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https://worksheetgenius.com/unit-converter/convert-9-mm-to-in/ | 1,659,988,161,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00262.warc.gz | 561,264,913 | 5,347 | # Convert 9 mm to in
In this article I will show you how to convert 9 millimeters into inches. Throughout the explanation below I might also call it 9 mm to in. They are the same thing!
## How to Convert Millimeters to Inches
A millimeter is smaller than a inch. I know that a mm is smaller than a in because of something called conversion factors.
Put very simply, a conversion factor is a number that can be used to change one set of units to another, by multiplying or dividing it. So when we need to convert 9 millimeters into inches, we use a conversion factor to get the answer.
The conversion factor for mm to in is:
1 mm = 0.039370078740157 in
Now that we know what the conversion factor is, we can easily calculate the conversion of 9 mm to in by multiplying `0.039370078740157` by the number of millimeters we have, which is 9.
9 x 0.039370078740157 = 0.35433070866142 in
So, the answer to the question "what is 9 millimeters in inches?" is 0.35433070866142 in.
## Millimeters to Inches Conversion Table
Below is a sample conversion table for mm to in:
Millimeters (mm) Inches (in)
0.010
0.10
10.04
20.08
30.12
50.2
100.39
200.79
501.97
1003.94
100039.37
## Best Conversion Unit for 9 mm
Sometimes when you work with conversions from one unit to another, the numbers can get a little confusing. Especially when dealing with really large numbers.
I've also calculated what the best unit of measurement is for 9 mm.
To determine which unit is best, I decided to define that as being the unit of measurement which is as low as possible, without going below 1. Smaller numbers are more easily understood and can make it easier for you to understand the measurement.
The best unit of measurement I have found for 9 mm is millimeters and the amount is 9 mm.
## Link to Us / Reference this Page
Please use the tool below to link back to this page or cite/reference us in anything you use the information for. Your support helps us to continue providing content!
• "Convert 9 mm to in". WorksheetGenius.com. Accessed on August 8, 2022. http://worksheetgenius.com/unit-converter/convert-9-mm-to-in/.
• "Convert 9 mm to in". WorksheetGenius.com, http://worksheetgenius.com/unit-converter/convert-9-mm-to-in/. Accessed 8 August, 2022
• Convert 9 mm to in. WorksheetGenius.com. Retrieved from http://worksheetgenius.com/unit-converter/convert-9-mm-to-in/. | 606 | 2,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-33 | latest | en | 0.918682 |
https://crypto.stackexchange.com/questions/101934/qr-pke-not-cca-secure | 1,709,033,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00661.warc.gz | 178,147,337 | 39,981 | # QR-PKE not CCA secure
Due to a comment stating "... QR-PKE is secure (CPA)..." I've been thinking of how to prove that it's not CCA secure, and would like to understand whether my proof is correct.
Here's the QR-PKE scheme:
• $$KeyGen(1^n)$$ produces:
$$sk = (p,q)$$ - two large n-bit primes
$$pk = (n, x)$$ where $$N=pq$$ and $$x$$ is a random $$QNR \in_R \mathbb{Z}^*_N$$.
• $$Enc_{pk}(b) = x^br^2 \mod N$$ , where $$r \in_R \mathbb{Z}^*_N$$ (random)
• $$Dec_{sk}(c)=0$$ iff $$c$$ is a $$QR$$
I thought of the following interaction in the CCA security game (key points) - the adversary sends the two challenge options: $$0,1$$. The challenger chooses one of them, $$b$$, and sends its encryption, $$c$$. The adversary randomly chooses $$r \in \mathbb{Z}^*_N$$ and asks to decrypt $$c x^0r^2$$. If the response is $$1$$ he knows that the second option (i.e. $$1$$) was chosen, otherwise $$0$$ was chosen, and responds accordingly. The correctness follows from the fact that the group product of two encryptions is the encryption of the XOR, in this scheme.
Have I got it right?
• "the XOR of two encryptions is the encryption of the XOR" seems incorrect. Can you show it?
– erth
Sep 18, 2022 at 9:11
• @erth Right - just fixed it now - I meant for it to be: "the group product of two encryptions is the encryption of the XOR" - this I proved separately, but stems from the fact that QRxQR is QR, QRxQNR is QNR and QNRxQNR is a QNR.
– Anon
Sep 18, 2022 at 9:36
• The last one should be QR (QNRxQNR=QR)
– Anon
Sep 18, 2022 at 9:43
• Yes, I think you can write an answer based on this. I think it is perhaps more to the point to phrase it as that multiplying by $r^2$ preserves the decryption result.
– erth
Sep 18, 2022 at 10:12
• The adversary sends the two challenge options: $$0,1$$.
• The challenger chooses one of them, $$b$$, and sends its encryption, $$c$$, to the adversary.
• The adversary randomly chooses $$r \in \mathbb{Z}^*_N$$ and asks to decrypt $$c x^0r^2$$. If the response is $$1$$ he knows that the second option (i.e. $$1$$) was chosen, otherwise $$0$$ was chosen, and responds accordingly.
The correctness follows from the fact that, in this scheme, the group product of two encryptions is the encryption of the XOR of the message (in other words, multiplying by $$r^2$$ preserves the decryption result). This is because it can be shown that: $$QR\times QR = QR$$, and $$QR \times QNR = QNR$$. | 740 | 2,426 | {"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": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-10 | longest | en | 0.918269 |
https://discourse.biologos.org/t/if-you-tied-a-string-around-the-earth-intuition-and-science-question-riddles/51366 | 1,685,339,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644683.18/warc/CC-MAIN-20230529042138-20230529072138-00354.warc.gz | 249,361,955 | 9,897 | If you tied a string around the earth.... [Intuition and Science Question/Riddles]
Pick one. No cheating.
[SRC: NSTA: Uncovering Student Ideas in Physical Science, Keeley & Harrington]
4 Likes
Are we restricted to choices a-e or are we allowed to reject them and suggest something single celled or even virus like?
I understand why you would want to go smaller but the correct answer is there!
Yep, d and r should increase proportionally to the circumfrence so I’ll go along with Tim.
Full adult, standard size? Either a cat or a goat, if the goat isn’t huge within goats.
It probably isn’t fair for math teachers to weigh in here; but I do have to say, I like the way this problem was stated! I think I’ll use that in the future. It’s better than the other way I’ve sometimes posed it to students: If you walked all the way around the world (my students can walk on water obviously!) … how much farther would your head go than your feet?
All I will say is this … remember your high school simple geometry formulas, people! You really don’t need a calculator for this one - AND - you don’t even need to know how big the earth is!
Treat it like a Fermi problem (meaning using simplified estimates in your head just to get to the rough magnitude of things - like just using three for pi because that gets you close enough.)
Or to assume most people know what a fermi problem is
2 Likes
Enrico Fermi was a famous physicist. But apparently he was also known for posing math problems that a person could solve with no calculator handy - or not even an envelope and pencil. Just head only - the goal being to at least get to the right magnitude of number or get your decimals in the right places.
So if I wanted to guestimate a reasonable approximation for how many calories were in the homemade peanut butter ball I just ate, I might work it thus (without looking up anything) … So I’ll guess it was about 30 grams (I know it was way more than 3g, and couldn’t have been close to 300 - since there’s no way three of them are anything close to as heavy as a kilogram). So already I’m only ball parking it here by calling it 30. And I also think I remember that every gram of carbs is about 4 food Calories and every gram of fats about 9. Since those balls had mostly peanut butter, oats, and honey as their primary ingredients, I’ll split the difference between sugars and fats and say roughly 7 Calories per gram - meaning I probably just had around a 200 Calorie snack.
Unless I was egregiously off on something there, then my answer should be reasonable. As long as the actual answer is closer in magnitude to 200 Cal than to 20 or 2000, then I was basically successful.
4 Likes
I don’t know why but questions like “how many golf balls fit on a school bus” interest me.
The Yale Physics Olympics gives a fermi quiz every year if you are ever bored:
Fermi
2 Likes
Dang! I got it. D. Does depend on the goat breed though.
I haven’t comepletely lost all reason.
Go with a beautiful little angora goat. You also get the lovely fiber and a fantastic warm sweater or socks or mittens.
1 Like
1. I did! At least the easiest ones, repeated over many classes and used in the kitchen.
2. I didn’t see that until you mentioned it. But there’s a reason the problem adds 6 meters. It did come from a text book, didn’t it?
1 Like
6.28 would work better but also make the problem odd/worse for most people. But yes, it is from a book.
They also write: “Because of the sophisticated nature of this probe, it is best used with middle or high school students who have the procedural and conceptual mathematical knowledge to per- form and understand the calculation. If stu- dents are unsure of the height of the animals, you might consider providing height data such as the following: ant is 1 mm or .001 m; mouse is 3 cm or .03 m; cat is 30 cm or .3 m; goat is 90 cm or .9 m; and horse is 2 m. Another alternative is to provide only two choices—the ant and the goat.”
I teach in the city. Most of my students will not generally not know about the many types and sizes of goats. They don’t get a lot of nature experience. I remember polling my students one year and a huge number of them have never gone fishing before. A goat will be somewhere in between a cat and horse to them.
The expectation is that most people will pick an ant because the earth is so big, its thought adding on 6m of string couldn’t possibly do much.
The gist is that it doesn’t. The earth has a radius of ~6.4 million meters. Adding 1m to it is not very significant but that is what 6m of extra string will do.
“The best answer is D: goat. The answer is highly counterintuitive. Even after performing the calculation, many students will still not believe the answer.” – NSTA
5 Likes
I do this for stuff all the time, I just thought it was called being bad at maths…
2 Likes
I just did the math for circum’s using 6.28. With a calculator.
I love that the problem (and hopefully the discussion that would follow (maybe many times) is intended to help kids connect the math and the physical concepts.
I have said before that my physics teacher expected us to take more on faith than my pastor. You either saw the connection or you didn’t. The kids that saw it were so much more interesting to him than those who didn’t. To the rest of us: trust me.
Why is d counterintuitive?
Quite the opposite actually! It’s called: being able to recognize that when a pencil is said or calculated to have 0.2 kg mass, that a mistake must have been made because there’s no way that five typical pencils are as heavy as a kilogram! Students need to have some concept of what a plausible sounding answer should be. Too often they don’t.
So it’s a good thing indeed if your fermi skills are well-honed!
3 Likes
I think it’s because our minds are conditioned to think that any problems involving huge things (like our whole planet!) - must also then result in extreme answers (either very BIG - as in - how much does the earth weigh? … or very tiny - as in how much does the ocean rise if I go for a swim in it?) We’re surprised when an operation on something so big ends up giving us some very ordinary change.
2 Likes
So true. This reminds me of the time Neil deGrasse Tyson blew my mind when he said the Earth in the hand of a giant would be smoother than a cue ball.
Because the earth is 24,000 miles in circumference. Most of my students would expect the string to not do much with only 6m added to that distance. Many of them would also not think to do a circumference calculation at least at first.
Not smoother than a professional grade cue-ball made for a tournament from what I remember but yes, as smooth as a well used one.
I do this demonstration with my class where we model out the earth on the floor. Its about 8,000mi in diameter so I work with the scale of 1 mi = 1mm. I refresh their memory on a meter stick (100 cm and 1000mm per stick) and it ends up that we have to lay 8 meter sticks on the ground to model the earth (8 meter sticks = 800cm = 8000 mm and 8000mm = 8000 mi on our scale).
We then draw a giant 8m circle (I had trouble erasing the dry erase from my floor this year!) on the floor. representing the earth. I ask students to draw a dot showing me how deep the oceans are and how tall the atmosphere is after that. The oceans have an average depth of 2mi so the correct answer is 2mm below the surface.
Technically I suppose the atmosphere never ends but:
~50% of mass of atmosphere lies below 3.5mi (= 3.5mm)
~90% is below 10 miles which his only 10mm or 1 cm.
~ 99% us below 20 miles (= 2cm)
~99.99997% lies below 60 miles (=6cm)
Seeing this modeled out live in front of them is always an aha moment.
I then mention the billiard ball thing. And how astronauts often point out the thin sliver of our atmosphere. I tell them the oceans are like puddles on the earth and the atmosphere like plastic wrap around a basketball or the skin on an apple.
I might just buy a 30ft by 30ft blue tarp and cut it into an 8m circle to preserve my classroom floor.
But its a good way of giving a sense of scale to the earth, a good way to talk about pollution and how finite our ocean and atmosphere is and also a good illustration of why earth is an inner ROCKY planet. Students think the oceans are really deep but usually don’t realize that the while the deepest part is 7 miles, it is about 4,000mi to the center of the earth.
Vinnie
5 Likes
This topic was automatically closed 6 days after the last reply. New replies are no longer allowed. | 2,097 | 8,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-23 | latest | en | 0.961594 |
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# Mrs E Teaches Math
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These logic puzzles with emojis are sure to be a hit! They would make a great Back to School activity or activity for a sub day! A digital version is included for distance learning.This product includes 32 logic puzzles. All of the puzzles are
Subjects:
Math, Other (Math)
7th, 8th, 9th, 10th
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250
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Math, Geometry, Word Problems
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329
In this NO-PREP activity, students will move around the room to practice multiplying decimals using word problems. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!In this activity, students will solve word
Subjects:
Math, Word Problems, Decimals
5th, 6th
Types:
CCSS:
5.NBT.B.7, 6.NS.B.3
\$3.50
264
These READY TO PRINT activities get students to actively discuss GEOMETRY PROOFS, without being afraid of making a mistake. These are perfect for PRACTICE or REVIEW!This is a bundle of my SEVEN proofs cut-out activities. Included are:1)
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Math, Geometry
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276
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In this NO-PREP activity, students will move around the room to practice writing the equations of parallel and perpendicular lines. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will write the
Subjects:
Algebra, Geometry
8th, 9th, 10th
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CCSS:
HSG-GPE.B.5
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235
This READY TO PRINT foldable for triangle congruence is a great addition to an interactive notebook or as a review foldable in your student's notes. MULTIPLE OPTIONS included.There are six flaps in the foldable. The flaps include: • SSS (Side
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Math, Geometry
8th, 9th, 10th
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Printables, Graphic Organizers, Interactive Notebooks
\$2.25
379
In this NO-PREP activity, students will move around the room to practice special right triangles. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will need to use 45-45-90 and 30-60-90 triangle rules
Subjects:
Math, Geometry
8th, 9th, 10th
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CCSS:
HSG-SRT.C.8
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147
This activity will keep your students active and engaged while visiting character education stations. You can extend parts of this activity into sub plans throughout the school year!Included:• Seven Stations - The stations included are Kindness
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For All Subject Areas, Character Education, Back to School
6th, 7th, 8th, 9th, 10th, 11th, 12th
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89
STOP THE SEARCH! You've just found the PERFECT back to school activities for your classes! These FIVE back to school activities are READY TO PRINT. Do all five on the first day of school or spread them throughout the first week.There are 5
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Back to School, Classroom Community
6th, 7th, 8th, 9th, 10th, 11th, 12th
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37
This NO PREP bundle is bursting with READY TO PRINT task cards. Your geometry proofs unit will be so much easier with your proofs activities DONE.This is a bundle of my four proofs task card activities. Included are:1) Algebraic Proofs Task
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\$16.00
\$12.50
109
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This is a fun activity that gets students working together and excited to practice trig ratios! Students won't mind practicing right triangle trig with this group activity.In this activity, students will use trig ratios to find a missing side or
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Math, Geometry, Trigonometry
8th, 9th, 10th, 11th
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\$4.00
193
In this NO-PREP activity, students will move around the room to practice interpreting slope and intercepts. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will read word problems to determine what
Subjects:
Math, Algebra, Graphing
8th, 9th, 10th
Types:
CCSS:
8.F.B.4
\$3.50
189
This NO PREP project will have students looking at the world around them to find geometric transformations. This is READY TO PRINT and will keep students engaged while working on rotations, reflections, translations, and glide reflections.In this
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8th, 9th, 10th
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Projects, Assessment
CCSS:
HSG-CO.A.2, HSG-CO.A.3, HSG-CO.A.5
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16
In this NO-PREP activity, students will move around the room to practice solving angle of depression and angle of elevation problems. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their trig skills!Students will solve angle
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Math, Geometry, PreCalculus
9th, 10th, 11th
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CCSS:
HSG-SRT.C.8
\$3.50
99
In this NO-PREP activity, students will move around the room to practice graphing and evaluating piecewise functions. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will solve problems related to
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10th, 11th, 12th
Types:
CCSS:
HSF-IF.C.7b
\$3.50
126
In this NO-PREP activity, students will move around the room to practice function operation and composition. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will use function operations to add,
Subjects:
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10th, 11th, 12th
Types:
\$3.50
158
In this NO-PREP activity, students will move around the room to practice using the midpoint and distance formulas. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will use the distance formula or
Subjects:
Math, Algebra, Geometry
8th, 9th, 10th
Types:
CCSS:
HSG-GPE.B.6
\$4.50
220
In this NO-PREP activity, students will move around the room to practice circle vocabulary. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their geometry skills!Students will need to recognize the vocabulary for radius,
Subjects:
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5th, 6th
Types:
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\$3.00
71
In this NO-PREP activity, students will move around the room to practice solving word problems with systems of equations. This is READY TO PRINT and will KEEP STUDENTS ENGAGED while working on their math skills!Students will use systems of equations
Subjects:
Math, Algebra, Algebra 2
8th, 9th, 10th
Types:
CCSS:
HSA-REI.C.6
\$3.50
200
Students will love working together to complete this mixed numbers and improper fractions puzzle! In this activity, students will match the improper fraction to its mixed number. Each pair of numbers matches to form a square.Two levels of the
Subjects:
Math, Fractions
4th, 5th, 6th
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CCSS:
4.NF.B.3c, 5.NF.A.1
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360
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TEACHING EXPERIENCE
I taught high school math for eight years. Before that, I spent two years teaching at the college level. I have experience teaching Statistics, Remedial Math, Algebra 1, Geometry, and Algebra 2. I have taught on-level and honors level students. I was on a team to completely re-write the mathematics standards for our private school and spearheaded the implementation of the new standards. Currently, I homeschool my young children. In our weekly co-op, I teach science classes to elementary school children.
MY TEACHING STYLE
I like to use various techniques in my classroom. I frequently use direct instruction with frequent breaks, authentic math tasks, foldables, and discovery activities.
HONORS/AWARDS/SHINING TEACHER MOMENT
I published a textbook ancillary while teaching a college-level remedial math class. I have been a team-lead for multiple courses and have been the technology mentor for my department.
MY OWN EDUCATIONAL HISTORY
Master of Science in Statistics
Bachelor of Arts in Mathematics
Check out my blog Mrs. E Teaches Math.
Reach me by email at mrseteachesmath@gmail.com
Logo design by Little Peace of Africa | 2,025 | 7,786 | {"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} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.906229 |
http://blog.csdn.net/Miracle_ma/article/details/51557673 | 1,516,640,819,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891485.97/warc/CC-MAIN-20180122153557-20180122173557-00703.warc.gz | 42,189,688 | 18,303 | # 离线专题学习
hdu 4288
1ikai[i%5=3]
sum[rt][i]=sum[lrt][i]+sum[rrt][5((num[lrt]i1)%5+5)%51]
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 100005
#define MAXN 500005
#define maxnode 105
#define sigma_size 2
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const ull mx = 1333333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
char op[MAX][10];
int a[MAX];
int b[MAX];
LL sum[MAX<<2][5];
int num[MAX<<2];
void build(int l,int r,int rt){
num[rt]=0;
for(int i=0;i<5;i++) sum[rt][i]=0;
if(l==r) return;
middle;
build(lson);
build(rson);
}
void pushup(int rt){
for(int i=0;i<5;i++)
sum[rt][i]=sum[lrt][i]+sum[rrt][5-((num[lrt]-i-1)%5+5)%5-1];
}
void update(int l,int r,int rt,int pos,int d,int v){
num[rt]+=v;
if(l==r){
sum[rt][0]+=d;
return;
}
middle;
if(pos<=m) update(lson,pos,d,v);
else update(rson,pos,d,v);
pushup(rt);
}
int main(){
//freopen("test.txt","r",stdin);
int n;
while(~scanf("%d",&n)){
int tmp=0;
for(int i=0;i<n;i++){
scanf("%s",op[i]);
if(op[i][0]!='s'){
scanf("%d",&a[i]);
b[tmp++]=a[i];
}
}
build(1,n,1);
sort(b,b+tmp);
int tot=unique(b,b+tmp)-b;
for(int i=0;i<n;i++){
int pos=lower_bound(b,b+tot,a[i])-b+1;
if(op[i][0]=='a') update(1,n,1,pos,a[i],1);
else if(op[i][0]=='d') update(1,n,1,pos,-a[i],-1);
else printf("%I64d\n",sum[1][2]);
}
}
return 0;
}
hdu 4417
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 100005
#define MAXN 500005
#define maxnode 105
#define sigma_size 2
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const ull mx = 1333333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
struct num{
int a,pos;
bool operator < (const num &b) const{
return a<b.a;
}
}pp[MAX];
struct Node{
int l,r,h,id,ans;
}p[MAX];
bool cmp1(Node a,Node b){
return a.h<b.h;
}
bool cmp2(Node a,Node b){
return a.id<b.id;
}
int sum[MAX<<2];
void build(int l,int r,int rt){
sum[rt]=0;
if(l==r) return ;
middle;
build(lson);
build(rson);
}
void pushup(int rt){
sum[rt]=sum[lrt]+sum[rrt];
}
void update(int l,int r,int rt,int pos){
if(l==r){
sum[rt]++;
return;
}
middle;
if(pos<=m) update(lson,pos);
else update(rson,pos);
pushup(rt);
}
int query(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R) return sum[rt];
middle;
int ans=0;
if(L<=m) ans+=query(lson, L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main(){
//freopen("test.txt","r",stdin);
int t,kase=0;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
for(int i=0;i<n;i++){
int a;
scanf("%d",&a);
pp[i]=(num){a,i+1};
}
sort(pp,pp+n);
for(int i=0;i<m;i++){
int l,r,h;
scanf("%d%d%d",&l,&r,&h);
p[i]=(Node){l+1,r+1,h,i,0};
}
sort(p,p+m,cmp1);
int ret=0;
build(1,n,1);
for(int i=0;i<m;i++){
while(pp[ret].a<=p[i].h&&ret<n){
update(1,n,1,pp[ret].pos);
ret++;
}
p[i].ans=query(1,n,1,p[i].l,p[i].r);
}
sort(p,p+m,cmp2);
kase++;
printf("Case %d:\n",kase);
for(int i=0;i<m;i++) printf("%d\n",p[i].ans);
}
return 0;
}
hdu 3874
http://acm.hdu.edu.cn/showproblem.php?pid=3874
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 50005
#define MAXN 6005
#define maxnode 15
#define sigma_size 30
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
//const double inf = 1e18;
const double eps = 1e-8;
const LL mod = 1e9+7;
const ull mx = 133333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
struct Node{
int l,r,id;
bool operator < (const Node&e)const{
return r>e.r;
}
}p[MAX*4];
LL ans[MAX*4];
int a[MAX];
int pos[1000005];
int pre[MAX];
LL sum[MAX<<2];
void pushup(int rt){
sum[rt]=sum[lrt]+sum[rrt];
}
void build(int l,int r,int rt){
sum[rt]=0;
if(l==r) return;
middle;
build(lson);
build(rson);
pushup(rt);
}
void update(int l,int r,int rt,int pos,int d){
if(l==r){
sum[rt]=d;
return;
}
middle;
if(pos<=m) update(lson,pos,d);
else update(rson,pos,d);
pushup(rt);
}
LL query(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R) return sum[rt];
middle;
LL ans=0;
if(L<=m) ans+=query(lson,L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
mem(pre,-1);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
pos[a[i]]=-1;
}
for(int i=1;i<=n;i++){
if(pos[a[i]]!=-1){
pre[i]=pos[a[i]];
pos[a[i]]=i;
}
else pos[a[i]]=i;
}
build(1,n,1);
int m;
cin>>m;
for(int i=0;i<m;i++){
int l,r;
scanf("%d%d",&l,&r);
p[i]=(Node){l,r,i};
}
sort(p,p+m);
for(int i=1;i<=n;i++){
if(pre[i]!=-1) update(1,n,1,pre[i],0);
update(1,n,1,i,a[i]);
}
int r=n;
for(int i=0;i<m;i++){
while(r>p[i].r){
if(pre[r]!=-1) update(1,n,1,pre[r],a[r]);
r--;
}
ans[p[i].id]=query(1,n,1,p[i].l,p[i].r);
}
for(int i=0;i<m;i++){
printf("%I64d\n",ans[i]);
}
}
return 0;
}
hdu 4638
http://acm.hdu.edu.cn/showproblem.php?pid=4638
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 100005
#define MAXN 6005
#define maxnode 15
#define sigma_size 30
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000
//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
//const double inf = 1e18;
const double eps = 1e-8;
const LL mod = 1e9+7;
const ull mx = 133333331;
/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
int pos[MAX];
int a[MAX];
struct que{
int l,r,id;
bool operator < (const que&e)const{
return r<e.r;
}
}p[MAX];
int ans[MAX];
int sum[MAX<<2];
void pushup(int rt){
sum[rt]=sum[lrt]+sum[rrt];
}
void build(int l,int r,int rt){
if(l==r){
sum[rt]=0;
return;
}
middle;
build(lson);
build(rson);
pushup(rt);
}
void update(int l,int r,int rt,int pos,int d){
if(l==r){
sum[rt]+=d;
return;
}
middle;
if(pos<=m) update(lson,pos,d);
else update(rson,pos,d);
pushup(rt);
}
int query(int l,int r,int rt,int L,int R){
if(L<=l&&r<=R) return sum[rt];
middle;
int ans=0;
if(L<=m) ans+=query(lson,L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main(){
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
pos[a[i]]=i;
}
for(int i=0;i<m;i++){
int l,r;
scanf("%d%d",&l,&r);
p[i]=(que){l,r,i};
}
sort(p,p+m);
build(1,n,1);
int tot=0;
for(int i=1;i<=n;i++){
update(1,n,1,i,1);
if(pos[a[i]-1]&&pos[a[i]-1]<=i) update(1,n,1,pos[a[i]-1],-1);
if(pos[a[i]+1]&&pos[a[i]+1]<=i) update(1,n,1,pos[a[i]+1],-1);
while(tot<m&&p[tot].r==i){
ans[p[tot].id]=query(1,n,1,p[tot].l,p[tot].r);
tot++;
}
}
for(int i=0;i<m;i++) printf("%d\n",ans[i]);
}
return 0;
}
• 本文已收录于以下专栏：
举报原因： 您举报文章：离线专题学习 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 3,575 | 10,358 | {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 2.703125 | 3 | CC-MAIN-2018-05 | latest | en | 0.133822 |
https://www.includehelp.com/python/numpy-set-diagonal-of-a-x-matrix-as-inf.aspx | 1,726,535,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00375.warc.gz | 756,103,812 | 35,589 | Python NumPy - Set Diagonal of a 3x3 Matrix as inf
By IncludeHelp Last updated : March 8, 2024
In NumPy, inf is a floating-point value that represents positive infinity. Whenever a calculation exceeds the possible limit of representation, we use inf to represent it.
Problem Statement
Given the NumPy matrix of shape 3X3, we need to set the diagonals of this matrix as inf values.
Setting Diagonal of a 3x3 Matrix as inf
To set the diagonal of a 3X3 matrix as inf, we first need to create a 3X3 matrix of zeros by using np.zeros() which will fill all the elements as 0 and then we can use np.fill_diagonals() function and pass np.inf() as parameter.
Let us understand with the help of an example,
Python Code to Set Diagonal of a 3x3 Matrix as inf
```# Importing numpy
import numpy as np
# Creating an array
arr = np.zeros((3, 3))
# Display original array
print("Original array:\n", arr, "\n")
# Assigning diagonal as inf
np.fill_diagonal(arr, np.inf)
# Display result
print("Result:\n", arr)
```
Output
The output of the above program is:
```Original array:
[[0. 0. 0.]
[0. 0. 0.]
[0. 0. 0.]]
Result:
[[inf 0. 0.]
[ 0. inf 0.]
[ 0. 0. inf]]
```
To understand the above program, you should have the basic knowledge of the following Python topics: | 363 | 1,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-38 | latest | en | 0.83179 |
https://domygreexam.com/the-order-of-operations-problem/ | 1,721,490,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00568.warc.gz | 182,569,929 | 22,679 | # The Order of Operations Problem
Most people need a fairly standard set of instructions for carrying out the computation. For some reason, however, most people have a very poor view of how this process actually works. It’s as if mathematics is a black box where no one can step inside and see the inner workings. In fact, many years ago, many math professors actually developed a rather common order of operations that essentially tells you what calculations to do in an equation with multiple operations.
This is a very common way to go about doing calculations, as it helps you get through the job much quicker. Unfortunately, most people are still doing things this way. For example, most students are taking classes where there is an equation and they will be required to work out the various factors that are involved in the equation. Often, there is not a standard procedure for doing this. In other words, they do it their own way and they are not particularly well aware of it.
It may not seem like it to people that don’t understand the standard procedures, but it actually has its advantages. First of all, the standard process makes sure that the student always gets everything that they are required to get. In fact, most people never get everything. That means that they get a lot more out of the equation when they use the standard procedure. When they don’t, the problem is compounded by having to work out every detail individually.
The second advantage of the most common procedure for getting through an equation is that it allows for an easier understanding. After all, it doesn’t matter how well you know the inner workings of the formula unless you understand the way that you are going to proceed with the next step in the process.
There are many different ways that people use in order to do this. For example, many people will begin with a very simple factor. In other words, they’ll simply add up one part of the equation with another part. This is quite often followed by adding two more factors together. Of course, you can also just add the three elements and multiply them to get the final number.
This can result in a number that is very difficult to understand. However, it can be relatively easy to grasp if you have a good idea of how this type of procedure works. It is a good idea to go over a few examples of these before beginning with the process.
A very popular way to do the whole process is by using a single factor. It works much the same way that this one does. Basically, you only need to include one element at a time, but there is a significant increase in the difficulty of computing. This is a good idea to do first, in fact, because it shows how many different ways there are for you to get to the final answer.
The third thing to remember is that the process is always changing. It’s always changing. As long as there are different variables, then the process changes.
The fourth method is to use a combination of various methods. It can be very hard to work out the order of operations with a single method. In addition, it can be very difficult to work out the order of operations with a mixture of several methods.
One thing to remember is that there are some cases where it is better to get the procedure right first. For example, if you are trying to compute the difference of the sales figures of two similar products.
The fifth method is to use a few different methods to get to the answer for the order of operation. This means that you will multiply your answers by at different rates. to get the answer that is the same.
