate what will happen in the two-storey house design described earlier. Each step is explained in some detail in order to make it easier to adapt the calculation to the different problem you may be trying to solve. Pipes often go round corners The pressure driving the water along the pipes is the head. For the bath, this is 3 metres (say), and for the kitchen sink on the floor below it's 5 metres (say). This pressure is opposed by the friction losses in the pipes, which can be thought of as the pressure-difference-per-metre needed to push the water along at the flow rate you want. The log-log graph can be used to find the flow rate in a pipe run when the head is known. There is one small difficulty. Real pipe goes round corners, and through tees, and valves, and other fittings. Each fitting creates its own bit of turbulence and absorbs some energy. How can this be taken into account? Quite easily, as it happens. In just the same way that a length of straight pipe needs a pressure difference to push water through it, so does an elbow, or a valve. The pressure difference required across a 15mm elbow to move water though it at, say, 0.2 litres/sec can be measured. Whatever this number is, it must be the same as the pressure difference required to move water through some length of straight 15mm pipe at the same speed. In fact, this equivalent length is about 0.4 metres for a 15mm elbow. So the pressure drop in the elbow can be included by pretending that the 15mm pipe is really straight, but 0.4 metres longer than it actually is. The "equivalent lengths" of some common fittings are listed below. Fittings: equivalent length Table 5: The equivalent lengths (in metres) of some standard Pipe size Elbow Tee: through Tee: into branch Tee: from branch 15 mm 0.4 0.05 0.7 0.6 22 mm 0.6 0.09 1.1 1.0 28 mm 0.9 0.12 1.6 1.4 One common fitting that doesn't appear in the table is the shower head. Its function is to take the stream of water flowing in a 15mm pipe and split it into many little streams, each about 1mm in diameter. This process takes a lot of energy. In terms of equivalent pipe length, a shower head might represent as much as 10-20 metres of 15mm, or even more, and this has a serious impact on flow-rate. That's why many people opt for a pumped shower, or one run directly from mains pressure via a combi boiler. Example - the kitchen sink feed from the bathroom The 15mm pipes run under the bathroom floor, then down to the ground floor, along to the sink, then up again to the taps. There are 5 elbows (right-angle bends) in each pipe. According to Table 5, each elbow causes the same pressure drop as 0.4 metres of 15mm pipe. So the elbows represent 5 x 0.4m = 2m of pipe. The pipes themselves are about 7m long, so the total equivalent length of each one is 7m + 2m = 9m of pipe. Then from the log-log graph, to achieve a water flow rate of 0.3 litres/second, the head would have to be 9m x 4,000 N/sq.m = 36,000 N/sq.m. Pipes are different sizes, too Suppose someone turns on the cold tap at the kitchen sink. What will happen? Water will begin to flow out of the header tank, down the 28mm pipe to the cylinder, along the 22mm pipe to the bath, then down the 15mm pipe to the sink. How fast it flows depends on the head and the opposing frictional pressure drop. The head is known to be 5m, but the opposing frictional loss must be calculated. The problem is that each different pipe size offers a different resistance to the same flow rate. What's needed is some way of expressing these different resistances in some common unit, so that they can be just added together. A clue comes from the log-log graph. The lines are (nearly) parallel. This means that the rate of pressure drop (RPD) for 22mm pipe (say) is always some fraction of that for 15mm pipe, at the same flow rate. At 0.05 litres/sec, 15mm pipe has a RPD of about 150 N/sq.m/m, whilst for 22mm RPD is just 20 N/sq.m/m - about seven times smaller. (RPD for 15mm pipe) / (RPD for 22mm pipe) = 7 / 1 And at 0.2 litres/sec the figures are 1900 and 270 - again, a ratio of about 7 to 1. So to get the same flow rate, 15mm pipe needs seven times the pressure difference that 22mm needs! 