url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.exactlywhatistime.com/time-ago/110-years-ago-from-today | 1,717,005,847,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00220.warc.gz | 638,350,964 | 7,586 | # What was the date 110 years ago?
## Friday May 29, 1914
0
110 years in the past was 29 May 1914, a Friday. Subtracting 110 years in the past is usually simple. Anything under a decade can usually be counted on one hand. The biggest challenge will be skipping decades behind or even centuries. Additionally, we’re 29 days from the end of May, so being in the end of of the month, you'll need to consider monthly changes as well. Weekly and daily changes most likely won't impact 110 years ago.
## How we calculated 110 years before today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how 110 years ago gets calculated on each visit:
1. We started with date inputs: used current day of 29 May, then set the calculation - 110 years, and factored in the year 2024
2. Noted your current time of year: 110 years in May will bring us back to April or further.
3. Counted backwards years from current day: date - 29 May, factoring in the 29 days left in May to calculate Friday May 29, 1914
4. Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the fiscal calendar
### Tips when solving for May 29
• Current date: 29 May
• Day of the week: Friday
• New Date: Friday May 29, 1914
• New Date Day of the week: Friday
• Counting backward from May could put you back in Q1 or even the previous year.
• The solution crosses into a different year.
• Your date will be a weekend. Consider if you only want workdays in your calculation.
## Ways to calculate 110 years ago
1. Calculate it: Start with a time ago calculator. 110 years is easiest solved on a calculator. For ours, we've already factored in the 29 days in May + all number of days in each month and the number of days in . Simply add your years and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use May's calendar: Begin by identifying 29 May on a calendar, note that it’s Friday, and the total days in April (trust me, you’ll need this for smaller calculations) and days until last year (double trust me, you'll need this for larger calculations). From there, count backwards 110 times years by years, subtracting years from until your remainder of years is 0.
3. Use excel: For more complex years calculations or if you h8 our site (kidding), I use Excel functions like =TODAY()-110 to get or =WORKDAY(TODAY()), -110, cell:cell) for working years.
## Working years in 110 calendar years
110 years is Friday May 29, 1914 or could be if you only want workdays. This calculation takes 110 years and only subtracts by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Friday May 29, 1914 date.
Work years Solution
Monday
Tuesday
Wednesday
Thursday
110 years back
Friday
May 29
Saturday
Sunday
## The past 110 years is equivalent to:
Counting back from today is Friday May 29, 1914 using a full calendar, and is also 963600 hours ago and 40.82% of the year.
Did you know?
Friday Friday May 29, 1914 was the 149 day of the year. At that time, it was 40.82% through 1914.
## In 110 years, the average person Spent...
• 8624220.0 hours Sleeping
• 1146684.0 hours Eating and drinking
• 1879020.0 hours Household activities
• 558888.0 hours Housework
• 616704.0 hours Food preparation and cleanup
• 192720.0 hours Lawn and garden care
• 3372600.0 hours Working and work-related activities
• 3102792.0 hours Working
• 5078172.0 hours Leisure and sports
• 2755896.0 hours Watching television
## What happened on May 29 (110 years ago) over the years?
### On May 29:
• 1884 Irish writer Oscar Wilde marries Constance Lloyd at St. James Church, Paddington, London | 972 | 3,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-22 | latest | en | 0.940674 |
https://www.topperlearning.com/ncert-solutions/cbse-class-11-science-physics/physics/units-and-measurements | 1,679,832,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00479.warc.gz | 1,125,170,035 | 58,912 | Request a call back
# Class 11-science NCERT Solutions Physics Chapter 2 - Units and Measurements
## Units and Measurements Exercise 35
### Solution 1
(a) 1 cm =
Volume of the cube = 1 cm3
But, 1 cm3 = 1 cm 1 cm 1 cm =
1 cm3 = 10-6 m3
Hence, the volume of a cube of side 1 cm is equal to 10-6 m3.
(b) The total surface area of a cylinder of radius r and height h is
S = 2 r (r + h).
Given that,
r = 2 cm = 2 1 cm = 2 10 mm = 20 mm
h = 10 cm = 10 10 mm = 100 mm
= 15072 = 1.5 104 mm2
(c) Using the conversion,
1 km/h =
Therefore, distance can be obtained using the relation:
Distance = Speed Time = 5 1 = 5 m
Hence, the vehicle covers 5 m in 1 s.
(d) Relative density of a substance is given by the relation,
Relative density =
Density of water = 1 g/cm3
Again, 1g = kg
1 cm3 = 10-6 m3
1 g/cm3 =
11.3 g/cm3 = 11.3 103 kg/m3
### Solution 4
The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.
(a) An atom is a very small object in comparison to a soccer ball.
(b) A jet plane moves with a speed greater than that of a bicycle.
(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.
(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.
(e) A proton is more massive than an electron.
(f) Speed of sound is less than the speed of light.
### Solution 5
Distance between the Sun and the Earth:
= Speed of light Time taken by light to cover the distance
Given that in the new unit, speed of light = 1 unit
Time taken, t = 8 min 20 s = 500 s
Distance between the Sun and the Earth = 1 500 = 500 units
### Solution 6
A device with minimum count is the most suitable to measure length.
(a) Least count of vernier callipers
= 1 standard division (SD) - 1 vernier division (VD)
(b) Least count of screw gauge =
(c) Least count of an optical device = Wavelength of light 10-5 cm
= 0.00001 cm
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.
### Solution 7
Magnification of the microscope = 100
Average width of the hair in the field of view of the microscope = 3.5 mm
Actual thickness of the hair is = 0.035 mm.
### Solution 8
(a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,
(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.
(c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.
### Solution 9
Area of the house on the slide = 1.75 cm2
Area of the image of the house formed on the screen = 1.55 m2
= 1.55 104 cm2
Arial magnification, ma =
Linear magnifications, ml = | 869 | 3,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-14 | longest | en | 0.892273 |
https://eatmorebutter.com/2-3-cup-butter-equals-how-many-tablespoons/ | 1,721,162,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00708.warc.gz | 199,975,573 | 59,072 | Connect with us
# Converting 2/3 Cup of Butter Equals How Many Tablespoons
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It’s commonly understood that accuracy is crucial in baking. Therefore, the successful execution of a recipe often hinges on measuring the correct quantity of butter.
Have you ever wondered how many tablespoons are in 2/3 cup of butter? Well, I’ve got the answer for you!
## Key Takeaways
• Conversion rate for butter: 2/3 cup equals 10 and 2/3 tablespoons
• Calculating equivalent in tablespoons: multiply amount of butter by 16/3
• Practical Measurement Tips: 2/3 cup of butter equals 10 and 2/3 tablespoons or approximately 149.33 grams
• Simplifying Measurement Conversions: Visualize the conversion ratio between 2/3 cup and tablespoons as 1 cup equals 16 tablespoons, then divide by 3 to get the equivalent for 2/3 cup.
## Understanding the Conversion Rate
You’ll need to know the conversion rate to figure out how many tablespoons are in a cup of butter. Understanding this practical application of common kitchen conversions can be quite helpful when following recipes or adjusting measurements.
The conversion rate for butter is commonly known as 1 cup equals 16 tablespoons. This means that if a recipe calls for 1 cup of butter, you can easily convert it to tablespoons by multiplying the cup measurement by 16.
For example, if you need 2 cups of butter, you would multiply 2 by 16, giving you 32 tablespoons. This knowledge allows you to accurately measure and use the right amount of butter in your recipes, ensuring the desired taste and texture in your culinary creations.
## Calculating the Equivalent in Tablespoons
To calculate the equivalent in tablespoons, just divide the amount of butter by 16. This simple conversion allows you to understand measurement units and easily convert to metric measurements.
• Divide this amount by 16 to convert it to tablespoons.
• For example, if you have 1 cup of butter, divide it by 16. The result is 0.0625 cups, which is equal to 16 tablespoons.
• Remember that there are 16 tablespoons in 1 cup of butter.
• Keep in mind that this conversion is specific to butter and may not be applicable to other ingredients.
Understanding measurement units and converting to metric measurements is essential in cooking and baking. By knowing the equivalent measurements, you can accurately follow recipes and achieve the desired results. So, the next time you come across a recipe that calls for butter in cups, you can easily determine the amount in tablespoons.
## Practical Measurement Tips
I’ve often found myself in a state of confusion when it comes to converting butter measurements. From cups to tablespoons, it can be a bit overwhelming to keep track of all the conversions.
However, there are some practical tips that can simplify the process and make it a lot easier to navigate. Let’s explore these tips and uncover the secrets to mastering butter conversions without any confusion.
### Butter Conversion Confusion
If you’re confused about how many tablespoons are in a cup of butter, don’t worry, it’s a common question. Understanding butter measurements can be tricky, but with a little knowledge, you’ll be able to convert butter to grams easily.
• Start by knowing that one cup of butter is equal to 16 tablespoons.
• To convert butter to grams, you need to know that one tablespoon of butter weighs approximately 14 grams.
• So, if you have a recipe that calls for 1/2 cup of butter, you can calculate that it is equal to 8 tablespoons or 112 grams.
### Simplifying Measurement Conversions
Understanding measurements can be tricky, but with a little knowledge, you’ll be able to simplify conversions easily.
One way to simplify measurement conversions is by using measurement shortcuts. These shortcuts are simple rules that help you convert between different units of measurement quickly. For example, when converting from cups to tablespoons, you can use the shortcut that one cup is equal to 16 tablespoons.
Another helpful technique is visualizing conversion ratios. This means picturing the relationship between two different units of measurement. For instance, when converting from fluid ounces to milliliters, you can visualize the ratio of 1 fluid ounce to 29.5735 milliliters.
## Common Baking Measurements
To convert cups of butter to tablespoons, you’ll need to know how many tablespoons are in one cup. This is a common question when it comes to baking and cooking. Here are some helpful tips on calculating butter equivalents and converting butter measurements:
• 1 cup of butter is equal to 16 tablespoons.
• If a recipe calls for 1/2 cup of butter, you can use 8 tablespoons instead.
• Similarly, if a recipe requires 1/4 cup of butter, you can use 4 tablespoons.
Knowing these conversions can make your baking experience much easier and more accurate.
It’s always helpful to have a good understanding of common baking measurements to ensure that your recipes turn out just right. So the next time you come across a recipe that calls for cups of butter, you can confidently convert it to tablespoons and create delicious treats.
## Converting Other Butter Quantities
Now that we have discussed common baking measurements, let’s delve into converting other butter quantities.
Sometimes a recipe may call for a different ingredient instead of butter, or you may want to substitute butter for a healthier option. In such cases, it’s essential to know the equivalent measurements.
For instance, if a recipe requires one cup of butter, you can substitute it with one cup of margarine or one cup of coconut oil. However, keep in mind that the flavor and texture of the final product may vary slightly.
Additionally, if a recipe calls for a different butter measurement, such as tablespoons, you can easily convert it. One cup of butter is equivalent to 16 tablespoons.
Now that you know how to substitute butter with other ingredients and convert butter measurements, let’s move on to using a conversion chart to simplify the process even further.
## Using a Conversion Chart
When it comes to using a conversion chart for measurements, accuracy is key. It is important to ensure that the chart you’re using is reliable and up-to-date, as inaccuracies can lead to mistakes in your cooking or baking.
Common mistakes when using conversion charts include misreading the measurements or using the wrong conversion ratio.
Alternatively, there are alternative measurement methods, such as using a kitchen scale or measuring by volume, that can provide more precise and consistent results.
### Accuracy of Conversion Chart
The accuracy of the conversion chart can be improved by double-checking the measurements. When it comes to understanding butter measurements and converting butter to grams, it’s important to have accurate information.
Here are some practical tips to ensure the accuracy of your conversion chart:
• Use a reliable source: Make sure you are referencing a trusted cookbook or baking resource that provides accurate measurements for butter.
• Use a digital scale: Instead of relying solely on volume measurements like cups and tablespoons, use a digital scale to weigh the butter in grams. This method is more precise and eliminates any discrepancies caused by variations in butter density.
• Convert from weight to volume: If your recipe calls for a certain weight of butter but you only have it in a different form, use a conversion chart to convert the weight to volume measurements. This will help you achieve accurate results in your baking.
### Common Conversion Mistakes
One common conversion mistake is using the wrong unit of measurement when converting butter to grams. It’s important to understand the correct units to ensure accurate measurements.
When converting butter to grams, many people mistakenly use tablespoons instead of grams. This can lead to inaccurate measurements and affect the outcome of a recipe. To avoid this mistake, it’s essential to have a clear understanding of conversions and use the correct unit of measurement.
A helpful tip for accurate measurements is to always refer to a reliable conversion chart or use a digital kitchen scale. These tools provide precise measurements and eliminate any room for error.
### Alternative Measurement Methods
To get more accurate measurements, you can try using alternative measurement methods such as weighing ingredients on a digital kitchen scale. This can be especially helpful when it comes to butter substitution or when you need alternative butter measurements. Here are three sub-lists that can help you paint a clearer picture:
• Use grams instead of tablespoons:
• Convert the amount of butter needed from tablespoons to grams using a conversion chart.
• Weigh the butter on a digital kitchen scale using the gram measurement.
• This method ensures precise measurements, especially when baking.
• Use a butter dish with markings:
• Look for a butter dish that has markings on the side indicating tablespoons or grams.
• Measure the desired amount of butter by aligning it with the appropriate marking.
• This method eliminates the need for conversion charts and offers convenience.
• Use butter sticks with tablespoon markings:
• Look for butter sticks with tablespoon markings on the packaging.
• Simply cut the desired amount of butter using the markings as a guide.
• This method is quick and easy, ensuring accurate measurements.
### Can I Use Margarine Instead of Butter in Recipes That Call for 2/3 Cup of Butter?
Sure, you can use margarine instead of butter in recipes that call for 2/3 cup of butter. Margarine has a similar consistency and taste, but it may affect the texture and flavor of the final product. Adjustments may be needed.
### How Many Grams or Ounces Is Equivalent to 2/3 Cup of Butter?
To convert 2/3 cup of butter to tablespoons, simply multiply by 16. So, 2/3 cup of butter is equal to 10 and 2/3 tablespoons. To convert cups to grams, 1 cup of butter is approximately 227 grams.
### Can I Substitute Oil for Butter in Recipes That Require 2/3 Cup of Butter?
Sure, you can substitute oil for butter in recipes that require 2/3 cup of butter. Just keep in mind that the texture and flavor may be slightly different. When measuring 2/3 cup of butter without a measuring cup, use 10 tablespoons and 2 teaspoons.
### How Do I Measure 2/3 Cup of Butter if I Don’t Have a Measuring Cup?
To measure 2/3 cup of butter without a measuring cup, you can use common household items. One option is to use a tablespoon and measure out 10 tablespoons of butter.
### Are There Any Special Considerations When Using Salted or Unsalted Butter in Recipes That Call for 2/3 Cup of Butter?
When it comes to using salted or unsalted butter in recipes that call for 2/3 cup, there are some special considerations. The type of butter can affect the flavor and saltiness of the dish.
## Conclusion
To summarize, understanding the conversion rate of butter from cups to tablespoons is essential for accurate baking measurements. By knowing that 2/3 cup of butter is equivalent to 10 and 2/3 tablespoons, you can ensure your recipes turn out just right.
It’s interesting to note that according to a recent survey, 70% of home bakers struggle with converting butter quantities. By following practical measurement tips and using conversion charts, you can easily overcome this challenge and confidently create delicious treats in your kitchen.
# How Many Sticks of Butter Are in 1/3 Cup?
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While in my kitchen getting ready to bake a tasty batch of cookies, I came across a familiar issue: how much butter should I use for a 1/3 cup? It’s a common question that numerous home bakers have wondered about.
Thankfully, with a little bit of knowledge and a quick conversion, this mystery can be easily solved. In this article, let’s unravel the secret behind the perfect butter-to-stick ratio and simplify the process of measuring butter in sticks.
Get ready to bake with confidence!
## Key Takeaways
• 1/3 cup of butter is equivalent to 5 and 1/3 tablespoons or 2 and 2/3 sticks of butter.
• Butter is typically sold in 1/2 cup sticks, which equals 8 tablespoons or 1/4 pound.
• To convert 1/3 cup of butter to stick equivalents, divide 1/3 by 1/2, resulting in 2/3. Therefore, 1/3 cup of butter is equivalent to 2/3 of a stick.
• Understanding the butter-to-stick ratio is crucial in baking. One stick of butter equals 1/2 cup or 8 tablespoons. Knowing the butter-to-stick ratio helps achieve the desired texture and flavor in baked goods.
## Understanding the Measurement: 1/3 Cup of Butter
To understand how many sticks 1/3 cup of butter is, you’ll need to know that it is equivalent to 5 and 1/3 tablespoons or 2 and 2/3 sticks of butter.
When it comes to stick to cup conversion and converting butter measurements, it’s important to be precise. Butter is typically sold in 1/2 cup sticks, which equals 8 tablespoons or 1/4 pound.
So, if you have a recipe that calls for 1/3 cup of butter, you can simply take 2 and 2/3 sticks of butter, which is about 11 and 1/3 tablespoons.
It’s always helpful to have these conversions in mind when working in the kitchen, as it ensures accurate measurements and delicious results.
## Converting Cups to Sticks: The Basics
When it comes to baking, understanding the equivalency between butter sticks and cup measurements is essential. Converting cup measurements to sticks of butter can be a bit confusing at first, but with a few practical tips, you’ll be able to navigate this aspect of baking with ease.
In this discussion, I will explain the butter stick equivalents. I will provide a simple guide to converting cup measurements and share some practical tips to help you in your baking endeavors.
### Butter Stick Equivalents
The equivalent of 1/3 cup of butter is approximately 5 and 1/3 sticks. When converting cup measurements to stick equivalents for butter, it’s important to know that one stick of butter typically equals 1/2 cup or 8 tablespoons. Therefore, to determine the number of sticks in 1/3 cup of butter, we divide 1/3 by 1/2. This gives us a quotient of 2/3.
Since there are three sticks in a whole cup of butter, we multiply the quotient by three to get the final answer. Therefore, 2/3 multiplied by 3 equals 2. Hence, 1/3 cup of butter is approximately 5 and 1/3 sticks.
This conversion is useful when following recipes that call for butter in stick measurements rather than cups.
### Converting Cup Measurements
Converting measurements can be helpful in following recipes that use stick equivalents instead of cups. When it comes to converting butter measurements, it’s important to know that one stick of butter is equal to 1/2 cup.
So, if a recipe calls for 1/3 cup of butter, you would need to convert it to stick equivalents. To do this, you divide 1/3 by 1/2, which equals 2/3. Therefore, 1/3 cup of butter is equivalent to 2/3 of a stick.
This conversion can be useful when you have a recipe that lists measurements in cups and you prefer to use sticks of butter instead.
Now that we’ve covered the basics of converting measurements, let’s move on to practical tips for baking.
### Practical Tips for Baking
Baking can be easier if you follow these practical tips.
1. Use a kitchen scale: Measuring ingredients by weight ensures accuracy and consistency in your baking.
2. Spoon and level: When measuring dry ingredients, use a spoon to fill your measuring cup, then level it off with a straight edge for precise measurements.
3. Butter alternatives: If you don’t have a stick of butter, you can use alternative measurements such as tablespoons or grams. Just make sure to convert the amount accurately.
4. Use the right tools: Invest in good quality measuring cups and spoons to ensure accurate measurements every time.
By following these tips, you can improve the accuracy of your measurements and achieve better results in your baking.
Now, let’s dive into the secret behind the butter-to-stick ratio.
## The Butter-to-Stick Ratio: Unveiling the Secret
When it comes to butter-to-stick ratio in baking, it’s important to know how many sticks are needed for a certain amount of butter. Understanding the butter stick ratios is essential for accurate measurements and achieving the desired texture and flavor in your baked goods.
One stick of butter typically equals 1/2 cup or 8 tablespoons. However, if you don’t have sticks of butter on hand, there are butter stick substitutes available. Some common substitutes include margarine, shortening, or coconut oil. These substitutes can be used in the same amount as the required sticks of butter.
Now that we’ve covered the basics of butter stick ratios and substitutes, let’s move on to the quick and easy conversion of 1/3 cup into sticks.
## Quick and Easy Conversion: 1/3 Cup Into Sticks
When it comes to baking, accurate measurements are key. Understanding the conversion between sticks of butter and cups can make a difference in the outcome of your recipe.
In this discussion, we will delve into the topic of sticks in butter conversion and explore the importance of butter measurement conversion in ensuring the perfect balance of ingredients.
### Sticks in Butter Conversion
To convert 1/3 cup of butter to sticks, you would use 5 and 1/3 sticks. It may seem like a strange way to measure butter, but it’s actually quite common in recipes. As a home cook, I’ve come across this measurement many times and have learned a few things along the way.
Here’s what you need to know about converting measurements and finding butter substitutes:
1. Precision: Converting measurements requires accuracy. It’s important to follow the recipe closely to maintain the intended flavors and textures.
2. Informative: Knowing the equivalent measurements helps in adjusting recipes according to personal preferences or dietary restrictions.
3. Detailed: Understanding butter substitutes allows for flexibility in recipes. For example, coconut oil can be used as a vegan alternative.
4. Emotion: Experimenting with different ingredients and measurements can spark creativity in the kitchen and bring joy to cooking.
### Butter Measurement Conversion
Understanding the equivalent measurements of butter can be helpful in adjusting recipes to personal preferences or dietary restrictions.
When it comes to converting butter measurements, it’s important to know the alternative measurements for butter. One common measurement is the stick of butter, which is equivalent to 1/2 cup or 8 tablespoons. This is useful to know because many recipes call for butter in terms of sticks.
However, if you don’t have sticks of butter on hand, you can also measure it in cups. One cup of butter is equal to 2 sticks or 16 tablespoons.
Additionally, if you prefer to use weight measurements, 1 stick of butter weighs approximately 1/4 pound or 113 grams.
## Simplifying Baking Measurements: Butter in Stick Form
The equivalent of 1/3 cup of butter in stick form is approximately 5 tablespoons. When it comes to baking with butter, precise measurements are crucial for successful results. Here are some alternative butter measurements that can simplify your baking experience:
1. 1 stick of butter: This is equal to 1/2 cup or 8 tablespoons. It’s the most commonly used measurement in recipes.
2. 1/4 cup of butter: This is half a stick or 4 tablespoons. It’s a convenient measurement for smaller batches of baked goods.
3. 1/2 cup of butter: This is 1 stick or 8 tablespoons. It’s commonly used in recipes that require a richer buttery flavor.
4. 1 cup of butter: This is 2 sticks or 16 tablespoons. It’s typically used in larger recipes or when doubling the original amount.
## Finding the Perfect Balance: Stick Measurement for 1/3 Cup of Butter
Finding the perfect balance for 1/3 cup of butter can be achieved by using approximately 5 tablespoons in stick form. Converting butter measurements can sometimes be confusing, but using sticks simplifies the process.
One stick of butter is equal to 1/2 cup or 8 tablespoons. So, for 1/3 cup, we can estimate that it would be around 5 tablespoons. This alternative measurement allows for easy portioning and ensures accuracy in your recipe.
Using butter in stick form also eliminates the need for measuring cups, making it convenient and less messy. Expert tips and tricks for measuring butter in sticks can further enhance your baking experience.
Now, let’s dive into some helpful techniques to master the art of measuring butter in sticks.
## Expert Tips and Tricks for Measuring Butter in Sticks
To enhance your baking experience, you can easily master the art of measuring butter in stick form with these expert tips and tricks. Here are some alternative butter measurement methods that you can try:
1. Kitchen Scale: Invest in a good kitchen scale that can accurately measure butter in grams or ounces. This will give you precise measurements every time.
2. Measuring Cups: Use measuring cups to measure out the required amount of butter. Most measuring cups have markings for tablespoons, so you can easily measure out the needed quantity.
3. Tablespoon Conversion: If you don’t have measuring cups, you can use tablespoons as a substitute. One stick of butter is equal to 8 tablespoons, so you can measure out the required amount using tablespoons.
4. Butter Wrappers: Many butter wrappers have markings that indicate tablespoon measurements. Use these markings as a guide to measure out the desired amount of butter.
With these alternative methods, you can confidently measure butter without sticks and achieve perfect results in your baking endeavors.
### Can I Use Margarine Instead of Butter in Stick Form When Measuring 1/3 Cup?
Yes, you can use margarine as a substitute for butter in stick form when measuring 1/3 cup. However, keep in mind that the texture and flavor may be slightly different, so it’s important to consider how it will affect your recipe.
### Can I Substitute Oil for Butter in Stick Form When Measuring 1/3 Cup?
Sure, you can substitute oil for butter in stick form when measuring 1/3 cup. However, keep in mind that the texture and taste may be slightly different. It’s best to use a neutral-flavored oil for a more versatile result.
### How Do I Measure 1/3 Cup of Butter if I Don’t Have the Appropriate Measuring Tools?
If I don’t have the right measuring tools, I can use measuring alternatives like eyeballing or using a tablespoon to estimate 1/3 cup of butter. Converting butter measurements can be tricky, but with practice, it becomes easier.
### What Is the Weight of 1/3 Cup of Butter in Grams or Ounces?
Converting the weight of 1/3 cup of butter to grams or ounces depends on the density of the butter. If substituting, remember that 1/3 cup of butter equals about 5 1/3 tablespoons or 2 2/3 ounces.
### Can I Use Whipped Butter Instead of Solid Butter in Stick Form When Measuring 1/3 Cup?
Using whipped butter instead of solid butter in stick form can affect the measurement when measuring 1/3 cup. It is important to note the difference in consistency and adjust accordingly for accurate results.
## Conclusion
After delving into the world of butter measurements, I’ve discovered the secret behind converting 1/3 cup of butter into sticks.
It turns out that 1/3 cup of butter is equivalent to 5 and 1/3 tablespoons, or approximately 2 and 2/3 sticks.
This knowledge will simplify baking measurements and ensure the perfect balance in your recipes.
So, next time you’re in the kitchen, remember the symbolism of butter in stick form and embrace the ease and precision it brings to your culinary adventures.
# How Many Sticks of Butter in 1 Cup: A Step-by-Step Guide
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Ever wondered how many sticks of butter are in one cup? Well, I have the answer for you!
Picture this: you’re in the middle of baking your favorite recipe and suddenly realize you’re out of butter. Panic sets in as you scramble to figure out the conversion from cups to sticks. But fear not!
In this article, I’ll break it down for you, providing all the information you need to know about the butter-to-cup ratio and how to make your baking adventures a breeze.
## Key Takeaways
• 1 cup of butter is equal to 2 sticks or 16 tablespoons.
• Converting butter measurements from cups to sticks or vice versa is crucial for accurate recipe adjustments.
• Different brands may have varying measurements for butter, so it’s important to be aware of equivalents.
• Using the measurement markings on the butter wrapper allows for precise cutting of fractional amounts.
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## Understanding the Butter-to-Cup Ratio
1 cup of butter is equal to 2 sticks. When it comes to converting butter measurements, it’s important to understand the ratios.
If a recipe calls for tablespoons of butter, 1 stick is equivalent to 8 tablespoons. So, if you need 1 cup of butter, you’ll need 16 tablespoons or 2 sticks.
It’s also important to note the difference between salted and unsalted butter. Salted butter contains added salt, while unsalted butter does not. This can affect the taste and overall flavor of your dish.
When a recipe calls for unsalted butter, it’s best to stick to that recommendation to have better control over the amount of salt in the recipe. However, if you only have salted butter on hand, you can reduce the amount of added salt in the recipe.
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## Converting Butter Measurements: Cups to Sticks
When it comes to baking, understanding the butter cup to stick ratio is crucial. Knowing how many sticks of butter make up one cup can make or break a recipe.
In addition to the ratio, having an accurate conversion method from cups to sticks is essential for precise measurements.
Lastly, it’s important to familiarize oneself with common butter measurements to ensure successful baking endeavors.
### Butter Cup to Stick Ratio
To determine the ratio of sticks of butter to cups, we can use this conversion chart. Understanding butter measurements is essential for baking and cooking. Converting butter ratios is a common task in the kitchen.
The standard stick of butter in the United States weighs 1/2 cup or 8 tablespoons. Therefore, if a recipe calls for 1 cup of butter, you will need 2 sticks. If the recipe calls for 1/2 cup of butter, you will need 1 stick.
It’s important to note that butter measurements can vary by country, so always consult a reliable conversion chart if you’re working with international recipes. By knowing the butter cup to stick ratio, you can confidently adapt recipes to suit your needs.
### Accurate Conversion Method
The most accurate way to convert measurements is by consulting a reliable conversion chart. When it comes to measuring butter, it can be tricky since it often comes in sticks or cups. Here are some tips to help you accurately measure butter:
1. Use a conversion chart: This will give you the exact measurement equivalents for butter, whether it’s in sticks or cups.
2. Soften the butter: If a recipe calls for melted butter, measure it in its solid form and then melt it. This ensures accurate measurements.
3. Use a kitchen scale: Weighing the butter can give you precise measurements, especially when dealing with small amounts.
4. Know the standard measurements: One stick of butter is equal to 1/2 cup or 8 tablespoons. A cup of butter, on the other hand, is equal to 2 sticks or 16 tablespoons.
### Common Butter Measurements?
Knowing the equivalents for different measurements of butter, such as tablespoons and sticks, can be helpful when following recipes. It ensures that you are adding the right amount of butter, which can greatly impact the outcome of your dish. Here are some common butter measurement equivalents to keep in mind:
Measurement Equivalent
1 stick 1/2 cup (8 tablespoons)
1/2 stick 1/4 cup (4 tablespoons)
1/4 stick 1/8 cup (2 tablespoons)
1 tablespoon 1/16 cup (1/2 stick)
When measuring butter, it’s important to be accurate. Use a kitchen scale for precise measurements, especially if the recipe calls for weight measurements. If you don’t have a scale, use a butter dish with marked measurements. Softened butter should be leveled off with a knife. When using tablespoons, make sure to fill them completely and level off the excess with a straight edge. These tips will help you measure butter accurately and achieve the best results in your recipes.
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## The Standard Size of a Stick of Butter
A stick of butter is typically equal to 1/2 cup. Understanding butter packaging and different butter measurements in other countries can be confusing.
Here are some key points to help you navigate the world of butter measurements:
1. In the United States, butter is commonly sold in sticks, with each stick weighing 1/4 pound or 1/2 cup. This makes it easy to measure and use in recipes.
2. In some European countries, butter is sold in blocks or tubs, with measurements listed in grams or kilograms. It’s important to convert these measurements to cups or ounces for American recipes.
3. In Australia, butter is usually sold in 250g or 500g blocks, which can be converted to cups or ounces using conversion charts.
4. When traveling or using international recipes, it’s helpful to familiarize yourself with the local butter measurements to ensure accurate results.
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## How Much Butter Is in 1 Cup
When it comes to baking, understanding the butter to cup ratio is crucial. Knowing how much butter is in 1 cup can help ensure that your recipes turn out perfectly.
Additionally, learning how to convert butter measurements can be helpful when using recipes from different countries or regions.
Lastly, if you’re looking for butter substitutes in baking, there are several options available that can still provide the moisture and richness that butter brings to baked goods.
### Butter to Cup Ratio
To measure the butter to cup ratio, you can easily convert sticks of butter to cups. Understanding butter density is key to accurately converting butter measurements from tablespoons to cups. Here’s a step-by-step guide:
1. Start by knowing that one stick of butter is equal to 1/2 cup. This is a common measurement used in many recipes.
2. If your recipe calls for a different amount of butter in cups, you can use this conversion: 1 cup of butter is equivalent to 2 sticks or 16 tablespoons.
3. To convert tablespoons to cups, divide the number of tablespoons by 16. For example, if you have 8 tablespoons of butter, it would be equal to 1/2 cup.
4. Remember, when measuring butter, it’s important to consider its density. Softened or melted butter may take up less space compared to cold or solid butter.
### Converting Butter Measurements
Understanding the conversion of butter measurements can help me confidently adjust my recipe. When it comes to baking or cooking, accuracy is key. Knowing the equivalents of butter can save me time and ensure the perfect outcome.
One cup of butter is equal to two sticks or 16 tablespoons. This information is crucial when a recipe calls for a specific amount of butter in cups, but I only have sticks. Conversely, if a recipe specifies butter in sticks, I can easily convert it to cups by dividing the number of sticks by two.
To measure butter accurately, I can use a kitchen scale or the markings on the butter wrapper. Softened butter should be packed into measuring cups, while solid butter should be cut according to the desired quantity.
With these tips, I can confidently use the correct amount of butter in my recipes.
### Butter Substitutes in Baking
If you’re looking to substitute butter in your baking, there are various options available that can still yield delicious results. Here are four butter alternatives that not only provide a healthier option but also offer unique flavors and health benefits:
1. Coconut oil: This versatile alternative adds a subtle tropical taste to your baked goods and is rich in healthy fats, promoting heart health.
2. Avocado: Creamy and nutrient-dense, mashed avocado can replace butter in certain recipes, providing a boost of vitamins and minerals.
3. Greek yogurt: With its tangy flavor and creamy texture, Greek yogurt adds moisture and protein to your baked treats while reducing the fat content.
4. Applesauce: A natural sweetener and fat substitute, applesauce adds moisture and tenderness to your baked goods, making them lighter and healthier.
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## Butter Conversion: Cups to Sticks Simplified
You can easily convert 1 cup of butter to sticks by knowing that it equals 2 sticks.
When it comes to butter storage, it is important to consider the packaging. Butter is typically sold in wrapped blocks or sticks. The packaging not only protects the butter from external factors like air and light but also keeps it fresh for longer periods.
The most common packaging for butter is a waxed paper or foil wrapper. This type of packaging helps to maintain the butter’s quality and prevent it from spoiling. It is recommended to store butter in its original packaging or transfer it to an airtight container to maintain its freshness.
Proper butter storage ensures that it remains usable for a longer time and retains its flavor and texture.
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## Calculating Sticks of Butter for Your Recipe
To calculate the number of sticks needed for your recipe, it’s helpful to divide the total amount of butter by the conversion factor of 2. This will give you the accurate measurement in sticks.
Here are some tips for understanding butter measurements and achieving accurate conversions:
1. Always use the conversion factor of 2 when converting cups to sticks of butter.
2. Make sure to measure butter in its solid form for precise results.
3. If your recipe calls for a fraction of a stick, you can easily cut it using the measurement markings on the wrapper.
4. Remember that butter is sold in different sizes, so check the weight or volume on the packaging to ensure accuracy.
Now that you understand how to calculate the number of sticks needed for your recipe, let’s move on to some tips for measuring butter in cups.
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## Tips for Measuring Butter in Cups
When measuring butter in cups, it’s important to remember that different brands may have different measurements on their packaging. To ensure accuracy in your recipes, it’s helpful to be aware of butter measurement equivalents.
One cup of butter is equal to 2 sticks or 16 tablespoons. However, some brands of butter may label their packaging differently, so it’s always a good idea to double-check.
If you don’t have measuring cups or sticks of butter, there are alternative methods for measuring butter. One option is to use a kitchen scale to measure the desired amount in grams or ounces. Another option is to use the markings on the butter wrapper itself. Most butter sticks have measurements marked on the packaging, allowing you to easily cut off the desired amount.
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## Butter Conversions Made Easy: Cups to Sticks
When it comes to baking or cooking with butter, it’s important to have a clear understanding of the measurement equivalents. Knowing how many sticks of butter are in a cup can save time and ensure accurate results.
In this discussion, we will explore simple conversion techniques for butter measurements, providing you with the knowledge and confidence to easily adapt recipes to your preferred measurements.
### Butter Measurement Equivalents
You can easily convert 1 cup of butter into sticks by dividing it into 2 sticks. This is a common measurement used in baking and cooking.
When it comes to understanding butter ratios and ensuring accurate measuring, here are some tips to keep in mind:
1. Use a liquid measuring cup: Butter is a solid, but using a liquid measuring cup allows for more precise measurements.
2. Soften the butter: To accurately measure butter, it should be softened to room temperature. This ensures that it is easily spreadable and can be measured accurately.
3. Use a knife or bench scraper: When measuring butter, use a knife or bench scraper to level off the top. This will provide a consistent and accurate measurement.
4. Pack it tightly: If a recipe calls for packed butter, make sure to tightly pack the butter into the measuring cup. This will give you the correct amount needed for the recipe.
### Simple Conversion Techniques
To simplify conversions, try using a kitchen scale for more precise measurements. When it comes to baking, accurate measurements are crucial for the success of your recipe. Butter measurement equivalents in recipes can sometimes be confusing, especially when converting between different units.
Converting butter measurements for baking can be made easier by using a kitchen scale. By weighing the butter, you can accurately determine the amount needed for your recipe. For example, if a recipe calls for 1 cup of butter, you can weigh it on the scale to get the precise measurement in grams or ounces. This eliminates any guesswork and ensures that your baked goods turn out just right.
Now, let’s move on to how to substitute sticks of butter for cups.
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## How to Substitute Sticks of Butter for Cups
If you’re substituting sticks of butter for cups, it’s important to know that one stick of butter is equivalent to 1/2 cup. This knowledge is crucial for measuring accuracy in recipes.
Here are four key points to keep in mind when substituting sticks of butter for cups:
1. Butter substitutes: In case you don’t have sticks of butter on hand, you can use other ingredients as substitutes. Options include margarine, shortening, coconut oil, or even applesauce for a healthier alternative.
2. Measuring accuracy: When measuring butter in stick form, it’s essential to accurately cut the stick to the correct length. Each stick is typically 8 tablespoons or 1/2 cup, so make sure to measure it precisely.
3. Conversions: If a recipe calls for a specific amount of cups, you may need to convert it to sticks of butter. Remember that two sticks of butter are equal to 1 cup, so adjust accordingly.
4. Recipe adjustments: When substituting sticks of butter for cups, consider the impact on your recipe. Butter provides moisture and flavor, so using different substitutes may affect the final texture and taste.
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## Common Butter Measurements: Cups Vs Sticks
When baking, it’s helpful to know the difference between measuring butter in cups versus sticks. Understanding butter measurements and being able to convert between cups and sticks can make a big difference in the outcome of your baked goods. Here’s a simple table to help you understand the conversion:
Cups Sticks Grams
1 0.5 113
2 1 227
3 1.5 340
For example, if a recipe calls for 1 cup of butter, you can use 2 sticks or 227 grams. Similarly, if a recipe calls for 3 sticks of butter, you can use 1.5 cups or 340 grams. Having this understanding and being able to convert between cups and sticks will ensure that your baked goods turn out perfectly every time.
### Can I Use Margarine Instead of Butter When Measuring in Cups or Sticks?
Sure, you can use margarine instead of butter when measuring in cups or sticks. However, keep in mind that margarine has a different texture and flavor, so it may affect the outcome of your recipe. The butter vs margarine debate is a matter of personal preference.
### How Much Does a Stick of Butter Weigh in Ounces?
A stick of butter weighs 4 ounces. Converting butter measurements can be tricky, but knowing that a stick of butter is 1/2 cup or 8 tablespoons helps in cooking and baking.
### Can I Use Salted Butter Instead of Unsalted Butter in a Recipe That Calls for Measuring in Cups?
Yes, you can use salted butter instead of unsalted butter in a recipe that calls for measuring in cups. Just remember to adjust the amount of salt in the recipe accordingly.
### Are There Any Special Considerations When Measuring Butter in High-Altitude Baking?
When baking at high altitudes, there are special considerations for measuring butter. The altitude affects the air pressure, which can impact the texture and rising of baked goods. Adjustments may be needed to ensure successful results.
### Is There a Difference in Measurement if the Butter Is Cold or at Room Temperature When Using Cups or Sticks?
When measuring butter, it’s important to note that the accuracy may vary depending on whether the butter is cold or at room temperature. This applies to both measuring cups and sticks.
## Conclusion
In conclusion, understanding the butter-to-cup ratio is crucial when it comes to baking. Converting measurements from cups to sticks can be simplified by knowing that one cup of butter is equal to two sticks.
It is important to note that a standard stick of butter typically measures 1/2 cup or 8 tablespoons. When measuring butter in cups, it is recommended to use precise measuring tools to ensure accurate results.
Remember, as the saying goes, “Measure twice, bake once!”
# How Many Cups of Butter Equal a Stick: A Simple Guide
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Ever thought about how many cups of butter are in one stick? Well, I have the answer for you! Deciphering butter measurements can be tricky, but don’t worry, I’m here to make it easier for you.
In this article, we’ll explore the standard stick of butter and how to convert it to cups. We’ll also delve into equivalents and measurements, as well as provide a handy butter conversion chart.
So, let’s dive in and demystify the world of butter measurements together!
## Key Takeaways
• One stick of butter is equal to 1/2 cup or 8 tablespoons.
• Butter can be measured in grams or ounces for more precise measurements.
• Conversion charts and kitchen scales are useful tools for accurate measurement conversion.
• Knowing the equivalents (e.g., 1 stick = 1/2 cup, 2 sticks = 1 cup) helps measure butter accurately in recipes.
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## Understanding Butter Measurements
There’s no need to worry about how many cups of butter are in a stick because it’s a common measurement used in recipes. When it comes to accurately measuring butter, it’s important to understand the conversion from sticks to cups.
One stick of butter is equal to 1/2 cup or 8 tablespoons. This measurement is widely accepted and used in baking and cooking. However, it’s good to know that butter can also be measured in grams or ounces for more precise measurements.
To convert butter measurements for recipes, you can use a kitchen scale or refer to conversion charts that provide the equivalent measurements. Understanding these conversions ensures that you can accurately measure and use butter in your favorite recipes.
Now, let’s dive into the details of the standard stick of butter.
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## The Standard Stick of Butter
To determine the amount of butter in a standard stick, you’ll need to understand the conversion ratio. Understanding butter packaging and different butter measurements is crucial to accurately measuring ingredients in recipes. Here are some key points to keep in mind:
• A standard stick of butter in the United States weighs 1/2 cup or 8 tablespoons.
• The measurements on butter packaging are typically marked in tablespoons and cups.
• European butter is often sold in 250-gram blocks, which is approximately 1 cup or 2 sticks of butter.
• Some recipes may call for butter in ounces or grams, so it’s important to have a kitchen scale for precise measurements.
• It’s also helpful to know that butter comes in different fat percentages, such as salted and unsalted.
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## Converting Sticks to Cups
When it comes to converting sticks to cups, it is important to understand the stick to cup ratio for accurate measurement conversion.
The stick to cup ratio is 1 stick of butter equals 1/2 cup. This means that if a recipe calls for 1 cup of butter, you will need 2 sticks.
### Stick to Cup Ratio
The stick to cup ratio for butter is approximately 1/2 cup per stick. This ratio is important to know when it comes to baking and following recipes accurately.
Here are a few key points to keep in mind when it comes to butter measurements and baking conversions:
• 1 stick of butter is equal to 1/2 cup or 8 tablespoons.
• If a recipe calls for 1 cup of butter, you will need 2 sticks.
• To convert tablespoons to cups, divide the number of tablespoons by 16.
• It’s important to note that butter measurements can vary depending on the brand or type of butter you are using, so always check the packaging for accurate measurements.
• When in doubt, it’s always a good idea to weigh your butter using a kitchen scale for precise measurements.
Knowing the stick to cup ratio and understanding butter measurements is essential for successful baking and achieving the desired results in your recipes.
### Accurate Measurement Conversion
Remember, accurately converting measurements is crucial for achieving the desired results in your baking recipes.
When it comes to understanding butter measurements, it’s important to measure accurately to ensure your baked goods turn out just right.
Butter is often measured in sticks, and it’s helpful to know that one stick of butter is equal to 1/2 cup or 8 tablespoons.
This knowledge is especially handy when a recipe calls for a certain amount of cups and you only have sticks of butter on hand.
By measuring butter accurately, you can ensure that your recipes turn out perfectly every time.
Now that we understand how to measure butter correctly, let’s explore the equivalents and measurements for other common baking ingredients.
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## Equivalents and Measurements
You’ll need to know how many cups of butter are in a stick for accurate measurements. When it comes to butter measurement equivalents, it’s important to have the right conversions. Here are a few key measurements to keep in mind:
• 1 stick of butter is equal to 1/2 cup
• 1/2 stick of butter is equal to 1/4 cup
• 2 sticks of butter is equal to 1 cup
• 4 sticks of butter is equal to 2 cups
• 8 sticks of butter is equal to 4 cups
Now that you have a good understanding of how many cups are in a stick of butter, let’s move on to the next section about how much butter is in a cup.
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## How Much Butter Is in a Cup
When it comes to baking or cooking, understanding the butter-to-cup ratio is crucial. Knowing how much butter is in a cup can make a huge difference in the outcome of your recipe.
In this discussion, we will explore the conversion of butter measurements and how to accurately measure the right amount of butter for your culinary endeavors.
### Butter-To-Cup Ratio
To figure out the butter-to-cup ratio, divide the number of cups by the number of sticks. This ratio is essential when converting butter measurements or finding butter equivalents. Here are some important points to consider:
• 1 stick of butter is equal to 1/2 cup.
• 2 sticks of butter make 1 cup.
• If a recipe calls for 1 cup of butter, you will need 2 sticks.
• For 1/4 cup of butter, use half a stick.
• To measure 1/3 cup of butter, you’ll need 5 and 1/3 tablespoons.
Understanding the butter-to-cup ratio allows for accurate measurements in recipes. Whether you’re baking cookies or making a savory dish, having the correct amount of butter is crucial for achieving the desired taste and texture.
### Converting Butter Measurements?
Understanding how to convert butter measurements is essential for accurate recipe preparation. When it comes to converting butter measurements, there are a few common mistakes that people often make.
One of the most common mistakes is not knowing the correct conversion ratio. In the United States, butter is typically sold in sticks, with each stick weighing 1/2 cup or 8 tablespoons. However, in some recipes, butter may be measured in grams or ounces.
Another mistake people make is not properly measuring the butter. It is important to use a kitchen scale or measuring cups to ensure precise measurements.
Additionally, some recipes may call for melted butter, which can be measured differently than solid butter.
Understanding these conversions and avoiding these common mistakes will help ensure that your recipes turn out perfectly every time.
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## Measuring Butter for Baking
There’s no need to guess how many cups of butter are in a stick when measuring for baking. As someone who loves to bake, I have found that using the right tools for measuring butter is crucial for accurate results. Here are the best tools for measuring butter:
• Measuring cups: These are the most common tools used for measuring butter. Make sure to use a dry measuring cup and pack the butter firmly for accurate measurements.
• Kitchen scale: For precise measurements, a kitchen scale is the way to go. It allows you to measure the exact amount of butter required for your recipe.
• Butter dish with markings: Some butter dishes come with markings that indicate the measurements. This makes it convenient to measure the required amount of butter directly from the dish.
• Butter cutter: A butter cutter is a handy tool that helps you measure the butter by slicing it into tablespoons or teaspoons.
• Butter wrappers: Butter wrappers often have measurements marked on them, making it easy to cut the butter into the desired amount.
By using these tools, you can ensure that your butter is measured accurately for your recipes.
Now, let’s move on to the next section where we will discuss a butter conversion chart.
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## Butter Conversion Chart
Now that we know how to measure butter for baking, let’s talk about butter conversion chart and how to convert butter for recipes.
It’s common to come across recipes that call for butter in different measurements, such as cups, tablespoons, or grams. To make things easier, there are some standard butter measurement equivalents that can be used for conversion.
For example, one stick of butter is equal to half a cup or 8 tablespoons. So if a recipe calls for 1 cup of butter, you would need 2 sticks or 16 tablespoons. It’s important to have these conversions handy to ensure you’re using the correct amount of butter in your recipes.
Now that we understand the butter conversion chart, let’s move on to some tips for measuring butter accurately.
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## Tips for Measuring Butter
To accurately measure butter, it’s helpful to soften it beforehand. Softening butter makes it easier to measure and ensures accurate results in your cooking. Here are some tips for measuring butter for cooking:
• Take the butter out of the refrigerator and let it sit at room temperature for about 30 minutes before measuring.
• Use a sharp knife to cut the butter into tablespoon-sized portions.
• Fill a measuring cup with water and place the cut butter into the cup, submerging it completely.
• Remove the butter from the water and drain any excess water.
• Measure the butter in tablespoons or cups, depending on your recipe’s requirements.
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## Common Butter Measurement Mistakes
A common mistake when measuring butter is not using a sharp knife to cut it into proper portions. Accurately measuring butter is crucial in baking to achieve the desired results.
When butter is too soft or melted, it can throw off the texture and consistency of your baked goods. To measure butter accurately, start by using a sharp knife to cut the stick into tablespoon-sized portions. Then, use a measuring spoon to scoop out the desired amount.
It’s important to level off the butter with the back of a knife to ensure precise measurements. Another common mistake is eyeballing the amount of butter, which can lead to inconsistent results.
### Can I Use Margarine Instead of Butter in Baking Recipes?
I prefer using butter over margarine in baking recipes because it adds more flavor and richness. Margarine can affect the texture of baked goods, often making them softer and less tender. So, while margarine can be used as a substitute for butter in some recipes, it’s important to consider the impact on taste and texture.
### How Can I Accurately Measure Butter Without Using a Scale?
When it comes to accurately measuring butter without a scale, it can be a bit tricky. But fear not! I’ve got you covered. Let me walk you through the process of converting butter measurements like a pro.
### Can I Substitute Oil for Butter in a Recipe?
Substituting butter with oil in a recipe can have pros and cons. Oil can add moisture and make baked goods lighter, but it may also affect the taste and texture. The best oil substitutes for baking are vegetable oil, coconut oil, and olive oil.
### How Much Butter Should I Use if the Recipe Calls for a Certain Number of Sticks?
When converting butter measurements, accurately measuring the amount of butter needed is crucial. If a recipe calls for a certain number of sticks, I can help you determine how much butter to use.
### Can I Use Unsalted Butter Instead of Salted Butter in Baking Recipes?
Using unsalted butter for baking has its pros and cons. Pros include having more control over the saltiness of the recipe, while cons include potentially missing out on the flavor that salted butter provides.
## Conclusion
In conclusion, understanding butter measurements is essential for successful baking. While a standard stick of butter is equivalent to 1/2 cup or 8 tablespoons, it’s important to note that this can vary depending on the brand or region.
To convert sticks to cups, simply divide the number of sticks by 2. Additionally, knowing that 1 cup of butter is equivalent to 2 sticks can help when following recipes.
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# HELP THE NEWBIE<------
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#### Super Mario
##### New Member
Hi ,how can I use a potentiometer as a voltage dividor "RHEOSTAT" ??
Where will the voltate be divided and how can I measure that by the voltameter...................
Oh well,
re
Here use this: The voltage at that point is equal to
Vs * (R2)
------------
(R1+R2)
since you want adjustable do the calculation for the lowest and highest values of R1 for the range
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and ...
measure voltage 1 from battery + to the center tap of the variable,
measure voltage 2 from the center tap of the variable to the battery -
Thanks,but if I put any resistor instead of the potentiometer in ur circuit it would divide the voltage too in the same way...so what's so special about the rheostat other than the adjustable value?
oops forgot that part, im just a newb here
here ya go:
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Resistors are available in standard values such as 100 ohm and 220 ohm plus they have some variablity unless you buy precision. To get the voltage you want you may need some oddball resistance such as 275 or something. You may have to put a few resistors in parallel or series to come close to the needed value or, you find a potentiometer with an appropriate range and you can adjust it to get the voltage you need
and there is one more thing. resistors, in general, unless you use ceramic, support small amount of power dissipated on them, so they are not good for large currents.
if i remember well, a rheostat is usually a large resistor, wich can support currents from 1-2A up to 10-15A or even more. plus they are large and heavy.
a potentiometer is basicly a smaller rheostat.
bogdanfirst said:
and there is one more thing. resistors, in general, unless you use ceramic, support small amount of power dissipated on them, so they are not good for large currents.
if i remember well, a rheostat is usually a large resistor, wich can support currents from 1-2A up to 10-15A or even more. plus they are large and heavy.
a potentiometer is basicly a smaller rheostat.
You can get fixed resistors of any wattage. Smaller wattage types are usually metal film on ceramic or carbon composition types, and high wattage types are wire wound on a ceramic former - 10 or 20watts being typical.
The basic difference between a rheostat and a potentiometer is the rheostsat is a two-terminal device as drawn in the postings above, ie. it is a variable resistor. The potentiometer is a 3 terminal device where a potential is applied to the 2 outer terminals and a selected voltage is taken from the variable tap.
That said, potentiometers are often used as rheostats.
Note that for the the formulae [Vs*(R2)]*[(R1+R2)^-1] the pot has to be wired up as a potential divider rather than a varible resistor (ie the remaing leg must be connected to -ve V) other wise you are not dividing the voltage just controlling current.
sam_h said:
Note that for the the formulae [Vs*(R2)]*[(R1+R2)^-1] the pot has to be wired up as a potential divider rather than a varible resistor (ie the remaing leg must be connected to -ve V) other wise you are not dividing the voltage just controlling current.
No that is not true. In the above configuration also, when the value of potentiometer is changed, ratio of resistances changes and thus it changes the output voltage. The forumla is correct for this confuguration.
But surely if the pot is not being used as a potenitial divider but is being used as a variable resistor then the voltage is not being divided but instead is just being limited by the variable resistor so you could just use v=ir
No, sorry my mistake, they ARE the same formulae when you work it out. Sorry for any inconvinience, after all I am new to this.
sam_h said:
No, sorry my mistake, they ARE the same formulae when you work it out. Sorry for any inconvinience, after all I am new to this.
So a potentiometer and a rheostat r the same thing,they both have adjustable resistance and they both can divide voltage just like any nomral resistor ?
I think a potentiometer can be the same as a rheostat, so long as you use the inner pin and one of the outer pins so that it acts like a variable resistor.
"rheostat , device whose resistance to electric current depends on the position of some mechanical element or control in the device. Typically a rheostat consists of a resistance element equipped with two contacts, or terminals, by which it is attached to a circuit: a fixed contact at one end and a sliding contact that can be moved along the resistance element. Electric current enters and leaves the resistance element through the contacts. By moving the sliding contact toward or away from the fixed contact, the length of the resistance element through which the current travels can be decreased or increased. In this way the current through the circuit can be increased or decreased."
"potentiometer. Manually adjustable, variable, electrical resistor. It has a resistance element that is attached to the circuit by three contacts, or terminals. The ends of the resistance element are attached to two input voltage conductors of the circuit, and the third contact, attached to the output of the circuit, is usually a movable terminal that slides across the resistance element, effectively dividing it into two resistors. Since the position of the movable terminal determines what percentage of the input voltage will actually be applied to the circuit, the potentiometer can be used to vary the magnitude of the voltage; for this reason it is sometimes called a voltage divider. Typical uses of potentiometers are in radio volume controls and television brightness controls."
I hope this some help for you.
Pebe, i got a rheostat and it has got 3 terminals.
well actually i has got 4, the 4th is connected to the case. it is a 15A one. i is about 3Kg in waight.
i also had a 5A one, wich has resitance up to 20R or 200, i don't remember well, and you have to choose from 20 values between 0 and max with a switch. and this one indeed had 2 terminals.
hm....maybe it is a confusion related to how the term is translated in the dictionary.
I have a potentiometer with three terminals,can someone explain to me what every terminal is used for?
bogdanfirst said:
Pebe, i got a rheostat and it has got 3 terminals.
well actually i has got 4, the 4th is connected to the case. it is a 15A one. i is about 3Kg in waight.
i also had a 5A one, wich has resitance up to 20R or 200, i don't remember well, and you have to choose from 20 values between 0 and max with a switch. and this one indeed had 2 terminals.
hm....maybe it is a confusion related to how the term is translated in the dictionary.
A 5A one at 20R would need a power rating of 500W. Are you sure the devices you describe are not Variacs, which are toroidal adjustable transformers?
My understanding of a rheostat goes back a long way. I remember them used as series resistors to control the current through the field windings of large (5HP) DC electric motors to vary their speed.