It can be very important to think about what the answers are before you use a procedure. Even though it may take you a while to make the decision, you should be able to get the answer for the whole process. | 767 | 3,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.977128 |
https://www.r-bloggers.com/2017/05/euler-problem-20-large-integer-factorials/ | 1,638,155,760,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00027.warc.gz | 1,072,134,445 | 24,039 | [This article was first published on The Devil is in the Data, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Euler Problem 20 is the third problem that requires special consideration for working with very large integers. In this problem, we look at factorials. These numbers are useful in combinatorics if, for example, you like to know in how many ways you can arrange a deck of cards.
The problem with computing factorials is that they are mostly very large numbers, beyond the generic capabilities of computers to process. This problem can be solved using a specialised R package and using only base-R code.
## Euler Problem 20 Definition
$n! = n \times (n - 1) \times (n-2) \times \ldots \times 3 \times 2 \times 1$.
For example: $10! = 10 \times 9 \times \ldots \times 3 \times 2 \times 1 = 3628800$.
The sum of the digits in the number $10!$ is $3 + 6 + 2 + 8 + 8 + 0 + 0 = 27$.
Find the sum of the digits in the number 100!
## Euler Problem 20 Solution
The factorial of the number 100 contains 158 digits, which is a lot more digits than a 64-bit operating system can accurately produce. Using the standard function: factorial(100) = 9.332622e+157. Without using a specialised algorithm, we cannot determine the sum of all digits. We need to deploy arbitrary-precision arithmetic to solve this problem.
Many computer languages, including R, have special libraries to deal with such large numbers. The GMP Multiple Precision Arithmetic package renders this problem almost trivial.
library(gmp)
digits <- factorialZ(100)
digits <- as.character(digits)
## Base-R Code
The problem becomes more interesting when only using basic R code. I developed the big.add function to solve Euler Problem 13 through the addition of very large integers. We can extend this function to also calculate factorials. A factorial can be replaced by a series of additions, for example:
$3! = 1 \times 2 \times 3 = (((1+1) + (1+1)) + (1+1))$
This can be mimicked in R using the Reduce function. This function reduces a vector to a single value by recursively calling a function. Reduce(“+”, rep(4, 5)) is the same as:
$4 \times 5 = ((((4 + 4) + 4) + 4) + 4) = 20$
Using a loop, we can use the Reduce function to calculate a factorial, using only additions:
fact <- 1
x <- 100
for (i in 2:x) {
fact <- Reduce("+", rep(fact, i))
}
print(fact)
The big.factorial function below implements this idea by combining the big.add and Reduce functions to calculate large integer factorials. The function returns a value of 1 when the factorial of 0 or 1 is requested. This function does not calculate the Gamma-function for fractions. For all other values, it goes through a loop from 2 to the requested factorial. The temporary values are stored in the bf variable. The code loops through the factorials by using the result of the previous Reduce call into the current one.
big.factorial <- function(x) {
x <- floor(x)
bf <- 1 if (x > 1) {
for (i in 2:x) {
}
}
return (bf)
}
digits <- big.factorial(100)
This function is most certainly not as fast as the GMP package but it was fun to write and to learn about the mechanics behind arbitrary precision arithmetic at work.
If you like to know how factorials can be used to determine the number of ways a deck can be shuffled the watch this video.
The post Euler Problem 20: Large Integer Factorials appeared first on The Devil is in the Data. | 897 | 3,535 | {"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": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | longest | en | 0.850336 |
https://math.answers.com/math-and-arithmetic/What_is_thirty_three_times_one_hundred_and_forty | 1,723,562,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00462.warc.gz | 305,291,258 | 49,415 | 0
What is thirty three times one hundred and forty?
Updated: 9/18/2023
Wiki User
β 13y ago
Best Answer
haha did that in my head it is 4620
Wiki User
β 13y ago
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Two hundred times eight equals sixteen hundred. Forty times eight equals three hundred twenty. Two times eight equals sixteen. Sixteen hundred plus three hundred twenty plus sixteen equals nineteen hundred thirty-six. Two hundred forty-two times eight equals nineteen hundred thirty-six.
What is thirty eight times forty?
one thousand five hundred twenty
what number times forty eight equals one hundred and eighty?
the answer to that would be one hundred and thirty two
How many times does 333 appear in the bible?
The phrases "three hundred thirty three" and "three hundred and thirty three" do not appear anywhere in the KJV bible.
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How much is twelve times two hundred and forty three times times twelve equals?
34, 992 is the answer.
In the number 3333333 how does the 3 in the hundreds place compare to the 3 in the place to its left?
it is 10 times smaller on the left is thousands 3,333,333 three million, three hundred and thirty-three thousand, three hundred and thirty-three
1200
What is three thousand three hundred and two times forty one equals?
3302 * 41 = 135,382
What does sixty-one times forty-three equal?
Two thousand six hundred and twenty-three
What is three times the square root of one hundred and forty four?
It is: 3 times 12 = 36
What is one hundred thirty three times eleven?
133 multiplied by 11 is 1,463. | 423 | 1,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.924029 |
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April 1st, 2013, 03:43 PM #1 Senior Member Joined: Jan 2012 Posts: 118 Thanks: 1 Recurrence Relations & Induction Hello there. Have a quick question. A problem says to : Show that: $$a_{1}= 3$$ $$a_{2}= 8$$ $$a_{n}= 2a_{n-1} + 2a_{n-2}$$ is a recurrence relation for an. This may be an extremely simple question but I barely understand this whole recurrence relation and induction stuff. Would I use induction here ?
April 1st, 2013, 07:38 PM #2
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Re: Recurrence Relations & Induction
Quote:
Originally Posted by Jet1045 Hello there. Have a quick question. A problem says to : Show that: $$a_{1}= 3$$ $$a_{2}= 8$$ $$a_{n}= 2a_{n-1} + 2a_{n-2}$$ is a recurrence relation for an. This may be an extremely simple question but I barely understand this whole recurrence relation and induction stuff. Would I use induction here ?
Ok, so A_3 = 2(A_1+A_2) = 2(3+ = 22
So what does A_4 equal?
April 2nd, 2013, 09:55 AM #3 Senior Member Joined: Jan 2012 Posts: 118 Thanks: 1 Re: Recurrence Relations & Induction thanks! ya i totally get that. Like i understand how to get all of the next terms in the sequence. But if its asking to show, is that all you would do then?!
April 2nd, 2013, 10:46 AM #4
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Re: Recurrence Relations & Induction
Quote:
Originally Posted by Jet1045 thanks! ya i totally get that. Like i understand how to get all of the next terms in the sequence. But if its asking to show, is that all you would do then?!
Yes, the value of every number depends on the values of the immediately prior two numbers and the formula for combining them.
These sequences are fun to work with and of some real importance in number theory. Happy adventuring!
April 2nd, 2013, 09:11 PM #5 Senior Member Joined: Jan 2012 Posts: 118 Thanks: 1 Re: Recurrence Relations & Induction Okay Great! Yeah I just expected that maybe he would want us to show more, but hopefully that's fine. I'll just show the relation but giving the next couple values. Heres one i'm having trouble with. Using strong induction show that: $a_{n}=\frac{(1 + \sqrt{3})^{n+2} - (1-\sqrt{3})^{n+2}}{4\sqrt{3}}$ is a solution to the recurrence relation from part a. Part a being the recurrence relation I posted above. I HAVE NO CLUE WHERE TO START, mainly because I don't understand strong induction because he barely explained it. So with strong induction you start with the inductive step opposed to the base case correct? For our example in class he showed that Pn was equal to P(n+4). So i'm confused, do we just pick a random integer. Could I insert n+4 into this equation? I just don't get it ;(
April 3rd, 2013, 04:41 AM #6 Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26 Re: Recurrence Relations & Induction I'm not sure this will help, but my first step confronted with that would be to pick a couple of terms that are in the series such as a_3 and a_4, then plug them into the equation to see how it works. Firstly this will confirm it does indeed work. Secondly it might give you a clue as to why it works, which could be a start.
April 3rd, 2013, 11:41 AM #7 Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Recurrence Relations & Induction You have to use strong induction to solve this. Strong induction allows you to assume that the statement is true for all n less than or equal to k. Now, the first step is to demonstrate that $a_1= 3$ and $a_2= 8$ according to the formula. Now, if this is true, the first part of the induction proof is done. Then all you have to do is show that the formula for $a_n$ satisfies the recurrence relation $a_n= 2a_{n-1} + 2a_{n-2}$. In this case, you must show: $\frac{(1 + \sqrt{3})^{n+2} - (1 - \sqrt{3})^{n+2}}{4 \sqrt{3}}= 2 \left(\frac{(1 + \sqrt{3})^{n+1} - (1 - \sqrt{3})^{n+1}}{4 \sqrt{3}}\right) + 2 \left(\frac{(1 + \sqrt{3})^{n} - (1 - \sqrt{3})^{n}}{4 \sqrt{3}}\right)$
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# A piece of fabric is cut into three sections so that the first is thre
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A piece of fabric is cut into three sections so that the first is thre [#permalink]
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25 Feb 2018, 11:21
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A piece of fabric is cut into three sections so that the first is three times as long as the second and the second section is three times as long as the third. What part of the entire piece is the smallest section?
(A) 1/12
(B) 1/9
(C) 1/3
(D) 1/7
(E) 1/13
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Re: A piece of fabric is cut into three sections so that the first is thre [#permalink]
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25 Feb 2018, 11:36
Bunuel wrote:
A piece of fabric is cut into three sections so that the first is three times as long as the second and the second section is three times as long as the third. What part of the entire piece is the smallest section?
(A) 1/12
(B) 1/9
(C) 1/3
(D) 1/7
(E) 1/13
start from the smallest and let the size = 1
3rd part= 1
2nd part = 3 ( thrice of 3rd part)
1st = 9
total = 9+3+1= 13
smallest = 1/13
(E) imo
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Re: A piece of fabric is cut into three sections so that the first is thre [#permalink]
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26 Feb 2018, 01:03
Let third part= x
Second part= 3x
First part= 3*3x= 9x
Total= 9x+3x+x= 13x
Third part (Smallest)= $$\frac{x}{13x}$$= $$\frac{1}{13}$$
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Re: A piece of fabric is cut into three sections so that the first is thre [#permalink]
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26 Feb 2018, 02:43
For these type of ratio questions, I would usually go for table method:
Name the 3 pieces of fabrics as A, B & C
A B C
3 1 -
- 3 1
Make B the same value, i.e. 3
A B C
9 3 -
- 3 1
Therefore, $$\frac{C}{A+B+C} = \frac{1}{9+3+1} = \frac{1}{13}$$
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Re: A piece of fabric is cut into three sections so that the first is thre [#permalink]
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27 Feb 2018, 10:25
Bunuel wrote:
A piece of fabric is cut into three sections so that the first is three times as long as the second and the second section is three times as long as the third. What part of the entire piece is the smallest section?
(A) 1/12
(B) 1/9
(C) 1/3
(D) 1/7
(E) 1/13
We can let f = the longest piece, s = the second longest piece, and t = the shortest piece and create the equations:
f = 3s
and
s = 3t
Thus, f = 3(3t) = 9t
The total piece of fabric is:
9t + 3t + t = 13t
Thus, the shortest piece is t/(13t) = 1/13 of the entire piece.
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Re: A piece of fabric is cut into three sections so that the first is thre [#permalink] 27 Feb 2018, 10:25
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- Aug 22, 2020-
Relationship between surface load and surface temperature of heating pipe in static air medium：
The surface temperature of the electric heating tube is determined by the electric power per unit area and the properties of the heating medium around the heating tube. Today, we first analyze the relationship between the surface load and the surface temperature of the heating tube in the static air medium.
Take an electric heating tube with an outer diameter of ψ 12mm, and use the surface temperature of No.10 steel to compare with Junjun's electric heating pipe to do the experiment. In order to ensure the accuracy of temperature measurement, nickel aluminum couple is used to weld directly on the surface of heating pipe (other experiments prove that the surface temperature of heating rod measured by this method is relatively correct). The heating rod is electrified and heated in the still air medium, and the power on the unit surface of the heating rod is 0.1-6w/cm2, and the temperature measured on the surface is 70-600 ℃.
Through the test, we can get that the relationship between the surface load and the surface temperature of the heating pipe in the static air medium is in direct proportion
In general static air medium,
The surface load is 1W / cm2, and the surface temperature of the heating rod reaches 300 ℃
The surface load is 2W / cm2, and the surface temperature of the heating rod reaches 400 ℃
The surface load is 3W / cm2, and the surface temperature of the heating rod reaches 500 ℃
The surface load is 4W / cm2, and the surface temperature of the heating rod reaches 580 ℃
The surface load is 5W / cm2, and the surface temperature of the heating rod reaches 630 ℃
The surface load is 6W / cm2, and the surface temperature of the heating rod reaches 680 ℃
The above values are only approximate temperature, for reference only.
According to the data obtained, the power required for the required temperature of the electric heating tube can be calculated in the still air medium. | 455 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-04 | latest | en | 0.902161 |
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Agglomerative clustering (AC) - PowerPoint PPT Presentation
Agglomerative clustering (AC). Clustering algorithms: Part 2c. Pasi Fränti 25.3.2014 Speech & Image Processing Unit School of Computing University of Eastern Finland Joensuu, FINLAND. Agglomerative clustering Categorization by cost function. Single link
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Agglomerative clustering (AC)
Clustering algorithms: Part 2c
• Pasi Fränti
• 25.3.2014
• Speech & Image Processing Unit
• School of Computing
• University of Eastern Finland
• Joensuu, FINLAND
Agglomerative clusteringCategorization by cost function
• Minimize distance of nearest vectors
• Minimize distance of two furthest vectors
Ward’s method
• Minimize mean square error
• In Vector Quantization, known as Pairwise Nearest Neighbor (PNN) method
We focus on this
Pseudo code
• PNN(X, M) → C, P
• FOR i←1 TO N DO
• p[i]←i; c[i]←x[i];
• REPEAT
• a,b ← FindSmallestMergeCost();
• MergeClusters(a,b);
• m←m-1;
• UNTIL m=M;
O(N)
O(N2)
N times
T(N) = O(N3)
Merge cost:
Local optimization strategy:
Nearest neighbor search:
• Find the cluster pair to be merged
• Update of NN pointers
Example of the overall process
M=5000
M=50
M=5000
M=4999
M=4998
.
.
.
M=50
.
.
M=16
M=15
M=16
M=15
Example - 25 Clusters
MSE ≈ 1.01*109
Example - 24 Clusters
MSE ≈ 1.03*109
Example - 23 Clusters
MSE ≈ 1.06*109
Example - 22 Clusters
MSE ≈ 1.09*109
Example - 21 Clusters
MSE ≈ 1.12*109
Example - 20 Clusters
MSE ≈ 1.16*109
Example - 19 Clusters
MSE ≈ 1.19*109
Example - 18 Clusters
MSE ≈ 1.23*109
Example - 17 Clusters
MSE ≈ 1.26*109
Example - 16 Clusters
MSE ≈ 1.30*109
Example - 15 Clusters
MSE ≈ 1.34*109
Storing distance matrix
• Maintain the distance matrix and update rows for the changed cluster only!
• Number of distance calculations reduces from O(N2) to O(N) for each step.
• Search of the minimum pair still requires O(N2) time still O(N3) in total.
• It also requires O(N2) memory.
• Search reduces O(N2) O(logN).
• In total: O(N2 logN)
Store nearest neighbor (NN) pointers[Fränti et al., 2000: IEEE Trans. Image Processing]
Time complexity reduces to O(N 3) Ω (N 2)
Pseudo code
• PNN(X, M) → C, P
• FOR i←1 TO N DO
• p[i]←i; c[i]←x[i];
• FOR i←1 TO N DO
• NN[i]← FindNearestCluster(i);
• REPEAT
• a ← SmallestMergeCost(NN);
• b ← NN[i];
• MergeClusters(C,P,NN,a,b,);
• UpdatePointers(C,NN);
• UNTIL m=M;
O(N)
O(N2)
O(N)
O(N)
http://cs.uef.fi/pages/franti/research/pnn.txt
Algorithm:Lazy-PNN
T. Kaukoranta, P. Fränti and O. Nevalainen, "Vector quantization by lazy pairwise nearest neighbor method", Optical Engineering, 38 (11), 1862-1868, November 1999
Monotony property of merge cost [Kaukoranta et al., Optical Engineering, 1999]
Merge costs values are monotonically increasing:
d(Sa, Sb) d(Sa, Sc) d(Sb, Sc)
d(Sa, Sc) d(Sa+b, Sc)
Lazy variant of the PNN
• Store merge costs in heap.
• Update merge cost value only when it appears at top of the heap.
• Processing time reduces about 35%.
Algorithm:Iterative shrinking
P. Fränti and O. Virmajoki “Iterative shrinking method for clustering problems“Pattern Recognition, 39 (5), 761-765, May 2006.
Agglomeration based on cluster removal[Fränti and Virmajoki, Pattern Recognition, 2006]
Cluster removal in practice
Find secondary cluster:
Calculate removal cost for every vector:
Complexity analysis
Number of vectors per cluster:
If we iterate until M=1:
Adding the processing time per vector:
Algorithm:PNN with kNN-graph
P. Fränti, O. Virmajoki and V. Hautamäki, "Fast agglomerative clustering using a k-nearest neighbor graph". IEEE Trans. on Pattern Analysis and Machine Intelligence, 28 (11), 1875-1881, November 2006
Time distortion comparison
-PNN (229 s)
Trivial-PNN (>9999 s)
Graph-PNN (1)
MSE = 5.36
Graph-PNN (2)
• Graph created by MSP
• Graph created by D-n-C
Conclusions
• Simple to implement, good clustering quality
• Straightforward algorithm slow O(N3)
• Fast exact (yet simple) algorithm O(τN2)
• Beyond this possible:
• O(τ∙N∙logN) complexity
• Complicated graph data structure
• Compromizes the exactness of the merge
Literature
• P. Fränti, T. Kaukoranta, D.-F. Shen and K.-S. Chang, "Fast and memory efficient implementation of the exact PNN", IEEE Trans. on Image Processing, 9 (5), 773-777, May 2000.
• P. Fränti, O. Virmajoki and V. Hautamäki, "Fast agglomerative clustering using a k-nearest neighbor graph". IEEE Trans. on Pattern Analysis and Machine Intelligence, 28 (11), 1875-1881, November 2006.
• P. Fränti and O. Virmajoki, "Iterative shrinking method for clustering problems", Pattern Recognition, 39 (5), 761-765, May 2006.
• T. Kaukoranta, P. Fränti and O. Nevalainen, "Vector quantization by lazy pairwise nearest neighbor method", Optical Engineering, 38 (11), 1862-1868, November 1999.
• T. Kurita, "An efficient agglomerative clustering algorithm using a heap", Pattern Recognition 24 (3) (1991) 205-209.
Literature
• J. Shanbehzadeh and P.O. Ogunbona, "On the computational complexity of the LBG and PNN algorithms". IEEE Transactions on Image Processing6 (4), 614‑616, April 1997.
• O. Virmajoki, P. Fränti and T. Kaukoranta, "Practical methods for speeding-up the pairwise nearest neighbor method ", Optical Engineering, 40 (11), 2495-2504, November 2001.
• O. Virmajoki and P. Fränti, "Fast pairwise nearest neighbor based algorithm for multilevel thresholding", Journal of Electronic Imaging, 12 (4), 648-659, October 2003.
• O. Virmajoki, Pairwise Nearest Neighbor Method Revisited, PhD thesis, Computer Science, University of Joensuu, 2004.
• J.H. Ward, Hierarchical grouping to optimize an objective function, J. Amer. Statist.Assoc. 58 (1963) 236-244. | 1,993 | 6,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.70559 |
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MPH1. Helen believes that the random variable C, representing cloud cover from the large dataset, can be modelled by a discrete uniform distribution.(1)Helen used all the data from the large data set for Hurn in 2015 and found that theproportion of days with cloud cover of less than 50% was 0.315(c) Comment on the suitability of Helen’s model in the light of this information.(d) Suggest an appropriate refinement to Helen’s model.(1)(1)DO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREA(b) Using this model, find the probability that cloud cover is less than 50%(2)DO NOT WRITE IN THIS AREA(a) Write down the probability distribution for C.DO NOT WRITE IN THIS AREAAnswer ALL questions. Write your answers in the spaces provided.DO NOT WRITE IN THIS AREASECTION A: STATISTICS2*P58350A0236*
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(a) Stating your hypotheses clearly and using a 5% level of significance, test whether ornot the correlation between sales figures and average weekly temperature is negative.(b) Suggest a possible reason for this correlation.(3)(1)Tessa suggests that a linear regression model could be used to model these data.(c) State, giving a reason, whether or not the correlation coefficient is consistent withTessa’s suggestion.(d) State, giving a reason, which variable would be the explanatory variable.DO NOT WRITE IN THIS AREA2. Tessa owns a small clothes shop in a seaside town. She records the weekly sales figures, w,and the average weekly temperature, t C, for 8 weeks during the summer.The product moment correlation coefficient for these data is 0.915DO NOT WRITE IN THIS AREAMPH(1)(1)DO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREA(1)DO NOT WRITE IN THIS AREA(e) Give an interpretation of the gradient of this regression equation.DO NOT WRITE IN THIS AREATessa calculated the linear regression equation as w 10 755 – 171t4*P58350A0436*
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Peta models H as B(10, 0.1)(a) State two assumptions Peta needs to make to use her model.(b) Using Peta’s model, find P(H 4)(2)(1)For each child the random variable F represents the number of the throw on which thedart first hits the target.DO NOT WRITE IN THIS AREA3. In an experiment a group of children each repeatedly throw a dart at a target.For each child, the random variable H represents the number of times the dart hits thetarget in the first 10 throws.DO NOT WRITE IN THIS AREAMPHUsing Peta’s assumptions about this experiment,(c) find P(F 5)(d) Find the value of α(4)(e) Using Thomas’ model, find P(F 5)(f) Explain how Peta’s and Thomas’ models differ in describing the probability that adart hits the target in this experiment.(1)DO NOT WRITE IN THIS AREAwhere α is a constant.DO NOT WRITE IN THIS AREAP(F n) 0.01 (n 1) αDO NOT WRITE IN THIS AREAThomas assumes that in this experiment no child will need more than 10 throws for thedart to hit the target for the first time. He models P(F n) asDO NOT WRITE IN THIS AREA(2)(1)6*P58350A0636*
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(a) State the sampling method Charlie used.(1)(b) State and briefly describe an alternative method of non-random sampling Charliecould have used to obtain a sample of 40 workers.(2)Taruni decided to ask every member of the company the time, x minutes, it takes them totravel to the office.(c) State the data selection process Taruni used.DO NOT WRITE IN THIS AREA4. Charlie is studying the time it takes members of his company to travel to the office.He stands by the door to the office from 08 40 to 08 50 one morning and asks workers, asthey arrive, how long their journey was.DO NOT WRITE IN THIS AREAMPH(1)405060708090100 110 120 130 140Journey time (minutes)n 95 x 4133 x2 202 294(d) Write down the interquartile range for these data.(e) Calculate the mean and the standard deviation for these data.(f) State, giving a reason, whether you would recommend using the mean and standarddeviation or the median and interquartile range to describe these data.(1)(3)(2)Rana and David both work for the company and have both moved house since Tarunicollected her data.Rana’s journey to work has changed from 75 minutes to 35 minutes and David’s journeyto work has changed from 60 minutes to 33 minutes.Taruni drew her box plot again and only had to change two values.(g) Explain which two values Taruni must have changed and whether each of thesevalues has increased or decreased.(3)10*P58350A01036*DO NOT WRITE IN THIS AREA30DO NOT WRITE IN THIS AREA20DO NOT WRITE IN THIS AREA10DO NOT WRITE IN THIS AREATaruni’s results are summarised by the box plot and summary statistics below.
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MPH(a) Find the probability that a randomly selected battery will last for longer than 16 hours.(1)At the start of her exams Alice put 4 new batteries in her calculator.She has used her calculator for 16 hours, but has another 4 hours of exams to sit.(b) Find the probability that her calculator will not stop working for Alice’s remaining exams.(5)DO NOT WRITE IN THIS AREAAlice’s calculator requires 4 batteries and will stop working when any one battery reachesthe end of its lifetime.DO NOT WRITE IN THIS AREA5. The lifetime, L hours, of a battery has a normal distribution with mean 18 hours andstandard deviation 4 hours.(d) Stating your hypotheses clearly and using a 5% level of significance, test Alice’s belief.(5)DO NOT WRITE IN THIS AREAAfter her exams, Alice believed that the lifetime of the batteries was more than 18 hours.She took a random sample of 20 of these batteries and found that their mean lifetime was19.2 hours.DO NOT WRITE IN THIS AREA(3)DO NOT WRITE IN THIS AREA(c) Show that the probability that her calculator will not stop working for the remainderof her exams is 0.199 to 3 significant figures.DO NOT WRITE IN THIS AREAAlice only has 2 new batteries so, after the first 16 hours of her exams, although hercalculator is still working, she randomly selects 2 of the batteries from her calculator andreplaces these with the 2 new batteries.14*P58350A01436*
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MPHAnswer ALL questions. Write your answers in the spaces provided.6. At time t seconds, where t 0, a particle P moves in the x-y plane in such a way that itsvelocity v m s 1 is given by 1v t 2 i 4tjWhen t 1, P is at the point A and when t 4, P is at the point B.Find the exact distance AB.DO NOT WRITE IN THIS AREAUnless otherwise stated, whenever a numerical value of g is required, take g 9.8 m s 2 and give youranswer to either 2 significant figures or 3 significant figures.DO NOT WRITE IN THIS AREASECTION B: MECHANICS(6)DO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREA18*P58350A01836*
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MPH7.DO NOT WRITE IN THIS AREAα20 kgFigure 1A wooden crate of mass 20 kg is pulled in a straight line along a rough horizontal floorusing a handle attached to the crate.3The handle is inclined at an angle α to the floor, as shown in Figure 1, where tan α 4The tension in the handle is 40 N.The coefficient of friction between the crate and the floor is 0.14The crate is modelled as a particle and the handle is modelled as a light rod.DO NOT WRITE IN THIS AREA40 Nα40 N20 kgDO NOT WRITE IN THIS AREAThe crate is now pushed along the same floor using the handle. The handle is againinclined at the same angle α to the floor, and the thrust in the handle is 40 N as shown inFigure 2 below.DO NOT WRITE IN THIS AREA(6)DO NOT WRITE IN THIS AREA(a) find the acceleration of the crate.DO NOT WRITE IN THIS AREAUsing the model,Figure 2(b) Explain briefly why the acceleration of the crate would now be less than theacceleration of the crate found in part (a).(2)20*P58350A02036*
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MPH(a) Show that the magnitude of the acceleration of P is 2.5 m s 2(4)At the instant when P leaves the point A, the acceleration of P changes so that P nowmoves with constant acceleration (4i 8.8j) m s 2At the instant when P reaches the point B, the direction of motion of P is north east.(b) Find the time it takes for P to travel from A to B.DO NOT WRITE IN THIS AREAA particle P moves with constant acceleration.At time t 0, the particle is at O and is moving with velocity (2i 3j) m s 1At time t 2 seconds, P is at the point A with position vector (7i 10j) m.DO NOT WRITE IN THIS AREA8. [In this question i and j are horizontal unit vectors due east and due north respectivelyand position vectors are given relative to the fixed point O.](4)DO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREA24*P58350A02436*
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MPHxPAαDO NOT WRITE IN THIS AREACB2aDO NOT WRITE IN THIS AREA9.3, as shown in Figure 3.4The plank is modelled as a uniform rod, the block is modelled as a particle and the ropeis modelled as a light inextensible string.(a) Using the model, show that the tension in the rope is5Mg (3 x a )6aDO NOT WRITE IN THIS AREAThe angle between the rope and the plank is α, where tan α DO NOT WRITE IN THIS AREAA small block of mass 3M is placed on the plank at the point P, where AP x.The plank is in equilibrium in a vertical plane which is perpendicular to the wall.DO NOT WRITE IN THIS AREAA plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall.The plank is held in a horizontal position by a rope. One end of the rope is attached to the plankat B and the other end is attached to the wall at the point C, which is vertically above A.DO NOT WRITE IN THIS AREAFigure 3(3)The magnitude of the horizontal component of the force exerted on the plank at A by thewall is 2 Mg.(b) Find x in terms of a.(2)The force exerted on the plank at A by the wall acts in a direction which makes anangle β with the horizontal.(c) Find the value of tan β(5)The rope will break if the tension in it exceeds 5 Mg.(d) Explain how this will restrict the possible positions of P. You must justify youranswer carefully.28*P58350A02836*(3)
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MPH10.DO NOT WRITE IN THIS AREAAα3m2mTFigure 4A boy throws a ball at a target. At the instant when the ball leaves the boy’s hand at thepoint A, the ball is 2 m above horizontal ground and is moving with speed U at an angle αabove the horizontal.DO NOT WRITE IN THIS AREAU(2)The point T is at a horizontal distance of 20 m from A and is at a height of 0.75 m abovethe ground. The ball reaches T without hitting the ground.(b) Find the size of the angle α(c) State one limitation of the model that could affect your answer to part (b).(d) Find the time taken for the ball to travel from A to T.(9)DO NOT WRITE IN THIS AREA2g.sin 2 αDO NOT WRITE IN THIS AREA(a) show that U 2 DO NOT WRITE IN THIS AREAUsing the model,DO NOT WRITE IN THIS AREAIn the subsequent motion, the highest point reached by the ball is 3 m above the ground.The target is modelled as being the point T, as shown in Figure 4.The ball is modelled as a particle moving freely under gravity.(1)(3)32*P58350A03236*
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MPHQuestion 10 continuedDO NOT WRITE IN THIS AREADO NOT WRITE IN THIS AREA(Total for Question 10 is 15 marks)TOTAL FOR SECTION B IS 50 MARKSTOTAL FOR PAPER IS 100 MARKS36*P58350A03636*DO NOT WRITE IN THIS AREA
Paper 3: Statistics and Mechanics Friday 15 June 2018 – Afternoon Time: 2 hours 9MA0/03 You must have: Mathematical Formulae and Statistical Tables, calculator Pearson Edexcel Level 3 GCE. MPH. marksphysicshelp. 2 *P58350A0236* T TE TS AEA T TE TS AEA T TE TS AEA T TE TS AEA T TE TS AEA T TE TS AEA
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The Pearson Edexcel Level 3 Advanced GCE in Economics A is designed for use in schools and colleges. It is part of a suite of GCE qualifications offered by Pearson. Purpose of the specification This specification sets out: the objectives of the qualification any other qualification | 5,034 | 18,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.848123 |
http://www.alternatievewiskunde.nl/article/SUNA11.txt | 1,553,267,146,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-13/segments/1552912202671.79/warc/CC-MAIN-20190322135230-20190322161230-00277.warc.gz | 229,221,742 | 2,665 | sci.math.research SUNA11, Theorem Preliminaries ============================= This article is part of the "Series on Unified Numerical Approximations" UNIFICATION OF FINITE VOLUME AND GALERKIN METHODS FOR CONSERVATION EQUATIONS, IN THE TWO-DIMENSIONAL CASE PART I : PRELIMINARIES Consider the following partial differential equation: dQx/dx + dQy/dy = 0 ("d" denotes partial differentiation) Here (Qx,Qy) can be interpreted as the heat-flow in a two-dimensional medium or as the velocities (u,v) in a 2-D flow field. According to the Finite Domain technique, a kind of Finite Difference Method, this equation is first integrated over a certain control volume/area: // || (dQx/dx + dQy/dy) dx.dy = 0 // Partial integration (Green's theorem) results in an equivalent formulation, which contains a line-integral: / O (Qx.dy - Qy.dx) (counterclockwise) / According to the (Galerkin) Finite Element Method, the conservation equation must first be multiplied by a weighting function F , and then integrated over a finite element area, according to: // || F.(dQx/dx + dQy/dy) dx.dy = 0 // Partial integration results in an expression with line-integrals, all of which become part of the boundary conditions, and a term for the bulk material: // - || (dF/dx.Qx + dF/dy.Qy) dx.dy (: mind the minus sign) // For the sake of generality, both the Finite Volume and the Galerkin formulation are assumed to be applicable at a curvilinear grid, consisting of quadrilateral shaped cells or elements. For the sake of simplicity, we wish to restrict ourselves to regular molecules and elements, such as the classical five point star, and square quadrilaterals. The general curvilinear grid is mapped upon a regular grid by a transformation of coordinates x(h,k) , y(h,k) , in such a way that the _whole_ problem may be assumed to be regular in the (h,k) plane. This is the Finite Volume point of view, essentially a _global_ transformation. From a Finite Element point of view, however, such coordinate transformations are always carried out _locally_/element-wise: by isoparametrics. Restrictions of orthogonality do not apply in a Finite Element context. In either case, integrands of the Finite Volume and the Galerkin formulations must be transformed first into such regular (h,k) coordinates. The vector (Qx,Qy) is expressed in components at the (h,k) coordinate system as follows: | Qx | | dx/dh | | dx/dk | | | = Qh.| | + Qk.| | | Qy | | dy/dh | | dy/dk | The inverse of this transformation is the one that's needed: J.Qh = + dy/dk.Qx - dx/dk.Qy J.Qk = - dy/dh.Qx + dx/dh.Qy Where: J = dx/dh.dy/dk - dx/dk.dy/dh (assume that J > 0 ). Differentials of the global coordinates are transformed as follows: dx = dx/dh.dh + dx/dk.dk (with proper interpretation of the dy = dy/dh.dh + dy/dk.dk partial and non-partial "d's") Last but not least: dT/dh = dT/dx.dx/dh + dT/dy.dy/dh dT/dk = dT/dx.dx/dk + dT/dy.dy/dk Solving for dT/d(x,y) by Cramer's rule: J.dT/dx = dy/dk.dT/dh - dy/dh.dT/dk J.dT/dy = dx/dh.dT/dk - dx/dk.dT/dh The integrand of the Finite Volume formulation can now be written as follows: Qx.dy - Qy.dx = Qx.[ dy/dh.dh + dy/dk.dk ] - Qy.[ dx/dh.dh + dx/dk.dk ] = [ Qx.dy/dh - Qy.dx/dh ].dh + [ Qx.dy/dk - Qy.dx/dk ].dk = J.[ Qh.dk - Qk.dh ] Which gives a remarkebly simple result: || / / || O (Qx.dy - Qy.dx) = O J.(Qh.dk - Qk.dh) || / / Same kind of procedure for the Galerkin integral: { dF/dx.Qx + dF/dy.Qy }.dx.dy = { J.dF/dx.Qx + J.dF/dy.Qy }.dh.dk = { [ dy/dk.dF/dh - dy/dh.dF/dk ].Qx + [ dx/dh.dF/dk - dx/dk.dF/dh ].Qy }.dh.dk = { dF/dh.[ dy/dk.Qx - dx/dk.Qy ] + dF/dk.[ dx/dh.Qy - dy/dh.Qx ] }.dh.dk = { dF/dh.J.Qh + dF/dk.J.Qk }.dh.dk Same kind of remarkebly simple result: || // // || - || (dF/dx.Qx + dF/dy.Qy) dx.dy = - || J.(dF/dh.Qh + dF/dk.Qk) dh.dk || // // In order to shorten notation, we shall make everywhere the substitution: Qh := J.Qh Qk := J.Qk Resulting in the two localized integrals: / // O (Qh.dk - Qk.dh) ; - || (dF/dh.Qh + dF/dk.Qk) dh.dk / // F.D. F.E. To be continued ... - * Han de Bruijn; Applications&Graphics | "A little bit of Physics * No * TUD Computing Centre; P.O. Box 354 | would be NO idleness in * Oil * 2600 AJ Delft; The Netherlands. | Mathematics" (HdB). * for * E-mail: Han.deBruijn@RC.TUDelft.NL --| Fax: +31 15 78 37 87 ----* Blood | 1,348 | 4,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-13 | latest | en | 0.859989 |
https://www.tutorialspoint.com/what-is-loop-testing-methodology-example | 1,660,758,308,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573029.81/warc/CC-MAIN-20220817153027-20220817183027-00622.warc.gz | 875,890,618 | 8,945 | # What is Loop Testing(Methodology, Example)?