1m of 15mm pipe behaves like 7m of 22mm pipe. These figures aren't exact, but they're near enough to be useful in the real world. 1m of 22mm pipe behaves like (1/7)m - 0.13m - of 15mm pipe. The idea can be extended to the other pipe sizes. The table below shows the length of each standard size pipe that is equivalent to a 1 metre length of 15mm pipe. It says, for example, that just 3.5cm of 28mm pipe has the same pressure drop as 1m of 15mm pipe. Table 6: The lengths (in metres) of standard pipe sizes equivalent to 1m of 15mm 10 mm 15 mm 22 mm 28 mm 35 mm 42 mm 54 mm 7 1.0 0.13 0.035 0.012 0.0047 0.0013 Flow rate calculations So, back at the sink... What is the flow rate out of the kitchen cold tap? The question was, how fast will water come out of the cold tap at the kitchen sink? 1. Work out the equivalent length of the 15mm section. 2. Work out the equivalent lengths of the 22mm and the 28mm sections. 3. Convert the 22mm and 28mm lengths to their equivalent 15mm length. 4. Add all the lengths of 15mm equivalent together. 5. Work out the total pressure drop (from head, ρ, g). 6. Find the average rate of pressure drop (divide by pipe length). 7. Look up the corresponding flow rate on the log-log graph. The 15mm section runs from the kitchen tap itself up to the tee with the bath tap. It is about 7m long with five elbows, so it has an equivalent length of [15mm actual] = 7.0m (the pipe) + (5 x 0.4m) (the elbows) = 9.0m. The 22mm section includes two tees, and the pipe itself. If the 22mm pipe is (say) 3.5m long, this represents an equivalent length of [22mm actual] = 3.5m (the pipe) + (0.09m + 1.1m) (the tees: 1 in, 1 through) = 4.7m. [Convert 22mm actual --> 15mm equivalent] = 4.7m x 0.13 = 0.6m. The 28mm pipe is 6m long, with two elbows, giving an equivalent length of [28mm actual] = 6.0m (the pipe) + (2 x 0.9m) (the elbows) = 7.8m. [Convert 28mm actual --> 15mm equivalent] = 7.8m x 0.035 = 0.3m. So the total equivalent length of 15mm pipe is: 9.0m (15mm) + 0.6m (22mm) + 0.3m (28mm) = 9.9m. Now, the head is 5m, and we know that: Pressure = Length x Density x g so putting in numbers for density and g, the pressure at the kitchen tap will be: 5 [m.of water] x 1,000 [kg/m^3] x 9.8 [m/sec^2] = 49,000 N/sq.m This pressure drop is shared out along the pipe run - that is, along the 9.9m equivalent length of 15mm - which means the average rate of pressure drop is 49,000 / 9.9 = 5,000 N/sq.m per metre more or less. From the log-log graph, 15mm pipe with a RPD of 5,000 N/sq.m per metre has a flow rate of about 0.35 litres/second. This is what will come out of the tap, and more by luck than by skilful design, it's close to the 0.3 litres/second that it should be. But is this figure true? Cross-check the result by working backwards. Breaking it down, the answer says that the 9m of real 15mm pipe accounts for (9 x 5,000) = 45,000 of the 49,000 N/sq.m of available pressure, the 22mm length takes (0.6 x 5,000) = 3,000 N/sq.m, and the 28mm needs (0.3 x 5,000) = 1,500 N/sq.m. This adds up to 49,500, which is close enough to the expected figure of 49,000. This is supposed to be engineering, not physics. Then the flow rate in the actual 4.6m of 22mm pipe at its RPD of (3,000 / 4.6m) = 652 N/sq.m per metre is, from the log-log graph, about 0.35 litres/second. And for the actual 7.8m of 28mm at its RPD of (1,500 / 7.8) = 192 N/sq.m per metre, the flow is once again 0.35 litres/second. Each pipe is carrying the same flow rate, as it should do. So the kitchen sink tap really will deliver 0.35 litres/ What if the pipes are too noisy? In a different design - perhaps one with with fatter pipes, or fewer elbows, or a larger head - the calculation might have predicted a much higher flow rate. In that case you would expect the pipes to be noisy when the water is running. To make them quieter, the water has to be slowed down, and this is actually very easy to do. Any competent plumber installing a system