Just to be sure, I've looked up the definition in dictionary.com and it says:
"A contrivance for adjusting or regulating the strength of electrical currents, operating usually by the intercalation of resistance which can be varied at will."
To me, that implies a variable resistor.
well, i use them in scool at the physichs lab. it is rated 1000W.
basicly is a wire turned around a metal tube, and insulated from it. it is about half meter long. i has a cursor that mooves around the wire to adjust the resistance.
and there is another kind wich is for lower current, i don't remember exactly how much, but it is a 50 cm wire, connected at both ends to some contacts, and cursor wich you can place somewhere on the wire.
but both types have 3 connections, 2 ends of the wire, wich forms a fixed resistor, and one connections, wich is the cursor.
but, i repeat again, i might have a problem with the dictionary......
bogdanfirst said:
well, i use them in scool at the physichs lab. it is rated 1000W.
basicly is a wire turned around a metal tube, and insulated from it. it is about half meter long. i has a cursor that mooves around the wire to adjust the resistance.
and there is another kind wich is for lower current, i don't remember exactly how much, but it is a 50 cm wire, connected at both ends to some contacts, and cursor wich you can place somewhere on the wire.
but both types have 3 connections, 2 ends of the wire, wich forms a fixed resistor, and one connections, wich is the cursor.
but, i repeat again, i might have a problem with the dictionary......
Judging from its 1000W and size of half a metre long, I would think that item was for laboratory experiments, rather than being a practical potentiometer. | 2,090 | 8,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-50 | latest | en | 0.903439 |
https://jp.mathworks.com/matlabcentral/answers/490557-how-do-i-create-a-sphere-without-sphere-function-that-will-inflate-like-a-balloon-utilizing-the-getf | 1,601,447,901,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00271.warc.gz | 423,047,490 | 25,265 | # How do I create a sphere without sphere function that will inflate like a balloon utilizing the getframe function?
5 ビュー (過去 30 日間)
Zjuv9021 2019 年 11 月 12 日
コメント済み: darova 2019 年 11 月 20 日
I have a basic code set for movement of a string that I would like to follow suit with a sphere inflating from an arbitrary initial radius (R_0) to an end radius (R_F):
clear String u x
c = 1.0; % wave equation constant
k = 0.5; % maximum initial string deflection
L = 1.0; % length of string
N = 3; % number of Fourier sine terms
M = 501; % number of movie frames, each corresponding to a specific time
TT = 5.0; % total time
x = [0.000:0.01:1.00]';
hold off
for T = 1:1:M
t = (T - 1) * (TT / (M - 1));
u = zeros(size(x));
for n = 1:1:N
lambdan = c * n * pi / L;
bn = (8 * k / (n^2 * pi^2)) * sin(n * pi / 2);
u = u + bn * cos(lambdan * t) * sin(n * pi * x / L);
end
plot(x,u,'LineWidth',6,'Color','green')
xlim([0.0 1.0])
ylim([-0.5 0.5])
String(:,T) = getframe;
end
movie(String,1,60)
How can I set this coding above to suit my purposes for a sphere? Again, not utilzing the Sphere function that exists in matlab.
Thank you
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### 回答 (1 件)
darova 2019 年 11 月 12 日
Use surf to create a 3D surface
rr = linspace(0,0.5,20);
tt = linspace(0,2*pi,40);
[R,T] = meshgrid(rr,tt);
[X,Y] = pol2cart(T,R);
for
% ...
z = interp1(x-0.5,u,rr);
[Z,~] = meshgrid(z,tt);
surf(X,Y,Z)
pause(0.05)
end
#### 4 件のコメント
darova 2019 年 11 月 16 日
You want to plot a sphere or plot your curve in 3D (surface)?
Zjuv9021 2019 年 11 月 20 日
I essentially would like to obtain a visualization of a sphere inflating in 3D space.
darova 2019 年 11 月 20 日
Can you make a simple drawing or something? How it should looks like?
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Translated by | 647 | 1,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.511544 |
https://discuss.codecademy.com/t/lesson-8-part-of-the-whole-what-am-i-doing-wrong/148657 | 1,537,855,720,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161098.75/warc/CC-MAIN-20180925044032-20180925064432-00494.warc.gz | 483,976,074 | 4,784 | # Lesson 8: part of the whole, what am I doing wrong?
#1
The posts here helped me fix a couple issues, but I am still getting this error:
Oops, try again. get_class_average([alice]) should return a number
Here is my code:
``````lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
def average(numbers):
total = sum(numbers)
total = float(total)
def get_average(student):
homework = average(student['homework'])
quizzes = average(student['quizzes'])
tests = average(student['tests'])
return (homework * .10 + quizzes * .3 + tests * .6)
if score >= 90.0:
return 'A'
elif score >= 80.0:
return 'B'
elif score >= 70.0:
return 'C'
elif score >= 60.0:
return 'D'
else:
return 'F';
def get_class_average(students):
results = []
for student in students:
x = get_average(student)
results.append(x)
y = (results)
return y``````
What am I doing wrong?
#2
This line should not be inside the loop.
#3
Thank You!
I actually kept digging and just figured that out from another post. I was coming back here to delete this one. Such a simple, easy thing and it's been giving me a headache for a few hours now.
#4
A post was split to a new topic: Why the below one is not working for me
#5 | 503 | 1,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-39 | latest | en | 0.860413 |
https://reason.town/introduction-to-tensor-decompositions-and-their-applications-in-machine-learning/ | 1,674,863,766,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00675.warc.gz | 494,941,698 | 26,691 | # Introduction to Tensor Decompositions and Their Applications in Machine Learning
A tensor decomposition is a type of matrix factorization that allows for the reduction of a high-dimensional tensor to a set of lower-dimensional matrices. In this blog post, we’ll introduce the different types of tensor decompositions and explore how they can be used in machine learning.
Checkout this video:
## Introduction to Tensor Decompositions
Tensor decompositions are a powerful tool for dimensionality reduction and data analysis in a variety of settings. In general, a tensor is an array of data with more than two dimensions. Decomposing a tensor into its constituent parts can often provide insights that are difficult to obtain from the original data.
There are many different types of tensor decompositions, but the most commonly used ones are the CP decomposition, Tucker decomposition, and PARAFAC decomposition. Each of these decompositions has different strengths and weaknesses, and so it is important to choose the right one for the problem at hand.
The CP decomposition is particularly well-suited for data that can be represented as a sum of rank-one tensors. This includes data such as images, video, and spectral data. The Tucker decomposition is best suited for data that contains redundant information, such as time series or text data. Finally, the PARAFAC decomposition is best suited for analyzing probabilistic graphical models.
Applications of tensor decompositions include document clustering, recommender systems, image processing, video analysis, and many other areas. In machine learning, tensor decompositions can be used for preprocessing data or as a tool for feature selection. Tensor decompositions can also be used to train deep neural networks faster and more effectively.
## Tensor Decompositions for Machine Learning
Decomposing a tensor is a way of reducing its dimensionality while still preserving its important features. This can be useful in machine learning, where lower-dimensional data is often easier to work with.
There are several different ways of decomposing a tensor, each with its own advantages and disadvantages. The most common methods are the CP decomposition and the Tucker decomposition.
The CP decomposition is often used for data that can be represented as a sum of rank-one tensors. This includes data such as images, video, and 3D objects. The Tucker decomposition is more general and can be used for any kind of data. However, it is more computationally expensive.
Both methods have applications in machine learning. The CP decomposition can be used for dimensionality reduction, feature extraction, and image classification. The Tucker decomposition can be used for feature selection and pattern recognition.
## Applications of Tensor Decompositions in Machine Learning
Tensor decompositions are a powerful tool for analyzing and representing data. They have been used in a variety of applications, including machine learning. In this article, we will review some of the most common applications of tensor decompositions in machine learning.
One of the most popular applications of tensor decompositions is latent variable analysis. Latent variable models are used to represent data in terms of hidden or latent variables. Tensor decompositions can be used to provide a low-dimensional representation of data, which can be used for dimensionality reduction or feature extraction.
Tensor decompositions have also been used for learning latent structures in data. For example, they have been used to learn the hidden layers in deep neural networks. Tensor decompositions can also be used to initialize the weights of neural networks, which can help improve the performance of the network.
In addition, tensor decompositions have been used for unsupervised learning tasks such as clustering and anomaly detection. Tensor decompositions can be used to find patterns in data that are not easily detected by other methods.
Finally, tensor decompositions can be used for reinforcement learning. Reinforcement learning is a type of machine learning that involves agents interacting with an environment in order to learn how to optimally complete tasks. Tensor decompositions can be used to represent the state space and action space in reinforcement learning problems, which can help make them more tractable and allow for more efficient learning algorithms.
## Benefits of Tensor Decompositions in Machine Learning
Tensor decompositions have many potential benefits for machine learning applications. For example, they can help reduce the dimensionality of data, improve the computational efficiency of algorithms, and enable new types of data analysis. In addition, tensor decompositions can provide insights into the structure of data and help identify meaningful patterns.
## Drawbacks of Tensor Decompositions in Machine Learning
Tensor decompositions are a powerful tool for dimensionality reduction and data analysis, but they have some drawbacks that can limit their usefulness in machine learning applications.
Firstly, tensor decompositions scale poorly with the size of the data set. For example, if you have a large image dataset with millions of images, it would be very difficult to apply a tensor decomposition to all of them.
Secondly, tensor decompositions can be sensitive to noise and outliers. This means that if your data is not perfectly clean, the results of the decomposition may be inaccurate.
Thirdly, most tensor decompositions require the entire data set to be present in memory in order to be computed. This can be a problem for very large data sets or for applications where data is constantly being added or updated (such as streaming video).
## Future Directions for Tensor Decompositions in Machine Learning
The application of tensor decompositions to machine learning tasks has shown great promise, and there is much excitement about their potential. In this article, we will review some of the recent progress in this area and discuss some future directions for research.
One promising direction for future work is the development of more efficient algorithms for computing tensor decompositions. Current methods are often prohibitively slow for large-scale data sets. Another direction is the incorporation of prior knowledge into tensor decomposition algorithms. For instance, many data sets have a known structure that can be exploited to speed up computation and improve the quality of the results.
Additionally, there is a need for more theoretical understanding of how different tensor decompositions relate to each other and how they can be applied to specific machine learning tasks. Finally, there is a need for more real-world applications of tensor decompositions in order to demonstrate their usefulness and advantages over traditional methods.
## Conclusion
In this paper, we have reviewed the fundamentals of tensor decompositions and their potential applications in machine learning. We have shown that tensor decompositions can be used to represent data in a concise and structured way, which can be beneficial for both supervised and unsupervised learning tasks. We have also discussed a number of challenges that need to be addressed in order to make full use of tensor decompositions in machine learning. Despite these challenges, we believe that tensor decompositions offer a promising direction for further research in this area.
Scroll to Top | 1,334 | 7,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | longest | en | 0.913048 |
https://fr.slideserve.com/jescie-moore/2013-treatment-episode-data-set-teds-reporting-supplementary-guidance | 1,618,926,332,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00255.warc.gz | 368,522,662 | 22,297 | # 2013 Treatment Episode Data Set (TEDS) Reporting Supplementary Guidance - PowerPoint PPT Presentation
2013 Treatment Episode Data Set (TEDS) Reporting Supplementary Guidance
1 / 28
2013 Treatment Episode Data Set (TEDS) Reporting Supplementary Guidance
## 2013 Treatment Episode Data Set (TEDS) Reporting Supplementary Guidance
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##### Presentation Transcript
1. 2013 Treatment Episode Data Set (TEDS) Reporting Supplementary Guidance
2. Purpose These slides are intended to boil down the 21-page coding instructions into bullet point statements. The selection and emphasis in this presentation will address only those areas where there has been confusion. Contact person at MDCH for questions or concerns is Phil Chvojka at chvojkap@michigan.gov
3. Overview Specific guidance on Each Field can be found in the CODING INSTRUCTIONSFor: Treatment Episode Data Set (TEDS) These instructions are on the Internet at http://www.michigan.gov/documents/mdch/Coding_Instructions_TEDS_398729_7.pdf
4. TEDS Admission and Discharge Records • TEDS uses a “capture--recapture” approach to data collection • Client data are collected on the first date of service (admission) at each licensed site... and then re-collected on the last date of service (discharge) • This allows tracking of changes during the course of treatment and allows us to demonstrate improved outcomes • It means that what is reported at admission WILL be compared to what is reported at discharge
5. Snapshot in Time • Admission field responses should reflect the known status on the reported date of admission in the record • Discharge field responses should reflect the known status on that reported date of discharge in the record • If there is no variation in what is reported from admission to discharge, it raises a red flag
6. Three Types of Fields Let’s borrow the concepts of Constants and Variables from algebra: • Constants do not change (like the number of hours in a day...always 24) • Variables do change (like the reported high temperature for each day) • There is also a special type of constant called the “Primary Key Fields”
7. TEDS Constants For TEDS, constants are collected only once, typically only at admission (examples) • Referral Source • County of Residence • Date of Birth (DOB) • Sex • Race • Ethnicity • Military Status
8. TEDS Constants (continued) Plus a few “honorary constants” are only collected at admission (examples) • Marital Status • Education (00 to 25 years) • Currently in Training • Number of Dependents • Total Annual Income • Primary Language Spoken
9. TEDS Variables The TEDS variables are collected first at admission and then at discharge. Of primary importance are the ones used to calculate National Outcome Measures (NOMS): • 30-day use of Primary Substance • Employment Status • Living Arrangement • Total Recent 30-Day Arrests • Attendance at Self-Help Groups
10. TEDS Variables (continued) • Variables are collected at both admission and discharge because the answers can and do change RULE: “Never pre-fill a discharge variable with what was reported at admission” Automatically reporting the same data for a field at discharge with what was collected at admission defeats the purpose for “capture-recapture”
11. TEDS Primary Key Fields These fields are in both the admission and discharge records...and are required to match for successful batch processing • CA Payer ID (CA only issue) • Provider License Number (CA only issue) • CA Client ID (CA only issue) • SSN (provider and CA) • Beneficiary ID (provider and CA issue) • Date of Admission (provider and CA issue)
12. Reporting the Admission Date in the Discharge Record Of those primary key fields listed, the one that’s crucial for data entry is the Month, Date, and Year of Admission. It’s reported both at admission and at discharge...and they must match exactly in both places for the discharge record to process
13. Beneficiary (Medicaid/ABW ID) • Ten Digit Beneficiary ID covers both Medicaid and ABW • Report the Beneficiary ID in the admission record, if known, even if Medicaid or ABW is not the actual known payer
14. County of Residence • Avoid using the 96 (Homeless) code, if at all possible (use “Living Situation” to report homelessness) • Even if the person has no fixed address, please code the county in which he/she is located (for example, the location of a shelter). • If the shelter/location is in a different county than the treatment facility, use the county where he/she lives. As a last resort, use the county of the facility.
15. Primary Substance of Abuse = 00 (None) • When you report that the person admitted had no substances involved in treatment, this must be defended • How can it be defended? With either “Co-dependent” = Yes, or with an “Other Factor” of Adult Child, Significant Other, or Gambling Addiction
16. Number of Prior Treatments • Include only treatment admissions and not assessment services • These should be episodes and not changes in levels of care • Answers the question, “How many times have you tried to address this problem via treatment at any provider?”
17. Employment Status at Admission and Discharge Reporting employment status for students who have no current job: • Under age 18 = Not Applicable • Over Age 18 = Not in the Labor Force
18. Detailed Not In Labor Force • Use only when Employment Status = 4 (Not in competitive labor force) Make sure if a student (college) is over the age of 18 that Detailed Not in the Labor Force is coded with 2 =Student This is being missed systematically. Feds use this for Employment/Education NOM
19. Number of Dependents • Number of dependents claimed on federal tax return • Only report zero if the client is reported as a dependent on another person’s federal tax return ( i.e. dependent minors)
20. Total Annual Income at Admission • This is estimate of income for 12-months prior to admission • Only legally earned taxable and non-taxable income should be reported • If client is single, indicate the total amount of gross income of the individual; if client is married, indicate the total of the client and spouse • For a minor client, indicate the 12-month income of the parent(s)
21. Opioid Replacement Part of Treatment at Admission • Only reply with “1= Methadone Yes” if you are an approved methadone dispensing site • The license number listed in the record MUST be an approved methadone provider; so only these sites should have the ability to code “yes”
22. TEDS Reporting Discharge Date • The reported date of discharge is also the “last date of contact or billable service” • When a case is administratively discharged after a period of inactivity, the reported discharge date reverts back to that last date of contact or the last date of a billable service
23. Discharge Reason • Discharge reason is unique in that it is only collected once at discharge “Death” means the death of the person in treatment....not the death of a relative or friend We should not see a re-admission for persons reporting the reason for discharge as “death”
24. A New Approach to Consider • In 2013, for the same six-digit provider license number...the reported service category at discharge does NOT have to match the service category at admission • Example: A person receives detox, short-term residential, and outpatient from the same provider under one license
25. Reporting 30-Day Windowat Discharge Only consider events that occurred since admission when reporting the “previous 30-day window” at discharge for: • PSA Frequency of Use • SSA Frequency of Use • TSA Frequency of Use • Recent (30-Day) Total Arrests • Recent (30-Day) Possession/Sales Arrests • Recent (30-Day) DUI/DWI • Attendance at Self Help Programs
26. Reporting 30-Day Windowat Discharge (continued) • Only take into consideration events that occurred after the admission date – Never count events in the discharge record that took place before treatment began (before the admission date) • Those pre-admission events would have been captured already at admission • Again, If the length of stay is less than 30 days, do not go back past the admission date to capture use or arrests or self-help group attendance
27. Other Considerations • There will be instances where you will have to use your judgment on how to code unique situations • Trust your judgment and select your best answer • The discharge record is there to track changes from admission and only looks at events that take place after the admission date
28. And Some Kudos Your efforts on TEDS reporting have allowed Michigan to show that, for persons in treatment, there are consistent improvements in use patterns, employment, homelessness, social support/self help, and arrests | 1,893 | 8,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | latest | en | 0.845355 |
https://www.esaral.com/q/let-p-h-k-be-a-point-on-the-curve-56981 | 1,722,797,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640408316.13/warc/CC-MAIN-20240804164455-20240804194455-00005.warc.gz | 614,615,264 | 11,588 | Let P(h, k) be a point on the curve
Question:
Let $P(h, k)$ be a point on the curve $y=x^{2}+7 x+2$, nearest to the line, $y=3 x-3$. Then the equation of the normal to the curve at $P$ is :
1. (1) $x+3 y+26=0$
2. (2) $x+3 y-62=0$
3. (3) $x-3 y-11=0$
4. (4) $x-3 y+22=0$
Correct Option: 1
Solution:
The given curve is, $y=x^{2}+7 x+2$
$\Rightarrow \frac{d y}{d x}=2 x+7$
$\left(\frac{d y}{d x}\right)_{(h, k)}=2 h+7$
The tangent at $P(h, k)$ will be parallel to given line
$2 h+7=3 \Rightarrow h=-2$
Point $P(h, k)$ lies on curve
$k=(-2)^{2}-7 \times 2+2=-8$
Slope of normal at point $P(-2,-8)=-\frac{1}{3}$
$\therefore$ The equation of normal to the cuve at $P$ is
$x+3 y+26=0$ | 301 | 695 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.549088 |
https://engineeringdaze.com/tag/baseball/ | 1,680,240,075,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00473.warc.gz | 279,827,303 | 16,252 | ## A Baseball Story for an Engineer
(engineeringdaze.com has some wrap-up thoughts on baseball, now that the World Series just finished the season.)
The best baseball story I have heard recently that most any engineer will appreciate is Moneyball. I have not seen the movie, but I went through the book a few years ago and I am still impressed by the logic and sheer emotionless decision-making that was introduced by the Oakland team. These are qualities that engineers admire.
One may say that these were really math nerds or statisticians, and to some extent that is correct. But considering the level of application of mathematic principles to solve a real-world problem, I would say that these people acted much more like engineers than mathematicians.
Whether a math guru or an engineer, it is good to point out that Brad Pitt is not typical of the way these numbers geeks look, but I’m not complaining. It’s good press.
## You Might Be an Engineer If…
– you watch the World Series, and may care about one team winning, but what you really watch for are the cool slow-motion shots that show the movement of the baseball and the angles of the bat-ball interaction. That was really neat.
(engineeringdaze.com has some wrap-up thoughts on baseball, now that the World Series just finished the season.)
## 13.3%
engineeringdaze.com has some wrap-up thoughts on baseball, now that the World Series just finished the season.
This number, 13.3% is not a number that represents anything to an engineer, except frustration regarding the pure lack of logic of some baseball stats. While watching a playoff game recently, a player was batting and a guy was on second base. The announcers stated that this player got hits 13.3% of the time that a runner was in scoring position. That is all well and good, and, as an engineer, I appreciate statistics which break down the game and explore different facets of the odds. They could have broken that statistic down for the player to dissect his hitting percent against left- and right-handed pitchers, outdoors versus domed stadiums, at night or in day games, in his first or second or third at bat, if he had a fielding error in the previous ten games or not, or whether they were playing on real or artificial turf. That kind of examination is fascinating to explore.
However, the guy at second only got there after the count was 2-1, by stealing second. The announcers didn’t talk about that. They didn’t state was this guy’s hitting percent was with men in scoring position but only getting there by stealing a base part-way through the batter’s at-bat which would obviously skew the hitting percent seeing as that he would have less opportunities to get a hit.
Here they are, and by “they” I mean the baseball “people”, with all these statistics and they fail to completely inform us of the true odds of what will about to take place. Disappointing? Most definitely. But, I will be OK. Baseball season is over and now I watch football. Which brings me to some illogical statistics from this past weekend…
## An ERA of what???
engineeringdaze.com has some wrap-up thoughts on baseball, now that the World Series just finished the season.
I was watching a baseball game recently and the team in the field brought in a relief pitcher. There was one or possibly two people on base, but he started pitching and no more outs were made when the first person he faced scored.
As an engineer, I am fascinated with baseball. Sure, the sport is mildly interesting, but the stats are what is really amazing. “This guy bats .322 with men on third base and when the pitcher he is facing is left-handed and has pitched more than once in the previous 3 days, as long as those appearances were further then 400 miles apart.”
Getting back to the game at hand, I did a quick calculation and figured, since there were no outs yet, something was terribly wrong. It looked like they were about to take this relief pitcher out of the game, resulting in him with an earned run counted against him, and no outs. That means his earned run average (ERA) for the day would have been either, impossible to calculate, because one would have to divide by zero, or, to make it a little more acceptable, infinity! Technically, it is impossible to calculate.
For the sake of logic, baseball should have the following rule:
“If a pitcher has an earned run counted against him, and he has not gotten any outs, then he will play until he gets an out, or, if he is leaves the game, an out will be charged to the other team.” In order for teams to not yank players just to get an out, then their team would start with two outs the next half inning.
Engineers can add so much logic to the game. Because, believe me, dividing by zero is not logical.
Thankfully, they let the pitcher get one out before taking him out of the game, and then with two runs against him, he ended the day with an ERA of 54.00. At least it was not infinity. | 1,063 | 4,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-14 | latest | en | 0.979707 |
http://econowaugh.blogspot.com/2017_02_18_archive.html | 1,532,229,344,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00440.warc.gz | 107,754,752 | 23,250 | ## Saturday, February 18, 2017
### 2012 Macro Multiple Choice (Unemployment, Inflation, Phillips Curve)
2012 Macro Multiple Choice (Unemployment, Inflation, Phillips Curve)
Cheat Sheet here
Hyperinflation Video
If you are so hopelessly out of work that you've stopped looking over the past four weeks -- the Department of Labor doesn't count you as unemployed. That's right. While you are as unemployed as one can possibly be, and tragically may never find work again, you are not counted in the figure we see relentlessly in the news.
From the cheat sheet
From the cheat sheet
### 2012 Macro Multiple Choice (Monetary Policy)
2012 Macro Multiple Choice (Monetary Policy)
Monetary Policy Cheat Sheet here.
### Transactions Demand
On a daily basis people need money on hand for the things that they routinely buy. You have to get a haircut or stop by the store on the way home from work to pick up some milk. You have transactions that you need to conduct, and therefore you have a demand for money. The transactions demand for money is using money as a medium of exchange. Notice in the graph below that the Transactions Demand for Money (DMT) is denoted as a vertical line when graphed against the interest rate. The demand for money as a medium of exchange is independent of the interest rate, because when you are on your way home from work and need to pick up milk, the interest rate does not affect how much milk you buy.
If interest rates increase, then citizens will get more money to have their cash in the bank. They will be paid more to keep their cash in the bank. In essence their opportunity cost of holding cash in their pockets has increased. They will hold less cash in their pockets if banks pay them to deposit it in the bank. How many more times can I say it????
From the cheat sheet -
Let me say tis another way - a decrease in inflationary expectations = reducing the price level, thus reducing AD or RGDP.
(A) decrease in MPS = increase in MPC, therefore people spend more and increase AD
(B) decrease in imports means an increase in exports = AD increases
(C) decrease in MS means an decrease in investment = AD decrease
(D) increase in deficit means increase in government spending = AD increase
(E) increase in the price of raw materials is an increase in AS = Cost push inflation
Economies operating below full-employment means that the economy is in recession. The FED would want to have an expansionary policy.
From the cheat sheet.
If investment is more responsive it means that as the money supply is increased and nominal interest rates fall, then investment will be stimulated to higher levels and therefore RGDP increases more.
FED wouldn't want to stimulate growth if the inflation rate is high.
Monetarists
Rational Expectation Theory
### 2012 Macro Multiple Choice (FOREX)
2012 Macro Multiple Choice (FOREX)
Forex Cheat Sheet here.
(A) Demand for a country's exports,, increases the demand in the FOREX for the currency
(B) Money supply increases, more money less value
(C) Travel abroad increases, citizens must exchange their currency for the currency of the country travelled to,, that would increase the supply of the currency in the FOREX which would cause the value to fall.
(D) Interest Rates decreasing means less foreigners would invest in our country, less demand for our currency.
(E) Tariffs decrease, we buy more imports and the supply of our currency in the FOREX increases driving down the value
If the US\$ value decreases, then our goods are relatively cheaper than the foreign goods, so our sports will increase. The foreigners purchasing power has increased because the value of the US\$ has fallen.
Country A has a higher inflation (PL, Price Level) than country B,,, if the PL in country A is rising, then their goods are getting more and more expensive. Country B is buying less and less of them. Therefore, there is less and less demand for country A's goods and less need to go to the FOREX and buy country A's currency. So less demand for country A's (Goods) currency means the value of A's currency falls.
Let me say this differently - The Indian Rupee will get stronger and the Japanese Yen will get weaker.
(A) India's inflation rate increases relative to Japans. - This means that India's goods will become more expensive leading to less demand in the FOREX, for India's currency and the value of India's currency will therefore decrease.
(B) India has a trade deficit with Japan. - This mean that India has more imports from Japan than exports, so more imports means that the supply of the rupee is increasing in the FOREX and therefore the value of the rupee is decreasing.
(C) Japan enters a recession - meaning that the PL of Japanese goods will fall leading to Indian citizens to import more of them and the rupee's value will decrease. (see above)
(D) India's money supply increases, and therefore nominal interest rates fall, investment and consumption increase, aggregate demand therefore increases along with Indian incomes (Y), and with higher incomes comes higher imports. (see above)
(E) Real interest rates in India increase, attracting Japanese investors to buy Rupees to invest in Indian interest assets (Bonds) etc,, In the FOREX the demand for Rupees increases driving up the value.
WE don't see these types of questions very often on the AP exams.
(A) they both do this
(B) they both do this
(C) Neither do this
(D) Neither do this
(E) Tariffs are a tax and the tax money goes to the government.
### FOREX Cheat Sheet
FOREX Cheat Sheet | 1,195 | 5,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.951466 |
https://au.mathworks.com/matlabcentral/cody/problems/31-remove-all-the-words-that-end-with-ain/solutions/252234 | 1,582,226,388,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00439.warc.gz | 274,297,730 | 15,735 | Cody
# Problem 31. Remove all the words that end with "ain"
Solution 252234
Submitted on 31 May 2013 by Marek Kuklis
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% s1 = 'The rain in Spain falls mainly on the plain'; s2 = 'The in falls mainly on the '; assert(strcmp(remAin(s1),s2));
2 Pass
%% s1 = 'The pain from my migraine makes me complain'; s2 = 'The from my migraine makes me '; assert(strcmp(remAin(s1),s2));
3 Pass
%% s1 = 'I had to explain that "ain" is not a word'; s2 = 'I had to that "" is not a word'; assert(strcmp(remAin(s1),s2)); | 204 | 677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.885136 |
https://kr.mathworks.com/matlabcentral/profile/authors/9598914-christopher-stapels?s_tid=maker_to_profile | 1,553,631,629,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206016.98/warc/CC-MAIN-20190326200359-20190326222359-00332.warc.gz | 527,555,490 | 20,301 | Community Profile
# Christopher Stapels
### MathWorks
33 total contributions since 2017
ThingSpeak Documentation Writer
and LCH enthusiast
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You should also be able to update the value of a ThingSpeak field using the REST API. The MQTT connection is intended to be ve...
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matlab and thingspeak manipulate data before plotting
I would recommend setting up a new channel. Then you can set a react to call a MATLAB analysis to write the data to the new cha...
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Battery Voltage
two six volt bettery packs in paralell in the basement
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Soil Monitor Data Summary
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5 Prime Numbers
Your function will be given lower and upper integer bounds. Your task is to return a vector containing the first five prime numb...
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Make an awesome ramp for a tiny motorcycle stuntman
Okay, given a vector, say v=[1 3 6 9 11], turn it into a matrix 'ramp' like so: m=[1 3 6 9 11; 3 6 9 11 0; 6 9 ...
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2년 이하 ago | 901 | 3,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-13 | longest | en | 0.781207 |
http://stackoverflow.com/questions/23554169/in-r-make-a-plot-ylim-is-a-logarithmic-scale-the-same-as-one-generated-by-exc | 1,469,294,517,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00100-ip-10-185-27-174.ec2.internal.warc.gz | 230,628,025 | 20,589 | Dismiss
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# In R: make a plot (ylim is a logarithmic scale) the same as one generated by excel
Here is the csv data: https://www.dropbox.com/s/87bwvhyo6i4f68u/test.csv , first column is the date and second column is the corresponding value.
I would like to plot such picture as made in excel: https://www.dropbox.com/s/13wyk1aeajgs6fq/test.png . Note: the ylim in the plot is changed to the logarithmic scale (base 10), which can the picture more better than the original one. I would like to this in R, especially using ggplot making it beautiful, but I can not. So I hope you can help.
``````ba <- as.data.frame(fread("test.csv"))
plot(as.factor(ba[, 1]), ba[, 2], log = "y", xaxt = "n", xlab = "")
idx <- seq(1, 100, length = 10)
labels <- ba[, 1]
labels <- labels[idx]
axis(1, seq(1, 100, length = 10), par("usr")[3], srt = 45, labels = labels,
adj = 1, xpd = TRUE)
``````
Thanks.
EDIT:
[Below is the answer from rawr]. That is perfect.
``````tmp <- read.csv('~/desktop/test.csv', stringsAsFactors = FALSE,
strip.white = TRUE, header = FALSE)
tmp\$V1 <- paste0(tmp\$V1, '-01')
tmp <- within(tmp, {
V1 <- as.Date(V1)
date <- format(as.Date(tmp\$V1, '%Y-%m-%d'), '%Y-%m')
stuff <- V2
})
par(tcl = -.1, xpd = FALSE)
with(tmp,
plot(V1, log(stuff), type = 'n', ylim = c(0,6),
col = 'royalblue1', lwd = 3, bty = 'l',
axes = FALSE, xlab = '', ylab = ''))
abline(h = 0:6, lwd = .5)
with(tmp, points(x = V1, y = log10(stuff), type = 'l',
col = 'royalblue1', lwd = 3))
par(xpd = TRUE)
axis(2, at = 0:6, cex.axis = .6,
labels = format(10 ** (0:6), scientific = FALSE, big.mark = ','), las = 2)
x <- with(tmp, seq(min(V1), max(V1), length = 12))
text(x = x, y = -.5, cex = .8, labels = format(x, '%Y-%m'), srt = 45)
``````
-
is ggplot a requirement? – MrFlick May 8 '14 at 23:19
plot is also OK. But for me, even using plot function in basic R, I can not duplicate the one produced by excel. can you help? thanks. – BioChemoinformatics May 8 '14 at 23:32
Please post the code you have tried so far. – josliber May 8 '14 at 23:43
@josilber, my code does not work well. – BioChemoinformatics May 9 '14 at 0:08
why you want to match what excel does is beyond me, but maybe this will get you started
``````set.seed(1618)
tmp <- data.frame(date = seq(as.Date("2000-1-1"),
by = "month", length.out = 12),
stuff = sort(rpois(12, 5)) * 10000)
par(tcl = -.1, xpd = NA)
with(tmp,
plot(date, stuff, type = 'l', ylim = c(0, max(stuff)),
col = 'royalblue1', lwd = 3, bty = 'l',
axes = FALSE, xlab = '', ylab = ''))
axis(2, at = pretty(seq(0, max(tmp\$stuff))),
labels = format(pretty(seq(0, max(tmp\$stuff), length = 5)),
scientific = FALSE), las = 2)
text(x = tmp\$date, y = -5000, labels = format(tmp\$date, '%Y-%m'), srt = 45)
par(xpd = FALSE)
abline(h = pretty(seq(0, max(tmp\$stuff))), lwd = .5)
``````
EDIT
Here is how to make your own breaks and make the commas using `prettyNum`
``````par(tcl = -.1, xpd = NA)
with(tmp,
plot(date, stuff, type = 'l', ylim = c(0, 100000),
col = 'royalblue1', lwd = 3, bty = 'l',
axes = FALSE, xlab = '', ylab = ''))
nums <- c(0,10000, 30000, 70000, 100000)
prettyNum(nums, big.mark = ',', scientific = FALSE)
axis(2, at = nums,
labels = prettyNum(nums, big.mark = ',', scientific = FALSE),
las = 2)
text(x = tmp\$date, y = -10000, labels = format(tmp\$date, '%Y-%m'), srt = 45)
par(xpd = FALSE)
abline(h = nums, lwd = .5)
``````
EDIT
``````tmp <- read.csv('~/desktop/test.csv', stringsAsFactors = FALSE,
strip.white = TRUE, header = FALSE)
tmp\$V1 <- paste0(tmp\$V1, '-01')
tmp <- within(tmp, {
V1 <- as.Date(V1)
date <- format(as.Date(tmp\$V1, '%Y-%m-%d'), '%Y-%m')
stuff <- V2
})
par(tcl = -.1, xpd = FALSE)
with(tmp,
plot(V1, log(stuff), type = 'n', ylim = c(0,6),
col = 'royalblue1', lwd = 3, bty = 'l',
axes = FALSE, xlab = '', ylab = ''))
abline(h = 0:6, lwd = .5)
with(tmp, points(x = V1, y = log10(stuff), type = 'l',
col = 'royalblue1', lwd = 3))
par(xpd = TRUE)
axis(2, at = 0:6, cex.axis = .6,
labels = format(10 ** (0:6), scientific = FALSE, big.mark = ','), las = 2)
x <- with(tmp, seq(min(V1), max(V1), length = 12))
text(x = x, y = -.5, cex = .8, labels = format(x, '%Y-%m'), srt = 45)
``````
-
Thank you. Since in the beginning I try to make the picture but I can not do that, since I do not know how the scale the ylim as dose in the excel, later, I use the excel to make a simple plot. But it looks not good as do in R. So I find the help here. I will learn your answer, which is more excel style. thank you very much. My concern is if you are using my data, how to deal with the ylim? I would like to the ylab is the same to excel one: 1 10 100 1,000 10,000 100,000 1,000,000. – BioChemoinformatics May 9 '14 at 0:39
I don't see how a log scale is misleading in this instance. They are pretty common. – thelatemail May 9 '14 at 0:58
oh it's a log scale. I thought it was just random (10, 1000, 100000000000), nevermind – rawr May 9 '14 at 0:58
@BioChemoinformatics you can make the labels in `axis` anything you want. You just need to put the `at` argument on a log scale, eg, `pretty(log10(data\$values))` or something similar. or set the breaks manually like I showed – rawr May 9 '14 at 1:14
@rawr: Thanks so much. I am trying your suggestion. Hope I can deal with it. – BioChemoinformatics May 9 '14 at 1:26
With `pkg:ggplot2` I would have thought you could just add `+scale_y_log10("log_10 Scale")`
`````` pl + geom_line() + scale_y_log10("log_10 Scale", limits=c(1, 1000000) )
perhaps you can change the `theme()` if you are interested in a more cosmetic aspect. For example try: `p1 + theme_bw()` docs.ggplot2.org/current/theme_bw.html – marbel May 9 '14 at 2:12 | 1,982 | 5,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-30 | latest | en | 0.803734 |
https://www.studyadda.com/question-bank/surface-area-and-volume-mensuration_q18/4628/365225 | 1,597,373,225,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00516.warc.gz | 846,558,627 | 18,063 | • # question_answer The dimensions of a field are 20 m by 9 m. A pit 10 m long, 4.5 m wide and 3 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field. What will be the rise in the height of field as a result of this operation? A) 1 mB) 2 m C) 3 m D) 4 m
(a): Volume of mud dug out $=10\times 4.5\times 3$$=135\,{{m}^{3}}$ Let the remaining ground rise by = ?h? m Then $\{(20\times 9)-(10\times 4.5)\}h=135$ $135\,h=135\Rightarrow h=1\text{ }m$ | 192 | 554 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-34 | latest | en | 0.776723 |
https://www.scribd.com/document/384821112/PE-Mannual-1 | 1,568,646,355,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572744.7/warc/CC-MAIN-20190916135948-20190916161948-00245.warc.gz | 1,016,532,962 | 60,563 | You are on page 1of 5
1
Reg. # 2014-EE-408,2014-EE-420,2014-EE-494
Marks
EXPERIMENT # 1
Design and simulation of a AC/DC converter
AC/DC Converter
A rectifier is an electrical device that converts alternating current (AC), which periodically
reverses direction, to direct current (DC), which flows in only one direction. The process is
known as rectification, since it "straightens" the direction of current.
Single Phase Rectifier:
Half-wave rectification:
In half-wave rectification of a single-phase supply, also called uncontrolled one-pulse
midpoint circuit, either the positive or negative half of the AC wave is passed, while the other
half is blocked. Mathematically, it is a ramp function (for positive pass, negative block):
passing positive corresponds to the ramp function being the identity on positive inputs,
blocking negative corresponds to being zero on negative inputs. Because only one half of the
input waveform reaches the output, mean voltage is lower. Half-wave rectification requires a
single diode in a single-phase supply.
Figure: 1
Full-wave rectification
A full-wave bridge rectifier converts the whole of the input waveform to one of constant
polarity (positive or negative) at its output. Mathematically, this corresponds to the absolute
value function. Full-wave rectification converts both polarities of the input waveform to
pulsating DC (direct current), and yields a higher average output voltage.
2
Figure: 2
Figure. 3
Figure. 4
3
Design the AC/DC converter in Matlab/Simulink.
Observe the output voltage waveform.
Observe the output current waveform.
4
5
I/O Voltages(three phase)
Results:
We have simulated single phase and three phase AC to Dc converters i-e full and half wave
rectifiers and observed the current and voltage characteristics. | 428 | 1,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-39 | latest | en | 0.851957 |
https://www.causeweb.org/cause/statistical-topic/multivariate-quantitative-relationships?page=1 | 1,726,863,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00134.warc.gz | 638,750,971 | 14,248 | Multivariate Quantitative Relationships
• Cartoon: Ice-cream sales and shark sightings
A cartoon to aid in discussing confounding and correlative versus causal relationships, for example by asking students to suggest an alternate reason for the relationship besides the one jokingly illustrated in the cartoon. The cartoon was created by English cartoonist John Landers in December, 2021 based on an idea by Dennis Pearl (Penn State University) and Larry Lesser (The University of Texas at El Paso).
• Dancing Correlation
A video using dance to teach about concepts involved with correlation. This 2013 video is from the “Dancing Statistics” series developed by Lucy Irving from Middlesex University (UK) funded by a BPS Public Engagement grant and additional funding from IdeasTap. Full credits are within the video. The Dancing Statistics project is described at https://www.frontiersin.org/articles/10.3389/fpsyg.2015.00050/full
The video also comes with teaching notes for viewing by instructors who are logged into CAUSEweb.org.
• Cartoon: Non-Constant Variance
A 2020 cartoon illustrating the idea of heteroscedasticity (non-constant variance) that might be used to start a discussion on the important of the constant variance of errors in making inferences from regression models. The cartoon was used in a 2021 Teaching Statistics paper "Statistical edutainment that lines up and fits," by Dennis Pearl from Penn State University and Larry Lesser from The University of Texas at El Paso.
• Hand Squeeze Activity
A quick "hands on" activity for an in-class experience of data collection as a simple linear regression example where students predict the time needed for a human chain of hand squeezes to make a full circuit as a function of number of people in the chain. The lesson plan secondary school lesson plan adapted from Cynthia Lanius’ hand squeeze activity by Bo Brawner at Tarleton State University.
• Cartoon: Lookout Groups
A cartoon designed to support a discussion of using dummy variables to code for categories of a categorical variable in a regression model (e.g. 5 are needed when there are 6 categories). The cartoon was used in the February 2020 CAUSE cartoon caption contest and the winning caption was written by Dominic Matriccino, a student at the University of Virginia. The cartoon was drawn by British cartoonist John Landers (www.landers.co.uk) based on an idea by Dennis Pearl from Penn State University. A second winner in the February 2020 contest was "The grass really is greener on the homogeneity side," written by Jennifer Ann Morrow, an instructor from University of Tennessee. Jennifer's cartoon caption can be used in discussing the importance of within-group variability in judging differences between groups and the difficulty when the groups being compared have different levels of variability.
• Cartoon: Constellations
A cartoon that can be used for discussing the traditional theme of "Correlation does not imply Causation" as well as what observational evidence does provide the most convincing evidence of a causal relationship. The cartoon was used in the June 2019 CAUSE cartoon caption contest. The cartoon was drawn by British cartoonist John Landers (www.landers.co.uk) based on an idea by Dennis Pearl from Penn State University.
• Cartoon: The Mars Orbiter
A cartoon to initiate discussions about how the correlation is a unitless number that does not change with changes in the units of the variables involved. The cartoon was created in February 2020 by British caetoonist John Landers based on an idea by Dennis Pearl (Penn State) and Larry Lesser (Univ of Texas at El Paso). An outline of a lesson plan for the use of the cartoon is given in a 2020 Teaching Statistics article by Dennis Pearl and Larry Lesser.
• Poem: This Function Dysfunction
A poem written in 2019 by Larry Lesser from The University of Texas at El Paso to discuss the simplest case of line of fit where the slope and correlation coefficients each have a value of 0. The poem is part of a collection of 8 poems published with commentary in the January 2020 issue of Journal of Humanistic Mathematics.
• Joke: Banana Peel
A joke to use in discussing the meaning of the slope in a linear trend. The joke was written in May 2019 by Larry Lesser, The University of Texas at El Paso, and Dennis Pearl, Penn State University.
• The Statistics of Visual Representation
This paper comes from researchers at the NASA Langley Research Center and College of William & Mary.
"The experience of retinex image processing has prompted us to reconsider fundamental aspects of imaging and image processing. Foremost is the idea that a good visual representation requires a non-linear transformation of the recorded (approximately linear) image data. Further, this transformation appears to converge on a specific distribution. Here we investigate the connection between numerical and visual phenomena. Specifically the questions explored are: (1) Is there a well-defined consistent statistical character associated with good visual representations? (2) Does there exist an ideal visual image? And (3) what are its statistical properties?" | 1,075 | 5,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.888125 |
http://powerpointpresentationon.blogspot.com/2013/08/ppt-on-solar-pv-system-design.html | 1,529,946,924,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00154.warc.gz | 254,970,533 | 20,514 | ## Thursday, August 8, 2013
### PPT On Solar PV System Design
Solar PV System Design Presentation Transcript:
1.Solar PV & Diesel GENERATOT Hybrid System Design
2.System Sizing
3.System Calculations:
Anatomy: 48 hrs (continued)
Whr = watt (load) x hrs
25000 x 48
= 120,000 x 1.3 (losses)
= 156,000
PV Panels Required:
Panel Size: = 156,000/5.5
= 28363.6363 Watts
= 28.36 KW or 29 KW
System available with Client is 35 KW
4.PV Panels Required for 35 KW systemSystem = 35000/ 230
= 152.17 or 153
Recommended 154 Panel
5.Battery Calculations
Battery Sizing :
0.85 for battery loss
0.8 for depth of discharge
Battery Capacity (Ah) = Total Watt-hours per day used by appliances x Days of autonomy (0.85 x 0.8 x nominal battery voltage)
= 60,000 x 2
32.64
= 3676.47 Ah
No. of Batteries Required = 24 batteries (2 volt and 4000 Ah)
6.Solar Charge Controller:
According to standard practice, the sizing of solar charge controller is to take the short circuit current (Isc) of the PV array, and multiply it by 1.1 for MPPT and 1.3 for PWM
Charge controller = 8 x 8.25 x 1.1
= 72.6 Ampere
Therefore 10 , 80A (T80 charge controller will be required for system to fulfill the 600 A) .
7.Inverter Sizing
An inverter is used in the system where AC power output is needed. The input rating of the inverter should never be lower than the total watt of appliances. The inverter must have the same nominal voltage as your battery.
For stand-alone systems, the inverter must be large enough to handle the total amount of Watts you will be using at one time. The inverter size should be 25-30% bigger than total Watts of appliances. In case of appliance type is motor or compressor then inverter size should be minimum 3 times the capacity of those appliances and must be added to the inverter capacity to handle surge current during starting.
Client –Energy has 20 KW inverter which is more than requirements.
8.Battery Selection:
Applications:
¦ Utility (stations and substations)
¦ Power Distribution
¦ Switchgear
¦ Telecom Equipment
¦ Solar/Photovoltaic
¦ UPS
24 batteries will be connected in series to obtain 48 volts and capacity will remains constant 4000 Ah which is Client requirement
9.Equipment Selection
10.Inverter:
Single Phase bidirectional dual mode hybrid inverter for mini grid system
Pure sine wave out put
Single phase configuration
High efficiency more than 94%
Automatic/manual generator control
11.System layout | 685 | 2,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.764801 |
https://tex.stackexchange.com/questions/263219/add-plot-to-node-in-tikz-matrix | 1,719,076,284,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00005.warc.gz | 480,732,828 | 35,789 | # Add plot to node in Tikz matrix
Using the code at the end, I was able to create this image.
I would like to add a plot (I'm plotting kernel densities) to each of the nodes on the edge, and move their labels below the plot, such that the plot and label are the same height as the center box.
This would mean, on the left would be a plot with y below the x-axis and on the right would be a plot with u below the x-axis.
Is this possible to do with the matrix structure, and if not, is there an alternate way to create this?
\documentclass[tikz,border=2]{standalone}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{shapes,snakes,calc}
\tikzstyle{connector} = [->,thick]
\begin{document}
\begin{tikzpicture}[x=3in,y=2in]
\tikzstyle{ann} = [draw=none,fill=none,right]
\matrix[nodes={draw, thick, fill=none},
row sep=0.3cm,column sep=0.5cm] {
\node[draw=none,fill=none] (N1) {$\mathbf{y}$}; &
\node[rectangle, minimum height=0.5in, minimum width = 0.75in] (N2) {Center}; &
\node[draw=none] (N3) {$\mathbf{u}$};\\
};
\draw [connector] (N1) -- node {} (N2);
\draw [connector] (N2) -- node {} (N3);
\end{tikzpicture}
\end{document}
Yes, this can be done (see example below). However, it's not clear that you really need a matrix here since all the required elements can be arranged as desired without using a matrix.
One option would be to use pgfplots to draw your plots; box them (a precaution to avoid the plots from inheriting settings from the "outer" tikzpicture) and then use the boxes in the \nodes of your matrix:
The code:
\documentclass[tikz,border=2]{standalone}
\usepackage{amsmath}
\usepackage{pgfplots}
\usetikzlibrary{shapes,snakes,calc}
\tikzstyle{connector} = [->,thick]
\newsavebox\myboxa
\newsavebox\myboxb
\savebox\myboxa{%
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
width=4cm,
]
\end{axis}
\end{tikzpicture}%
}
\savebox\myboxb{%
\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
width=4cm
]
\end{axis}
\end{tikzpicture}%
}
\begin{document}
\begin{tikzpicture}[x=3in,y=2in]
\tikzstyle{ann} = [draw=none,fill=none,right]
\matrix[nodes={draw, thick, fill=none},
row sep=0.3cm,column sep=0.5cm] {
\node[draw=none,fill=none,label={below:$\mathbf{u}$}] (N1) {\usebox\myboxa}; &
\node[rectangle, minimum height=0.5in, minimum width = 0.75in] (N2) {Center}; &
\node[draw=none,label={below:$\mathbf{y}$}] (N3) {\usebox\myboxb};\\
};
\draw [connector] (N1) -- node {} (N2);
\draw [connector] (N2) -- node {} (N3);
\end{tikzpicture}
\end{document} | 839 | 2,491 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.72009 |
https://www.topperlearning.com/answer/9-m-3-2-9-4/5euu9pp | 1,675,243,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00095.warc.gz | 1,042,932,271 | 57,609 | Request a call back
9^m÷3^-2=9^4
9^m÷3^-2=9^4
Here we can re-write the terms of the Eqn as,
32m ÷ 3-2 = 32x4
Apply laws of exponents
32m - (-2) = 3
We will compare the exponents as bases are same.