Software TestingTesting ToolsAutomation Testing
Loop Testing is a kind of software testing that focuses exclusively on the correctness of loop constructions. It is a component of Control Structure Testing (path testing, data validation testing, condition testing).
Loop testing is an example of white-box testing. This approach is used to test software loops.
## Types of Loop Tests
The following are some examples of loop tests −
• Simple loop
• Nested loop
• Concatenated loop
• Unstructured loop
## Why Loop Testing?
Following are some of the reasons why loop testing is performed −
• Testing can help to resolve loop recurrence concerns.
• Loop testing can indicate constraints in efficiency and operations.
• The loop's uninitialized variables can be identified by testing loops.
• It aids in the identification of loop initiation issues.
## How to do Loop Testing − Complete Methodology?
It must be tested at three separate stages within the testing loop −
• When the loop is activated.
• When the loop is executed.
• When the loop is terminated.
The following is the testing technique for all of these loops −
### Simple Loop
The following is how a simple loop is tested −
• Ignore the entire loop.
• Make a single pass across the loop.
• Make a number of passes through the loop where a<b, n is the maximum limit of passes.
• Make b, b-1; b+1 passes through the loop, where "b" is the highest amount of passes through the loop allowed.
### Nested Loop
You must perform the following steps to create a nested loop −
• Adjust all the other loops to their smallest value and begin with the innermost loop.
• Initiate a simple loop test on the innermost loop and keep the outside loops at their smallest iteration parameter value.
• Conduct the test for the following loop and make your way outwards.
• Keep testing till the outermost loop is reached.
## Concatenated Loops
If two loops in a chained loop are free of one other, they are checked as simple loops; otherwise, they are tested as nested loops.
However, if the loop counter for one loop is utilized as the starting value for the others, the loops are no longer considered separate.
### Unstructured Loops
For unstructured loops, the architecture must be restructured to represent the use of structured programming techniques.
## Limitation in Loop testing
• Loop issues are especially common in low-level applications.
• The flaws discovered during loop testing are not significant.
• Numerous defects may be identified by the operating system, resulting in storage boundary breaches, identifiable pointer failures, and so on.
## Summary
• Loop testing is a type of White Box testing in software engineering. This approach is used to evaluate loops in the program.
• Loop testing can indicate constraints in functionality and operations.
• Loop issues are especially common in low-level applications.
Updated on 09-Jun-2021 12:49:22 | 605 | 2,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | latest | en | 0.88766 |
https://magoosh.com/gre/gre-math-a-rectangle-parallelogram-with-equal-area/ | 1,713,708,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00203.warc.gz | 322,626,119 | 73,473 | # GRE Math: A Rectangle & Parallelogram with Equal Area
Today, we’ll quickly look at how a rectangle and parallelogram can have the same area.
In the diagram to the above, the parallelogram and rectangle share a vertex (D), one vertex of the rectangle (E) is on a side of the parallelogram, and one vertex of the parallelogram (C) is on a side of the rectangle. That is enough information to guarantee that the rectangle and parallelogram have equal area.
Here’s an argument why. In the diagram below, notice I have constructed segment EQ, which is perpendicular to CD. This segment is the height of the parallelogram, so that times the length of CD would be the area of the parallelogram.
Look at ΔDGC and ΔEQD. Those two triangles are similar. Why?
Well, first of all, ∠QDG and ∠EDQ are complementary: they both add up to the 90° angle of ∠EDG. Also, ∠QDG and ∠QCG are complementary, because they are the acute angles of a right triangle. Since ∠EDQ and ∠QCG are both complementary to the same angle (∠QDG), they are congruent: ∠EDQ≅ ∠QCG.
Since we know ∠EDQ≅ ∠QCG and we know ∠EQD≅ ∠G (both right angles), we know two angles in ΔDGC are congruent to two angles in ΔEQD. By the AA Similarity Theorem, they must be similar triangles.
ΔDGC ~ ΔEQD
Similar triangles have proportional sides. In particular, we can set up a proportion:
After cross-multiplying, we get two equal products. (ED)*(DG) = the area of the rectangle. (EQ)*(DC) = the area of the parallelogram. Therefore, those two areas are equal.
That’s a more equation-based way of proving the areas equal. Here’s another HUGE idea, which is much more appealing for visual thinkers. Imagine extending one pair of sides in a parallelogram like railroad rails:
If we slide either side along its “rail”, the shape of the parallelogram will change, but the area will stay the same, because the base (the length on the rail) doesn’t change, the height (the distance between the parallel lines) doesn’t change.
The purple and the orange parallelograms must have exactly the same area. That’s a HUGE geometry idea. | 533 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-18 | latest | en | 0.895368 |
https://fifthgradewinterhaven.org/2010/01/10/winterhaven-school-5th-grade-student-web-page-2-7/ | 1,638,703,818,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363157.32/warc/CC-MAIN-20211205100135-20211205130135-00139.warc.gz | 324,752,179 | 20,146 | ## Winterhaven School 5th Grade Student Web Page (2.7)
IMPORTANT DATES:
January
18- Martin Luther King, Jr. Day- NO SCHOOL
February
1- Teacher Planning Day- NO SCHOOL
4- Field Trip OMSI (Subject to approval)
Classroom topics for week beginning January 11, 2010 ( 二〇十年 一月 十一日
Chautauqua:
The Medicine Wheel
Math: Investigation 2-Polygons and Angles
Developing Angle Benchmarks– Students play a game on circular grids.This helps them to develop angle sense using rotations of 30 degree and 45 degree and their multiples.
Using an Angle Ruler- Students practice using an angle ruler (a movable protractor) to measure angles of some of the shapes in the Shapes Set.
Analyzing Measurement Errors-
Students analyze the effects of measurement error in computing the angle Amelia Earhart needed to fly to reach her next destination.
Angles and Parallel Line-
Students explore some interesting patterns among the angles created when two or more parallel lines are cut by a line.
Humanities:
Civics and Government-The founding of America continued.
Socratic Seminar: Analysis and significance of I Have A Dream Speech, by Dr,. Martin Luther King, Jr.
Systems Thinking:
Government as a system
Science: Constructing a light bulb model and understanding electrical resistance
Japanese: Lesson 15 Hiragana, Lessons 3 Conversation.
HOMEWORK
Math Investigation 2:
ACE, 1–3 required, 27-29 extra credit, due Tuesday
ACE, 4-9 required, 30-36, 41 extra credit, due Wednesday
ACE, 10-17 required, 37 extra credit, due Thursday
Japanese: Hiragana Lesson 15, study for Friday’s test.
Pen pal letter:
Respond to what you learned about your pen-pal in the last letter. Mention things you have in common as well as differences. Tell them how you celebrated the holidays. Any trips, sledding, family gatherings and traditions, etc… Due Thursday
Japanese Practice From Chapter 3
## いま ８じです。 ひるやすみは １２じはんから １じはんまでです。
Japanese Practice From Chapter 15
## １。えきへ いきます。 ２。がっこうへ いきます。 ３。きょうとへ いきました。 ４。わたしは うちへ かえります。 ５。がっこうから こうえんへ きました。
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 557 | 2,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-49 | latest | en | 0.74479 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2001/Hays/WriteUp01/solution.html | 1,539,956,372,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512400.59/warc/CC-MAIN-20181019124748-20181019150248-00270.warc.gz | 190,631,456 | 2,200 | ## is tangent to each of f(x) and g(x) at two distinct points.
(This is problem 3 of Assignment 1.)
The solution will look something like this.
To make things simple, let's try to make a parabola that has a vertex on the y-axis. This means that the parabola will be symmetric about the y-axis. To make the graphs of our linear functions look like the lines in the picture above, the slopes of our lines need to have equal magnitudes but opposite signs. Therefore, we know that a = -c. Thus, we also know that the parabola opens down.
In the picture above, we also see that the two lines cross directly above the vertex of the parabola. Since we are choosing the vertex to be on the y-axis, we know that these two lines will cross each other on the y-axis. Therefore, their y-intercepts must be the same, or b = d.
Since we know a = -c and b = d, we can how write
We can differentiate h to find the slope of the tangent lines.
Now we know the slope of the tangent lines from taking the derivative of h, but we also know that the slope of the two tangent lines are a and -a, respectively. Thus, we now have
-2a2x1 = a,
where x1 is the x value of the point where the parabola and one of the lines intersect. Dividing through by a yields
-2ax1 = 1 (1)
We can find x1 by setting the equation for the parabola and the equation for the line equal to each other and solving for x1.
(ax1 + b) (-ax1 + b) = ax1 + b
Dividing through by ax1 + b yields
-ax1 + b = 1
-ax1 = 1 - b
x1 = (b-1)/a
Substituting this into equation (1), we see
-2a(b-1)/a = 1
-2a(b-1)/a = 1
-2ab + 2 = 1
b = 1/2
Now we have a family of functions that satisfy the conditions listed above.
f(x) = a x + 1/2
and
g(x) = - a x + 1/2.
An Example
Here is an example when a = 1.
These two lines appear to be tangent to the parabola, but let's check.
f(x) = x + 1/2, and g(x) = - x + 1/2. Therefore, h(x) = - x2 + 1/4.
The graphs of the functions f and h intersect at the point (-1/2, 0). The derivative of h at x = -1/2 is 1, which is the same as the coefficient of x in f.
The graphs of the functions g and h intersect at the point (1/2, 0). The derivative of h at x = 1/2 is -1, which is the same as the coefficient of x in g.
Thus, our functions satisfy the conditions listed above.
A Few More Examples...
Let a = 1/2. Then
Let a = 2. Then
Extensions
I will not answer these questions in this write-up, but they are good extensions to this problem.
1. Must the tangent lines have slopes of equal magnitude but opposite sign?
2. Must the tangent lines cross directly above the vertex of the parabola?
3. Must the parabola open down?
4. Must the parabola be symmetric about the y-axis?
Write-Up 1 | Alison's web page | Student Web Pages | EMAT 6680 Page | 793 | 2,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2018-43 | latest | en | 0.935309 |
https://puzzling.stackexchange.com/questions/67646/me-i-am-circumference/67656 | 1,563,429,433,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00096.warc.gz | 504,636,942 | 36,056 | # Me, I am circumference
Here is a puzzle in the same form of my previous puzzles, here and here.
Me, I am circumference,
My anagram is what you should do,
My anagram's homophone is what you do in school,
My anagram's other homophone is a ceremony, often in religion.
You are
GIRTH - the measurement around the middle of something, AKA circumference
My anagram is what you should do,
You should do what is RIGHT
My anagram's homophone is what you do in school,
In school, you WRITE a lot of things
My anagrams other homophone is a ceremony, often in religion
A religious ceremony can be called a RITE
@QuantumTwinkie (using your anagram's other homophone's anagram):
I hope you never TIRE of making these puzzles! =D
• That is the correct answer :D. – QuantumTwinkie Jun 28 '18 at 16:21
• @QuantumTwinkie Nice, thank you for the puzzle! – hagfy Jun 28 '18 at 16:48
• Ha ha ha, I like the edit! @hagfy – QuantumTwinkie Jun 28 '18 at 16:53
Totally sacrificing the appropriate meaning of "homophone", this was my first idea, heh.
You are
pi*d, circumference = pi * diameter
My anagram is what you should do,
You should dip—your chips. You're bland if you don't.
My anagram's homophone is what you do in school,
Dipping class means ditching class.
My anagrams other homophone is a ceremony, often in religion
Baptism is getting dipped in water.
• This is a very clever answer! ha ha ha :D – QuantumTwinkie Jun 29 '18 at 0:00
• Nice. +1, and welcome to Puzzling.SE. – Chowzen Jun 29 '18 at 0:27
• So, if it were dπ, would it be "de apple pie", or "de steak 'n' kidn'y pie"? – Chronocidal Jun 29 '18 at 7:51 | 475 | 1,627 | {"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} | 2.65625 | 3 | CC-MAIN-2019-30 | longest | en | 0.911666 |
https://www.bestfreenow.com/article/1927605.html | 1,675,714,659,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00737.warc.gz | 663,644,594 | 7,476 | # opticampus thickness
## OptiCampus OptiCampus Optical Calculators
OptiCampus Optical Calculators. Lens Tilt & Wrap Compensation. Crossed Cylinders Calculation. Vertex Distance Compensation. Spectacle Magnification Calculation. Lens Thickness Calculation. Vertical Imbalance Compensation. Single Vision Blank Size Calculation. Compounding Prisms Calculation.
## OptiCampus Continuing Education Course
Typical minimum thickness guidelines for traditional ophthalmic lenses range from a minimum edge thickness of 1 mm to a minimum center thickness of 2 mm.
## OptiCampus OptiCampus Optical Calculators
OptiCampus Online Optical Continuing Education. Lens Tilt & Wrap Compensation. Fill in the input fields and then press the Calculate button in order to compute the approximate prescription necessary to compensate for the effects of pantoscopic tilt and facial wrap. Note that these fields represent the tilt of the actual lens relative to the line of sight, not the tilt of the frame.
## OptiCampus OptiCampus Optical Calculators
OptiCampus Online Optical Continuing Education. Crossed Cylinders Calculation. Fill in the input fields and then press the Calculate button in order to compute the resultant spherocylindrical prescription that is produced by combining—or crossing—two individual prescriptions (such as for an overrefraction).
## OptiCampus Continuing Education Courses
OptiCampus Online Optical Continuing Education. Choose a Continuing Education Course. Select a continuing education course to review from the list below.
## Opticampus OptiBoard
Through his OptiCampus website, which was free and open to all to use, he shared information on ophthalmic lens design, progressive lens design, the nature of light and optics, wavefront aberrations, and a host of other technical lens subjects.
## Opticare, OptiEdge Edge Thickness Estimation tool
LEFT. Note: The calculator results are estimates only. For minus lens the result is for edge thickness. For plus lens the result is for centre thickness. For minus stock lens use CT =
## The Lens Formula 20/20 Mag
Added thickness means added weight in all lenses. For strongest meridians that fall close to 90° or straight up and down, avoid deep (i.e. large B measurement) frames. The effective diameter (ED) of a lens shape is defined as twice the longest radius of the shape. | 489 | 2,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.785371 |
https://www.scienceforums.net/profile/45964-mathematic/ | 1,670,554,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00452.warc.gz | 1,031,785,880 | 14,587 | mathematic
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107
1. Mathematics is Inconsistent!
The paradox is in pure mathematics. A physical object cannot be decomposed into non-measurable pieces.
2. Million equivalent in different !base-10 system ?
The base doesn't matter. It'a \\;s a language issue. What is big?
4. Distance conversion
Look for 'distance converter on line. Pick one.
F=1.8C+32.
C=5(F-32)/9
8. wormholes and warp drives
Where did you get these ideas?
9. Applications of 4 fundamental forces
Strong and weak forces are microscopic - no control? Gravity is affected by moving stuff around.
10. Is math really just counting one and two?
You seem to be jumping from math ro physics. Math starts with arithmetic, next elementary algebra, plane geometry, etc. From your description you need a start at the very beginning.
11. How to use the change of variables technique here?
Change (x,y) to polar coordinates.
12. Binary numbers--- what does "base-2" mean?
base digits 2......0,1 3......0,1,2 4......0,1,2,3 etc.
13. The Rewriting of Particle Physics....possibly:
Disagreement with other measurements. Error here?
14. The speed of gravity = the speed of light. Does that mean gravity and light are the same thing?
Incorrect assertion: "The only thing that can travel at the speed of light is light, and if gravity can travel at the speed of light, then it's almost definitely the same thing as light. Also, there's no gravity or light in outer space. Almost certainly a sign that gravity and light are the same particle. " "it's almost definitely the same thing as light." Where did you get this? "there's no gravity or light in outer space" False
15. Alone again?
It is too soon to say. There are about 100 billion stars in the milky way. Only an infinitesimal number has been explored to determine whether life supporting planets exist.
×
×
• Create New... | 477 | 2,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.915158 |
http://www.warren-wilson.edu/~dcollins/PhysPhotOfWeek/2013PPOW/20130405CrossCheckRows/index.html | 1,406,630,580,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510267075.55/warc/CC-MAIN-20140728011747-00227-ip-10-146-231-18.ec2.internal.warc.gz | 933,793,779 | 3,734 | # Physics Photo of the Week
## April 5, 2013
Cross Checked Rows
California grows extensive row crops (grapes, peaches, almonds) on large plots of level land. The rows and columns of fruit and nut plantings are planted and pruned to very even tolerances to enable mechanized harvesting, trellising, and even irrigation (notice the drip hose also suspended by the trellis). With cross-checked rows one notices many different diagonals. If the space between rows and columns are equal, the most obvious diagonal is 45 degrees - the pathway leading away from the camera in the right center. The diagonals are even given names for their application in crystallography - the three-dimensional structure of atoms arranged in a regular lattice. This main 45 degree diagonal is called the (1,1) direction because the diagonal is formed by "sideways one unit, depthways one unit". The diagonal in the center of the image is the (2,1)
diagonal (sideways two, depthways one). The next diagonals in the series are (3,1), (4,1), etc. In between some of these (n,1) series of diagonals are higher order diagonals such as (3,2) or (5, 3). If the size of the field of the row crop were infinite, and the plants or posts were perfectly straight and narrow, there would be no limit to the number of diagonals visible. One sees the same effect in empty theater seats, military grave plots, or any regular array of objects.
What is amazing about viewing the cross checked rows viewed from a moving car window (at 60 miles/hour) is that although the individual plants or posts are a blur, an infinite array of diagonals are still visible in the second image, and even more pronounced. This blurred image was taken from a long-enough time exposure such that all the individual trunks of this nut orchard are just a blur, yet many diagonal paths are visible - similar to interference of waves creating standing patterns as the waves are moving. No matter where the camera is positioned, the camera will always see the (3,2) diagonal in the same direction. One series of diagonals has been labeled beginning with (2,1) on the left and progressing with (n, n+1) series which eventually becomes the (1,1) diagonal - the convergence of the series for very large n. The identity of the diagonals was verified by carefully measuring the position angles of these diagonals within the digital image with geometry. Of course many more series are visible in the blurred image.
Physics Photo of the Week is published weekly during the academic year on Fridays by the Warren Wilson College Physics Department. These photos feature interesting phenomena in the world around us. Students, faculty, and others are invited to submit digital (or film) photographs for publication and explanation. Atmospheric phenomena are especially welcome. Please send any photos to dcollins@warren-wilson.edu.
All photos and discussions are copyright by Donald Collins or by the person credited for the photo and/or discussion. These photos and discussions may be used for private individual use or educational use. Any commercial use without written permission of the photoprovider is forbidden.
Observers are invited to submit digital photos to: | 697 | 3,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.928977 |
https://www.coursehero.com/file/6346569/OPS-571-wk2The-principle-of-the-learning-curve-theory-is-process-improvement/ | 1,490,374,935,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188213.41/warc/CC-MAIN-20170322212948-00164-ip-10-233-31-227.ec2.internal.warc.gz | 901,767,284 | 94,780 | OPS 571 wk2The principle of the learning curve theory is process improvement
# OPS 571 wk2The principle of the learning curve theory is process improvement
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Pizza Store Layout Simulation 1 Pizza Store Learning Curve Theory Carl R. Foster November 2, 2010 University of Phoenix
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Pizza Store Layout Simulation 2 Pizza Store Layout The learning curve theory is process improvement and concerned with that when a new job, process or activity starts for the first time, the probability of achieving maximum operation efficiency immediately is remote. In pizza operations, process improvement occurs as a result of the repetitive nature of the tasks and as confidence builds, the result is a more efficient, and productive operation. The improvements occur on an individual and organizational level. “Learning curve theory is based on three assumptions: 1. The amount of time required to complete a given task or unit of a product will be less each time the task is undertaken. 2. The unit time will decrease at a decreasing rate. 3. The reduction in time will follow a predictable pattern” (Chase, Jacobs, Aquilano, 2006). The Pizza Store Layout performance data is as follows: Weeks: No. of Customers Group of 2 No. of Customers Group of 4 Avg. Wait Time (Min) Avg. Queue Length Profit (\$)
Pizza Store Layout Simulation 3 3-4 71 105 4.96 2.50 1,477 5-6 67 109 4.32 2.26 1,695 7-8 96 143 3.32 2.72 2,040 As indicated above, definite improvement in average waiting time for customers from
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## This note was uploaded on 07/28/2011 for the course OPS 571 taught by Professor Conques during the Spring '10 term at University of Phoenix.
### Page1 / 5
OPS 571 wk2The principle of the learning curve theory is process improvement
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Ask a homework question - tutors are online | 511 | 2,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.895954 |
http://mathhelpforum.com/advanced-applied-math/4942-possible-system-have-negative-potential-energy.html | 1,495,672,221,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607871.54/warc/CC-MAIN-20170524230905-20170525010905-00343.warc.gz | 220,134,976 | 10,187 | Thread: Is it possible for a system to have negative potential energy?
1. Is it possible for a system to have negative potential energy?
Is it possible for a system to have negative potential energy?
2. Originally Posted by Candy
Is it possible for a system to have negative potential energy?
Yes. All potential energies have an arbitrary "zero point:" the point we designate a zero value for the potential energy. Consider gravitational potential energy. What is the GPE of a ball on the ground? We need to set a coordinate system, so +y upward and the origin at the ground. The GPE of the ball is 0 J. Now let's take our origin to be the top of a tall building. Since upward is positive the y coordinate of the ball is negative. Thus the GPE is also negative.
We generally need a zero point for any PE, but note that all Physics really cares about is changes in the PE as an object moves from one point to another. So ultimately our choice of origin is whatever we feel like...the choice of origin doesn't affect the problem.
-Dan | 224 | 1,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-22 | longest | en | 0.941536 |
https://www.emathhelp.net/notes/pre-algebra/factors-and-multiples/ | 1,718,662,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00395.warc.gz | 671,370,668 | 5,321 | Category: Factors and Multiples
Divisibility of Integers
When we talked about division of integers, we assumed that result of the division is an integer number.
From another side, when we talked about division with remainder, we discovered that result of the division is not always an integer number.
Even Numbers (Integers)
A number (integer) is even if it is divisible by 2.
Example 1. Determine whether 10 is an even number.
10 is even because $\frac{{10}}{{2}}=5$ and 5 is an integer number.
Another example.
Example 2. Determine whether 15 is even.
Odd Numbers (Integers)
Number (integer) is odd if it is NOT divisible by 2.
Alternatively, number is odd if it is not even.
Example 1. Determine whether 24 is odd number.
24 is even because $\frac{{24}}{{2}}={12}$ and 12 is integer number. Thus, 24 is not odd number.
Divisibility Rules
In general it is hard to determine from first glance whether one number is divisible by another. We need to perform division.
However, in some cases, we can determine whether one number divides another without dividing them by applying divisibility rules.
What are Factors and Multiples
Let's see what are factors and multiples.
When we talked about divisibility of integers, we said that number ${a}$ is divisible ${b}$ if ${c}=\frac{{a}}{{b}}$ is integer number.
If ${a}$ is divisible by ${b}$, then ${b}$ is a factor of ${a}$ and ${a}$ is multiple of ${b}$.
Integer Factorization
When we talked about factors and multiples we learned how to find factors of number.
This allows to write number as product of its factors.
Recall that factors of 12 are 1,2,3,4,6,12.
So, we can write that ${12}={1}\cdot{12}$, ${12}={2}\cdot{6}$ and ${12}={3}\cdot{4}$.
What is a Prime Number
Prime number is an integer number greater than 1 that has only two factors: 1 and itself.
For example, 5 is prime number because it has no other factors except 1 and 5.
By convention 1 is not prime, so prime numbers start from 2.
Composite Numbers
Composite number is a number, that is not prime.
For example, 6 is composite number because its has other factors except 1 and 6, namely, 2 and 3.
By convention 1 is not composite.
This means that 1 is neither prime, nor composite.
How do you do Prime Factorization
When we talked about integer factorization we noticed that integer number can have more than one factorization.
For example, ${18}={1}\cdot{18}={2}\cdot{9}={3}\cdot{6}={2}\cdot{{3}}^{{2}}$.
However, if we use only prime numbers as factors than such factorization is unique (ignoring order of factors).
Greatest Common Divisor (GCD)
Suppose we are given two numbers 18 and 24.
Let's find their factors.
18: 1,2,3,6,9,18.
24: 1,2,3,4,6,8,12,24.
As can be seen some factors are same for both numbers (they are in bold). These numbers are called common factors (divisors) of 18 and 24.
Least Common Multiple (LCM)
Suppose we are given two numbers 18 and 24.
Let's find some of their multiples.
18: 18,36,54,72,90,108,126,144,...
24: 24,48,72,96,120,144,168,...
As can be seen some factors are same for both numbers (they are in bold: 72 and 144). These numbers are called common multiples of 18 and 24. | 847 | 3,180 | {"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} | 4.875 | 5 | CC-MAIN-2024-26 | latest | en | 0.918531 |
http://sigmaquality.pl/data-plots/perfect-plots_-matrix-of-corelation-281120190000/ | 1,722,644,971,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00644.warc.gz | 30,842,619 | 234,126 | # Perfect Plots_ Matrix of corelation
## Combined Cycle Power Plant Data Set¶
#### Data Set Information:¶
The dataset contains 9568 data points collected from a Combined Cycle Power Plant over 6 years (2006-2011), when the power plant was set to work with full load. Features consist of hourly average ambient variables Temperature (T), Ambient Pressure (AP), Relative Humidity (RH) and Exhaust Vacuum (V) to predict the net hourly electrical energy output (EP) of the plant.