will have included valves at strategic points, so that sections of the system can be isolated - when, for example, you need to change a tap washer. All you have to do is find the right valve and turn it down a bit. The extra resistance this adds will reduce the flow rate to a more sensible value. Halving the flow rate would reduce the noise by a factor of four. Running a bath What is the flow rate out of the bath cold tap? This calculation is a bit more complicated, because it involves both the hot and cold water pipes in the two-storey house sketched above. The approach is exactly the same: find the equivalent lengths , convert them to the same size pipe, add them up, find the pressure drop per metre, look up the corresponding flow rate. Cold feed only: Think about the cold water first. The 22mm pipe from the tap is 3.5m long and includes two tees. It has an apparent length of: [22mm actual] = 3.5m + (1.1m + 1.0m) = 5.6m. Similarly, the apparent length of the 28mm pipe is: [28mm actual] = 6.0m + (2 x 0.9m) = 7.8m. Since there is no 15mm pipe involved in the runs to the bath, it seems silly to convert these lengths to their equivalent 15mm lengths, then add them together, then convert them back again to 22mm. Instead, I'll simply convert the 28mm length to its equivalent 22mm value, using the figures in Table 6: [28mm actual --> 22mm equivalent] = 7.8m x (0.035 / 0.13) = 2.1m. Then the total equivalent length of 22mm is: 5.6m + 2.1m = 7.7m. The head is 3m, which corresponds to a pressure of: 3 [m.of water] x 1,000 [kg/m^3] x 9.8 [m/sec^2] = 29,400 N/sq.m So the average rate of pressure drop is: 29,400 / 7.7 = 3,800 N/sq.m per metre which according to the log-log graph means a (rather noisy) flow rate of close to 0.9 litres/second for a cold bath, rather than the 0.5 litres/second one might have hoped for. Still, things will change when the hot tap is running too. Hot feed only: Now for the hot water. The hot pipe is all 22mm, which makes it slightly easier. The pipe run to the top of the cylinder is (let's say) 6m long, and includes two tees and three elbows. [Hot: 22mm actual] = 6m + (1.1m + 1.0m + [3 x 0.6m]) = 9.9m. However, the hot water leaving the cylinder is replaced by cold water flowing from the header tank. The cylinder itself is only a kind of fitting, and it too has resistance, just like an elbow. The resistance of the whole circuit must be calculated. The 22mm run is only 1m or so, plus a tee and an elbow. The cylinder's resistance is equivalent to about 1.6m of 22mm pipe. Adding these up gives: [Cold: 22mm actual] = 1m + (1.0m + 0.6m + 1.6m) = 4.2m. Finally, there's the 28mm pipe from the header tank. I've already calculated that this is 7.8m (actual) and 2.1m (22mm equivalent), so the total equivalent length of 22mm pipe in this circuit is: 9.9m + 4.2m + 2.1m = 16.2m. The head is still 3m, or 29,400 N/sq.m, so the average rate of pressure drop is 29,400 / 16.2 = 1,800 N/sq.m per metre which the log-log graph says represents close to 0.6 litres/second for a hot bath - pretty much what it should be. The flow rate from the hot tap is less than from the cold tap because of the resistance of all the extra pipe this water has to flow through. Both hot and cold: Most people turn on both taps when they are running a bath. What happens then? It's a more difficult problem, because now the 28mm pipe from the header tank is carrying cold water both to the bath and to the bottom of the cylinder. A higher flow rate means a greater resistance. How much greater? That depends on the flow rate it's carrying, and that in turn depends on its Breaking this circle demands a little algebra, since there are now two unknown (and inter-dependent) quantities: the flow rates from each of the bath taps. I don't know yet what they are, so I'll call the flow rate out of the hot tap H litres/sec, and that from the cold tap C litres/sec. Now, the hot water circuit runs from the tee (with the 28mm pipe) up through the cylinder, down and along to the hot tap. It has an equivalent length of (9.9m + 4.2m) = 