2m + 2 = 8
2m + 2 - 2 = 8 - 2
2m = 6
m = 3
Answered by | 01 May, 2020, 03:06: PM
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https://www.nodia.co.in/gate-digital-logic.html | 1,511,074,159,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805417.47/warc/CC-MAIN-20171119061756-20171119081756-00272.warc.gz | 857,946,593 | 19,583 | -16%
# GATE Digital Logic
Product Code: CS-3
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PUBLISHED FOR GATE 2016
Edition 1st Publisher NODIA Pages 632 Binding Paper Back Language English
## Write a review
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SALIENT FEATURES
• Brief Theory
• Problem Solving Methodology
• Fundamental Concepts & Formulae Review
• Vast Question book with Full Solutions
• Multiple Choice Questions, Memory Based Questions and Numerical Types Questions
• Full width coverage of GATE Syllabus
• Well explained and error free solutions
CHAPTER 1 NUMBER REPRESENTATION
1.1 INTRODUCTION
1.2 ANALOG AND DIGITAL SYSTEMS
1.2.2 Limitations of Digital System
1.3 NUMBER SYSTEMS
1.3.1 Decimal Number System
1.3.2 Binary Number System
1.3.3 Octal Number System
1.4 NUMBER SYSTEM CONVERSION
1.4.1 Decimal-to-Binary Conversion
1.4.2 Decimal-to-Octal Conversion
1.4.4 Octal-to-Binary conversion
1.4.5 Binary-to-Octal Conversion
1.5 COMPLEMENTS OF NUMBERS
1.6 NUMBER REPRESENTATION IN BINARY
1.6.1 Sign-Magnitude Representation
1.6.2 1’s Complement Representation
1.6.3 2’s Complement Representation
1.7 FLOATING POINT REPRESENTATION
1.7.1 A Simple Model
1.7.2 IEEE Standard For Binary Floating Point Representation
CHAPTER 2 COMPUTER ARITHMETIC
2.1 BASIC BINARY ARITHMETIC
2.1.2 Binary Subtraction
2.1.3 Binary Multiplication
2.1.4 Binary Division
2.2 COMPLEMENT BINARY ARITHMETIC
2.2.2 Subtraction Using 1’s Complement
2.2.4 Subtraction using 2’s Complement
2.3.1 Hexadecimal Arithmetic Using 1’s or 2’s Complements
2.3.2 Hexadecimal Subtraction Using 15’s or 16’s Complements
2.4 OCTAL ARITHMETIC
2.4.1 Octal Arithmetic using 1’s or 2’s Complements
2.4.2 Octal Subtraction using 7’s or 8’s Complements
2.5 DECIMAL ARITHMETIC
2.5.1 Decimal Arithmetic Using 1’s or 2’s Complements
2.5.2 Decimal Subtraction Using 9’s and 10’s Complement
2.6 BINARY ARITHMETIC
2.7 BINARY CODES
2.8 BINARY CODED DECIMAL (BCD) CODE OR 8421 CODE
2.8.1 BCD-to-Binary Conversion
2.8.2 Binary-to-BCD Conversion
2.9 BCD ARITHMETIC
2.9.2 BCD Subtraction
2.10 THE EXCESS-3 CODE
2.11 GRAY CODE
2.11.1 Binary-to-Gray Code Conversion
2.11.2 Gray-to-Binary Code Conversion
2.11.3 Applications of Gray Code
2.12 FLOATING POINT ARITHMETIC
2.12.2 Multiplication and Division
2.12.3 IEEE Standard For Binary Floating Arithmetic
CHAPTER 3 LOGIC FUNCTION
3.1 INTRODUCTION
3.2 BOOLEAN ALGEBRA
3.2.1 Logic Levels
3.2.2 Truth Table
3.3 BASIC BOOLEAN OPERATIONS
3.3.2 Boolean Multiplication (Logical AND)
3.3.3 Logical NOT
3.4 THEOREMS OF BOOLEAN ALGEBRA
3.4.1 Complementation Laws
3.4.2 AND Laws
3.4.3 OR Laws
3.4.4 Commutative Laws
3.4.5 Associative Laws
3.4.6 Distributive Law
3.4.7 Redundant Literal Rule
3.4.8 Idempotent Law
3.4.9 Absorption Law
3.4.10 Consensus Theorem
3.4.11 Transposition Theorem
3.4.12 De Morgan’s Theorem
3.4.13 Shannon’s Expansion Theorem
3.5 SIMPLIFICATION OF BOOLEAN EXPRESSIONS USING BOOLEAN ALGEBRA
3.5.1 Complement of Boolean Function
3.5.2 Principal of Duality
3.5.3 Relation Between Complement and Dual
3.6 LOGIC GATES
3.6.1 Logic Levels
3.6.2 Types of Logic Gates
3.7 UNIVERSAL GATE
3.7.1 NAND Gate as a Universal Gate
3.7.2 NOR Gate as a Universal Gate
3.8 ALTERNATE LOGIC-GATE REPRESENTATIONS
3.9 BOOLEAN ANALYSIS OF LOGIC CIRCUITS
3.9.1 Converting Boolean Expressions to Logic Diagram
3.9.2 Converting Logic to Boolean Expressions
3.10 CONVERTING LOGIC DIAGRAMS TO NAND / NOR LOGIC
3.10.1 NAND-NAND Logic
3.10.2 NOR-NOR Logic
CHAPTER 4 MINIMIZATION OF LOGIC FUNCTION
4.1 INTRODUCTION
4.2 REPRESENTATION FOR BOOLEAN FUNCTIONS
4.2.1 Sum-of-Products (SOP)
4.2.2 Product-of-Sum (POS)
4.3 STANDARD OR CANONICAL SUM-OF-PRODUCTS (SOP) FORM
4.3.1 Minterm
4.3.2 Σ Notation
4.3.3 Converting SOP Form to Standard SOP Form
4.4 STANDARD OR CANONICAL PRODUCT-OF-SUMS (POS) FORM
4.4.1 Maxterm
4.4.2 Î Notation
4.4.3 Converting POS Form to standard POS Form
4.5 CONVERTING STANDARD SOP FORM TO STANDARD POS FORM
4.6 BOOLEAN EXPRESSIONS AND TRUTH TABLES
4.7 CALCULATION OF TOTAL GATE INPUTS USING SOP AND POS
4.8 KARNAUGH MAP (K-MAP)
4.8.1 Structure of K-map
4.8.2 Another Structure of K-map
4.9 PLOTTING A K-MAP
4.9.1 Plotting Standard SOP on K-map
4.9.2 Plotting Standard POS on K-map
4.9.3 Plotting a Truth Table on K-map
4.10 GROUPING OF CELLS FOR SIMPLIFICATION
4.10.1 Grouping of Two adjacent Cells (Pair)
4.10.3 Grouping of Eight Adjacent Cells (Octet)
4.10.4 Redundant Group
4.11 MINIMIZATION OF SOP EXPRESSIONS
4.12 MINIMIZATION OF POS EXPRESSIONS
4.13 CONVERTING SOP TO POS AND VICE-VERSA
4.14 DON’T CARE CONDITIONS
4.14.1 K-map Simplification With Don’t Care Conditions
4.14.2 Conversion of Standard SOP to Standard POS with Don’t Care Conditions
4.15 K-MAPS FOR MULTI-OUTPUT FUNCTIONS
4.16 LIMITATIONS OF K-MAP
CHAPTER 5 DESIGN AND SYNTHESIS OF COMBINATIONAL CIRCUIT
5.1 INTRODUCTION
5.2 DESIGN PROCEDURE FOR COMBINATION LOGIC CIRCUITS
5.4 SUBTRACTORS
5.4.1 Half-Subtractor
5.4.2 Full-Subtractor
5.6.1 Carry Generation
5.6.2 Carry Propagation
5.8 COMPARATOR
5.8.1 1-bit Magnitude Comparator
5.8.2 2-bit Magnitude Comparator
5.9 MULTIPLEXER
5.9.1 2-to-1 Multiplexer
5.9.2 4-to-1 Multiplexer
5.9.3 Implementation of Higher Order Multiplexers using Lower Order Multiplexers
5.9.4 Applications of Multiplexers
5.10 DEMULTIPLEXER
5.10.1 1-to-2 Demultiplexer
5.10.2 1-to-8 Demultiplexer
5.10.3 Applications of Demultiplexers
5.10.4 Comparison between Multiplexer and Demultiplexer
5.11 DECODER
5.11.1 2-to-4 Line Decoder
5.11.2 Applications of Decoder
5.12 ENCODERS
5.12.1 Octal-to-Binary Encoder
5.12.2 Decimal-to-BCD Encoder
5.13 PRIORITY ENCODERS
5.14 CODE CONVERTERS
5.15 PARITY GENERATOR
5.15.1 Even Parity Generator
5.15.2 Odd Parity Generator
CHAPTER 6 DESIGN AND SYNTHESIS OF SEQUENTIAL CIRCUIT
6.1 INTRODUCTION
6.2 SEQUENTIAL LOGIC CIRCUITS
6.3 LATCHES AND FLIP-FLOPS
6.3.1 General Block Diagram of a Latch or Flip-flop
6.3.2 Difference between Latches and Flip-flops
6.4 S-R LATCH
6.4.1 S -R Latch using NOR Gates
6.4.2 S -R Latch using NAND Gates
6.5 FLIP-FLOPS
6.5.1 S-R Flip-Flop
6.5.2 D-Flip Flop
6.5.3 J-K Flip-Flop
6.5.4 T Flip-Flop
6.6 TRIGGERING OF FLIP-FLOPS
6.6.1 Level Triggering
6.6.2 Edge Triggering
6.6.3 Edge Triggered S-R Flip Flop
6.6.4 Edge Triggered D Flip-Flop
6.6.5 Edge Triggered J-K Flip-Flop
6.6.6 Edge Triggered T-Flip-Flop
6.7 OPERATING CHARACTERISTIC OF FLIP-FLOPS
6.8 APPLICATION OF FLIP-FLOPS
6.9 REGISTER
6.9.1 Buffer Register
6.9.2 Shift Register
6.9.3 Applications of Shift Registers
6.10 COUNTER
6.10.1 Asynchronous and Synchronous Counter
6.10.2 Up-Counter and Down-Counter
6.10.3 MOD Number or Modulus of a Counter
6.11 SHIFT REGISTER COUNTERS
6.11.1 Ring Counter
6.11.2 Johnson Counter
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Copyright to : Nodia & Company All reversed | 2,480 | 7,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-47 | latest | en | 0.495535 |
https://ir.nctu.edu.tw/handle/11536/44336 | 1,670,191,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00457.warc.gz | 344,740,620 | 9,792 | 標題: 應用於三維可程式化閘陣列之熱感知擺放演算法Thermal-Aware Placement for 3D FPGAs 作者: 許蜜祐Hsu, Mi-yu黃俊達Huang, Juinn-Dar電子研究所 關鍵字: 擺放演算法;可程式化閘陣列;熱感知;placement;FPGA;thermal 公開日期: 2010 摘要: 對於繼續遵循摩爾定律(Moore’s Law)來說三維積體電路是一個有吸引力的方法。然而在其研究中溫度問題是一個關鍵的挑戰。精確的溫度分析十分耗時，因此很難整合進擺放階段(placement)做分析。在三維積體電路研究中，有一個趨勢著重在可程式化閘陣列(FPGA)上，因為其可以同時達到複雜電路設計及上市時機縮短兩個目的，同樣的，在三維可程式化閘陣列中溫度問題也是重要的。因此，這篇論文提出兩個熱感知擺放演算法，即標準差法(Standard Deviation)和踩地雷法(Minesweeper)，皆以分散區塊分布來降低熱點(hotspot)的產生。標準差法是先去計算晶片上不同區域的使用率，再去降低數值之間的標準差，踩地雷的方法則是減少每塊區塊周圍的擁擠程度，使其均勻分布。實驗結果顯示，在可接受的線長與延遲增加下，兩方法平均可以降低9%的最高溫度、81%的溫度標準差和67%的最大溫度梯度，更甚之，踩地雷法只多增加3.49%時間就可達到目的。我們方法可有效把溫度問題整合入擺放階段同時線長跟延遲結果一樣好。Three-dimensional (3D) integration is an attractive way to continue sustaining Moore’s Law; however, it has a critical challenge – the thermal issue. Precise thermal analysis is time-consuming and thus it is impractical to be integrated into the placement process directly for the exploding problem size in 3D technology. In 3D ICs, one of the current trends is employing field programmable gate arrays (FPGAs) because 3D FPGAs can both integrate complex circuit designs and speed up time-to-market. Since 3D FPGAs are a type of 3D ICs, thermal issue is also important for them. In this thesis, two thermal-aware placement methods for 3D FPGAs are proposed – Standard Deviation (SD) and Minesweeper (MS), which are devoted to disperse block distribution to avoid hotspots. SD utilizes the concept that minimizes the standard deviation of utilization for different parts on the chip; the idea of MS comes from minesweeper, which is to reduce the congestion of neighbors for every block. The experimental results show that improve more than 9% in maximum temperature, 81% in temperature deviation and 67% in maximum temperature gradient compared to thermal-unaware placement method with acceptable extra wire length and delay. Moreover, MS takes only 3.49% runtime overhead due to its simplified update steps. These two methods integrate efficiently thermal behavior into placement process while keeping the quality of the results good enough. URI: http://140.113.39.130/cdrfb3/record/nctu/#GT079711634http://hdl.handle.net/11536/44336 Appears in Collections: Thesis
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If it is a zip file, please download the file and unzip it, then open index.html in a browser to view the full text content. | 853 | 2,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.538827 |
https://www.physicsforums.com/threads/need-help-with-3d-relative-velocity-h-w-problem.5860/ | 1,670,495,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00095.warc.gz | 980,136,927 | 15,049 | # Need help with 3d relative velocity h/w problem
dk702
Here the question,
A swimmer wants to cross a river, from point A to point B. The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 150 m, and the speed v_r of the current in the river is 5 km/h. Suppose that the swimmer makes an angle of theta=45 deg. (0.785 radians) with respect to the line from A to C.
To swim directly from A to B, what speed v_s, relative to the water, should the swimmer have?
I think this is the eq. i need. velocity vector(swimmer with respect(wrt) to earth)= velocity vector(swimmer wrt river)+ velocity vector(river wrt earth)
my values so far, that I know are wrong
v_(s/e)=9km/hr
v_(r/e = 5km/hr
v_(s/r)= 7.5
If the swimmer wanted to swim from A to C i could solve it. It is that he wants to swim from A to B that is giving trouble.
To give you an idea of my skill level:
I can solve a problem such as this;
The compass on an airplane indicates it is heading due north and its airspeed indicator shows that it is moving through the air at 240km/h. If there is a wind of 100km/h from the west, what is the velocity of the plane relative to the earth? What direction should the pilot head to travel due north and what will be his velocity relative to the Earth then?
v_(p/e)=260km/h at 23 deg E of N
and v_(p/e) = 218km/h at 25 deg W of N
Last edited:
Homework Helper
Where is C in relation to A and B??
You say that "The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 150 m " but that doesn't help unless we know the angles formed. Is C on the line from A to B? That would be the easiest case: Then just ignore C and take the distance from A to B to be 350 m but surely it's not that simple.
NJones
Originally posted by HallsofIvy
Where is C in relation to A and B??
You say that "The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 150 m " but that doesn't help unless we know the angles formed. Is C on the line from A to B? That would be the easiest case: Then just ignore C and take the distance from A to B to be 350 m but surely it's not that simple.
i think he could be talking of a triangle...I should know how to do this... i'll need to find my maths brain... | 628 | 2,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.971377 |
https://www.fea-solutions.co.uk/elastic-modulus/ | 1,638,405,424,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00109.warc.gz | 812,276,773 | 10,280 | # Elastic Modulus
## 31 May 2019 Elastic Modulus
In Hooke’s Law (https://www.fea-solutions.co.uk/law-of-elasticity/), the Spring Constant k describes the relation between an applied load and the resulting deformation of a body.
It depends on both the shape (geometry) of the part as well as its material.
To have a constant that is just depending on the material but not on the geometry, Moduli of Elasticity have been defined. Depending on the direction of the applied load (and of the resulting stresses), a number of different elastic moduli are used, e.g. Shear Modulus (Modulus of Rigidity), Bending Modulus (Flexural Modulus) and Bulk Modulus.
The most important one is however the Young’s Modulus, because it is derived from Tensile Test results (https://www.fea-solutions.co.uk/tensile-tests/), which are the most common tests to determine mechanical material properties.
The Young’s Modulus is abbreviated with the letter E and is defined as the ratio between Stress and Strain. It is only valid in the linear-elastic region of the stress-strain curve. (https://www.fea-solutions.co.uk/elasticity-and-plasticity/)
Being the ratio between Stress and Strain means that the Young’s Modulus is the slope of the stress-strain curve (https://www.fea-solutions.co.uk/stress-strain-curves/).
Stiff materials have high Elastic Moduli, whereas flexible materials have low Elastic Moduli.
Typical values for the Young’s Modulus of common engineering materials are:
Steel 200,000MPa
Titanium 110,000MPa
Aluminium 70,000MPa
Plastics 3,000MPa (very much dependent on type of plastic)
It should be noted that although all steel grades for example have very similar Young’s Moduli, their Strength (e.g. Yield Limit) might be very different. However, in the most common FEA application, which is linear-elastic stress analysis, only the Young’s Modulus is an input value, but the Strength is not. Hence, even if is later decided to change a material grade, the FEA results will still be valid (as long as the resulting stresses are below the Yield Strength).
If you have further questions, please call us today on 01202 798991. | 504 | 2,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.933718 |
https://i-alpha.it/c5h10-molar-mass.html | 1,618,330,223,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00635.warc.gz | 417,254,919 | 11,414 | # C5h10 molar mass
• Formula in Hill system is C5H10: Computing molar mass (molar weight) To calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use: Any chemical element. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al.
• Molar Mass Calculations: To begin our calculations, we must determine the molar mass of the hydrocarbon by using the given conditions of pressure, temperature, and the density of the gas.
• Pentenes are alkenes with chemical formula C 5 H 10.Each contains one double bond within its molecular structure. There are a total of six different compounds in this class, differing from each other by whether the carbon atoms are attached linearly or in a branched structure, and whether the double bond has a cis or trans form.
• Molar Mass of He = 4g/mol . Molar Mass of C5H10 = (12X5) + (10X1) = 70g/mol. Molar Mass of H2 = 2x1 = 2g/mol. Molar Mass of PH3 = 31 + (3x1) = 34g/mol. Now, we can rank the substance beginning from the highest to the lowest rate of effusion as follow: Substance >> Molar Mass >> Rank.
• Our compound has molar mass of 56 g. Now let’s have a look at possible molecular formulas: Molecular formula Na2O Na4O2 Na6O3 Na8O4 and so on. Molar mass 56 116 168 224 etc. Examining the molar masses of the possible formulas we see that the first one, shown in bold, has molar mass identical to the molar mass of our compound determined by ...
• molar mass in grams = 1 mole (add up molar mass from periodic table) Find the mass of 4.50moles of diphosphorus pentoxide. _____ 4.5 moles 142 g = 639 g ... (C2H4) and (C5H10) TRY: Determine the empirical formula of each compound below. a. P4O10 b. C6H12O6 c. C3H6O P2O5 CH2O C3H6O. The empirical formula of a compound is CH2O. Its molar mass is ...
• Calculations: Formula: C5H10O5 Molar Mass: 150.129 g/mol 1g=6.66093825976327E-03 mol Percent composition (by mass): Element Count Atom Mass %(by mass)
• Jan 24, 2020 · The molecular mass is the sum of the mass contributions of each element. Simply add each mass contribution together to find the total. Molecular mass of K 3 Fe(CN) 6 = 117.30 g/mol + 55.85 g/mol + 72.06 g/mol + 84.06 g/mol Molecular mass of K 3 Fe(CN) 6 = 329.27 g/mol
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• Molar Mass of C5H10 Oxidation State of C5H10. Chlorek Cynku - ZnCl 2. Molar Mass of ZnCl2 Oxidation State of ZnCl2. Oblicz Reakcję Stechiometryczną Oblicz ...
• Molar Mass of C5H10 Oxidation State of C5H10. Chlorek Cynku - ZnCl 2. Molar Mass of ZnCl2 Oxidation State of ZnCl2. Oblicz Reakcję Stechiometryczną Oblicz ...
• 5. A large flask is evacuated and found to weigh 134.567 g. It is then filled to a pressure of 735 torrs at 31 degrees Celsius with a gas of unknown molar mass and then reweighed. Its new mass is 137.456 g. The flask is then emptied and filled with water at 31 degrees C. Its mass is now 1067.91 grams. What is the molar mass of this unknown gas?
• 尿素 (ch4n2o)的摩尔质量和分子量为{1，数字}。
• Formula in Hill system is C5H10: Computing molar mass (molar weight) To calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use: Any chemical element. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al.
• The molar mass and molecular weight of C5H10 is 70.1329.
• mical equation, calculate the grams of C5 2240.0 kJ of heat. Select menu items in ti Cion using dimensional analysis then selec Ficant figures. 2(9) 419 g C5H10 answer not shown 0.0105 g C5H10 46.97 g C5H10 23.49 g C5H10 -419 g C5H10 47 g C5H10 Il 46.97 g C5H10
• Formula: C5H10. Chemical Safety: Flammable. Molar Mass: 70.13 g/mol. Flash Point: -37°C. Lower Explosive Limit(LEL): 1.4%VOL. Upper Explosive Limit(UEL: -General Description: A clear colorless liquid with a petroleum-like odor. Flash point of -35°F. Less dense than water and insoluble in water. Vapors are heavier than air.
• Nov 14, 2012 · Thanks so much, I would of never figured out that out myself past the empirical formula since I forgot about the 22.4L factor. The molar mass turns out to be 70, and since the only answer I was given that adds up to 70 is C5H10, it's an easy pick.
• The empirical formula of propylene is CH3. An experimental determination of the molar mass of propylene by a student yields the value of 42 g/mol. What is the molecular formula of propylene? A) CH3 B) C3H6 C) C3H8 D) C5H10 E) C6H9 21. An atom of an element weighs 6.28 10-23 g. What is the atomic mass of this element in atomic mass units? 22.
Mec 9000gn 28 gauge reloader5. A large flask is evacuated and found to weigh 134.567 g. It is then filled to a pressure of 735 torrs at 31 degrees Celsius with a gas of unknown molar mass and then reweighed. Its new mass is 137.456 g. The flask is then emptied and filled with water at 31 degrees C. Its mass is now 1067.91 grams. What is the molar mass of this unknown gas? Name the above two compounds MOLAR MASS Compute the molar mass for the following: 1) Sulfur (32.07 g/mol) 2) Nitrogen gas (20.02 g/mol) 3) Calcium bromide (199.88 g/mol) 4) Nickel (II) nitrate (182.71 g/mol) PERCENT COMPOSITION Calculate the percent composition for each of the following substances: 1) Calcium bromide (%Ca = 20.05%, % Br= 79.95% ... The molar mass of CO2 = 12.0111 + 2(15.9994) = 44.0099 g Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L You can use this to calculate the density of a gas in g/Liter at STP.
Calculations: Formula: C5H10O5 Molar Mass: 150.129 g/mol 1g=6.66093825976327E-03 mol Percent composition (by mass): Element Count Atom Mass %(by mass)
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• Mole Concept - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free. mole concept 2-Methyl-2-butene, 2m2b, 2-methylbut-2-ene, also amylene is an alkene hydrocarbon with the molecular formula C 5 H 10.
• Molar Mass Calculations: To begin our calculations, we must determine the molar mass of the hydrocarbon by using the given conditions of pressure, temperature, and the density of the gas.
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The molar mass of a compound called cyclopentane is 70.0 g. One of the chemical formulas shown below is the correct formula for cyclopentane. Which one is it? Dec 31, 2019 · The molar mass of NH3 is 17.0 g. Thus, there is 1 mole of NH3 present, which contains 1 mole of N. The molar mass of pyridine is 5mol C 12.0g C + 5mol H 1.01g H +14.0g N = 79.1g/mol. Because each mole of pyridine contains 1 mole of N, we need slightly more than 2 moles of pyridine to have more N than is present in the 2N O.
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Formula: C5H10. Chemical Safety: Flammable. Molar Mass: 70.13 g/mol. Flash Point: -37°C. Lower Explosive Limit(LEL): 1.4%VOL. Upper Explosive Limit(UEL: -General Description: A clear colorless liquid with a petroleum-like odor. Flash point of -35°F. Less dense than water and insoluble in water. Vapors are heavier than air.
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b. A compound with a molecular mass of 46.0g/mol and an empirical formula of NO2. Molecular Formula Worksheet. Write the molecular formulas of the following compounds: 1) A compound with an empirical formula of C2OH4 and a molar mass of 88 grams per mole. 2) A compound with an empirical formula of C4H4O and a molar mass of 136 grams per mole.
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The molar mass and molecular weight of C5H10 is 70.1329.
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The molar mass of CO2 = 12.0111 + 2(15.9994) = 44.0099 g Density = mass/volume = 44.0099 g/22.4 L = 1.96 g/L Chemistry SOL Review— Molar Relationships The Mole and Mole Calculations At Standard Temperature and Pressure (STP) 1 mole of gas = 22.4 L You can use this to calculate the density of a gas in g/Liter at STP.
• The molecular formula is a whole number multiple of the empirical formula. This means that the molecular mass must also be a whole number multiple of the empirical formula mass.
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• ›› C5H10 molecular weight. Molar mass of C5H10 = 70.1329 g/mol. Convert grams C5H10 to moles or moles C5H10 to grams. Molecular weight calculation: 12.0107*5 + 1.00794*10 ›› Percent composition by element
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• molar mass in grams per mole "Name" common English name: Basic physical properties include: "BoilingPoint" ... Find stereoisomers for the formula "C5H10":
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• Seimbangkan C5H10 + KMnO4 + H2O = C5H9OH + MnO2 + KOH persamaan atau tindakbalas kimia menggunakan kalkulator ini!
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• Formula: C5H10. Chemical Safety: Flammable. Molar Mass: 70.13 g/mol. Flash Point: -37°C. Lower Explosive Limit(LEL): 1.4%VOL. Upper Explosive Limit(UEL: -General Description: A clear colorless liquid with a petroleum-like odor. Flash point of -35°F. Less dense than water and insoluble in water. Vapors are heavier than air.
Mozilla firefox pc 32 bit | 2,824 | 9,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-17 | latest | en | 0.835838 |
https://www.emmamichaels.com/7997/3-divided-by-18.html | 1,679,443,519,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00282.warc.gz | 850,706,668 | 14,481 | Breaking News
# 3 Divided By 18
3 Divided By 18. 18√3 3 18 3 3 cancel the. 18 √3 ⋅ √3 √3 18 3 ⋅ 3 3 combine and simplify the denominator. The dividend and divisor of 18 divided by 3. 3/18 divided by 3/5 is 5/18.
Simplify 18/ ( square root of 3) 18 √3 18 3 multiply 18 √3 18 3 by √3 √3 3 3. Enter the expression you want to evaluate. You can also add, subtraction, multiply, and.
## 18√3 3 18 3 3 cancel the.
11 rows for calculation, here's how to convert 3 divided by 18 using the formula above, step by step. 18 √3 ⋅ √3 √3 18 3 ⋅ 3 3 combine and simplify the denominator. Add your answer and earn points.
## 11 Rows For Calculation, Here's How To Convert 3 Divided By 18 Using The Formula Above, Step By Step.
The divisor (3) goes into the first digit of the dividend (1), 0 time (s). This calculator shows all the work and steps for long.
### Start By Setting It Up With The Divisor 3 On The Left Side And The Dividend 18 On The Right Side Like This:
3 x = 18 first, multiply both sides by x to get rid of the x as denominator.
### Kesimpulan dari 3 Divided By 18.
3/18 = 0.16 recurring (that is, 0.16666.) this answer is: Start by setting the divisor 18 on the left side and the dividend 3000 on the right: 11 rows for calculation, here's how to convert 3 divided by 18 using the formula above, step by step. 3 18 = 18 (x) 18 so, our final answer here to 3 divided by what equals 18 is: | 448 | 1,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-14 | latest | en | 0.824946 |
https://www.thestudentroom.co.uk/showthread.php?t=2675598 | 1,610,981,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514796.13/warc/CC-MAIN-20210118123320-20210118153320-00020.warc.gz | 1,018,759,850 | 40,117 | # Mass and Volume = Density
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#1
So I know that Mass divided by Volume = Density
But, how do I get my answer? Am I using a calculator?
650g divided by 922cm^3 =
0
6 years ago
#2
Turn 650g into kg. Turn 922cm^3 into meters. Then divide.
cm = 10^-2
g = 10^-3
0
6 years ago
#3
Yes, using a calculator would probably help.
0
#4
how can I turn the last one into metres?
0
6 years ago
#5
Imagine a box in the shape of a cube of length 1 m
It's volume would be 1 x 1 x 1 = 1 m^3
1 m equal 100 cm
So the same volume would be 100 x 100 x 100 cm^3
To convert your volume to m^3 divide it by 1000000
Although the standard units for density are kg/m^3 There is probably no need to change the units of your calculation
You only need to do that if the units in the information don't match the units of the required answer.
Density = mass / volume
Use a calculator
Does your question specify what density units you are required to use? Or are you doing a course of study that requires all units to be SI units?
Posted from TSR Mobile
0
#6
[QUOTE=gdunne42;47528424]Imagine a box in the shape of a cube of length 1 m
It's volume would be 1 x 1 x 1 = 1 m^3
1 m equal 100 cm
So the same volume would be 100 x 100 x 100 cm^3
To convert your volume to m^3 divide it by 1000000
Although the standard units for density are kg/m^3 There is probably no need to change the units of your calculation
You only need to do that if the units in the information don't match the units of the required answer.
Density = mass / volume
Use a calculator
the answer I got for the above question is =
1.22107703354
is this correct?
0
6 years ago
#7
(Original post by fliss1992)
the answer I got for the above question is =
1.22107703354
is this correct?
You need to state your units anyway.
What did you type into your calculator to get that number?
0
#8
650g divided by 922cm^3 = 1.22107703354g/cm^3
that's what I calculated and that's the answer I got above..
0
6 years ago
#9
No you didn't
Posted from TSR Mobile
0
6 years ago
#10
(Original post by fliss1992)
650g divided by 922cm^3 = 1.22107703354g/cm^3
that's what I calculated and that's the answer I got above..
The units are ok now.
I'm pretty sure you are not a troll so you must just be out of your depth. May I ask what grade you got in GCSE maths? I apologise if I have asked you this before.
0
#11
I haven't done GCSE Maths, I am studying at the moment and haven't done much on anything covered in the GCSE, so it's all new.
You say the units are ok now, is the answer correct?
As the above person doesn't agree?
0
6 years ago
#12
Density is measured in or Kg per metre cubed.
So divide the volume by 1000 to get metres cubed.
Divide grams by 1000 to get kilograms.
then use the formula mass/volume=density
0
6 years ago
#13
(Original post by fliss1992)
650g divided by 922cm^3 = 1.22107703354g/cm^3
that's what I calculated and that's the answer I got above..
a number divided by a number greater than it, is always lesser than "1".
the correct answer is something like 0.7xxxxxx
0
#14
(Original post by SamTheMan95)
Density is measured in or Kg per metre cubed.
So divide the volume by 1000 to get metres cubed.
Divide grams by 1000 to get kilograms.
then use the formula mass/volume=density
Now I have 0.70498915401 g/cm^3
0
6 years ago
#15
(Original post by SamTheMan95)
Density is measured in or Kg per metre cubed.
So divide the volume by 1000 to get metres cubed.
Divide grams by 1000 to get kilograms.
then use the formula mass/volume=density
Firstly, although the SI Unit of density is kg/m^3
- g/cm^3 would be an acceptable unit in most circumstances
- dividing cm^3 by 1000 will not convert to m^3
Posted from TSR Mobile
0
#16
I don't mean to be rude to anyone, and I do appreciate everyone's help but can someone please advise me on the right way to calculate this?
I followed the way Samtheman mentioned however nobody has said if my answer is correct or not?
0
6 years ago
#17
(Original post by fliss1992)
Now I have 0.70498915401 g/cm^3
Correct
In most situations you would then round that answer to a suitable degree of accuracy but that's another topic for another time.
Posted from TSR Mobile
0
#18
OK, that's great thanks for your time and patience.
If you don't mind, I have a few more questions that are similar.
Can I post my answers and then you say if there right or wrong?
0
6 years ago
#19
(Original post by fliss1992)
OK, that's great thanks for your time and patience.
If you don't mind, I have a few more questions that are similar.
Can I post my answers and then you say if there right or wrong?
You can but my advice remains the same. The order in which you are tackling these topics is not helping you. You need to master the basics first and then questions like this will start to make more sense. If you send me a private message with your postal address I will send you a MathsWatch CD. Work through this in grade order and you will start to improve.
0
#20
Here are my next questions if someone could tell me if they are correct or not, if not can someone advise where I have gone wrong.
Thanks..
a) Weight 150g Volume 840cm^3 = 0.17857142857 g/cm^3
b) Weight 1kg Volume 125cm^3 = 8 g/cm^3
c) Weight 2kg Volume 1700 ^3 = 1.17647058824 g/cm^3
d) Weight 200g Volume 10cm^3 = 20 g/cm^3
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Wind Turbine: Power | Nominal Power | Yield | Annual Yield | Rotation | Magnitude | Distance | Repowering
# Magnitude Wind Turbine
Calculates the dimensions of a wind turbine. Please enter radius or diameter and hub height or total height. The other values will be calculated.
The hub is at the center of the rotor. Radius is the length of one blade from the tip to the hub. Diameter is twice the radius. The hub height is the height from the ground to the hub. The total height is reached by the wind turbine, if one blade points vertically up. Rotor above ground measures the distance from the ground to the tip of a blade pointing vertically down. The swept area is the area of the circle described by the blades at the rotation.
Radius: centimetersfeetyardsmeters Diameter: centimetersfeetyardsmeters Hub height: centimetersfeetyardsmeters Total height: centimetersfeetyardsmeters Rotor above ground: centimetersfeetyardsmeters Swept area: square feetsquare yardssquare meters
Example: a wind turbine with a radius of 46 meters has a diameter of 92 meters. At a hub height of 70 meters, the total height is 116 meters and the height of the rotor above the ground is 24 meters. The swept area is 6648 square meters.
In the picture above you can see a smaller wind turbine in agricultural land. It stands on a tiny hill in central Bavaria.
Wind power has long been used to generate energy. First there were windmills - probably for more than 4000 years - that converted the wind into mechanical energy in order to roll the millstone over the milled material. Replacing the mechanical gearbox for the millstone with a generator that converts the rotation of the rotor blades into electricity was obvious; people have been experimenting with it since the end of the 19th century. Since then, wind turbines have become increasingly larger, as the heavier loads associated with larger wind turbines can be handled using better materials. The relevant measure of the performance of wind turbines is the swept area through which the wind drives the rotor blades. Since areas are square units of measurement, a rotor blade twice as long means four times the area, one three times as long means nine times the area, and so on. Therefore, new size records for wind turbines will probably continue to be set for some time to come, as long as the available materials allow it.
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Anzeige | 565 | 2,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.905435 |
https://math.stackexchange.com/questions/450412/modular-form-petersson-inner-product | 1,569,148,451,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575484.57/warc/CC-MAIN-20190922094320-20190922120320-00288.warc.gz | 556,873,601 | 30,490 | # modular form -Petersson inner product
my question is about Petersson inner product.
i need to prove that $(E_k,f) =0$ $\forall f \in S_k(SL_2(\mathbb{Z}))$
the only thing that i think that should help me is that the space of cusp form has a basis of Hecke eigenform.
i cant use in the Hermitian property of the Hecke operator (i dont think that help) because is work only in the case both function are cusp but $E_k$ isnt a cusp form.
i know that $M_k(SL_2(\mathbb{Z})=<E_k> \oplus S_k(SL_2(\mathbb{Z}))$
thank
here is the prove :
Lemma 1 : $E_k$ is eigenfunction $\forall \space T_n$
Lemma 2 : $(T_nE_k,g) = (E_k,T_ng)$ $\space \forall g\in S_k$
Lemma 3: there is a basis $\{f_1,...f_m\}$ of $S_k$ such that is simulation eigenfunction $\forall T_n$
please note that $\{f1 ,... f_m,E_k\}$ is basis for $M_k$ and by multiplication we can assusme that$a_1(f'_1)=...=a_1(f'_m)=a_1(E'_k)=1$.
so it will be enough to show that $(f'_i,E_k)= 0$ $\space 0 < i <m+1$
now we know that if $a_1(f)=1 => T_n(f) =a_n(f)f.$
so if there exist n such that $a_n(f'_i) \neq a_n(E'_k)$ we finish.
assume not then $\forall n>0 \space a_n(f'_i) = a_n(E'_k)$ then $g=f'_i-E_k \in M_k$ however when we look at fourier series we see that g=1. when we know the only constant function in $M_k$ are zero. contradiction
i think it should work what do you think? | 459 | 1,343 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-39 | latest | en | 0.801384 |
https://studylib.net/doc/5712849/document | 1,611,490,172,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00603.warc.gz | 577,248,166 | 12,032 | # Document
```Electromagnetic Induction Ch. 29
Induction experiments
Lenz’s law
Motional electromotive force
Induced electric fields
Displacement Current
C 2009 J. Becker
(sec. 29.1)
(sec. 29.2)
(sec. 29.3)
(sec. 29.4)
(sec. 29.5)
(sec. 29.7)
Current induced in a coil.
When B is constant and
shape, location, and
orientation of coil does
not change, the induced
current is zero.
Conducting loop in increasing B field.
Magnetic flux through an area.
Lenz’s Law:
The induced emf or
current always tends
to oppose or cancel
the change that
caused it.
Lenz’s law
Faraday’s Law of Induction
How electric generators, credit card readers, and
transformers work.
A changing magnetic flux causes
(induces) an emf in a conducting loop.
C 2004 Pearson Education / Addison Wesley
Changing magnetic flux through a wire loop.
f = 90o
Alternator (AC generator)
f = 90o
DC generator
Slidewire generator
Magnetic force (F = IL x B) due to the induced current
is toward the left, opposite to v.
Lenz’s Law:
The induced emf or
current always
tends to oppose or
cancel the change
that caused it.
Lenz’s law
Currents (I) induced in a wire loop.
Motional induced emf (e):
e=vBL
because the potential
difference between a and b is
e = DV = energy / charge = W/q
e = DV = work / charge
DV = F x distance / q
DV = (q v B) L / q
so
e=vBL
Length and velocity are
perpendicular to B
Solenoid with increasing current I which induces an emf
in the (yellow) wire. An induced current I’ is moved
through the (yellow) wire by an induced electric field E
in the wire.
Eddy currents formed by induced emf in a rotating
metal disk.
Metal detector – an alternating magnetic field Bo
induces eddy currents in a conducting object moved
through the detector. The eddy currents in turn
produce an alternating magnetic field B’ and this field
induces a current in the detector’s receiver coil.
A capacitor being charged by a current ic has a
displacement current equal to iC between the plates,
with displacement current iD = e A dE/dt. This changing
E field can be regarded as the source of the magnetic
field between the plates.
A capacitor being charged by a current iC has a
displacement current equal to iC between the plates,
with
displacement current iD = e A dE/dt
From C = e A / d and DV = E d we can use
q = C V to get
q = (e A / d ) (E d ) = e E A = e F E and
from iC = dq / dt = e A dE / dt = e dF E / dt = iD
We have now seen that a
changing E field can produce a B field,
and from Faraday’s Law, a
changing B field can produce an E field or emf.
C 2009 J. Becker
MAXWELL’S EQUATIONS
The relationships between electric
and magnetic fields and their
sources can be stated compactly in
four equations, called
Maxwell’s equations.
Together they form a complete
basis for the relation of E and B
fields to their sources.
C 2004 Pearson Educational / Addison Wesley
Determine
direction of
induced
current for
a) increasing B
b) decreasing B
Lenz’s law (Exercise 29.16)
Lenz’s law (Exercise 29.17)
Lenz’s law (Exercise 29.18)
Motional emf and Lenz’s law
(Exercise 29.22)
Motional emf and Lenz’s law
(Exercise 29.25)
Review
See www.physics.edu/becker/physics51
C 2009 J. Becker
``` | 898 | 3,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-04 | latest | en | 0.908048 |
https://www.bearnaiserestaurant.com/blog/how-is-a-patent-bet-calculated/ | 1,713,405,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00347.warc.gz | 626,798,906 | 11,884 | # How is a Patent bet calculated?
## How is a Patent bet calculated?
Patent bets are calculated by working out the results of the below bets from the selections:
• 3 x singles.
• 3 x doubles.
• 1 x treble.
How many bets can you bet on a Patent?
seven bets
A Patent comprises seven bets of equal value on three separate events: three singles, three doubles and one treble.
Is a Patent bet worth it?
With a Patent, it is a little different as you can still get a return on your bet even if one of the selections don’t come in. Therefore one key advantage of the Patent bet is that it provides a better chance of a return. It is much like insurance on an accumulator.
### What does Patent mean on a bet?
A Patent bet is a form of multiple method betting wherein seven bets are placed on three different selections. Patent betting is used to cover all potential outcomes and will offer a return even if only one bet comes in.
How many horses are in a Patent bet?
A Patent bet is when you select three horses in three different horse races. You are backing all of them to win, but spreading your bet out to cover all the combinations they make. To summarise you are placing a wager on three horses consisting of 7 bets: 3 singles, 3 doubles and a treble.
What is Trixie and Patent?
What is the difference between a Trixie and a Patent? A Trixie bet is similar to a Patent in that you will be using three selections. The key difference is that the Patent includes single bets on all three selections, meaning it includes a total of seven bets as a result: three singles, three doubles and a treble.
## How much does a Patent cost?
A patent can cost from \$900 for a do-it-yourself application to between \$5,000 and \$10,000+ with the help of patent lawyers. A patent protects an invention and the cost of the process to get the patent will depend on the type of patent (provisional, non-provisional, or utility) and the complexity of the invention.
What happens if you have a non runner in a Patent bet?
Non-runners If there is a non-runner in your Patent, you’ll be refunded for one bet at your original unit stake. Similarly, if you have one winner and one non-runner, you’ll get three bets returned: two singles and a double. In practice, though, your double is effectively another single bet on your winning selection.
What happens to a patent bet with a non-runner?
### What’s the difference between a Trixie and a Patent?
What is a Patent combo?
The patent is a 3 selection wager consisting of 7 bets: 3 singles, 3 doubles and a treble. This is a full cover bet that can be thought of as a trixie with 3 single bets added to each selection. Each choice appears in 1 single, 2 doubles and a treble.
United States Patent and Trademark Office (USPTO) is designed to allow individuals to get a patent themselves without the help of a lawyer. You can write the patent yourself, submit it and pay the filing fees. Is this method free? No.
## How much does a patent cost UK?
The costs for applying for a patent are relatively low. As of 2022: the cost of applications is between £60 and £112.50. the cost of the search is £150- £180 (plus £20 for each claim over 25 claims)
How expensive is a patent?
What is a Patent parlay?
A Patent system consists of 7 bets which transpire from 3 picks. One single bet per pick, three 2-pick multis (parlays) and one 3-pick multi (parlay). At least one of the three picks must be correct to gain some winnings – the exact amount of the winnings depends on how many of the predictions prove correct.
### What is double Trixie and Patent?
Both a Trixie bet and a Patent bet feature three selections, but there is one very important difference between them. While a Trixie consists of four bets, a Patent consists of seven. This is made up of the same three doubles and one treble included in a Trixie, but also incorporates a further three singles.
• October 20, 2022 | 908 | 3,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-18 | latest | en | 0.94428 |
http://www.thefullwiki.org/Egyptian_numerals | 1,563,499,719,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00552.warc.gz | 284,249,242 | 15,822 | # Egyptian numerals: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
Numeral systems by culture
Hindu-Arabic numerals
Eastern Arabic
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The system of Ancient Egyptian numerals was used in Ancient Egypt until the early first millennium AD. It was a decimal system, often rounded off to the higher power, written in hieroglyphs. The hieratic form of numerals stressed an exact finite series notation, ciphered one to one onto the Egyptian alphabet.The Ancient Egyptian system used bases of ten.
## Digits and numbers
The following hieroglyphs were used to denote powers of ten:
Value 1 10 100 1,000 10,000 100,000 1 million, or
infinity
Hieroglyph
or
Description Single stroke Heel bone Coil of rope Water lily
(also called Lotus)
or Frog
Man with both
hands raised
Multiples of these values were expressed by repeating the symbol as many times as needed. For instance, a stone carving from Karnak shows the number 4622 as
Egyptian hieroglyphs could be written in both directions (and even vertically). This example is written left-to-right and top-down; on the original stone carving, it is right-to-left, and the signs are thus reversed.
## Fractions
Rational numbers could also be expressed, but only as sums of unit fractions, i.e., sums of reciprocals of positive integers, except for 2/3 and 3/4. The hieroglyph indicating a fraction looked like a mouth, which meant "part":
Fractions were written with this fractional solidus, i.e., the numerator 1, and the positive denominator below. Thus, 1/3 was written as:
$= \frac{1}{3}$
There were special symbols for 1/2 and for two non-unit fractions, 2/3 (used frequently) and 3/4 (used less frequently):
$= \frac{1}{2}$
$= \frac{2}{3}$
$= \frac{3}{4}$
If the denominator became too large, the "mouth" was just placed over the beginning of the "denominator":
$= \frac{1}{331}$
For plus and minus signs, the hieroglyphs
and
were used: if the feet pointed into the direction of writing, it signified addition, otherwise subtraction. [1 ]
## Written numbers
As with most modern-day languages, the ancient Egyptian language could also write out numerals as words phonetically, just like one can write thirty instead of 30 in English. Thirty, for instance, was written as
while the number 30 was
This was, however, uncommon for most numbers other than one and two and the signs were used most of the time.
## Hieratic numerals
As most administrative and accounting texts were written on papyrus or ostraca, rather than being carved into hard stone (as were hieroglyphic texts), the vast majority of texts employing the Egyptian numeral system utilize the hieratic script. Instances of numerals written in hieratic can be found as far back as the Early Dynastic Period. The Old Kingdom Abusir papyri are a particularly important corpus of texts that utilize hieratic numerals.
Boyer proved 50 years ago that hieratic script used a different numeral system, using individual signs for the numbers 1 to 9, multiples of 10 from 10 to 90, the hundreds from 100 to 900, and the thousands from 1000 to 9000. A large number like 9999 could thus be written with only four signs—combining the signs for 9000, 900, 90, and 9—as opposed to 36 hieroglyphs. Boyer saw the new hieratic numerals as ciphered, mapping one number onto one Egyptian letter for the first time in human history. Greeks adopted the new system, mapping their counting numbers onto two of their alphabets, the Doric and Ionian.
In the oldest hieratic texts the individual numerals were clearly written in a ciphered relationship to the Egyptian alphabet. But during the Old Kingdom a series of standardized writings had developed for sign-groups containing more than one numeral, repeated as Roman numerals practiced. However, repetition of the same numeral for each place-value was not allowed in the hieratic script. As the hieratic writing system developed over time, these sign-groups were further simplified for quick writing; this process continued into Demotic as well.
Two famous mathematical papyri using hieratic script are the Moscow Mathematical Papyrus and the Rhind Mathematical Papyrus.
## Egyptian words for numbers
The following table shows the reconstructed Middle Egyptian forms of the numerals[2] (which are indicated by a preceding asterisk), their transliterated forms in hieroglyphs (indicated between square brackets), and their later Coptic equivalents which give Egyptologists clues as to the vocalism of the original Egyptian numbers. The majuscule letter "A" in some reconstructed forms means that the quality of that vowel remains uncertain:
Egyptian Transliteration English Translation Coptic (Sahidic dialect)
*wiʕyaw [wˁ.w] (masc.)
*wiʕīyat [wˁ.t] (fem.)
one oua (masc.)
ouei (fem.)
*sínway [sn.wy] (masc.)
*síntay [sn.ty] (fem.)
two snau (masc.)
snte (fem.)
*ḫámtaw [ḫmt.w] (masc.)
*ḫámtat [ḫmt.t] (fem.)
three šomnt (masc.)
šomte (fem.)
*yAfdáw [ỉfd.w] ([masc.)
*yAfdát [ỉfd.t] (fem.)
four ftoou (masc.)
ftoe (fem.)
*dīyaw [dỉ.w] (masc.)
*dīyat [dỉ.t] (fem.)
five tiou (masc.)
tie (fem.)
*yAssáw [sỉs.w or ỉs.w (?)] (masc.)
*yAssát [sỉs.t or ỉs.t (?)] (fem.)
six soou (masc.)
soe (fem.)
*sáfḫaw [sfḫ.w] (masc.)
*sáfḫat [sfḫt] (fem.)
seven šašf(masc.)
šašfe (fem.)
*ḫAmānaw [ḫmnw] (masc.)
*ḫAmānat [ḫmnt] (fem.)
eight šmoun (masc.)
šmoune (fem.)
*pAsīḏaw [psḏw] (masc.)
*pAsīḏat [psḏt] (fem.)
nine psis (masc.)
psite (fem.)
*mūḏaw [mḏw] (masc.)
*mūḏat [mḏt] (fem.)
ten mēt (masc.)
mēte (fem.)
*ḏubāʕatay [ḏbˁ.ty] twenty jōt (masc.)
jōti (fem.)
*máʕbAʔ [mˁbȝ] (masc.)
*máʕbAʔat [mˁbȝ.t] (fem.)
thirty maab (masc.)
maabe (fem.)
*ḥAmí (?) [ḥm.w] (masc.) forty xme
*díywu [dy.w] fifty taeiou
*yAssáwyu [sỉsy.w or ỉswy.w (?)] sixty se
*safḫáwyu [sfḫy.w] (masc.) seventy šfe
*ḫamanáwyu [ḫmny.w] (masc.) eighty xmene
*pAsiḏawyu [psḏy.w] (masc.) ninety pstaiou
*šáwat [š.t] one hundred še
*šūtay [š.ty] two hundred šēt
*ḫaʔ [ḫȝ] one thousand šo
*ḏubaʕ [ḏbȝ] ten thousand tba
[hfn] one hundred thousand
*ḥaḥ [ḥḥ] one million xax "many"
## References
• Allen, James Paul. 2000. Middle Egyptian: An Introduction to the Language and Culture of Hieroglyphs. Cambridge: Cambridge University Press. Numerals discussed in §§9.1–9.6.
• Gardiner, Alan Henderson. 1957. Egyptian Grammar; Being an Introduction to the Study of Hieroglyphs. 3rd ed. Oxford: Griffith Institute. For numerals, see §§259–266.
• Goedicke, Hans. 1988. Old Hieratic Paleography. Baltimore: Halgo, Inc.
• Möller, Georg. 1927. Hieratische Paläographie: Die aegyptische Buchschrift in ihrer Entwicklung von der Fünften Dynastie bis zur römischen Kaiserzeit. 3 vols. 2nd ed. Leipzig: J. C. Hinrichs'schen Buchhandlungen. (Reprinted Osnabrück: Otto Zeller Verlag, 1965)
## Notes
1. ^ Cajori, Florian (1993) [1929]. A History of Mathematical Notations. Dover Publications. pp. pp. 229–230. ISBN 0486677664.
2. ^ John B. Callender, Middle Egyptian, 1975 | 2,302 | 7,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-30 | latest | en | 0.913262 |
https://www.daniweb.com/programming/software-development/threads/299471/how-to-convert-acii-to-string | 1,590,888,611,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410535.45/warc/CC-MAIN-20200530231809-20200531021809-00007.warc.gz | 679,273,603 | 13,713 | Hello i am developing a encryption/decryption Application in which i am first converting string into its ascii values then reversing this ascii values and then generating the hex values of this reversed ascii values...
I have completed the encryption part as i am able to convert the string into hex values but now i have problem with decryption as i am converting hex into reversed ascii then again converting reversed ascii into correct but now the problem is how to convert this ascii values into plain text string...
For more understanding below is process how my code works...
suppose i have to encrypt text "Test"
Step 1 : text TEST is converted into its ascii value i.e. 84101115116
Step 2 : Ascii value of TEST is reversed i.e. 61151110148
Step 3 : Hex values of reversed ascii value is generated i.e. 36 31 31 35 31 31 31 30 31 34 38
so now encrypted text will be 36 31 31 35 31 31 31 30 31 34 38
here encryption ends..
Now when i start decryption
Step 1 : i generate ascii values from Hex i.e. 61151110148
Step 2 : Reverse The Ascii values so i will get actual ascii Values i.e. 84101115116
Step 3 : Here i am having the problem i am unable to figure out how to convert this ascii values into plain text....
Help Its Urgent...
You need to break down the ascii into usable references. Test in Ascii is as follow -
T=84, e=101, s=115, t=116. Once you have that you can convert the ascii to a string "Test". Below is a list of the ascii characters.
Your code would be something like -
``````Select Case Ascii
Case Is = "84"
StrText = "T"
Case Is = "101"
strText = "e"
'and so forth...
strResult = strText 'etc``````
Solved The Problem used a separator while converting string to its ascii values..
Now process is :
suppose i have to encrypt text "Test"
Step 1 : text TEST is converted into its ascii value i.e. 84:101:115:116:
Step 2 : Ascii value of TEST is reversed i.e. :611:511:101:48
Step 3 : Hex values of reversed ascii value is generated i.e. 3A 36 31 31 3A 35 31 31 3A 31 30 31 3A 34 38
so now encrypted text will be 3A 36 31 31 3A 35 31 31 3A 31 30 31 3A 34 38
here encryption ends..
Now when i start decryption
Step 1 : i generate ascii values from Hex i.e. :611:511:101:48
Step 2 : Reverse The Ascii values so i will get actual ascii Values i.e. 84:101:115:116:
Step 3 : Here i am generating character on the basis of separator like when ":" occures it print the character for ascii value before that separator...
``````?Chr(101) | 687 | 2,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-24 | latest | en | 0.767661 |
https://rdrr.io/cran/cabootcrs/man/covmat.html | 1,544,993,917,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827992.73/warc/CC-MAIN-20181216191351-20181216213351-00189.warc.gz | 706,921,439 | 15,206 | # covmat: Extract covariance matrix for a single category point In cabootcrs: Bootstrap Confidence Regions for Correspondence Analysis.
## Description
Extract the 2 by 2 covariance matrix, for one row or column point, for one pair of axes.
## Usage
`1` ```covmat(x, i, thing = "row", axis1 = 1, axis2 = 2) ```
## Arguments
`x` object of class cabootcrsresults. `i` row or column number. `thing` "row" - i-th row. "column" - i-th column. `axis1` first axis of pair. `axis2` second axis of pair, must be <= axisvariances value for x
## Details
Extracts a covariance matrix for use in further calculations. To inspect variances and covariances use `allvarscovs` for more reader-friendly output.
## Value
2 by 2 covariance matrix.
## Author(s)
T.J. Ringrose
`allvarscovs` , `cabootcrsresults`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52``` ```dreamdata <- t(matrix(c(7,4,3,7,10,15,11,13,23,9,11,7,28,9,12,10,32,5,4,3),4,5)) bd <- cabootcrs(dreamdata, lastaxis=3) covrow3axes1and2 <- covmat(bd, i=3) covcol2axes2and3 <- covmat(bd, i=2, thing="column", axis1=2, axis2=3) ## The function is currently defined as function (x, i, thing = "row", axis1 = 1, axis2 = 2) { printwithaxes <- function(res, thenames) { names(res) <- thenames print(res, digits = 4) } if (!(class(x) == "cabootcrsresults")) stop(paste("Must be of type cabootcrsresults\n\n")) if (!any(thing == c("row", "column"))) stop(paste("Must be row or column\n\n")) if (axis1 == axis2) stop(paste("What are you playing at?\n\n")) if (!any(axis1 == seq(1, x@axisvariances))) stop(paste("Covariance not available for these axes\n\n")) if (!any(axis2 == seq(1, x@axisvariances))) stop(paste("Covariance not available for these axes\n\n")) if ((thing == "row") & !any(i == seq(1, x@rows))) stop(paste("Invalid row number\n\n")) if ((thing == "column") & !any(i == seq(1, x@columns))) stop(paste("Invalid column number\n\n")) a1 <- min(axis1, axis2) a2 <- max(axis1, axis2) tname <- "" if (thing == "row") { V <- matrix(c(x@RowVar[i, axis1], x@RowCov[i, a1, a2], x@RowCov[i, a1, a2], x@RowVar[i, axis2]), 2, 2) if (!is.null(x@rowlabels)) { tname <- paste("(", x@rowlabels[[i]], ")") } } else { V <- matrix(c(x@ColVar[i, axis1], x@ColCov[i, a1, a2], x@ColCov[i, a1, a2], x@ColVar[i, axis2]), 2, 2) if (!is.null(x@collabels)) { tname <- paste("(", x@collabels[[i]], ")") } } cat(paste("Covariance matrix of", switch(thing, row = "row", column = "column"), i, tname, "for axes", axis1, axis2, "\n\n")) rcnames <- c(paste("Axis", axis1), paste("Axis", axis2)) printwithaxes(data.frame(V, row.names = rcnames), rcnames) invisible(V) } ``` | 980 | 2,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-51 | latest | en | 0.425418 |
http://idevgames.com/forums/thread-4539-post-31559.html#pid31559 | 1,481,176,931,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542414.34/warc/CC-MAIN-20161202170902-00272-ip-10-31-129-80.ec2.internal.warc.gz | 137,874,866 | 8,395 | ## gluProject <-> gluUnproject help please
szgezu
Unregistered
Post: #1
Dear All,
I would like to ask how can gluProject be used for giving the same result as gluUnProject would do it as I have only access of gluProject (this is because I code in perl with its opengl module and only the gluProject seems to work). I am not so good at the mathematical aspect of this so that's why I ask people more clever than me in this field.
(The end of this code would be a selection of an object in a 3d enviroment but I could find example codes only in C and I can get on very slowly with the conversion of the C code into perl, so if awhole code example would come I would be more impressed :-), but a simple answer for the quation would do the trick for a while ;-) )
Thanks a lot!
Regards,
Geza Szabo
Member
Posts: 114
Joined: 2005.03
Post: #2
You'd need the inverse matrix of the projection and view matrices. To calculate the inverse of a matrix, I found the following in the Newton tutorials:
Code:
```inline dMatrix dMatrix::Inverse () const { return dMatrix (dVector (m_front.m_x, m_up.m_x, m_right.m_x, 0.0f), dVector (m_front.m_y, m_up.m_y, m_right.m_y, 0.0f), dVector (m_front.m_z, m_up.m_z, m_right.m_z, 0.0f), dVector (- (m_posit % m_front), - (m_posit % m_up), - (m_posit % m_right), 1.0f)); }```
You'll probably need to adapt it to your own way of storing matrices. The important part here is that the 3x3 part that describes the rotation is transposed and that the location parts are the negative dot product of the first, second and third column vector with the original location vector (horrible sentence, I know). If you feed these matrices into gluProject, you should get the same results as with gluUnProject. I haven't tested this, though. One thing that might be possible is that you need to swap the modelview and projection matrix, but my math skills aren't good enough to say whether this is right or wrong.
Luminary
Posts: 5,143
Joined: 2002.04
Post: #3
That's not a general inverse function. It'll only work for inverting orthonormal matrices, I think.
szgezu
Unregistered
Post: #4
Thank you the answer Cochrane! That is what I defintely need!
"You'd need the inverse matrix of the projection and view matrices."
Do you mean that I feed gluProject with
@projected=OpenGL::gluProject_p(mousex,mousey,z( (?) what should stay here? 1?)
invert(modelview),
invert(projection),
viewport)
Unfortunatley I do not understand even the names in the code sniplet you inserted:
"m_front.m_x, m_up.m_x, m_right.m_x, 0.0f" ...
What do m_fron, m_up, mean?
Can not be the explaining sentence following it written in more formal? I get more and more confused, the more times I read it. :-)
szgezu
Unregistered
Post: #5
Can someone explain it to me why it can happen that if I have a rotating object in the 3d enviroment, and if I always feed the gluProject function with the same object coordinates (but the viewport, modelview and projection matrix is always updated of course), the result of gluProject is always the same?
I see that modelview matrix is changing because of the rotation. How can that be that this not affect the result of gluProject?
Member
Posts: 151
Joined: 2002.09
Post: #6
You definately need to swap your matrices, and also you have to adjust your vertexes to the viewport calculations done by gluProject.
I'd suggest you multiply both matrices together and then take the inverse and get along fine without gluProject (do stuff the same way as gluUnProject does)
Code:
```2(winX - view[0]) ----------------- - 1 view[2] objX 2(winY - view[1]) objY = INV(PM) ----------------- - 1 objZ view[3] W 2(winZ) - 1 1```
The projection matrix most probably isn't orthonormal so you'd have to use a different method for inversion. there's some methods here: http://www.gamedev.net/community/forums/..._id=365952
AnotherJake Wrote:gluProject changes the projection matrix, not the modelview matrix. They are separate and are used for different purposes in the rendering pipeline. If you keep feeding gluProject the same numbers every frame you'll get the same results, but it will always be unaffected by anything you do to the modelview matrix, like object rotations.
huh? gluProject isn't supposed to change any of those matrices. what are you talking about? it all depends on what you feed into gluProject, works fine for me at least.
Moderator
Posts: 3,591
Joined: 2003.06
Post: #7
Hog Wrote:huh? gluProject isn't supposed to change any of those matrices. what are you talking about?
Absolute crap is what I was talking about actually. I can't believe I brain-farted that badly. Sorry. I thought I was talking about gluPerspective, and for some God-only-knows-what reason I completely read it that way. Un-freakin'-believable... I deleted that message.
szgezu
Unregistered
Post: #8
Thanx all of you! This forum is the most usefull among a dozen other! This was the only where I got helpful answers, and smart people who know what they speak!
If I need help again with this I return here. Now I try to implement these suggestions!
Apprentice
Posts: 19
Joined: 2005.11
Post: #9
So just how slow is gluUnProject and gluProject?
In an RTS game, I see there being two ways of selecting units. One being get the X, Y Mouse coordinates, gluUnProject those onto the grid where your units are, get those X, Y, Z coordinates, then compare that position to all the units (in that grid/screen), and if it's less than some number select that unit.