A combined cycle power plant (CCPP) is composed of gas turbines (GT), steam turbines (ST) and heat recovery steam generators. In a CCPP, the electricity is generated by gas and steam turbines, which are combined in one cycle, and is transferred from one turbine to another. While the Vacuum is colected from and has effect on the Steam Turbine, he other three of the ambient variables effect the GT performance.
For comparability with our baseline studies, and to allow 5×2 fold statistical tests be carried out, we provide the data shuffled five times. For each shuffling 2-fold CV is carried out and the resulting 10 measurements are used for statistical testing.
We provide the data both in .ods and in .xlsx formats.
#### Attribute Information:¶
Features consist of hourly average ambient variables
• Temperature (T) in the range 1.81°C and 37.11°C,
• Ambient Pressure (AP) in the range 992.89-1033.30 milibar,
• Relative Humidity (RH) in the range 25.56% to 100.16%
• Exhaust Vacuum (V) in teh range 25.36-81.56 cm Hg
• Net hourly electrical energy output (EP) 420.26-495.76 MW
The averages are taken from various sensors located around the plant that record the ambient variables every second. The variables are given without normalization.
In [1]:
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
del df['Unnamed: 0']
df.columns = ['Temperature', 'Exhaust_Vacuum', 'Ambient_Pressure', 'Relative_Humidity', 'Energy_output']
df.sample(3)
Out[1]:
Temperature Exhaust_Vacuum Ambient_Pressure Relative_Humidity Energy_output
7071 21.98 59.39 1015.25 84.52 446.79
1815 14.12 42.86 1011.84 88.29 471.86
5227 23.14 58.18 1008.89 81.82 444.51
In [2]:
sns.set(style="ticks")
corr = df.corr()
f, ax = plt.subplots(figsize=(12, 6))
cmap = sns.diverging_palette(180, 90, as_cmap=True)
square=True, linewidths=.9, cbar_kws={"shrink": .9})
Out[2]:
<matplotlib.axes._subplots.AxesSubplot at 0x7fd267636d10>
In [3]:
del df2['Unnamed: 0']
del df2['Unnamed: 0.1']
Out[3]:
age job marital education default housing loan contact month day_of_week campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
0 44 blue-collar married basic.4y unknown yes no cellular aug thu 1 999 0 nonexistent 1.4 93.444 -36.1 4.963 5228.1 0
1 53 technician married unknown no no no cellular nov fri 1 999 0 nonexistent -0.1 93.200 -42.0 4.021 5195.8 0
2 28 management single university.degree no yes no cellular jun thu 3 6 2 success -1.7 94.055 -39.8 0.729 4991.6 1
3 39 services married high.school no no no cellular apr fri 2 999 0 nonexistent -1.8 93.075 -47.1 1.405 5099.1 0
4 55 retired married basic.4y no yes no cellular aug fri 1 3 1 success -2.9 92.201 -31.4 0.869 5076.2 1
5 rows × 21 columns
In [4]:
sns.set(style="ticks")
corr = df2.corr()
f, ax = plt.subplots(figsize=(22, 10))
cmap = sns.diverging_palette(580, 10, as_cmap=True)
square=True, linewidths=.9, cbar_kws={"shrink": 0.8})
Out[4]:
<matplotlib.axes._subplots.AxesSubplot at 0x7fd2910ce590>
### Definition¶
In [5]:
def matrix_plot(df,title):
sns.set(style="ticks")
corr = df2.corr()
f, ax = plt.subplots(figsize=(22, 10))
#cmap = sns.diverging_palette(580, 10, as_cmap=True)
cmap = sns.diverging_palette(180, 90, as_cmap=True) #Inna paleta barw
square=True, linewidths=.9, cbar_kws={"shrink": 0.8})
plt.xticks(rotation=90)
plt.title(title,fontsize=22,color='#0c343d',alpha=0.5)
plt.show
In [6]:
matrix_plot(df2, 'Perfect Plots: Matrix of corelation')
## Definition by class¶
In [7]:
class mx_plot:
def __init__(self,df,title):
self.df = df
self.title = title
def matrix(self):
sns.set(style="ticks")
corr = df2.corr()
f, ax = plt.subplots(figsize=(22, 10))
#cmap = sns.diverging_palette(580, 10, as_cmap=True)
cmap = sns.diverging_palette(580, 10, as_cmap=True) #Inna paleta barw | 1,310 | 4,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.844179 |
https://corbettmaths.com/2016/07/16/textbook-answers-lcmhcf-using-product-of-primes/ | 1,696,015,120,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00557.warc.gz | 204,427,797 | 12,547 | # LCM/HCF using Product of Primes Textbook Answers
Workout
Question 1:
(a) 105
(b) 154
(c) 105
(d) 99
(e) 60
(f) 90
(g) 80
(h) 120
(i) 144
(j) 78
(k) 150
(l) 144
(m) 168
(n) 180
(o) 360
(p) 630
Question 2:
(a) 7
(b) 5
(c) 6
(d) 9
(e) 15
(f) 14
(g) 30
(h) 16
(i) 8
(j) 1
(k) 12
(l) 19
(m) 11
(n) 28
(o) 45
(p) 32
Apply
Question 1(a) 420
1(b) 12
Question 2: 300
Question 3: 168 seconds
Question 4: 1pm or 13:00
Question 5: 6200
Question 6: 26 and 65
Question 7: The highest common factor of 20 and 21 is 1.
Question 8: 1200 coins each, so they have 2400 in total.
Question 9: It should be 4, since 2 x 2 = 4. Perhaps he just put the two 2’s beside each other to make 22 instead of multiplying them. | 365 | 789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.519932 |
http://tylerneylon.com/blog/2007_03_01_archive.html | 1,556,296,554,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578841544.98/warc/CC-MAIN-20190426153423-20190426175423-00406.warc.gz | 165,266,446 | 13,871 | thinking
Name: Tyler
Location: Mountain View, California, United States
thinking := [life, games, movies, philosophy, math, coding, pizza, &c.]
Monday, March 26, 2007
angle between subspaces
Someone asked me this question today, which turns out to have a very elegant solution if you know a "decent amount" of linear algebra (say, at least one advanced class about matrix computations):
Given two subspaces of Rn, what is the angle between them?
To be rigorous, we could state this as: how can you efficiently find the minimum angle between any two nonzero vectors, one in subspace A, and the other in subspace B? I choose the minimum here because I feel that it corresponds most closely with our intuition in 3D in the case of a plane vs. a line. Given a line that intersects a plane at a single point (maybe picture a sundial in your mind), we usually think of "the angle" between the two as the smallest angle. It also seems more distance-like than the maximum angle since, for example, we expect the "distance" between a subspace and itself to be zero, which is true for the min angle, but not necessarily the max.
I'll post what I think is a nice solution in a little bit. In the meantime, can anyone think of how to answer this?
Friday, March 16, 2007
Strings and knots
A friend of mine at work recently asked me the following problem, which was supposedly given as an interview question (not at my company):
Suppose n strings (physical strings, not character arrays!) are placed in a bowl so that we only see the ends of each string, and do not know which ends belong to the same string. Next we proceed by picking two ends at random and tying them together, and continue to do so until all the ends are tied. On average, how many loops will you end up with?
As an example of how it might work: if n=2, then there are 4 string ends left hanging out of the bowl. When we pick the first two ends and tie them together, either we are making a loop out of a single string, or we are tying two different strings together. Next we tie the last ends together (there are only two ends left so there is no choice). If our first knot was a self-knot that made a loop out of a single string, then there will end up being two loops in total. But if our first knot was between two different strings, then the remaining ends joined will result in a single long loop containing both strings. So the average in this case is somewhere between 1 and 2 (it's a fraction).
I will not give away the answer just yet! If anyone would like to know, just post a comment, and I'll put it up.
Update: I've written up the answer. If you've thought at least a little bit about the problem, then you have permission to check out the answer :)
:)
Monday, March 12, 2007
the subjective nature of causality
I would like to argue that the idea of something "causing" another thing is not an intrinsic property of the universe -- rather, it is a subjective property which is somewhat artificially assigned to the world by our own minds. It is a "useful fiction", much like the idea of a center of gravity in physics. So I do not mean to say that causality is nothing, but rather that it is an idea we use to simplify our understanding of the world, and which does not inherently exist outside our interpretations.
This is why: we feel that event x causes event y when
1. y would have not happened without x, and
2. we feel that "x almost didn't happen".
Both of these statements are purposefully vague, because the idea of causality is itself vague. The first point means that, if we imagine the world exactly the same, except that x did not happen, then it seems to us that y would also have (most likely) not happened. The second point means that, along the chain of events leading up to y, x was somehow among the least predictable. If the second point holds for several events as a possible cause (I'll give an example in a second), then we often feel that each of these events was also a cause of the ultimate event y.
In a moment I will argue that these points are dependent in a fundamental way on the ignorance of the one making the causality judgement, because if we "knew everything" then it would be hard (perhaps impossible) to imagine a world in which everything else was the same except that x didn't happen, and, further, it would be truly impossible to distinguish some event x which more "almost didn't happen" than any other past event. (I am assuming a deterministic world here -- and that is a matter for a different posting!)
Here are a couple examples to help get our mind around this intuitive "definition" of causality:
Example 1. Suppose someone named Boberta shoots an apple. In this very straightforward scenario, we might say that the bullet rather speedily touring the apple's interior is a cause of the apple's demise. But what is the "real reason" the apple was destroyed? I think most people would be more satisfied with saying that "Boberta shot the apple" is more of a "real" cause. Why is this a better answer? By the second point above, it is easier to imagine that Boberta didn't shoot the apple than it is to imagine that Boberta did shoot the apple yet the bullet did not destroy it. (Let's think of the shooting as being at very close range, so that aim is not a question.) Out of those two events, it would be less surprising if Boberta never shot the gun than if Boberta did shoot the gun and the bullet did not destroy the apple. This illustrates how the second point is important for our idea of causality. (By the way, if you are wondering why Boberta shot the apple, it's because the apple killed Boberta's father. It was a bad apple.)
Example 2. Now let us suppose that both Alfonzo and Boberta, with a merciless thirst for applejuice, simultaneous shoot the apple. In this case it intuitively feels as if either "Alfonzo shot the apple" or "Boberta shot the apple" alone are each not quite the cause of the apple's demise. In some sense, we feel a bit cheating if we do not mention both shootings. This is meant to illustrate the importance of point one -- if only one of these events didn't happen, the apple would still be maimed. Only when we imagine both shootings not taking place do we imagine the apple remaining, at least for a short while, peacefully intact.
So far I have offered some defense of my definition of causality. Now let me proffer the notion that "x caused y" is inevitably a subjective idea. This follows because our ability to imagine a slightly different world, in which either x did not happen and all else remained the same, or further our ability to estimate the probability with which x almost didn't happen, could never exist in the presence of a perfect knowledge of the world. That is, our ignorance is an integral part of our causality judgement.
Example 3. This one is a little more far-fetched in order to really capture ignorance as an almost essential property of someone in the situation. We've all seen The Matrix, our pop culture modern version of Plato's allegory of the cave. Suppose that in the future we are running a highly complex simulation of an entire world on a vast system of computers. This world is completely deterministic. And the creators of this simulated world have set it up so that at some point in time, the simulated planet of all these simulated people will be hit by a (simulated) comet, leading to the deaths of many (simulated) people. The question is: what is the cause of all these deaths? Within the simulated world, as far as these people ever know, the comet has caused this disaster. Certainly if the comet were not present (or had missed the planet), it would not have happened (point one of causality). And, for most people, it seems very easy to imagine a world in which a particular comet did not exist, or at least whose trajectory was a little bit different (point two of causality). Now let us recall that we know the entire world to be a simulation. In this example, we could also pretend that the entire simulation was set up by a rather sadistic fellow who enjoyed the idea of a planet being devastated by a comet. Now that we know more about the world, we can give a better-informed "real cause" of the destruction to be this programmer's decision to set up the simulated world in such a way.
The point of this example is to illustrate the role of ignorance in causality. No one in the simulated world could ever possibly know about the sadistic programmer unless someone in the real, non-simulated world decided to somehow interfere (and for the sake of the thought experiment we may suppose this does not happen). Hence within the simulated world, there is simply no way to ever know a better "cause" of the comet hit beyond the mere laws of physics within that world.
As far as we may ever know, we might as well be these simulated people, never knowing the complete story, but filling in those gaps in our minds by guessing at the small differences in which the world may evolve, and basing our causal links on our own internal model of many possibilities and our own idea of their probabilities. Yet these possibilities and probabilities could never exist with complete knowledge of the world, for there is but one world; no other possibilities, and no question of probability.
By the way, I suspect there's a lot of pre-existing philosophical work discussing similar ideas, and I'd love to hear about them! Also, thanks to my friend Lara for some useful brainstorming while thinking about this, and Rebecca for reminding me of Plato's somewhat-Matrixy allegory :)
Friday, March 09, 2007
the mean and the nice
The term "expected value" has a very precise meaning in probability theory, but I think it often clashes with what we as everyday humans would really "expect" from certain probabilistic situations.
To avoid confusion, I will refer to the traditional definition of expected value as the mean (this is also standard terminology [wiki]) and contrast it with a new idea which I will here refer to as the nice.
Let's see why the mean is not always what's expected. Consider a game of chance which you can play for free. Here's how it works: you flip 20 coins. If any of them are heads, you win a million dollars. If they're all tails, you owe 10 trillion dollars (yes, you're really screwed if you lose). Would you play this game? I would, and I think that many people would happily do the same. The chances you'll lose are less than one in a million (since (1/2)^(20) < 1/million). Yet the mean value is negative, since the loss of -10 trillion outweighs the far-more-likely win of a million. Intuitively speaking, what value do you really expect? I think most people would agree that they are least surprised by the outcome of winning a million dollars.
Hence I am hoping to mathematically capture this notion of the "least surprising" outcome, which we will call the nice value. Here are some properties the nice value should have:
• The values close to the nice value should be far more likely than those far away from the nice value.
• The nice value should be an actually possible value in the probability distribution.
• It should be understood that some nice values are "expected" with much less confidence than others.
I mention the last point because the above game is relatively easy to predict. Consider rolling a six-sided die. What should be the nice value? I think in that case, intuitively speaking, there is no best value. We expect any of the 6 numbers to show up with equal probability. So there are many probability distributions in which we have low confidence about the outcome. The motivation for a nice value is targeted toward probability distributions which are concentrated around a particular value.
a gambling strategy
Part of my motivation for thinking about the nice value is the following old "guaranteed-to-work" gambling strategy. Suppose you're playing a double-or-nothing game with 50/50 odds. In other words, you bet a certain amount, flip a coin, and if it's heads you get twice your money, tails you lose all your money. (So the mean value is zero, since half the time you lose x dollars and the other half you gain as much.) Here is the strategy: first you bet x. If you win, you stop, and you've successfully won x dollars. If you lose, you bet 2x. If you win, your net gain is now 2x-x = x, and you stop. If you lose, you're down by 3x. So bet 4x. If you win, you're up by x, and you stop. Et cetera. You just keep doubling your bet until you win, and you're "guaranteed" to end up winning x dollars.
What's the catch? The problem is that you could run out of money before you ever win. Suppose you start with 6x dollars. Then there's a 25% chance that you'll lose twice in a row, in which case you won't be able to afford doubling your bet again. If you play this strategy long enough, you're essentially guaranteed to go broke.
However, what this strategy does achieve is a way to effectively turn this essentially unpredictable game into a (possibly longer) game with a single very likely outcome -- that you'll end up winning x dollars. Specifically, let's say you start with 63x dollars. Then you can afford to lose up to 5 times in a row and still ultimately win overall in the last game (with a bet of 32x on that last game). This means you'll net x dollars profit with probability 63/64 = 98.4375% chance. So I would call x the nice value of that game.
By the way, you may be wondering why people don't use this strategy all the time if it sounds so tempting? Well, in the long run, it's actually not a great strategy. I think it's a good one-time strategy, but not something you'd like to repeat. For example, suppose you go to a casino and use this strategy once a month. By the end of 4 years, it's most likely that you'll have lost at least once. Even if you only lose once, and win the other 47 times, you've still lost overall since a single loss means losing 6 times in a row, or a loss of 63x. This perspective shows that the mean value is the "right value" for long-term additive behavior, while the nice value is meant to express a one-time value which is the least surprising outcome.
toward a rigorous definition of the nice
Suppose we have a metric space X and a probability distribution P() on X. (I also assume we have a measure space on X including all spheres generated by the distance function of the metric.) Given any x in X, we can measure its niceness via the function f_x:R → [0,1] defined by
f_x(z) = P({y: dist(x,y) < z}).
Basically, f_x() is an increasing function which (if no distance in X is infinite) asymptotically approaches 1. The faster f_x() approaches 1, the nicer x is -- that is, the less surprising it would be that x is the outcome. The trick is that functions are not easily put into a linear ordering, so it is hard to say which value of x is the nicest. Beyond that, we can easily imagine situations in which many values of x have the same function f_x() -- for example, rolling a die. But when there is a value x in X which gives the greatest function f_x(), we can safely say that that value is the nicest.
Future work: try to define the nicest value in other cases!
Sunday, March 04, 2007
corner bistro
To those in search of an excellent new york city cheeseburger: may I recommend to you the Corner Bistro, happily located a mere two blocks from my place of abode (not that this proximity affects in any way the taste of said burger).
Vegetarians, lovers of healthifulnessity, deniers of hedonism: avert your fair untainted eyes.
Meandering the serpentine streets of the west village, your first hint of grilled-bovine salvation is a concise neon appellation: "Corner Bistro", hold the serifs. Once inside, this sans frills style saturates the scene. A lazy haze of smoke once permeated the very soul of this cozy joint -- haze illegal now, its ghost remains in the nostalgic distance which, in your mind alone, separates your suddenly-anachronistic chic from the grimy aged authenticity of the regulars around you. The bar's cash register, a metalic curve of plastic mushroom buttons, is easily a century old if a day. A shaded flourescent bulb dimly illuminates a ten-line "menu" on the wall, consisting primarily of four varieties of burger. The missing 'i' in "chiliburger" imparts a lacunal singularity of character. An impressively laconic waitstaff waves your thankfully small party down a narrow coat-racked hall, adorned most curiously by a row of Heinz ketchup bottles -- each, label facing out, standing at attention, patiently awaiting its ephemeral tabletop role.
Be seated, or rather scrunched, at your table, and choose your fare; a 'bistro' means baconcheeseburger, or hold the pork, or the cheese or both or add some chili. Curiously baconic fries? Grilled cheese? A drink or two or so away, and you may wonder for a moment if that last fleeting gesticulation of your server was acknowledgement of an order soon fulfilled or was it just your hope assigning undue meaning to a twitch as they flurried away, order unheeded. But here! Yes, they heard! Half a pound of freshly-grilled, juice-imbued, cheese-melting, tomato-strewn, barely-bun-encompassed, beautiful beef sirloin sits steaming snuggly under your overwhelmed olfactory-orifacing salivation-inducing proboscis dinostrilonimus (don't worry I completely made up that last word). Now here we find a burger which, viewed in any cross section of significance, is still beef by vast majority: in volume, density, height, diameter, and sheer incorrigible flavorocity. Every bite produces from its opposite edge another egress of warm oily beef-nectar, which in turn deposits itself, filtered lingeringly along your fingertips, into a congealing puddle of wisely ignored goop atop your paper plate.
Upon completion of this iniquitous consumption, it will eventually occur to you that you must now open your eyes, pay the bill, and peacefully resume your normally scheduled life. | 4,038 | 18,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-18 | latest | en | 0.963507 |
https://www.lmfdb.org/ModularForm/GL2/ImaginaryQuadratic/2.0.11.1/5184.3/k/ | 1,623,522,805,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00372.warc.gz | 798,790,413 | 5,412 | # Properties
Label 2.0.11.1-5184.3-k Base field $$\Q(\sqrt{-11})$$ Weight $2$ Level norm $5184$ Level $$\left(72\right)$$ Dimension $1$ CM no Base change yes Sign $+1$ Analytic rank $$0$$
# Related objects
## Base field: $$\Q(\sqrt{-11})$$
Generator $$a$$, with minimal polynomial $$x^2 - x + 3$$; class number $$1$$.
## Form
Weight: 2 Level: 5184.3 = $$\left(72\right)$$ Level norm: 5184 Dimension: 1 CM: no Base change: yes 72.2.a.a , 8712.2.a.u Newspace: 2.0.11.1-5184.3 (dimension 11) Sign of functional equation: $+1$ Analytic rank: $$0$$ L-ratio: 2
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$$4$$ 4.1 = ($$2$$) $$-1$$
$$3$$ 3.1 = ($$-a$$) $$-1$$
$$3$$ 3.2 = ($$a - 1$$) $$-1$$
## Hecke eigenvalues
The Hecke eigenvalue field is $\Q$. The eigenvalue of the Hecke operator $T_{\mathfrak{p}}$ is $a_{\mathfrak{p}}$. The database contains 200 eigenvalues, of which 20 are currently shown below. We only show the eigenvalues $a_{\mathfrak{p}}$ for primes $\mathfrak{p}$ which do not divide the level.
$N(\mathfrak{p})$ $\mathfrak{p}$ $a_{\mathfrak{p}}$
$$5$$ 5.1 = ($$-a - 1$$) $$2$$
$$5$$ 5.2 = ($$a - 2$$) $$2$$
$$11$$ 11.1 = ($$-2 a + 1$$) $$-4$$
$$23$$ 23.1 = ($$a + 4$$) $$8$$
$$23$$ 23.2 = ($$a - 5$$) $$8$$
$$31$$ 31.1 = ($$-3 a + 4$$) $$8$$
$$31$$ 31.2 = ($$3 a + 1$$) $$8$$
$$37$$ 37.1 = ($$-3 a - 2$$) $$6$$
$$37$$ 37.2 = ($$3 a - 5$$) $$6$$
$$47$$ 47.1 = ($$-2 a + 7$$) $$0$$
$$47$$ 47.2 = ($$2 a + 5$$) $$0$$
$$49$$ 49.1 = ($$7$$) $$-14$$
$$53$$ 53.1 = ($$-4 a + 5$$) $$2$$
$$53$$ 53.2 = ($$4 a + 1$$) $$2$$
$$59$$ 59.1 = ($$a + 7$$) $$-4$$
$$59$$ 59.2 = ($$a - 8$$) $$-4$$
$$67$$ 67.1 = ($$-3 a - 5$$) $$-4$$
$$67$$ 67.2 = ($$3 a - 8$$) $$-4$$
$$71$$ 71.1 = ($$-5 a + 1$$) $$-8$$
$$71$$ 71.2 = ($$5 a - 4$$) $$-8$$
Display number of eigenvalues | 841 | 1,780 | {"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} | 3.078125 | 3 | CC-MAIN-2021-25 | latest | en | 0.426045 |
https://math.stackexchange.com/questions/2076034/graph-homomorphisms-allowing-contractions-of-edges | 1,721,690,065,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00177.warc.gz | 332,829,800 | 35,536 | # Graph homomorphisms allowing contractions of edges
Let $G,H$ be graphs with at most one edge between vertices, with loops allowed. A homomorphism of graphs from $G$ to $H$ is a vertex mapping $f\colon V(G)\to V(H)$ such that if $\{u,v\}$ is an edge of $G$, then $\{f(u),f(v)\}$ is an edge of $H$.
I my research, I need a name for a somewhat more general notion: a vertex mapping $f\colon V(G)\to V(H)$ such that if $\{u,v\}$ is an edge of $G$, then either $f(u)=f(v)$ or $\{f(u),f(v)\}$ is an edge. So every edge of $G$ is either mapped to an edge or contracted to a vertex.
Question: Is there (at least a semi-standard) name of such mappings between graphs?
I am aware that this terminological problem commonly solved/handwaved by putting loops on every vertex. However, in my case I need to consider both types of morphisms at the same time, so I cannot do this.
(2) Another option might be to embed the category of graphs without loops $$\mathbb G$$ to the category $$\mathbb G^+$$ of graphs which possibly can have loops (but do not have to). Then, you have a monad $$T\colon \mathbb G^+ \to \mathbb G^+$$ which adds all missing loops (I hope that this is a monad :-)). Then your morphisms are just Kleisli morphisms $$A \to T(B)$$ where $$A$$ and $$B$$ are without loops.
The advantage of the second option is that you can still recover the original subcategory $$\mathbb G$$ within $$\mathbb G^+$$. | 396 | 1,411 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.901092 |
http://www.fixya.com/support/t792091-log_reciprocal_functions | 1,529,675,899,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864482.90/warc/CC-MAIN-20180622123642-20180622143642-00057.warc.gz | 410,243,291 | 33,952 | Question about Texas Instruments TI-89 Calculator
# Log & reciprocal functions
I can't find lo and reciprocal functions tabs.
Posted by on
• Anonymous Oct 27, 2008
i dont not know where to access the reciprocal function on the TI-89
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• Contributor
There is no log (base 10) button on a TI-89. It's buried in the CATALOG as log(. As a shortcut, you can do ln(#)/ln(base), where # and base are replaced with what you want. For the inverse of a log, do the base raised to the part on the other side of the equals sign. So, log(x) = 2 would become 10^2 = x.
Posted on Oct 04, 2009
Hi there,
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## Related Questions:
### Cannot find my factorial button
It's the shifted function of the x^-1 key, located just below and right of the big round cursor pad, above the log key.
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### Im pressing the buttom that everyone is saying is the inverse key but it is saying error. i really need help with this.
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This is just a quick overview.
Feb 16, 2012 | Texas Instruments TI-84 Plus Silver...
### Where is the inverse button on the casio fx-82ES
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Jan 09, 2012 | Casio FX-115ES Scientific Calculator
### How do i change blade on dewalt dw303 reciprocating saw
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http://www.ereplacementparts.com/bosch-rs15-060164e064-114-reciprocating-saw-parts-c-128_1131_7890.html
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### 0.5=log Io/It find Io/It
Hello
A set a parentheses would sure help remove any ambiguities.
I assume your equation is 0.5 =log(lo/lt)
There is an equivalence
y=log(x) is equivalent to x=10^y
You can write
10^(0.5)=10^(log(lo/lt))
But 10^(log(x))=x, thus
10^(0.5)= lo/lt
and lo/lt=square root of 10
Here is a screen capture to convince you.
Hope it helps
Oct 25, 2009 | Casio FX-300MS Calculator
### Trigonometric function
The 3 trig functions most commonly
used in triangle calculations are
sine = Opposite/Hypotenuse Co secant = Hypotenuse/Opposite
Hope this helps Loringh
Oct 12, 2008 | Measuring Tools & Sensors
## Open Questions:
#### Related Topics:
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Level 2 Expert | 1,403 | 5,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-26 | latest | en | 0.938173 |
https://en.doc.boardgamearena.com/Gamehelpcrazyeights | 1,726,651,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00296.warc.gz | 208,211,283 | 6,320 | This is a documentation for Board Game Arena: play board games online !
# Gamehelpcrazyeights
Copied from ' https://en.wikipedia.org/wiki/Crazy_Eights '
## Basic play
Five cards are dealt to each player (or seven in a two-player game). The remaining cards of the deck are placed face down at the center of the table as the stock pile. The top card is then turned face up to start the game as the first card in the discard pile.
Players discard by matching rank or suit with the top card of the discard pile, starting with the player left of the dealer. They can also play any 8 at any time, which allows them to declare the suit that the next player is to play; that player must then follow the named suit or play another 8. If a player is unable to play, that player draws cards from the stock pile until a play can be made, or until the stock pile is exhausted. If the player cannot play when the stock pile is exhausted, that player must pass the turn to the player on the left. A player may draw from the stock pile at any time, even when holding one or more playable cards.
As an example: Once 6♣ is played the next player:
• can play 6♦, 6♥ or 6♠
• can play any club
• can play any 8 (then must declare a suit)
• can draw from the stockpile and continue their turn
If the stock pile runs out, all played cards except for the top one are reshuffled to form a new stock.
The game ends as soon as one player has emptied their hand. That player collects a payment from each opponent equal to the point score of the cards remaining in that opponent's hand. 8s score 50, court cards 10 and all other cards face value. If the players run out of cards in the deck, the player with the lowest point score in their hand scores the difference between that hand and each opponent's hand.
## Variations
Card game historian John McLeod describes Crazy Eights as "one of the easiest games to modify by adding variations", and many variant rules exist. Common rules applied to cards include:
• First player to reach specific number of points - The winner of the game is the first player to reach a specific number of points. For two players it is 100 points, three players 150, four 200, five 250, six 300 and for seven players 350.
• Maximum pickup Cards - the maximum number of cards a player must pick up is 5 if they cannot play a card
• Queens skip - Playing a Queen causes the next player to miss their turn.
• Aces reverse direction - Playing an Ace reverses the direction of play.
• Draw 2 - Playing a two forces the next player to draw two cards. Multiple twos "stack"; if a two is played in response to a two, the next player must draw four. | 613 | 2,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.96645 |
http://www.mathworks.fr/fr/help/matlab/ref/isrow.html?action=changeCountry | 1,386,960,716,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164977792/warc/CC-MAIN-20131204134937-00029-ip-10-33-133-15.ec2.internal.warc.gz | 445,275,150 | 7,451 | Accelerating the pace of engineering and science
# Documentation Center
• Trials
• Product Updates
# isrow
Determine whether input is row vector
isrow(V)
## Description
isrow(V) returns logical 1 (true) if size(V) returns [1 n] with a nonnegative integer value n, and logical 0 (false) otherwise.
## Examples
Determine if a vector is a row. This example is a column so isrow returns 0:
V = rand(5,1);
isrow(V)
ans =
0
Transpose the vector to make it a row. isrow returns 1:
V1 = V';
isrow(V1)
ans =
1
## See Also
Was this topic helpful? | 155 | 550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-48 | latest | en | 0.602829 |
https://itunes.apple.com/us/app/id843388724?mt=8 | 1,534,372,192,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210362.19/warc/CC-MAIN-20180815220136-20180816000136-00416.warc.gz | 728,381,626 | 36,364 | ## Description
7th Grade Math Learning Games is a year’s worth of math lessons, worksheets, and problems for seventh graders. Covering 19 common core subjects with a near endless amount of questions, it is perfect for teachers or parents who want to give their students a fun way to practice and learn.
Track your student’s progress with detailed reports and multiple profiles. And students will want to keep playing as they earn tokens to spend in the arcade!
This free version allows for up to 20 questions per day, 1 arcade game, and 1 profile. Get the full version through a simple in-app purchase to unlock the entire game.