14.1m. This pipe run is carrying H litres/sec. Similarly, the effective length of the cold water circuit, from the cold tap to the same junction, is 5.6m. This pipe run is carrying C litres/sec. And the 28mm pipe, with an effective length of 7.8m (or 2.1m of 22mm equivalent), has to carry (H + C) litres/sec. I know that H and C must be less than 0.6 and 0.9 litres/sec respectively, because those are the flow rates with only one tap open. The flow rates with both taps open must be smaller, because the hot and cold flows share space in the 28mm pipe, and it will offer greater resistance to the flow, so (for now) guess that H = 0.5 litres/sec. From the graph, this implies a Rate of Pressure Drop (RPD) of 1,400 N/sq.m per metre. The effective length of the pipes carrying just hot water is 14.1m. The total pressure drop along these pipes would then be (14.1 x 1,400) = 19,700 N/sq.m. The head is 29,400 N/sq.m, so the pressure difference between the water surface in the header tank and the junction of the hot and cold circuits - at the tee near the bottom of the cylinder - would be (29,400 N/sq.m - 19,700 N/sq.m) = 9,700 N /sq.m. I'll come back to this figure in a moment. But the same pressure of 19,700 N/sq.m that drives the hot water flow is driving the cold water flow too. The effective length of the pipes carrying just cold water is 5.6m, so the RPD for the cold-water pipes is (19,700 / 5.6) = 3,500 N/sq.m per metre, and the graph says that this implies a flow rate of about C = 0.82 litres/sec. My original guess was that H was 0.5 litres/sec, and this guess resulted in a predicted value for C of 0.82 litres/sec. In other words, C is 1.64 times bigger than H. But this ratio depends only on the pipe layout. It's independent of the actual values of H and C. Whatever the real figures are, this ratio will stay the same. If my original guess that H was 0.5 litres/sec had been correct, then the combined flow in the 28mm pipe would have been (0.5 + 0.82) = 1.32 litres/sec. The graph says that the RPD of 28mm pipe carrying 1.32 litres/sec is about 2,700 N/sq.m per metre, so the total pressure drop along its effective length of 7.8m is (2,700 x 7.8) = 21,000 N/sq.m. But I have already calculated that if H really had been 0.5 litres/sec, the pressure drop along the 28mm pipe would have been 9,700 N/sq.m - only half as much. The original guess was plainly wrong! So how can the problem be solved? A Useful Approximation The straight-line log-log graph could also be written as a power law. For 15mm pipe, it would be: RPD = 35,000 x FR^1.83 For 22mm pipe, it would be: RPD = 5,000 x FR^1.85 The relationship between the two quantities of interest - flow rate and pressure drop - is extremely complex, but fortunately it can be approximated by a rather simple formula: Rate of Pressure Drop (RPD) = A x (Flow Rate)^2 + B - where A and B are constants that depend only on pipe size. I give values for A and B in the table below. Table 7: Values of constants A and B in the Useful Pipe size 10mm 15mm 22mm 28mm 35mm 42mm 54mm A 400,000 44,000 5,300 1,400 450 160 40 B 100 70 40 30 18 20 14 The approximation is accurate when the pipe is carrying a flow rate of between 30% and 100% of its maximum capacity. Running the bath Here is a simplified diagram showing only the pipe runs to the hot and cold bathtaps. Water is flowing from both taps. The cold water feed pipe is 5.6 metres long and carrying C litres/second. The hot water pipe is 14.1 metres long and carries H litres/sec. The common feed, carrying cold water to the tap and also into the bottom of the cylinder - that is, (C + H) litres/sec - is 7.8 metres long, from the header tank to the tee. Now, from the Useful Approximation, the total pressure difference between the ends of a pipe is Pressure drop in pipe = Length x [A x (Flow Rate)^2 + B] The hot and cold pipes are both fed from the common pipe, and both end in open taps. The pressure difference between the common point and each tap must be the same. So by applying the formula, doing a bit of algebra, and discarding terms