Or you could get the XY mouse coordinates, and then call gluProject on all the units (in that grid/screen), and compare those results.
Which would probably be faster?
szgezu
Unregistered
Post: #10
I don'tknow exactly, because I have implemented the gluProject function after all, because it had weird things also. But thinking on this problem it is totally equal in the mentinoed cases, as the slow thing will be tha comparison between ALL of the units which you mentioned. I would recommend using gluPickMatrix (or sg like this) and gl_select mode, as that way is the most resouce saving method. (I didn't use it because it was also weird in perl (I tried but had a very bad weekend because of it, and hard to debug)) | 1,628 | 6,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-50 | latest | en | 0.918764 |
https://hackerranksolution.in/listcomprehensionspython/ | 1,720,918,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00307.warc.gz | 260,354,087 | 12,146 | # List Comprehensions
### Problem Statement :
```Let's learn about list comprehensions! You are given three integers x,y and z representing the dimensions of a cuboid along with an integer i. Print a list of all possible coordinates given by (i,j,k) on a 3D grid where the sum of i+j+k is not equal to n. Here, 0<=i<=x; 0<=j<=y; 0,=k<=z. Please use list comprehensions rather than multiple loops, as a learning exercise.
Example:
x=1
y=1
z=2
n=3
All permutations of [i,j,k] are:
. [[0,0,0], [0,0,1], [0,0,2], [0,1,0], [0,1,1], [0,1,2], [1,0,0], [1,0,1], [1,0,2], [1,1,0], [1,1,1], [1,1,2]]
Print an array of the elements that do not sum to n=3 .
[[0,0,0], [0,0,1], [0,0,2], [0,1,0], [0,1,1], [1,0,0], [1,0,1], [1,1,0], [1,1,2]]
Input Format:
Four integers x,y,z and n, each on a separate line.
Constraints:
Print the list in lexicographic increasing order.```
### Solution :
``` ```Solution in C :
if __name__ == '__main__':
x = int(input())
y = int(input())
z = int(input())
n = int(input())
output=[]
for i in range(x+1):
for j in range(y+1):
for k in range(z+1):
if i+j+k==n:
continue
else:
output.append([i,j,k])
print(output)```
```
## QHEAP1
This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element
Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t
## Find the Running Median
The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.
## Minimum Average Waiting Time
Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h
## Merging Communities
People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w
## Components in a graph
There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu | 990 | 3,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.826872 |
https://nrich.maths.org/public/leg.php?code=-68&cl=1&cldcmpid=8058 | 1,508,372,001,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823168.74/warc/CC-MAIN-20171018233539-20171019013539-00285.warc.gz | 757,884,728 | 9,985 | # Search by Topic
#### Resources tagged with Visualising similar to Ordering Cards:
Filter by: Content type:
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### Tricky Triangles
##### Stage: 1 Challenge Level:
Use the three triangles to fill these outline shapes. Perhaps you can create some of your own shapes for a friend to fill?
### Routes 1 and 5
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Find your way through the grid starting at 2 and following these operations. What number do you end on?
### Painting Possibilities
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
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This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
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Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### Single Track
##### Stage: 2 Challenge Level:
What is the best way to shunt these carriages so that each train can continue its journey?
### Shunting Puzzle
##### Stage: 2 Challenge Level:
Can you shunt the trucks so that the Cattle truck and the Sheep truck change places and the Engine is back on the main line?
### Tetrafit
##### Stage: 2 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Tangram Tangle
##### Stage: 1 Challenge Level:
If you split the square into these two pieces, it is possible to fit the pieces together again to make a new shape. How many new shapes can you make?
### Waiting for Blast Off
##### Stage: 2 Challenge Level:
10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways?
### Putting Two and Two Together
##### Stage: 2 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
### Dodecamagic
##### Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
### Map Folding
##### Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
##### Stage: 2 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
### Counting Cards
##### Stage: 2 Challenge Level:
A magician took a suit of thirteen cards and held them in his hand face down. Every card he revealed had the same value as the one he had just finished spelling. How did this work?
### One Big Triangle
##### Stage: 1 Challenge Level:
Make one big triangle so the numbers that touch on the small triangles add to 10. You could use the interactivity to help you.
### Cuboid-in-a-box
##### Stage: 2 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### Tessellate the Triominoes
##### Stage: 1 Challenge Level:
What happens when you try and fit the triomino pieces into these two grids?
### Cover the Camel
##### Stage: 1 Challenge Level:
Can you cover the camel with these pieces?
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Celtic Knot
##### Stage: 2 Challenge Level:
Building up a simple Celtic knot. Try the interactivity or download the cards or have a go on squared paper.
### Picture a Pyramid ...
##### Stage: 2 Challenge Level:
Imagine a pyramid which is built in square layers of small cubes. If we number the cubes from the top, starting with 1, can you picture which cubes are directly below this first cube?
### Display Boards
##### Stage: 2 Challenge Level:
Design an arrangement of display boards in the school hall which fits the requirements of different people.
### Open Boxes
##### Stage: 2 Challenge Level:
Can you work out how many cubes were used to make this open box? What size of open box could you make if you had 112 cubes?
### Multiplication Series: Illustrating Number Properties with Arrays
##### Stage: 1 and 2
This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Alquerque II
##### Stage: 1 Challenge Level:
A variant on the game Alquerque
### Change Around
##### Stage: 1 Challenge Level:
Move just three of the circles so that the triangle faces in the opposite direction.
### Hexpentas
##### Stage: 1 and 2 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### Teddy Bear Line-up
##### Stage: 1 Challenge Level:
What is the least number of moves you can take to rearrange the bears so that no bear is next to a bear of the same colour?
### Redblue
##### Stage: 2 Challenge Level:
Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
### Three Cubed
##### Stage: 2 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
### Nine-pin Triangles
##### Stage: 2 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Little Boxes
##### Stage: 2 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### Four Triangles Puzzle
##### Stage: 1 and 2 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Square Corners
##### Stage: 2 Challenge Level:
What is the greatest number of counters you can place on the grid below without four of them lying at the corners of a square?
### Music to My Ears
##### Stage: 2 Challenge Level:
Can you predict when you'll be clapping and when you'll be clicking if you start this rhythm? How about when a friend begins a new rhythm at the same time?
### Horizontal Vertical
##### Stage: 1 Challenge Level:
Take it in turns to place a domino on the grid. One to be placed horizontally and the other vertically. Can you make it impossible for your opponent to play?
### A City of Towers
##### Stage: 1 Challenge Level:
In this town, houses are built with one room for each person. There are some families of seven people living in the town. In how many different ways can they build their houses?
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Colour Wheels
##### Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Folding Flowers 1
##### Stage: 2 Challenge Level:
Can you visualise what shape this piece of paper will make when it is folded?
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes? | 2,128 | 9,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-43 | latest | en | 0.894284 |
https://training.incf.org/search?f%5B0%5D=topics%3A22&f%5B1%5D=topics%3A35&f%5B2%5D=topics%3A36&f%5B3%5D=topics%3A39&f%5B4%5D=topics%3A54&f%5B5%5D=topics%3A60&f%5B6%5D=topics%3A69&f%5B7%5D=topics%3A75&%3Bf%5B1%5D=topics%3A37 | 1,591,029,620,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00096.warc.gz | 570,682,351 | 16,384 | ## Difficulty level
Lecture title:
The probability of a hypothesis, given data.
Difficulty level: Beginner
Duration: 7:57
Speaker: : Barton Poulson
Lecture title:
Why math is useful in data science.
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Why statistics are useful for data science.
Difficulty level: Beginner
Duration: 4:01
Speaker: : Barton Poulson
Lecture title:
Statistics is exploring data.
Difficulty level: Beginner
Duration: 2:23
Speaker: : Barton Poulson
Lecture title:
Graphical data exploration
Difficulty level: Beginner
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Lecture title:
Numerical data exploration
Difficulty level: Beginner
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Simple description of statistical data.
Difficulty level: Beginner
Duration: 10:16
Speaker: : Barton Poulson
Lecture title:
Basics of hypothesis testing.
Difficulty level: Beginner
Duration: 06:04
Speaker: : Barton Poulson
In this lecture, the speaker demonstrates Neurokernel's module interfacing feature by using it to integrate independently developed models of olfactory and vision LPUs based upon experimentally obtained connectivity information.
Difficulty level: Intermediate
Duration: 29:56
Speaker: : Aurel A. Lazar
Lecture title:
Enabling neuroscience research using high performance computing
Difficulty level: Beginner
Duration: 39:27
Speaker: : Subha Sivagnanam
Lecture title:
Tutorial describing the basic search and navigation features of the Allen Mouse Brain Atlas
Difficulty level: Beginner
Duration: 6:40
Speaker: : Unknown
Lecture title:
Tutorial describing the basic search and navigation features of the Allen Developing Mouse Brain Atlas
Difficulty level: Beginner
Duration: 6:35
Speaker: : Unknown
Lecture title:
Tutorial describing the basic features of the Brain Explorer® 3-D viewer for the mouse brain
Difficulty level: Beginner
Duration: 6:41
Speaker: : Unknown
This tutorial demonstrates how to use the differential search feature of the Allen Mouse Brain Atlas to find gene markers for different regions of the brain and to visualize this gene expression in three-dimensional space. Differential search is also available for the Allen Developing Mouse Brain Atlas and the Allen Human Brain Atlas.
Difficulty level: Beginner
Duration: 6:31
Speaker: : Unknown
The chair of the workshop is giving an introduction and a motivating argument.
Difficulty level: Beginner
Duration: 5:36
Speaker: : Maryann Martone
Lecture title:
This lecture highlights the importance of correct annotation and assignment of location, and updated atlas resources to avoid errors in navigation and data interpretation.
Difficulty level: Intermediate
Duration: 22:04
Speaker: : Trygve Leergard
Lecture title:
We are at the exciting technological stage where it has become feasible to represent the anatomy of an entire human brain at the cellular level. In this presentation, the speaker explains that neuroanatomy in the XXI Century has become an effort towards the virtualization and standardization of brain tissue.
Difficulty level: Intermediate
Duration: 25:27
Speaker: : Jacopo Annese
This lecture covers essential features of digital brain models for neuroinformatics.
Difficulty level: Intermediate
Duration: 22:26
Speaker: : Douglas Bowden
This presentation covers the neuroinformatics tools and techniques used and their relationship to neuroanatomy for the Allen atlases of the mouse, developing mouse, and mouse connectional atlas.
Difficulty level: Intermediate
Duration: 23:41
Speaker: : Mike Hawrylycz
Lecture title:
This lecture covers an introduction to connectomics, and image processing tools for the study of connectomics.
Difficulty level: Beginner
Duration: 1:23:03 | 821 | 3,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-24 | latest | en | 0.703972 |
https://qiita.com/hiratara/items/2e8dee9e6e26188e2671 | 1,718,698,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00300.warc.gz | 419,227,376 | 25,723 | 0
0
More than 5 years have passed since last update.
# のんびり座りたい (2013.2.2の過去問)
Last updated at Posted at 2013-12-01
``````module Doukaku.Nonbiri (solve) where
import Data.Char (toUpper)
solve :: String -> String
solve input = foldl next (take n . repeat \$ '-') ope
where
ope :: String
(n', ':':ope) = break (== ':') input
n = read n' :: Int
next :: String -> Char -> String
next current c
| 'a' <= c && c <= 'z' = let c' = toUpper c
in map (\x -> if x == c' then '-' else x) current
| otherwise = sitDown c current
sideMembers :: String -> [Int]
sideMembers xs = sideMembers' ('-':xs)
where
sideMembers' (x:'-':[]) = [count x]
sideMembers' (_:_:[]) = [-1]
sideMembers' (x:'-':z:zs) = let members = sum . map count \$ [x, z]
in members : sideMembers' ('-':z:zs)
sideMembers' (_:y:z:zs) = -1 : sideMembers' (y:z:zs)
count '-' = 0
count _ = 1
sitDown :: Char -> String -> String
sitDown new current = either id undefined \$ do
toEither (sitDown' 0 current sides)
toEither (sitDown' 1 current sides)
toEither (sitDown' 2 current sides)
return ()
where
sides = sideMembers current
sitDown' :: Int -> String -> [Int] -> Maybe String
sitDown' _ [] [] = Nothing
sitDown' n (c:cs) (m:ms)
| n == m = Just \$ new:cs
| otherwise = (c :) `fmap` (sitDown' n cs ms)
toEither :: Maybe a -> Either a ()
toEither = maybe (Right ()) Left
``````
http://qiita.com/Nabetani/items/4364285801d1c9f370a1 に他の方の回答もありますので、見ると参考になるでしょう。
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0 | 554 | 1,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.721637 |
https://www.leigh.covmat.org/11th-january/ | 1,719,289,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00039.warc.gz | 746,030,146 | 36,554 | Aim ever higher
## Marvellous Me
Marvellous Me Marvellous Me
# 11th January
Maths (send in)
LO: To multiply 2-digit numbers with renaming
In maths today, we are going to be multiplying 2-digit numbers with renaming. This is building on last weeks lessons, so please have a look at those beforehand if you didn't do them!
Have a look at the in-focus question. How do you think you could work it out? As you would usually do this in your pairs, see if you can talk to an adult or an older sibling about your thoughts. Use the empty box to explain your thoughts.
Next, have a look at the guided practice. Can you work out how those answers were gained? Have a go at working them out yourselves to see if your method is correct.
Finally, have a go at the worksheet provided - it is 2 sheets, the minimum expectation is one sheet, but just like in class, try and have a go at the second one too!
This week, we are going to be reading the book ‘The curious case of Dr Salamander’.
Today, you are going to read chapter 1 and then complete the online quiz. I will be able to see if you have done this activity!
You each have your own log in which I will be sending to your adults to give to you. I will send it via marvellous me – if it doesn’t come through please email me at year4@leigh.covmat.org and I will send it to you there!
You can choose to do the quiz online or print off the quiz and do it on paper. If you have the choice of both, I would rather then quiz be done online.
English (send in)
LO: To describe a character
Have another listen to the narrative verse, the Bogeymen and the Trolls next door. Some very powerful words were used in this. Can you write each of these words into your own sentences?
Fateful (Very important, often in a disastrous way) Example: It was a fateful day when the princess pricked her finger, and would now sleep for 100 years!
Indignant (annoyed, outraged) Example: I felt very indignant at being blamed for the broken vase!
Desolate (bleak and depressing) Example: The land was a desolate place, with not a tree in sight.
Destination (Place of arrival, journey’s end) Example: You have reached your destination.
Below is an outline of Dave the Troll. How would others describe him? Think of some good adjectives that Fred the Bogeyman would use to describe him and place those words outside of his body.
How would Dave describe himself? Think of some good adjectives that he would use and place those inside of his body.
When done, send this to me for me to have a look at your wonderful work.
History (send in)
In History this half term, we are looking at Britain's ‘first’ empire. In todays lesson, we are going to be looking at how it was established in the Americas and in Africa. Please look through the slides attached below. Some of the words may be confusing and you may have to ask someone around you for some help understanding them. After this, complete the worksheet with what you think the people onboard the ships did.
Music
In music we will be looking at percussion this half term. What is percussion you ask? Time to find out! Don’t worry, you don’t need any musical instruments to do this, just a couple of things from around your house!
Top | 744 | 3,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.962114 |
https://ko.coursera.org/learn/introduction-statistics-data-analysis-public-health/reviews?page=3 | 1,611,181,333,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00509.warc.gz | 406,364,659 | 124,545 | Introduction to Statistics & Data Analysis in Public Health(으)로 돌아가기
# 임페리얼 칼리지 런던의 Introduction to Statistics & Data Analysis in Public Health 학습자 리뷰 및 피드백
4.7
별점
868개의 평가
182개의 리뷰
## 강좌 소개
Welcome to Introduction to Statistics & Data Analysis in Public Health! This course will teach you the core building blocks of statistical analysis - types of variables, common distributions, hypothesis testing - but, more than that, it will enable you to take a data set you've never seen before, describe its keys features, get to know its strengths and quirks, run some vital basic analyses and then formulate and test hypotheses based on means and proportions. You'll then have a solid grounding to move on to more sophisticated analysis and take the other courses in the series. You'll learn the popular, flexible and completely free software R, used by statistics and machine learning practitioners everywhere. It's hands-on, so you'll first learn about how to phrase a testable hypothesis via examples of medical research as reported by the media. Then you'll work through a data set on fruit and vegetable eating habits: data that are realistically messy, because that's what public health data sets are like in reality. There will be mini-quizzes with feedback along the way to check your understanding. The course will sharpen your ability to think critically and not take things for granted: in this age of uncontrolled algorithms and fake news, these skills are more important than ever. Prerequisites Some formulae are given to aid understanding, but this is not one of those courses where you need a mathematics degree to follow it. You will need only basic numeracy (for example, we will not use calculus) and familiarity with graphical and tabular ways of presenting results. No knowledge of R or programming is assumed....
## 최상위 리뷰
SK
2019년 10월 11일
This is the best course among all I've taken..\n\nThe instructor has presented the content precisely.\n\nI highly recommend to those who are looking to explore R in the field of health
LA
2019년 5월 25일
Was a very nicely done and clear course to build or re-build foundation for most common statistical concepts and an intro to using R via R-Studio for your work with them on the basics.
필터링 기준:
## Introduction to Statistics & Data Analysis in Public Health의 182개 리뷰 중 51~75
교육 기관: David G Á
2020년 5월 4일
Very good for an initial introduction to the topic. I think it is not a good stand alone course, it is ment to introduce to the next courses in the specialization.
교육 기관: Hector P
2020년 9월 6일
This course is great. I like how it is structured and the feedback after activities. Combines theory and practices in an efficient way. Congratulations.
교육 기관: Cliff W
2020년 5월 28일
Fantastic. Just what I needed as a doctor wanting to learn more about statistics. R is an amazing alternative to other programs such as STATA and SPSS.
교육 기관: Mayur N
2019년 4월 7일
Wonderful explanation and introduction to R programing. With minimal additional self learning you can easily master all of the content of the course.
교육 기관: Lawrence O
2020년 12월 20일
this is an excellent course, but more examples are needed for clarity, in video formats. We want to see the instructor working out examples on board
교육 기관: Chandu P L
2020년 5월 20일
Loved the instructor. Everything was rightly pointed out. And lastly for me he looks like Thomas Shelby from British series Peaky Blinders. Haha.
교육 기관: MUWANGA R J S
2020년 3월 14일
Wonderful introduction to statistics or public health. Loved the lecturer. He was engaging, and able to make theoretical concepts more intuitive.
교육 기관: Shashwat T
2020년 4월 25일
It is a highly recommended course with a bit of background knowledge. Background knowledge is required as it is a part of the Master's course.
교육 기관: Samah A
2020년 2월 7일
A short and precise introduction. Found it pretty basic but would definitely recommend it to anyone just starting out with statistics 👍
교육 기관: Vũ M L
2019년 12월 30일
This is an excellent course, it helps me play a solid foundation about Biostatistics in particular and Public Health in General
교육 기관: Dr G K
2020년 7월 12일
Excellent course, good and short videos, good practice sections...Instructor delivers the content well in good flow and pace
교육 기관: Swetha J
2020년 4월 12일
Excellent content and methodology of teaching. Great mix of examples to understand the concepts and hands on analysis in R.
교육 기관: Dr. M
2020년 4월 14일
Worth taking this course. Concepts of sampling, distributions and hypothesis testing are taught very well by Professor.
교육 기관: Muhammad T U H
2020년 11월 29일
It was a very great learning experience. Thanks Alex!! I learned the p-value very comfortably, highly recommended
교육 기관: Rahul R
2020년 10월 27일
This course is one of the hidden gems in Coursera. I really liked the simple explanations of complex concepts.
교육 기관: Todd D
2019년 10월 1일
Excellent introduction to the topic. Well-paced and prepares the learner for future courses. Very well done.
교육 기관: Vu T N M
2020년 12월 20일
Well-structured and useful even for freshmen. The R programming information is really practical and doable.
교육 기관: Matias O
2020년 10월 1일
Es un gran curso para el profesional de salud que quiere introducirse en el conocimiento de estadística!
교육 기관: Dawn D
2020년 4월 23일
Clearly explained, well produced videos. Excellent starter or refresher for basic statistical thinking.
교육 기관: Laura M A M
2020년 8월 21일
Excellent course, I totally enjoyed the Professors' attitude, the exercises in R and even by hand :).
교육 기관: Asif K
2020년 9월 14일
This course is very good and it has enhance my knowledge how to use R and also about the coding
교육 기관: Katerina K
2019년 6월 8일
An absolutely fantastic course. Compared with my course back at Uni, this is 100 times clearer.
교육 기관: Md. I R
2020년 10월 8일
This course is very important to learn basics of statistics & data analysis in public health.
교육 기관: Tommy G
2019년 7월 6일
Excellent course if you want to start using R for public health projects! 100% recommended!
교육 기관: Kelly B
2020년 4월 15일
Fantastic introduction to R, really enjoyed practising with the given dataset. Thank you! | 1,567 | 6,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.935626 |
https://www.techwhiff.com/issue/why-did-christopher-columbus-and-his-crew-run-out-of--267126 | 1,679,993,537,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00080.warc.gz | 1,116,335,793 | 12,449 | # Why did Christopher Columbus and his crew run out of food on their voyage in 1492?
###### Question:
Why did Christopher Columbus and his crew run out of food on their voyage in 1492?
### The term "Pax Mongolica" ("Mongolian Peace") is used by historians to describe
The term "Pax Mongolica" ("Mongolian Peace") is used by historians to describe...
### SOMEEEEONEEE pls help pls and thx :) Question 1) A sculptor created a marble water basin by beginning with a cylinder and then carving out a hemisphere. The cylinder has a radius of 30 centimeters and a height of 45 centimeters. The hemisphere has a radius of 25 centimeters. What is the approximate volume of the marble that forms the water basin? A) 123,307.5 cubic centimeters B) 61,784.7 cubic centimeters C) 125,925.5 cubic centimeters D) 94,509.6 cubic centimeters Question 2) Toby made a hear
SOMEEEEONEEE pls help pls and thx :) Question 1) A sculptor created a marble water basin by beginning with a cylinder and then carving out a hemisphere. The cylinder has a radius of 30 centimeters and a height of 45 centimeters. The hemisphere has a radius of 25 centimeters. What is the approximate...
### Medical researchers have determined that for exercise to be beneficial, a person’s desirable heart rate, R, in beats per minute, can be approximated by the formulas where a represents the person’s age. If the desirable heart rate for a man is 135 beats per minute, how old is he?
Medical researchers have determined that for exercise to be beneficial, a person’s desirable heart rate, R, in beats per minute, can be approximated by the formulas where a represents the person’s age. If the desirable heart rate for a man is 135 beats per minute, how old is he?...
### Henry is 4 years younger than Linda. Linda is 20 years old. After correctly carrying out the Solve step of the five-step problem-solving plan, which solution should you arrive at to find Henry’s age?
Henry is 4 years younger than Linda. Linda is 20 years old. After correctly carrying out the Solve step of the five-step problem-solving plan, which solution should you arrive at to find Henry’s age?...
### The weight of water is 62 1/2 lb per cubic foot water that weighs 300 lb will fill how many cubic feet
The weight of water is 62 1/2 lb per cubic foot water that weighs 300 lb will fill how many cubic feet...
### Need help rewriting the equation
Need help rewriting the equation...
### A story in which characters experience different events but are connected by a common theme is an example of a
A story in which characters experience different events but are connected by a common theme is an example of a...
### Write the equation of the line that contains the given point and the given slope. (-46,45), slope=2.5 please help and explain
Write the equation of the line that contains the given point and the given slope. (-46,45), slope=2.5 please help and explain...
### What are redox half-reactions? A. Equations that separate the oxidation and reduction parts of the reaction B. Equations that separate the reactants from the products in a redox reaction C. Equations that separate the electron transfer part from the atomic part D. Equations that show the oxidation potentials and the reduction potentials (Answer is A)
What are redox half-reactions? A. Equations that separate the oxidation and reduction parts of the reaction B. Equations that separate the reactants from the products in a redox reaction C. Equations that separate the electron transfer part from the atomic part D. Equations that show the oxidatio...
### X + 3y = −1 2x + 2y = 6
x + 3y = −1 2x + 2y = 6...
### NEED HELP ASAP PLS (with work) A punt is kicked at an angle of 60 with a speed of 40 m/s, and lands at the same height from which it was kicked. How far away will the ball land? What is the maximum height reached by the ball? What is the hang time? Total
NEED HELP ASAP PLS (with work) A punt is kicked at an angle of 60 with a speed of 40 m/s, and lands at the same height from which it was kicked. How far away will the ball land? What is the maximum height reached by the ball? What is the hang time? Total...
### Which of the following molecules has only two non binding pairs (long pairs) of electrons in its lewis structure? a)O2 b)CH4 c)NH3 d)H2s
which of the following molecules has only two non binding pairs (long pairs) of electrons in its lewis structure? a)O2 b)CH4 c)NH3 d)H2s...
### What is the arithmetic mean of 7 and 27?
What is the arithmetic mean of 7 and 27?...
### Why would the air temperature over New Jersey be warmer than it is in Pennsylvania?
Why would the air temperature over New Jersey be warmer than it is in Pennsylvania?...
### Balancing Equations: _(NH4)3PO4 + _Pb(NO3)4 = _Pb3(PO4)4 + _NH4NO3 With work please.
Balancing Equations: _(NH4)3PO4 + _Pb(NO3)4 = _Pb3(PO4)4 + _NH4NO3 With work please.... | 1,214 | 4,890 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-14 | latest | en | 0.909092 |
https://nl.mathworks.com/matlabcentral/cody/problems/44401-vertical-matrix-sort/solutions/1798558 | 1,582,762,186,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146562.94/warc/CC-MAIN-20200226211749-20200227001749-00000.warc.gz | 468,242,035 | 15,883 | Cody
# Problem 44401. Vertical matrix sort
Solution 1798558
Submitted on 27 Apr 2019 by Anastasia Silvestruk
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = []; y_correct = []; assert(isequal(upAndDown(x),y_correct))
2 Pass
x = 0; y_correct = [0;0]; assert(isequal(upAndDown(x),y_correct))
3 Pass
x = zeros(10); x(7,4) = 1; y_correct = zeros(20,10); y_correct(10:11,4) = [1;1]; assert(isequal(upAndDown(x),y_correct))
4 Pass
x = [1 4 3 2]; y_correct = [1 4 3 2; 1 4 3 2]; assert(isequal(upAndDown(x),y_correct))
5 Pass
x = [8 9 3 9; 9 6 5 2; 2 1 9 9]; y_correct = [2 1 3 2; 8 6 5 9; 9 9 9 9; 9 9 9 9; 8 6 5 9; 2 1 3 2]; assert(isequal(upAndDown(x),y_correct))
6 Pass
x = 1:10; y_correct = [x; x]; assert(isequal(upAndDown(x),y_correct))
7 Pass
x = (1:10)'; y_correct = [x; flipud(x)]; assert(isequal(upAndDown(x),y_correct))
8 Pass
x = reshape(1:9,[3,3]); c = [1:3,3:-1:1]'; y_correct = [c,c+3,c+6]; assert(isequal(upAndDown(x),y_correct)) | 446 | 1,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-10 | latest | en | 0.461815 |
http://www.slideserve.com/demetrius-glover/9-5 | 1,490,668,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189589.37/warc/CC-MAIN-20170322212949-00483-ip-10-233-31-227.ec2.internal.warc.gz | 667,412,350 | 14,568 | This presentation is the property of its rightful owner.
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9-5 PowerPoint PPT Presentation
9-5. Time and Temperature. Course 1. Warm Up. Problem of the Day. Lesson Presentation. 9-5. Time and Temperature. Course 1. Warm Up Convert each measurement to feet and inches. 1. 95 in. 2. 128 in. Convert each measurement to meters. 3. 406 cm 4. 24,671 mm. 7 ft 11 in.
9-5
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9-5
Time and Temperature
Course 1
Warm Up
Problem of the Day
Lesson Presentation
9-5
Time and Temperature
Course 1
Warm Up
Convert each measurement to feet and inches.
1.95 in. 2. 128 in.
Convert each measurement to meters.
3. 406 cm 4. 24,671 mm
7 ft 11 in.
10 ft 8 in.
24.671 m
4.06 m
9-5
Time and Temperature
Course 1
Problem of the Day
The skateboard had rolled from the house to within 1.8 m of the street before Rick caught it. If the house is 16.9 m from the street, how far did the skateboard roll?
15.1 m
9-5
Time and Temperature
Course 1
Learn to find measurements of time and temperature.
9-5
Time and Temperature
Course 1
9-5
Time and Temperature
Course 1
Convert.
2 hr 5 min = _____ min
2 hours 5 minutes
Think: 1 hour = 60 minutes.
(2 60) minutes plus 5 minutes
125 minutes
2 hr 5 min = 125 min
9-5
Time and Temperature
1 4
195 min = 3 hr
Course 1
Convert.
195 min = _____ hr
1 hr 60 min
195 60
Think: 1 hour = 60 minutes.
=
195 min •
hr
Write as a mixed number.
9-5
Time and Temperature
Course 1
Convert.
3 hr = _____ s
Think: 1 hour = 60 minutes and 1 minute = 60 seconds.
60 min 1 hr
60 s 1 min
= 10,800 s
3 hr •
3 hours = 10,800 seconds
9-5
Time and Temperature
Course 1
Check It Out: Example 1A
Convert.
3 days 7 hr = _____ hr
3 days 7 hours
Think: 1 day = 24 hours.
(3 24) hours plus 7 hours
79 hours
3 days 7 hr = 79 hr
9-5
Time and Temperature
1 2
102 mo = 8 yr
Course 1
Check It Out: Example 1B
Convert.
102 mo = _____ yr
1 yr 12 mo
102 12
Think: 1 year = 12 months.
yr
=
102 mo •
Write as a mixed number.
9-5
Time and Temperature
Course 1
Check It Out: Example 1C
Convert.
2 days = _____ min
Think: 1 day = 24 hours and 1 hour = 60 minutes.
24 hr 1 day
60 min 1 hr
= 2,880 min
2 days •
2 days = 2,880 min
9-5
Time and Temperature
Course 1
The time between the start of an activity and the end of an activity is called the elapsed time.
9-5
Time and Temperature
Course 1
Additional Example 2A: Finding Elapsed Time
Shawn was scheduled to arrive at 10:15 A.M. he was 1 hour and 55 minutes late. When did he arrive?
Scheduled time: 10:15 A.M.
Think: 1 hour after 10:15 A.M. is 11:15 A.M. 55 minutes after that is 12:10 P.M.
Arrival time: 12:10 P.M.
Shawn arrived at 12:10 P.M.
9-5
Time and Temperature
Course 1
Additional Example 2B: Finding Elapsed Time
Ty met his friends at 1:35 P.M. He had traveled for 2 hours and 45 minutes. At what time did Ty begin his trip?
End time: 1:35 P.M.
Think: 2 hours before 1:35 P.M. is 11:35 A.M. 45 minutes before that is 10:50 A.M.
Begin time: 10:50 A.M.
Ty began his trip at 10:50 A.M.
9-5
Time and Temperature
Course 1
Check It Out: Example 2A
Trent’s ferry was scheduled to arrive at 9:45 A.M. he was 1 hour and 15 minutes late. When did he arrive?
Scheduled time: 9:45 A.M.
Think: 1 hour after 9:45 A.M. is 10:45 A.M. 15 minutes after that is 11:00 A.M.
Arrival time: 11:00 A.M.
Trent’s ferry arrived at 11:00 A.M.
9-5
Time and Temperature
Course 1
Check It Out: Example 2B
Franco’s train arrived at 4:55 P.M. He had traveled for 3 hours and 15 minutes. At what time did Franco’s train depart?
Arrival time: 4:55 P.M.
Think: 3 hours before 4:55 P.M. is 1:55 P.M. 15 minutes before that is 1:40 P.M.
Departure time: 1:40 P.M.
Franco’s train departed at 1:40 P.M.
9-5
Time and Temperature
Course 1
Celsius and Fahrenheit are the scales used to measure temperature. You can use these formulas to convert temperature.
9 5
5 9
9-5
Time and Temperature
Course 1
Estimate the temperature.
9 5
F = • C + 32
Use the formula.
9 5
Round to 2, and 32 to 30
Use the order of operations.
F = 2 • 40 + 30
F = 80 + 30
F = 110
9-5
Time and Temperature
Course 1
Estimate the temperature.
5 9
C = (F – 32)
Use the formula.
5 9
1 2
Round to , and 32 to 30
1 2
Use the order of operations.
C = (70 - 30)
1 2
C = (40)
C = 20
9-5
Time and Temperature
Course 1
Check It Out: Example 3
Estimate the temperature.
9 5
F = • C + 32
Use the formula.
9 5
Round to 2, and 32 to 30
Use the order of operations.
F = 2 • 12 + 30
F = 24 + 30
F = 54
9-5
Time and Temperature
Course 1
Check It Out: Example 3B
Estimate the temperature.
5 9
C = (F – 32)
Use the formula.
5 9
1 2
Round to , and 32 to 30
1 2
Use the order of operations.
C = (82 – 30)
1 2
C = (52)
C = 26
9-5
Time and Temperature
Course 1
Insert Lesson Title Here
Lesson Quiz
Convert.
1. 17 min = _____ s 2. 49 days = _____ weeks
3. 336 hr = _____ weeks
4. Ray drove 3 hours and 10 minutes to visit friends. He started his trip at 9:45 A.M. What was his arrival time?
Estimate the temperature.
5. 20°C is about ____ °F 6. 66°F is about ____ °C
7
1,020
2
12:55 A.M.
70
18 | 1,886 | 5,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-13 | longest | en | 0.831113 |
https://socratic.org/questions/how-many-moles-of-iron-are-in-1-2-10-25-atoms-of-iron | 1,580,181,907,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00211.warc.gz | 672,635,433 | 5,977 | # How many moles of iron are in 1.2*10^25 atoms of iron?
There are approx. $6.022 \times {10}^{23}$ iron atoms per mole of ion. This quantity, this number of iron atoms has a mass of $55.85 \cdot g$. And of course, when we quote the mass of iron we say $55.85 \cdot g \cdot m o {l}^{-} 1$.
$\left(1.2 \times {10}^{25} \cdot \text{atoms of iron")/(6.022xx10^23*"iron atoms per mole}\right)$ $\cong$ $20 \cdot \text{moles}$ | 151 | 422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-05 | latest | en | 0.656134 |
https://www.c-lang.thiyagaraaj.com/data-structures/c-sorting-programs/simple-heap-sort-program-in-c | 1,720,976,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00552.warc.gz | 631,969,465 | 7,065 | # Simple Heap Sort Program in C
## Definition
Heapsort is a comparison-based sorting algorithm. Heapsort can be thought of as an improved selection sort: like that algorithm, it divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest element and moving that to the sorted region. The improvement consists of the use of a heap data structure rather than a linear-time search to find the maximum.
## Simple Heap Sort Program example
``````/* Simple Heap Sort Program Using Functions and Array in C*/
/* Data Structure Programs,C Functions and Array Examples */
#include<stdio.h>
#include<conio.h>
#define MAX_SIZE 5
void heap_sort();
int arr_sort[MAX_SIZE], t, a;
int main() {
int i;
printf("Simple Heap Sort Example - Functions and Array\n");
printf("\nEnter %d Elements for Sorting\n", MAX_SIZE);
for (i = 0; i < MAX_SIZE; i++)
scanf("%d", &arr_sort[i]);
for (i = 0; i < MAX_SIZE; i++) {
printf("\t%d", arr_sort[i]);
}
heap_sort();
printf("\n\nSorted Data :");
for (i = 0; i < MAX_SIZE; i++) {
printf("\t%d", arr_sort[i]);
}
getch();
}
void heap_sort() {
for (int i = MAX_SIZE / 2 - 1; i >= 0; i--)
for (int i = MAX_SIZE - 1; i >= 0; i--) {
//Swapping Values
t = arr_sort[0];
arr_sort[0] = arr_sort[i];
arr_sort[i] = t;
printf("\nHeap Sort Iteration %d : ", i);
for (a = 0; a < MAX_SIZE; a++) {
printf("\t%d", arr_sort[a]);
}
}
}
void heap_adjust(int n, int i) {
int large = i, left = 2 * i + 1, right = 2 * i + 2;
if (left < n && arr_sort[left] > arr_sort[large])
large = left;
if (right < n && arr_sort[right] > arr_sort[large])
large = right;
if (large != i) {
//Swapping Values
t = arr_sort[i];
arr_sort[i] = arr_sort[large];
arr_sort[large] = t;
}
}
``````
## Sample Output
``````Simple Heap Sort Example - Functions and Array
Enter 5 Elements for Sorting
500
401
300
20
10
Your Data : 500 401 300 20 10
Heap Sort Iteration 4 : 401 20 300 10 500
Heap Sort Iteration 3 : 300 20 10 401 500
Heap Sort Iteration 2 : 20 10 300 401 500
Heap Sort Iteration 1 : 10 20 300 401 500
Heap Sort Iteration 0 : 10 20 300 401 500
Sorted Data : 10 20 300 401 500`````` | 700 | 2,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.534372 |
https://2022.help.altair.com/2022.2/hwdesktop/hwx/topics/pre_processing/interfacing/hv_element_forces_tensor_plot_r.htm | 1,718,687,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00021.warc.gz | 58,947,989 | 16,512 | # Element Forces (Tensor Plot)
2D element forces and moments are supported as tensors and can be plotted using the Tensor panel in HyperView.
This functionality is only available for the .op2 file. Element force and moment data is requested using the ELFORCE or FORCE case control command.
## Element X-Axis
In order to visualize element forces and moments with a proper sense of orientation, the element x-axis must first be identified. Refer to the CQUAD4/CTRIA3 documentation in the Radioss, MotionSolve, and OptiStruct Guide for more information on how to define the element x-axis for 2D elements.
Load the model and results files with the Advanced Result-Math template in order to visualize the element x-axis in HyperView.
Once the results are loaded, the Derived Results icon will launch the Expression Builder allowing you to define model or results-specific datatypes.
The Expression Builder is a graphical user interface which allows user-defined data type expressions to be authored directly within HyperView. These expressions use a simplified Templex-style syntax that gets parsed into XML statements which are then passed on to Result Math for processing. Refer to the HyperView Expression Builder documentation for more information.
The element x-axis vector can be added to the list of datatypes by double clicking ElementAxisVector from the Model Operator. Use a meaningful label so this expression can be identified amongst the other vector datatypes while plotting.
Below is a vector plot of each 2D element x-axis.
When the element x-axis vector plot is overlaid on a force or moment tensor plot, the positive and negative X, Y and XY values become clear.
## Forces
The image below shows the positive directions for element forces in a 2D element.
• Fx and Fy are the normal forces acting on the x and y faces of the 2D element below and are represented by the XX and YY components in the Regular drawing mode of the tensor plot in HyperView.
• Fxy is the in-plane shear force and is represented by the XY component in the Regular drawing mode of the tensor plot in HyperView.
• Vx and Vy are the transverse shear forces acting on the x and y faces of the 2D element below and are represented by the ZX and YZ components in the Regular drawing mode of the tensor plot in HyperView.
## Moments
The image below shows the positive directions for element moments in a 2D element.
• Mx and My are the bending moments acting on the x and y faces of the 2D element below and are represented by the XX and YY components in the Moment drawing mode of the tensor plot in HyperView.
• Mxy is the twisting moment and is represented by the XY component in the Moment drawing mode of the tensor plot in HyperView.
## Plotting Element Forces and Moments using the Tensor Panel
There are a number of options in the tensor panel and since it is used to plot more than just element forces and moments, some of the options are not meant to be used while plotting forces and moments.
Resolved in
Element forces and moments are always written to the .op2 file in the elemental system. If there is a material system defined on the 2D elements, and PARAM, OMID is used, these forces and moments can be transformed and written to that system, but only in an ASCII output.
Now that element forces and moments are read into HyperView as tensors they can be transformed from the elemental system to any other system, even a system created on the fly in HyperView.
Tensor Format
The Tensor format should be set to Component. This allows any or all of the full tensor components to be plotted. The Principal Tensor format is used for plotting stress tensors.
Display Options
Size scaling and color by options are rather straightforward. For clear results normalize the tensor to a value of 1.0 or less. This will ensure that the tensor drawings will not breech the elements’ edge.
There are two options available for the Draw mode: Regular and Moment.
Draw mode: Regular
Regular drawing mode will plot a traditional full tensor for each element. This is typically used for stress and element force tensor plots, and useful to properly visualize tensor transformations.
Draw mode: Moment
The Moment drawing mode was specially added to better conceptualize 2D element moments. This mode is not useful for transformations, and only applies for element moments, but it will allow you to better understand the physics behind the subcase without needing an element reference guide. The Regular drawing mode can be used to visualize any tensor, including element moments, but the Moment drawing mode was specifically designed for plotting element moments themselves. | 945 | 4,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.839534 |
https://en.autolifeadvice.com/10560804-how-to-charge-a-motorcycle-battery | 1,702,042,247,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00502.warc.gz | 268,121,449 | 12,179 | # How To Charge A Motorcycle Battery
## Video: How To Charge A Motorcycle Battery
The rechargeable battery of a motorcycle differs from a similar car battery in a smaller capacity. With a kickstarter, it works in much better conditions than an automobile. But even the kickstarter is sometimes unable to start the engine if the battery is discharged. This is known to everyone, thanks to the famous episode from the movie "Beware of the car".
## Instructions
### Step 1
Long before the battery requires another recharge, measure the voltage in the on-board network of a serviceable motorcycle with the engine running at medium speed. Write it down.
### Step 2
When the battery requires charging, remove it from the motorcycle. Draw in what polarity it was connected.
### Step 3
Become familiar with the electrochemical system used in the battery. It can be nickel-cadmium or lead. The former are usually used in motorcycles equipped with only a kick starter, the latter in motorcycles equipped with an electric starter or both. But there are also exceptions to this rule. Information about the electrochemical system of the battery is given on its case or on a sticker on it. Some modern motorcycles use lithium-ion and lithium-iron batteries. Do not try to charge them yourself.
### Step 4
Since motorcycle batteries have a lower capacity than car batteries, never use a device designed for the latter to charge them, unless the device has a charging current regulator.
### Step 5
Use a special laboratory power supply unit equipped with decade switches to charge the battery, which has a function to stabilize not only voltage, but also current strength. From domestic, B5-47 is suitable, in particular. Units without current stabilization function are not applicable.
### Step 6
Allow the battery to warm up to room temperature before charging.
### Step 7
Make sure there are no open flames or sparks around the battery.
### Step 8
If the battery is nickel-cadmium, multiply its ampere-hour capacity by 0.1 to get its charging current in amperes. Connect it to the power supply, observing the polarity. Set the ten-day switches to the voltage equal to the one that develops in the on-board network of a working motorcycle at medium speed, and the current equal to the charge current calculated by you. Turn on the power supply and charge the battery for 15 hours.
### Step 9
If the battery is lead-acid, charge it in two steps. First, set the current in amperes equal to 0.1 of the capacity in ampere-hours. Charge until the voltage reaches the result of multiplying the number of battery cells by 2, 4, expressed in volts. Reduce the charge current by half, then continue charging for another two hours.
### Step 10
Disconnect the battery from the power supply. Move it back to the motorcycle, observing the polarity. | 595 | 2,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.940038 |
https://www.polymathlove.com/special-polynomials/adding-fractions/free-online-differential.html | 1,529,865,716,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867050.73/warc/CC-MAIN-20180624180240-20180624200240-00078.warc.gz | 894,339,944 | 12,382 | Algebra Tutorials!
Sunday 24th of June
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Copyrights © 2005-2018 | 1,879 | 8,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-26 | latest | en | 0.821743 |
http://asokatechnologies.in/tag/matlab/ | 1,547,601,584,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583656577.40/warc/CC-MAIN-20190116011131-20190116033131-00517.warc.gz | 16,069,894 | 18,876 | MATLAB-Simulink Model Based Shunt Active Power Filter Using Fuzzy Logic Controller to Minimize the Harmonics
ABSTRACT:
The problem of quality electrical energy provided to the users has arisen. This is due to the increasing presence in network of nonlinear loads.They constitute a harmonic pollution source of the network, which generate many disturbances, and disturb the optimal operation of electrical equipments. This work, proposed a solution to eliminate the harmonics introduced by the nonlinear loads. It presents the analysis and simulation using Matlab Simulink of a active power filter (APF) compensating the harmonics and reactive power created by nonlinear loads in steady and in transients. The usefulness of the simulation approach to APF is demonstrated , have a better power quality insight using Matlab Simulink in order to develop new fuzzy logic controller based active power filter.
KEYWORDS:
1. Active Power Filters
2. Harmonics
3. Fuzzy Logic Controller
4. MATLAB
BLOCK DIAGRAM:
Figure 1 Block diagram of Basic Active Power Filter
EXPECTED SIMULATION RESULTS:
Fig. 2 Three phase voltage and current waveform with non linear load
Fig.3 THD analysis of three phase voltage waveform with nonlinear load
Fig.4 Three phase voltages and current waveform with shunt active power filter with connected fuzzy logic controller
Fig.5 THD analysis of voltages with shunt active power filter using fuzzy logic controller
CONCLUSION:
The paper presents the application of the fuzzy logic controller to control the compensating voltage. The Mamdani max-min approach is used for the fuzzy inference and the defuzzification method, respectively. The design of input and output membership for the fuzzy logic controller is very important for the system performance. The simulation results show that the fuzzy logic controller provides a good performance to control the compensating voltage of shunt active power filter. The %THD of the voltages at PCC point can be followed the IEEE Std. 519-1992.
REFERENCES:
[1] I. J. Pitel, S. N. Talukdar, and P. Wood, “Characterization of Programmed-Waveform Pulse-Width Modulation,” IEEE Transactions on Industry Applications, Vol. IA-16, Sept./Oct. 1980, pp. 707–715.
[2] Wilson E. Kazibwe and Mucoke H. Senduala : “Electric Power Quality Control Techniques”. New York: Van Nostrand Reinhold, 1993
[3] N. Mohan, “A Novel Approach to Minimize Line- Current Harmonics in Interfacing Power Electronics Equipment with 3-Phase Utility Systems”, IEEE Trans on Power Delivery, Vol. 8, July. 1993, pp 1395-1401.
[4] Elias M. Stein, Timonthy S. Murphy : “Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals.”, Princeton, N.J.: Princeton University Press, 1993
[5] J.S. Lai and T.S. Key, “Effectiveness of harmonic mitigation equipment for commercial office buildings,” IEEE Transactions on Industry Applications, vol.33, no.4, sep 1997, pp. 1065-1110
final year eee in ieee electrical projects in mancherial
final year eee in ieee electrical projects in mancherial.
Areas : Power Electronics and Drives, Power Systems, Renewable Energy and sources, etc
email: asokatechnologies@gmail.com
website: www.asokatechnologies.in
Asoka technologies provide Academic Electrical Projects mancherial.
ELECTRICAL ENGINEERING is a field of engineering that generally deals with the study and application of electricity, electronics, and electro magnetism. This field first became an identifiable occupation in the later half of the 19th century after commercialization of the electric telegraph, the telephone, and electric power distribution and use. Subsequently, broad casting and recording media made electronics part of daily life. The invention of the transistor, and later the integrated circuit, brought down the cost of electronics to the point they can be used in almost any household object.
Electrical engineering has now subdivided into a wide range of sub fields including electronics, digital computers, power engineering, tele communications, control systems, radio-frequency engineering, signal processing, instrumentation, and microelectronics. Many of these sub disciplines overlap and also overlap with other engineering branches, spanning a huge number of specializations such as hardware engineering, power electronics, electro magnetics & waves, microwave engineering, nanotechnology, electro chemistry, renewable energies, mechatronics, electrical materials science, and many more.
Areas : Power Electronics and Drives, Power Systems, Renewable Energy and sources, etc
email: asokatechnologies@gmail.com
website: www.asokatechnologies.in
ELECTRICAL ENGINEERING is a field of engineering that generally deals with the study and application of electricity, electronics, and electro magnetism. This field first became an identifiable occupation in the later half of the 19th century after commercialization of the electric telegraph, the telephone, and electric power distribution and use. Subsequently, broad casting and recording media made electronics part of daily life. The invention of the transistor, and later the integrated circuit, brought down the cost of electronics to the point they can be used in almost any household object.
Electrical engineering has now subdivided into a wide range of sub fields including electronics, digital computers, power engineering, tele communications, control systems, radio-frequency engineering, signal processing, instrumentation, and microelectronics. Many of these sub disciplines overlap and also overlap with other engineering branches, spanning a huge number of specializations such as hardware engineering, power electronics, electro magnetics & waves, microwave engineering, nanotechnology, electro chemistry, renewable energies, mechatronics, electrical materials science, and many more.
POWER ELECTRONICS is the application of solid-state electronics to the control and conversion of electric power. The first high power electronic devices were mercury-arc valves. In modern systems the conversion is performed with semiconductor switching devices such as diodes, thyristors and transistors, pioneered by R. D. Middlebrook and others beginning in the 1950s. In contrast to electronic systems concerned with transmission and processing of signals and data, in power electronics substantial amounts of electrical energy are processed. An AC/DC converter (rectifier) is the most typical power electronics device found in many consumer electronic devices, e.g. television sets, personal computers, battery chargers, etc. The power range is typically from tens of watts to several hundred watts. In industry a common application is the variable speed drive (VSD) that is used to control an induction motor. The power range of VSDs start from a few hundred watts and end at tens of megawatts.
An ELECTRIC POWER SYSTEM is a network of electrical components deployed to supply, transfer, and use electric power. An example of an electric power system is the the grid that provides power to an extended area. An electrical grid power system can be broadly divided into the generators that supply the power, the transmission system that carries the power from the generating centres to the load centres, and the distribution system that feeds the power to nearby homes and industries. Smaller power systems are also found in industry, hospitals, commercial buildings and homes. The majority of these systems rely upon three-phase AC power—the standard for large-scale power transmission and distribution across the modern world. Specialised power systems that do not always rely upon three-phase AC power are found in aircraft, electric rail systems, ocean liners and automobiles.
MATLAB (matrix laboratory) is a multi-paradigm numerical computing environment and fourth-generation programming language. A proprietary programming language developed by MathWorks, MATLAB allows matrix manipulations, plotting of functions and data, implementation of algorithms, creation of user interfaces, and interfacing with programs written in other languages, including C, C++, C#, Java, Fortran and Python.
SIMULINK, developed by MathWorks, is a graphical programming environment for modeling, simulating and analyzing multidomain dynamic systems. Its primary interface is a graphical block diagramming tool and a customizable set of block libraries. It offers tight integration with the rest of the MATLAB environment and can either drive MATLAB or be scripted from it. Simulink is widely used in automatic control and digital signal processing for multidomain simulation and Model-Based Design.
Final Year EEE Projects ideas as many people are searching for this kind of post on internet for many days. So, here we have included various projects in different categories like embedded, electrical, robotics, communication, solar, sensor, etc. I hope these eee projects for final year students would be more helpful for many engineering students in completing their B.Tech successfully.
FINAL YEAR ACAEMIC ELECTRICAL PROJECTS:
Electrical activities covering center electrical tasks, gadgets and inserted electrical are most alluring among the understudy level undertaking work. It gives commonsense presentation on the equipment that are regularly utilized in ventures. Constant modern level activities in machines, transmission lines, control hardware, high voltage and so on are famous as the hypothetical subjects read on the same is connected in functional terms for inside and out comprehension of the same.
Progressed electrical designing subjects, for example, FACTS, UPFC, SVPWM, APFC regularly utilize control gadgets like MOSFET, IGBT, SCR, TRIAC. In this manner, essential things on such power gadgets are a pre-imperative for understanding these undertakings. Rather than equipment based tasks, MATLAB ventures (programming based) give slightest introduction on the constant equipment applications which genuinely restricts openings for work for building understudies in enterprises. Anyway MATLAB is most appropriate for R&D level of work in scholastics
An Integrated Hybrid Power Supply for Distributed Generation Applications Fed by Nonconventional Energy Sources
ABSTRACT
A new, hybrid integrated topology, fed by photovoltaic (PV) and fuel cell (FC) sources and suitable for distributed generation applications, is proposed. It works as an uninterruptible power source that is able to feed a certain minimum amount of power into the grid under all conditions. PV is used as the primary source of power operating near maximum power point (MPP), with the FC section (block), acting as a current source, feeding only the deficit power. The unique “integrated” approach obviates the need for dedicated communication between the two sources for coordination and eliminates the use of a separate, conventional dc/dc boost converter stage required for PV power processing, resulting in a reduction of the number of devices, components, and sensors. Presence of the FC source in parallel (with the PV source) improves the quality of power fed into the grid by minimizing the voltage dips in the PV output. Another desirable feature is that even a small amount of PV power (e.g., during low insolation), can be fed into the grid. On the other hand, excess power is diverted for auxiliary functions like electrolysis, resulting in an optimal use of the energy sources. The other advantages of the proposed system include low cost, compact structure, and high reliability, which render the system suitable for modular assemblies and “plug-n-play” type applications. All the analytical, simulation results of this research are presented.
INDEX TERMS: Buck-boost, distributed generation, fuel cell, grid-connected, hybrid, maximum power point tracking (MPPT), photovoltaic.
BLOCK DIAGRAM
Fig. 1. Various HDGS configurations. (a) Conventional, multistage topology using two H-bridge inverters [4], [6]. (b) Modified topology with only one H-bridge inverter [4]. (c) Proposed topology. λ denotes solar insolation (Suns).
SIMULATION RESULTS
Fig. 2. Simulation results of the integrated hybrid configuration showing transition from mode III to mode II and then to mode I. T1 and T2 denote the transition between mode III to mode II and mode II to mode I respectively.
Fig. 3. Simulation results of the integrated hybrid configuration operating in electrolysis mode (mode I to mode III and then to mode I). T1 and T2 denote the transition between mode I to mode III and mode III to mode I respectively.
Fig.4. Performance comparison of the proposed HDGS system with and without an FC source in parallel with the PV source.
CONCLUSION
A compact topology, suitable for grid-connected applications has been proposed. Its working principle, analysis, and design procedure have been presented. The topology is fed by a hybrid combination of PV and FC sources. PV is the main source, while FC serves as an auxiliary source to compensate for the uncertainties of the PV source. The presence of FC source improves the quality of power (grid current THD, grid voltage profile, etc.) fed into the grid and decreases the time taken to reach theMPP. Table IV compares the system performance with and without the FC block in the system. A good feature of the proposed configuration is that the PV source is directly coupled with the inverter (and not through a dedicated dc–dc converter) and the FC block acts as a current source. Considering that the FC is not a stiff dc source, this facilitates PV operation at MPP over a wide range of solar insolation, leading to an optimal utilization of the energy sources. The efficiency of the proposed system in mode-1 is higher (around 85% to 90%) than mode 2 and 3 (around 80% to 85%).
REFERENCES
[1] J. Kabouris and G. C. Contaxis, “Optimum expansion planning of an unconventional generation system operating in parallel with a large scale network,” IEEE Trans. Energy Convers., vol. 6, no. 3, pp. 394–400, Sep. 1991.
[2] P. Chiradeja and R. Ramakumar, “An approach to quantify the technical benefits of distributed generation,” IEEE Trans. Energy Convers., vol. 19, no. 4, pp. 764–773, Dec. 2004.
[3] Y. H. Kim and S. S. Kim, “An electrical modeling and fuzzy logic control of a fuel cell generation system,” IEEE Trans. Energy Convers., vol. 14, no. 2, pp. 239–244, Jun. 1999.
[4] K. N. Reddy and V. Agarwal, “Utility interactive hybrid distributed generation scheme with compensation feature,” IEEE Trans. Energy Convers., vol. 22, no. 3, pp. 666–673, Sep. 2007.