*** Features ***
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+ Thousands of Questions - The problems are randomly generated, so every round will be different
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+ Fun Way To Learn - An arcade with 5 fun games makes students want to keep playing and earning more tokens
*** Topics ***
+ Ratios and Proportions
-- Compute Ratios; Proportional Relationships; Multi-Step Ratio and Percent Problems
+ The Number System
-- Add and Subtract Rational Numbers; Multiply and Divide Rational Numbers; Real-World Problems with Rational Numbers
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-- Properties of Operations / Linear Expressions; Rewriting Expressions, Multi-Step Problems with Rational Numbers; Simple Equations and Inequalities
+ Geometry
-- Scale Drawings; 2-D Figures From 3-D Figures; Circles; Angles; Area, Volume, and Surface Area
+ Statistics and Probability
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## What's New
Version 2.2
Huge new update!
• Support for iPhone X!
• Bug fixes and UI improvements
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If you're having any trouble with our games, please email us at help@rosimosi.com and we'll get back to you ASAP. And if you love the games then be sure to leave us a review, it really helps us out!
## Ratings and Reviews
### Works we’ll!!! But should tell the person how to solve the problems after the get it wrong
#### Jul 9, 2018
This app is very great 7th graders but, it would be more helpful
If they tell you what you did wrong. Overall I would only get this game if you are already in 7th grade and you need more practice.
### This is really amazing
#### Nov 29, 2016
Harkinzus
It has everything you could possibly want for your students! Very well designed and easy for parents or students to navigate and play with.
### Geat appp
#### Nov 30, 2016
ArmyTea
This amazing app. it helped me entertain after a stressful working hours. My family is very commonly used applications.
## Information
Seller
RosiMosi LLC
Size
19.8 MB
Category
Education
Compatibility
Requires iOS 8.0 or later. Compatible with iPhone, iPad, and iPod touch.
Languages
English
Age Rating
Rated 4+
Price
Free
In-App Purchases
1. 7th Grade Math - Full Version \$2.99 | 709 | 3,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-34 | latest | en | 0.894773 |
http://www.jiskha.com/display.cgi?id=1347411089 | 1,498,183,902,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319943.55/warc/CC-MAIN-20170623012730-20170623032730-00218.warc.gz | 567,572,272 | 3,720 | math
posted by on .
A cylindrical pipe has a radius of 8 inches and a height of 100 inches. About how many cubic inches of water will fill the pipe?
• math - ,
Volume = pi * r^2 * h
V = 3.14 * 64 * 100 | 63 | 206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.855241 |
http://physics.qandaexchange.com/?qa=2855/toppling-of-cube-balanced-on-a-hemisphere | 1,566,633,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00084.warc.gz | 146,375,702 | 8,148 | # Toppling of cube balanced on a hemisphere
1 vote
63 views
Find the max edge length $a$ for which cube will remain in stable equilibrium on hemispherical surface. Consider no sliding of the base of cube and hemisphere.
edited Jul 8, 2018
I'm getting $a=2R$ but not aware of the correct answer.
Draw a diagram of the cube after it has rocked (not slid) through some small angle. Determine whether the position of the cube is stable - ie whether the moment of the weight of the cube about the contact point is turning the cube back towards its equilibrium position or turning it further away from equilibrium.
The difficult part is to calculate the position of the COM of the cube relative to the contact point when the cube has rocked through an angle.
1 vote
In the diagram I have drawn only one quarter of the cube, which has centre C and side $a=2b$.
Initially the cube is balanced in equilibrium with point O below C in contact with the hemisphere. After it has rolled through some angle $\theta$ contact has shifted to point P. If C is to the left of the vertical through P ($b \lt h$) then the cube will roll back to the equilibrium position shown on the left, whereas if C is to the right of the vertical through P ($b \gt h$) then the cube will continue rolling and will topple.
$\tan\theta=\frac{s}{h}=\frac{R\theta}{h}$
For small angles $\tan\theta \approx \theta$ so for stable equilibrium
$b \le h \approx R$
$a = 2b \le 2 R$
Comment :
As $\theta$ increases the limit for $a$ gets smaller :
$h=R\theta \cot\theta=R\theta (\frac{1}{\theta}-\frac{\theta}{3}-\frac{\theta^3}{45}-...)$
$a \le 2R(1-\frac{\theta^2}{3}-\frac{\theta^4}{45}-...)$
For a cube the maximum angle $\theta$, beyond which it can no longer roll, is given by $R\theta=b$ so the maximum angle is $\theta_m=\frac{b}{R}=\frac{a}{2R}$ hence
$\frac{a}{2R} = \theta_m \approx 1-\frac13 \theta_m^2-\frac{1}{45} \theta_m^4-...$
$\theta_m^2+3\theta_m -3 \approx 0$
$\theta_m \approx \frac12 (-3 + \sqrt{21}) = 0.791^c \approx 45^{\circ}$
The problem assumes that the coefficient of friction is sufficient to avoid sliding. The condition for no sliding on an inclined plane applies :
$\mu \ge \tan\theta$
answered Jul 9, 2018 by (28,746 points)
edited Jul 9, 2018 | 626 | 2,247 | {"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-2019-35 | longest | en | 0.7846 |
http://mathhelpforum.com/algebra/174960-factorise.html | 1,481,124,975,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00121-ip-10-31-129-80.ec2.internal.warc.gz | 181,478,754 | 10,397 | 1. ## factorise
hi,
i nned to factorise this:
3x^2+5x-2=0
i know how to do these but im stuck on this one!
so if someone could lend some advise i would be very grateful!
thanks!
2. 5x = 6x-x
So,
$3x^2+5x-2$
$=3x^2+6x-x-2$
3. .
4. i don't understand, wouldn't you have 2 brackets one starting with 3x and the other x.
5. Originally Posted by andyboy179
i don't understand, wouldn't you have 2 brackets one starting with 3x and the other x.
Hi andyboy,
You are correct, but harish21 was showing you a neat little way of factoring besides the "guess and check" method.
Here is harish21's method.
First, multiply the leading coefficient times the constant: 3 times -2 = -6
Next, determine two numbers whose product is -6 and whose sum is the coefficient of the linear term which is +5.
You should easily see that -1 and 6 meet that criteria.
$3x^2+6x-1x-2$
Group the first two terms and factor out 3x.
$3x(x+2)-1x-2$
Group the last two terms and factor out -1.
$3x(x+2)-1(x+2)$
It's easy to see the two binomial factors are $(x+2)(3x-1)$
It takes a lot longer to explain it than it does to actually do it. | 353 | 1,126 | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.936864 |
https://yourprofoundinfo.com/what-is-the-laplace-transform-of-cos-3t/ | 1,702,204,193,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00758.warc.gz | 1,194,654,085 | 14,557 | # What is the Laplace transform of cos 3t?
## What is the Laplace transform of cos 3t?
From the formulas, L[cos(3t)] = s s2 + 9 . So, by s-shift, F(s) = L[e-t cos(3t)] = s + 1 (s + 1)2 + 9 .
### How do you find the Laplace transform of cos t?
Let cos be the real cosine function. Let L{f} denote the Laplace transform of the real function f. Then: L{cosat}=ss2+a2.
What is the Laplace transform of cosh?
Let L{f} denote the Laplace transform of the real function f. Then: L{coshat}=ss2−a2.
What is the formula of cos 2 t?
cos 2t = 1 – 2 sin2 t.
## What is the formula of Cos Cube T?
Cosine cubed formula is given as cos3α = (3cosα + cos3α) / 4, where cos = cosine, α = Angle. Through this cos cube formula, you can find the value of cosine cube which the anti derivative involves sin3 and sin. The functions sine, cos and tan are three important functions of trigonometry.
READ: How long does it take to pay off \$30000 student loans?
### How do you calculate Laplace transform?
Method of Laplace Transform
1. First multiply f(t) by e-st, s being a complex number (s = σ + j ω).
2. Integrate this product w.r.t time with limits as zero and infinity. This integration results in Laplace transformation of f(t), which is denoted by F(s).
What is formula of Cos?
The cosine formulas talk about the cosine (cos) function. Then the cosine formula is, cos x = (adjacent side) / (hypotenuse), where “adjacent side” is the side adjacent to the angle x, and “hypotenuse” is the longest side (the side opposite to the right angle) of the triangle.
What does cosh mean in math?
hyperbolic cosine
The Math.cosh() function returns the hyperbolic cosine of a number, that can be expressed using the constant e: Math.cosh(x) = e x + e – x 2.
## What is the formula of cos t?
Truly obscure identities
sin s cos t = sin (s + t) + sin (s – t) 2
cos s cos t = cos (s + t) + cos (s – t) 2
sin s sin t = cos (s – t) – cos (s + t) 2
READ: Where does the Bible say God cares?
### What is the Laplace of T 3 2?
What is the Laplace of t to the power of -3/2? – Quora. Hence L{t^(-3/2)} = -2 √π /s^(-1/2)= -2√(πs).
What is the formula for COS 3 Theta?
Cos 3θ = 4cos3θ – 3cos θ
What is the Laplace transform of f(t) = sin t?
The laplace transform of f(t) = sin t is L{sin t} = 1/(s^2 + 1). As we know that the Laplace transform of sin at = a/(s^2 + a^2).
## How to find inverse Laplace transform of a function?
If a unique function is continuous on o to ∞ limit and have the property of Laplace Transform, F(s) = L {f (t)} (s); is said to be Inverse laplace transform of F(s). It can be written as, L -1 [f(s)] (t).
### What is Laplace transformation in control systems?
Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. The best way to convert differential equations into algebraic equations is the use of Laplace transformation. Laplace transformation plays a major role in control system engineering. | 877 | 2,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-50 | longest | en | 0.799691 |
http://forum.allaboutcircuits.com/threads/design-troubles-this-really-shouldnt-be-that-hard.25767/ | 1,484,967,606,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280899.42/warc/CC-MAIN-20170116095120-00072-ip-10-171-10-70.ec2.internal.warc.gz | 104,668,313 | 16,791 | # Design troubles. This really shouldn't be that hard...
Discussion in 'General Electronics Chat' started by melmac82, Jul 18, 2009.
1. ### melmac82 Thread Starter New Member
Jul 18, 2009
9
0
Ok, trying to design a relatively simple logic circuit with 3 inputs and 3 outputs. (A, B, C and a, b, c) When A is high, a is high and all the rest are low. Then when B goes high a drops out and b goes high even if A is still high. If C goes high, c is high and both a and b are low even if A and B are high. I've tried experimenting with various latches and other combinations. Just when I think I'm getting close I get oscillation issues. Any ideas or help would be great.
2. ### SgtWookie Expert
Jul 17, 2007
22,183
1,728
Is this a homework assignment?
3. ### millwood Guest
that's easy.
c=C;
b=(!C) & B
a=((!C) & (!B)) & A
where !X = not X.
4. ### melmac82 Thread Starter New Member
Jul 18, 2009
9
0
Nope its not a homework problem... just something that got me thinking and now I have to figure it out, heh.
And I should have elaborated more but it can't just be c=C because it has to go backwards too. So if C and c are both high and then A goes high with C still high, c would go low and a high. So no matter what inputs are high, when a new input goes high, all outputs go low except the one associated with the new high input.
5. ### hgmjr Moderator
Jan 28, 2005
9,030
214
A good place to start is to build a truth table relating the inputs to the outputs.
hgmjr
6. ### millwood Guest
that will make it become an oscillator - the circuit will never have a steady state.
7. ### melmac82 Thread Starter New Member
Jul 18, 2009
9
0
I was beginning to think that it was just going to make an oscillator because everything I had tried resulted in that happening. I was just hoping that someone else with a little more experience might be able to tell me otherwise, but oh well, at least it looks like I'm getting good at creating a lot of different circuits that oscillate.
8. ### hobbyist Distinguished Member
Aug 10, 2008
773
62
If there manual inputs,
couldn't you latch the outputs, then use steering diodes connected to the input switchs to reset the other 2 latches.
switch A would connect to S on it's latched output while simultaneously switch A could steer a signal to the R of the other 2 latches. Thereby trigering the opposite outputs.
So when A or B or C switch is inputed the other 2 latches will change to an off state every time a switch is hit.
and do the same for the other 2 switches
However it would have to be the same thing with the input switchs themselves steering diodes and some gating so that 2 switchs on at the same time would feedback a shutoff signal to the other switchs. etc...
Switch A and B and C would be wired to some gating so that A switch blocks B and C switch inputs and vice versa.
I think you know what i mean.
Something like that...If there manual switch inputs.
just an idea, probably not what your looking for..
Last edited: Jul 18, 2009
9. ### melmac82 Thread Starter New Member
Jul 18, 2009
9
0
I'm having some problems with latches though because all my inputs and outputs will be low to start. I've found that latches tend to either oscillate or give high outputs with starting low inputs. I can get low inputs to have low outputs when I use XOR gates instead of the usual NOR and NAND gates, but then I run into other oscillation problems later.
I was going to try some stuff with the manual switches, latches, and diodes, but I don't know what software to use. Right now I am using just a basic program that only includes inputs/outputs, gates, plexers, arithmetic, and some memory modules. Any suggestions for some better free software for simulations?
10. ### melmac82 Thread Starter New Member
Jul 18, 2009
9
0
Ok, so now I've been thinking about this way to go. I'm not very knowledgeable about latches, so are we talking NAND latches or NOR latches?
Also, when you are saying to have a steering diode go from one switch to the R input of the other latches, am I correct in assuming then that there would be 2 signals with steering diodes going to each R input of each latch?
And I got a little confused when you were talking about having to do the same thing with the input switches. Does that mean that there would be steering diodes directing the inputs from the two other switches to the S input of the remaining latch?
I've never done anything like this before and it is pretty interesting and its hard to stop thinking about it, lol. And thanks for all the help so far everyone.
11. ### hobbyist Distinguished Member
Aug 10, 2008
773
62
sorry for the delay, I forgot to look at this post.
I'll see if I can come up with a logic diagram.
What I think your asking about.
Something like this.
I never worked with IC's so can't help you there but this logic diagram might be what your asking about.
There RS latchs, when S goes high its output Q goes high
when R goes high the output goes off.
So A switch connects to "S" on "a" output latch and simultaneously connects to R inputs of the other 2 latchs to turn them off.
This only works for inputs that are momentary, if you want to cancel outputs when more than one input is already prisent it will take some elaborate circuitry to cancel input switchs as well.
If you want the switchs to remain on and then they get canceled with another switch input, that's a whole new ballgame, which would take some real heavy thinking and a lot of extra circuitry.
Last edited: Jul 21, 2009
12. ### melmac82 Thread Starter New Member
Jul 18, 2009
9
0
Thanks a lot for posting this. It's pretty much what i was thinking that you were saying and I'll have to look at some more stuff to see if it is more complicated. I did another post that is along the same lines as this. I was trying to explain some stuff better and put a better title to the post to see if I could get some other people to respond. Again, thanks very much for your help | 1,480 | 5,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-04 | latest | en | 0.949363 |
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### What really makes this educational discipline so integral to our learning as humans?
Math, as a whole, has become such an intrinsic part of comprehensive education.
Without the calculated use of numbers, shapes, equations, and so much more- we wouldn’t have made nearly as many advancements we use in our daily lives today! Isn’t that crazy?
Although math is a subject many people have a hard time grasping, it’s greatly applicable to assure our general wellbeing. Those who are more adept in the field have the power to alter the course of our evolution- FOREVER. And even those who don’t choose to make a career out of it can still possess applicable life skills to make their daily living that much easier.
Math was originally discovered and used to construct empires and their systems of living in ancient times. Nowadays, it’s applied to nearly everything from cooking to architecture to dance choreography. We measure and calculate and articulate so many things in our habitual lifestyle it’s easy to overlook how much math we exercise consistently.
#### What makes math SO important, you ask?
Well, here are six core reasons why Math is incredibly important to include in a diversified model of education:
1. Math promotes critical thinking skills: Critical thinking is an important skill covering a breadth of other cognitive abilities such as analysis, interpretation, inference, explanation, self-regulation, open-mindedness, and problem-solving. Math utilizes its kind of critical thinking to get students (or anyone using it) to brainstorm reasoned decisions to approach particular problems. Rather than guessing or applying a rule without testing its validity, they have to be able to concoct educated judgments that best suit the necessary solution to the problem.
2. Truth be told, STEM is the future: STEM includes varying math-oriented disciplines such as science, engineering, and technology. These are fields able to help our technological evolution progress, so it’s incredibly important to implement the fundamental learnings that provide the solid basis for pursuing and studying such prominent endeavors.
3. Logical thinking is a communicable life skill: Unlike critical thinking, logical thinking is how to evaluate the processes behind the reasoning. Critical thinking uses logic and its core principles to separate truth from falsehood. However, it doesn’t focus on said reasoning specifically. Solidifying a strong math education, especially for those at a young age, can empower them to make creative yet careful decisions.
4. Math productively teaches about failure: Look, we’ve all experienced an excruciatingly difficult math problem we SWORE we would be unable to solve. But sure enough, that’s one of the most crucial lessons math teaches us- failure is the opportunity to learn even more. When someone fails to solve a problem, they have the option to give up or attempt the problem again. If they choose to try again, this strengthens their learning process; if not, then they’re stuck in a limiting belief they can’t achieve a resolution. Math subtly emphasizes that failure is a part of reality, and we have the choice to turn it into something greater.
5. Math ALSO exemplifies the importance of a growth mindset: Math has the power to teach us that competence is an ever-evolving process- NOT a limited set of skills. Combining a growth mindset with the reality of failure prepares young students to attack their decision-making with patience and an awareness that they have the power to reach their end goal! These are two very essential skills in optimal social, emotional, and cognitive development.
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Mathematics is so deeply incorporated into our daily functions, and the mechanic behind our most used tools, that it’s nearly impossible to avoid. Our societal standard of success is based greatly upon our inherent ability to make informed, sound, logical decisions; we don’t even realize how much math helps us to condition this technique!
If you encounter a young student with a difficulty in learning math, now is the time to encourage these key principles. Math can be taught in environments outside of the traditional classroom, such as the kitchen or certain extracurriculars. Although that young child may not NEED math for their future professional endeavors, it’s still going to be a functional skill once they reach adulthood.
Even though math is a meticulous discipline, it has the power to better our means of thriving. And without it, who knows where we’d all be! | 954 | 5,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.960503 |
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# The efficiency of microwave ovens in destroying the harmful
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09 May 2005, 17:15
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19. The efficiency of microwave ovens in destroying the harmful bacteria frequently found in common foods is diminished by the presence of salt in the food being cooked. When heated in a microwave oven, the interior of unsalted food reaches temperatures high enough to kill bacteria that cause food poisoning, but the interior of salted food does not. Scientists theorize that salt effectively blocks the microwaves from heating the interior.
Which one of the following conclusions is most supported by the information above?
(A) The kinds of bacteria that cause food poisoning are more likely to be found on the exterior of food than in the interior of food.
(B) The incidence of serious food poisoning would be significantly reduced if microwave ovens were not used by consumers to cook or reheat food.
(C) The addition of salt to food that has been cooked or reheated in a microwave oven can increase the danger of food poisoning.
(D) The danger of food poisoning can be lessened if salt is not used to prepare foods that are to be cooked in a microwave oven.
(E) Salt is the primary cause of food poisoning resulting from food that is heated in microwave ovens.
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09 May 2005, 17:51
D is the answer. Since salt blocks microwave heat and therefore will not kill bacteria, avoiding salt will solve this issue
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09 May 2005, 18:46
(A) The kinds of bacteria that cause food poisoning are more likely
to be found on the exterior of food than in the interior of food.
- If so, then it does not matter if the food is salted or not.
(B) The incidence of serious food poisoning would be significantly reduced if microwave ovens were not used by consumers to cook or reheat food.
- This goes to say microwave ovens are not effective in removing bacteria, and it has nothing to do with salt content in the food
(C) The addition of salt to food that has been cooked or reheated in a microwave oven can increase the danger of food poisoning.
- Not supported for in passage.
(D) The danger of food poisoning can be lessened if salt is not used to prepare foods that are to be cooked in a microwave oven.
(E) Salt is the primary cause of food poisoning resulting from food that is heated in microwave ovens.
- If salt is the primary cause, then it does not matter if the interior is heated or not so long as salt is added to the food.
D is the best choice. The danger of food poisoning is lessened if salt is not used to prepare food that are to be cooked in microwave ovens. This is because we're not told if salt is the only cause of food poisoning.
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09 May 2005, 18:53
D
the conclusion of the Argument :" The efficiency of microwave ovens in destroying the harmful bacteria frequently found in common foods is diminished by the presence of salt in the food being cooked"
--> only D support this
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10 May 2005, 03:49
any of you can explain why C is wrong?
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10 May 2005, 04:48
wunderbar03 wrote:
any of you can explain why C is wrong?
C says "C) The addition of salt to food that has been cooked or reheated in a microwave oven can increase the danger of food poisoning. "
While the passage says is talking about the salt being already in the food before reheating/cooking it in the oven.[/b]
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10 May 2005, 05:04
(D) .... (D) ... (D) .. for obvious reasons
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16 May 2005, 14:33
The OA is D.
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My math professor tends to write $\exists x\in\mathbf{X} \ni P(x)$. Is this a correct use of the such that symbol $\ni$? If not, what is the use of that symbol? Isn't it better to write $\exists x\in\mathbf{X} (P(x))$ assuming $P(x)$ is a complex proposition? Also, while we're at it, is $\forall x\in\mathbf{X} : P(x)$ an acceptable notation or should it really be $\forall x\in\mathbf{X} (P(x))$ for precision?
• I'm guessing he means it as "such that." Often we just use a colon: $\exists x\in\mathbf{X}: P(x)$. You could put parentheses around it, as you do. Incidentally, even that is shorthand, since what it really means is: $\exists x(x\in\mathbf{X} \text{ and } P(x))$. Jan 27, 2012 at 13:54
• This is semi-formal notation, at some distance from the standard ways of expressing the idea in a formal language. Jan 27, 2012 at 15:15
• I would describe $\exists x\in\mathbf{X}\ni P(x)$, $\exists x\in\mathbf{X}: P(x)$, and $\exists x\in\mathbf{X}$ s.t. $P(x)$ as shorthand for an English sentence, There is an x in ... . $\exists x\in\mathbf{X}\big(P(x)\big)$ and $(\exists x\in\mathbf{X}) P(x)$, on the other hand, are expressions in slightly different formal languages, abbreviating the even more formal $\exists x\big(x\in\mathbf{X}\land P(x)\big)$ or $(\exists x)\big(x\in\mathbf{X}\land P(x)\big)$. Jan 27, 2012 at 15:29
The backwards epsilon means "such that", but in this context it's slightly bizarre since the usual set membership symbol appears symmetrically before the $\mathbf{X}$ and of course means something quite different. After all, if $P$ were a function symbol rather than a relation symbol then we might parse the statement quite differently, as $\exists{x} (x \in \mathbf{X} \wedge P(x) \in \mathbf{X})$.
We can't see inside your professor's mind, but with usual usage in mind, your expansion is correct: it abbreviates $\exists{x \in \mathbf{X}} (P(x))$, which is in turn an abbreviation for $\exists{x} (x \in \mathbf{X} \wedge P(x))$.
Universal quantifiers expand similarly: $\forall{y \in Y} \; \varphi(y)$ is short for $\forall{y} (y \in Y \rightarrow \varphi(y))$. Both colons and full stops (periods) are acceptable separators: $\exists{y \in Y} : \varphi(y)$ and $\forall{z \in Z} \; . \; \psi(z)$ are reasonably common. | 711 | 2,320 | {"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} | 3.71875 | 4 | CC-MAIN-2024-18 | latest | en | 0.844918 |
https://www.teacherspayteachers.com/Browse/Search:rally%20coach%20math/Type-of-Resource/Computation?ref=filter/resourcetype/all | 1,568,665,256,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572934.73/warc/CC-MAIN-20190916200355-20190916222355-00126.warc.gz | 1,012,689,812 | 32,657 | # Results forrally coach math
Rally Coach is a great cooperative learning strategy which allows students to learn from each other. Because working with others and communicating learning is so vital to learning and the Common Core Standards, this activity assists students with these challenges. Eight different 2 digit by 2 digi
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Use the Kagan structure, Rally Coach, to have your students practice teaching long division to their peers.
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You will LOVE this great Partner Practice Activity, created by Tricks and Treats for Teaching. Add and Subtract One-Step Equations Rally Coach is a worksheet for Partner A and Partner B. After you partner up the students, cut down the middle and give each partner their problems to show their work an
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You will LOVE this great Partner Practice Activity, created by Tricks and Treats for Teaching. Multiply or Divide One-Step Equations Rally Coach is a worksheet for Partner A and Partner B. After you partner up the students, cut down the middle and give each partner their problems to show their work
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This is a great cooperative learning activity that allows the students to practice their skills in finding volume and dimensions. It is aligned to Common Core State Standards 5.MD.4 and 5.MD.5 . Students will be in partners. Together they will share one paper and once pencil. Students take turns ei
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Use this flip chart to introduce/review order of operations using whole numbers. Also included in the flip chart is the opportunity to use the Kagan structure: Rally Coach. *Does not include exponents.
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This is a practice page designed to get students working with using volume and mass measurements to calculate density of an object. This should be done prior to the use of "Identifying Substances By Density" lab included in the store. There are two versions with identical problems that can be used.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 564 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.908518 |
https://fr.mathworks.com/matlabcentral/cody/problems/45970-vector-of-digits/solutions/2652788 | 1,611,022,619,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00360.warc.gz | 347,657,172 | 16,954 | Cody
# Problem 45970. Vector of digits (★★)
Solution 2652788
Submitted on 4 Jul 2020 by Athi
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 = 172850; y_correct = [1 7 2 8 5 0]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 1455; y_correct = [1 4 5 5]; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 1; y_correct = [ 1 ]; assert(isequal(your_fcn_name(x),y_correct))
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## How can i count unique text in rows which occur more than 1 time
Hello,
I have a question and i seems to figure this one out, in my database i have two rows. 1 is a Stamordernumber row and the other a ordernumber row when there is a project which has no additional cost the stamorder row and the ordernumber row have the same number. However when i make a addional order which is link to the Stamorder number the stamorder number will stay the same and the ordernumber will change. See below axample.
Stamorder number - Ordernumber
SO32100060 SO32200060 (This is a order with no additional cost on is, so this is the main and only order)
SO32100105 SO32200105 (This is a order with additional cost on is, so this is the main order)
SO32100105 SO32200106 (and this is the additonal order with the addition cost on it)
Now i would lik to have a colum count in front of the stamorder number which counts the amount of rows which has the same stamorder number, see below example.
Count - Stamorder number - Ordernumber
1 SO32100060 SO32200060 (This is a order with no additional cost on is, so this is the main and only order)
2 SO32100105 SO32200105 (This is a order with additional cost on is, so this is the main order)
2 SO32100105 SO32200106 (and this is the additonal order with the addition cost on it)
I tried everthing with CALCULATE and DISTINCTCOUNT but i just don't seem to get a correct count, could anyone help me please.
The table with the stamordernumber in it is a text table.
Thank You,
Harry
1 ACCEPTED SOLUTION
Community Support
Hi @HarryHKE ,
From what I am understanding, you want to count group by Stamorder number.
``Count = CALCULATE(COUNT('Table'[Stamorder number]),ALLEXCEPT('Table','Table'[Stamorder number]))``
Best Regards,
Stephen Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 REPLIES 2
Frequent Visitor
Tank you !!
Community Support
Hi @HarryHKE ,
From what I am understanding, you want to count group by Stamorder number.
``Count = CALCULATE(COUNT('Table'[Stamorder number]),ALLEXCEPT('Table','Table'[Stamorder number]))``
Best Regards,
Stephen Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly. | 653 | 2,668 | {"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} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.841481 |
https://it.mathworks.com/matlabcentral/cody/problems/2576-sum-of-series-ii/solutions/1210409 | 1,610,878,761,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00669.warc.gz | 402,348,213 | 16,972 | Cody
# Problem 2576. Sum of series II
Solution 1210409
Submitted on 12 Jun 2017 by Chris Cleveland
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### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; s_correct = 1; assert(isequal(sumOfSeriesII(n),s_correct))
2 Pass
n = 2; s_correct = 10; assert(isequal(sumOfSeriesII(n),s_correct))
3 Pass
n = 3; s_correct = 35; assert(isequal(sumOfSeriesII(n),s_correct))
4 Pass
n = 4; s_correct = 84; assert(isequal(sumOfSeriesII(n),s_correct))
5 Pass
n = 5; s_correct = 165; assert(isequal(sumOfSeriesII(n),s_correct))
### Community Treasure Hunt
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Start Hunting! | 244 | 835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.681084 |
https://metanumbers.com/100442600 | 1,638,816,325,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00270.warc.gz | 483,700,595 | 7,734 | # 100442600 (number)
100,442,600 (one hundred million four hundred forty-two thousand six hundred) is an even nine-digits composite number following 100442599 and preceding 100442601. In scientific notation, it is written as 1.004426 × 108. The sum of its digits is 17. It has a total of 7 prime factors and 48 positive divisors. There are 39,513,600 positive integers (up to 100442600) that are relatively prime to 100442600.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 9
• Sum of Digits 17
• Digital Root 8
## Name
Short name 100 million 442 thousand 600 one hundred million four hundred forty-two thousand six hundred
## Notation
Scientific notation 1.004426 × 108 100.4426 × 106
## Prime Factorization of 100442600
Prime Factorization 23 × 52 × 61 × 8233
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 5022130 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 100,442,600 is 23 × 52 × 61 × 8233. Since it has a total of 7 prime factors, 100,442,600 is a composite number.
## Divisors of 100442600
48 divisors
Even divisors 36 12 12 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 48 Total number of the positive divisors of n σ(n) 2.37386e+08 Sum of all the positive divisors of n s(n) 1.36944e+08 Sum of the proper positive divisors of n A(n) 4.94555e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 10022.1 Returns the nth root of the product of n divisors H(n) 20.3097 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 100,442,600 can be divided by 48 positive divisors (out of which 36 are even, and 12 are odd). The sum of these divisors (counting 100,442,600) is 237,386,220, the average is 4,945,546,.25.
## Other Arithmetic Functions (n = 100442600)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 39513600 Total number of positive integers not greater than n that are coprime to n λ(n) 41160 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5778114 Total number of primes less than or equal to n r2(n) 48 The number of ways n can be represented as the sum of 2 squares
There are 39,513,600 positive integers (less than 100,442,600) that are coprime with 100,442,600. And there are approximately 5,778,114 prime numbers less than or equal to 100,442,600.