that are too small to matter, we get a relationship between the flow rates that just depends on pipe lengths: This is really just a more formal way of expressing the idea that the hot and cold flow rates will always bear the same ratio to each other. But we also know that, for the whole system: Head = (Pressure drop in common pipe) + (Pressure drop in hot [or cold] pipe) and this, with a bit of algebra, can be made to yield an expression for the actual hot or cold flow rate in terms of numbers we already know! To make the equation as general as possible, I have used the symbols L[c] and L[h] to stand for the lengths of the cold and hot pipe runs respectively, and L[28] to mean the length of the common 28mm pipe. For the cold flow rate, C, the equation is: This equation looks forbiddingly complex, but finding a value for C is simply a matter of substituting known numbers for all the variables and calculating the answer. The head is 29,400 N/sq.m., L[c] is 5.6m., and L[h] is 14.1m. It's important that all the lengths be expressed in the same units, so L[28] is 2.1m (of 22mm equivalent) rather than the actual figure of 7.8m. Finally, from Table 7, the constants for 22mm pipe are A = 5,300 and B = 40. The answer I got was C^2 = 0.49 litres/second, so C = 0.7 litres/second. And since (C / H)^2 = (14.1 / 5.6), I calculate that H = 0.44 litres/second. The answer can be checked by working out the individual pressure drops using the Useful Approximation. In the hot and cold pipes: Cold pipe: Pressure drop = 5.6m x [5,300 x (0.7)^2 + 40] = 14,800 N/sq.m Hot pipe: Pressure drop = 14.1m x [5,300 x (0.44)^2 + 40] = 15,000 N/sq.m which is near enough the same, as it should be, and in the common pipe, Common pipe: Pressure drop = 2.1m x [5,300 x (0.7 + 0.44)^2 + 40] = 14,500 N/sq.m making a total of about 29,500 N/sq.m. The actual head is 29,400 N/sq.m. I think the conclusion is that the sums really do add up. The method works. If you want another look at the theoretical background to all this, you'll find it here in Part 1.	{"url":"http://www.johnhearfield.com/Water/Water_in_pipes2.htm","timestamp":"2024-11-02T23:16:09Z","content_type":"application/xhtml+xml","content_length":"45100","record_id":"<urn:uuid:aa393e36-7cc4-4726-a689-0798f575fcd2>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00059.warc.gz"}
NCERT Solutions Std 4 Maths Long and Short - MathswallahPadhai NCERT Solutions Std 4 Maths Long and Short Long and Short Questions with () sign are open ended questions. Guess the distance between any two dots. How many centimetres is it? Now measure it with the help of a scale. Did you guess right? Which two dots do you think are the farthest from each other? Check your answer. The dots M and O are farthest to each other. Which two dots do you think are nearest to each other? Check your answer. The dots D and O are nearest to each other. Look at the picture and explain how Birbal made Akbar’s line shorter. Birbal drew a line longer than the Akbar’s line. Now can you be as smart as Birbal? Make his line shorter without erasing it. Make her right arm 1 cm longer than the left arm. Draw a cup 1 cm shorter than this cup. Draw a broom half of double the length. Draw another hair of double the length. Do you remember that in class 3 you measured your height? Do you think you have grown taller? Yes, I have grown taller. How much? About 5 cm. Have your friends also grown taller? Find out and fill the table below. Friend’s name Height (last year) Height (this year) How much grown? Ayush 1 m 20 cm 1 m 28 cm 8 cm Vinay 1 m 45 cm 1 m 50 cm 5 cm Ajay 1 m 22 cm 1 m 26 cm 4 cm Nilesh 1 m 15 cm 1 m 22 cm 7 cm Jhumpa once read a list of the tallest people in the world. One of them was 272 cm tall! That is just double of Jhumpa’s height. How tall is Jhumpa? Height of the tallest person = 272 cm It is double Jhumpa’s height. Therefore, Jhumpa’s height = half of 272 = 136 cm. Could that person pass through the door of your classroom without bending? No, he cannot pass through the classroom door without bending. Will his head touch the roof of your house if he stands straight? Yes, his head touches the roof. () Who is the tallest in your family? My father is the tallest in my family. His height is 168 cm. () Who is the shortest in your family? My younger sister is the shortest in my family. Her height is 102 cm. What is the difference between their heights? 