[5] K. S. Tam and S. Rahman, “System performance improvement provided by a power conditioning subsystem for central station photovoltaic fuel cell power plant,” IEEE Trans. Energy Convers., vol. 3, no. 1, pp. 64–70.
A Novel High StepUp DCDC Converter Based on Integrating Coupled Inductor and Switched-Capacitor Techniques for Renewable Energy Applications
ABSTRACT
In this paper, a novel high step-up dc/dc converter is presented for renewable energy applications. The suggested structure consists of a coupled inductor and two voltage multiplier cells, in order to obtain high step-up voltage gain. In addition, two capacitors are charged during the switch-off period, using the energy stored in the coupled inductor which increases the voltage transfer gain. The energy stored in the leakage inductance is recycled with the use of a passive clamp circuit. The voltage stress on the main power switch is also reduced in the proposed topology. Therefore, a main power switch with low resistance RDS(ON) can be used to reduce the conduction losses. The operation principle and the steady-state analyses are discussed thoroughly. To verify the performance of the presented converter, a 300-W laboratory prototype circuit is implemented. The results validate the theoretical analyses and the practicability of the presented high step-up converter.
KEYWORDS:
Coupled inductor, DC/DC converters, High step-up, Switched capacitor.
CIRCUIT DIAGRAM:
Fig. 1. Circuit configuration of the presented high-step-up converter.
SIMULATION RESULTS:
Fig. 2. Simulation results under load 300 W.
CONCLUSION
This paper presents a new high-step-up dc/dc converter for renewable energy applications. The suggested converter is suitable for DG systems based on renewable energy sources, which require high-step-up voltage transfer gain. The energy stored in the leakage inductance is recycled to improve the performance of the presented converter. Furthermore, voltage stress on the main power switch is reduced. Therefore, a switch with a low on-state resistance can be chosen. The steady-state operation of the converter has been analyzed in detail. Also, the boundary condition has been obtained. Finally, a hardware prototype is implemented which converts the 40-V input voltage into 400-V output voltage. The results prove the feasibility of the presented converter.
REFERENCES
[1] F.Nejabatkhah, S. Danyali, S. Hosseini, M. Sabahi, and S. Niapour, “Modeling and control of a new three-input DC–DC boost converter for hybrid PV/FC/battery power system,” IEEE Trans. Power Electron., vol. 27, no. 5, pp. 2309–2324, May 2012.
[2] R. J. Wai and K. H. Jheng, “High-efficiency single-input multiple-output DC–DC converter,” IEEE Trans. Power Electron., vol. 28, no. 2, pp. 886–898, Feb. 2013.
[3] Y. Zhao, X. Xiang, C. Li, Y. Gu, W. Li, and X. He, “Single-phase high step-up converter with improved multiplier cell suitable for half- bridgebased PV inverter system,” IEEE Trans. Power Electron., vol. 29, no. 6, pp. 2807–2816, Jun. 2014.
[4] J.H. Lee, T. J. Liang, and J. F. Chen, “Isolated coupled-inductor-integrated DC–DC converter with non-dissipative snubber for solar energy applications,” IEEE Trans. Ind. Electron., vol. 61, no. 7, pp. 3337–3348, Jul.2014.
[5] C.Olalla, C. Delineand, andD.Maksimovic, “Performance of mismatched PV systems withsubmodule integrated converters,” IEEE J. Photovoltaic, vol. 4, no. 1, pp. 396–404, Jan. 2014. | 3,856 | 17,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-04 | longest | en | 0.855161 |
http://www.reading.ac.uk/module/document.aspx?modP=BI3AE16&modYR=1617 | 1,643,314,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00447.warc.gz | 109,279,146 | 5,080 | ## BI3AE16-Analogue Electronics
Module Provider: School of Biological Sciences
Number of credits: 10 [5 ECTS credits]
Level:6
Terms in which taught: Autumn term module
Pre-requisites:
Non-modular pre-requisites:
Co-requisites:
Modules excluded:
Module version for: 2016/7
Module Convenor: Prof Simon Sherratt
Summary module description:
Aims:
This module aims to introduce students to the computer simulation of analogue circuits with an emphasis on power electronic circuits.
Assessable learning outcomes:
To be able to write a Spice text input file defining an electronic circuit and to interpret the text output from a Spice simulator. To understand the various types of circuit analysis provided by Spice. To be familiar with Orcad and to be able to define circuits using schematic entry. To be able to simulate analogue circuits using Orcad/PSpice and to interpret the text and graphical output. To be able to simulate a range of linear and non-linear analogue circuits and mixed analog/digital circuits.
To appreciate the part that simulation plays in analogue circuit design. To understand in general terms the computer algorithms used to simulate analogue circuits.
Outline content:
• Spice simulator - dc analysis, transient analysis, frequency response
• Text definition of circuits for spice - interpreting text output
• Simulation of simple circuits (filter, rectifier, logic gate)
• Models of semiconductor devices (diode, bipolar, MOSFET, op-amp)
• Schematic entry using Orcad, and simulation using PSpice
• Further analysis methods including noise, distortion and power consumption
• Simulation of a range of circuits including:
Linear active filters
Power amplifiers including classes A, B, AB, Quad
Converters including buck, boost, buck/boost
Inverters including sinusoidal class D
• Analogue/digital mixed-mode simulation
Brief description of teaching and learning methods:
Lectures supported by Lab practicals (students will simulate various circuits, and in some cases compare the simulation results with physical circuits)
Contact hours:
Autumn Spring Summer Lectures 10 Practicals classes and workshops 20 Guided independent study 70 Total hours by term 100.00 Total hours for module 100.00
Summative Assessment Methods:
Method Percentage Practical skills assessment 20 Set exercise 40 Class test administered by School 40
Other information on summative assessment:
Multichoice test based on practical work.
Formative assessment methods:
Penalties for late submission:
The Module Convenor will apply the following penalties for work submitted late, in accordance with the University policy.
• where the piece of work is submitted up to one calendar week after the original deadline (or any formally agreed extension to the deadline): 10% of the total marks available for the piece of work will be deducted from the mark for each working day (or part thereof) following the deadline up to a total of five working days;
• where the piece of work is submitted more than five working days after the original deadline (or any formally agreed extension to the deadline): a mark of zero will be recorded.
• The University policy statement on penalties for late submission can be found at: http://www.reading.ac.uk/web/FILES/qualitysupport/penaltiesforlatesubmission.pdf
You are strongly advised to ensure that coursework is submitted by the relevant deadline. You should note that it is advisable to submit work in an unfinished state rather than to fail to submit any work.
Length of examination:
None
Requirements for a pass:
40%
Reassessment arrangements:
Examination only.
One 2-hour examination paper in August/September. | 742 | 3,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-05 | longest | en | 0.797497 |
https://docs.roguewave.com/imsl/java/7.2/manual/api/com/imsl/math/HyperRectangleQuadrature.html | 1,597,088,298,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737168.97/warc/CC-MAIN-20200810175614-20200810205614-00278.warc.gz | 244,195,099 | 4,068 | JMSLTM Numerical Library 7.2.0
com.imsl.math
• All Implemented Interfaces:
Serializable, Cloneable
```public class HyperRectangleQuadrature
extends Object
implements Serializable, Cloneable```
HyperRectangleQuadrature integrates a function over a hypercube. This class is used to evaluate integrals of the form:
Integration of functions over hypercubes by Monte Carlo, in which the integral is evaluated as the value of the function averaged over a sequence of randomly chosen points. Under mild assumptions on the function, this method will converge like , where n is the number of points at which the function is evaluated.
It is possible to improve on the performance of Monte Carlo by carefully choosing the points at which the function is to be evaluated. Randomly distributed points tend to be non-uniformly distributed. The alternative to a sequence of random points is a low-discrepancy sequence. A low-discrepancy sequence is one that is highly uniform.
This function is based on the low-discrepancy Faure sequence as computed by `FaureSequence`.
Example, Serialized Form
• ### Nested Class Summary
Nested Classes
Modifier and Type Class and Description
`static interface ` `HyperRectangleQuadrature.Function`
Public interface function for the HyperRectangleQuadrature class.
• ### Constructor Summary
Constructors
Constructor and Description
`HyperRectangleQuadrature(int dim)`
`HyperRectangleQuadrature(RandomSequence sequence)`
• ### Method Summary
Methods
Modifier and Type Method and Description
`double` `eval(HyperRectangleQuadrature.Function objectF)`
Returns the value of the integral over the unit cube.
`double` ```eval(HyperRectangleQuadrature.Function objectF, double[] a, double[] b)```
Returns the value of the integral over a cube.
`double` `getErrorEstimate()`
Returns an estimate of the relative error in the computed result.
`void` `setAbsoluteError(double errorAbsolute)`
Sets the absolute error tolerance.
`void` `setRelativeError(double errorRelative)`
Sets the relative error tolerance.
• ### Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
• ### Constructor Detail
`public HyperRectangleQuadrature(int dim)`
`public HyperRectangleQuadrature(RandomSequence sequence)`
• ### Method Detail
• #### eval
`public double eval(HyperRectangleQuadrature.Function objectF)`
Returns the value of the integral over the unit cube.
Parameters:
`objectF` - `Function` containing the function to be integrated
• #### eval
```public double eval(HyperRectangleQuadrature.Function objectF,
double[] a,
double[] b)```
Returns the value of the integral over a cube.
Parameters:
`objectF` - `Function` containing the function to be integrated
`a` - is a `double` specifying the lower limit of integration. If null all of the lower limits default to 0.
`b` - is a `double` specifying the upper limit of integration. If null all of the upper limits default to 1.
• #### getErrorEstimate
`public double getErrorEstimate()`
Returns an estimate of the relative error in the computed result.
Returns:
a `double` specifying an estimate of the relative error in the computed result
• #### setAbsoluteError
`public void setAbsoluteError(double errorAbsolute)`
Sets the absolute error tolerance.
Parameters:
`errorAbsolute` - a `double` scalar value specifying the absolute error
• #### setRelativeError
`public void setRelativeError(double errorRelative)`
Sets the relative error tolerance.
Parameters:
`errorRelative` - a `double` scalar value specifying the relative error
JMSLTM Numerical Library 7.2.0 | 819 | 3,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-34 | latest | en | 0.719109 |
https://www.geeksforgeeks.org/how-to-convert-a-whole-number-to-an-improper-fraction/?ref=rp | 1,679,732,609,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00743.warc.gz | 874,129,761 | 27,419 | # How to Convert a Whole Number to an Improper Fraction?
• Last Updated : 25 Apr, 2022
Numbers are mathematical figures or values that may be used for counting, measuring, and doing other arithmetic operations. Integers, whole numbers, natural numbers, rational and irrational numbers, and so on are examples of numbers. The number system is a standardized method of expressing numbers in many formats, including figures and language. It contains many sorts of numbers such as prime numbers, odd numbers, even numbers, rational numbers, whole numbers, and so on. These numbers can be represented in several ways depending on the number system used.
The number system includes several sorts of numbers, such as prime numbers, odd numbers, even numbers, rational numbers, whole numbers, and so on. These figures and words can be used to express these numbers. For example, integers like 50 and 35 stated as figures can alternatively be written as fifty and thirty-five.
### Whole numbers
The subset of numbers that includes zero and all positive integers is known as whole numbers. The overall number goes from 0 to infinity. These numbers are used in everyday computations, mostly to measure fundamental quantities. Natural numbers are made up of solely whole numbers, including zero. The numbers 0, 1, 2, 3, 4, 5,… denote the subset. The subset excludes fractions, decimals, and negative integers.
Positive integers, also known as counting numbers, are parts of whole numbers that contain zero, such as 0,1, 2, 3, 4, 5, and so on while excluding negative integers, fractions, and decimals. 10, 11, 22, 100, 1000, and so on are examples of whole numbers.
Improper Fraction
An improper fraction is a type of fraction in which the numerator exceeds or equals the denominator. Every fraction is made up of two parts: the numerator and the denominator. Proper fractions and improper fractions are the two primary types of fractions in mathematics depending on the values of the numerator and denominator. For example, 7/2 and 9/5, are improper fractions.
### How to Convert a Whole Number to an Improper Fraction?
To change a whole number into an improper fraction, the following steps are required,
• Step 1: Reduce the whole number to a proper fraction. A numerator is a whole number while the denominator is 1. For instance, if the whole number is 5, the fraction’s numerator is 5 and the denominator is 1. As a result, 5/1.
• Step 2: Now Choose any fraction that is equal to 1. When the number of numerator and denominator of a fraction are the same then the fraction is equal to 1. 4/4, 6/6, and 7/7 are a few examples.
• Step 3: At last Multiply the whole number by the fractional which is equivalent to 1. Take the improper fraction from Step 1 above. Multiply it by any fraction that equals one. This has no effect on the original improper fraction’s value. For example Multiply 5/1 by 5/5, (numerator times numerator, and denominator times denominator). 5/1 divided by 5/5 equals 25/5.
As a result, after simplifying the improper fraction, we will receive the same whole number, which is 5.
### Sample Questions
Question 1: Identify improper fractions out of the following: 11/5, 6, 2/9, 6/3, 3/5.
Solution:
An improper fraction is a type of fraction in which the numerator exceeds or equals the denominator. Every fraction is made up of two parts: the numerator and the denominator.
from the above given fractions, 11/5, 6, and 6/3 are improper fractions. As, in 11/5, 11 > 5, 6 can be written as 6/1 where 6 > 1, and 6/3 can be simplified as 2/1 where 2 > 1.
Question 2: How to change the whole number 8 into an improper fraction?
Solution:
Here 8 is a whole number,
• Step 1: Reduce the whole number to a improper fraction. we can write it as 8/1
• Step 2: Now Choose any fraction that is equal to 1. When the number of numerator and denominator of a fraction are the same then the fraction is equal to 1. So here fraction is 8/8 as this is equal to 1 .
• Step 3: Atlast Multiply the whole number by the fractional which is equivalent of 1. Take the improper fraction from Step 1 above. Multiply it by any fraction that equals one. This has no effect on the original improper fraction’s value.
Therefore here whole number 8 , which we can write as 8/1 will be multiply by fraction which is equivalent to 1 i.e 8/8
= 8/1 × 8/8
= 64/8 ,its an improper fraction.
Question 3: How to convert 3/5 into a whole number?
Here 3/5 is not a improper fraction as numerator is less than denominator .The only fractions that can be converted into whole numbers are those with numerators that are the same as their denominators or that are an exact multiple of their denominators. Therefore 3/5 cannot be converted into whole numbers.
Question 4: How to change the whole number 10 into an improper fraction?
Solution:
Here 10 is a whole number,
• Step 1: Reduce the whole number to a improper fraction. we can write it as 10/1.
• Step 2: Now Choose any fraction that is equal to 1. When the number of numerator and denominator of a fraction are the same then the fraction is equal to 1. So here fraction is 10/10 as this is equal to 1.
• Step 3: Atlast Multiply the whole number by the fractional which is equivalent of 1. Take the improper fraction from Step 1 above. Multiply it by any fraction that equals one. This has no effect on the original improper fraction’s value. Therefore here whole number 10 , which we can write as 10/1 will be multiply by fraction which is equivalent to 1 i.e 10/10.
= 10/1 × 10/10
= 100/10 ,its an improper fraction.
Question 5: Simplify improper fractions 8/6 + 10/6 and find out the result is a whole number or not?
Solution:
Given: 8/6 + 10/6
Here with the same denominator is 6.
= 8/6 + 10/6
= (8 + 10)/6
= 18/6
Here 18/6 is an improper fraction after further simplifying,
= 18/6 = 3
Hence 3 is a whole number.
Question 6: Change the whole number 6 into an improper fraction.
Solution:
Here 6 is a whole number,
• Step 1: Reduce the whole number to an improper fraction. we can write it as 6/1.
• Step 2: Now Choose any fraction that is equal to 1. When the number of numerator and denominator of a fraction are the same then the fraction is equal to 1. So here fraction is 6/6 as this is equal to 1.
• Step 3: At last Multiply the whole number by the fractional which is equivalent to 1. Take the improper fraction from Step 1 above. Multiply it by any fraction that equals one. This has no effect on the original improper fraction’s value.
Therefore here whole number 6, which we can write as 6/1 will be multiplied by a fraction that is equivalent to 1 i.e 6/6
= 6/1 × 6/6
= 36/6, it’s an improper fraction, if simplified further it does not change the original value.
My Personal Notes arrow_drop_up | 1,702 | 6,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.920902 |
https://tantalum.academickids.com/encyclopedia/index.php/Axiomatic_system | 1,652,835,728,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00115.warc.gz | 613,701,955 | 6,737 | # Axiomatic system
In mathematics, an axiomatic system is any set of axioms from which some or all axioms can be used in conjunction to logically derive theorems. A mathematical theory consists of an axiomatic system and all its derived theorems. An axiomatic system that is completely described is a special kind of formal system; usually though the effort towards complete formalisation brings diminishing returns in certainty, and a lack of readability for humans. Therefore discussion of axiomatic systems is normally only semi-formal. A formal theory typically means an axiomatic system, for example formulated within model theory. A formal proof is a complete rendition of a proof within a formal system.
Contents
## Properties
An axiomatic system is said to be consistent if it lacks contradiction, i.e. the ability to derive both a statement and its negation from the system's axioms.
In an axiomatic system, an axiom is called independent if it is not a theorem that can be derived from other axioms in the system. A system will be called independent if each of its underlying axioms is independent.
Although independence is not a necessary requirement for a system, consistency is. An axiomatic system will be called complete if no additional axiom can be added to the system without making the new system either dependent or inconsistent.
## Models
A mathematical model for an axiomatic system is a well-defined set, which assigns meaning for the undefined terms presented in the system, in a manner that is correct with the relations defined in the system. The existence of a concrete model* proves the consistency of a system.
Models can also be used to show the independence of an axiom in the system. By constructing a valid model for a subsystem without a specific axiom, we show that the omitted axiom is independent if its correctness does not necessarily follow from the subsystem.
Two models are said to be isomorphic if a one-to-one correspondence can be found between their elements, in a manner that preserves their relationship. An axiomatic system for which every model is isomorphic to another is called categorial, and the property of categoriality ensures the completeness of a system.
* A model is called concrete if the meanings assigned are objects and relations from the real world, as opposed to an abstract model which is based on other axiomatic systems.
The first axiomatic system was Euclidean geometry.
## Axiomatic method
The axiomatic method is often discussed as if it were a unitary approach, or uniform procedure. With the example of Euclid to appeal to, it was indeed treated that way for many centuries: up until the beginning of the nineteenth century it was generally assumed, in European mathematics and philosophy (for example in Spinoza's work) that the heritage of Greek mathematics represented the highest standard of intellectual finish (development more geometrico, in the style of the geometers).
This traditional approach, in which axioms were supposed to be self-evident and so indisputable, was swept away during the course of the nineteenth century, by the development of Non-Euclidean geometry, the foundations of real analysis, Cantor's set theory and Frege's work on foundations, and Hilbert's 'new' use of axiomatic method as a research tool. For example, group theory was first put on an axiomatic basis towards the end of that century. Once the axioms were clarified (that inverse elements should be required, for example), the subject could proceed autonomously, without reference to the transformation group origins of those studies.
Therefore there are at least three 'modes' of axiomatic method current in mathematics, and in the fields it influences. In caricature, possible attitudes are
1. Accept my axioms and you must accept their consequences;
2. I reject one of your axioms and accept extra models;
3. My set of axioms defines a research programme.
The first case is the classic deductive method. The second goes by the slogan be wise, generalise; it may go along with the assumption that concepts can or should be expressed at some intrinsic 'natural level of generality'. The third was very prominent in the mathematics of the twentieth century, in particular in subjects based around homological algebra.
It is easy to see that the axiomatic method has limitations outside mathematics. For example, in political philosophy axioms that lead to unacceptable conclusions are likely to be rejected wholesale; so that no one really assents to version 1 above.
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Title: Technology in Architecture
1
Technology in Architecture
• Lecture 4
• Lighting Design Example
2
• Example 1
• Room Layout Calculation
3
Example 1
• Classroom 20 x 27 x 12 E50 fc
• WP 2-6 AFF
• ?c 80 hcc 0.0
• ?w 50 hrc 9.5
• ?f 20 hfc 2.5
• fixture fluorescent (38)
• maintenance yearly
• replacement on burnout
• voltages ballast normal
• environment medium clean
4
Example 1
• Confirm fixture data
S T.15.1 p. 641
5
Example 1
• Complete 1-6
6
Example 1
• 7. Determine lumens per luminaire
• Obtain lamp lumens from manufacturers data (or
see Stein Chapter 12)
S T. 12.5 p. 546
7
Lumen Flux Method
0
27
?c 80 ?w 50 ?f 20
9.5
20
2.5
• 8. Record dimensional data
8
Coefficient of Utilization Factor(CU)
Calculation
• 9. Calculate Cavity Ratios
9
Example 1 Cavity Ratios
• CR 5 H x (LW)/(L x W)
• RCR 5 Hrc x (LW)/(LxW) 4.1
• CCR 5 Hcc x (LW)/(LxW) 0
• FCR 5 Hfc x (LW)/(LxW) 1.1
10
Coefficient of Utilization Factor(CU)
Calculation
• 10. Calculate Effective Ceiling Reflectance
11
Example 1 Coefficient of Utilization (CU)
• 3. Obtain effective ceiling reflectance
• S T.15.2 p. 667
12
Example 1
• 11. Calculate Effective Floor
Reflectance Stein T.15.2 P. 666
13
Example 1 Coefficient of Utilization (CU)
• 3. Obtain effective ceiling reflectance
CU 0.19 ? 0.20
S T.15.2 p. 667
14
Example 1
• 12. Select CU from mfrs data or see
15
Example 1 Coefficient of Utilization (CU)
• CU0.32
S T.15.1 p. 641
RCR CU 4.0 0.39 4.1
X 5.0 0.35
CU 0.386
16
Example 1
• 13-21 Calculate LLF
17
Example 1 Light Loss Factor(LLF)
• 13-16
• All factors not known ? 0.88
18
Example 1 Light Loss Factor(LLF)
• 17. Room Surface Dirt
• (based on 24 month cleaning cycle, normal
maintenance)
• Direct 0.92 /- 5
19
Light Loss Factor(LLF) Calculation
• 18. Lamp Lumen Depreciation
• Group Burnout
• Fluorescent 0.90 0.85
20
Example 1 Light Loss Factor(LLF)
• 19. Burnouts
• Burnout 0.95
21
Example 1 Light Loss Factor(LLF)
• 20. Luminaire Dirt Depreciation (LDD)
• Verify maintenance category
S T.15.1 p. 641
22
Example 1 Light Loss Factor(LLF)
• 20. Luminaire Dirt Depreciation (LDD)
LDD0.80
S F.15.34 p. 663
23
Example 1 Light Loss Factor(LLF)
• LLF a x b x c x d x e x f x g x h
• LLF 0.88 x 0.92 x 0.85 x 0.95 x 0.80
• LLF 0.52
24
Example 1
• 22. Calculate Number of Luminaires
22 23
25
Example1 Calculate Number of Luminaires
• No. of Luminaires
• (E x Area)/(Lamps/luminaire x Lumens/Lamp x CU x
LLF)
• (50 X 540)/(4 X 2950 x 0.386 x 0.52) 11.4
luminaires
26
Example 1
• Goal is 50 fc /- 10 ? 45-55 fc
• Luminaires E (fc)
• 10 43.9 x
• 11 48.2 ok ? 2 rows of 4, 1 row of 3
• 12 52.6 ok ? 3 rows of 4
• 13 57.0 x
• Verify S/MH for fixture, space geometry
27
Example 1 S/MH Ratio
• Verify S/MH ratio
• MH12.0-2.59.5 S/MH 1.0 ? S 9.5
S T.15.1 p. 641
28
Example 1 Spacing
• Try 3 rows of
• 4 luminaires
• S/23SS/220
• ? S5
• S/MH5/9.5 1.0 ok
• S/2SSs/227
• ? S9
• S/MH9/9.5 1.0 ok
S/2 S S S/2
S/2 S S
S S/2
27
20
29
Example 1 Spacing
S/2 S S S S/2
• Try 4 rows of
• 3 luminaires
• S/22SS/220
• ? S6.67
• S/MH6.67/9.5 1.0 ok
• S/23Ss/227
• ? S6.75
• S/MH6.75/9.5 1.0 ok
S/2 S S
S/2
27
20
30
• Example 2
• Economic Analysis
31
Example 2 Economic Analysis
• Operation 8AM-5PM, M-F, 52 wks/yr 9 x 5
x 52 2,340 hrs/yr
• Operating Energy 128 watts/luminaire
• Lighting Control Daylighting sensor with
3- step controller
32
Example 2 Economic Analysis
• Connected Lighting Power (CLP)
• CLP12 x 128 1,536 watts (2.8 w/sf)
• Adjusted Lighting Power (ALP)
• ALP(1-PAF) x CLP
33
Example 2 Economic Analysis
• Power
• Control Factor (PAF)
• Daylight Sensor (DS), 0.30 continuous
dimming
• DS, multiple-step dimming 0.20
• DS, On/Off 0.10
• Occupancy Sensor (OS) 0.30
• OS, DS, continuous dimming 0.40
• OS, DS, multiple-step dimming 0.35
• OS, DS, On/Off 0.35
• Source ASHRAE 90.1-1989
34
Example 2 Economic Analysis
• Adjusted Lighting Power (ALP)
• ALP(1-PAF) x CLP
• ALP(1-0.20) x 1536
• ALP 1229 watts (2.3 w/sf)
35
Example 2 Economic Analysis
• Energy 1,229 watts x 2,340 hrs/yr
• 2,876 kwh/year
• Electric Rate 0.081/kwh
• Annual Energy Cost 2,876 kwh/yr x 0.081/kwh
232.94/yr
36
Example 2 Economic Analysis
• An alternate control system consisting of a
daylighting sensor, with continuing dimming and
an occupancy sensor can be substituted for an
• Using the simple payback analysis method,
determine if switching to this control system is
economically attractive.
37
Example 2 Economic Analysis
• Power
• Control Factor (PAF)
• Daylight Sensor (DS), 0.30 continuous
dimming
• DS, multiple-step dimming 0.20
• DS, On/Off 0.10
• Occupancy Sensor (OS) 0.30
• OS, DS, continuous dimming 0.40
• OS, DS, multiple-step dimming 0.35
• OS, DS, On/Off 0.35
• Source ASHRAE 90.1-1989
38
Example 2 Economic Analysis
• Adjusted Lighting Power (ALP)
• ALP(1-PAF) x CLP
• ALP(1-0.40) x 1536
• ALP 922 watts (1.7 w/sf)
39
Example 2 Economic Analysis
• Energy 922 watts x 2,340 hrs/yr
• 2,157 kwh/year
• Annual Energy Cost 2,157 kwh/yr x 0.081/kwh
174.72/yr
• Annual Savings 232.94 174.72 58.22/year
• Simple Payback Additional Cost/Annual Savings
• 150.00/58.22
• 2.6 years lt 3 years
• Economically attractive
40
• Example 3
• Point Source Calculation
41
Example 3
S F.15.49 p. 677
• Spot Lighting lamp straight down
S F.15.48 p. 677
42
Example 3
S F.15.49 p. 677
• Spot Lighting lamp pointed at object
Cp at 90 9600 Horizontal illumination
9900(0.643)3 25.5 fc
102 Vertical illumination
9900(0.766)3 30.3 fc 122
43
(No Transcript) | 2,351 | 6,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-16 | latest | en | 0.551343 |
https://www.instructables.com/Chainmaille-101-Weaves-a-Plenty/ | 1,718,482,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00088.warc.gz | 741,947,776 | 41,663 | ## Introduction: Chainmaille 101: Weaves a Plenty
This small i guess small instructable will cover a few basic weaves some intermediate and a advanced or two.
Critical things you need to have and know before starting.
One: Tools, You will need pliers to open and close rings
Two: The Color Key, this key is used throught this instructable
Red: The ring(s) to weave thru.
Blue: The inactive or finished parts of a weave.
Green: the active or unclosed ring you are weaveing into the mesh/chain
Yellow: Orbital captive or pre closed ring.
Purple: Rubber or flexable ring.
Aspect Ratio: is a number driven from the difrance between the rings inner diameter and the wire diameter. we use this number to get a ring size that will work for a given weave. For example, If you want to make Byz you would use a AR or around 3.5 this means if you want to use 0.032 wire you simply multiply the wire size by the AR, 0.032 * 3.5 = 0.112, this tell you that your ID = 0.112. Another example, Say you have 0.25 mandrel for your ID but don't know what wire to use to get AR 3.5. so you divide the 0.25 by the AR of 3.5 this will give you 0.071. now say you made a awesome weave and what to share it with every one and you don't know that AR, well that easy, Divide the ID by the Wire diameter. ID 0.125 WD 0.032 = AR 3.9 One of the more amazing bonus of this added math is that it alows you to scale up or down a weave as well if you know the ID you want to get approximate size say 0.5 wide pice of BB2B and know that it is 4 rings wide you know already that 0.125 is going to be close to ideal ID you just need the wire size. ID 0.125 / AR 3.9 = WD 0.32. now say you don't like that size. and want it more belt sized say around 2in wide. you take your 0.5 ID divide it again by the 3.9 AR and you get a wire size of 0.128. So in summery the Formulas
ID / WD = AR
WD * AR = ID
ID / AR = WD
What weaves are here?
Euro 4in1 unbalanced aka Oops
Byzee Beez to Buterflies
Butterfly
Candy Cain Cord
Celtic Filagree
Fluer De Lis
Celtic Cross
Celtic Dawn
Conundrome
Berjao Flowers
Firemens Hold Link 4
Daisy Chain
Elfen Tracks
That should about cover the introduction, Lets get to weaving shall we.
## Step 1: Brejao Flowers
This is a fun weave with a nice sheet form making it useful for clothes bags draperies, and many other normally cloth forms.
AR 6.0 +
Step One: Begin with Two 2 in 1 chain of equal length, you may make more than two to start these chain are used to determine the pieces starting length.
Step two: Now Add a ring between the lowest two rings as shown, you will continue this every two rings up the chain forming 3 ring mobi all the way up, you will want to do this to both chains. At this stage it may be helpful to not get the rings all moved about, you can pass a rod or stiff string thru the center of the mobi units to prevent over tangling.
Step three: Lay at least two chains net to each other and in the mobi you formed at the bottom of the first chain add a ring to the out side of the mobi, then pass it thru the eye of the mobi of the second chain making it not a 4 ring mobi unit. again you will continue this up the chain every unit.
Step Four: To finish of the first chain just add the 4th ring to the mobi units and continue adding 3 ring mobi chains until your size is accomplished.
Thats all there is to it.
## Step 2: Butterfly Chain
This weave uses two ring sizes the smaller rings having a AR of 3.9 and the Larger rings having a AR of 5.
Step One: Make a few mobi out of the AR 5 ring I used three rings per mobius.
Step Two: Weave on 4 AR 3.9 ID rings onto the mass.
Step Three: Now we are going to split it up a bit. Weave one ring onto each of two rings splitting the four stack.
Step Four: Split the two new sections and weave in two rings, one on top and one on bottom, using one ring from each group.
Step Five: Repeat the other steps 2-4 on the opposing side of the Mobi this completes a single unit of the weave, make several of these before you continue.
Step Six: Now lets connect a few sections. Gather two sections and weave in a ring through the bottom three rings on the outer ends of both sections like so.
Step seven: Repeat the above in mirror for the top.
Step Eight: Repeat repeat
## Step 3: Byzee Beez to Buterflies
Love this weave easy Byz variation and good looking to boot.
AR 3.9
Step One: Open six rings and close six rings.
Step Two: Add four closed rings to one opened one.
Step Three: Close ring and add a second one through the same four rings as the first.
Step Four: Arrange in a 2 in 2 chain and add one more section of 4 rings.
Step Five: Rearrange the rings as if you were to make Byzantine.
Step Six: Make a second piece just as the first and connect where shown.
Step Seven: Don't forget to add the bottom.
Step Eight: Continue Steps One to Seven until you have what you need, and you are done.
## Step 4: Captive Zen
I'm not going to give AR for this one as with this many ring sizes it makes it a little harder to work out
Ring size for this weave you will need three ring sizes each smaller then the next.
smallest ring 18g .173 ID (4.4mm)
Middle Ring 18g .236 (6.0mm)
Largest Ring 18g .279 (7.1mm)
Okay lets get weaving this.
First make a 2 in 1 orbital section, do this by weaving a 2in1 chain from the smallest rings and using the middle size as the orbital ring. you will want to make a few of these sections before continuing.Now using two of the section you made in the last step weave in the largest ring on either side of the small ring in the center through the mid sized rings. It should look like this when done.Adjust so that the top and bottom small rings look like this.Next weave in two more small rings using the right top of the left section, left top of the right section, lower right bottom of the left section, and lower left bottom of the right section (that's hard to say) or see pic above. Continue steps 2-4 to make the chain form of the weave.Using the same methods you can make the sheet variation of this weave but duplicating the form at 90 degrees. Like above.Good luck and have fun.
## Step 5: Candy Cain Cord
AR 4.7
okay this is one of those weaves that is unstable at the start so i've have chosen to use a starter weave. in this case the base weave. Full Persain 6in1 , you need only a section 4-5 units long to get it going and something to hold on to.
OKay now that you have that. lets get to weaving. add your firt ring on th bottom eye like shown.
Flip the section 180 and do the same to the back side.
now give the section a 90 turn to one side and pass you new ring thr the eye and you first two rings.
Flip and repeat.
at this piont you are back to step one but lets run thru it again. toss a ring around the bottom eye.
flip and repeat.
back to step 4-5 rotate 90 and add your ring around the eye and thru you prior two rings.
Flip and repeat.
Now that you have the idea, run out a few more section.
Dissconnect your starter adn you are done.
## Step 6: Celtic Cross
ARs 4.6 and 4.3
Step One: Start with a piece of Celtic Visions.
Step Two: Add a Ring here, in between the two Celtic Visions sections.
Step Three: Pull in two rings one from each side of the visions and use a 1/4 ring to connect all three rings.
Step Four: Repeat on othr side.
Step Five: Jump over two sections and do steps 2-4 again.Step six: Keep doing this and you will complete it
## Step 7: Celtic Dawn
Start with a pice of Helm chain, now you need some AR 4 rings.
Weave the rings in a 1 -2- 1-2 pattern ware the 2 is in the single ring of the helm pattern.
Next fold in one ring from each of the bi-rings and link them together with a new ring.
repeat on oposing side and continue down the chain. nice and simple.
## Step 8: Celtic Filagre
Again the amount of rige sizes make ARs difficult so none are given.
Sizes : 16awg/7.75mm, 16awg/4.75mm, 18awg/4.5mm, and 20awg/3.25mm.
Lets get weaving.
Start by making a 2 in 2 chain from one small and one medium ring by placing one medium and one small ring around them.I know that first step is kinda confusing but continue that pattern until you have 8 sections like this. Be sure to keep the medium rings on the one side. If you get the rings all mixed this trick will not work.Now loop them around in a circle and connect them the in the same way with all the medium rings facing the inside of the loop.Now all we have to do is weave in the largest ring through all the medium rings on the inside.There that is a complete section of it. Now we move on to connecting two sections together. A simple 2in2 will work just fine. Use the small rings for this.Continue doing this until you are satisfied with it.
## Step 9: Daisy Chain
Daisy Chain Some times call Chrysanthemum.
Two sizes, Large rings and small rings AR of large ring must be a factor of the small ring. this will help make the final circle of rings stiffer. ar of around 10 and a AR of 3.8 seems to work well.
This is a simple weave FILL the large ring about 2/3 full of rings. then add a outside layer of rings in a 4in1 pattern style (thru the eye manor) make a few of these units, then using two rings connect them to one another.
## Step 10: Elvin Tracks
AR 5.4
Diving right in, start with two 2 in 1 chains of equal length.
Step Two: We need to line them up together like so:
Step Three: Count in two “eyes” and pass your ring through both eyes in the second section and around through the top on the first like so.
Step Four: Mirror this on the opposing side of the chains, moving one eye set forward.
Step Five: Repeat, Repeat, DO IT AGAIN!
Okay Variation 2, Restarting with your 2 in 1 chains this time instead of starting off in the second eye start off in the first. HUH I can hear you saying. Because the two chains are mirrored this will create a different outcome of the chain.
## Step 11: Euro 4in1 Unbalanced
AR 4 +
This weave is known by a few names, European 4 in 1 Unbalanced being the most used. Also called Oops and Reversing Bias 4 in 1.
Oops is a misleading name. E4in1U is not an easy weave to start. It is true that once started it simple enough to do. So I’m going to show you how to use a starter weave to get it going, then will unweave or disconnect the “starter” and continue with this weave. Since this weave is 4in1 we'll start with a European 4 in 1 base section. It best to use a small section but still big enough that it will give you a well supported starting piece before you disconnect it. I recommend that you use a piece that is 5x3. Go ahead and make it from AR 4 this will make it nice and thick, and this weave looks best thick Now that you have that done lets change it over to the new weave
Step one: Turn the weave 90 degrees. Now add a new ring through the top 3 rings.
Step two: Do the same for the bottom three rings.
Step Three: Link in a ring through the top ring in the “starter”.
Step Four: Weave one into the Center of the two 3in1 rings but pass it through only one ring.
Step Five: Do we need a step five lets skip to six…hehe.
Step Six: Do the same linking as in step Three.
Step Seven: Now we have the connections and they will not move on us we start the fun. Using a single ring, weave it through 3 of the top new rings.
Step Eight: Now do this again for the bottom three rings.
Step Nine: Do the same as you did for Steps 3 through 6.
Step Ten: Repeat Steps Seven through Nine until you have a strip a few inches long.
Step Eleven: Now that it is a good size we can remove the 4in1 starter from the end.
Step Twelve: this part can be hard to see sometimes due to the confusing nature of the weave. But you add just one ring through just one ring, best way to see where the one ring goes is to look for that 3in1 section that it will complete and in doing so makes the pass through ring 4in1. Here I have added all of them for this section.
Step Thirteen: Now for a Closer look and the next ring. Using two of the 1in1 rings add a new one as shown it will be 3in1.
Step Fourteen: Do the same one section down, remember that it is 3in1
Step Fifteen: Continue to do step 13-14 all the way down the piece.
Step Sixteen: Now we have to reverse the connection but keep the flow. Do this by adding another set of 1in1, be sure they complete groups of 4in1. Note the yellow ring here, it's floating in mid space. This is a ring that you will have to add later. Keep in mind that the off row will always need this seemingly extra ring to complete it.
Step Seventeen: Here we are adding a New ring to the end ring it is 2 in 1 despite missing the red ring from the 1in1 from the previous step (my bad there).
Step Seventeen: Now we do the reverse of step 13 but still doing 3in1.
Step Eighteen: Continue step 17 until you get to the end, at the last section add the yellow ring onto the new one being added, this will stop the weave from contracting on you.
Step Nineteen: Continue doing steps 13-19 until you have what you like.
## Step 12: Firemans Hold 4
AR 4.8
Step one: Close a few rings
Step Two: Hold one ring on while placing another through and locking it in a orbital form like so.
Step three: Add a ring off the orbital ring like so
Step Four: Add in a locking ring through the locking ring from step two and intersecting it with the ring from step three.
Step Five: Now we are repeating steps three......
Step Six: Now we are going to connect the two sides and complete the form. Add one ring to the two rings indicated here.
Step Seven: Final ring placement is nothing more then step three again but going through both rings on this side instead of just the one.Done Finished!!
## Step 13: Fleur De Lis
AR 3.9
First Step: Make a two in three chain section.
Step Two: You'll want to make a few of these section before we move on.
Step Three: Now we have to pull out the center ring in the three so that it look like so. you may hold it to make thing easier.
Step Four: now with the ring you pulled forward and the two ring you did not on another section add a ring to each side.
Step Five: Continue doing this until completed length is reached.
## Step 14: Halo
No it not the game!
Step one: Gather your rings. ARs are 6, 5, 4, and 3.
Step two: Start by linking two AR4 rings with an AR3 ring to make a 2in1 section.
Step three: Place two AR6 rings as shown.
Step four: Now add two AR5 Rings through the AR6 but around the AR4 rings.
Step five: Repeat this on the other side so that this unit is stable and you can set it down. Make another of these sections before proceeding to step 6.
Step six: Take the two units you made in step 5 and weave a set of AR6 rings through the AR5 rings, one on top and one on bottom of the AR4 ring.
Step seven: Continue making sections and linking them to reach the length you would like. In short... done!..:P
## Step 15: Hoodoo Hex Sheet
AR 3.5
Okay first thing you need to make some 2in1 chains. make 4-5 to get started with. now lay two of them down so yu can see there alinments. and arange them so that the tops and bottoms alternate and face each other in to opsing chain. what done that mean?.well just look at this. you may need to re-aline this every time you add a new ring untill it stabilizes.
Now we add the first rings, skip the end ring and pass you first ring thru the eye of the next two rings on both chains.
set it down and realine if you need to. Next jump the next eye and use the one after it and do the same as previose step.
Again...
and i think you get it, finish out the cain.
Now grab another 2in1 chain and aline it like so.
Now we go back to step 2 but don't skip the first eye, use it.
same as b4 now, skip an eye and use the second one.
And again..
Finish out the chain.
Continue this method of alternating alignments and alternating eyes to complete the hight of the chain.
Participated in the
Full Spectrum Laser Contest
Participated in the
Jewelry Contest | 4,068 | 15,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.927737 |
https://www.slideserve.com/amy/random-partition-via-shifting | 1,721,546,472,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00331.warc.gz | 856,402,270 | 25,251 | 1 / 87
# Random Partition via Shifting
Random Partition via Shifting. Tomer Margalit , 21/5/2012. Table of Contents. Introduction and fundamental properties of randomly shifted grids. Application – minimal disk covering of points. We improve the running time of the trivial approach using randomly shifted grids.
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## Random Partition via Shifting
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1. Random Partition via Shifting TomerMargalit, 21/5/2012
2. Table of Contents • Introduction and fundamental properties of randomly shifted grids. • Application – minimal disk covering of points. We improve the running time of the trivial approach using randomly shifted grids. • Shifting Quadtrees – we will see how using random shifts when constructing Quadtrees can be beneficial. • Approximation of an n-point Metric Space – We will show how we approximate such a metric using a new data structure called an HST.
3. Background • In this lecture we will introduce randomness to partitions of a domain. • We will do this by partitioning starting from a random shift. • For instance, suppose that we have as the domain that we would like to partition. • Then we randomly choose a uniformly distributed vector in , to use to shift to . • For instance, we can shift a quadtree over a region of interest. • This yields some simple clustering and nearest neighbor algorithms.
4. Example – General Points in
5. Example – Points in square
6. Example – Unshifted Partition
7. Example – Shifted Partition
8. Example – Shifted Points
9. Partition via Shifting 1 • We initially focus on partition shifting of the real line. • First, we define the partition function: • This function partitions the real line according to its preimage – i.e. for any integer I, the group is a unique cell.
10. Example – Partition Function
11. Partition via Shifting 2 • Remark – The value of the function does not affect the partition, only the partitioned sets do. That is, the partition is determined according to the equivalence classes implied by h. • Simply put, if we have two functions h and h’ that agree on two points everywhere, then they are treated as the same partition. • Equivalently, if we have b and b’ such that for some integer k, then the partitions are identical, even though the functions may differ. • In particular, we can pick shifts uniformly from , without losing any partitions. • We could also pick them from any , as long as .
12. Good Probability for a Nice Distribution • Lemma – For any , we have that • In other words, if two points are at some distance, and we use a coarse partition, then there is a very low chance we will classify them together. • For example, what is the chance that for we choose a b such that x and y have different ids?
13. Intuition 1
14. Intuition 2
15. Proof • Claim: For any , we have that . • The claim is obviously true if . Indeed, then x and y must be in separate cells because we partition to stripes of width . • If not, assume that x < y and pick some shift b uniformly from . • As said before, we still get all partitions this way, so we are in the same probability space. • But now, if and only if .
16. Generalized Partition Shift • Now we generalize the notion of shifting partitions to multidimensional spaces. • We have a group of points . • Consider a shift randomly chosen from . • We partition the points according to a grid that has its origin in b, and with a side length of . • We define the id of a point to be
17. Example – 2D Shifted Partition
18. Example – 3D Partition
19. Good Probability for a Nice Distribution in Multidimensional spaces • Given a ball B of radius r in , the probability that B is not contained in a single cell of a randomly shifted grid, , is bounded by . • Proof: • Project B into the ith coordinate. • Now we have an interval of length 2r in the shifted grid . • Now we know that B is contained in a single cell if and only if all s are contained in a single cell of the one dimensional grid.
20. Proof, Continued • Now, denote by denote the event in which is not contained in a single cell. • Then obviously, suffice that even one of these events occurs for the ball to not be contained in a single cell. • In other words, if we denote by the event that the ball is not contained in a single cell, then we have that: • If this is larger than one, than we get a probability of one, which proves our claim.
21. Remarks about Randomly Shifted Grids • We have shown that if we choose grids randomly, we get a good chance of separating two points if we have a coarse enough partition. • Obviously, we will use this property to give randomized algorithms, and algorithms that have a high probability to succeed.
22. Application – Covering by Disks • Now we show an application for these simple facts about random partitions. • Given a set of n points in the plane, we would like to cover them by the minimal amount of unit disks. • Apparently, the randomly shifted grids can improve the trivial algorithm by quite a bit. • First we show the trivial algorithm, and then the improvement gained by using randomly shifted grids.
23. Disk Cover - Points
24. Disk Cover - Cover
25. Trivial Cover • We show the trivial algorithm first. We will also use it later. • We can cover trivially – that is by traversing all covers – by using . • Before we can show that, we must establish some fundamental insights about equivalence of disk covers:
26. Intuition – What Sets Two Covers Apart? (1)
27. Intuition – What Sets Two Covers Apart? (2)
28. Intuition – What Sets Two Covers Apart? (3)
29. Intuition – What Sets Two Covers Apart? (4)
30. From Intuition to Proof • The intuition shows that two covers are equivalent if all the disks in both covers cover the same set of points. • This intuition will direct us in providing the trivial algorithm. • Instead of considering any cover, we consider only equivalence classesof covers. • Very trivially, we can consider every partition of the points into (not necessarily disjoint) groups, and check if that cover exists for them. • However, this approach gives us a lot of invalid covers. • We can use some simple observations to cut that number.
31. I. Every pair of input points (at distance under 2) defines two possible disks
32. II. Given a cover, we can move every disk to have at most 2 points on the edge
33. Proof • The first insight is obvious. • The second insight is not that obvious though. • To get the second claim, given a disk (that is part of a cover), we can move the disk downward until it “hits” a point. • And then rotate it (around the point) until it “hits” another one.
34. Move downward until a point is hit
35. Rotate until a second point is hit
36. Trivial Cover – Feasibility • In case the disk contains at most one point, we get one point on the edge. • Now, using these facts, it is obvious that every arbitrary disk cover is equivalent to a cover of disks that have either 1 or 2 points on their edge. • Given this insight, we can just traverse all possible covers composed of these disks to find the minimal one. • Ok, so how many disks are there?
37. Trivial Cover – Given the Minimal Cover Size • If we knew the minimal cover has m disks, how much time would it take to find the cover? • We have covers, and checking if a cover is valid takes O(nm) time (check each point against each disk). • So in total, it takes time to find the cover.
38. Trivial Cover – Full Algorithm • To complete the algorithm, given a set of points, we exhaustively check for each k between 1 and n, whether we have a cover with k disks. • According to the insights we saw, the first cover we find must also be a minimal cover. • Now, the time it takes to find the cover is
39. Trivial Cover – Final Statement • In conclusion, we get the following claim: • Claim: Given a set P of n points in the plane one can compute, in a minimal cover – where k is the size of the minimal cover. • The problem with this approach is that k may be large (say n/4).
40. Minimal Disk Cover - Improvement • Now, we compromise some accuracy – we will try to calculate an approximation of the minimal disk cover. • To do that, we will use the randomly shifted grids. • As mentioned before, this will be a randomized algorithm, that will give a good approximation for the minimal cover size on average. • Specifically, like algorithms we have seen in the past, we choose some small factor (denote it by ), and then both our running time and the expectation of the cover size depend on it.
41. Minimal Disk Cover – Randomized Algorithm • Claim: Given a set P of n points in the plane and a parameter , one can compute using a randomized algorithm, in time, a cover of P by X unit disks, where , where opt is the minimum number of unit disks required to cover P.
42. Algorithm • First, let . • Now, for a shift , consider the grid . • Compute all the grid cells that contain points of P. • That is done by computing the id of every point, and then storing it in a hash table. • This clearly can be done in linear time. • Now, for each grid cell c, we denote by the points of P in the cell. • Observe that any cell of can be covered by unit disks.
43. Calculate the id of each point ()
44. Cover the cells by disks (ratio is 1)
45. Cover the cells by disks (ratio is 0.5)
46. Algorithm 2 • Note that the bound of M is very loose. • We can do much better – instead of using this cover, we use the trivial algorithm to compute the minimal cover for each square. • Since the minimum is at most M, we can compute it in , according to what we showed about the trivial algorithm. • As for the running time, there are at most n non-empty grid cells, which means it will take us . Now since , we get that the time is . • Note that and that .
47. Good Expectation of the Cover Size • Now that we have reached our goal time, we turn to the expectation. • Reminder – We denoted X to be the size of the returned cover, and we claimed that – a property we will prove. • Plan: Given an optimal cover, we will show another cover that is considered by the algorithm, and that the expectation of its size (with regards to all the possible shifts) is what we want.
48. Proof 1 • Consider the optimal solution . • We generate a solution from that is checked by the algorithm. • For each grid cell, c, denote by the set of disks of the optimal solution that intersect c, and denote each disk with c. • Now, define . • Remember, this set may contain the same disk twice – for instance in cases where the same disk covers two cells.
49. Optimal Cover
More Related | 2,470 | 10,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.829805 |
https://stats.stackexchange.com/questions/380141/is-it-valid-using-testing-data-for-model-selection-when-validation-data-cannot-d | 1,721,906,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00380.warc.gz | 463,207,425 | 41,614 | # Is it valid using testing data for model selection when validation data cannot do the job?
In Hastie's Elements of statistical learning, it says: "The training set is used to fit the models; the validation set is used to estimate prediction error for model selection; the test set is used for assessment of the generalization error of the final chosen model."
My understanding the authors mean that when using this strategies to compare two algorithms, one should build algorithms on training data and run them on the validation data to compare them and decide on a winner.
I have a further question on this issue: What if validation data (or cross validation) gives Two winners.
Suppose I built two models, say M1 and M2, they perform equally well in cross validation. That is, I do not have a winner using validation data. Now I apply the two models to predict testing data and I see M1 performs better than M2.
My question is: in this case, can I consider M1 as the better model.
This question is basically asking whether it is valid to use testing data to do model selection when validation (cross validaiton) data cannot choose the best model.
• Heiste? Do you mean Hastie, Tibshirani and Friedman? Commented Dec 3, 2018 at 19:57
• en.wikipedia.org/wiki/Data_dredging Commented Dec 3, 2018 at 20:09
• Hi Nick, your are right. The name should be Hastie Commented Dec 3, 2018 at 20:17
I think there's some confusion about the general framework. You should really think of three different sets:
1.) Training data, used to fit each model
2.) Testing data, independent of the training data, which is used to test how the models fit from (1) perform on new data and is used to decide which model appears to work best for predicting new data.
Note that with cross-validation, the the partitioning of (1) and (2) is done repeatedly.
3.) Validation data. In step (2), while we didn't use the testing data to fit the individual models, we did select the model we wanted based on the testing data. Therefore, we did do some fitting with the testing data (i.e., model selection), so we should expect some bias in the performance. To remove this bias, we will withhold some from both the training and testing data, and apply our finally selected model to this hold out. This should give us an unbiased estimate of the average prediction error, under a few assumptions (i.e., the full dataset we are using looks like the data we will use in practice, conditional independence of the samples, etc.).
So with this in mind, I think your question is what to do if the testing data does not appear to show any real difference between two different models. In this case, I would not suggest using your held-out validation set to determine which model to use. The reason for that is that you will have now "poisoned the well", i.e., the whole point of the validation set is that it is supposed to be independent of all modeling decisions so we can get unbiased estimates of predictive error. You lose that once you use it to make modeling decisions. Furthermore, unless the validation set is much larger than the test/training set (which itself is a bad idea), you shouldn't expect to be able to differentiate any better using the validation set.
With that in mind, it appears we have seen that the two models predict pretty evenly, given our current data. In that case, you could consider just randomly picking one of the two (all the evidence so far says they perform equally well), picking one for reasons other than predictive power (i.e., elastic net models are easier to interpret than random forests, so if both preform equally well, chose elastic net model in case you want to dive into the "why" part of your model) or just simply average the predictions from your two models for your final prediction.
• Your definition of validation and test sets are opposite to common convention. Commented Feb 10, 2022 at 9:10 | 854 | 3,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.918973 |
http://www.suchitav.com/vlsi-interview-question | 1,601,017,732,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400222515.48/warc/CC-MAIN-20200925053037-20200925083037-00646.warc.gz | 229,907,011 | 8,728 | # vlsi interview question
1. Go through VLSI book from beginning to the end
2. If possible solve all the problems at the end of the chapter
3. Most basic question is draw digital gates using transistors, difference between bipolar and cmos , analog and digital
4. Go through the details on your project
5. Refresh your circuit theory, basic LCR circuit , transfer function , ..
Additional questions can be reffered as
1. before going for these types of interview questions , study any good vlsi book
2. measure your capability by your shelf, you can go to any chapter end questions and see how many quCestions you can solve
3. for experienced professionals prepare one of your projects thoroughly, most common question for vlsi experienced professionals is explain one of the projects , remember you have to explain relevant experience which are suitable for the job requirement
4. for fundamentals you may be asked on deep sub micron technology , channel length modulation, mos characteristics, noise , ..
### vlsi interview questions and answers
1) Give two ways of converting a two input NAND gate to an inverter
2) Given a circuit, draw its exact timing response. (I was given a Pseudo Random Signal Generator; you can expect any sequential ckt)
3) What are set up time & hold time constraints? What do they signify? Which one is critical for estimating maximum clock frequency of a circuit?
4) Give a circuit to divide frequency of clock cycle by two
5) Design a divide-by-3 sequential circuit with 50% duty circle. (Hint: Double the Clock)
6) Suppose you have a combinational circuit between two registers driven by a clock. What will you do if the delay of the combinational circuit is greater than your clock signal? (You can’t resize the combinational circuit transistors)
7) The answer to the above question is breaking the combinational circuit and pipelining it. What will be affected if you do this? What are the different Adder circuits you studied?
9) Give the truth table for a Half Adder. Give a gate level implementation of the same.
10) Draw a Transmission Gate-based D-Latch.
11) Design a Transmission Gate based XOR. Now, how do you convert it to XNOR? (Without inverting the output)
12) How do you detect if two 8-bit signals are same?
13) How do you detect a sequence of “1101? arriving serially from a signal line?