## Divisibility of 100442600
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 6 0 8
The number 100,442,600 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (100442600)
Base System Value
2 Binary 101111111001010000111101000
3 Ternary 21000000000100022
4 Quaternary 11333022013220
5 Quinary 201203130400
6 Senary 13544500012
8 Octal 577120750
10 Decimal 100442600
12 Duodecimal 2977a608
20 Vigesimal 1b7f6a0
36 Base36 1nsu08
## Basic calculations (n = 100442600)
### Multiplication
n×y
n×2 200885200 301327800 401770400 502213000
### Division
n÷y
n÷2 5.02213e+07 3.34809e+07 2.51106e+07 2.00885e+07
### Exponentiation
ny
n2 10088715894760000 1013336855131020776000000 101782188405183067395457600000000 10223267637106440765174989533760000000000
### Nth Root
y√n
2√n 10022.1 464.843 100.11 39.8459
## 100442600 as geometric shapes
### Circle
Diameter 2.00885e+08 6.31099e+08 3.16946e+16
### Sphere
Volume 4.24466e+24 1.26779e+17 6.31099e+08
### Square
Length = n
Perimeter 4.0177e+08 1.00887e+16 1.42047e+08
### Cube
Length = n
Surface area 6.05323e+16 1.01334e+24 1.73972e+08
### Equilateral Triangle
Length = n
Perimeter 3.01328e+08 4.36854e+15 8.69858e+07
### Triangular Pyramid
Length = n
Surface area 1.74742e+16 1.19423e+23 8.2011e+07
## Cryptographic Hash Functions
md5 34ee198c302a81ee56c36f73e923f78b fdb8f91f8e43e60aea804f87b09ee57e7c3f280a d46df448c12208c7b1bdff388badaf1ebb0ec510651180bca74bd73fa80a67d8 a15ff9918618ae07a11c27285600f81345b53905edf294555ac6a02d775b411ee731d02807d00f9d7c005c9d63f951a0e7e81eea9be0a8ff5fc928a875178513 e3d0d96a5403d25a70cfa0e37c2fdf3bbdd7f973 | 1,600 | 4,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-49 | latest | en | 0.776319 |
https://guia.unl.pt/en/2019/novaims/program/9155/course/100053 | 1,701,504,066,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00096.warc.gz | 331,182,528 | 7,053 | # Statistics I
## Objectives
This unit allows the acquisition of competences related to the most the important techniques of description, summarization and exploration of data. Furthermore, it allows the acquisition of competences related to a set of concepts and methods of probability theory. The emphasis is especially on such topics which will be necessary for understanding statistical inference methods that are presented in other units. The aim is to deepen knowledge of descriptive statistics and probability theory, including the concepts of descriptive measures, conditional probabilities, independence, random variables and distribution functions, expectation, variance and moment generating function, joint probability distributions, and the most important probability distributions and their applications.
At the end of this course, students should be able to:
- Organize information on charts and graphs
- Construct and interpret frequency tables
- Calculate and interpret descriptive measures
- Calculate probabilities by classical definition
- Calculate probabilities by the axiomatic and conditional probabilities definition
- Verify that two events are independent
- Determine and characterize the distribution function (df)
- Calculate probabilities based on the probability (density) function (pdf) and df
- Calculate the mean and variance and apply their properties
- Calculate the moment generating function and obtain the moments
- Indicate the characteristics of families of distributions and identify the distribution of concrete phenomena
- Calculate probabilities and percentiles of the Normal, Student-t, Chi-square, F
- Obtain joint probabilities and marginal distributions
- Check if two random variables are independent
- Calculate the covariance and correlation coefficient
- Calculate joint probabilities and conditional probabilities of the bivariate Normal distribution
## General characterization
100053
6.0
##### Responsible teacher
Ana Cristina Marinho da Costa
##### Hours
Weekly - Available soon
Total - Available soon
##### Teaching language
Portuguese. If there are Erasmus students, classes will be taught in English
### Prerequisites
In order to meet the leaning objectives successfully, students must possess knowledge of Math I.
### Bibliography
• Pedrosa, A. C. e Gama, S. M. A. (2016). Introdução Computacional à Probabilidade e Estatística. 3ª edição, Porto Editora, reimpressão 07-2018.
• Afonso, A. e Nunes, C. (2011). Probabilidades e Estatística. Aplicações e Soluções em SPSS. Escolar Editora, Lisboa.
• Mood, A. M., Graybill, F. A. e Boes, D. C. (1974). Introduction to the Theory of Statistics.3rd Edition, McGraw?Hill.
• Murteira, B., Ribeiro, C. S., Silva, J. A. e Pimenta, C. (2002). Introdução à Estatística. McGraw Hill.
• Reis, E. (1996). Estatística Descritiva. 3ª Edição, Edições Sílabo, Lisboa.
### Teaching method
The curricular unit is based on theoretical and practical lessons. A variety of instructional strategies will be applied, including lectures, slide show demonstrations, step-by-step applications, questions and answers. The sessions include presentation of concepts and methodologies, solving examples, discussion and interpretation of results. The practical component is geared towards solving problems and exercises, including discussion and interpretation of results. A set of exercises to be completed independently in extra-classroom context is also disclosed.
### Evaluation method
1st call: three tests (35%, 20% and 45%, respectively). It is required a minimum of 8 points in the 3rd test for approval.
2nd call: final exam (100%).
## Subject matter
The curricular unit is organized in seven Learning Units (LU):
LU0. Descriptive statistics
• Introduction to statistics
• Organization of the information
• Frequency distributions
• Descriptive measures
LU1. Introduction to probability theory
• History
• Combinatorial models
• Probability definitions
LU2. Probability axioms
• Probability measure
• Conditional probability and independence
• Bayes’s theorem
LU3. Random variables and distribution functions
• Random variable concept
• Distribution function
• Discrete random variables
• Continuous random variables
LU4. Mathematical expectation and moments
• Mathematical expectation
• Variance
• Random variables’ moments
• Moment generating function
LU5. Specific probability distributions
• Uniform (discrete and continuous)
• Bernoulli
• Binomial
• Hypergeometric
• Negative Binomial
• Geometric
• Poisson
• Exponential
• Normal
• Chi-squared
• t-Student
• F-Fisher-Snedecor
LU6. Joint distributions
• Random vectors
• Bivariate discrete probability distributions
• Bivariate continuous probability distributions
• Bivariate Normal distribution
## Programs
Programs where the course is taught: | 1,015 | 4,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-50 | latest | en | 0.861177 |
https://www.jiskha.com/display.cgi?id=1265674350 | 1,503,183,326,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00210.warc.gz | 921,172,467 | 3,541 | # Math I
posted by .
How do you find whether a function is even or odd?
Identify the odd function.
A)f(x) = 5
B)f(x) = 8x
C)f(x) = |x3|
D)f(x) = x2 + 7
## Similar Questions
1. ### odd and even functions
Using f is odd if f(-x) = -f(x) or even if f(-x) = f(x) for all real x, how do I 1)show that a polynomial P(x) that contains only odd powers of x is an odd function 2)show that if a polynomial P(x) contains both odd and even powders …
2. ### math
I know It's probably an easy question but I don't know remember how to do it. Show the work to determine if the relation is even, odd, or neither. a ) f(x) = 2x^2 - 7 b) f(x) = -4x^3 - 2x c) f(x) = 4x^2 - 4x + 4 If f(-x) = f(x), for …
3. ### college algebra
Let f denote an odd function and g an odd function. Decide whether the function h(x)=g(x) f(x) is even or odd.
4. ### Algebra
Let f(x) = 1 – 3x^2. Which of the following is true?
5. ### Algebra
Let f(x) = 1 – 3x^2. Which of the following is true?
7. ### Algebra
Let f(x) = 1 – 3x^2. Which of the following is true? | 360 | 1,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-34 | latest | en | 0.83734 |
https://nbgrader.readthedocs.io/en/stable/user_guide/feedback/hacker/ps1/problem2.html | 1,582,599,783,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00311.warc.gz | 485,340,308 | 46,778 | #### problem2 (Score: 0.0 / 3.0)
1. Written response (Score: 0.0 / 1.0)
2. Written response (Score: 0.0 / 2.0)
Before you turn this problem in, make sure everything runs as expected. First, restart the kernel (in the menubar, select Kernel$\rightarrow$Restart) and then run all cells (in the menubar, select Cell$\rightarrow$Run All).
Make sure you fill in any place that says YOUR CODE HERE or "YOUR ANSWER HERE", as well as your name and collaborators below:
In [1]:
NAME = "Alyssa P. Hacker"
COLLABORATORS = "Ben Bitdiddle"
Consider the following piece of code:
def f(x):
if x == 0 or x == 1:
return x
return f(x - 1) + f(x - 2)
## Part A (1 point)¶
Describe, in words, what this code does, and how it does it.
Student's answer Score: 0.0 / 1.0 (Top)
It computes the fibonacci sequence using recursion, with base cases of $x=0$ and $x=1$.
## Part B (2 points)¶
For what inputs will this function not behave as expected? What will happen?
Student's answer Score: 0.0 / 2.0 (Top)
It computes the fibonacci sequence using recursion, with base cases of $x=0$ and $x=1$. | 325 | 1,084 | {"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} | 2.953125 | 3 | CC-MAIN-2020-10 | latest | en | 0.825231 |
https://certifiedcalculator.com/standard-goldendoodle-weight-calculator/ | 1,721,388,063,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00186.warc.gz | 135,266,540 | 14,629 | # Standard Goldendoodle Weight Calculator
The estimated weight of your Goldendoodle is:
Introduction: Goldendoodles are popular and lovable crossbreeds known for their friendly nature and hypoallergenic fur. As Goldendoodle owners, it’s important to have a general idea of how much your furry friend might weigh as they grow. The Standard Goldendoodle Weight Calculator is a handy tool that helps you estimate your Goldendoodle’s weight based on their age and gender. This information can be helpful for planning their diet, choosing the right collar size, and understanding their development.
Formula: The estimated weight of a standard Goldendoodle is calculated using a simple formula based on age and gender. Here’s the formula in plain text:
• For Male Goldendoodles: Estimated Weight (in pounds) = 50.5 + 5.2 * Age (in months)
• For Female Goldendoodles: Estimated Weight (in pounds) = 48.5 + 4.7 * Age (in months)
This formula provides a close approximation of your Goldendoodle’s weight at a particular age.
How to Use: Using the Standard Goldendoodle Weight Calculator is straightforward and user-friendly. Here are the steps to estimate your Goldendoodle’s weight:
1. Enter the age of your Goldendoodle in months.
2. Select the gender of your Goldendoodle (Male or Female) from the dropdown menu.
3. Click the “Calculate” button.
The calculator will instantly provide you with the estimated weight of your Goldendoodle in pounds.
Example: Suppose you have a Male Goldendoodle that is 9 months old. By entering the age and selecting “Male” from the gender options, clicking “Calculate” will yield an estimated weight of approximately 99.7 pounds.
FAQs:
1. Q: How accurate is the estimated weight provided by the calculator? A: The calculator provides a close approximation of your Goldendoodle’s weight based on standard growth patterns. However, individual variations may occur.
2. Q: Can I use this calculator for Mini or Toy Goldendoodles? A: This calculator is specifically designed for estimating the weight of standard-sized Goldendoodles. It may not provide accurate results for Mini or Toy Goldendoodles, which have different growth patterns.
3. Q: What if my Goldendoodle is a mix of different breeds? A: The calculator assumes a standard Goldendoodle, which is a mix of Golden Retriever and Poodle. For other mixes, the accuracy of the estimate may vary.
4. Q: Is the estimated weight useful for planning my Goldendoodle’s diet? A: Yes, knowing the estimated weight of your Goldendoodle can help you determine appropriate portion sizes and nutritional needs.
5. Q: How often should I use the calculator to estimate my Goldendoodle’s weight? A: It’s recommended to use the calculator periodically as your Goldendoodle grows. For puppies, estimating every month can be helpful.
6. Q: Does the calculator consider the Goldendoodle’s parents’ weights? A: The calculator does not consider parental weights, but it provides a general estimate based on age and gender.
7. Q: Are there any other factors that can influence a Goldendoodle’s weight? A: Diet, exercise, genetics, and overall health can influence a Goldendoodle’s weight. This calculator offers a baseline estimate.
8. Q: Is this calculator suitable for Goldendoodle puppies of all generations (F1, F2, etc.)? A: The calculator can be used for Goldendoodle puppies of different generations, but individual growth rates may vary.
9. Q: Can I use this calculator for older Goldendoodles, or is it just for puppies? A: The calculator is primarily designed for puppies, but you can use it for older Goldendoodles to get a rough estimate.
10. Q: What if my Goldendoodle is underweight or overweight compared to the estimate? A: If you have concerns about your Goldendoodle’s weight, it’s best to consult with a veterinarian for personalized guidance.
Conclusion: The Standard Goldendoodle Weight Calculator is a useful tool for Goldendoodle owners to estimate their pet’s weight as they grow. While it provides a close approximation based on age and gender, individual variations may occur. This information can be valuable for various aspects of Goldendoodle care, from choosing the right-sized accessories to planning their nutrition. Remember to consult with your veterinarian for personalized advice on your Goldendoodle’s health and development. | 925 | 4,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.883259 |
https://www.careercup.com/question?id=5748062545772544 | 1,555,662,279,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527518.38/warc/CC-MAIN-20190419081303-20190419103303-00527.warc.gz | 632,935,517 | 8,431 | ## Amazon Interview Question for Software Engineer / Developers
Country: United States
Interview Type: In-Person
Comment hidden because of low score. Click to expand.
1
of 3 vote
``geeksforgeeks.org/kth-largest-element-in-a-stream/``
Edit:
Although as pointed out k is not the index but the number of elements, we can apply the same concept:
``````Solution #1 O(k)
- Keep sorted array of size K descending
- on every new element
- check if larger than last
- if so pop last( smallest ) and insert new item
Solution #2 O(logk)
- Use Binary Search Tree
- on insert check if item is larger than smallest
- if so remove smallest and insert new item
Solution #3 O(logk) but faster min lookup
- Use MinHeap
- on insert check if item is larger than smallest
- if so remove smallest and insert new item``````
Comment hidden because of low score. Click to expand.
1
of 1 vote
K is not an index here, it's the number of elements 3,5,10 etc.
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````import java.util.PriorityQueue;
import java.util.ArrayList;
import java.util.List;
public class KMaximumNumbers
{
PriorityQueue<Integer> pq;
int capacity;
KMaximumNumbers(int k)
{
pq = new PriorityQueue(k);
this.capacity = k;
}
public static void main(String[] args)
{
KMaximumNumbers k = new KMaximumNumbers(3);
int[] a = {0, 10, 7, 1, 3, 4, 9, 5};
for(int i = 0;i<a.length;i++)
{
System.out.println(k.findKMax());
}
}
{
if(pq.size()>=this.capacity && pq.peek() < a)
{
pq.poll();
}
else if(pq.size() < this.capacity)
}
public List<Integer> findKMax()
{
List<Integer> list = new ArrayList<>(pq);
return list;
}
}``````
Comment hidden because of low score. Click to expand.
0
of 0 vote
``````#include <iostream>
#include <vector>
#include <queue>
using namespace std;
class Tracker {
public:
Tracker(int k)
{
k_ = k;
}
{
if (pq_.size() >= k_ &&
n > pq_.top())
{
pq_.pop();
}
if (pq_.size() < k_) {
pq_.push(n);
}
}
vector<int> GetMaxNumbers()
{
vector<int> out;
while (!pq_.empty()) {
out.push_back(pq_.top());
pq_.pop();
}
for (int n : out) {
pq_.push(n);
}
reverse(out.begin(), out.end());
return out;
}
private:
int k_;
priority_queue<int, vector<int>, greater<int>> pq_;
};
int main(int argc, char const **argv)
{
vector<int> in = {4, 5, 2, 1, 7, 3, 6, 20, 0, -5, 1, 100};
Tracker tr(3);
for (int n : in) {
auto max_numbers = tr.GetMaxNumbers();
for (int m : max_numbers) {
cout << m << ", ";
}
cout << "\n";
}
}``````
Name:
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# Homework 4.pdf - Heather Macgregor Assignment Homework 4...
• Homework Help
• kanraorange
• 4
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Heather Macgregor MATH6A-03-S18-Martinez Assignment Homework 4 due 05/30/2018 at 11:30pm PDT 1. (1 pt) Find the maximum and minimum values of f ( x , y ) = 4 x + y on the ellipse x 2 + 4 y 2 = 1 maximum value: minimum value: Answer(s) submitted: 4(8/sqrt(65)) + (1/(2*sqrt(65))) -4(8/sqrt(65)) - (1/(2*sqrt(65))) (correct) 2. (1 pt) Find the maximum and minimum values of the func- tion f ( x , y , z ) = x + 2 y subject to the constraints y 2 + z 2 = 144 and x + y + z = 10. Maximum value is , occuring at ( , , ). Minimum value is , occuring at ( , , ). Answer(s) submitted: 10+12*sqrt(2) 10 6sqrt2 -6sqrt2 10-12sqrt2 10 -6sqrt2 6sqrt2 (correct) 3. (1 pt) Find the absolute maximum and minimum of the function f ( x , y ) = x 2 - y 2 subject to the constraint x 2 + y 2 = 256. As usual, ignore unneeded answer blanks, and list points in lexicographic order. Absolute minimum value: attained at ( , ) and ( , ). Absolute maximum value: attained at ( , ) and ( , ). Answer(s) submitted: -256 0 -16 0 16 256 -16 0 16 0 (correct) 4. (1 pt) Find the arclength of the curve r ( t ) = h 9 2 t , e 9 t , e - 9 t i , 0 t 1 Answer(s) submitted: eˆ9 - (1/eˆ9) (correct) 5. (1 pt) Consider the path r ( t ) = ( 12 t , 6 t 2 , 6ln t ) defined for t > 0. Find the length of the curve between the points ( 12 , 6 , 0 ) and ( 60 , 150 , 6ln ( 5 )) . Answer(s) submitted: 153.66 (correct) 6. (1 pt) Find the length of the given curve: r ( t ) = ( 5 t , - 4sin t , - 4cos t ) where - 2 t 5. Answer(s) submitted: 7sqrt(41) (correct) 7. (1 pt) Find the arclength of the curve r ( t ) = h 10sin t , - 9 t , 10cos t i , - 3 t 3 Answer(s) submitted: 6sqrt(181) (correct) 8.
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http://mymathforum.com/algebra/16081-upper-lower-bound-theorems.html | 1,495,892,849,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608954.74/warc/CC-MAIN-20170527133144-20170527153144-00012.warc.gz | 323,272,169 | 10,487 | My Math Forum Upper and Lower Bound Theorems
Algebra Pre-Algebra and Basic Algebra Math Forum
November 29th, 2010, 02:17 PM #1 Member Joined: Nov 2009 Posts: 67 Thanks: 0 Upper and Lower Bound Theorems I ran across an exercise that has me a bit puzzled. Consider: $f(x)=x^3+3x^2-5x-10$ Find an integral upper bound using the Upper Bound Theorem and find an integral lower bound using the Lower Bound Theorem. An upper bound is an integer greater than or equal to the greatest real zero. Upper Bound Theorem: If you divide a polynomial function f(x) by (x - c), where c > 0, using synthetic division and this yields all non-negative numbers, then c is an upper bound to the real roots of the equation f(x) = 0. Using synthetic division, I found the integral upper bound to be 2. All the coefficients and the remainder are non-negative. Now on to the lower bound. A lower bound is an integer less than or equal to the least real zero. Lower Bound Theorem: If you divide a polynomial function f(x) by (x - c), where c < 0, using synthetic division and this yields alternating signs, then c is a lower bound to the real roots of the equation f(x) = 0. Special note that zeros can be either positive or negative. Using synthetic division again, I tested -1, -2, -3, -4, and -5. Only -5 yielded alternating signs in the coefficients and remainder of the quotient. This says that -5 is a lower bound and all the real zeros of $f(x)=x^3+3x^2-5x-10$ can be found in the interval $-5 \leq x \leq 2$ Now, -4 failed the lower bound theorem test because the quotient did not produce alternating signs. This would suggest that -4 is not a lower bound, right? However, upon further inspection, the actual zeros are {-3.62, -1.38, 2) This would seem to indicate that a lower bound (in fact the greatest lower bound) should have been -4. Why did -4 fail the Lower Bound Theorem test?
November 29th, 2010, 07:54 PM #2 Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Upper and Lower Bound Theorems If P, then Q. Not all bounds pass the test. The problem is just in the logic of the theorem.
November 29th, 2010, 10:37 PM #3 Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Upper and Lower Bound Theorems The point here is that this is not exclusive. The theorem does not say that, to use Chaz's notation, if not P, then not Q. Instead, this is just a one-sided logic test. This is similar to the idea that if a function is differentiable at a point, then the function is continuous at that point. You can of course have functions that are continuous somewhere but not differentiable there (like a cusp, an angle, etc.). But we do not say, if it is not differentiable then it is not continuous.
November 30th, 2010, 10:36 AM #4
Member
Joined: Nov 2009
Posts: 67
Thanks: 0
Re: Upper and Lower Bound Theorems
Quote:
Originally Posted by DLowry The point here is that this is not exclusive. The theorem does not say that, to use Chaz's notation, if not P, then not Q. Instead, this is just a one-sided logic test. This is similar to the idea that if a function is differentiable at a point, then the function is continuous at that point. You can of course have functions that are continuous somewhere but not differentiable there (like a cusp, an angle, etc.). But we do not say, if it is not differentiable then it is not continuous.
This makes perfect sense to me. If the question were asked this way:
Given the function $x^3+3x^2-5x-10$, is -4 a lower bound?
Even though the answer is "yes", this cannot be shown using the Lower Bound Theorem.
December 23rd, 2015, 06:49 AM #5 Newbie Joined: Dec 2015 From: USA Posts: 4 Thanks: 0 Is it true that the theorem only provides conditions in cases where the polynomial has at least one positive root and at least one negative root? I.e., if a polynomial has at least one positive root, then the upper bound for the roots of the polynomial will be the number c, such that when the polynomial is divided by c the coefficients of the quotient and the remainder will be nonnegative, and also the analogous conditions will apply to the lower bound. And if the polynomial has two positive roots then the theorem can only be used to show an upper bound, not a lower bound? (Similarly in the case of two negative roots the theorem can be used to find a lower bound.) Is this accurate?
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# upper and lower bounds of the zeros
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Contact - Home - Forums - Cryptocurrency Forum - Top | 1,285 | 4,940 | {"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": 4, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-22 | longest | en | 0.889977 |
https://mathematica.stackexchange.com/questions/105896/randompoint-works-with-a-2d-implicitregion-but-fails-for-any-higher-dimension?noredirect=1 | 1,582,232,716,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145282.57/warc/CC-MAIN-20200220193228-20200220223228-00113.warc.gz | 465,772,097 | 31,586 | # RandomPoint works with a 2D ImplicitRegion but fails for any higher dimension
This is inspired by this answer which features code that I cannot get to work at all, and I tried to figure out why.
It boils down to using the function RandomPoint, which is new and a little bit tricky to use (see here and here), in combination with ImplicitRegion.
Here are two and three dimensional regions,
twoDRegion =
ImplicitRegion[{0 <= a <= 1, 0 <= b <= 1, a + b == 1}, {a, b}];
threeDRegion =
ImplicitRegion[{0 <= a <= 1, 0 <= b <= 1, 0 <= c <= 1,
a + b + c == 1}, {a, b, c}];
These regions are the set of all positive tuples that sum to 1,
RandomPoint works on the 2D version, but not the 3D version,
RandomPoint /@ {twoDRegion, threeDRegion}
(* {{0.146978, 0.853022},
RandomPoint[
ImplicitRegion[
0 <= a <= 1 && 0 <= b <= 1 && 0 <= c <= 1 && a + b + c == 1, {a, b,
c}]]} *)
Interestingly, RegionPlot also has trouble with the 3D version,
RegionPlot /@ {twoDRegion, threeDRegion}
but DiscretizeRegion can handle both
DiscretizeRegion /@ {twoDRegion, threeDRegion}
This led me to think that I just need to discretize before selecting the random point, and this in fact works for the 3D case
RandomPoint@DiscretizeRegion@# & /@ {twoDRegion, threeDRegion}
(* {{0.579605, 0.420395}, {0.154819, 0.0404491, 0.804732}} *)
But it won't work for higher dimensions since DiscretizeRegion is limited to 3 dimensions or lower.
Is this a bug?
• I have run into the problem you describe here myself and have accepted the behavior as either an incomplete implementation or a design choice by the developers, not a bug. That is only a speculation on my part, of course. I hope you get an informed answer, perhaps from someone connected to WRI. – m_goldberg Feb 8 '16 at 12:03
• Dear OP, did you find any solution for this issue? I also have similar requirement to use RandomPoint in higher dimension region. Thanks – Prashanth Nov 7 '16 at 2:57
• Any word on a fix for this? I can't sample from BooleanRegions of with n>3. – Edmund May 18 '18 at 23:27 | 593 | 2,047 | {"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} | 2.953125 | 3 | CC-MAIN-2020-10 | latest | en | 0.910049 |
http://www.jiskha.com/display.cgi?id=1318427148 | 1,462,499,672,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861700326.64/warc/CC-MAIN-20160428164140-00043-ip-10-239-7-51.ec2.internal.warc.gz | 604,505,144 | 3,662 | Thursday
May 5, 2016
# Homework Help: math
Posted by unknown on Wednesday, October 12, 2011 at 9:45am.
If the twelfth term of an arithmetic sequence is 58 and the constant difference is 2, what is the first term of the sequence?
• math - Reiny, Wednesday, October 12, 2011 at 9:56am
easy.
d = 2
12 term is 58 ---> a+11d = 58
a+22=58
a = 36 | 119 | 344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2016-18 | longest | en | 0.95901 |
https://socratic.org/questions/how-do-you-find-the-derivative-for-2x-4-1 | 1,716,389,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00535.warc.gz | 466,343,165 | 5,826 | How do you find the derivative for 2x^4-1?
$f \left(x\right) = 2 {x}^{4} - 1 = g \left(x\right) + h \left(x\right)$, where $g \left(x\right) = 2 {x}^{4}$ and $h \left(x\right) = - 1$
$f ' \left(x\right) = g ' \left(x\right) + h ' \left(x\right) = \left(2 {x}^{4}\right) ' + \left(- 1\right) ' = 4 \cdot 2 {x}^{4 - 1} + 0 = 8 {x}^{3}$. | 163 | 335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "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} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.29363 |
https://trustconverter.com/en/weight-conversion/pounds/pounds-to-nanograms.html | 1,719,262,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00647.warc.gz | 516,751,249 | 9,723 | # Pounds to Nanograms Conversion
Pound to nanogram conversion allow you make a conversion between pound and nanogram easily. You can find the tool in the following.