168 – 102 = 66 The difference between their heights = 66 cm This is a 100 metre race for girls. Arundhati is nearest the finishing line. She is about 6 metres from it. Behind her is Rehana. Konkana and Uma are running behind Rehana. Look at the picture. To answer the questions below choose from these distances. -3 metres -6 metres -10 metres -15 metres a). How far is Rehana from Arundhati? -3 metres b). How far ahead is Rehana from Konkana and Uma? 6 metres. c). How far are Konkana and Uma from the finishing line? 15 metres. Have you heard about a 1500m or 3000m race? (you remember that 1000 metres make 1 kilometre and 500 metres make half a kilometer). So you can say- In a 1500 metres race people run 1 and half km. In a 3000 metres race people run 3 km. Have you heard about Marathon races in which people have to run about 40 kilometres? People run marathons on roads because the track of a stadium is only 400 metres. 10 rounds of a stadium track = ________ km So, if you run a marathon on a stadium track, you will have to complete ___________ rounds. Length of the stadium track = 400 m 10 rounds of stadium track = 400 X 10 = 4000 m (1 km = 1000 m) = 4000 ÷ 1000 = 4 km 10 rounds of a stadium track = 4 km Length of marathon race = 40 km = 40 X 1000 (1 km = 1000 m) = 40000 m Length of the stadium track = 400 m Number of rounds in stadium track = 40000 ÷ 400 = 100 So, if you run a marathon on a stadium track, you will have to complete 100 rounds. Dhanu has the longest jump of 3 metres 40cm. Gurjeet is second. His jump is 20 cm less than Dhanu’s. Gopi comes third. His jump is only 5 cm less than Gurjeet’s jump. How long are Gurjeet’s and Gopi’s jump? Gurjeet’s jump: Gurjeet’s jump is = 20 cm less than Dhanu’s jump. = 3m 40cm – 20 cm = 3m 20cm Gopi’s jump: Gopi’s jump = 5 cm less than Gurjeet’s jump = 3m 20cm – 5 cm = 3m 15cm () Try and see how far you can jump. I can jump 1m (*) How far can you throw a ball? 7 to 8 metres. Look for a big ball, like a football or volleyball. How far can you kick it? About 20 metres Sports World Record Indian Record High jump (Men) Javier S. (2m 45cm) Chandra Pal (2m 17cm) Long jump (Men) Mike P (8m 95cm) Amrit Pal (8m 8cm) High jump (Women) Stefka K (2m 9cm) Bobby A (1m 91cm) Long jump (Women)) Galina C (7m 52cm) Anju G (6m 83cm) Find out from the table: 1). How many centimetres more should Chandra Pal jump to equal the Men’s World Record for the high jump? Men’s World Record for high jump = 2m 45cm Chandra Pal’s Record for high jump = 2m 17 cm 2m 45cm – 2m 17cm = 28 cm More centimetres required by Chandra Pal to equal the Men’s World Record for high jump is 28cm. 2). How many centimetres high should Bobby A. jump to reach 2 metres? Remember that 1m = 100cm Half metre = ? Record of Bobby A. for high jump = 1m 91 cm Centimetres required to reach 2 m = 2m – 1m 91cm = 9 cm 3). Galina’s long jump is nearly a). 7 metres b). 7 and a half metres c). 8 metres Option b. (7 and a half metres). 4). Look at the Women’s World Records. What is the difference between the longest jump and the highest jump? Women’s World Record for long jump = 7m 52cm Women’s World Record for high jump = 2m 9cm Difference = 7m 52 cm – 2m 9 cm = 5 m 43 cm 5). If Mike P. could jump _____ centimetres longer, his jump would be full 9 metres. Record of Mike P’s long jump = 8m 95 cm Centimetres required to reach 9 metres = 9m – 8m 95cm = 5 cm 6). Whose high jump is very close to two and a half metres? a). Stefla K. b). Chandra Pal c). Javier S d). Bobby A. The jump of Javier S. is very close to two and a half metres. The doctor has told Devi Prasad to run 2 km everyday to stay fit. He took one round of this field. How far did he run? Length of the field = 500 + 500 + 500 + 500 = 2000 m (1000 m = 1 km) = 2 km He ran 2 km. The field was very far from his home. So he chose a park nearby. The boundary of the park was about 400 metres long. How many rounds of the park must Devi Prasad run to complete 2 km? The distance Devi Prasad run every day = 2 km = 2000 m The boundary of the park = 400 m Number of rounds = 2000 ÷ 400 = 5 Devi Prasad has to run 5 rounds to complete 2 km. One day the weather was very good and a cool breeze was blowing. He felt so good that he kept jogging till he got tired after 8 rounds. That day he ran ___ km and ________ metres! The boundary of the park = 400 m Number of rounds = 8 Distance ran by Devi Prasad = 400 X 8 = 3200 m = 3000 + 200 = 3 km 200m The Qutub Minar is 72 metres high. About how many metres high is your classroom? Height of my classroom = 3 m Guess how many rooms, one top of the other, will be equal to the Qutub Minar. Number of rooms = 21 Subodh is going to Kozhikode which is 24 kilometres (km) away. Manjani is going to Thalassery which is 46 km away in the opposite direction. How far is Kozhikode from Thalassery? Distance of Kozhikode from Thalassery = 24 + 46 = 70 km Momun comes to school from very far. He first walks about 400 metres to the pond. With slippers in his hands, he then walks 150 metres through the pond. Next, he runs across the 350 metres wide green field. Then he carefully crosses the 40m wide road to reach his school. How much does Momun walk every day to reach shool = 400 + 150 + 350 + 40 = 940 m Is it more than 1 km? No, it is less than 1 km. Find out how far your friends live from school and fill the table. Friend’s Name Distance of home from the school Ayush 400 m Vinay 2 km 600 m Ajay 6 km 300 m Nilesh 1 km 200 m Amit 2 km 800 m Write in metres or kilometers. Who among you lives nearest to the school? Ayush lives nearest to the school. Who lives farthest from the school? Ajay lives farthest to the school. How many children live less than 1 kilometre away from your school? Only one student lives less than 1 kilometre away from my school. Is there anyone who lives more than 5 km away from the school? Yes, Ajay lives more than 5 km away from the school. How did they come to school? He comes to school by rickshaw. 1). How long is the thread in a reel? It may be 50m or 100 m. 2). How long is the string of a kite reel? Can it be more than a kilometre long? The length of the string of a kite reel comes in different measurements. It may be 1000m 1500m and 2000m. Yes, it can be more than a kilometer long. 3). If a handkerchief is made out of a single thread, how long would that thread be? If a handkerchief is made out of a single thread, then the length of the handkerchief may be about 4 to 5 thousands metre. 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[Solved] Question 9 12 pts Each year about 1500 st \| SolutionInn Answered step by step Verified Expert Solution Question 9 12 pts Each year about 1500 students take the Introductory statistics course at a large university. This year scores on the final Question 9 12 pts Each year about 1500 students take the Introductory statistics course at a large university. This year scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) What is the probability that the average score for a random sample of 46 students is above 75? P(xbar > 75)= (please round to four decimal places). (b) What is the probability that the average score for a random sample of 46 students is below 67? P(xbar There are 3 Steps involved in it Step: 1 Get Instant Access to Expert-Tailored Solutions See step-by-step solutions with expert insights and AI powered tools for academic success Ace Your Homework with AI Get the answers you need in no time with our AI-driven, step-by-step assistance Get Started	{"url":"https://www.solutioninn.com/study-help/questions/question-9-12-pts-each-year-about-1500-students-take-1026005","timestamp":"2024-11-05T12:05:02Z","content_type":"text/html","content_length":"104748","record_id":"<urn:uuid:0ff1ab87-c971-4882-bc89-9be992f6649b>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00636.warc.gz"}