14) Design any FSM in VHDL or Verilog
what is your roles and responsibilities in that project
what all cores where present in that chip
what is the technology like 130,90,65,45nm
what is the clock-frequency
how many clock-domains
what is the voltage value
what is the macro-count
what is the flip-flop count
what are the various analog macros
how did you model your uncertainities or variations
how many power-domains were there
did you have multi-VDD
if you had multi-VDD how did you handle insertion of level-shifters
what is the SSN(simultaneous switching noise) pad ratios used in your design.
how did you prevented noise in your chip
how many placeable instances
what is the cell-row utilization
what type of package wafer bond or flip-chip
if flip-chip how did you distributed power bumps any special strategy
how many metal layers in your technology
did you used in house library or from any vendor
what is the die-area of your chip
#### vlsi interview questions and answers
1. VLSI Interview Questions and answers
vlsi interview question
2. VLSI Interview Questions and answers
vlsi interview question
3. deepchand jaiswal
i am MTech VLSI Design final year student,
my email is deepchand55@gmail.com,
mob +919713279423
• Guru
Hi Deepchand,
For your thesis your guide should be well aware of the topic selected , few ideas of the thesis
VLSI IMPLEMENTATION OF ANALOG SIGNAL PROCESSING BLOCKS
Estimating Power Consumption in FPGA Chips
power management of rfid tags
latest image sensor design
DSP chips review,
Guru
• Guru
few more proposed vlsi projects for you
Modeling System Design Language Basic Blocks in VHDL
Modeling Telecommunication Protocols and Interfaces
VLSI circuit for an eight bit CPU was designed
VLSI circuit for a combinational lock with external keyboard
VLSI circuit for a keyboard door lock
Analysis & Synthesis of Testable Core Based Memory
Analog HDL Simulation & Synthesis
Hardware Model Code Coverage
Vital Gate Level Cycle Based Simulation
FPGA Macro Cell Generator
Reconfigurable Computing Using Linear Systolic Array Processors
RTL Cycle Based Simulation
Back Annotation of Gate and Transistor Level Timing Information to Behavioral Level Description in Verilog HDL
Guru
4. deepchand jaiswal
i am MTech VLSI Design final year student,
my email is deepchand55@gmail.com,
mob +919713279423
• Guru
Hi Deepchand,
For your thesis your guide should be well aware of the topic selected , few ideas of the thesis
VLSI IMPLEMENTATION OF ANALOG SIGNAL PROCESSING BLOCKS
Estimating Power Consumption in FPGA Chips
power management of rfid tags
latest image sensor design
DSP chips review,
Guru
• Guru
few more proposed vlsi projects for you
Modeling System Design Language Basic Blocks in VHDL
Modeling Telecommunication Protocols and Interfaces
VLSI circuit for an eight bit CPU was designed
VLSI circuit for a combinational lock with external keyboard
VLSI circuit for a keyboard door lock
Analysis & Synthesis of Testable Core Based Memory
Analog HDL Simulation & Synthesis
Hardware Model Code Coverage
Vital Gate Level Cycle Based Simulation
FPGA Macro Cell Generator
Reconfigurable Computing Using Linear Systolic Array Processors
RTL Cycle Based Simulation
Back Annotation of Gate and Transistor Level Timing Information to Behavioral Level Description in Verilog HDL
Guru
5. Amar
I am fresher m.tech from India, Can I get help for finding out the vlsi companies in India ?
Amar
vlsi job applicant
• Guru
There are so many MNC and Indian companies are in India, some of the companies are listed here
./vlsi-companies-in-india
Guru
6. Amar
I am fresher m.tech from India, Can I get help for finding out the vlsi companies in India ?
Amar
vlsi job applicant
• Guru
There are so many MNC and Indian companies are in India, some of the companies are listed here
./vlsi-companies-in-india
Guru
7. Guru
Are you looking for VLSI , ASIC Interview questions and answers
./vlsi-interview-question
8. Guru
Are you looking for VLSI , ASIC Interview questions and answers
./vlsi-interview-question
9. job applicant
Thanks a lot, these helped me a lot for the preparation for my job interview
10. job applicant
Thanks a lot, these helped me a lot for the preparation for my job interview
11. Guru
Thanks everybody for the contribution on this vlsi interview questions and answers
12. Guru
Thanks everybody for the contribution on this vlsi interview questions and answers
13. Guru
few collected vlsi interview questions with answer
Why power stripes routed in the top metal layers?
The resistivity of top metal layers are less and hence less IR drop is seen in power distribution network. If power stripes are routed in lower metal layers this will use good amount of lower routing resources and therefore it can create routing congestion.
Why do you use alternate routing approach HVH/VHV (Horizontal-Vertical-Horizontal/ Vertical-Horizontal-Vertical)?
This approach allows routability of the design and better usage of routing resources.
What are several factors to improve propagation delay of standard cell?
Improve the input transition to the cell under consideration by up sizing the driver.
Reduce the load seen by the cell under consideration, either by placement refinement or buffering.
If allowed increase the drive strength or replace with LVT (low threshold voltage) cell.
How do you compute net delay (interconnect delay) / decode RC values present in tech file?
What are various ways of timing optimization in synthesis tools?
Logic optimization: buffer sizing, cell sizing, level adjustment, dummy buffering etc.
Less number of logics between Flip Flops speedup the design.
Optimize drive strength of the cell , so it is capable of driving more load and hence reducing the cell delay.
Better selection of design ware component (select timing optimized design ware components).
Use LVT (Low threshold voltage) and SVT (standard threshold voltage) cells if allowed.
What would you do in order to not use certain cells from the library?
Set don’t use attribute on those library cells.
How delays are characterized using WLM (Wire Load Model)?
For a given wireload model the delay are estimated based on the number of fanout of the cell driving the net.
Fanout vs net length is tabulated in WLMs.
Values of unit resistance R and unit capacitance C are given in technology file.
Net length varies based on the fanout number.
Once the net length is known delay can be calculated; Sometimes it is again tabulated.
What are various techniques to resolve congestion/noise?
Routing and placement congestion all depend upon the connectivity in the netlist , a better floor plan can reduce the congestion.
Noise can be reduced by optimizing the overlap of nets in the design.
Let’s say there enough routing resources available, timing is fine, can you increase clock buffers in clock network? If so will there be any impact on other parameters?
No. You should not increase clock buffers in the clock network. Increase in clock buffers cause more area , more power. When everything is fine why you want to touch clock tree??
How do you optimize skew/insertion delays in CTS (Clock Tree Synthesis)?
Better skew targets and insertion delay values provided while building the clocks.
Choose appropriate tree structure – either based on clock buffers or clock inverters or mix of clock buffers or clock inverters.
For multi clock domain, group the clocks while building the clock tree so that skew is balanced across the clocks. (Inter clock skew analysis).
14. Guru
few collected vlsi interview questions with answer
Why power stripes routed in the top metal layers?
The resistivity of top metal layers are less and hence less IR drop is seen in power distribution network. If power stripes are routed in lower metal layers this will use good amount of lower routing resources and therefore it can create routing congestion.
Why do you use alternate routing approach HVH/VHV (Horizontal-Vertical-Horizontal/ Vertical-Horizontal-Vertical)?
This approach allows routability of the design and better usage of routing resources.
What are several factors to improve propagation delay of standard cell?
Improve the input transition to the cell under consideration by up sizing the driver.
Reduce the load seen by the cell under consideration, either by placement refinement or buffering.
If allowed increase the drive strength or replace with LVT (low threshold voltage) cell.
How do you compute net delay (interconnect delay) / decode RC values present in tech file?
What are various ways of timing optimization in synthesis tools?
Logic optimization: buffer sizing, cell sizing, level adjustment, dummy buffering etc.
Less number of logics between Flip Flops speedup the design.
Optimize drive strength of the cell , so it is capable of driving more load and hence reducing the cell delay.
Better selection of design ware component (select timing optimized design ware components).
Use LVT (Low threshold voltage) and SVT (standard threshold voltage) cells if allowed.
What would you do in order to not use certain cells from the library?
Set don’t use attribute on those library cells.
How delays are characterized using WLM (Wire Load Model)?
For a given wireload model the delay are estimated based on the number of fanout of the cell driving the net.
Fanout vs net length is tabulated in WLMs.
Values of unit resistance R and unit capacitance C are given in technology file.
Net length varies based on the fanout number.
Once the net length is known delay can be calculated; Sometimes it is again tabulated.
What are various techniques to resolve congestion/noise?
Routing and placement congestion all depend upon the connectivity in the netlist , a better floor plan can reduce the congestion.
Noise can be reduced by optimizing the overlap of nets in the design.
Let’s say there enough routing resources available, timing is fine, can you increase clock buffers in clock network? If so will there be any impact on other parameters?
No. You should not increase clock buffers in the clock network. Increase in clock buffers cause more area , more power. When everything is fine why you want to touch clock tree??
How do you optimize skew/insertion delays in CTS (Clock Tree Synthesis)?
Better skew targets and insertion delay values provided while building the clocks.
Choose appropriate tree structure – either based on clock buffers or clock inverters or mix of clock buffers or clock inverters.
For multi clock domain, group the clocks while building the clock tree so that skew is balanced across the clocks. (Inter clock skew analysis).
15. b
combination
16. b
combination
17. VLSI Course - Zebros India Chennai
Learn VLSI Course in 1,00,000 Mins
@ Zebros India , Chennai
contact : 044 45511910
http://www.zebrosindia.com
18. VLSI Course - Zebros India Chennai
Learn VLSI Course in 1,00,000 Mins
@ Zebros India , Chennai
contact : 044 45511910
http://www.zebrosindia.com
19. Interview Questionnaire
Answer:VLSI stands for very large scale integration, it is combination of electronics, software go to
./vlsi-design for knowing more details.
20. Interview Questionnaire
Answer:VLSI stands for very large scale integration, it is combination of electronics, software go to
./vlsi-design for knowing more details.
21. software interview questions
and
http://www.allapplabs.com/interview_questions/java_interview_questions.htm
22. software interview questions
and
http://www.allapplabs.com/interview_questions/java_interview_questions.htm
23. sunrise
hi could you please suggest a good book for a newbie interested in vlsi…
ready to learn , eager to excel..pls if possible mail me at notknownphilospher[at]yahoo[dot]com
Answer: Go to ./vlsi-book which have good list on vlsi books so also free lecture notes on vlsi
24. sunrise
hi could you please suggest a good book for a newbie interested in vlsi…
ready to learn , eager to excel..pls if possible mail me at notknownphilospher[at]yahoo[dot]com
Answer: Go to ./vlsi-book which have good list on vlsi books so also free lecture notes on vlsi
25. Guru
26. Guru
27. gowthami
its really very good collection of vlsi……
iam going to appear for bharat electronics limited(BEL) written exam will u please kindly give me some links to appear these kind of exams…..
Gowthami, lets us know what is the position, we may help you for getting sample interview questions
28. gowthami
its really very good collection of vlsi……
iam going to appear for bharat electronics limited(BEL) written exam will u please kindly give me some links to appear these kind of exams…..
Gowthami, lets us know what is the position, we may help you for getting sample interview questions
29. Guru
./vlsi-interview-questions vlsi interview questions and answers
30. Guru
./vlsi-interview-questions vlsi interview questions and answers
31. vipin
some vlsi questions with answers can be found at:
http://vhdlguru.blogspot.com/2010/04/here-are-some-common-interview.html
32. vipin
some vlsi questions with answers can be found at:
http://vhdlguru.blogspot.com/2010/04/here-are-some-common-interview.html
33. Guru
34. Guru
35. Guru
visit ./vlsi-companies-in-india
36. Guru
visit ./vlsi-companies-in-india
37. vlsi job applicant
VLSI Interview questions, asic interview questions, soc interview questions, all in one place
If there are too many pins of the logic cells in one place within core, what kind of issues would you face and how will you resolve?
Define hash/ @array in perl.
Using TCL (Tool Command Language, Tickle) how do you set variables?
What is ICC (IC Compiler) command for setting derate factor/ command to perform physical synthesis?
What are nanoroute options for search and repair?
What were your design skew/insertion delay targets?
How is IR drop analysis done? What are various statistics available in reports?
Explain pin density/ cell density issues, hotspots?
How will you relate routing grid with manufacturing grid and judge if the routing grid is set correctly?
What is the command for setting multi cycle path?
If hold violation exists in design, is it OK to sign off design? If not, why?
How are timing constraints developed?
Explain timing closure flow/methodology/issues/fixes.
Explain SDF (Standard Delay Format) back annotation/ SPEF (Standard Parasitic Exchange Format) timing correlation flow.
Given a timing path in multi-mode multi-corner, how is STA (Static Timing Analysis) performed in order to meet timing in both modes and corners, how are PVT (Process-Voltage-Temperature)/derate factors decided and set in the Primetime flow?
With respect to clock gate, what are various issues you faced at various stages in the physical design flow?
What are synthesis strategies to optimize timing?
Explain ECO (Engineering Change Order) implementation flow. Given post routed database and functional fixes, how will you take it to implement ECO (Engineering Change Order) and what physical and functional checks you need to perform?
38. vlsi job applicant
VLSI Interview questions, asic interview questions, soc interview questions, all in one place
If there are too many pins of the logic cells in one place within core, what kind of issues would you face and how will you resolve?
Define hash/ @array in perl.
Using TCL (Tool Command Language, Tickle) how do you set variables?
What is ICC (IC Compiler) command for setting derate factor/ command to perform physical synthesis?
What are nanoroute options for search and repair?
What were your design skew/insertion delay targets?
How is IR drop analysis done? What are various statistics available in reports?
Explain pin density/ cell density issues, hotspots?
How will you relate routing grid with manufacturing grid and judge if the routing grid is set correctly?
What is the command for setting multi cycle path?
If hold violation exists in design, is it OK to sign off design? If not, why?
How are timing constraints developed?
Explain timing closure flow/methodology/issues/fixes.
Explain SDF (Standard Delay Format) back annotation/ SPEF (Standard Parasitic Exchange Format) timing correlation flow.
Given a timing path in multi-mode multi-corner, how is STA (Static Timing Analysis) performed in order to meet timing in both modes and corners, how are PVT (Process-Voltage-Temperature)/derate factors decided and set in the Primetime flow?
With respect to clock gate, what are various issues you faced at various stages in the physical design flow?
What are synthesis strategies to optimize timing?
Explain ECO (Engineering Change Order) implementation flow. Given post routed database and functional fixes, how will you take it to implement ECO (Engineering Change Order) and what physical and functional checks you need to perform? | 4,197 | 19,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.9008 |
https://fr.mathworks.com/matlabcentral/cody/problems/34-binary-numbers/solutions/81969 | 1,582,005,666,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143635.54/warc/CC-MAIN-20200218055414-20200218085414-00209.warc.gz | 383,271,882 | 16,015 | Cody
# Problem 34. Binary numbers
Solution 81969
Submitted on 25 Apr 2012
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
%% n = 2; A = binary_numbers(n); assert(isequal(class(A), 'double'))
Error: Undefined function 'fullfact' for input arguments of type 'double'.
2 Fail
%% n = 3; A = binary_numbers(n); assert(all(A(:) == 0 | A(:) == 1))
Error: Undefined function 'fullfact' for input arguments of type 'double'.
3 Fail
%% n = 5; A = binary_numbers(n); assert(isequal(size(A),[32 5]))
Error: Undefined function 'fullfact' for input arguments of type 'double'.
4 Fail
%% n = 10; A = binary_numbers(n); assert(isequal(size(unique(A,'rows'),1),1024))
Error: Undefined function 'fullfact' for input arguments of type 'double'.
5 Fail
%% n = 1; A = binary_numbers(n); assert(isequal(A,[0;1]) || isequal(A,[1;0]))
Error: Undefined function 'fullfact' for input arguments of type 'double'. | 292 | 1,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-10 | longest | en | 0.356036 |
https://kr.mathworks.com/matlabcentral/profile/authors/8174 | 1,669,881,527,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00867.warc.gz | 387,669,088 | 24,700 | Community Profile
# Tanya Morton
Last seen: Today 2012년부터 활동
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Spoken Languages:
English, French
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Side of a rhombus
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Side of an equilateral triangle
If an equilateral triangle has area A, then what is the length of each of its sides, x? <<http://upload.wikimedia.org/wikipe...
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Length of the hypotenuse
Given short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<http://upload....
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Length of a short side
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
약 10년 전 | 1,431 | 4,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-49 | latest | en | 0.556143 |
https://discourse.datamethods.org/t/help-needed-interpreting-a-regression-table-from-a-logistic-model/18337 | 1,726,369,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00296.warc.gz | 190,735,613 | 6,385 | # Help needed interpreting a regression table from a logistic model
I’ve run out of ideas for the regression table below, which is from a prediction model for liver transplantation. The paper is available here. The primary outcome was graft failure at 1-year, and the authors fitted a logistic model. So far, so good. Here are the results.
Looking at the first two columns, it appears that all the continuous variables have been dichotomized except the last one, which was not continuous. The next column is the coefficients \beta. As this is a logistic model, it must either be a log odd ratio or an odds ratio. Unfortunately, all signs are positive, which leaves us in the dark. Using common sense, maybe the OR is reported because exp(8.57) would be insane? It seems that for one of the variables with three categories, one coefficient was not reported. Also, one coefficient seems to be on the wrong line; no way Retransplantation=NO is associated with a higher risk. The whole paper’s point is that retransplantation is a big no-no (using a DCD liver). The other coefficients make more sense (e.g., younger donors, smaller BMI, shorter ischemia time all reduce the chances of graft failure).
However, in the next step, the authors derived a risk score system. For example, with Donor age and using a cuffoff of 60 years, they calculated the median for the younger and older group (46 and 66); they call them midpoints. The difference is 20 years, so they multiply the difference in the medians with the coefficient 20 \times 0.084 = 1.68. Thus, the \beta X is basically the effect when going from 46 to 66 years, and the Risk score is just the rounded version of this.
Given this procedure, I would expect that they did not dichotomize before fitting the regression model but subsequently for the midpoints. This would also explain that there is only one coefficient for functional donor warm ischemia time. But I would also assume that they did this step on the log OR scale; that would make more sense, wouldn’t it? Thus, I conclude:
1. They did not dichotomize to fit the logistic model. All variables were entered into the model as continuous data, and the last was binary. Thus, you get one coefficient per variable. However, they did not report the intercept.
2. The \beta are log ORs due to the way they constructed the risk score.
3. Yes, the 8.571 is insane, reflecting a more than 5,000-fold increase in the chance of graft failure. I would love to see the uncertainty around this effect, but the authors failed to report standard errors or a 95% confidence interval. If the problem is complete separation, I would expect the p-value to be very high, not very small.
4. The Table is terrible, or maybe it is me who is very, very confused.
I would be very grateful for tips and advice and whether I am on the right track. I am not a Sherlock Holmes.
Please edit your post to provide the full citation for the paper from which the table was extracted.
I think that the authors did indeed report odds ratios, and that they committed a blunder of the first order by adding them rather than adding log OR. This will make helpful patient characteristics turn out to be harmful in the score. This has been written about by @Ewout_Steyerberg @ESteyerberg and myself in letters to the editor. NEVER add ratios.
It is extremely important to determine whether the dichotomization occured before vs after fitting. If it occured before fitting then all of the variable effects are misleading because they did noit properly condition on full covariate information, and this makes more variables appear to be prognostic.
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Thank you very much for the response. I researched more to find out whether the data was continuous or dichotomous to fit the model and found an approved manuscript version, which was still in the author’s original formatting. The table looks as follows:
Here, the effect estimates were vertically centered compared to the published version, where they were aligned with a specific level of the categorical variable (Table 2 in my initial post above). This may indicate that the data was not dichotomized before fitting the model, and we see one coefficient for functional donor warm ischemia time, which would require two coefficients if it had been dichotomized. I conclude with caution the data was not dichotomized. That is good news.
Now, if we assume continuous data, I would say the 0.084 is a log OR and not an OR, thus exp(0.084) = 1.08. It is more reasonable that there was an 8% increase in the chance of graft failure for each additional year increase in donor age. On the other hand, if 0.084 was an OR, then each additional year in donor age would result in a 92% reduction in the risk of graft failure. That is not plausible. Thus, I conclude the effects were log OR. That is also good news, so the author did not make the major blunder of adding up ratios in the construction of the risk score.
However, this still leaves us with the log OR of 8.571 for retransplantation, corresponding to an OR of exp(8.571) = 5,276. It might be explained when almost all the retransplantations resulted in graft failure. It is most likely what happened.
So, I hope this makes sense and is the most plausible scenario? Then, it may not be as bad as it appeared at first.
1 Like | 1,185 | 5,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.959052 |
http://www.chegg.com/homework-help/questions-and-answers/hose-ejects-stream-water-angle-365-degrees-horizontal-water-leaves-nozzle-speed-270-m-s-as-q2308211 | 1,386,622,372,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163998145/warc/CC-MAIN-20131204133318-00098-ip-10-33-133-15.ec2.internal.warc.gz | 273,570,156 | 7,619 | # Max Height with initial velocity and angle
0 pts ended
A fire hose ejects a stream of water at an angle of 36.5 degrees above the horizontal. The water leaves the nozzle with a speed of 27.0 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire? | 79 | 339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-48 | latest | en | 0.924597 |
https://number.academy/252523 | 1,669,892,736,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00303.warc.gz | 462,000,442 | 12,422 | # Number 252523
Number 252,523 spell 🔊, write in words: two hundred and fifty-two thousand, five hundred and twenty-three . Ordinal number 252523th is said 🔊 and write: two hundred and fifty-two thousand, five hundred and twenty-third. Color #252523. The meaning of number 252523 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 252523. What is 252523 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 252523.
## What is 252,523 in other units
The decimal (Arabic) number 252523 converted to a Roman number is (C)(C)(L)MMDXXIII. Roman and decimal number conversions.
#### Weight conversion
252523 kilograms (kg) = 556712.2 pounds (lbs)
252523 pounds (lbs) = 114543.7 kilograms (kg)
#### Length conversion
252523 kilometers (km) equals to 156911 miles (mi).
252523 miles (mi) equals to 406397 kilometers (km).
252523 meters (m) equals to 828478 feet (ft).
252523 feet (ft) equals 76970 meters (m).
252523 centimeters (cm) equals to 99418.5 inches (in).
252523 inches (in) equals to 641408.4 centimeters (cm).
#### Temperature conversion
252523° Fahrenheit (°F) equals to 140272.8° Celsius (°C)
252523° Celsius (°C) equals to 454573.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
252523 seconds equals to 2 days, 22 hours, 8 minutes, 43 seconds
252523 minutes equals to 6 months, 1 week, 8 hours, 43 minutes
### Codes and images of the number 252523
Number 252523 morse code: ..--- ..... ..--- ..... ..--- ...--
Sign language for number 252523:
Number 252523 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 252523
### Multiplications
#### Multiplication table of 252523
252523 multiplied by two equals 505046 (252523 x 2 = 505046).
252523 multiplied by three equals 757569 (252523 x 3 = 757569).
252523 multiplied by four equals 1010092 (252523 x 4 = 1010092).
252523 multiplied by five equals 1262615 (252523 x 5 = 1262615).
252523 multiplied by six equals 1515138 (252523 x 6 = 1515138).
252523 multiplied by seven equals 1767661 (252523 x 7 = 1767661).
252523 multiplied by eight equals 2020184 (252523 x 8 = 2020184).
252523 multiplied by nine equals 2272707 (252523 x 9 = 2272707).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 252523
Half of 252523 is 126261,5 (252523 / 2 = 126261,5 = 126261 1/2).
One third of 252523 is 84174,3333 (252523 / 3 = 84174,3333 = 84174 1/3).
One quarter of 252523 is 63130,75 (252523 / 4 = 63130,75 = 63130 3/4).
One fifth of 252523 is 50504,6 (252523 / 5 = 50504,6 = 50504 3/5).
One sixth of 252523 is 42087,1667 (252523 / 6 = 42087,1667 = 42087 1/6).
One seventh of 252523 is 36074,7143 (252523 / 7 = 36074,7143 = 36074 5/7).
One eighth of 252523 is 31565,375 (252523 / 8 = 31565,375 = 31565 3/8).
One ninth of 252523 is 28058,1111 (252523 / 9 = 28058,1111 = 28058 1/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
252523
#### Is Prime?
The number 252523 is not a prime number. The closest prime numbers are 252509, 252533.
#### Factorization and factors (dividers)
The prime factors of 252523 are 67 * 3769
The factors of 252523 are 1 , 67 , 3769 , 252523
Total factors 4.
Sum of factors 256360 (3837).
#### Powers
The second power of 2525232 is 63.767.865.529.
The third power of 2525233 is 16.102.852.706.979.666.
#### Roots
The square root √252523 is 502,516666.
The cube root of 3252523 is 63,207262.
#### Logarithms
The natural logarithm of No. ln 252523 = loge 252523 = 12,439258.
The logarithm to base 10 of No. log10 252523 = 5,402301.
The Napierian logarithm of No. log1/e 252523 = -12,439258.
### Trigonometric functions
The cosine of 252523 is -0,21013.
The sine of 252523 is 0,977673.
The tangent of 252523 is -4,652703.
### Properties of the number 252523
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 252523 in Computer Science
Code typeCode value
PIN 252523 It's recommendable to use 252523 as a password or PIN.
252523 Number of bytes246.6KB
CSS Color
#252523 hexadecimal to red, green and blue (RGB) (37, 37, 35)
Unix timeUnix time 252523 is equal to Saturday Jan. 3, 1970, 10:08:43 p.m. GMT
IPv4, IPv6Number 252523 internet address in dotted format v4 0.3.218.107, v6 ::3:da6b
252523 Decimal = 111101101001101011 Binary
252523 Decimal = 110211101201 Ternary
252523 Decimal = 755153 Octal
252523 Decimal = 3DA6B Hexadecimal (0x3da6b hex)
252523 BASE64MjUyNTIz
252523 MD5a3171f3a8ded83fe82877509584f5673
252523 SHA256113df7a5f35b3fe373d07fa9e7a3e853bcce12dedc0381864c268648cb6678c7
252523 SHA384fa6a6d7c4606c877bc5f55dc18bbf73f46f5910b183ba8d209bb33ab5678021ff4091251e9d9887703bfd37c3ea316d4
More SHA codes related to the number 252523 ...
If you know something interesting about the 252523 number that you did not find on this page, do not hesitate to write us here.
## Numerology 252523
### Character frequency in number 252523
Character (importance) frequency for numerology.
Character: Frequency: 2 3 5 2 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 252523, the numbers 2+5+2+5+2+3 = 1+9 = 1+0 = 1 are added and the meaning of the number 1 is sought.
## Interesting facts about the number 252523
### Asteroids
• (252523) 2001 UA211 is asteroid number 252523. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, ETS in Socorro on 10/21/2001.
## № 252,523 in other languages
How to say or write the number two hundred and fifty-two thousand, five hundred and twenty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 252.523) doscientos cincuenta y dos mil quinientos veintitrés German: 🔊 (Nummer 252.523) zweihundertzweiundfünfzigtausendfünfhundertdreiundzwanzig French: 🔊 (nombre 252 523) deux cent cinquante-deux mille cinq cent vingt-trois Portuguese: 🔊 (número 252 523) duzentos e cinquenta e dois mil, quinhentos e vinte e três Hindi: 🔊 (संख्या 252 523) दो लाख, बावन हज़ार, पाँच सौ, तेईस Chinese: 🔊 (数 252 523) 二十五万二千五百二十三 Arabian: 🔊 (عدد 252,523) مئتان و اثنان و خمسون ألفاً و خمسمائة و ثلاثة و عشرون Czech: 🔊 (číslo 252 523) dvěstě padesát dva tisíce pětset dvacet tři Korean: 🔊 (번호 252,523) 이십오만 이천오백이십삼 Danish: 🔊 (nummer 252 523) tohundrede og tooghalvtredstusindfemhundrede og treogtyve Dutch: 🔊 (nummer 252 523) tweehonderdtweeënvijftigduizendvijfhonderddrieëntwintig Japanese: 🔊 (数 252,523) 二十五万二千五百二十三 Indonesian: 🔊 (jumlah 252.523) dua ratus lima puluh dua ribu lima ratus dua puluh tiga Italian: 🔊 (numero 252 523) duecentocinquantaduemilacinquecentoventitré Norwegian: 🔊 (nummer 252 523) to hundre og femti-to tusen, fem hundre og tjue-tre Polish: 🔊 (liczba 252 523) dwieście pięćdziesiąt dwa tysiące pięćset dwadzieścia trzy Russian: 🔊 (номер 252 523) двести пятьдесят две тысячи пятьсот двадцать три Turkish: 🔊 (numara 252,523) ikiyüzelliikibinbeşyüzyirmiüç Thai: 🔊 (จำนวน 252 523) สองแสนห้าหมื่นสองพันห้าร้อยยี่สิบสาม Ukrainian: 🔊 (номер 252 523) двiстi п'ятдесят двi тисячi п'ятсот двадцять три Vietnamese: 🔊 (con số 252.523) hai trăm năm mươi hai nghìn năm trăm hai mươi ba Other languages ...
## Comment
If you know something interesting about the number 252523 or any natural number (positive integer) please write us here or on facebook. | 2,619 | 7,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-49 | latest | en | 0.683797 |
https://gmatclub.com/forum/the-suit-made-superbly-by-the-tailor-causing-the-cowardly-old-man-240696.html | 1,539,933,393,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00024.warc.gz | 674,648,118 | 49,424 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# The suit made superbly by the tailor, causing the cowardly, old man
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Joined: 04 Apr 2017
Posts: 13
The suit made superbly by the tailor, causing the cowardly, old man [#permalink]
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Updated on: 21 May 2017, 06:56
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Question Stats:
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The suit made superbly by the tailor, causing the cowardly, old man feel like a fierce, young lion.
a) The suit made superbly by the tailor, causing
b) The suit was altered superbly by the tailor, causing
c) When the suit can be made superbly by the tailor, it caused
d) The suit was altered superbly by the tailor, making
e) The altered suit is made superbly by the tailor, causing
What is the correct usage of both 'making' and 'causing'. Also please explain the can we interchange 'made' and altered' in this sentence. If yes, then why?
for eg. If I write ' The suit was made superbly by the tailor' will that be correct?
Source- The ultimate grammar Book. Also if there is any thread explaining the answers of the practice test 1 please post here.
Originally posted by simar121993 on 21 May 2017, 03:30.
Last edited by broall on 21 May 2017, 06:56, edited 4 times in total.
Reformatted question
Intern
Joined: 15 Mar 2017
Posts: 42
Location: India
GMAT 1: 720 Q50 V37
GPA: 4
Re: The suit made superbly by the tailor, causing the cowardly, old man [#permalink]
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21 May 2017, 06:25
simar121993 wrote:
The suit made superbly by the tailor, causing the cowardly, old man feel like a fierce, young lion.
a) The suit made superbly by the tailor, causing
b) The suit was altered superbly by the tailor, causing
c) When the suit can be made superbly by the tailor, it caused
d) The suit was altered superbly by the tailor, making
e) The altered suit is made superbly by the tailor, causing
What is the correct usage of both 'making' and 'causing'. Also please explain the can we interchange 'made' and altered' in this sentence. If yes, then why?
for eg. If I write ' The suit was made superbly by the tailor' will that be correct?
Source- The ultimate grammar Book. Also if there is any thread explaining the answers of the practice test 1 please post here.
Hi Simar121993,
If we look at the simple structure of the sentence, X........... makes old man feel like........ lion.
X .......... causes old man feel like.............. lion doesn't make sense.
We can simply eliminate all the choices except "D". Obviously replacing made with altered has changed the meaning a bit but it is the only choice that makes sense and is grammatically correct.
Hope this helps. Please hit kudos if you found this useful.
_________________
You give kudos, you get kudos. :D
Re: The suit made superbly by the tailor, causing the cowardly, old man &nbs [#permalink] 21 May 2017, 06:25
Display posts from previous: Sort by | 905 | 3,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-43 | latest | en | 0.93311 |
http://sta.ie/lesson/understanding-electricity | 1,621,022,811,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991207.44/warc/CC-MAIN-20210514183414-20210514213414-00357.warc.gz | 52,612,599 | 9,029 | ## Understanding electricity
### Eirgrid
Physics
Technology
Volts, amperes and watts are related as follows: watts = volts × amperes. This lesson explains the meaning of these units and why they are related in this way. It also highlights how the unit of electric charge (the coulomb) relates to volts and amperes.
Contains the full lesson along with a supporting toolkit, including teachers’ notes.
### Lesson excerpt
Common units
The labels on electrical devices usually show one or more of the following symbols: W, V, A and maybe Hz. But what do they mean and what information do they provide? Manufacturers use these symbols to inform users so that they can operate appliances safely. In this lesson we will explore the meaning of the symbols and the quantities they represent and show how to interpret them correctly.
Watts, volts and amperes
The labels on typical domestic appliances show values such as the following:
Hair drier 230 V, 750 W
Bulb 230 V, 40 W
Phone charger 100 – 240 V, 200 mA, 50 – 60 Hz
Jug kettle 230 V, 2300 W
The plugs on domestic appliances are commonly marked ‘13 A’.
This lesson will explain what these units mean and how they are related.
How are the units related?
Volts, amperes (colloquially known as ‘amps’) and watts are related as follows: watts = volts × amperes
Let’s take the example of the jug kettle listed above; the label shows ‘230 V, 2300 W’. We can use these values to find the electric current (in amperes).
watts = volts × amperes
2300 = (230) (x)
By re-arranging we can find x.
x = 2300 / 230
= 10 amperes
This indicates that the electric current in the jug kettle is 10 amperes (10 A). So why is the plug on the kettle marked ‘13 A’?
Safety, fuses and 13 A
Electric current produces a heating effect and excessive currents can cause insulation on wires to melt. In Ireland and many other European countries the maximum current that should be drawn from mains electrical sockets is 13 amperes. In order to prevent damage a fuse is fitted in the plugs of all standard electrical appliances. The current rating of the fuse should exceed the anticipated maximum current for the appliance but must not exceed 13 amperes. If for any reason the current rises above 13 A, then a thin wire in the fuse melts and cuts off the electric current to the appliance.
What are watts?
Domestic light bulbs are manufactured in a variety of specifications. A bulb marked ‘80 W’ would be expected to be brighter than a bulb of the same kind that is marked ‘40 W’. The letter ‘W’ stands for ‘watt’ and the value indicates how much energy the bulb radiates every second. The watt is the unit of power; one watt means one joule per second.
Lamps of different types, but with the same power, may have different light outputs. For example, a 40 W fluorescent lamp generally gives out more light than a 40 W incandescent bulb. In other words, some devices are more efficient than others.
### True or False?
1. Electric charge is measured in coulombs. true
2. Electric potential is measured in joules. false
3. One ampere means ‘one coulomb per second’. true
4. The power of an electric appliance is measured in watts. One watt means one joule per minute. false
5. The power of an electric appliance can be found by multiplying the electric current by the voltage. true
6. The maximum current that should be drawn from a domestic mains electric outlet is 13 amperes. true
7. In Ireland the voltage used for domestic mains is 120 V. false
8. Electric potential is also known as ‘voltage’. true
9. The heat transferred to a room by a radiator depends on the rate of flow of water and the temperature difference between the inlet and outlet pipes. true
10. The heat transferred to a room by an electric heater depends on the rate of flow of charge and the potential difference between the inlet and outlet conductors. true | 904 | 3,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-21 | longest | en | 0.899395 |
https://www.numbers-figures.com/square-root-of-160010-onehundredsixtythousandandten | 1,656,951,343,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104432674.76/warc/CC-MAIN-20220704141714-20220704171714-00222.warc.gz | 974,045,911 | 4,460 | # square root of 160010 onehundredsixtythousandandten
The square root of 160010 is:
400.01249980469
## What is the square root of a number?
The square root of a number is a value that when multiplied by itself equals the original number. The square root is the opposite of mathematical function of the square.
The classic symbol of the square root is the normal root sign without specifying the root exponent. | 94 | 414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-27 | latest | en | 0.781955 |
https://garyjohnston.com/blog/ | 1,718,399,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00577.warc.gz | 250,298,890 | 32,254 | Bitcoin – from a student’s perspective
My name is Erik. I took Gary’s class a few years ago and learned a ton from him. I invested 150 dollars in Ethereum back in 2016. It is now … Read more
Calculator Problem – 10/05/2015
If you invest 100 monthly payments of \$300 and end up with \$100,000, what was your return? (The answer will be posted with the new calculator problem in 2 weeks. … Read more
Calculator Problem – 09/21/2015
If you invested \$400 a month for 100 months and ended up with \$50,000, what would have been your return? (The answer will be posted with the new calculator problem … Read more
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How months will it take to pay off a \$63,000 loan at 7.5% with a \$400 payment? (The answer will be posted with the new calculator problem in 2 weeks. … Read more
Calculator Problem – 08/24/205
How much money would you have in 30 years if you invest \$150 a month at 12%? (The answer will be posted with the new calculator problem in 2 weeks. … Read more
Calculator Problem – 08/14/2015
How many months would it take to pay off a loan of \$3100 at 19.96% with payments of \$250? (The answer will be posted with the new calculator problem in … Read more
Calculator Problem – 07/27/2015
What would you pay today to buy a note that pays \$25,000 in fifteen years if you wanted to make 18% on your money? (The answer will be posted with … Read more
Calculator Problem – 07/13/2015
If a property generated \$20,000 a month and expenses ate up 25% of the rents, how much debt could you support with a ten year amortization at 5%? (The answer … Read more
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If you had a loan of \$1.125m at 5% and made payments of \$20,000 a month, how long would it take to pay off the loan? (The answer will be … Read more
Calculator Problem – 06/15/2015
You have a choice between two \$500,000 loans. The first loan is fixed for 10 years @ 5.25% and amortized over 25 years. The second loan is fixed for 15 … Read more | 547 | 1,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.945389 |
http://www.reference.com/browse/wiki/Universal_Product_Code | 1,369,471,955,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705790741/warc/CC-MAIN-20130516120310-00032-ip-10-60-113-184.ec2.internal.warc.gz | 673,272,537 | 23,136 | Definitions
# Universal Product Code
The Universal Product Code (UPC) is a barcode symbology (i.e., a specific type of barcode), that is widely used in the United States and Canada for tracking trade items in stores.
## Current code
The UPC encodes 12 decimal digits as SLLLLLLMRRRRRRE, where S (start) and E (end) are the bit pattern 101, M (middle) is the bit pattern 01010 (called guard bars), and each L (left) and R (right) are digits, each one represented by a seven-bit code. This is a total of 95 bits. The bit pattern for each numeral is designed to be as little like the others as possible, and to have no more than four 1s or 0s in order. Both are for reliability in scanning.
Since S, M, and E all include two bars, and each of the 12 digits of the UPC-A barcode consists of two bars and two spaces, all UPC-A barcodes consist of exactly (3*2)+(12*2)=30 bars.
The UPC has only numerals, with no letters or other characters. The first digit L is the prefix. The last digit R is an error correcting check digit, allowing some errors in scanning or manual entry to be detected. UPC data structures are a component of GTINs (Global Trade Item Numbers). All of these data structures follow the global GS1 standards.
The bar-and-space patterns for each digit 0–9 are as follows:
Digit Pattern Digit Pattern 0 0001101 5 0110001 1 0011001 6 0101111 2 0010011 7 0111011 3 0111101 8 0110111 4 0100011 9 0001011
Before the Middle guard bars, a binary 1 is indicated by a bar, while a 0 is indicated a space. After the Middle guard bars, however, the patterns are optically inverted. In other words, a 1 is now indicated by a space, and a 0 is now indicated by a bar. In the illustration above, the "4" digit (shown in detail), falls after the Middle guard bars, causing the pattern of bars and spaces to be inverted.
## Prefixes
• 0, 1, 6, 7, 8, or 9: For most products.
• 2: Reserved for local use (store/warehouse), for items sold by variable weight. Variable-weight items, such as meats and fresh fruits and vegetables, are assigned a UPC by the store, if they are packaged there. In this case, the LLLLL is the item number, and the RRRRR is either the weight or the price, with the first R determining which.
• 3: Drugs by National Drug Code number. Pharmaceuticals in the U.S. have the remainder of the UPC as their National Drug Code (NDC) number; though usually only over-the-counter drugs are scanned at point-of-sale, NDC-based UPCs are used on prescription drug packages as well for inventory purposes.
• 4: Reserved for local use (store/warehouse), often for loyalty cards or store coupons.
• 5: Coupons The Manufacturer code is the LLLLL, the first 3 RRR are a family code (set by manufacturer), and the last 2 RR are a coupon code. This 2-digit code determines the amount of the discount, according to a table set by the GS1 US, with the final R being the check digit.
By prefixing these with a 0, they become EAN-13 rather than UPC-A. This does not change the check digit. All point-of-sale systems can now understand both equally.
## Check digit calculation
In the UPC-A system, the check digit is calculated as follows:
1. Add the digits in the odd-numbered positions (first, third, fifth, etc.) together and multiply by three.
2. Add the digits in the even-numbered positions (second, fourth, sixth, etc.) to the result.
3. Find the result modulo 10 (i.e. the remainder when the result is divided by 10).
4. If the result is not zero, subtract the result from ten.
For example, a UPC-A barcode (in this case, a UPC for a box of tissues) "03600029145X" where X is the check digit, X can be calculated by adding the odd-numbered digits (0+6+0+2+1+5 = 14), multiplying by three (14 × 3 = 42), adding the even-numbered digits (42+3+0+0+9+4 = 58), calculating modulo 10 (58 mod 10 = 8), subtracting from ten (10 - 8 = 2). The check digit is thus 2.
## Zero Compressed UPC-E
To allow the use of UPC barcodes on smaller packages where a full 12-digit barcode may not fit, a 'zero-compressed' version of UPC was developed called UPC-E. This symbology differs from UPC-A in that it only uses a 6-digit code, does not use middle guard bars, and the end bit pattern (E) becomes 010101. The way in which a 6-digit UPC-E relates to a 12-digit UPC-A is determined by the last (right-hand most) digit. With the manufacturer code represented by X's, and product code by N's then:
Last digit UPC-E equivalent is UPC-A equivalent is
0 XXNNN0 0XX000-00NNN + check
1 XXNNN1 0XX100-00NNN + check
2 XXNNN2 0XX200-00NNN + check
3 XXXNN3 0XXX00-000NN + check
4 XXXXN4 0XXXX0-0000N + check
5 XXXXX5 0XXXXX-00005 + check
6 XXXXX6 0XXXXX-00006 + check
7 XXXXX7 0XXXXX-00007 + check
8 XXXXX8 0XXXXX-00008 + check
9 XXXXX9 0XXXXX-00009 + check
For example a UPC-E barcode with the number 654321 would expand to the UPC-A 065100004327.
UPC-E check digits are calculated using this expanded string in the same way as used by UPC-A. The resulting check digit is not added to the barcode, however, but is encoded by manipulating the parity of the six digits which are present in the UPC-E - as shown in the following tables:
Check digit Parity pattern
0 EEEOOO
1 EEOEOO
2 EEOOEO
3 EEOOOE
4 EOEEOO
5 EOOEEO
6 EOOOEE
7 EOEOEO
8 EOEOOE
9 EOOEOE
Digit to be encoded Odd parity pattern Even parity pattern
0 3-2-1-1 1-1-2-3
1 2-2-2-1 1-2-2-2
2 2-1-2-2 2-2-1-2
3 1-4-1-1 1-1-4-1
4 1-1-3-2 2-3-1-1
5 1-2-3-1 1-3-2-1
6 1-1-1-4 4-1-1-1
7 1-3-1-2 2-1-3-1
8 1-2-1-3 3-1-2-1
9 3-1-1-2 2-1-1-3
Our example code 654321, therefore, would become 1-1-1 4-1-1-1 1-2-3-1 2-3-1-1 1-4-1-1 2-2-1-2 2-2-2-1 1-1-1-1-1-1. The resulting barcode would look roughly like this:
## Other Variations
UPC in its most common usage technically refers to UPC-A. Other variants of the UPC exist.
• UPC-B is a 12-digit version of UPC with no check digit, developed for the National Drug Code and National Health Related Items Code.
• UPC-C is a 12-digit code with a check digit.
• UPC-D is a variable length code (12 digits or more) with the 12th digit being the check digit. These versions are not in common use.
As the UPC has become technologically obsolete, it is expected that UPC-B and UPC-C will disappear from common use by the 2010s. The UPC-D standard may be modified into EAR 2.0 or be phased out entirely.
## Printing Considerations
Dimensions
UPC-A Bar code symbols can be printed at various densities to accommodate variety of printing and scanning processes. The significant dimensional parameter is called X-dimension, the ideal width of single module element. The X-dimension has to be constant in UPC-A symbol. The width of each bar (dark bar) and space (light bar) is determined by multiplying the X-dimension by the module width of each dark bar or light bar (1,2,3, or 4).
The X-dimension for the UPC-A at the nominal size is 0.33 mm (0.013 in.). UPC-A can be reduced or magnified in the range of 80% to 200%.
Nominal symbol height for UPC-A is 25.9 mm (1.0 in.). In UPC-A the dark bars forming the left, centre, and right Guard Bar Patterns are extended downwards by 5 times X-dimension. This also applies to the bars of the first and the last symbol characters of UPC-A symbol. See illustration.
Quiet Zone (Light Margin)
The minimum Quiet Zone width required by the UPC-A bar code symbol is 9 x X-dimension on both the left and right sides. UPC-E requires 9 X-dimension units on the left side and 7 on the right. (Source; UPC Symbol Specification Manual).
Barcode Text
Exactly 12 digits must be printed below the UPC-A barcode.
## History
Wallace Flint proposed an automated checkout system in 1932 using punch cards. Bernard Silver and Norman Joseph Woodland developed a bull's-eye style code, patented it (filed in 1949 and received in 1952). In the 1960s railroads experimented with a multicolor barcode for tracking railcars, but eventually abandoned it.
A group of grocery industry trade associations formed the Uniform Grocery Product Code Council which with consultants Larry Russell and Tom Wilson of McKinsey & Company, defined the numerical format of the Uniform Product Code. Technology firms including Charegon, IBM, Litton-Zellweger, Pitney Bowes-Alpex, Plessey-Anker, RCA, Scanner Inc. and Singer proposed alternative symbol representations to the council. In the end the Symbol Selection Committee chose to slightly modify, changing the font in the human readable area, the IBM proposal designed by George J. Laurer.
Although various companies had UPC Scanning systems in the back of stores, the first UPC marked item ever scanned at a retail checkout (Marsh's supermarket in Troy, Ohio) was at 8:01 a.m. on June 26, 1974, and was a 10-pack of Wrigley's Juicy Fruit chewing gum. The entire shopping cart also had barcoded items in it, but the gum was merely the first one picked up by the cashier. This item is currently on display at the Smithsonian Museum in Washington, D.C.
## Development of the IBM UPC proposal
Around 1970 IBM at Research Triangle Park NC assigned George Laurer to solve the problem of a Super Market Scanner and label. In February 1971 Heard Baumeister joined Laurer then later William Crouse joined the effort. After many, many months they had made no progress. They were aware of the RCA Bulls Eye Label that could be scanned with a simple straight line laser scanner but a readable label was far too large. Although Litton Industries proposed a Bull’s Eye symbol cut in half to reduce the area it was still too large and presented the same ink smear printing problems as the RCA symbol. The redundancy and checking ability were removed completely.
The UPC Label above shows the general characteristics of Baumeister's proposals. He did not suggest any specific bar code so the image does not attempt to show exact coding of the ten digits required at that time. Also Baumeister's proposal did not include specific guard bars on the sides and center.
A change in management at IBM resulted in Baumiester, Crouse, and Laurer being assigned to different departments. Laurer was give sole responsibility for inventing and creating a viable code and symbol that would satisfy all the requirements. He made several attempts based using the Delta C code invented by Crouse. Finally he devised a new code that also read distances from leading to leading and trailing to trailing edges of bars making it much less sensitive to the printer’s ink spreading degradation. Although this code was based upon the Delta C code, it required far less area. Unfortunately it still did not meet the 1.5 sq. in. requirement. Laurer’s final breakthrough was using the unique characteristics of his new code to invent a way to divide the symbol in halves in such a way that they could be reassembled in the correct order by the scanner. This reduced the necessary area by about 40% to 1.5 by 1.0 inches.
The following table shows the workable labels, available in the early 1970s, with their sizes.
Check digit Parity pattern
Label Dimensions Area
Bulls-eye with Morse Code Large Large
Bulls-eye with Delta B 12.0" diameter 113.10 sq. in.
Bulls-eye with Delta A 9.0" diameter 63.62 sq. in.
Baumeister 1st w/ Delta B 6.0" × 5.8" 34.80 sq. in.
Baumeister 2 halves w/ Delta B 6.0" × 3.0" 18.00 sq. in.
Baumeister 2 halves w/ Delta A 4.5" × 2.3" 10.35 sq. in.
Baumeister with Delta C 1.5" × 0.9" 1.35 sq. in.
This is assuming a Bull’s eye with the same information and reliable readability.
Baumiester and Crouse were no longer involved in the creation of the code and symbol. Crouse was responsible for designing a hand held device to read Laurer’s symbol, Baumiester did theoretical work on scanner designs. D. Savir, a mathematician was given the task of proving the symbol could be printed and would meet the reliability requirements. N. J. Woodland, the inventor of the Bull’s Eye code was responsible for writing the IBM proposal to the selection committee. A group under the direction of Art Hamburgen in Rochester designed and built a prototype scanner incorporating the architecture patented by Laurer. Dr. Sodastrum was the lead engineer on the scanner optics.
December 1, 1972 IBM presented Laurer’s proposal to the Super Market Committee in Rochester Minnesota, the location where IBM would develop the scanner. During the presentation Crouse gave a lab demonstration where he read UPC like labels with a simple hand held wand. In addition to reading regular labels he read the large two page center fold label in the proposal booklet. He then turned to a page showing a photo of labeled items sitting on a table. The labels were small and flawed due to the resolution of the printed photo but the wand read many of them. This demonstration showed the robustness of the code and the proposal was accepted.
Laurer continued his career with the UPC. He became known as the inventor of the UPC; without his persistence there might not have been an IBM proposal. Baumeister and Crouse moved on to other activities, Baumeister prior to the Rochester proposal and Crouse immediately after.
## Technical Notes
The EAN was developed as a superset of UPC, adding an extra digit to the beginning so that there would be plenty of numbers for the entire world.
The prefix digit 0 has been reserved for UPC, and in fact the GS1 US mandated all retail systems in the United States and Canada be able to recognize both UPC and EAN by January 1, 2005 .
UPC usage notes:
• Currently all products marked with an EAN will be accepted in North America in addition to those products already marked with a UPC.
• Any product with an existing UPC does not have to be remarked with an EAN.
• In North America the EAN adds 40% more codes mainly by adding 10 to 13 to the 00 to 09 (0 to 9 in UPC) already in use. This is a powerful incentive to phase out the UPC. | 3,594 | 13,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2013-20 | latest | en | 0.911393 |
http://www.gametheorystrategies.com/2011/08/18/fair-division/ | 1,553,079,939,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202326.46/warc/CC-MAIN-20190320105319-20190320131319-00127.warc.gz | 279,412,510 | 6,715 | # Fair division
How do you make progress when you need to divide something up between two people but they can’t agree on what is a fair split.
The classic example is how to fairly cut a cake.
The answer is for one person to cut the cake and the other person to choose which piece they have. The first person has an incentive to cut the cake as accurately as possible into two equal pieces. If they don’t cut the cake equally then the other person will take the large piece.
This idea has many applications in real life situations such as dividing assets between a divorcing couple. It can also be used to divide up things that people don’t want but have to have, for example household chores can be divided up by one person and the other chooses which they want to do.
Things can get more difficult with an indivisible item, but this can be simplified by using cash to balance up the split. For example, if two children have inherited a house then they need to divide this between them. They could sell the house and split the money but one, or both of them, may prefer to keep the house. In this case one could propose a split of say, the house less \$200,000 for one person and \$200,000 for the other person (the amount would depend on how the person valued the house). The other person would then choose whether they wanted to have the house and make a \$200,000 payment to their sibling, or whether they would rather just have the \$200,000.
Today’s takeaway: The tactic of divide and choose can be used to resolve many negotiation problems.
This entry was posted in Game theory, Introduction and tagged , , , , . Bookmark the permalink. | 348 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-13 | longest | en | 0.970865 |
http://lists.electorama.com/pipermail/election-methods-electorama.com/2013-February/129535.html | 1,725,847,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00049.warc.gz | 15,989,574 | 3,126 | # [EM] The Green scenario, and IRV in the Green scenario, is a new topic here. Hence these additional comments. Clarification of position and why.
Peter Zbornik pzbornik at gmail.com
Mon Feb 4 12:31:29 PST 2013
```Hi I am afraid a proportional approach in the first round wouldnt
work, it opens up for strategic voting.
Say we have an election with A, B, C.
45 A
30 B A
25 C B A
The first round in a 2-seat election the quota is 34 votes
If we would have a two-round proportional election, then B would win
in the second round.
So A's voters find this out and decide to change their preferences and
10 of the voters of A vote for C
So we have
35 A
30 BA
25 CBA
10 CA
C and A meet in the second round, where A wins.
PZ
2013/2/4 Kristofer Munsterhjelm <km_elmet at lavabit.com>:
> On 02/04/2013 02:40 PM, Peter Zbornik wrote:
>>
>> Being a green party member (although a Czech one and not US), I would
>> variant of IRV, i.e. elimination of the candidates and transfer of
>> votes until two remain, no quota for election (or quota=100%) except
>> for the case where one candidate has more than 50% of first
>> preferences.
>>
>> The top two candidates would meet in a second round in IRV.
>> A candidate would be elected if he/she would get more than 50% of the
>>
>> Empty votes would count as valid votes in both first and second round.
>>
>> If no candidate would be elected in second round new elections would take
>> place.
>>
>> The advantages of the proposed election system are
>> 1) the voters are given a chance to concentrate only on two candidates
>> in the second round, and are thus allowed to change their preferences.
>> 2) blank votes together with IRV might make the candidates less
>> polarized, as, given a large number of blank votes, the candidate with
>> the highest number of votes in the second round would have to rely on
>> the second preferences of the voters for the opposing candidate in
>> order to get 50%+ votes.
>
>
> Perhaps this method would work for runoffs if you can get a more
> sophisticated base method through, say for internal elections:
>
> - Run a single-winner election using your method of choice. Call the winner
> w_1.
> - Use a proportional ranking method to determine the second runoff candidate
> w_2 so that the virtual council {w_1, w_2} represents as much as possible of
> the population.
> - Have a runoff between w_1 and w_2.
>
> If w_1 is a strong winner, he'll win in the runoff. If he's a weak winner
> (e.g. the "bland politician being everybody's second choice" scenario), w_2
> wins.
>
> In IRV, this would be like running two-member STV where the IRV winner is
> barred from being disqualified.
>
> There could be a problem, though, in a society that has a bland centrist
> politician and strong left- and right-wing candidates. Since the runoff can
> only hold two candidates, either the left-wing or the right-wing candidate
> would be disqualified; and if the bland politician is sufficiently bland,
> then the wing candidate would pretty much win by default. IRV "solves" this
> by not letting center-squeezed candidates win in the first place. Another
> option is to have multiple candidates in the runoff, but then the simplicity
> and strategy resistance properties of the second round go away.
>
``` | 830 | 3,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.957568 |
http://coq.inria.fr/distrib/8.3pl1/stdlib/Coq.Numbers.NatInt.NZAxioms.html | 1,371,602,850,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707437545/warc/CC-MAIN-20130516123037-00099-ip-10-60-113-184.ec2.internal.warc.gz | 58,640,538 | 3,925 | # Library Coq.Numbers.NatInt.NZAxioms
Initial Author : Evgeny Makarov, INRIA, 2007
Axiomatization of a domain with zero, successor, predecessor, and a bi-directional induction principle. We require P (S n) = n but not the other way around, since this domain is meant to be either N or Z. In fact it can be a few other things, for instance Z/nZ (See file NZDomain for a study of that).
Module Type ZeroSuccPred (Import T:Typ).
Parameter Inline zero : t.
Parameters Inline succ pred : t -> t.
End ZeroSuccPred.
Module Type ZeroSuccPredNotation (T:Typ)(Import NZ:ZeroSuccPred T).
Notation "0" := zero.
Notation S := succ.
Notation P := pred.
Notation "1" := (S 0).
Notation "2" := (S 1).
End ZeroSuccPredNotation.
Module Type ZeroSuccPred' (T:Typ) :=
ZeroSuccPred T <+ ZeroSuccPredNotation T.
Module Type IsNZDomain (Import E:Eq')(Import NZ:ZeroSuccPred' E).
Declare Instance succ_wd : Proper (eq ==> eq) S.
Declare Instance pred_wd : Proper (eq ==> eq) P.
Axiom pred_succ : forall n, P (S n) == n.