### Weight Conversion
to
input
= 453,592,370,000.00006104
= 4.53592 × 1011
= 4.53592E+11
= 4.53592e+11
= 1,133,980,925,000.00024414
= 1.13398 × 1012
= 1.13398E+12
= 1.13398e+12
= 1,814,369,480,000.00024414
= 1.81437 × 1012
= 1.81437E+12
= 1.81437e+12
= 2,494,758,035,000.00048828
= 2.49476 × 1012
= 2.49476E+12
= 2.49476e+12
= 3,175,146,590,000.00048828
= 3.17515 × 1012
= 3.17515E+12
= 3.17515e+12
= 3,855,535,145,000.00000000
= 3.85554 × 1012
= 3.85554E+12
= 3.85554e+12
= 4,535,923,700,000.00097656
= 4.53592 × 1012
= 4.53592E+12
= 4.53592e+12
### Quick Look: pounds to nanograms
pound 1 lb 2 lb 3 lb 4 lb 5 lb 6 lb 7 lb 8 lb 9 lb 10 lb 11 lb 12 lb 13 lb 14 lb 15 lb 16 lb 17 lb 18 lb 19 lb 20 lb 21 lb 22 lb 23 lb 24 lb 25 lb 26 lb 27 lb 28 lb 29 lb 30 lb 31 lb 32 lb 33 lb 34 lb 35 lb 36 lb 37 lb 38 lb 39 lb 40 lb 41 lb 42 lb 43 lb 44 lb 45 lb 46 lb 47 lb 48 lb 49 lb 50 lb 51 lb 52 lb 53 lb 54 lb 55 lb 56 lb 57 lb 58 lb 59 lb 60 lb 61 lb 62 lb 63 lb 64 lb 65 lb 66 lb 67 lb 68 lb 69 lb 70 lb 71 lb 72 lb 73 lb 74 lb 75 lb 76 lb 77 lb 78 lb 79 lb 80 lb 81 lb 82 lb 83 lb 84 lb 85 lb 86 lb 87 lb 88 lb 89 lb 90 lb 91 lb 92 lb 93 lb 94 lb 95 lb 96 lb 97 lb 98 lb 99 lb 100 lb nanogram 453 592 370 000 ng 907 184 740 000 ng 1 360 777 110 000 ng 1 814 369 480 000 ng 2 267 961 850 000 ng 2 721 554 220 000 ng 3 175 146 590 000 ng 3 628 738 960 000 ng 4 082 331 330 000 ng 4 535 923 700 000 ng 4 989 516 070 000 ng 5 443 108 440 000 ng 5 896 700 810 000 ng 6 350 293 180 000 ng 6 803 885 550 000 ng 7 257 477 920 000 ng 7 711 070 290 000 ng 8 164 662 660 000 ng 8 618 255 030 000 ng 9 071 847 400 000 ng 9 525 439 770 000 ng 9 979 032 140 000 ng 10 432 624 510 000 ng 10 886 216 880 000 ng 11 339 809 250 000 ng 11 793 401 620 000 ng 12 246 993 990 000 ng 12 700 586 360 000 ng 13 154 178 730 000 ng 13 607 771 100 000 ng 14 061 363 470 000 ng 14 514 955 840 000 ng 14 968 548 210 000 ng 15 422 140 580 000 ng 15 875 732 950 000 ng 16 329 325 320 000 ng 16 782 917 690 000 ng 17 236 510 060 000 ng 17 690 102 430 000 ng 18 143 694 800 000 ng 18 597 287 170 000 ng 19 050 879 540 000 ng 19 504 471 910 000 ng 19 958 064 280 000 ng 20 411 656 650 000 ng 20 865 249 020 000 ng 21 318 841 390 000 ng 21 772 433 760 000 ng 22 226 026 130 000 ng 22 679 618 500 000 ng 23 133 210 870 000 ng 23 586 803 240 000 ng 24 040 395 610 000 ng 24 493 987 980 000 ng 24 947 580 350 000 ng 25 401 172 720 000 ng 25 854 765 090 000 ng 26 308 357 460 000 ng 26 761 949 830 000 ng 27 215 542 200 000 ng 27 669 134 570 000 ng 28 122 726 940 000 ng 28 576 319 310 000 ng 29 029 911 680 000 ng 29 483 504 050 000 ng 29 937 096 420 000 ng 30 390 688 790 000 ng 30 844 281 160 000 ng 31 297 873 530 000 ng 31 751 465 900 000 ng 32 205 058 270 000 ng 32 658 650 640 000 ng 33 112 243 010 000 ng 33 565 835 380 000 ng 34 019 427 750 000 ng 34 473 020 120 000 ng 34 926 612 490 000 ng 35 380 204 860 000 ng 35 833 797 230 000 ng 36 287 389 600 000 ng 36 740 981 970 000 ng 37 194 574 340 000 ng 37 648 166 710 000 ng 38 101 759 080 000 ng 38 555 351 450 000 ng 39 008 943 820 000 ng 39 462 536 190 000 ng 39 916 128 560 000 ng 40 369 720 930 000 ng 40 823 313 300 000 ng 41 276 905 670 000 ng 41 730 498 040 000 ng 42 184 090 410 000 ng 42 637 682 780 000 ng 43 091 275 150 000 ng 43 544 867 520 000 ng 43 998 459 890 000 ng 44 452 052 260 000 ng 44 905 644 630 000 ng 45 359 237 000 000 ng
The pound or pound-mass is a unit of mass used in the imperial, United States customary and other systems of measurement.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
poundlb
0.45359237 kg = 7000 grains
≡ 0.45359237 kg
Imperial/US
#### conversion table
poundsnanogramspoundsnanograms
1≈ 45359237000011≈ 4989516070000
2.5≈ 113398092500012.5≈ 5669904625000
4≈ 181436948000014≈ 6350293180000
5.5≈ 249475803500015.5≈ 7030681735000
7≈ 317514659000017≈ 7711070290000
8.5≈ 385553514500018.5≈ 8391458845000
10≈ 453592370000020≈ 9071847400000
nanogram or nanogramme is equal to 10-9 gram (unit of mass), comes from a combination of the metric prefix nano (n) with the gram (g). Plural name is nanograms.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
nanogramng≡ 10-9 g≡ 10-12 kgMetric system SI
### conversion table
nanogramspoundsnanogramspounds
1≈ 2.2046226218488E-1211≈ 2.4250848840337E-11
2.5≈ 5.5115565546219E-1212.5≈ 2.755778277311E-11
4≈ 8.8184904873951E-1214≈ 3.0864716705883E-11
5.5≈ 1.2125424420168E-1115.5≈ 3.4171650638656E-11
7≈ 1.5432358352941E-1117≈ 3.7478584571429E-11
8.5≈ 1.8739292285715E-1118.5≈ 4.0785518504202E-11
10≈ 2.2046226218488E-1120≈ 4.4092452436976E-11
### Conversion table
poundsnanograms
1≈ 453 592 370 000
2.2046226218488 × 10-12≈ 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …) | 2,412 | 5,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.177032 |
https://www.coursehero.com/file/6085293/Chapter-6/ | 1,495,785,830,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608648.25/warc/CC-MAIN-20170526071051-20170526091051-00390.warc.gz | 1,046,457,871 | 24,284 | Chapter 6
# Chapter 6 - MAT2378 Rafal Kulik Version 2009/Oct/17 Rafal...
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Unformatted text preview: MAT2378 Rafal Kulik Version 2009/Oct/17 Rafal Kulik MAT2378 Probability and Statistics for the Natural Sciences Chapter 6 Comments • These notes cover material from Chapter 6, Sections 6.1-6.4, 6.6. They are not complete . I will do a lot of calculations on blackboard. • I’m planning to spend two lectures on this material. • MINITAB will be used to illustrate confidence intervals. Rafal Kulik 1 MAT2378 Probability and Statistics for the Natural Sciences Chapter 6 Point estimation Statistical inference consists of methods used to make conclusions about a population based on a random sample . In particular, we want to estimate an unknown parameter , say θ , using a single number called point estimate . This point estimate is obtained using a statistics , which is simply a function of a random sample. The probability distribution of statistics is called sampling distribution . Rafal Kulik 2 MAT2378 Probability and Statistics for the Natural Sciences Chapter 6 Example: If we want to estimate the parameter μ (the population mean), we may take a random sample Y 1 ,...,Y n and compute a statistics Y . In other words, ¯ Y is the point estimate of μ . We have learned that Y has normal distribution if the population is normal, or approximately normal distribution, if n is big. Therefore, the sampling distribution of ¯ Y is normal. Example: If we want to estimate parameter σ 2 (the population variance), we may take a random sample Y 1 ,...,Y n and compute a statistics S 2 = 1 n- 1 ∑ n i =1 ( Y i- ¯ Y ) (the sample variance). In other words, S 2 is the point estimate of σ 2 . Example: Soybean example p. 179. Rafal Kulik 3 MAT2378 Probability and Statistics for the Natural Sciences Chapter 6 Variance of the estimator and standard error If data Y 1 ,...,Y n come from a population with mean μ and variance σ 2 , then Var( Y ) = σ 2 /n . Thus Standard error of the Mean : σ Y = σ √ n ....
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Ask a homework question - tutors are online | 658 | 2,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-22 | longest | en | 0.832203 |
https://gis.stackexchange.com/questions/290586/how-to-create-a-polygon-from-excel-coordinate-file/290614 | 1,569,010,441,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574077.39/warc/CC-MAIN-20190920200607-20190920222607-00199.warc.gz | 516,288,308 | 31,053 | # How to create a polygon from Excel coordinate file
Currently I have an Excel file with 2 columns.
• area number.
• coordinates.
The coordinate column contains several X and Y points those are seperated with a comma. Next an example how the points are noted:
5.72567034 -3.61719203,5.72567034 -3.61719203,5.72565126 -3.61720133,5.72562551 -3.61721635,5.72560167 -3.61723351,5.72557211 -3.61725235, etc.
point 1: 5.72567034 (longitude) & -3.61719203 (longitude) {,}
point 2: 5.72567034 -3.61719203,
point 3: 5.72565126 -3.61720133,
I want to extract this data to ARCMap 10.6 and make polygon shapefiles out of it. The idea is to extract a polygon from points (points to polygon). but I am not able to create points out of these file. Because all the points of one area are in the same cell. When I try to put every point in one cell by doing in Excel "text to column" it exceeds the width of the worksheet (meaning >256 points per area). And thus by doing this I would lose some data. Then there is this problem that there are over 1000 areas of which this process should be done so manually is not an option.
Does anyone have a solution?
• Can you copy and paste from the excel file to notepad. Save text file. If Then in a new excel sheet import from text file and use delimited as comma separated? It will separate the data out by comma separation. Then you'll have three columns, area, x , y. – enolan Jul 25 '18 at 15:09
You can just edit your coordinate field by putting `POLYGON((` at the beginning and `))` to the end of each entry. Afterwards you can save it as *.txt file and use the 'import layer from textfile' function of QGIS to import your polygons as Well Known Text (WKT). From there you can save them as Shapefiles etc. You have to make shure, that the last and the first coordinate pair are the same to generate a valid polygon. | 502 | 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} | 2.796875 | 3 | CC-MAIN-2019-39 | latest | en | 0.892479 |
https://physics.stackexchange.com/questions/588761/geometry-of-deformation-of-axial-members-connected-to-a-rotating-rigid-element | 1,611,824,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704839214.97/warc/CC-MAIN-20210128071759-20210128101759-00709.warc.gz | 487,179,699 | 32,373 | # Geometry of deformation of Axial members connected to a rotating rigid element
On axial members connected to a rotating rigid element,
we have the following relationship $$\frac{-\delta_1}{a}=\frac{\delta_2}{b},$$
which is based on the concept of similar triangles.
In the relationship, what is $$a$$ and $$b$$?
$$a$$ and $$b$$ refers to the lengths of the rotating rigid member corresponding to the deformed member that is being attached. | 104 | 445 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.963281 |
https://chemistry.stackexchange.com/questions/173953/a-reaction-based-question-from-stoichiometry | 1,685,598,672,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00039.warc.gz | 196,661,001 | 35,302 | # A reaction based question from stoichiometry [closed]
A 110.0g sample of a mixture of CaCl2 and NaCl is treated with Na2CO3 to precipitate the Calcium as Calcium Carbonate. This CaCO3 is heated to convert all the Calcium into CaO and the final mass of CaO is 11.62 grams. The % by mass of the CaCl2 in the original mixture has to be found out..
i've tried out finding the no of moles of CaCO3 which which i got as 0.2075 and assuming that the CaCO3 and the no of moles of CaCl2 should be equal i tried finding out the % which came out to be 20.75% in this method and if i separately found the masses of Ca, Cl2 and summed them up to find the percentage over 110 the % turned out to be 20.09% need some suggestions to get the % anywhere near 15.2% in this process taking all of the compounds to be anhydrous
New contributor
kekule is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers with units. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. You may find useful formatting mathematical/chemical expressions/formulas. May 25 at 18:31
• Assuming no waste during the reaction process, amount of $\ce{CaO}$ isolated in $\pu{mol}$ is same as amount of $\ce{CaCl2}$ in $\pu{mol}$ in original mixture. You also have to assume $\ce{CaCl2}$ in anhydrous form during the calculation. May 25 at 20:19
There is no way to produce $$11.62 \pu{g}$$ of $$\ce{CaO}$$ from a $$\ce{CaCl2 + NaCl}$$ sample weighing $$110.0 \pu{g}$$ which is $$15.2\%$$ $$\ce{CaCl2}$$ by weight. Your calculations are correct.*
Assuming $$100\%$$ yield, and, as mentioned, no hydration whatsoever, you need at least $$20.96\%$$ $$\ce{CaCl2}$$ in a $$110.0 \pu{g}$$ $$\ce{CaCl2 + NaCl}$$ sample to produce $$11.62 \pu{g}$$ of $$\ce{CaO}$$.
*However, if the final product is hydrated, this is possible. Let me know if you need help with calculating percentage hydration.
### Underlying Principle
This is an application of the well-known law of conservation of matter, also known as the principle of atom conservation (POAC). You may refer to R. C. Mukherjee (2004), Modern Approach to Chemical Calculations: An Introduction to the Mole Concept, 7$$^\text{th}$$ edition for a thorough reading with lots of practice problems.
### Calculation
$$\ce{CaCl2 -> -> CaO}$$
We simply apply POAC:
$$\text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} \tag{1}$$
$$\text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \dfrac{m_\ce{CaO}}{M_\ce{CaO}}M_\ce{Ca} = \dfrac{11.62}{56.077}40.078 \pu{g} = 8.30\pu {g} \tag{2}$$
$$\text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} = \dfrac{m_\ce{CaCl2}}{M_\ce{CaCl2}}M_\ce{Ca} = \dfrac{m_\ce{CaCl2}}{110.98}40.078 = 0.36m_\ce{CaCl2} \tag{3}$$
where, in Equations (2) and (3), $$M_i$$ is the molar mass of $$i$$ and $$m_i$$ is the mass of $$i$$ in sample of.
Equating Equations (2) and (3) in accordance with Equation (1):
$$8.30\pu {g} = 0.36m_\ce{CaCl2} \implies m_\ce{CaCl2} = 23.06 \pu{g}$$
Thus, if the original $$\ce{CaCl2 + NaCl}$$ sample weighed $$110.0 \pu{g}$$, $$\ce{CaCl2}$$ was present in $$\dfrac{23.06}{110.0}\times100 \%$$ weight ratio, which is $$20.96\%$$
Note: All calculations have been rounded off to the second decimal place. | 1,112 | 3,484 | {"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": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-23 | latest | en | 0.938504 |
http://www.docstoc.com/docs/68168592/MATH-ACTIVITY-MATH-ACTIVITY--Derivative | 1,416,723,211,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400379190.66/warc/CC-MAIN-20141119123259-00241-ip-10-235-23-156.ec2.internal.warc.gz | 539,787,446 | 15,023 | # MATH ACTIVITY MATH ACTIVITY Derivative
Document Sample
MATH 116 ACTIVITY 1: Review of derivative and antiderivative concepts and techniques
WHY: The work of the second semester relies on concepts and techniques from the first semester; most heavily on the idea
and interpretation of derivative and antiderivative. We will be making extensive and frequent use of the ideas and
calculations for derivatives and antiderivatives in extending the formulas - particularly for calculating integrals and
for extending our antiderivative rules.
LEARNING OBJECTIVES:
1. Begin to know your team for the second semester.
2. Review the important concepts involving the antiderivative
3. Work as a team, using the team roles.
CRITERIA:
1. Success in completing the exercises.
2. Success in working as a team and in filling the team roles.
RESOURCES:
1. Your text - especially sections 4.1-4.6 and 7.1-7.2
2. The review sheet for material from the first semester
3. The team role desk markers (handed out in class for use during the semester)
4. 40 minutes
PLAN:
1. Select roles, if you have not already done so, and decide how you will carry out steps 2 through 5
2. Get to know the members of your new team - name, residence, other interests
3. Complete the exercises given here - be sure all members of the team understand and agree with all the results in the recorder's
report.
4. Assess the team's work and roles performances and prepare the Reflector's and Recorder's reports including team grade .
EXERCISES:
1. For each member of the team, write down her name, local address, and one interesting fact that other students (or faculty)
would not know .
2. Derivatives
5t 2
a.) Give Dt (t 1)e
4
b.) Give the slope of the graph of xy 2x y at the point (1, -1)
2
Dx y [requires implicit differentiation to find Dx y ]
3. General antiderivatives
x x 8 dx
2 3 5
a.)
cos5x
b.) dx
sin5x 9
ln(4 x)
c.) x
dx
4. Particular anitiderivative
3x3
If H(x) = 4 and H(0) = 2 + .75 ln 2 , what is the value of H(5) ?
x +2
SKILL EXERCISES
(Hand in the next Thursday that we meet) Text p. 387, # 6-7, 10-11, 16-17, 22, 28-29, 37-38
CRITICAL THINKING QUESTIONS:(answer individually in your journal)
1. Why is it not enough to know the formula for the derivative to find the formula for a function f(x)? [That is, if f(x) = 3x + 2,
why can’t we calculate an exact value for f(5), for example?] What else is needed? [Compare exercises 3 & 4 for a hint]
2. Did working the review exercises in a group bring up any facts or ideas that you had forgotten from your previous work?
3. What do you expect you can do to help your teammates during this semester? What help do you expect you will receive from
them?
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views: 10 posted: 1/1/2011 language: English pages: 2 | 867 | 3,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2014-49 | longest | en | 0.886684 |
https://www.teamstudy.com/resources/chapter-46-trigonometric-integrals-rtV8syrz | 1,660,129,839,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00171.warc.gz | 905,371,837 | 22,853 | # Chapter 46: Trigonometric Integrals
We introduce a variety of techniques for dealing with trigonometric functions in integrals
Example - $\int sin^{5}(x)cos(x)dx$
Apply Substitution
$u=sin(x) \Rightarrow du=cos(x)dx$
$\int sin^{5}(x)cos(x)dx=\int u^{5}du$
$=\frac{1}{6}u^{6}+C$
$=\frac{1}{6}sin(6)(x)+C$
Example - $\int sin^{3}(x)cos^{3}(x)dx$
$=\int sin^{3}(x)\cdot cos^{2}(x)\cdot cos(x)dx$
$=\int sin^{3}(x)[1-sin^{2}(x)]cos(x)dx$
Apply Substitution
$u=sin(x) \Rightarrow du=cos(x)$
$=\int u^{3}(1-u^{2})du = \int u^{3}-u^{5}du$
$=\frac{1}{4}u^{4}-\frac{1}{6}u^{6}+C=\frac{1}{4}sin^{4}(x)-\frac{1}{6}sin^{6}(x)+C$
Use one trig function for the substitution. Convert extra cos terms to sin terms using $cos^{2}(x)+sin^{2}(x)=1^{2}$
Example - $\int sin^{5}(x)cos^{4}(x)dx$
$=\int sin(x)[sin^{4}(x)]cos^{4}(x)dx$
$=\int sin(x)[1-cos^{2}(x)]^{2}cos^{4}(x)dx$
Apply substitution
$u=cos(x) \Rightarrow du=-sin(x)dx$
$=-\int [1-\underbrace{cos^{2}}(x)]^{2}\underbrace{cos^{4}}(x)\underbrace{(-sinx)dx}$
$u$ $u$ $du$
$=-\int [1-u^{2}]^{2}u^{4}du=-\int (1-2u^{2}+u^{4})u^{4}du$
$=-\int u^{4}-2u^{6}+u^{8}du=-\frac{1}{5}u^{5}-\frac{2}{7}u^{7}+\frac{1}{9}u^{9}+C$
Use the odd power to produce $du$
#### Key Fact
$sin^{2}(x)=\frac{1-cos(2x)}{2}$
$cos^{2}(x)=\frac{1+cos(2x)}{2}$
Example - $\int sin^{2}(x)cos^{2}(x)dx$
$=\int (\frac{1-cos2x}{2})(\frac{1+cos(2x)}{2})dx$
$=\int \frac{1-cos^{2}(2x)}{4}dx= \int \frac{1}{4}-\frac{1}{4}cos^{2}(2x)dx$
$=\int \frac{1}{4}-\frac{1}{8}(1+cos(4x))dx$
$=\frac{1}{4}x-\frac{1}{8}x-\frac{1}{32}sin(4x)+C$
Example - Calculate $\int_{0}^{\frac{\pi}{4}}\sqrt{1+cos(2x)}dx$
$\frac{1+cos(2x)}{2}=cos^{2}(x)\Rightarrow 1+cos(2x)= 2cos^{2}(x)$
Thus, $\int_{0}^{\frac{\pi}{4}}\sqrt{1+cos(2x)}dx$
The absolute value appears since $\sqrt{x^{2}}=|x|$
$=\int_{0}^{\frac{\pi}{4}}\sqrt{2 cos^{2}(x)}dx$
$=\int_{0}^{\frac{\pi}{4}}\sqrt{2}|cos(2x)|dx$
$=\int_{0}^{\frac{\pi}{4}}\sqrt{2}cos(2x)dx$
$=[\frac{\sqrt{2}}{2}sin(2x)]_{0}^{\frac{\pi}{4}}$
$=\frac{\sqrt{2}}{2}sin(\frac{\pi}{2})-\frac{\sqrt{2}}{2}sin(0)=\frac{\sqrt{2}}{2}$
Discuss - $\int_{0}^{\frac{\pi}{16}}\sqrt{1-cos(4x)}dx$
$\frac{1-cos(4x)}{2}=sin^{2}(2x) \Rightarrow 1- cos(4x)=2sin^{2}(2x)$
Thus, $\int_{0}^{\frac{\pi}{16}}\sqrt{1-cos(4x)}dx=\int_{0}^{\frac{\pi}{16}}\sqrt{2sin^{2}(2x)}dx$
$=\int_{0}^{\frac{\pi}{16}}\sqrt{2}sin(2x)dx$
$=[-\frac{\sqrt{2}}{2}cos(2x)]_{0}^{\frac{\pi}{16}}$
#### Key Facts
$sin^{2}(x)+cos^{2}(x)=1\Leftrightarrow tan^{2}(x)+1=sec^{2}(x)$
Divide both sides by $cos^{2}(x)$
$\int sec(x)tan(x)dx=sec(x)$ $\int sec^{2}(x)dx=tan(x)$
Example - $\int tan^{4}(x)dx$
$=\int tan^{2}(x)[sec^{2}(x)-1]dx$
$=\int tan^{2}(x)sec^{2}(x)-tan^{2}(x)dx$
$=\int \underbrace{ tan^{2}}(x)\underbrace{sec^{2}}(x)-\underbrace{sec^{2}(x)+1dx}$
$u$ $du$ we can integrate these
$=\int u^{2}du - \int sec^{2}(x)dx + \int 1dx$
$=\frac{1}{3}(tan(x))^{3}-tan(x)+x+C$
Discuss - $\int tan^{4}(x)sec^{4}(x)dx$
$=\int tan^{4}(x)\cdot (1+tan^{2}(x))sec^{2}(x)dx$
$=\int tan^{4}(x)sec^{2}(x)dx + \int tan^{6}(x)sec^{2}(x)dx$
$=\frac{1}{5}tan^{5}(x)+\frac{1}{7}(x)+C$ | 1,604 | 3,224 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 60, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-33 | latest | en | 0.283337 |
http://www.chegg.com/homework-help/when-will-i-ever-teach-this-an-activities-manual-for-mathematics-for-elementary-teachers-1st-edition-chapter-12.1-solutions-9780321237170 | 1,469,850,506,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257832939.48/warc/CC-MAIN-20160723071032-00198-ip-10-185-27-174.ec2.internal.warc.gz | 368,432,427 | 17,099 | View more editions
TEXTBOOK SOLUTIONS FOR A Problem Solving Approach to Mathematics for Elementary School Teachers 1st Edition
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Chapter: Problem:
Given three points in the plane, is it always possible to find a point equidistant from the three points?
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
Here we have to take any 3 points in the plane
Let the 3 points be
If we take the three points on a same line, that is if are collinear, we cannot find a point equidistant to the 3 points
Example: If take in a same line , suppose take a point any where in the plane, we observe that the distance between not same. That is they are not equidistant to the point .
So, when it is possible means if we take a circle and the 3 points are not collinear then the point is the center of the circum circle
The distance of is same that is the point is equidistant to the 3 points
Corresponding Textbook
A Problem Solving Approach to Mathematics for Elementary School Teachers | 1st Edition
9780321237170ISBN-13: 032123717XISBN: Authors: | 305 | 1,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.863356 |
https://tantalum.academickids.com/encyclopedia/index.php/Bayes_factor | 1,638,892,446,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00072.warc.gz | 614,469,507 | 6,009 | # Bayes factor
In statistics, the use of Bayes factors is a Bayesian alternative to classical hypothesis testing.
Given a model selection problem in which we have to choose between two models M1 and M2, on the basis of a data vector x. The Bayes factor K is given by
[itex]K = \frac{p(x|M_1)}{p(x|M_2)}.[itex]
This is similar to a likelihood-ratio test, but instead of maximising the likelihood Bayesians average it over the parameters. Generally, the models M1 and M2 will be parametrised by vectors of parameters θ1 and θ2; thus K is given by
[itex]K = \frac{p(x|M_1)}{p(x|M_2)} = \frac{\int \,p(\theta_1|M_1)p(x|\theta_1, M_1)d\theta_1}{\int \,p(\theta_2|M_2)p(x|\theta_2, M_2)d\theta_2}.[itex]
A value of K > 1 means that the data indicate that M1 is more likely than M2 and vice versa. Note that classical hypothesis testing gives one hypothesis (or model) preferred status (the 'null hypothesis'), and only considers evidence against it. Harold Jeffreys gave a scale for interpretation of K:
K Strength of evidence
< 1 Negative (supports M2)
1 to 3 Barely worth mentioning
3 to 12 Positive
12 to 150 Strong
> 150 Very strong
Many Bayesian statisticians would use a Bayes factor as part of making a choice, but would also combine it with their estimates of the prior probability of each of the models and the loss functions associated with making the wrong choice.
## Example
Suppose we have a random variable which produces either a success or a failure. We want to consider a model M1 where the probability of success is q=½, and another model M2 where q is completely unknown and we take a prior distribution for q which is uniform on [0,1]. We take a sample of 200, and find 115 success and 85 failures. The likelihood is:
[itex]{200 \choose 115}q^{115}(1-q)^{85}[itex]
So we have
[itex]P(X=115|M_1)={200 \choose 115}\left({1 \over 2}\right)^{200}=0.00595...[itex]
but
[itex]P(X=115|M_2)=\int_{q=0}^1 1{200 \choose 115}q^{115}(1-q)^{85}dq = {1 \over 201} = 0.00497...[itex]
The ratio is then 1.197..., which is "barely worth mentioning" even if it points very slightly towards M1.
This is not the same as a classical likelihood ratio test, which would have found the maximum likelihood estimate for q, namely 115200=0.575, and from that get a ratio of 0.1045..., and so pointing towards M2. A frequentist hypothesis test would have produced an even more dramatic result, saying that that M1 could be rejected at the 5% confidence level, since the probability of getting 115 or more successes from a sample of 200 if q=½ is 0.0200..., and as a two-tailed test of getting a figure as extreme as or more extreme than 115 is 0.0400... Note that 115 is more than two standard deviations away from 100.
M2 is a more complex model than M1 because it has a free parameter which allows it to model the data more closely. The ability of Bayes factors to take this into account is a reason why Bayesian inference has been put forward as a theoretical justification for and generalisation of Occam's razor, reducing Type I errors.
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• Space and Astronomy | 867 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.901997 |
http://www.physicsforums.com/showthread.php?p=2395995 | 1,411,069,756,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657129229.10/warc/CC-MAIN-20140914011209-00292-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 734,303,934 | 10,594 | # Parallel transport and geodesics
by dianaj
Tags: geodesics, parallel, transport
P: 16 A vector field is parallel transported along a curve if and only if the the corariant derivative of the vector field along the path is 0. That is $$\frac{d}{d\lambda} V^\mu + \Gamma^\mu_{\sigma \rho} \frac{dx^\sigma}{d\lambda} V^\rho = 0$$ This is basically what every book says. But what exactly does it mean? 1) if you have a vector (field), there is a specific set of paths, that - when you transport the vector along them - will leave the vector 'unchanged'. That means that there are certain paths that change the vector i.e. stretches it or turns it or whatever. 2) the equation describes how the vector (field) should change when you move it along a certain path (any path you choose) in order to keep it constant with respect to the connection. I can see arguments for both interpretations and it's driving me crazy. ;) In flat space space you can parallel transport a vector in any direction you want - there are no right or wrong paths. This speaks for 2). A geodesic is a path that parallel transports a vector that is the tangent vector to the path itself. This speaks for 1) (i.e. for a vector there is a specific parallel transport path, for the tangent vector to the path this is the path itself) If you have a sphere and parallel transport the vector $$V = (1,0)$$ around a circle of constant $$\theta$$ (altitude) the vector changes like $$V(\theta,\phi) = \left(\cos(\phi \cos\theta), \frac{-1}{\sin \theta}\sin(\phi \cos\theta)\right)$$ This speaks for 2) as you can clearly choose any path (any altitude (except the north/south pole of course)). It also shows the vector is not constant with respect to the path (the path parameter being $$\lambda = \phi$$) as it changes when you go round. So...which one is it? What point am I missing? ;)
HW Helper PF Gold P: 1,961 Vectors can be parallel transported in any direction. There are no preferred directions for parallel transport.
HW Helper PF Gold P: 1,961 In (1), you seem to be implying that somehow it is that path that changes the vector. The vector is defined in the tangent space at a point p. If you have a path through p, this vector may be parallel transported along it to another point. You also refer to vector fields instead of vectors, so again I don't quite understand what you're implying.
Emeritus Sci Advisor P: 7,662 Parallel transport and geodesics The formal definitions of parallel transport are very good for calculation, but for understanding the idea I prefer geometrical constructs, like Schild's ladder. http://en.wikipedia.org/wiki/Schild%27s_ladder Consider a parallelogram - if opposing sides are of equal lengths, the opposing sides are parallel. And we know how to calculate lengths if we have a metric, so there isn't any problem in the notion of "sides of equal length". So if we have a metric, the idea that the sides of a parallelogram "should be" parallel if the parallelogram is small enough gives a unique idea of parallel transport. As long as you don't mind representing vectors as line segments, that is.....
P: 707
Quote by dianaj A vector field is parallel transported along a curve if and only if the the corariant derivative of the vector field along the path is 0. That is $$\frac{d}{d\lambda} V^\mu + \Gamma^\mu_{\sigma \rho} \frac{dx^\sigma}{d\lambda} V^\rho = 0$$ This is basically what every book says. But what exactly does it mean? 1) if you have a vector (field), there is a specific set of paths, that - when you transport the vector along them - will leave the vector 'unchanged'. That means that there are certain paths that change the vector i.e. stretches it or turns it or whatever. 2) the equation describes how the vector (field) should change when you move it along a certain path (any path you choose) in order to keep it constant with respect to the connection. I can see arguments for both interpretations and it's driving me crazy. ;) In flat space space you can parallel transport a vector in any direction you want - there are no right or wrong paths. This speaks for 2). A geodesic is a path that parallel transports a vector that is the tangent vector to the path itself. This speaks for 1) (i.e. for a vector there is a specific parallel transport path, for the tangent vector to the path this is the path itself) If you have a sphere and parallel transport the vector $$V = (1,0)$$ around a circle of constant $$\theta$$ (altitude) the vector changes like $$V(\theta,\phi) = \left(\cos(\phi \cos\theta), \frac{-1}{\sin \theta}\sin(\phi \cos\theta)\right)$$ This speaks for 2) as you can clearly choose any path (any altitude (except the north/south pole of course)). It also shows the vector is not constant with respect to the path (the path parameter being $$\lambda = \phi$$) as it changes when you go round. So...which one is it? What point am I missing? ;)
When you have a Riemannian connection a vector parallel transported along a curve keeps its length and its angle to the tangent direction of the curve. If two vectors are parallel transported along the same curve, the angle between them remains constant.
If you parallel transport a vector around a closed curve then the vector may come back to a different vector even in a flat space. Also the change in the vector when it returns to the starting point may depend on the curve, even in a flat space.
P: 3,967
Quote by dianaj If you have a sphere and parallel transport the vector $$V = (1,0)$$ around a circle of constant $$\theta$$ (altitude) the vector changes like $$V(\theta,\phi) = \left(\cos(\phi \cos\theta), \frac{-1}{\sin \theta}\sin(\phi \cos\theta)\right)$$ ... This speaks for 2) as you can clearly choose any path (any altitude (except the north/south pole of course)). It also shows the vector is not constant with respect to the path (the path parameter being $$\lambda = \phi$$) as it changes when you go round.
Did you mean latitude?
Quote by dianaj A vector field is parallel transported along a curve if and only if the corariant derivative of the vector field along the path is 0. That is $$\frac{d}{d\lambda} V^\mu + \Gamma^\mu_{\sigma \rho} \frac{dx^\sigma}{d\lambda} V^\rho = 0$$
I think a better description might be "A path is a geodesic (the equivalent of a straight line in curved space) if and only if the covariant derivative of the vector field along the path is 0." I am not an expert on this subject so I welcome the real experts to correct the terminology and semantics.
Quote by dianaj This is basically what every book says. But what exactly does it mean? 1) if you have a vector (field), there is a specific set of paths, that - when you transport the vector along them - will leave the vector 'unchanged'. That means that there are certain paths that change the vector i.e. stretches it or turns it or whatever.
The specific set of paths that leave the vector unchanged are called geodesics.
Quote by dianaj 2) the equation describes how the vector (field) should change when you move it along a certain path (any path you choose) in order to keep it constant with respect to the connection.