Inverse Cotangent cot^−1 Calculator x= adjacent_side/opposite_side Calculating Inverse Cotangent of a Value The inverse cotangent (cot⁻¹) function, also known as arccotangent, is used to find the angle whose cotangent is a given value. This function is useful in trigonometry when you need to determine the angle based on the ratio of the adjacent side to the opposite side in a right-angled triangle. The inverse cotangent of a value $ x $ is defined as: $ \theta = \cot^{-1}(x) $ • θ is the angle corresponding to the cotangent value. • x is the cotangent value, representing the ratio of the adjacent side to the opposite side. Consider the following triangle ABC, right-angled at vertex B. For the angle $ \theta $ at vertex A: • Adjacent side = AB • Opposite side = BC $ \cot^{-1}(x) = \theta $ where $ x = \dfrac{\text{length of side AB}}{\text{length of side BC}} $ The following examples demonstrate how to use the inverse cotangent function to find the angle when the cotangent value is known. 1. A surveyor is measuring the angle of depression from the top of a hill to a point on the ground 100 meters away horizontally. If the height of the hill is 50 meters, what is the angle of • Length of the adjacent side (horizontal distance) = 100 meters • Length of the opposite side (height of the hill) = 50 meters First, calculate the cotangent of the angle: $ \cot(\theta) = \dfrac{\text{horizontal distance}}{\text{height of the hill}} = \dfrac{100}{50} $ Simplify the expression: $ \cot(\theta) = 2 $ Now, find the angle using the inverse cotangent function: $ \theta = \cot^{-1}(2) $ Using a calculator or reference table, the angle is: ∴ θ ≈ 26.6° 2. A ship is anchored 150 meters offshore. The distance from the base of a cliff to the ship is 120 meters along the water. If the height of the cliff is 30 meters, what is the angle of elevation from the ship to the top of the cliff? • Length of the adjacent side (distance along the water) = 120 meters • Length of the opposite side (height of the cliff) = 30 meters First, calculate the cotangent of the angle: $ \cot(\theta) = \dfrac{\text{distance along the water}}{\text{height of the cliff}} = \dfrac{120}{30} $ Simplify the expression: $ \cot(\theta) = 4 $ Now, find the angle using the inverse cotangent function: $ \theta = \cot^{-1}(4) $ Using a calculator or reference table, the angle is: ∴ θ ≈ 14.0° 3. A ski slope descends 60 meters over a horizontal distance of 180 meters. What is the angle of the slope with respect to the horizontal? • Length of the adjacent side (horizontal distance) = 180 meters • Length of the opposite side (vertical descent) = 60 meters First, calculate the cotangent of the angle: $ \cot(\theta) = \dfrac{\text{horizontal distance}}{\text{vertical descent}} = \dfrac{180}{60} $ Simplify the expression: $ \cot(\theta) = 3 $ Now, find the angle using the inverse cotangent function: $ \theta = \cot^{-1}(3) $ Using a calculator or reference table, the angle is: ∴ θ ≈ 18.4° 4. A photographer is setting up to take a picture of a tall building. The distance from the photographer to the building is 50 meters, and the height of the building is 150 meters. What is the angle of elevation of the camera to the top of the building? • Length of the adjacent side (distance from the building) = 50 meters • Length of the opposite side (height of the building) = 150 meters First, calculate the cotangent of the angle: $ \cot(\theta) = \dfrac{\text{distance from the building}}{\text{height of the building}} = \dfrac{50}{150} $ Simplify the expression: $ \cot(\theta) \approx 0.333 $ Now, find the angle using the inverse cotangent function: $ \theta = \cot^{-1}(0.333) $ Using a calculator or reference table, the angle is: ∴ θ ≈ 71.6°	{"url":"https://convertonline.org/mathematics/?topic=inverse-cot","timestamp":"2024-11-02T17:47:56Z","content_type":"text/html","content_length":"49711","record_id":"<urn:uuid:9e974c67-eefc-4378-8d4a-4454fd195fe7>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00450.warc.gz"}