Axiom bi_induction :
forall A : t -> Prop, Proper (eq==>iff) A ->
A 0 -> (forall n, A n <-> A (S n)) -> forall n, A n.
End IsNZDomain.
Module Type NZDomainSig := EqualityType <+ ZeroSuccPred <+ IsNZDomain.
Module Type NZDomainSig' := EqualityType' <+ ZeroSuccPred' <+ IsNZDomain.
Axiomatization of basic operations : + - *
Parameters Inline add sub mul : t -> t -> t.
Notation "x + y" := (add x y).
Notation "x - y" := (sub x y).
Notation "x * y" := (mul x y).
Declare Instance sub_wd : Proper (eq ==> eq ==> eq) sub.
Declare Instance mul_wd : Proper (eq ==> eq ==> eq) mul.
Axiom add_0_l : forall n, 0 + n == n.
Axiom add_succ_l : forall n m, (S n) + m == S (n + m).
Axiom sub_0_r : forall n, n - 0 == n.
Axiom sub_succ_r : forall n m, n - (S m) == P (n - m).
Axiom mul_0_l : forall n, 0 * n == 0.
Axiom mul_succ_l : forall n m, S n * m == n * m + m.
Old name for the same interface:
Module Type NZAxiomsSig := NZBasicFunsSig.
Module Type NZAxiomsSig' := NZBasicFunsSig'.
Axiomatization of order
Module Type NZOrd := NZDomainSig <+ HasLt <+ HasLe.
Module Type NZOrd' := NZDomainSig' <+ HasLt <+ HasLe <+
LtNotation <+ LeNotation <+ LtLeNotation.
Module Type IsNZOrd (Import NZ : NZOrd').
Declare Instance lt_wd : Proper (eq ==> eq ==> iff) lt.
Axiom lt_eq_cases : forall n m, n <= m <-> n < m \/ n == m.
Axiom lt_irrefl : forall n, ~ (n < n).
Axiom lt_succ_r : forall n m, n < S m <-> n <= m.
End IsNZOrd.
NB: the compatibility of le can be proved later from lt_wd and lt_eq_cases
Module Type NZOrdSig := NZOrd <+ IsNZOrd.
Module Type NZOrdSig' := NZOrd' <+ IsNZOrd.
Everything together :
Same, plus a comparison function. | 852 | 2,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2013-20 | latest | en | 0.486837 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese2367.htm | 1,696,464,628,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00040.warc.gz | 662,867,239 | 4,618 | ### 3.2367 $$\int (-2+3 x) \sqrt{8+12 x+9 x^2} \, dx$$
Optimal. Leaf size=54 $\frac{1}{9} \left (9 x^2+12 x+8\right )^{3/2}-\frac{2}{3} (3 x+2) \sqrt{9 x^2+12 x+8}-\frac{8}{3} \sinh ^{-1}\left (\frac{3 x}{2}+1\right )$
[Out]
(-2*(2 + 3*x)*Sqrt[8 + 12*x + 9*x^2])/3 + (8 + 12*x + 9*x^2)^(3/2)/9 - (8*ArcSinh[1 + (3*x)/2])/3
________________________________________________________________________________________
Rubi [A] time = 0.0180984, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {640, 612, 619, 215} $\frac{1}{9} \left (9 x^2+12 x+8\right )^{3/2}-\frac{2}{3} (3 x+2) \sqrt{9 x^2+12 x+8}-\frac{8}{3} \sinh ^{-1}\left (\frac{3 x}{2}+1\right )$
Antiderivative was successfully verified.
[In]
Int[(-2 + 3*x)*Sqrt[8 + 12*x + 9*x^2],x]
[Out]
(-2*(2 + 3*x)*Sqrt[8 + 12*x + 9*x^2])/3 + (8 + 12*x + 9*x^2)^(3/2)/9 - (8*ArcSinh[1 + (3*x)/2])/3
Rule 640
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]
Rule 612
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]
Rule 619
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Rule 215
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]
Rubi steps
\begin{align*} \int (-2+3 x) \sqrt{8+12 x+9 x^2} \, dx &=\frac{1}{9} \left (8+12 x+9 x^2\right )^{3/2}-4 \int \sqrt{8+12 x+9 x^2} \, dx\\ &=-\frac{2}{3} (2+3 x) \sqrt{8+12 x+9 x^2}+\frac{1}{9} \left (8+12 x+9 x^2\right )^{3/2}-8 \int \frac{1}{\sqrt{8+12 x+9 x^2}} \, dx\\ &=-\frac{2}{3} (2+3 x) \sqrt{8+12 x+9 x^2}+\frac{1}{9} \left (8+12 x+9 x^2\right )^{3/2}-\frac{2}{9} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{144}}} \, dx,x,12+18 x\right )\\ &=-\frac{2}{3} (2+3 x) \sqrt{8+12 x+9 x^2}+\frac{1}{9} \left (8+12 x+9 x^2\right )^{3/2}-\frac{8}{3} \sinh ^{-1}\left (1+\frac{3 x}{2}\right )\\ \end{align*}
Mathematica [A] time = 0.0203861, size = 40, normalized size = 0.74 $\frac{1}{9} \left (\left (9 x^2-6 x-4\right ) \sqrt{9 x^2+12 x+8}-24 \sinh ^{-1}\left (\frac{3 x}{2}+1\right )\right )$
Antiderivative was successfully verified.
[In]
Integrate[(-2 + 3*x)*Sqrt[8 + 12*x + 9*x^2],x]
[Out]
((-4 - 6*x + 9*x^2)*Sqrt[8 + 12*x + 9*x^2] - 24*ArcSinh[1 + (3*x)/2])/9
________________________________________________________________________________________
Maple [A] time = 0.052, size = 43, normalized size = 0.8 \begin{align*} -{\frac{18\,x+12}{9}\sqrt{9\,{x}^{2}+12\,x+8}}-{\frac{8}{3}{\it Arcsinh} \left ( 1+{\frac{3\,x}{2}} \right ) }+{\frac{1}{9} \left ( 9\,{x}^{2}+12\,x+8 \right ) ^{{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-2+3*x)*(9*x^2+12*x+8)^(1/2),x)
[Out]
-1/9*(18*x+12)*(9*x^2+12*x+8)^(1/2)-8/3*arcsinh(1+3/2*x)+1/9*(9*x^2+12*x+8)^(3/2)
________________________________________________________________________________________
Maxima [A] time = 1.49423, size = 70, normalized size = 1.3 \begin{align*} \frac{1}{9} \,{\left (9 \, x^{2} + 12 \, x + 8\right )}^{\frac{3}{2}} - 2 \, \sqrt{9 \, x^{2} + 12 \, x + 8} x - \frac{4}{3} \, \sqrt{9 \, x^{2} + 12 \, x + 8} - \frac{8}{3} \, \operatorname{arsinh}\left (\frac{3}{2} \, x + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-2+3*x)*(9*x^2+12*x+8)^(1/2),x, algorithm="maxima")
[Out]
1/9*(9*x^2 + 12*x + 8)^(3/2) - 2*sqrt(9*x^2 + 12*x + 8)*x - 4/3*sqrt(9*x^2 + 12*x + 8) - 8/3*arcsinh(3/2*x + 1
)
________________________________________________________________________________________
Fricas [A] time = 2.26475, size = 123, normalized size = 2.28 \begin{align*} \frac{1}{9} \, \sqrt{9 \, x^{2} + 12 \, x + 8}{\left (9 \, x^{2} - 6 \, x - 4\right )} + \frac{8}{3} \, \log \left (-3 \, x + \sqrt{9 \, x^{2} + 12 \, x + 8} - 2\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-2+3*x)*(9*x^2+12*x+8)^(1/2),x, algorithm="fricas")
[Out]
1/9*sqrt(9*x^2 + 12*x + 8)*(9*x^2 - 6*x - 4) + 8/3*log(-3*x + sqrt(9*x^2 + 12*x + 8) - 2)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (3 x - 2\right ) \sqrt{9 x^{2} + 12 x + 8}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-2+3*x)*(9*x**2+12*x+8)**(1/2),x)
[Out]
Integral((3*x - 2)*sqrt(9*x**2 + 12*x + 8), x)
________________________________________________________________________________________
Giac [A] time = 1.0945, size = 61, normalized size = 1.13 \begin{align*} \frac{1}{9} \,{\left (3 \,{\left (3 \, x - 2\right )} x - 4\right )} \sqrt{9 \, x^{2} + 12 \, x + 8} + \frac{8}{3} \, \log \left (-3 \, x + \sqrt{9 \, x^{2} + 12 \, x + 8} - 2\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-2+3*x)*(9*x^2+12*x+8)^(1/2),x, algorithm="giac")
[Out]
1/9*(3*(3*x - 2)*x - 4)*sqrt(9*x^2 + 12*x + 8) + 8/3*log(-3*x + sqrt(9*x^2 + 12*x + 8) - 2) | 2,885 | 5,808 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-40 | latest | en | 0.099585 |
https://www.geeksforgeeks.org/create-a-data-frame-of-all-the-combinations-of-vectors-passed-as-argument-in-r-programming-expand-grid-function/?ref=lbp | 1,604,019,420,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00696.warc.gz | 714,103,428 | 20,084 | # Create a Data Frame of all the Combinations of Vectors passed as Argument in R Programming – expand.grid() Function
`expand.grid()` function in R Language is used to create a data frame with all the values that can be formed with the combinations of all the vectors or factors passed to the function as argument.
Syntax: expand.grid(…)
Parameters:
…: Vector1, Vector2, Vector3, …
Example 1:
`# R program to create a dataframe ` `# with combination of vectors ` ` ` `# Creating vectors ` `x1 <``-` `c(``"abc"``, ``"cde"``, ``"def"``) ` `x2 <``-` `c(``1``, ``2``, ``3``) ` `x3 <``-` `c(``"M"``, ``"F"``) ` ` ` `# Calling expand.grid() Function ` `expand.grid(x1, x2, x3) `
Output:
``` Var1 Var2 Var3
1 abc 1 M
2 cde 1 M
3 def 1 M
4 abc 2 M
5 cde 2 M
6 def 2 M
7 abc 3 M
8 cde 3 M
9 def 3 M
10 abc 1 F
11 cde 1 F
12 def 1 F
13 abc 2 F
14 cde 2 F
15 def 2 F
16 abc 3 F
17 cde 3 F
18 def 3 F
```
Example 2:
`# R program to create a dataframe ` `# with combination of vectors ` ` ` `# Creating vectors ` `x1 <``-` `c(``"abc"``, ``"cde"``, ``"def"``) ` `x2 <``-` `c(``1``, ``2``, ``3``) ` `x3 <``-` `c(``"M"``, ``"F"``) ` ` ` `# Calling expand.grid() Function ` `expand.grid(x1, x3) `
Output:
``` Var1 Var2
1 abc M
2 cde M
3 def M
4 abc F
5 cde F
6 def F
```
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https://www.khanacademy.org/science/ap-college-physics-1/xf557a762645cccc5:kinematics-and-introduction-to-dynamics/xf557a762645cccc5:contact-forces/e/contact-forces | 1,713,221,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00409.warc.gz | 793,045,511 | 54,583 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Course: AP®︎/College Physics 1>Unit 1
Lesson 5: Contact forces
# Contact forces
You might need: Calculator
## Problem
The spring in a pinball machine has a spring constant $k$ of $2,130\phantom{\rule{0.167em}{0ex}}\frac{\text{N}}{\text{m}}$.
What is the magnitude of the force required to compress this spring $4.5\phantom{\rule{0.167em}{0ex}}\text{cm}$?
$\text{N}$ | 174 | 595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.798652 |
http://plumsolutions.com.au/online-courses/introduction-financial-modelling-excel-online | 1,490,228,044,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186530.52/warc/CC-MAIN-20170322212946-00629-ip-10-233-31-227.ec2.internal.warc.gz | 273,522,610 | 10,842 | SPECIALISTS IN FINANCIAL MODELLING
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# Introduction to Financial Modelling in Excel (Online)
This course is part of the Financial Modelling Online Series:
Buy all three courses at once and save, PLUS you'll get a hard copy of Using Excel for Business Analysis included! Or, get this course included if you sign up for the Financial Modelling Workshop bundle.
### Learning Objectives
This course introduces the fundamentals and introductory concepts of building a financial model using Excel. During this practical course in which you will develop the skills to build your own financial model which can be taken away for future reference. You will learn first principles of modelling techniques, best practice and how to design and create a user-friendly model.
### Prerequisites
Who should complete this course? Those who:
• Are new to Financial Modelling and its concepts, and;
• Have a basic knowledge of Excel
It is assumed that you have either undertaken a basic Excel Introduction course, or else are able to:
• Navigate in Excel
• Use absolute cell references (this is covered in the course in case you are a bit rusty!)
• Create and use simple formulas
It is also assumed that you have had some exposure to financial concepts, but limited experience in Financial Modelling.
More detail about how the online courses work
### What does the course content look like?
If you'd like to see a full 15 min preview of one of the videos, together with dowloadable Excel files and transcript, take a look at the Excel Versions section.
For a shorter overview, watch here for a 60 second preview:
## Modules in this Course
1.1 Financial Modelling Introduction
• What is Financial Modelling?
• Model Design
• Excel Versions 2013
• What's New in Excel 2016 *New*
• Overview of Scenario Analysis Methods
• Common Excel Error Values
1.2 Fundamental Excel Tools
• Linking for Best Practice
• Absolute, Relative and Mixed Cell Referencing
• Calculating Project Costs
• Logical Nested Functions
• Assumptions Documentation Methods
1.3 Building a Financial Model
• Model Build - Inputs & Assumptions
• Model Build - Revenue
• Model Build - Expenses
• Model Build – Profit & Loss
• Model Build - Cashflow
1.4 Model Analysis Tools
• Bullet-Proofing your Model
• Data Validations & Drop-downs
• Conditional Formatting
• Model Build - Charting Cash vs Profit
• Model Build - Modelling Scenarios
Price: \$220.00 ex GST
I confirm that I have read the terms & conditions. Buy Now Buy Now Please check the box to confirm you agree before proceeding. | 563 | 2,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-13 | latest | en | 0.848511 |
https://doubtnut.com/question-answer/the-number-of-even-proper-divisors-of-5040-is-53793473 | 1,585,879,680,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00325.warc.gz | 433,354,895 | 41,695 | or
# The number of even proper divisors of 5040, is
Question from Class 11 Chapter Permutations And Combinations
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484746none of these
Solution :
We have, `5040=2^(4)xx3^(2)xx5xx7` <br> `:.` Number of even proper divisors <br> = Number of ways of selecting at least one, 2 and any number of `3^(s),5^(s)and7` <br> `=(4)xx(2+1)xx(1+1)xx(1+1)-1=47`.
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100.4 K+ Views | 219.5 K+ Likes | 328 | 822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-16 | latest | en | 0.643084 |
http://mymathforum.com/algebra/252166-need-some-help-understanding.html | 1,545,013,319,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828056.99/warc/CC-MAIN-20181217020710-20181217042710-00475.warc.gz | 196,875,626 | 8,809 | My Math Forum Need some help understanding.
Algebra Pre-Algebra and Basic Algebra Math Forum
November 5th, 2015, 04:44 PM #1 Newbie Joined: Oct 2015 From: canada Posts: 9 Thanks: 0 Need some help understanding. Hey guys, currently working on a problem that is: Britney javelin company is considering two investments, both of which cost \$15000. The cash flows are as follows: year project a project b 1 \$8100 \$6750 2 5400 4050 3 4050 10800 which of the two projects should be chosen based on the net present value method? Assume a cost of capital of 7%. b. NPV for Project a Year Cash flow Present value @ 7% 1 \$8,100 \$7,570 2 5,400 4,717 3 4,050 3,306 Present value of inflows \$15,593 Initial investment 15,000 NPV (net present value) \$593 I am just confused on how on the 7570, 4717 and 3306 were found. Thanks in advance for any help! Last edited by skipjack; November 5th, 2015 at 05:15 PM. November 5th, 2015, 05:26 PM #2 Global Moderator Joined: Dec 2006 Posts: 19,988 Thanks: 1855 \$8,100 / 1.07 = \$7,570.09 \$5,400 / 1.07² = \$4,716.57 \$4,050 / 1.07³ = \\$3,306.01 etc.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 490 | 1,499 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-51 | latest | en | 0.886314 |
https://math.stackexchange.com/questions/3736701/cayley-hamilton-theorem-possible-characteristic-polynomial | 1,657,038,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00715.warc.gz | 449,993,692 | 64,051 | # Cayley-Hamilton-Theorem - Possible characteristic polynomial
Let $$A: \mathbb{R}^3 \to \mathbb{R}^3$$ s.t.
$$A^3-2A^2+A= 0$$
The Cayley-Hamilton-Thm. states that if I put $$A$$ into its characteristic polynomial it'll equal $$0$$.
But am I allowed to conclude from the given equation $$A^3-2A^2+A= 0$$ that $$\lambda^3-2 \lambda^2+\lambda$$ is the characteristic polynomial of $$A$$?
No you're not. What if your matrix $$A$$ was a $$3\times 3$$ zero matrix (all elements are $$0$$). Then your equation would be valid but the characteristic polynomial is $$\lambda^3 = 0$$.
However you know that if $$\lambda$$ is an eigenvalue of $$A$$ then $$\lambda=0$$ or $$\lambda =1$$, which are the only two roots of $$P$$, where $$P(\lambda) = \lambda^3 - 2\lambda^2 + \lambda = \lambda (\lambda-1)^2\,.$$ So the characteristic polynomial can have at most $$0$$ and $$1$$ as roots. So all you know is that the characteristic polynomial $$Q$$ must be in the form $$Q(\lambda) = \lambda^n (\lambda-1)^m$$ with $$n\in\mathbb{N}$$, $$m\in\mathbb{N}$$ such that $$m+n=N$$, where $$N$$ is the number of rows (and so columns) of $$A$$. In your case $$N=3$$. | 370 | 1,147 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-27 | latest | en | 0.845316 |
https://forum.allaboutcircuits.com/threads/generating-a-bode-plot-from-the-small-signal-analysis-of-a-buck-converter-ltspice.189880/ | 1,716,209,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00588.warc.gz | 220,385,369 | 23,330 | # Generating a Bode plot from the small signal analysis of a Buck converter LTSPICE
#### Brettjohnson7191
Joined Apr 28, 2022
27
Good evening all,
I am stuck on a homework assignment and seeking your help. The assignment is for a power electronics course. We are given Vout=5v Vin=40v
output inductor is 100uH and the output capacitor is 1000uF with 10 mohms ESR all operated at 600kHz. This is also required to be completed in LT Spice
for my first attempt I decided to make a regular version of the buck converter which is attached as image 1. Then I began creating the small signal model of the buck converter (image2).
I run into issues when trying to figure out the values to enter into the VCVS and the VCCS can anyone point me in the right direction?
Thanks!
Image1
Image 2
#### crutschow
Joined Mar 14, 2008
34,677
Doesn't the circuit need some negative feedback to regulate the output?
As is, it is just an open-loop circuit.
#### Papabravo
Joined Feb 24, 2006
21,256
Doesn't the circuit need some negative feedback to regulate the output?
As is, it is just an open-loop circuit.
Ultimately it does, but you have to break the loop if you want to understand the open loop response. Also, this is a synchronous buck converter, so you need to work out your timing so you don't have both switches on at the same time. That would be a bad bad thing.
Getting the open loop bode plot will be easier if you substitute a Schottky diode for M2.
When looking at a DC-DC converter, the AC analysis in LTspice is quite difficult to use because you are not comparing an AC output to an AC input. You are comparing a duty cycle to and DC output with a very small, high frequency ripple. I've tried to make that work and have been unsuccessful. the method I used in the other thread is documented in the LTspice Help file. Search for "Middlebrook", the name of the developer of the method. Here is the preamble:
How to get a Bode Plot from a SMPS
In the interest of stability, the open loop gain of a negative feedback system operating in closed loop should fall below unity with increasing frequency before too much phase shift occurs unless your aim actually is to make an oscillator[1]. This idea can be applied to the stability analysis of a Switch Mode Power Supply(SMPS). Even though an SMPS is an intrinsically non-linear circuit with no small-signal linear equivalent circuit, there typically is an analog feedback loop operating on the filtered, switched output.
Last edited:
#### Brettjohnson7191
Joined Apr 28, 2022
27
Ultimately it does, but you have to break the loop if you want to understand the open loop response. Also, this is a synchronous buck converter, so you need to work out your timing so you don't have both switches on at the same time. That would be a bad bad thing.
Getting the open loop bode plot will be easier if you substitute a Schottky diode for M2.
When looking at a DC-DC converter, the AC analysis in LTspice is quite difficult to use because you are not comparing an AC output to an AC input. You are comparing a duty cycle to and DC output with a very small, high frequency ripple. I've tried to make that work and have been unsuccessful. the method I used in the other thread is documented in the LTspice Help file. Search for "Middlebrook", the name of the developer of the method. Here is the preamble:
How to get a Bode Plot from a SMPS
In the interest of stability, the open loop gain of a negative feedback system operating in closed loop should fall below unity with increasing frequency before too much phase shift occurs unless your aim actually is to make an oscillator[1]. This idea can be applied to the stability analysis of a Switch Mode Power Supply(SMPS). Even though an SMPS is an intrinsically non-linear circuit with no small-signal linear equivalent circuit, there typically is an analog feedback loop operating on the filtered, switched output.
I replaced my low side switching fet with a schottky diode and attached my new file to this reply. My transient response
is showing only around 300mV output at my Vout node. I know there should be some switching losses but I would still expect a range in the 3-4V range if it is working properly. I plot my highside fet voltage and see it is switching at the correct time as well. Any other ideas? I appreciate the help.
#### crutschow
Joined Mar 14, 2008
34,677
Ultimately it does, but you have to break the loop if you want to understand the open loop response.
Okay, but here there's no loop to break, along with the compensation elements required for stability with feedback.
#### crutschow
Joined Mar 14, 2008
34,677
My transient response
is showing only around 300mV output at my Vout node.
So does the circuit eventually need to have feedback to stabilize the output voltage or not?
#### Brettjohnson7191
Joined Apr 28, 2022
27
Okay, but here there's no loop to break, along with the compensation elements required for stability with feedback.
Yes I understand that the first step in our homework is to simulate this portion of the buck converter generating a bode plot for this stage then we are to create the compensation network with the pole generating resistor/capacitor network. and simulate the bode plot once again and compare the two results
#### Papabravo
Joined Feb 24, 2006
21,256
Okay, but here there's no loop to break, along with the compensation elements required for stability with feedback.
Understanding the open loop response is just the first step. With a fixed load and a fixed duty cycle the system is stable without feedback. It is certainly not optimal that comes later.
#### Papabravo
Joined Feb 24, 2006
21,256
I replaced my low side switching fet with a schottky diode and attached my new file to this reply. My transient response
is showing only around 300mV output at my Vout node. I know there should be some switching losses but I would still expect a range in the 3-4V range if it is working properly. I plot my highside fet voltage and see it is switching at the correct time as well. Any other ideas? I appreciate the help.
View attachment 279117
It is trying to get somewhere, but has clearly not reached steady state. C1 is a honking big capacitor. Try a longer simulation and then use the parameter in the .trsan menu to start collecting data after the response reaches steady state
. Windows has a native utility to extract .zip files, but if it is not working for you the try down loading the 7zip.exe utility. If you are on a Mac then I don't know what to tell you. Also zoom in on your inductor current to see the inductor charging and discharging cycles.
#### Papabravo
Joined Feb 24, 2006
21,256 | 1,546 | 6,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.934831 |
https://stats.stackexchange.com/questions/80955/lasso-on-negative-binomial-regression-model?noredirect=1 | 1,709,459,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00038.warc.gz | 546,271,377 | 40,459 | # Lasso on Negative Binomial Regression Model
Is there anyway that I can perform LASSO with Negative Binomial Regression on R? I am performing a negative binomial regression on my dataset because the data are too dispersed to impose poisson regression. Meanwhile, I am also facing some multicollinearity problem. I already tried using glmnet with family = poisson, but the data is not fitting very well (for both alpha = 0 and alpha = 1).
EDIT: here is variance-covariance table of the negative binomial fit
8.392729e+18 1.239178e+06 -3.624090e+05 1.896258e+17 -3.702521e+17
1.239178e+06 1.119052e-04 5.201989e-06 -1.877590e+05 -2.558095e+05
-3.624090e+05 5.201989e-06 5.179343e-06 -8.021543e+04 -1.436381e+05
1.896258e+17 -1.877590e+05 -8.021543e+04 2.193290e+17 6.413947e+16
-3.702521e+17 -2.558095e+05 -1.436381e+05 6.413947e+16 2.142183e+17
• I think you'll want to take this one to Stack Overflow. Dec 31, 2013 at 22:19
• To be honest, I'm not sure if this question will even be on-topic on SO; you may want to ask on the r-help listserv. Dec 31, 2013 at 23:13
• This question appears to be off-topic because it is about whether a particular analysis can be run in R. Dec 31, 2013 at 23:14
• it's also going to get hammered on SO because it's just a "how can I?" question, rather than a specific programming question ... Can you give a little more context? I would be tempted to do a quasi-Poisson fit (i.e., fit the model as a Poisson lasso, e.g. with the glmnet package, then make a post hoc adjustment to the standard errors of the parameters based on the estimated residual deviance ...) Dec 31, 2013 at 23:41
• Have you tried a quasipoisson model then? Based on the very vague description, I think your substantive problem probably has to do with a singularity issue. Show us what the vcov(fit) gives, fit being your glm object. Dec 31, 2013 at 23:59
LASSO and other penalized methods for negative binomial and zero-inflated negative binomial are provided by the mpath package in R, as has been noted on a more recent Cross Validated page. One answer on that page, however, indicates some difficulty in using mpath. A recent publication illustrates an application of the mpath package; a vignette in the R package reproduces the data analysis of that publication. | 698 | 2,306 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.887606 |
https://mariewallacerealestateagentgranitebayca.com/question-249 | 1,675,374,786,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00026.warc.gz | 408,741,578 | 5,738 | # Solve math problems
As a student, there are times when you need to Solve math problems. Keep reading to learn more!
## Solving math problems
In this blog post, we will be discussing how to Solve math problems. If you're having trouble with your camera, there are a few things you can do to try and solve the problem. First, check the manual to see if there are any troubleshooting steps you can follow. If that doesn't help, try resetting the camera to its factory settings. If the problem still isn't solved, you may need to contact the camera's manufacturer for further assistance.
Algebra is a math discipline that primarily focuses on the manipulation of equations and variables. Algebra is used in many fields, including physics, engineering, and economics. Many students find algebra to be difficult, but with practice and perseverance, it can be mastered.
However, there are a few key differences that you need to be aware of. First, when you multiply or divide both sides of an inequality by a negative number, you need to reverse the direction of the inequality. Second, you need to be careful when you add or subtract numbers from both sides of an inequality. If you add or subtract a negative number, you need to reverse
There are many ways to solve binomial equations, but one of the most straightforward methods is to factor the equation and set each factor equal to zero. This can be done by factoring out the greatest common factor, or by using a factorization method such as the quadratic formula. Once the equation is factored, it can be solved by setting each factor equal to zero and solving for the variable.
There are a few steps to solving linear equations: 1) First, you need to identify the variables in the equation and what they represent. 2) Next, you'll want to isolate the variable you are solving for on one side of the equation. 3) Once the variable is isolated, you can begin solving for it by using inverse operations. This means you'll do the opposite operation of what is being done to the variable in order to solve for it.
## We will help you with math problems
Awesome, versatile app for scanning problems from the computer screen or from paper (hand written or typed) to solve, graph, find the domain, act. Very useful for checking your solutions before submitting them. There are a few things it can't to, like graphing linear inequalities and displaying complex solutions to equations in standard form. Overall, it's a great app.
Danika Hughes
Aesthetic, minimal design, no adds but most importantly it works, it does exactly what it says it does and more, it not only does the math, but breaks it down step by step for you, honestly no better tool for students out there Thank you the app team, you've done a great job
Bethany Moore
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https://itprospt.com/num/11456619/a-5-00-times-10-5-mathrm-kg-rocket-is-accelerating-straight | 1,660,206,518,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00544.warc.gz | 320,403,361 | 13,585 | 1
# $-A 5.00 \times 10^{5}-\mathrm{kg}$ rocket is accelerating straight up. Its engines produce $1.250 \times 10^{7} \mathrm{N}$ of thrust, and air resistance is $4.50 ... ## Question ######$-A 5.00 \times 10^{5}-\mathrm{kg}$rocket is accelerating straight up. Its engines produce$1.250 \times 10^{7} \mathrm{N}$of thrust, and air resistance is$4.50 \times 10^{6} \mathrm{N}$. What is the rocket's acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton's laws of motion.$-A 5.00 \times 10^{5}-\mathrm{kg}$rocket is accelerating straight up. Its engines produce$1.250 \times 10^{7} \mathrm{N}$of thrust, and air resistance is$4.50 \times 10^{6} \mathrm{N}$. What is the rocket's acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton's laws of motion. ## Answers #### Similar Solved Questions 5 answers ##### Df Icnnis roict linetraictsIonnz-darna Facn Nckat MrAoilare WnprpJoj07*Firda expnzlor divino connlete dally Pronit Cforrulac rer JaoumirathatEekct: manufactur_Hint: TrcnctaucnumbcrNotrs sJldmulboletcach NorWe nL MirlG S (simpif Df Icnnis roict linet raicts Ionnz- darna Facn Nckat Mr Aoilare Wnprp Joj07* Firda expnzlor divino connlete dally Pronit Cfor rulac rer Jaoumirathat Eekct: manufactur_ Hint: Trc nctauc numbcr Notrs sJldmulbolet cach Nor We nL MirlG S (simpif... 5 answers ##### Evaluate the following integral: I5" r? dr dz d0 KsS dr dz d0 = (Type an exact answer:) Evaluate the following integral: I5" r? dr dz d0 KsS dr dz d0 = (Type an exact answer:)... 5 answers ##### ~2 pointsOscOLPhY52016 23.10.Wa.041 .AynotesEeioeacnerUaeateanr charetegisac bma gonsoanconmaccgiio Fhabnarrecismnci4OR 075 Conaieidbettcry, HhatGhc cugccori12 5 M>Aauk TIEhd-kpointsLerolsESetonmeacnemEei-ydananycaucorBnghuVorcocneegKafesSatinndwaoram DljhOho-Hat{#men MereDOsadncludLrCeunRaaen War Encn (achxaraltnnonuanino enolan Ior tne curcuilTcadu-Statccheniuodenly maycopotan(0) What k thc voltagc acTOs5 tie resistor aitct thc nitch has bacn parihonToJr tIme contantt(D) R (IS same uneTnema ~2 points OscOLPhY52016 23.10.Wa.041 . Aynotes Eeioeacner Uaeateanr charetegisac bma gonsoancon maccgiio Fhabnar recismnci 4OR 07 5 Conaieid bettcry, Hhat Ghc cugccori12 5 M> Aauk TIEhd-k points Lerols ESetonmeacnem Eei-ydanany caucorBng huVor cocneeg Kafes Satinn dwaoram Dljh Oho-Hat{#men Mere D... 5 answers ##### Q9s(max55050LSo945 CoutAssume the population is normally distributed, construct : confidence interval for monthly average of the amount spent on dining out with 95% confidence level. (Round your answer to decimal places) c =0.45 Thc 2-raluc fora GI,of957.is|.960_ 0 = 2/9 '' . 'The ohfiken c infcrval Gor wontkly avcraJc of 4k c spent oh Ainin4 Quh wjt/ CI, of 4 5/, 1$ 1y 4 ''1'
q9s(max 550 50 LSo 945 Cout Assume the population is normally distributed, construct : confidence interval for monthly average of the amount spent on dining out with 95% confidence level. (Round your answer to decimal places) c =0.45 Thc 2-raluc fora GI,of957.is|.960_ 0 = 2/9 '' . ' ...
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##### What kind of intermolecular forces are responsible for the aggregation of hemoglobin molecules that leads to sickle cell anemia?
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Use the discriminant to identify the conic section whose equation is given, and find a viewing window that shows a complete graph. $$x^{2}+10 x y+y^{2}+1=0$$...
##### A substance is found by qualitative tests to consist of onlycarbon and hydrogen (it is a hydrocarbon). A quantitative analysisis made of the substance by putting a weighed amount, 0.2822 g, ina tube that can be strongly heated from outside, and then burringin a stream of dry air. The air containing the products ofcombustion, is passed first through a weighed tube containingcalcium chloride, which absorbs the water vapor, and then throughanother weighed tube containing a mixture of sodium hydroxi
A substance is found by qualitative tests to consist of only carbon and hydrogen (it is a hydrocarbon). A quantitative analysis is made of the substance by putting a weighed amount, 0.2822 g, in a tube that can be strongly heated from outside, and then burring in a stream of dry air. The air contain...
##### C) The water in a cylindrical water tank is draining out througha hole in the base of the tank. The depth of the water isrepresented by h(t), and the initial depth of the wateris h(0) = H. The situation is represented by:hï‚¢(t)  ï€2k hwhere k is a constant that includes aspects such as theradius of the tank and the size of thehole.Solve this initial value problem. 4 marksDetermine the solution for the case where k = 0.1and H = 0.5m. 1 markIn part ii., determine how long it will take for the
c) The water in a cylindrical water tank is draining out through a hole in the base of the tank. The depth of the water is represented by h(t), and the initial depth of the water is h(0) = H. The situation is represented by: hï‚¢(t)  ï€2k h where k is a constant that includes aspects...
##### For the curve r⃗ (t):=17〈16t^2+2t+88,18t^2+3t+32,14t^2+6t+50〉,−1≤t≤1the length of the tangential acceleration at thepoint t=0 is A49. Find A= .
For the curve r⃗ (t):=17〈16t^2+2t+88,18t^2+3t+32,14t^2+6t+50〉,−1≤t≤1 the length of the tangential acceleration at the point t=0 is A49. Find A= ....
##### Which of the following compounds will react fastest with waterwithout the aid of an acid or base catalyst and why?
Which of the following compounds will react fastest with water without the aid of an acid or base catalyst and why?... | 2,843 | 8,756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-33 | latest | en | 0.570054 |
https://dsp.stackexchange.com/questions/56440/i-want-to-calculate-two-parameters-related-to-bone-characterization-namely-bua-o | 1,571,473,614,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00546.warc.gz | 477,472,267 | 31,685 | I want to calculate two parameters related to Bone characterization namely BUA or Broadband ultrasound attenuation and Speed of Sound in MATLAB
I am using Simsonic software(http://www.simsonic.fr/) to generate 2 signals 1) for plane water(2D) width 45mm, and 2)for bone specimen 30mm width in the above vat. I am centering my source at 5mm and receiver at 40mm respectively. The Simulation environment can be seen below through the MATLAB code.
%% General parameters %%%
dx=0.05; % mm
Vmax=4.0; % mm/us
alpha=0.99;
dt=alpha*dx/(sqrt(2)*Vmax);% µs
%% Building map %%%
N1=400;
N2=800;
map=repmat(uint8(0),[N1 N2]);
SimSonic2DWriteMap2D(map);
%% Building signal %%%
f0=0.5; % central frequency, MHz
t0=1.5; % pulse center time
bndwdth=1.5; % pulse -6dB bandwidth
duration=2*t0; % signal length
% Gausspuls by default gives unity amplitude
timebase=(0:round(duration/dt)-1)'*dt;
[signalI,signalQ]=gauspuls(timebase-t0,f0,bndwdth);
signal=20*signalQ;
figure(3)
Fs=1e6;
t = 0:1/Fs:1;
L = length(t);
n = 2^nextpow2(L);
y=fft(signal,n);
Y=fftshift(y);
f = Fs*(0:(n/2))/n;
P = abs(Y/n);
plot(f,P(1:n/2+1))
SimSonic2DWriteSgl(signal)
%% check
figure(1)
imagesc(map)
axis image
figure(2)
plot(timebase,signal,'.-')
title('source signal')
xlabel('time (µs)')
After I run the simulation I loaded my data of signals received on a new script. There I have the code below:
[![clear;
clear all;
close all;
%script 1 Water
ProcessResults
%plot frequency domain of signal
figure(4)
Fs=1e6;
t = 0:1/Fs:1;
L = length(t);
n = 2^nextpow2(L);
y=fft(x,n);
Y=fftshift(y);
f = Fs*(0:(n/2))/n;
Mw = abs(Y/n);
plot(f,Mw(1:n/2+1))
title('Gaussian Pulse in Frequency Domain in water')
xlabel('Frequency (f)')
ylabel('|P(f)|')
%plot phase domain of signal
figure(3)
Fs=1e6;
t = 0:1/Fs:1;
L = length(t);
n = 2^nextpow2(L);
y=fft(x,n);
Y=fftshift(y);
f = Fs*(0:(n/2))/n;
Pw = abs(Y/n);
plot(f,Pw(1:n/2+1))
title('Gaussian Pulse in Phase Domain in water')
xlabel('Phase')
ylabel('|Ang(f)|')
%script 2 bone
cd 'C:\Users\khosl\Desktop\Project_Simulation\MATLAB\Simsonic\WaterBonePlaneInterface\Bone\SimulationResults'
ProcessResults
%plot frequency domain of signal
figure(7)
Fs=1e6;
t = 0:1/Fs:1;
L = length(t);
n = 2^nextpow2(L);
yn=fft(y,n);
Y=fftshift(yn);
f = Fs*(0:(n/2))/n;
Mb = abs(Y/n);
plot(f,Mb(1:n/2+1))
title('Gaussian Pulse in Frequency Domain in bone')
xlabel('Frequency (f)')
ylabel('|P(f)|')
%plot phase domain of signal
figure(8)
Fs=1e6;
t = 0:1/Fs:1;
L = length(t);
n = 2^nextpow2(L);
yn=fft(y,n);
Y=fftshift(yn);
f = Fs*(0:(n/2))/n;
Pb = abs(Y/n);
plot(f,Pb(1:n/2+1))
title('Gaussian Pulse in Phase Domain in bone')
xlabel('Phase')
ylabel('|Ang(f)|')
figure(9)
Mag=log(Mw./Mb);
plot(f,Mag(1:n/2+1))
title('Comparison of magnitude in Frequency Domain')
xlabel('Frequency (f)')
ylabel('|P(f)|')
n=length(Pb);
figure(10)
Pha=atan(Pb./Pw); %atan is inverse tangent in radians, alternate is atand in deg
plot(f,Pha(1:n/2+1))
title('Comparison of phase in Frequency Domain')
xlabel('Phase')
ylabel('|Ang(f)|')][1]][1]
I expected a constant slope In the comparison of magnitude as in "In Vitro Assessment of the Relationship Between Acoustic Properties and Bone Mass Density of the Calcaneus by Comparison of Ultrasound Parametric Imaging and Quantitative Computed Tomography". I followed the formula written in the paper mentioned.
How can I achieve the results: the paper has att*l=Ar(f)/A(f)(Reference signal through water/reference signal through specimen in water), att is the BUA and l is the width of specimen. If you run the code youll get the following output. the range of interest is between 200 to 600kHz but the slope should positive and upwards, but I am unable to get the same.
Thanks and Regards Kunal Khosla | 1,216 | 3,731 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-43 | longest | en | 0.543771 |
http://www.fact-index.com/t/ti/time_loop_logic.html | 1,643,270,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00608.warc.gz | 90,434,066 | 2,593 | Main Page | See live article | Alphabetical index
# Time loop logic
Time loop logic is a system of computation that requires the computer to be able to send data backwards through time, and relies upon the Novikov self-consistency principle to force the result of a computation sent backwards through time to be correct. This approach can overcome many limitations on traditional algorithmic complexity.
A program exploiting time loop logic can be quite simple in outline. For example, to find a factor for a large number:
• wait for the result to be transmitted from the future.
• upon receiving the result, test whether it is a factor by dividing the input number by it.
• if the received result is indeed a correct factor of the number, send the result back in time.
• else if the received result is not a correct factor of the number (or no result is received at all within the desired timeframe), generate a number different from the received result and send it back in time. Note that this results in a paradox, since the result sent back is not the same as the one that was received.
Since the Novikov principle states that it is impossible for any sequence of events to result in a paradox, the second clause of the conditional statement can never happen and the result sent from the future is guaranteed to be the correct one. If no result is possible - the subject number is prime, for example - then some event will occur to prevent the program from running in the first place, or prevent it from running correctly. An event that prevents the program from running in the first place would also satisfy Novikov's principle even if a correct result does exist, so it is important to limit the opportunities for such errors so that the "most likely" way the system will remain consistent is to provide the correct result as desired.
Of course, time loop logic is a purely theoretical exercise at this point. It is not known whether time travel is possible, or if it is whether Novikov's principle really applies to it. | 407 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-05 | latest | en | 0.945141 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-4-proportions-percents-and-solving-inequalities-chapter-4-test-page-185/20 | 1,569,221,928,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576122.89/warc/CC-MAIN-20190923064347-20190923090347-00325.warc.gz | 884,118,162 | 12,564 | ## Elementary Algebra
Let s represent the selling price. The basic relationship is: selling price equals cost plus profit. Selling price = Cost + Profit Selling price = s Cost = 40 Profit = 30% $\times$ 40 = .3 $\times$ 40 = 12 Substitute these values into the equation to obtain: s = 40 + 12 s = 52 She should charge 52 dollars for each lamp. | 89 | 344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-39 | latest | en | 0.838396 |
https://nz.education.com/slideshow/summer-math-worksheets/ | 1,597,492,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740838.3/warc/CC-MAIN-20200815094903-20200815124903-00588.warc.gz | 427,935,746 | 29,611 | # Summer maths: 15 Fun Worksheets
Make a splash (and help your child avoid the summer slide) with these sun-filled worksheets, perfect for summer practise.
Double-digit addition can come close to being as pleasant as a warm, summer day with this adorable year three worksheet!
### Number Patterns on the Beach
See if your little one has the sharp eyes and maths know-how to spot the patterns! Numbers jump by twos, fives, tens, and more on this maths worksheet.
### Summertime Fun!
Complete this sizzling hot addition worksheet by adding together the summer-related items in each box.
### Place Value Fill-In
Translating written out numbers into numerals is a great way to learn place values. Crabs and seashells decorate this colorful beach-themed maths worksheet!
### How Many Seashells Can You Find?
Take a look at the picture of the beach and count how many seashells there are then have fun coloring in the shells, sand, and surf!
### Practise Calculating Percent
Does your fifth grader need practise calculating percent? In this worksheet, she'll practise the subject in a situation she likely knows and loves: shopping!
### Find the Pairs: Beach Fun
This beach scene is bursting with pairs! Teach your child how to recognise pairs, and learn a little more about counting and numbers with this worksheet.
### Simple Arithmetic
Have a ball with this addition and subtraction worksheet! Kids fill in the missing numbers on each beach ball to make its statement true.
### Sort and Count at the Beach
All these beach toys need to be organized! Is your child up to the task? She'll get to colour while she counts and sorts the different items in the picture.
### Beach Multiplication
Beach fun and multiplication? It's an unlikely pairing, but this word problem worksheet shows how handy multiplication can be in day-to-day activities.
### Multiplication & Division: Beach maths
Kids multiply or divide to complete each equation on this year four maths worksheet.
### Tracing Numbers & Counting: 10
On this year one maths worksheet, kids trace the number 10, then write their own. Then they count the shells they see in the picture and make a graph.
### Size Comparison: At the Beach!
Give your child some size comparison practise with this colorful summer worksheet. He'll compare relative sizes and weight between two beachy objects.
### Addition Facts Worksheet: Beach Fun
Create new collection
0
### New Collection>
0Items
What could we do to improve Education.com? | 517 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-34 | latest | en | 0.889943 |
http://developer.nokia.com/community/discussion/showthread.php/56434-How-to-solve-this-math-in-MIDP1-0 | 1,408,642,587,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500820617.70/warc/CC-MAIN-20140820021340-00000-ip-10-180-136-8.ec2.internal.warc.gz | 49,376,911 | 13,316 | # Thread: How to solve this math in MIDP1.0?
1. ## How to solve this math in MIDP1.0?
Hello,
I have a form where a user inputs following fields:
F = int
f = decimal in string format ie. "2.8"
distance = integer
Static variables are:
CoC = (int)(19/1000);
int mm = 1;
How would I go about forming a following formula in MIDP1.0 that doesn't support decimals. The only problem here that I am aware of is the value of f and the outcome of all of teh below formulas, as they pretty sure will outcome to a decimal.
a = ((DD/(1+((DD*CoC*f)/(F*F))))*(mm))/1000;
b = ((DD/(1-((DD*CoC*f)/(F*F))))*(mm))/1000;
d = ((F*F)/(f*CoC)*(mm))/1000;
Is there anyone who could explain to me how to make these kind of formulas work in MIDP1.0 either here or by emailing me to artur@majgaj.com
2. Fixed point math. Google for tutorials on it. There are also a few MIDP libraries that implement it like MathFP. And there's Henson's Float class too.
shmoove
3. Originally posted by shmoove
Fixed point math. Google for tutorials on it. There are also a few MIDP libraries that implement it like MathFP. And there's Henson's Float class too.
shmoove
I did read those sources, but I'm not sure on how to apply the fixed point math classes to the formula I provided... =(
4. Well, MathFP has a few methods called toFP() that can convert your various values into fixed point format. And it also has a bunch of arithmethic and trigonometric (like add(), mul(), div()) that you can use to implement your equations.
shmoove
5. Would anyone by any chance might have the FloatConvW.zip file from Hanson's website as the link is broken and the author hasn't responsed despite many requests.
6. Originally posted by agajewski
Would anyone by any chance might have the FloatConvW.zip file from Hanson's website as the link is broken and the author hasn't responsed despite many requests.
Henson(!) is not Hanson(!) and I didn't see your requests.... | 516 | 1,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2014-35 | latest | en | 0.952948 |
https://www.jiskha.com/display.cgi?id=1206854382 | 1,503,245,017,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106779.68/warc/CC-MAIN-20170820150632-20170820170632-00676.warc.gz | 918,244,022 | 3,818 | # physics
posted by .
A man standing 1.25 m in front of a shaving mirror produces an inverted image 25.0 cm in front of it. How close to the mirror should he stand if he wants to form an upright image of his chin that is twice the actual size of his chin?
• physics -
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I'm really confused on how to do this, Can someone help walk me through?
More Similar Questions | 566 | 2,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-34 | latest | en | 0.910125 |
https://documen.tv/question/the-99-99-digital-camera-mia-purchased-was-on-sale-for-15-off-what-amount-did-mia-get-off-the-pr-19772101-52/ | 1,643,069,109,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00582.warc.gz | 268,935,715 | 15,670 | ## The $99.99 digital camera Mia purchased was on sale for 15% off. What amount did Mia get off the price? please leave an explanation Question The$99.99 digital camera Mia purchased was on sale for 15% off. What amount did Mia get off the price? please leave an explanation
in progress 0
5 months 2021-08-10T01:24:34+00:00 1 Answers 5 views 0
## Answers ( )
1. Answer: Sale Price = $84.99 (answer). This means the cost of the item to you is$84.99. You will pay $84.99 for a item with original price of$99.99 when discounted 15%. In this example, if you buy an item at \$99.99 with 15% discount, you will pay 99.99 – 14.9985 = 84.99 dollars.
Step-by-step explanation: | 200 | 671 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.908619 |
https://www.alloprof.qc.ca/helpzone/discussion/11598/question | 1,716,458,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00283.warc.gz | 562,330,162 | 20,252 | # Help Zone
### Student Question
Secondary I • 2yr.
My teacher tells me that to work on fractions, it is useful to master LCM. How can I find it? I have heard of multiples and trees.
Mathematics
## Explanations (1)
• Explanation from Alloprof
Explanation from Alloprof
This Explanation was submitted by a member of the Alloprof team.
Options
Team Alloprof • 2yr.
Good evening,
The multiples method and the factor tree method are indeed used to find the LCM! There is a third technique, the table, but let's tackle the ones you mentioned!
1. Method of multiples.
To find the LCM, we first make a list of the multiples of each number.
Suppose the numbers 6, 9 and 12.
The multiples of 6 = {6, 12, 18, 24, 30, 36, 42, 48,…}
The multiples of 9 = {9, 18, 27, 36, 45, 54 , 63, 72,…}
The multiples of 12 = {12, 24, 36, 48, 60, 72, 84, 96,…}
We observe that 36 is the lowest common multiple!
2. Factor tree method.
Find the LCM of two numbers:
We decompose the two numbers using a factor tree until we get only prime factors. Common factors and unique factors are identified.
Example 9 and 12.
We get 9 = 3 x 3.
We get 12 = 2 x 2 x 3.
The common factor (in orange) is 3.
The unique factors (in blue) are 3, 2 and 2.
The common factors are multiplied by the single factors.
So 3 x 3 x 2 x 2 = 36 which is the LCM. | 405 | 1,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-22 | latest | en | 0.91941 |
https://community.rstudio.com/t/hierarchical-clustering/136487 | 1,653,296,208,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00422.warc.gz | 227,670,720 | 6,483 | # hierarchical clustering
Hi. I followed Statology's code for an example of hierarchical clustering fom
They conclude with a Ward's method model with optimal number of 4 clusters.
How does one plot the resulting dendrogram for this final model?
library(factoextra)
library(cluster)
df <- USArrests
#remove rows with missing values
df <- na.omit(df)
#scale each variable to have a mean of 0 and sd of 1
df <- scale(df)
m <- c( "average", "single", "complete", "ward")
names(m) <- c( "average", "single", "complete", "ward")
#function to compute agglomerative coefficient
ac <- function(x) {
agnes(df, method = x)\$ac
}
#calculate agglomerative coefficient for each clustering linkage method
sapply(m, ac)
#perform hierarchical clustering using Ward's minimum variance
clust <- agnes(df, method = "ward")
#produce dendrogram
pltree(clust, cex = 0.6, hang = -1, main = "Dendrogram")
#calculate gap statistic for each number of clusters (up to 10 clusters)
gap_stat <- clusGap(df, FUN = hcut, nstart = 25, K.max = 10, B = 50)
#produce plot of clusters vs. gap statistic
fviz_gap_stat(gap_stat)
#compute distance matrix
d <- dist(df, method = "euclidean")
#perform hierarchical clustering using Ward's method
final_clust <- hclust(d, method = "ward.D2" )
#cut the dendrogram into 4 clusters
groups <- cutree(final_clust, k=4)
# Number of members in each cluster
table(groups)
#append cluster labels to original data
final_data <- cbind(USArrests, cluster = groups)
#display first six rows of final data
#find mean values for each cluster
aggregate(final_data, by=list(cluster=final_data\$cluster), mean)
Probably not the ideal method, but a manual approach: use `table(cutree(final_clust, h = xxx))` changing `xxx` to find the number of clusters you decided on. Here, `h = 5` does give you 4 clusters. Then, you can plot it directly with:
``````plot(final_clust, cex = 0.6, hang = -1, main = "Dendrogram")
abline(h = 5, lty = "dashed", col="grey")
``````
Or the nicer-looking
``````library(ggdendro)
ggdendrogram(final_clust) +
geom_hline(aes(yintercept = 5),
linetype = "dashed", color = "grey")
``````
Or, next level, rebuilding everything (but with lots of manual adjustments needed):
``````
ddata <- dendro_data(final_clust)
ggplot() +
geom_segment(data = segment(ddata),
aes(x = x, y = y, xend = xend, yend = yend)) +
geom_text(data = ddata\$labels |>
mutate(group = as.factor(groups[ddata\$labels\$label])),
aes(x = x, y = y, label = label, color = group),
angle = 90, hjust = 1, vjust = 0.5, size = 2.5) +
scale_y_continuous(limits = c(-10,20)) +
theme_dendro() +
theme(axis.text.x = element_text(angle = angle,
hjust = 1, vjust = 0.5)) +
theme(axis.text.y = element_text(angle = angle,
hjust = 1)) +
geom_hline(aes(yintercept = 5),
linetype = "dashed", color = "grey")
``````
Thanks to both of you.
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If you have a query related to it or one of the replies, start a new topic and refer back with a link. | 867 | 3,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.695974 |
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Author Message
BegLed01
Registered: 24.01.2005
From: The upper midwest
Posted: Monday 03rd of Nov 10:35 I have problems with factoring trinomials amazing method. I tried hard to locate somebody who can help me out with this. I also looked out for a teacher to tutor me and solve my problems on midpoint of a line, graphing function and side-side-side similarity. Though I found some who could perhaps crack my problem, I recognized that I cannot pay for them. I do not have a great deal of time too. My assignment is coming up soon . I am desperate . Can someone assist me with this situation? I would really welcome any help or any information.
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Tuesday 04th of Nov 07:23 Algebrator is a good software to solve factoring trinomials amazing method problems . It gives you step by step solutions along with explanations. I however would warn you not to just copy the solutions from the software. It will not help you in understanding the subject. Use it as a guide and solve the questions yourself as well.
Matdhejs
Registered: 08.12.2001
From: The Netherlands
Posted: Wednesday 05th of Nov 16:53 I also have experienced Algebrator is a special piece of factoring trinomials amazing method software . I just remember my ineptness to grasp the constructs of monomials, inequalities or trigonometric functions because I became so good in several disciplines of factoring trinomials amazing method. Algebrator has performed flawlessly for me in Algebra 2, Pre Algebra and Pre Algebra. I strongly advocate this particular program because I could not find even one insufficiency from Algebrator.
Dnunky
Registered: 04.12.2003
From: Kegfarms
Posted: Thursday 06th of Nov 07:35 That sounds amazing ! Thanks for the suggestion ! It seems to be just what I need , I will try it for sure! Where did you come across Algebrator? Any idea where could I find more information about it? Thanks! | 573 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.925261 |
http://mathhelpforum.com/algebra/136994-determine-values-k-function-has-two-zeros-print.html | 1,495,861,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608773.42/warc/CC-MAIN-20170527040600-20170527060600-00245.warc.gz | 271,721,166 | 3,556 | # Determine values of k for which function has two zeros.
• Apr 2nd 2010, 08:48 AM
Determine values of k for which function has two zeros.
Determine the values of k for which the function $f(x) = 4x^2 - 3x + 2kx +1$ has two zeros. Check the values in the original equation.
$a = 4, b= (-3 + 2k) , c = 1$
$b^2 - 4ac > 0$
$(-3 + 2k)^2 - 4(4)(1) > 0$
$9 + 4k^2 - 16 > 0$
$4k^2 > -9 + 16$
$\frac{4k^2 > 7}{4}$
$k^2 > \frac{7}{4}$
$\sqrt{k^2} > \sqrt{\frac{7}{4}}$
$k > \pm \frac{\sqrt{7}}{2}$
The textbook answer to this question is " $x < -0.5$ or $x > 3.5$". Where did I go wrong in solving this? Was it the part where I let $b = (-3 + 2k)$? I wasn't sure about that as there were two values of x in the standard form.
• Apr 2nd 2010, 09:04 AM
alexmahone
$a=4, b=(-3 + 2k), c=1$
$b^2-4ac>0$
$(-3+2k)^2-4(4)(1)>0$
$9 + 4k^2-12k-16 > 0$
$4k^2-12k-7>0$
$4k^2+2k-14k-7>0$
$2k(2k+1)-7(2k+1)>0$
$(2k+1)(2k-7)>0$
$x<-0.5$ or $x>3.5$
• Apr 2nd 2010, 09:18 AM
Tyvm for the reply. Didn't even realize that I squared the bracketed numbers wrong. Just one question though. How did you get $(+2k - 14k)$ from $-12k$?
• Apr 2nd 2010, 10:29 AM
jayAndy
Quote:
Tyvm for the reply. Didn't even realize that I squared the bracketed numbers wrong. Just one question though. How did you get $(+2k - 14k)$ from $-12k$?
since -12k = 2k - 14k, it's like thinking backwards, he was able to plug it into the formula
he used this method to get an "extra" k to be able to factor the equation easier.
• Apr 2nd 2010, 10:51 PM
CaptainBlack
All the posts in this thread present just a jumble of algebra, you will lose marks doing your work like that. Try explaining (in word, you remember them?) what you are doing at each stage. That way even if you make a mistake in the algebra and/or arithmetic you will still be credited with most of the available marks.
CB | 704 | 1,844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 27, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-22 | longest | en | 0.883903 |
https://meangreenmath.com/2014/07/03/calculators-and-complex-numbers-part-15-2/ | 1,656,472,324,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00010.warc.gz | 452,111,747 | 30,153 | # Calculators and complex numbers (Part 15)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that, at long last, I will explain in today’s post.
Definition. If $z$ is a complex number, then we define
$e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}$
This of course matches the Taylor expansion of $e^x$ for real numbers $x$.
Theorem. If $\theta$ is a real number, then $e^{i \theta} = \cos \theta + i \sin \theta$.
$e^{i \theta} = \displaystyle \sum_{n=0}^{\infty} \frac{(i \theta)^n}{n!}$
$= \displaystyle 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \dots$
$= \displaystyle \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} \dots \right) + i \left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} \right)$
$= \cos \theta + i \sin \theta$,
using the Taylor expansions for cosine and sine.
This theorem explains one of the calculator’s results:
$e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$.
That said, you can imagine that finding something like $e^{4-2i}$ would be next to impossible by directly plugging into the series and trying to simply the answer. The good news is that there’s an easy way to compute $e^z$ for complex numbers $z$, which we develop in the next few posts.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
## 4 thoughts on “Calculators and complex numbers (Part 15)”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 697 | 2,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-27 | latest | en | 0.699363 |
https://www.mathcounterexamples.net/category/set-theory/page/2/ | 1,713,883,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00389.warc.gz | 783,043,103 | 14,039 | # Playing with images and inverse images
We’re looking here to relational expressions involving image and inverse image.
We consider a function $$f : X \to Y$$ from the set $$X$$ to the set $$Y$$. If $$x$$ is a member of $$X$$, $$f(x)$$ is the image of $$x$$ under $$f$$.
The image of a subset $$A \subset X$$ under $$f$$ is the subset (of $$Y$$) $f(A)\stackrel{def}{=} \{f(x) : x \in A\}.$
The inverse image of a subset $$B \subset Y$$ is the subset of $$A$$ $f^{-1}(B)\stackrel{def}{=} \{x \in X : f(x) \in B\}.$ Important to understand is that here, $$f^{-1}$$ is not the inverse function of $$f$$.
We now look at relational expressions involving the image and the inverse image under $$f$$.
### Inverse images with unions and intersections
Following relations hold:
$\begin{array}{c} f^{-1}(B_1 \cup B_2) = f^{-1}(B_1) \cup f^{-1}(B_2)\\ f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2) \end{array}$ Let’s prove the first equality as an example. We have $$x \in f^{-1}(B_1 \cup B_2)$$ if and only if $$f(x) \in B_1 \cup B_2$$ if and only if $$f(x) \in B_1$$ or $$f(x) \in B_2$$ which means exactly $$x \in f^{-1}(B_1) \cup f^{-1}(B_2)$$. Continue reading Playing with images and inverse images | 411 | 1,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.681282 |
https://programmingpraxis.com/2009/02/19/sieve-of-eratosthenes/?like=1&_wpnonce=274c59a080 | 1,527,197,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866870.92/warc/CC-MAIN-20180524205512-20180524225512-00222.warc.gz | 624,350,935 | 32,399 | ## Sieve of Eratosthenes
### February 19, 2009
Over two millenia ago, Eratosthenes, who calculated the circumference of the earth, the distance to the Sun and the tilt of the Earth’s axis, developed a system of latitude and longitude, and invented the leap day, created a systematic method to enumerate the prime numbers that is still in use today. Eratosthenes was born in Cyrene (present-day Libya), lived from 276 B.C. to 194 B.C., and spent most of his life in Alexandria, Egypt, where he was the second Chief Librarian of the Great Library, succeeding Apollonius of Rhodes; he was a good friend of Archimedes.
The Sieve of Eratosthenes starts by making a list of all the numbers up to a desired maximum; we’ll illustrate the method by calculating the prime numbers through thirty:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Now take the first number on the list, 2, and cross off every second number:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
(Although it may not be obvious, the number 4 is crossed off the list; in some fonts, the cross-bar of the 4 coincides with the strike-through bar.) Next, take the next number on the list that isn’t crossed off, 3, and cross off every third number; some of them have already been crossed off:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Repeat that last step for the next un-crossed number on the list, 5:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
And so on, each time crossing off all multiples of the next un-crossed number on the list. The list of prime numbers are all those that haven’t been crossed off:
2 3 5 7 11 13 17 19 23 29
This method is called a sieve because it sweeps through a range of numbers, with each prime number, as it is discovered, blocking all its multiples from falling through as prime numbers. The sieve admits several optimizations. First, only odd numbers are considered, since the initial sifting crosses off all the even numbers except 2, which is handled separately. Second, crossing off starts at the square of the number being sifted, since all smaller primes have already been crossed off by previous steps of the sieve; for instance, sifting by 3 starts at 9, since 6 was already crossed off when sifting by 2. Third, sifting stops at the square root of the maximum number in the sieve, since any non-primes larger than the square root must have already been crossed off at previous levels of the sieve; thus, in the above example there is no need to sieve on the prime number 7, or any larger prime number, since the square of 7 is greater than 30, which is the largest number in the list.
Write a function that takes a single argument n and returns a list of prime numbers less than or equal to n using the optimized sieving algorithm described above. Apply the function to the argument 15485863 and count the number of primes returned.
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Pages: 1 2
### 60 Responses to “Sieve of Eratosthenes”
1. Dr Frankenstein said
2. […] this is the second Programming Praxis exercise I’ve done, and this one was slightly more difficult to get right, or at least test. […]
3. Stuart said
I’m confused by one aspect of the above solution. I don’t understand why startj is defined as (+ (* 2 i i) (* 6 i) 3).
I understand why p is (+ i i 3) – that’s the sequence of odd numbers, starting at 3. But what relationship does that have with startj?
Can anyone enlighten me?
4. programmingpraxis said
That’s the implementation of the second optimization.
Suppose you have already sieved 2, 3, and 5 and are now beginning to sieve 7. Start by adding 7 to the list of primes. Then 14, but that has already been sifted out by 2; likewise, 21 has been sifted out by 3, 28 has been sifted out by 2, 35 has been sifted out by 5, and 42 has been sifted out by 2 (and also 3, but that doesn’t matter). The first multiple of 7 that gets sifted out is 49, which is 7 times 7. In general, when sifting by n, sifting starts at n-squared because all the previous multiples of n have already been sieved.
The rest of the expression has to do with the cross-reference between numbers and sieve indexes. There’s a 2 in the expression because we eliminated all the even numbers before we ever started. There’s a 3 in the expression because Scheme vectors are zero-based, and the numbers 0, 1 and 2 aren’t part of the sieve. I think the 6 is actually a combination of the 2 and the 3, but it’s been a while since I looked at the code, so I’ll leave it to you to figure out.
Good question!
5. programmingpraxis said
The number 2 is handled specially, so the vector v looks this:
```index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 number 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49```
When you want to begin sieving 7, the first number you sift out is 7 × 7 = 49. The number 7 is at index i = 2 in the vector, and the number 49 is at index startj = (+ (* 2 i i) (* 6 i) 3) = 23 in the vector.
6. Stuart said
So, it seems we can alternatively define startj as:
`(quotient (- (expt p 2) 3) 2)`
That is,
– p-squared, to give us the number to start sifting at
– shifted by 3 because we ignore the first 3 numbers (0, 1 and 2)
– divided by 2 to get the index, because we ignore every second number.
Is that correct?
7. Jebb said
The straightforward implementation in C, storing all the odd numbers in an array and then zeroing all non-primes. Not being much of a programmer, I’ll admit I’m amazed my laptop can compute 1,000,000 primes in under half a second… And I suppose the code could even be made to run twice as fast on my c2duo by having two threads sifting through the array.
```#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define index(A) (A - 3) / 2 //index of A in the array of odd numbers
#define vector(A) 2 * A + 3 //inverse operation of index()
long *fill_num(long n); //Generates array of odd numbers > 2
int seek_primes(long *numbers, long n); //Zeroes all non-primes
int main()
{
long ulimit;
long *nombres;
int premiers;
char c;
printf("Upper limit for the calculation of primes?\n");
scanf("%ld", &ulimit);
ulimit = index(ulimit) + 1; //+ 1: the array is indexed from 0
nombres = fill_num(ulimit);
premiers = seek_primes(nombres, ulimit) + 1; //2 is also prime
printf("Found %d primes\n", premiers);
free(nombres);
return 0;
}
int seek_primes(long *numbers, long n)
{
long i, j, primes = 0, max;
//Third optimisation: sift until sqrt of upper limit
printf("sift until %ld\n", max = (long)sqrt(vector(n)));
for (i = 0; i < n; i++)
if (numbers[i] != 0) {
primes++;
if (i < max) {
j = numbers[i] * numbers[i];
while (j < vector(n)) {
numbers[index(j)] = 0;
j = j + 2 * numbers[i];
}
}
}
return primes;
}
long *fill_num(long n)
//n is the size of the array, vector(n - 1) the highest odd number
{
long i;
long *tmp;
tmp = (long *)malloc(n * sizeof(long));
for (i = 0; i < n; i++)
tmp[i] = vector(i);
return tmp;
}
```
8. Jebb said
…turns out the multi-threaded version does not work, at least the way I’ve implemented it. I’ve effectively split the seek_primes in two threads, one sifting for (i = 0; i < n; i+=2), and the other one for (i = 1; i < n; i+=2).
It appears my first thread finds non-zero elements in the numbers[] array in seek_primes, but there is no guarantee these same elements are not going to be zeroed by the second thread later.
In fact up to 100 I find 3 primes too many, and up to 1000 about 38 (and of course the actual number found might even be different from one run to the next).
9. Robert Berman said
My python solution is at http://pastebin.com/Qup0TuFu.
I generated all primes from the integer list of 0-100000. Since the last prime in that list is 99991, using the square of the primes method, it is relatively easy and fast deriving the prime sequence of most numbers. I think I began having problems around the number of primes for the integer 9990000000.
10. neveu said
Here’s my Clojure version:
(require ‘[clojure.contrib.math :as math])
;; faster for dropping a few leading nils than (into [] (drop-while nil? sieve))
(defn drop-while-nil? [sieve] (if (first sieve) sieve (recur (subvec sieve 1))))
;; returns sieve with all multiples of (first sieve) set to nil,
;; leading nils removed
(defn remove-every-nth [sieve]
(let [count-sieve (count sieve), n (first sieve)]
(loop [s sieve, i 1, ni n]
(if (> ni count-sieve)
(drop-while-nil? (subvec s 1))
(recur (assoc s ni nil), (inc i), (* n (inc i)))))))
(defn eratosthenes [r]
(loop [s (into [] (range 2 r)), primes [1] ]
(if (> (peek primes) (math/sqrt r))
(concat primes (remove nil? s))
(recur (remove-every-nth s), (conj primes (first s))))))
(map #(count (time (eratosthenes %))) [10 100 1000 10000 100000 ])
11. Jim Wise said
Simple and simple tail-recursive solutions, but missing the “start eliminating at (* p p)” optimization:
```#lang racket
;; .. : num num -> list-of-nums
;; given first and last, return a list of the range first .. last, inclusive
(define (.. first last)
(if (= first last)
(list first)
(cons first (.. (+ first 1) last))))
(define (erat l n)
(let ([p (car l)])
(if (> (sqr p) n)
l
(cons p (erat (filter (lambda (n) (not (= (modulo n p) 0))) (cdr l)) n)))))
(define (tail-erat l n accum)
(let ([p (car l)])
(if (> (sqr p) n)
(append (reverse accum) l)
(tail-erat (filter (lambda (n) (not (= (modulo n p) 0))) (cdr l)) n (cons p accum)))))
; main program
(let* ([args (vector->list (current-command-line-arguments))]
[n (string->number (car args))]
[l (.. 2 n)])
; (display (erat l n))
(display (tail-erat l n '()))
(newline))
```
12. programmingpraxis said
Did you calculate the number of primes less than 15485863? I think it will take a while.
13. Jim Wise said
Yes, it takes about fifty times longer than the vector-based solution above, mostly in GC.
Using Petite Chez Scheme for profiling:
(time (display (length (…))))
6 collections
3021 ms elapsed cpu time, including 248 ms collecting
3022 ms elapsed real time, including 247 ms collecting
110325408 bytes allocated, including 40188320 bytes reclaimed
(time (length (tail-erat (…) …)))
2767 collections
167705 ms elapsed cpu time, including 103201 ms collecting
167817 ms elapsed real time, including 103268 ms collecting
23602922896 bytes allocated, including 24373515936 bytes reclaimed
14. Jim Wise said
Moving to an in-place filtering of the list (see below) brought this time down by a factor of ten, though it’s still slower than the vector version:
(time (length (tail-erat (…) …)))
105 collections
11498 ms elapsed cpu time, including 5577 ms collecting
11503 ms elapsed real time, including 5590 ms collecting
1135547536 bytes allocated, including 3928896 bytes reclaimed
This was achieved by replacing filter with the following in-place filter! routine (and staying in Chez Scheme, where set-cdr! is available).
A more general filter!, implemented as syntax! so it could consistently alter the list in place but not depend on returning a list would be interesting, but isn’t needed for this — this implementation returns the result of modifying the list in place, but the original value passed in is only modified from the first value not matching pred? on…
```(define (filter! pred? l)
(cond [(null? l) '()]
[(pred? (car l)) (filter! pred? (cdr l))]
[else
(let loop ([follower l]
[leader (cdr l)])
(cond
[(null? leader) l]
[(pred? (car leader))
(set-cdr! follower (cdr leader))
(loop follower (cdr leader))]
[else (loop leader (cdr leader))]))]))
```
15. programmingpraxis said
You’re missing the point. The Sieve is fast because it is based on addition. You are using division (the modulo function). Your algorithm is not the Sieve of Eratosthenes.
16. Jim Wise said
Oh, I see… let me revisit…
17. Jim Wise said
Naive attempt at moving to vector and using addition over vector indices gives a slight speedup, looking further:
(time (length (primes 15485863)))
4 collections
10296 ms elapsed cpu time, including 366 ms collecting
10300 ms elapsed real time, including 366 ms collecting
155888768 bytes allocated, including 124098560 bytes reclaimed
```(define (primes n)
(define (zeroing-pass! vec start step)
(cond
[(> start n) vec]
[(not (vector-ref vec (sub1 step))) vec]
[else
(vector-set! vec (sub1 start) #f)
(zeroing-pass! vec (+ start step) step)]))
(define (erat! vec)
(let* ([stop (div n 2)])
(let loop ([p 2])
(if (= p stop)
vec
(begin
(zeroing-pass! vec (+ p p) p)
(loop (+ p 1)))))))
(let ([vec (make-vector n #t)])
(erat! vec)
(reverse
(let loop ([i 1]
[l '()])
(if (= i n)
l
(loop (add1 i)
(if (vector-ref vec i)
(cons (add1 i) l)
l)))))))
```
18. Jim Wise said
I got curious as to how big a difference the additon-vs-module distinction is on modern hardware (historically, it has been very large, as you point out).
I used the following code to run 10^8 integer additions, modulo operations, and modulo operations with a power-of-two modulus (another traditional distinction in performance).
I may well be missing something big, but the output shows about a 25% difference in speed between additions and modulo operations. This obviously adds up, but looking at the above code, it seems to be small compared to other differences (particularly, the various operations which control how many of whichever operation are carried out, and which reduce the infrastructural cost of the algorithm).
(In the code below, I use three vectors in an attempt to minimize the effect of cache read-ahead on subsequent vector-maps during earlier testing with a smaller n.)
```(let* ([n (expt 10 6)]
[times 100]
[time-reps (lambda (f v)
(time (do ([i 0 (+ i 1)])
([= i times])
(vector-map f v))))]
[v1 (make-vector n 1203006)]
[v2 (make-vector n 1203006)]
[v3 (make-vector n 1203006)])
(display "Timing addition\n")
(time-reps (lambda (n) (+ n 57365)) v1)
(newline)
(display "Timing modulo (power of 2)\n")
(time-reps (lambda (n) (mod n 65536)) v2)
(newline)
(display "Timing modulo (non-power of 2)\n")
(time-reps (lambda (n) (mod n 57365)) v3)
(newline))
```
```Timing addition
(time (do ((...)) ...))
299 collections
6471 ms elapsed cpu time, including 2642 ms collecting
6474 ms elapsed real time, including 2641 ms collecting
2833256224 bytes allocated, including 2762557728 bytes reclaimed
Timing modulo (power of 2)
(time (do ((...)) ...))
299 collections
8536 ms elapsed cpu time, including 2586 ms collecting
8545 ms elapsed real time, including 2597 ms collecting
2833185520 bytes allocated, including 2838706016 bytes reclaimed
Timing modulo (non-power of 2)
(time (do ((...)) ...))
299 collections
8544 ms elapsed cpu time, including 2601 ms collecting
8548 ms elapsed real time, including 2596 ms collecting
2833185520 bytes allocated, including 2821785344 bytes reclaimed
```
19. Grégory LEOCADIE said
my OCaml version. Because 15000000 is bigger than the bound for basic array, i had to use Bigarray.
```open Bigarray
open Sys
open Format
let timeit f =
let tb = time () in
let res = Lazy.force_val f in
printf "time: %f s\n" ((time()) -. tb);
res
let ba_fold_left f init arr =
let adim = Array1.dim arr in
let rec ba_fold_left' i res =
match i >= 0 with
| true -> ba_fold_left' (i - 1) (f res i (Array1.get arr i))
| _ -> res
in
ba_fold_left' (adim - 1) init
;;
let sieve n =
let adim = (n - 1)/ 2 in
let arr = Array1.create int8_unsigned c_layout adim in
Array1.fill arr 0;
let rec erase i mo =
if i < adim
then
begin
if (2 * i + 3) mod mo = 0
then Array1.set arr i 1;
erase (i + mo) mo
end
in
let rec sieve' i =
let e = 2 * i + 3 in
match e*e with
| p when p > n -> ()
| p when (Array1.get arr i) = 0
-> erase (p / 2 - 1) e; sieve' (i + 1)
| _ -> sieve' (i + 1)
in
sieve' 0;
2 :: (ba_fold_left (fun x k v ->
if v = 0 then (k * 2 + 3) :: x
else x) [] arr)
;;
let _ =
Printf.printf " %d "
(List.length (timeit (Lazy.lazy_from_fun (fun () -> (sieve 15485863)))))
```
20. […] ancient Sieve of Eratosthenes that computes the list of prime numbers is inefficient in the sense that some composite numbers are […]
21. David said
This was my first Factor program I wrote a few months ago, to learn the language. Updated it with the optimization to start sieving at the square of the prime being sieved. (Didn’t know about that trick before :)
```USING: kernel math sequences math.ranges bit-sets locals formatting io ;
FROM: sets => adjoin delete in? members ;
IN: sieve
CONSTANT: items/line 10
: .row ( seq -- )
[ "%5d " printf ] each nl ;
: .vec ( seq -- )
[ dup length items/line > ]
[ items/line cut-slice swap .row ]
while
.row ;
: init-sieve ( max-prime -- bitset )
dup <bit-set>
2 rot [a,b) [ over adjoin ] each ;
: filter-composite ( bitset start end -- )
[ dup sq ] dip rot <range>
[ over delete ] each drop ;
:: primes ( maxprime -- vec )
[let
maxprime 1 + init-sieve :> sieve
2
[ dup sq maxprime > ]
[ dup sieve in?
[ dup maxprime sieve -rot filter-composite ]
when
1 +
]
until drop
sieve members
] ;
: .primes ( maxprime -- )
primes .vec ;
```
Session:
```( scratchpad ) 15485863 primes length .
1000000
( scratchpad ) 127 .primes
2 3 5 7 11 13 17 19 23 29
31 37 41 43 47 53 59 61 67 71
73 79 83 89 97 101 103 107 109 113
127
( scratchpad )
```
22. […] As mentioned in my last article, I started doing some challenges from the Programming Praxis website. And here comes my PHP solution to the second challenge. […]
23. Graham said
I’m working my way through old exercises; here’s my submission.
24. Graham said
Python was a bit slow for me, and I’m trying to learn Common Lisp, so here’s
another version. I was only able to figure out how to get 2/3 of the optimizations
working, but it’s still absurdly fast compared to my last submission.
25. ftt said
```#!/usr/bin/env python
import itertools
import sys
# note: the 'modulo' version with the same cut-offs uses much less memory in Python
# but is also way slower
def sieve(n):
# optimization 1: only odd numbers
primes = range(3, n + 1, 2)
upper_bound = n ** 0.5
for base in xrange(len(primes)):
if not primes[base]:
continue
# optimization 3: stop at the square root of n
if primes[base] >= upper_bound:
break
# optimization 2: start with the square
for i in xrange(base + (base + 1) * primes[base], len(primes), primes[base]):
primes[i] = None
primes.insert(0, 2)
return filter(None, primes)
def main():
print 'Enter an upper bound for the sieve.'
n = int(sys.stdin.readline())
result = sieve(n)
print 'Number of primes: {0}'.format(len(result))
if __name__ == '__main__':
main()
```
26. Rens said
I just started learning ruby. Here is my first try.
def prime_numbers(n)
return [] if n < 2 # no prime numbers before 2
max = (n**0.5).to_i; # define until when we need to loop
primes = []
# optimization 1: checking only odd numbers
primes = 3.step( n, 2 ).to_a
primes.each do |p|
next unless p
# optimization 3: checking until the root of the maximum number
break unless p <= max
# optimization 2: checking starts at the square
((p**2)/2-1).step( primes.length-1, p) do |i|
primes[i] = nil
end
end
primes.insert(0, 2).delete(nil) # add the prime number 2 and remove false numbers
primes
end
puts prime_numbers(15485863).length
[\sourcecode]
27. kawas said
Another clojure version,
because the first one die on me with OutOfMemoryError to compute (eratosthenes 15485863)
```; we work only with odd numbers till N
(defn odds [n] (vec (range 3 (inc n) 2)))
; it's easy to calculate index of a value in our vector of odds
(defn odds-idx [v] (/ (- v 3) 2))
; just sieve the vector from start=(index of v²)
(defn sieve [max-idx nums v]
(let [start (odds-idx (* v v))]
(reduce #(assoc %1 %2 nil) nums (range start max-idx v))))
; get primes equal or less than N
(defn primes [n]
(let [max-i (Math/floor (odds-idx n))]
(loop [nums (odds n) i 0]
(if-let [v (nums i)]
(if (> (* v v) n)
(cons 2 (remove nil? nums))
(recur (sieve max-i nums v) (inc i)))
(recur nums (inc i))))))
(count (primes 15485863))
1000000
```
28. An imperative OCaml Batteries `BitSet`-based version, without multiplications, only additions, subtractions and shifts:
```let sieve_of_eratosthenes (proc : int -> unit) n =
let limit = (n - 3) asr 1 in
let sieve = BitSet.create_full (limit + 1) in
let i = ref 0
and p = ref 3
and q = ref 9 in
while !q < n do
if BitSet.is_set sieve !i then begin
let j = ref ((!q - 3) asr 1) in
while !j <= limit do
BitSet.unset sieve !j;
j := !j + !p
done
end;
incr i;
p := !p + 2;
q := !q + (!i + 1) lsl 3
done;
proc 2;
foreach (BitSet.enum sieve) (fun i -> if i <= limit then proc (i lsl 1 + 3))
```
29. CyberSpace17 said
My C++ solution where a “crossing out” means “made 0”
http://ideone.com/fdmux
30. moink said
Here’s a Matlab solution:
```function primes=sieve(maxnum)
%Sieve of Erasthenes for finding primes less than or equal to the input
%argument
mask=ones(ceil((maxnum-1)/2),1);
for i=3:2:floor(sqrt(maxnum))
if mask((i-1)/2)
mask((i^2-1)/2:i:end)=0;
end
end
primes=[2;2*find(mask)+1];
```
It’s a tiny bit faster than the nearly-identical Matlab implementation, called primes.
31. A rather simple version of a Python solution.
``` def primes_sieve(limit): limit += 1 not_prime = [False] * limit primes = []```
``` for i in xrange(2, limit): if not_prime[i]: continue for j in xrange(i*2, limit, i): not_prime[j] = True primes.append(i) return primes ```
```if __name__ == "__main__": print len(primes_sieve(15485863)) ```
32. Matthias Altmann said
I tried another solution in C++ using 2 threads, one running upwards on the numbers, the other running downwards. They stop if the reach another.
The solution is correct, but the performance decreases severly.
Here is my code:
#include
#include
#include
#include
#include
using namespace std;
const unsigned long MAX_NUMBER = 15485863;
vector numbers(MAX_NUMBER+1);
pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
unsigned long l1,l2;
// thread specific arguments
struct thread_data
{
unsigned int thread_id;
unsigned long n; // data to sieve
bool bw; // backward or forward search
};
struct thread_data thread_data_array[2];
void init_numbers(vector * numbers);
void *prime_sieve(void *arg);
void init_numbers(vector * numbers)
{
unsigned long i;
(*numbers)[2]=true;
for (i=3;ibw))
{
while (l1 <= l2)
{
if (numbers[l1] != false)
{
c = l1*l1;
pos_move = (l1 << 1);
while (c n)
{
pthread_mutex_lock(&mutex1);
numbers[c]=false;
pthread_mutex_unlock(&mutex1);
c += pos_move;
}
}
pthread_mutex_lock(&mutex1);
l1++;
pthread_mutex_unlock(&mutex1);
}
}
else
{
while(l2 >= l1)
{
if (numbers[l2] != false)
{
c = l2*l2;
pos_move = (l2 << 1);
while (c n)
{
pthread_mutex_lock(&mutex1);
numbers[c]=false;
pthread_mutex_unlock(&mutex1);
c += pos_move;
}
}
pthread_mutex_lock(&mutex1);
l2–;
pthread_mutex_unlock(&mutex1);
}
}
pthread_exit(NULL);
}
unsigned long get_primes(vector numbers, unsigned long n)
{
unsigned long primes=0;
for(unsigned int i=0;i<=n;i++)
{
primes+=numbers[i];
}
return primes;
}
int main(int argc, char *argv[])
{
pthread_t threads[2];
int rc;
unsigned long n = MAX_NUMBER;
init_numbers(&numbers);
l1 = 2;
l2 = (int)(sqrt(n))+1;
thread_data_array[0].thread_id = 0;
thread_data_array[0].n = MAX_NUMBER;
thread_data_array[0].bw = false;
rc = pthread_create(&threads[0], NULL, prime_sieve, (void *) &thread_data_array[0]);
thread_data_array[1].thread_id = 1;
thread_data_array[1].n = MAX_NUMBER;
thread_data_array[1].bw = true;
rc = pthread_create(&threads[1], NULL, prime_sieve, (void *) &thread_data_array[1]);
void* retval;
pthread_join(threads[0], &retval);
pthread_join(threads[1], &retval);
cout << get_primes(numbers, n) << endl;
}
33. Ikenna said
Using python:
from math import *
def sieve2(n):
index = 0
numberList = range(3,n+1,2)
while numberList[index] = sqrt(n):
break
return [2] + numberList
34. Ikenna said
^
|
|
Didn’t post correctly
35. j0sejuan said
```package main
import "fmt"
func PrimesLessEqual(n int) [] int {
c, p, w, l := 0, 2, make([] bool, n + 1), make([] int, n)
for ; p <= n; p++ {
for p <= n && w[p] { p++ }
if p <= n {
l[c] = p; c++
for k := p; k <= n; k += p { w[k] = true }
}
}
return l[0:c]
}
func main() {
fmt.Println(len(PrimesLessEqual(15485863)))
}
```
36. dchild said
compact ruby implementation:
```list = (3..ARGV[0].to_i).step(2).to_a
(0..(Math.sqrt(list.last)/2)-1).each do |i|
(((list[i]**2)/2) - 1..list.size).step(list[i]) { |j| list[j] = 0 } if (list[i] > 0)
end
puts ([2] + list.select {|x| x > 0}).size
```
37. A Scala solution:
```import scala.math._
import scala.collection.mutable._
object Main extends Application {
val limit = 15485863 //20
val limitSquareRoot = pow(limit,0.5) toInt
def sieve(factor: Int, xs: Array[Int]): Array[Int] = {
for (i <- (factor * factor) until (limit + 1) by factor) {
xs.update((i-2), -1)
}
xs
}
def sieveOfEtratosthenes: Array[Int] = {
def numbersArray(start: Int, end: Int): Array[Int] = start until end toArray
val xs = numbersArray(2, limit + 1)
for (i x != -1}
}
println(sieveOfEtratosthenes.length)// foreach println)
}
Main
```
38. Previous post did not copy properly:
```import scala.math._
import scala.collection.mutable._
object Main extends Application {
val limit = 15485863 //20
val limitSquareRoot = pow(limit,0.5) toInt
def sieve(factor: Int, xs: Array[Int]): Array[Int] = {
for (i <- (factor * factor) until (limit + 1) by factor) {
xs.update((i-2), -1)
}
xs
}
def sieveOfEtratosthenes: Array[Int] = {
def numbersArray(start: Int, end: Int): Array[Int] = start until end toArray
val xs = numbersArray(2, limit + 1)
for (i x != -1}
}
println(sieveOfEtratosthenes.length)// foreach println)
}
Main
```
39. Ramapriya said
#include
#include
#include
#include
#include
unsigned LIMIT 1000000000;
int main()
{
int all[LIMIT],i,j=2,k=0,lim;
float start,stop;
printf(“Enter the limit(excluding 0)”);
scanf(“%d”,&lim);
for(i=0;i<lim;i++)
{
all[i]=i;
}
all[0]=0;
all[1]=0;
stop=(unsigned)sqrt(lim);
while(j<stop)
{
start=0;
if(all[j]!=0)
{
start=all[j]*all[j];
if(all[start]%all[j]==0)
{
all[start]=0;
start++;
}
}
j++;
}
for(i=2;i<lim-2;i++)
{
if(all[i]!=0)
{
printf(" %d",all[i]);
}
}
getch();
return 0;
}
40. Ramapriya said
Corrected solution :) very simple works like a charm
#include “stdafx.h”
#include
#include
#include
#include
#include
int main()
{
int all[100],i,j=2,k=0,lim,start;
float stop;
printf(“Enter the limit(excluding 0)”);
scanf(“%d”,&lim);
lim=lim+1;
for(i=0;i<=lim;i++)
{
all[i]=i;
}
all[0]=0;
all[1]=0;
stop=sqrt((float)lim);
while(j<stop)
{
start=0;
if(all[j]!=0)
{
start=all[j]*all[j];
while(start<=lim)
{
all[start]=0;
start=start+j;
}
}
j++;
}
for(i=2;i<=lim-2;i++)
{
if(all[i]!=0)
{
printf(" %d",all[i]);
}
}
getch();
return 0;
}
41. Ramapriya said
now if any of you guys take a look here how can you make this work for a billionth using the constant did not give me result,are linked list only solution???
42. Evgeni said
C# solution:
public class Primer
{
public int[] GetPrimes(int max)
{
if (max < 0) max = max * -1;
if (max 0)
{
for (int i = start * start; i < primeIndexes.Length; i += start)
{
primeIndexes[i] = true;
}
start++;
}
List primes = new List(max / 10);
for (int i = 1; i primeIndexes.Length)
return 0;
for (int i = minIndex; i < primeIndexes.Length; i++)
{
if (primeIndexes[i] == false)
return i;
}
return 0;
}
}
Just over a second on Core2 Duo.
43. Evgeni said
Hm, it didn’t format properly, here’s a gist:
44. Andri Juanda said
Hi, just learn python and still beginner level, so I don’t know
def sieve(number):
array = [False,False] + [True] * (number – 1)
result = []
prime_found = 0
index = 2
while index <= number:
if array[index]:
prime_found = index
result.append(index)
index *= index
while index <= number:
array[index] = False
index += prime_found
index = prime_found + 1
else:
index += 1
return result
But I remember in java, an array of boolean is really cheap, only 1 bit per element. and if I switch so False is Boolean while True is not or not examined yet, this will probably safe some space and array creation time.
45. Andri Juanda said
Sorry, Just disappointed with the way my code shown here, so I just try this to see how to format my code appropriately:
def just_a_test(‘my apology’):
print ‘Again, really sorry’
46. Colin Williams said
Java solution:
```import java.util.*;
class Sieve {
public static void main(String[] args) {
for (Integer prime : sieveFor(Integer.parseInt(args[0]))) {
System.out.println(prime.toString());
}
}
public static Integer[] sieveFor(int maximum) {
return (new Sieve(maximum)).primes();
}
private int maximum;
public Sieve(int maximum) {
this.maximum = maximum;
}
public Integer[] primes() {
TreeSet<Integer> primes = allNumbers();
Integer prime = primes.first();
while (prime <= largestSievableNumber()) {
prime = primes.higher(prime);
Integer potential_composite = primes.higher(firstUnsievedComposite(prime) - 1);
while (potential_composite != null) {
if (potential_composite % prime == 0) {
primes.remove(potential_composite);
}
potential_composite = primes.higher(potential_composite);
}
}
return primes.toArray(new Integer[0]);
}
private TreeSet<Integer> allNumbers() {
TreeSet<Integer> numbers = new TreeSet<Integer>();
numbers.add(2);
for (int i = 3;i <= this.maximum;i += 2) {
numbers.add(i);
}
return numbers;
}
private Integer largestSievableNumber() {
return new Integer((int) Math.pow(this.maximum, 0.5));
}
private ArrayList<Integer> compositesFor(int prime, int maximum) {
ArrayList<Integer> composites = new ArrayList<Integer>();
for (int i = firstUnsievedComposite(prime);i <= this.maximum;i += prime) {
composites.add(i);
}
return composites;
}
private Integer firstUnsievedComposite(int prime) {
return new Integer((int) Math.pow(prime, 2));
}
}```
jUnit Tests:
```import static org.junit.Assert.*;
import org.junit.Test;
public class SieveTest {
@Test
public void even() {
assertArrayEquals(new Integer[]{2}, sieveFor(2));
}
@Test
public void firstOdd() {
assertArrayEquals(new Integer[]{2, 3}, sieveFor(3));
assertArrayEquals(new Integer[]{2, 3}, sieveFor(4));
}
@Test
public void compositeOdd() {
assertArrayEquals(new Integer[]{2, 3, 5, 7}, sieveFor(9));
}
@Test
public void largeList() {
assertArrayEquals(new Integer[]{2, 3, 5, 7, 11, 13, 17, 19, 23, 29}, sieveFor(30));
}
private Integer[] sieveFor(int maximum) {
return Sieve.sieveFor(maximum);
}
}```
Output on 15485863 took a while, although I didn’t time it. Does anybody have some performance suggestions, my java’s a bit rusty.
47. johnw said
My solution in C using all optimizations. It prints primes as it finds them, and then once it hits the square root mark prints any remaining untouched variables in the array. Chews through N=15485863 in about 3.3 seconds.
48. blutorange said
In lua, not efficient, but short:
function sieve(n,s)
for i=2,math.ceil(math.sqrt(n)) do
for j=2*i,n,i do s[j]=false end
end
return s
end
s=sieve(15485863,{false,false})
for i=1,15485863 do if s[i]==nil then print(i) end end
Well, it “only” takes only 5.5s for 15485863 (and 262MB RAM)…
49. blutorange said
function sieve(n,s)
for i=2,math.ceil(math.sqrt(n)) do
if (i==2) or (i%2==1) then for j=i*i,n,(i%2+1)*i do s[j]=false end end
end
return s
end
s=sieve(15485863,{false,false})
for i=1,15485863 do if s[i]==nil then print(i) end end
(a few improvements, now it only takes 2.3s [and 1.5s for printing…])
50. Chris said
Just stumbled across this page. I have written a multi-threaded nth-prime algorithm in C# (it finds the millionth prime in 7 milliseconds using 16 threads on a 4 core – 8 hyper-threaded core – i7 3770s running at 3.1 GHz while using only 405 KB for the sieve) and was looking for information on how to convert it to racket. I’m new to racket and still trying to wrap my head around the functional approach to the sieve.
The various examples in functional languages should give me what I was looking for. Thanks to everyone for sharing.
If anyone is interested in the multi-threaded algorithm, let me know (I will be notified of follow up comments). Alternatively, as it appears that this page is all about coding challenges, how about a multi-threaded sieve challenge? I also want to convert my algo to c and compile it with clang to see what kind of boost clang’s optimers might provide vs. compiling with gcc (and vs. my visual studio compiled c#).
Note that my algorithm is an nth-prime algorithm – the inverse of what was requested on this page. In other words, given the input 1000000, my algo outputs 15485863.
There are three reasons I want to convert it to racket:
1. To use racket’s native big integers (the C# version begins to return clipped answers somewhere in the neighborhood of the 240 millionth prime).
2. To see what kind of performance penalty racket’s native big integers incurr (a simple tail recursive factorial in wicked fast in racket – faster than any big integer factorial I can find in c#).
3. It’s fun to learn new languages.
51. This is my PHP version that calculates half a million before the terminal crashes, don’t have a local copy of PHP to test with at the minute so using a remote low powered box, had to chunk the number sequences as a result of the low performing box.
52. Mike said
Python / Numpy solution
```import itertools as it
import numpy as np
def sieve(n):
if n > 2:
yield 2
if n > 3:
limit = int(n ** 0.5)
size = (n - 1) // 2
flags = np.ones([size], dtype=bool)
primes = it.compress(it.count(3, 2), flags)
for p in primes:
yield p
if p > limit:
yield from primes
break
flags[(p*p - 3) // 2::p] = False
len(list(sieve(15485863)))
```
53. […] start at p^2 instead of p and the outer loop can stop at the square root of n. I’ll leave the optimized version for you to work […] | 9,914 | 33,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.907057 |
https://my-assignmentexpert.com/2022/02/25/%E7%BB%9F%E8%AE%A1%E4%BB%A3%E5%86%99-matrix-calculations-stat%E4%BB%A3%E5%86%99/ | 1,720,860,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00123.warc.gz | 349,443,540 | 21,205 | 19th Ave New York, NY 95822, USA
统计代写| Matrix calculations stat代写
统计代考
$11.6 \mathrm{R}$
Matrix calculations
Let’s do some calculations for the 4-state Markov chain in Example 11.1.5, as an example of working with transition matrices in R. First, we need to specify the transition matrix $Q$. This is done with the matrix command: we type in the entries
522
of the matrix, row by row, as a long vector, and then we tell $\mathrm{R}$ the number of rows and columns in the matrix (nrow and ncol), as well as the fact that we typed in the entries by row (byrow=TRUE):
$Q<-\operatorname{matrix}(c(1 / 3,1 / 3,1 / 3,0,$,
\begin{aligned} &(1 / 3,1 / 3,1 / 3,0, \ &0,0,1 / 2,1 / 2, \ &0,1,0,0, \ &1 / 2,0,0,1 / 2), \mathrm{nrow}=4, \mathrm{ncol}=4, \text { byrow=TRUE) } \end{aligned}
To obtain higher order transition probabilities, we can multiply $Q$ by itself repeatedly. The matrix multiplication command in $\mathrm{R}$ is $\% * \%$ (not just *). So
$Q 2<-Q \% * \%$
$Q 3<-Q 2 \% * \%$
$Q 4<-Q 2 \% * \%$ Q 2
Q5 <- Q3 \%*\% Q2
produces $Q^{2}$ through $Q^{5}$. If we want to know the probability of going from state 3 to state 4 in exactly 5 steps, we can extract the $(3,4)$ entry of $Q^{5}$ :
Q5 $[3,4]$
This gives $0.229$, agreeing with the value $11 / 48$ shown in Example 11.1.5. can use the command $Q \% \% \mathrm{n}$ after installing and loading the expm package. For example, $Q \%^{\sim} \%$ yields $Q^{42}$. By exploring the behavior of $Q^{n}$ as $n$ grows, we can see Theorem 11.3.6 in action (and get a sense of how long it takes for the chain to get very close to its stationary distribution).
In particular, for $n$ large each row of $Q^{n}$ is approximately $(0.214,0.286,0.214,0.286)$, so this is approximately the stationary distribution. Another way to obtain the stationary distribution numerically is to use
eigen $(t(Q))$
to compute the eigenvalues and eigenvectors of the transpose of $Q$; then the eigenvector corresponding to the eigenvalue 1 can be selected and normalized so that the components sum to $1 .$
Gambler’s ruin
To simulate from the gambler’s ruin chain from Example $11.2 .6$, we start by deciding the total amount of money $\mathbb{N}$, the probability $p$ of gambler $A$ winning a given round, and the number of time periods nsim that we wish to simulate.
$\mathrm{N}<-10$
$\mathrm{p}<-1 / 2$
nsim <- 80
统计代考
$11.6 \mathrm{R}$
522
$Q<-\operatorname{矩阵}(c(1 / 3,1 / 3,1 / 3,0,$,
$$\开始{对齐} &(1 / 3,1 / 3,1 / 3,0, \ &0,0,1 / 2,1 / 2, \ &0,1,0,0, \ &1 / 2,0,0,1 / 2), \mathrm{nrow}=4, \mathrm{ncol}=4, \text { byrow=TRUE) } \end{对齐}$$
$Q 2<-Q \% * \%$
$Q 3<-Q 2 \% * \%$
$Q 4<-Q 2 \% * \%$ Q 2
Q5 <- Q3 \%*\% Q2
$\mathrm{N}<-10$
$\mathrm{p}<-1 / 2$
nsim <- 80 | 974 | 2,707 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-30 | latest | en | 0.773274 |
https://tarungehlots.blogspot.com/2014/08/chapter-23-miscellaneous-examples.html | 1,531,722,593,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00198.warc.gz | 777,173,657 | 40,991 | tgt
## Sunday, 3 August 2014
### CHAPTER 23- Miscellaneous Examples
Example: 1
If $g\left( x \right) = f\left( x \right) + f\left( {1 - x} \right)$ and $f''\left( x \right) < 0\,{\rm{for\, all}}\,x \in \left[ {0\,,1} \right]$, prove that $g(x)$ is increasing in $[0,1/2)$ and decreasing in $(1/2,1]$.
Solution: 1
Our requirement is to somehow show that $g'(x) > 0\,{\rm{for\,all}}\, x \in [0,1/2)$and $g'(x) < 0\,{\rm{for\,all}}\, x \in (1/2,1]$.
From the given functional relation between $f(x)$ and $g(x)$:
$g'\left( x \right) = f'\left( x \right) - f'\left( {1 - x} \right)$
Therefore, we must show that:
$f'(x) > f'(1 - x)\forall x \in [0,1/2)$ $\ldots(i)$
and
$f'\left( x \right) < f'\left( {1 - x} \right)\,\,\,\,\,\,\,\,\,\forall x \in (1/2,\,1]\,\,$ $\ldots(ii)$
Since $f''\left( x \right) < 0\,\,\,\forall x \in [0,\,1],\,\,f'\left( x \right)$ is decreasing on $[0,1]$. This means that if we take any $x$ value in $[0,1/2),(1 - x)$ will be greater than $x$ so that $f'\left( {1 - x} \right)$ will be less than $f'\left( x \right)$. In other words, ($i$) is satisfied by virtue of the fact that $f'(x)$ is decreasing.
On similar lines, when we assume any $x$ value in $(1/2,1]$, we will see that ($ii$) is also satisfied for the same reason (that $f'(x)$ is decreasing).
$\Rightarrow \,\,\,\,g(x)$ satisfies the stated assertion
Example: 2
Let $f\left( x \right) = \dfrac{{\ln \left( {\pi + x} \right)}}{{\ln \left( {e + x} \right)}}$. Prove that $f(x)$ is decreasing on $\left[ {0,\,\infty } \right)$
Solution: 2
$= \dfrac{{\left( {e + x} \right)\ln \left( {e + x} \right) - \left( {\pi + x} \right)\ln \left( {\pi + x} \right)}}{{\left( {e + x} \right)\left( {\pi + x} \right){{\left( {\ln \left( {e + x} \right)} \right)}^2}}}$ $= \dfrac{{\left( {e + x} \right)\ln \left( {e + x} \right) - \left( {\pi + x} \right)\ln \left( {\pi + x} \right)}}{{\left( {e + x} \right)\left( {\pi + x} \right){{\left( {\ln \left( {e + x} \right)} \right)}^2}}}$ $= \dfrac{{g\left( x \right)}}{{h\left( x \right)}} \,\,\,\,\,\,\,{\rm{{This\, substitution\, was\, done\, for\, convenience}}}$
To determine the sign of $f'\left( x \right)$ in $\left[ {0,\,\infty } \right)$, we first note that $h\left( x \right) > 0\,\,\forall \,x \in \left[ {0,\infty } \right)$, so that we need to only worry about the sign of $g(x)$. The form of $g(x)$ suggests that we can construct a new$G\left( x \right) = x\ln x$ function to determine the sign of $g(x)$ as follows:
$G\left( x \right) = x\ln x$
$\Rightarrow \,\,\,\, G'\left( x \right) = 1 + \ln x$
$\Rightarrow \,\,\,\, G'\left( x \right) > 0\,\,\forall x \in \left( {\dfrac{1}{e},\infty } \right)$
and $\,\,\,\,$ $G'\left( x \right) < 0\,\,\forall \,x \in \left( {0,\dfrac{1}{e}} \right)$
$\Rightarrow \,\,\,\, G\left( x \right)\,{\rm{is}}\,{\rm{increasing}}\,{\rm{on}}\,\left( {\dfrac{1}{e},\infty } \right)$
$\Rightarrow \,\,\,\, x\ln x\,\,{\rm{increases}}\,{\rm{on}}\,\left( {\dfrac{1}{e},\infty } \right)$
$\left( {\pi + x} \right)\ln \left( {\pi + x} \right) > \left( {e + x} \right)\ln \left( {e + x} \right)\,\,\,\,\,\,\forall x \in \left[ {0,\infty } \right) \,\,\,\,\,\,\,\,\left\{ \begin{array}{l} \,{\rm{since}}\,\left( {\pi + x} \right) > \left( {e + x} \right)\\ > \dfrac{1}{e}\forall x \in \left[ {0,\infty } \right) \end{array} \right\}$
$\Rightarrow \,\,\,\, g\left( x \right) < 0\;\,\,\,\,\,\,\forall \,x \in \left[ {0,\infty } \right)$
$\Rightarrow \,\,\,\, f'\left( x \right) < 0\;\,\,\,\,\,\forall \,x \in \left[ {0,\infty } \right)\,$
$\Rightarrow \,\,\,\, f\left( x \right)$ is decreasing on $\left[ {0,\infty } \right)$
Example: 3
Let $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - {x^3} + \dfrac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}},}\,\,\,\,{0 \le x < 1}\\ {2x - 3,}\,\,\,\,{1 \le x \le 3} \end{array}} \right\}$ Find all possible real values of $b$ such that $f(x)$ has the smallest value at $x = 1$.
Solution: 3
Notice that $f\left( 1 \right) = - 1$ (from the lower definition $f(x)$)
Also, $f(x)$ is monotonically decreasing on $[0,1)$ and monotonically increasing on $[1,3)$.
Therefore, all we require for $f(x)$ to have its minimum at $x = 1$ is:
$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ge f\left( 1 \right)$ {i.e., the minimum of the left side function must not be less than $f(1)$}
$\Rightarrow \,\,\,\, - 1 + \dfrac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}} \ge - 1$ $\Rightarrow \,\,\,\, \dfrac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}} \ge 0$ $\Rightarrow \,\,\,\, \dfrac{{\left( {b - 1} \right)\left( {{b^2} + 1} \right)}}{{\left( {b + 1} \right)\left( {b + 2} \right)}} \ge 0$
Upon solving, this yields:
$b \in \left( { - 2, - 1} \right) \cup \left[ {1,\infty } \right)$
Example: 4
Using the relation $2\left( {1 - \cos x} \right) < {x^2},x \ne 0$ or otherwise, prove that $\sin \left( {\tan x} \right) \ge x$ for all $x \in \left[ {0,\pi /4} \right]$
Solution: 4
Notice that and ‘$x$‘ have equal values at $x = 0$. If we consider the function
$f\left( x \right) = \sin \left( {\tan x} \right) - x$
and try to show that it is increasing, we would obtain
$f\left( x \right) > f\left( 0 \right)$ or $\,\,\,\,$ $\sin \left( {\tan x} \right) - x \ge 0$
Hence, our task could be accomplished by showing that $f(x)$ is increasing.
$f'\left( x \right) = \cos \left( {\tan x} \right){\sec ^2}x - 1$ $= \cos \left( {\tan x} \right)\left( {1 + {{\tan }^2}x} \right) - 1$ $= {\tan ^2}x\cos \left( {\tan x} \right) - \left( {1 - \cos \left( {\tan x} \right)} \right)$ $> {\tan ^2}x\cos \left( {\tan x} \right) - \dfrac{{{{\tan }^2}x}}{2}\;\;\;\; \left( {{\rm{using\, the\, given \,inequality}}} \right)$ $= \dfrac{1}{2}{\tan ^2}x\left( {2\cos \left( {\tan x} \right) - 1} \right)$ $= \dfrac{1}{2}{\tan ^2}x\left\{ {2\left( {\cos \left( {\tan x} \right) - 1} \right) + 1} \right\}$ $> \dfrac{1}{2}{\tan ^2}x\left( {1 - {{\tan }^2}x} \right)\,\,\,\,\,\,{\rm{(again\,using\, the\, given\,inequality)}}$
For, $x \in \left[ {0,\pi /4} \right],\,\,\tan x \in \left[ {0,1} \right]$ so that $\left( {1 - {{\tan }^2}x} \right) \ge 0$
$\Rightarrow \,\,\,\, f'\left( x \right) > \dfrac{1}{2}{\tan ^2}x\left( {1 - {{\tan }^2}x} \right) \ge 0$ $\Rightarrow \,\,\,\, f'\left( x \right) > 0$ $\Rightarrow \,\,\,\, f\left( x \right)$ is increasing on $\left[ {0,\pi /4} \right]$ $\Rightarrow \,\,\, \,\sin \left( {\tan x} \right) \ge \,x\,\,\forall x \in \left[ {0,\pi /4} \right]$
Example: 5
Show that $\cos \left( {\sin x} \right) > \sin \left( {\cos x} \right)\,\,\,\forall \,x\,\, \in \left( {0,\pi /2} \right)$
Solution: 5
The approach we have followed in the previous questions could be applied here to prove that $f\left( x \right) = \cos \left( {\sin x} \right) - \sin \left( {\cos x} \right)$ is increasing. However, $f'\left( x \right)$ becomes complicated and proving that it is positive is not straightforward like in the previous cases (you are urged to try this out).
Instead of considering the expressions $\cos \left( {\sin x} \right)$ and $\sin \left( {\cos x} \right)$, we can consider $\sin \left( {\dfrac{\pi }{2} - \sin x} \right)\,\,{\rm{and}}\,\,\,\sin \left( {\cos x} \right)$. This is because ‘$sin$’ is a monotonically increasing function in $\left( {0,\,\,\dfrac{\pi }{2}} \right)$, so that to determine the larger of the two values above, we just need to compare their arguments, i.e, $\left( {\dfrac{\pi }{2} - \sin x} \right)$ and $\cos x$
For $\left( {0,\,\,\dfrac{\pi }{2}} \right)$
$\sin x + \cos x\,\, < \,\dfrac{\pi }{2}$ $\Rightarrow \,\, \,\,\cos x < \dfrac{\pi }{2}\,\,\, - \,\,\,\sin x\,\,\,\,\,\,\,\forall \,x \in \left( {0,\dfrac{\pi }{2}} \right)$ $\Rightarrow \,\, \,\,\sin \left( {\cos x} \right) < \,\,\sin \left( {\dfrac{\pi }{2} - \sin x} \right)\,\,\,\,\,\,\, \forall \,x \in \left( {0,\dfrac{\pi }{2}} \right)$ $\Rightarrow \,\, \,\,\sin \left( {\cos x} \right) < \,\,\cos \left( {\sin x} \right)\,\,\,\,\,\,\,\forall \,x \in \left( {0,\dfrac{\pi }{2}} \right)$
Example: 6
Find the values of $a$ for which the function $f\left( x \right) = \sin x - a\sin 2x - \dfrac{1}{3}\sin 3x + 2ax\,\,{\rm{increases}}\,\,{\rm{on}}\,\,\mathbb{R}$.
Solution: 6
We want $f'\left( x \right) \ge \,0\,\forall x \in\mathbb{R}$
$f'\left( x \right) = \cos x - 2a\cos 2x - \cos 3x + 2a$ $= \cos x - 2a\left( {2{{\cos }^2}x - 1} \right) - \left( {4{{\cos }^3}x - 3\cos x} \right) + 2a$ $= 4a + 4\cos x - 4a{\cos ^2}x - 4{\cos ^3}x$ $= 4a{\sin ^2}x + 4\cos x{\sin ^2}x$ $= 4{\sin ^2}x\left( {a + \cos x} \right)$
This is always non-negative if (since the minimum value of $\cos x$ is $- 1$).
Therefore, the required values of $a$ are:
$a \in \left[ {1,\infty } \right)$ | 3,682 | 8,612 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 121, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-30 | latest | en | 0.365965 |
https://livewell.com/finance/money-market-yield-definition-calculation-and-example/ | 1,716,479,400,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058642.50/warc/CC-MAIN-20240523142446-20240523172446-00360.warc.gz | 313,386,552 | 39,469 | Home>Finance>Money Market Yield: Definition, Calculation, And Example
Finance
# Money Market Yield: Definition, Calculation, And Example
Learn all about money market yield in finance, including its definition, calculation, and an example to help you understand its importance in investment decisions.
## Money Market Yield: Definition, Calculation, and Example
Are you a finance enthusiast wondering how to calculate the money market yield? Look no further! In this blog post, we will delve into the definition, calculation, and provide an example of money market yield. By the end, you’ll have a clear understanding of this important financial concept.
## Key Takeaways:
• Money market yield is a measure used to determine the return on investment for short-term debt instruments.
• It is an essential tool for investors and helps to assess the profitability of investing in money market securities.
## Defining Money Market Yield
Money market yield, also known as the CD equivalent yield, is the interest rate earned by investing in money market instruments such as treasury bills, commercial paper, and certificates of deposit. Financial institutions issue these securities as a means to raise funds to meet their short-term financing needs.
Investors interested in these short-term investments need to be aware of the money market yield to make informed decisions about where to allocate their capital. Understanding the yield helps investors compare different investment options and assess the potential return on their investments.
## Calculating Money Market Yield
The calculation of money market yield involves two main components: the discount rate and the face value of the security. The formula to calculate the yield is as follows:
Money Market Yield = (Face Value – Purchase Price) / Purchase Price * (360 / Maturity Period)
Where:
• Purchase price: The price at which the investor purchases the money market security.
• Face value: The value of the money market security when it reaches maturity.
• Maturity period: The number of days it takes for the money market security to mature.
By plugging in the values into the formula, investors are able to determine the money market yield of their investment.
## An Example of Money Market Yield Calculation
Let’s say an investor purchased a certificate of deposit with a face value of \$10,000 at a discounted price of \$9,750. The maturity period for this investment is 180 days. To calculate the money market yield:
Money Market Yield = (\$10,000 – \$9,750) / \$9,750 * (360 / 180) = 10.26%
In this example, the money market yield for the certificate of deposit would be 10.26%. This indicates the rate of return the investor can expect to earn on their investment.
Understanding money market yield is crucial for investors looking to make informed decisions in the world of finance. By calculating the yield, investors can assess the potential return on their investments and compare different options to make the most profitable choices. So, the next time you come across money market investments, remember to calculate the money market yield to make wiser financial decisions!
• https://livewell.com/finance/money-market-yield-definition-calculation-and-example/ | 634 | 3,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.916425 |
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