This is probably the better definition because a vector can be parallel transported along a path without the covariant derivative remaining zero. For example a vector can be parallel transported along a line of constant latitude on the globe and not necessarily still be parallel to itself on returning to the start point. This is because a line of latitude (other than the equator) is not on a great circle and is not a geodesic. This deviation or precession of the vector when parallel transported along a path (the non zero covariant derivative value) is a measure of how much the path deviates from the equivalent of a straight line in curved space (a geodesic).
It might help to think of a imaginary parallel transport vehicle. This vehicle is designed so that whenever the vehicle turns x degrees clockwise, the vector it is transporting turns x degrees anticlockwise (or vice versa) so that on a two dimensional plane the transported vector always remains parallel to its starting position. Now imagine the following journey. The vehicle starts at the equator and heads to the North pole. At the North pole it makes a 90 degree turn clockwise and heads South back towards the equator. At the equator it makes another 90 degree turn clockwise and heads back to its starting point and makes a final 90 degree turn so that the vehicle is aligned with its original heading. The vector it was transporting is now rotated 270 degrees anticlockwise relative to its starting position. The internal angles of the triangle formed by this journey add up to 270 degrees rather than the usual 180 degrees we would normally expect for a triangle in flat space. If this vehicle has its wheels locked in the straight ahead position it will be forced to follow a great circle or geodesic on the globe and its transported vector will not precess from its original direction when it returns to its starting point. Now if the vehicle follows a line of latitude it has to turn its front wheels to stay on the line of latitude because if it tries to keep pointing straight ahead it will be forced to deviate from the line of latitude. This turning of the front wheels of the vehicle to stay on the line of latitude, causes the vector it is transporting to precess. In a nutshell, deviating from a geodesic path causes precession of the parallel transported vector.
I think the source of your confusion is the use of the term "parallel transport" for something that is not necessarily parallel transported in the casual sense.
Page 26 of this document
http://www.shef.ac.uk/physics/teachi..._lecture_8.pdf shows a nice trick with a cone on sphere that may help visualise the situation.
P: 16
Quote by kev It might help to think of a imaginary parallel transport vehicle. This vehicle is designed so that whenever the vehicle turns x degrees clockwise, the vector it is transporting turns x degrees anticlockwise (or vice versa) so that on a two dimensional plane the transported vector always remains parallel to its starting position. Now imagine the following journey. The vehicle starts at the equator and heads to the North pole. At the North pole it makes a 90 degree turn clockwise and heads South back towards the equator. At the equator it makes another 90 degree turn clockwise and heads back to its starting point and makes a final 90 degree turn so that the vehicle is aligned with its original heading. The vector it was transporting is now rotated 270 degrees anticlockwise relative to its starting position. The internal angles of the triangle formed by this journey add up to 270 degrees rather than the usual 180 degrees we would normally expect for a triangle in flat space. If this vehicle has its wheels locked in the straight ahead position it will be forced to follow a great circle or geodesic on the globe and its transported vector will not precess from its original direction when it returns to its starting point. Now if the vehicle follows a line of latitude it has to turn its front wheels to stay on the line of latitude because if it tries to keep pointing straight ahead it will be forced to deviate from the line of latitude. This turning of the front wheels of the vehicle to stay on the line of latitude, causes the vector it is transporting to precess. In a nutshell, deviating from a geodesic path causes precession of the parallel transported vector.
That's a nice way to think of it. It kind of makes sense. :)
Quote by kev Page 26 of this document http://www.shef.ac.uk/physics/teachi..._lecture_8.pdf shows a nice trick with a cone on sphere that may help visualise the situation.
Thank you for the link, it looks good. Before I look at it more closely I just have a few questions.
The change of coordinates that they calculate, describe how the vector will look, when it has been transported to a new point, from the new point's view, right?
Also, the figure at page 24 is sort of how I thought a tangent vector to the equator would be parallel transported around the equator - i.e. along its own geodesic. I thought that transport around a circle in flat space (which is sort of like the top part of the sphere) would keep the vector pointing 'up' (whith up I here mean at 12 o'clock in the drawing).
P: 3,967
Quote by dianaj Also, the figure at page 24 is sort of how I thought a tangent vector to the equator would be parallel transported around the equator - i.e. along its own geodesic. I thought that transport around a circle in flat space (which is sort of like the top part of the sphere) would keep the vector pointing 'up' (whith up I here mean at 12 o'clock in the drawing).
I think the diagram you refer to is a little misleading. Starting at the 3 o'clock position with the vector pointing in the 12 o'clock direction, the transported vector should continue to point roughly in the 12 o'clock direction.
The equation that they give for the precession of a vector transported along a line of latitude is:
$$\alpha = \phi*\cos(\theta)$$
If the vector is transported half a circle around a line of latitude the equation becomes:
$$\alpha = \pi*\cos(\theta)$$
At the equator $\theta$ is 2*pi so $\alpha = 0$ so there is no precession and the vector is pointing in the same direction as the vehicle.
Very near the North pole $\theta$ is close to zero (say 0.1 radians) so for a half circle path $\alpha = \pi*\cos(0.1) = 0.995*\pi$ radians, so by the time the vehicle has travelled to the 9 o'clock position in the diagram the vector will have been rotated nearly half a turn clockwise relative to the vehicle (and in the plane tangential to the surface the vehicle is on) and still pointing roughly in the 12 o'clock direction. (The vector is now pointing roughly to the rear of our imaginary vector transportation vehicle.)
Quote by dianaj The change of coordinates that they calculate, describe how the vector will look, when it has been transported to a new point, from the new point's view, right?
I think that is correct.
P: 16 Thank you for your answers - I think my understanding has improved. :)
Related Discussions Calculus & Beyond Homework 1 Special & General Relativity 20 Special & General Relativity 1 Introductory Physics Homework 0 Differential Geometry 10 | 3,286 | 14,363 | {"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} | 3.796875 | 4 | CC-MAIN-2014-41 | latest | en | 0.908712 |
http://www.britannica.com/topic/harmonic-construction | 1,435,817,245,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095404.78/warc/CC-MAIN-20150627031815-00096-ip-10-179-60-89.ec2.internal.warc.gz | 337,037,895 | 13,135 | # Harmonic construction
Mathematics
Alternative title: harmonic-pair division
harmonic construction, in projective geometry, determination of a pair of points C and D that divides a line segment AB harmonically (see Figure), that is, internally and externally in the same ratio, the internal ratio CA/CB being equal to the negative of the external ratio DA/DB on the extended line. The theorem of harmonicity states that if the external point of division of a line segment is given, then the internal point can be constructed by a purely projective technique; that is, by using only intersections of straight lines. To accomplish this, an arbitrary triangle is drawn on the base AB, followed by an arbitrary line from the external point D cutting this triangle in two. The corners of the quadrilateral formed thus joined and the point determined by the intersection of these diagonals together with the point at the vertex of the triangle determine a line that cuts AB in the proper ratio.
This construction is of interest in projective geometry because the location of the fourth point is independent of the choice of the first three lines in the construction, and the harmonic relationship of the four points is preserved if the line is projected onto another line.
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http://topdrawer.aamt.edu.au/Mental-computation/Activities/How-many-possible-ways | 1,495,899,911,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608956.34/warc/CC-MAIN-20170527152350-20170527172350-00154.warc.gz | 454,374,871 | 6,554 | Home > Mental computation > Activities > How many possible ways?
# How many possible ways?
Factors are pairs of counting numbers that multiply to give another counting number.
Constraining the number of tiles and building different rectangular arrays develops the concept of factors and the commutative property of multiplication.
Pose the following situation:
I made an array with 24 tiles. What might it look like? Close your eyes and picture it. Open your eyes and make what you saw.
Note the way the students work out the different arrays. Do they recognise that by rotating the structure they generate another arrangement?
Students share the different possibilities and draw them on the board or interactive whiteboard.
• What is the same about all the arrays?
• What is different?
• What do you notice about the number of rows and number of tiles in each row?
Pose extension problems such as:
I wonder if another number such as 36 would have as many different possible arrays as 24?
Year 4: Recall multiplication facts up to 10 x 10 and related division facts
Year 3: Recall multiplication facts of two, three, five and ten and related division facts
Source | 234 | 1,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.923317 |
https://www.gamedev.net/forums/topic/301362-rolling-motion/ | 1,506,352,752,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691977.66/warc/CC-MAIN-20170925145232-20170925165232-00056.warc.gz | 762,677,479 | 30,892 | # Rolling Motion
## Recommended Posts
yaroslavd 150
Hi, guys. I'm having problems understanding rolling motion. Specifically, I am trying to simulate physically accurate pool ball movement. I'll tell you what I understand so far and what I don't. Hopefully, some of you can clear up my misconceptions and explain the answers to my questions: If the ball is hit hard enough, it slides at first. Because of kinetic friction, w gets bigger in the direction of motion, v gets smaller. When v = w x r, rolling motion now starts. From what I read, this type of motion is identical to sliding except with a much smaller coefficient of rolling friction. However, this doesn't make sense to me. I mean once v = w x r, the ball's gonna be rolling, but if v keeps decreasing and w keeps increasing, this condition is gonna break the next frame. And if w always increases, how is the ball gonna stop? Basically, I'm very confused about the transition from sliding motion to rolling motion, the conditions for each motion, and the movement of the ball during each type of motion. I'm very sorry, as this is a lot to explain, but if someone can even "common sensically" explain to me what's going on, I'd really appreciate it because I've read many tutorials and can't understand how this works.
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JeffLander 660
Once the ball is rolling true, there is no motion of the point making contact with the table relative to the ball. That is, if you take the velocity of the point of contact, it is 0. So as far as the standard friction model is concerned, the object is at rest so no frictional force is applied. As the velocity of the COM slows due to viscous drag, w x r will be larger than v and a the relative velocity of contact is > 0 so a small frictional force is applied to w in the direction of v. This slows w and increases v restoring the balance.
These forces by themselves aren't really enough to stop the ball rolling true realistically. In practice, you need to add some rolling friction to slow the object rolling true.
Th ebest explaination I have seen of how this works in life, is that the compression of the rolling object and the surface it is rolling on leads to the object actually continuously rolling up a small hill.
I googled a site that has a good set of diagrams.
http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html
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yaroslavd 150
So how would I incorporate this into the equation. Like can someone post short pseudocode like:
if(rolling()) v = ...; w = ...;else v = ...; w = ...;bool rolling(){ ...}
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yaroslavd 150
Please, guys. Even something common sense would do. Like "the ball starts rolling when... At this point, v =...." Please, I don't expect you to type up a thesis on rolling motion, but I would really appreciate ANY type of help.
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JeffLander 660
You need to just make the coefficient of rolling friction another variable. It is much less than kinetic friction though. For example if the kinetic friction is something like 0.8, the rolling friction would be more like 0.01 or less. I just apply that as a damping on the angular velocity just as viscous drag acts on the linear velocity.
Don't set up special cases like you have in your code sample. Just have a general model that applies kinetic friction and rolling friction when the surfaces are in contact. Viscous drag would always be applied even if not in contact with the ground.
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BobMcGee 102
There is no 'real way' of solving this problem. I have solve it however. It takes into account "real physics" but because there's no exact formula for how to do it, this is considered an ad-hoc method.
So, you already established that friction in itself won't bring the ball to rest. That's because it's a mathematically perfect sphere, and it 'sits' on a single point on the table. This means that the normal force (the force of the table pushing up on the ball) doesn't exert a torque on the ball, correct? This should make sense to you because the normal to the surface always points through the ball's center of mass.
Now, the "real" reason spheres come to rest is because spheres in real life, no matter how rigid you think they are, aren't 100% rigid and they never sit on a single point, they always rest over an area. The same is true for the plane that they sit on. Say you roll a billiard ball across the pool table. The fabric of the pool table bunches up in front of the billiard ball producing another torque which bleeds kinetic energy from the system and eventually makes it come to rest. This is what someone earlier said, it's like it's perpetually rolling up a hill when it's moving (even when it's on an incline).
So, in practice, I have solved it a different way (in code). My implementation may not make sense, but when there is a friction force to be applied, I change the direction of the normal force such that it points against the direction of the object's velocity. And, the effect is that the ball rolls for a while, taking into account kinetnc and static friction where applicable, but eventually comes to rest. Here's my implementation:
/* December 4, 2004*/void Physics::WorldImpulseOnSphere(PhysicsObject*a,InteractionData*b){ Vector3D ContactPointVelocity = a->mLinearVelocity + CrossProduct(&a->mAngularVelocity,&b->Position); double NormalSpeed = DotProduct(&ContactPointVelocity,&b->Normal); Vector3D NormalComponent = b->Normal * NormalSpeed; Vector3D TangentComponent = ContactPointVelocity - NormalComponent; double TangentSpeed = TangentComponent.GetLength(); Vector3D TangentDir(0,0,0); if(TangentSpeed > 0) { TangentDir = TangentComponent / TangentSpeed; //normalizes the direction } Vector3D CorrectedNormal = b->Normal - (a->mNormalizedLinearVelocity.Dir * .1); CorrectedNormal.Normalize(); double CorrectedNormalSpeed = DotProduct(&ContactPointVelocity,&CorrectedNormal); //Must do this, else the impulse won't support the object enough in the //geometric normal direction and the object could fall through the world double NormalForce = WorldGetForceLinearOnSphere(a,b,CorrectedNormalSpeed); // a->mLinearVelocity += (b->Normal * (NormalForce / a->mMass)); a->mLinearVelocity += (CorrectedNormal * (NormalForce / a->mMass)); Vector3D ActualFrictionForce(0,0,0); double ImpulseToStop(0); double ImpulseToStopNumerator = TangentSpeed; double InverseInertia = 1/(.4 * a->mMass * a->mRadius * a->mRadius); Vector3D RCrossNOverI = CrossProduct(&b->Position,&TangentDir) * InverseInertia; double ImpulseToStopDenominator = ((1/a->mMass) + DotProduct(&CrossProduct(&RCrossNOverI, &b->Position),&TangentDir) /* n * ((r x n)/i x r) */ ); ImpulseToStop = ImpulseToStopNumerator / ImpulseToStopDenominator; if(ImpulseToStop < (NormalForce * a->mStaticStickiness)) { ActualFrictionForce = TangentDir * -1 * ImpulseToStop; /* As an extra source of resistance, the surface normal bunches up in front of the ball inducing an extra source of resistance to rotational motion. It can be approximately equal to a fourth of the friction force applied at a fourth of the distance away, but there's no way to really know for sure */ // a->mLinearVelocity *= 1-(.1*mFrameTime); } else { ActualFrictionForce = TangentDir * -1 * (NormalForce * a->mDynamicStickiness); } a->mLinearVelocity += ActualFrictionForce / a->mMass; ApplyTorqueOnSphere(a,&b->Position,&ActualFrictionForce); this->UpdateNormalizedAngularVel(a); this->UpdateNormalizedLinearVel(a);}
EDIT:
also, there are no amateur problems with the concept. For instance, the normal force in the original normal direction doesn't decrease in magnitude (otherwise it could fall through the plane) I have taken things like that into account, and I typically know what I'm doign (although the next post willb e somebody pointing out a math mistake knowing mein luck lol)
Also ask me to in detail better explain what I am talking about. It's really just waht the others were speaking of however.
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yaroslavd 150
The thing is, I understand that there's kinetic friction and rolling friction and that the former is a lot bigger than the former. However, I am confused about when to use which. Also, the coefficient just affects the magnitude of the friction vector. As far as I can see, friction will still increase w and decrease v. So how does it all fit together?
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Fil 168
Sorry I've not been of help. :(
Maybe someone else can explain better my solution (if it really was a "solution" [grin]).
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Squirm 481
There is no when to use each - use both - the rolling friction is so small that you won't notice it when you are sliding, and when you are rolling the normal friction is so small that the rolling friction makes a difference, and when you are putting spin on a snooker ball they both apply in different directions, and it all just balances out.
TotalForce = NormalFriction + RollingFriction
There is no if statement :o) | 2,279 | 9,119 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-39 | longest | en | 0.939844 |
https://essayassignmentanswers.com/electrical-engineering-homework-help-4-2/ | 1,660,492,766,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00771.warc.gz | 250,702,292 | 19,344 | ECE505 Digital Signal Processing
Laboratory Project – Equalizer Design via the Windowing Method
1 Problem Description:
In this project we are going to design and construct a digital 5-channel equalizer for processing audio
files. Equalizers are used for adjusting the gain of an audio signal in different frequency bands.
2 Project Definition:
The equalizer design will have 5 bandpass filters in parallel, each multiplied by a gain constant (i.e.,
the band weight). The overall equalizer system can be drawn in block diagram form as follows:
So we have a parallel group of bandpass filters. The equivalent system transfer function can be written
as
H(z) = a1H1(z) + a2H2(z) + a3H3(z) + a4H4(z) + a5H5(z),
which leads to the total system impulse response,
h[n] = a1h1[n] + a2h2[n] + a3h3[n] + a4h4[n] + a5h5[n].
We are going to design each bandpass filter as an FIR using the windowing method. To begin, we
start with an ideal frequency response for a bandpass filter with lower and upper cutoff frequencies,
ωl and ωu, respectively, i.e.,
|H(e
jω)| =
(
1, if − ωu ≤ n ≤ −ωl
, ωl ≤ n ≤ ωu
0, otherwise.
We can then obtain the impulse response hi
[n] for this filter by applying the definition of the inverse
DTFT. The resulting impulse response will exist from (−∞, ∞) and we thus must window and shift
hi
[n] to make it a causal FIR filter for implementation. For each filter stage we have
hk[n] = hi
[n + (M − 1)/2]w[n + (M − 1)/2],
1
where k ∈ (1, 5) and w[n] is a windowing function of length M, with M odd. Once these filters are
combined to produce the overall impulse response h[n], we can process our audio file via convolution,
i.e.,
y[n] = x[n] ∗ h[n].
We will assume the input audio files are sampled at a rate of 44.1kHz, which is the rate used on audio
CDs. With our sampling rate fs = 44.1kHz (T = 1/44100 seconds), the effective continuous frequency
responses of the five filters should have a combined bandwidth ranging from 0 to 22.5kHz. I suggest
you divide the combined bandwidth up into five equal bands on a logarithmic scale. The reason for
such a selection is due to the logarithmic way in which humans perceive frequency.
I have provided a MATLAB function, bpw, that implements a causal FIR bandpass filter using the
windowing method. To use it, all you need to specify are the lower and upper digital cutoff frequencies
(in radians per sample), the length of the filter M, and the type of windowing function to apply (i.e.,
I’ve done the hard part). For additional help with the function open the .m file and read the help
section at the beginning of the file.
Your task is to design and create the 5 bandpass filters, which simply involves determining the cutoff
frequency for each filter and choosing a window type. For each filter use a length of M = 501. Also,
note that h1[n] and h5[n] will be lowpass and highpass filters, respectively. This is easily accounted
for by setting ωl = 0 for h1[n] and ωu = π for h5[n]. Once you have created each filter in MATLAB,
you simply need to multiply each filter by its respective gain factor (which are tuning parameters)
and sum the scaled impulse responses together. Then, apply the filter via convolution, i.e., using the
MATLAB command conv(x, h). Once processed, the reconstruction process is to be done using the
sound() command in MATLAB so that you can listen to the results of applying your filter.
To design your equalizer perform the following design tasks. Include any results generated in this
section in the Design section of the document your submit for this project.
1. Use the inverse DTFT to derive the impulse response hi
[n] for an ideal bandpass filter with
lower cutoff ωl and upper cutoff ωh.
2. Determine the upper and lower analog cutoff frequencies for each bandpass filter in radians per
second. I suggest using a logarithmic scale and dividing the frequency axis into five nearly equal
segments from the range of 22.5Hz to 22.05kHz. Once you determine these analog frequencies,
map each one to their corresponding digital frequency in radians per sample.
3. Create the five filters in MATLAB using the bpw function with the digital frequencies you
determined above. Create h3[n] using a rectangular window. Then generate h3[n] again using
a Blackman window. For the other four filters select windows of your choosing. Create the five
gain factors as configurable parameters. Now, multiply each impulse response by its respective
filter gain and sum them all together to obtain the total impulse response.
4. Let’s also generate 10th order IIR bandpass filters using ωl and ωu from h3[n] based upon
Butterworth, Chebychev, and Elliptic filters. Create each filter using these commands:
[b1, a1] = butter(5, [wl/pi wu/pi])
[b2, a2] = cheby1(5, 1, [wl/pi wu/pi])
2
[b3, a3] = ellip(5, 1, 80, [wl/pi wu/pi])
In each case, notice that the digital cutoff frequencies are normalized by π. The reason for this is
because MATLAB expects the [0, π] range to be normalized to the [0, 1] range. These functions
return the numerator transfer function coefficients in b and the denominator transfer function
coefficients in a that can then be used to generate the frequency response and corresponding
difference equation.
5. Now, read in an audio file using the audioread command (there are several music clips on the
class website that have been sampled at 44.1kHz). Convolve the audio file with the total impulse
response. Listen to the result using the sound command. Now, tune the gain parameters until
you find a set of values that you like.
4 Questions:
Prepare a document for online submission that includes all requested figures, information, and answers
to the questions below.
1. Show the results of your analytic derivation for the impulse response of the ideal bandpass filter.
Can you infer any meaning from this result?
2. Explain how you determined the analog filter cutoff frequencies and create a table in your
document that contains your designed analog cutoff frequencies in radians per second:
h1[n]
h2[n]
h3[n]
h4[n]
h5[n]
3. Explain how you obtained the digital cutoff frequencies and create a table in your document
that contains the digital cutoff frequencies in radians per sample corresponding to the analog
cutoff frequencies specified above in the following format:
h1[n]
h2[n]
h3[n]
h4[n]
h5[n]
4. Plot the middle bandpass filter impulse response, h3[n], using both the rectangular and Blackman
windows. Use the stem command and plot them on two separate subplots. Include this plot as
Figure 1. Discuss the differences you observe when comparing the impulse responses.
3
5. Next, plot the magnitude of the frequency response |H3(e
jω)| for the middle bandpass filter using
the rectangular and Blackman windows on the same plot. Use N = 4096 points in the fft. Also,
generate the magnitude frequency response for an ideal bandpass filter (i.e., a rect function) at
the corresponding digital cutoff frequencies and plot it on the same plot for comparison. I would
also suggest you zoom your plot to best see the details of the filters using the axis command,
e.g., axis([−.5 .5 0 1.2]) or whatever range best displays your filters. Use digital frequency
(radians/sample) for the frequency scale, i.e., [−π, π]. Include this plot as Figure 2. Compare
the plots. Discuss the differences you observe between the two filters and explain how they each
compare to the ideal bandpass filter.
6. Next, use the transfer function coefficients to generate the magnitude frequency response for
each 10th order IIR filter and plot them on the same plot, similar the Figure 2. Again, plot
the magnitude response for the ideal bandpass filter at the given cutoff frequencies on the same
plot for comparison. Use digital frequency (radians/sample) for the frequency scale, i.e., [−π, π].
Include this plot as Figure 3. Compare the plots. Discuss the differences you observe between
the three filters and explain how they each compare to the ideal bandpass filter. How do these
filters compare with the FIR counterparts? Comment on the performance in regards to the order
of the FIR versus IIR filters.
7. Create a table in your document similar to that below that specifies the filter gain factors you
selected for your equalizer and explain why you chose those values for modifying the sound clip.
Also, provide the values in dB (i.e., ˆak = 20 log10(ak) dB).
Filter Gain (V/V) Gain (dB)
h1[n]
h2[n]
h3[n]
h4[n]
h5[n]
8. Using these gain values, plot the magnitude frequency response for each individual filter, i.e.,
|Hk(e
jω)| for k = 1, 2, . . . , 5 (again use N = 4096 points in the fft). You should plot all five
filter responses on the same plot. Plot these as a function of Hertz rather than digital frequency
(f = fs
ω
) and use a logarithmic scale for the x-axis. To do this, instead of using the plot
command, use the semilogx command. Include this plot as Figure 4. Explain the characteristics
you observe with each individual filter. Does each filter passband appear to be nearly the same
width on a logarithmic scale per our design guidelines? Explain.
9. Now, repeat the previous step for the total system magnitude frequency response |H(e
jω)| and
include this result as Figure 5. Discuss the plot as compared with the individual frequency
responses.
10. Plot 2000 samples of the input and output audio sequence, x[n] and y[n], on the same plot.
Include this plot as Figure 6. Explain any differences that you observe and relate it to the shape
of the total frequency response and the gains you selected.
11. Plot the frequency spectrum of the input and output audio signals, X(e
jω) and Y (e
jω), on two
separate subplots again using a logarithmic frequency scale (Hertz) for the x-axis. In this case,
do not specify N for the fft so that it will set N equal to the length of the sequence. Include
these plots as Figure 7. Based upon your chosen band gains, does the output signal indicate
that these gains have been properly applied? Explain.
5 | 2,414 | 9,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.899702 |
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Is the 4 x 100m relay the same as the 400m?
Updated: 12/13/2022
Wiki User
14y ago
the total distance you run is 400m. so it is the same thing. all 4x100 means is that there are 4 people running 100 meters.
Wiki User
14y ago
Earn +20 pts
Q: Is the 4 x 100m relay the same as the 400m?
Submit
Still have questions?
Related questions
Which althlete won 4 gold medals in los angeles Olympics for the 100m 200m 400m relay and the long jump?
Carl Lewis of the United States.
What events did Michael Phelps win his 8 gold medals in?
200m Freestyle 100m Butterfly 200m Butterfly 200m Individual Medley 400m Individual Medley 4 x 100m Freestyle Reley (First Leg) 4 x 200m Freestyle Relay (First Leg) 4 x 100m Medley Relay (Butterfly Leg/Third Leg)
What 5 sports has been in every modern Olympics?
Athletics, cycling, fencing, gymnastics, and swimming have been competed at all Summer Olympics since the 1896 Games in Athens.
How many lengths of the pool is 400m?
4 lengths of 25m (not 25km) is 100m So 4x4=16 lengths to make 400m
What is the perimeter of a square with an area of 10000m2?
If the area is 10000m2, each side has length 100m Therefore the perimeter is 4*100m = 400m
What is the difference between a relay and a dash?
A relay involves 4 people passing on a baton for the race (4X100, 4X400 etc.) A dash is like a sprint and is run individually (100m 200m 400m etc.) The difference is that a relay is run with more than one person.
What is the difference between a dash and a relay?
A relay involves 4 people passing on a baton for the race (4X100, 4X400 etc.) A dash is like a sprint and is run individually (100m 200m 400m etc.) The difference is that a relay is run with more than one person.
What races did Michael Phelps enter in the Beijing games?
400m Individual medley 4x100m freestle relay 200m freestyle 200m butterfly 4 x 200m freestyle relay 200m individual medley 100m butterfly 4x100m medley relay. Winning gold in 8 out of 8.
What runner runs then longest leg of the 800 meters x 4 relay?
They all run 400m depending on where they get the handoff
What is a mile relay track?
The mile relay is another way of saying the 1600m relay or 4x400m relay. This event involves 4 runners each running and handing off a baton after completing 400m
How many times around an outdoor track is a 4x100meter relay?
Well if its your typical 400m track, then 4 times.
How many Olympic medals has Eamon Sullivan won?
Two - Bronze in the 2008 4 x 100m Freestyle Relay Silver in the 2008 100m Freestyle | 723 | 2,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.942013 |
https://capitalsportsbets.com/ZuluCodeBets4/sports-betting-explained-up-games.html | 1,575,929,546,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00280.warc.gz | 306,754,478 | 6,715 | There are many ways that you can bet on football and I will go over all of the different options below. The first and most common bet is the straight bet, which simply means the team wagered on must win by the point spread given at the time of the wager. Most of the time the odds on a straight bet is -110, which means for every \$110 bet, you win a \$100 back. An example of a straight bet is if the team you bet on is a 6 point (-6) favorite, also known as giving up 6 points. Then in order for you to win, the team taken in the wager must win by more than 6 points in order to collect your winnings. If they win by 6, then this will be called a push and your original bet will be refunded to you. You can also bet on the totals, which is also known as the over/under bet. The totals bet is The combined score of both teams for games wagered on, all totals pay out at -110, which is \$110 dollar bet makes you \$100, no matter if you bet on the over or the under. If the total score equals the line, then the bet is a push and you get your original bet back. Another type of bet is the money line bet. This bet is simply that team wagered on just has to win the game. The odds on these games vary depending on the disparity between the favorite and the underdog. You obviously will win more betting on the underdog; however there is a reason why these bets pay more because the odds of the underdog winning are decreased. Although betting the moneyline can sometimes return a big sum, it is wise to know that the bookmakers hold percentage is much larger than a typical straight bet on the side or total of a game. A classic straight wager on the point spread is almost always dealt at -110 on both sides which is 20 cents of vigorish. Compare that to a double digit favorite in college football with a money line of -600/+450 and you can see the extra 130 cents of vig added in.
If you've never set foot in an actual sportsbook before or logged into an online sportsbook, the chances of you getting overwhelmed when you actually do is very high. In an actual Las Vegas sportsbook, there is typically a lot of commotion and the odds and lines are displayed on a massive digital board for everyone to see. When a novice sports bettor looks at the massive digital signage, they will see a bunch of numbers, both positive and negative, some two digits, some three digits. They also won't have a clue what any of it means. The same can be said for the online sportsbooks. It looks like a massive spreadsheet with negative and positive numbers beside each teams' name. Read More >>
Sports betting would be easy — or maybe just easier — if all that was required was to correctly pick the winning team. Gambling institutions, sportsbooks and bookies fall back on point spreads to make the process a little more difficult and to create the ultimate wagering challenge. You'll need a solid understanding of the point spread system if you hope to have a profitable season.
Football betting is by far the most popular form of sports betting by far. Most professional gamblers only bet for 5 months out of the year because of this. When you include College and NFL together there are almost a hundred games to bet on during a 2-day span. This is what is so appealing to the professional gambler; the action packed two days that comes every fall for five months. The gambler has so much to watch in a 48 hour span that the rush alone makes these bets fun. Now I do not expect you to go out and bet on every game played on a Saturday or Sunday. However, I will outline below how you can bet on football also how to become more successful at football betting
```In this instance, the Dodgers are the favored team, as signified by the negative numeral. It would cost you \$130 in order to collect a \$100 payout on a Dodgers victory (plus the original wager of \$130). But if you bet \$100 on the Cubs, you'd collect \$120 if they win (plus the original wager). In other words, you'll have to wager more money on Los Angeles than you would Chicago in order collect \$100 on a bet.
```
Betting on sporting events has long been the most popular form of spread betting. Whilst most bets the casino offers to players have a built in house edge, betting on the spread offers an opportunity for the astute gambler. When a casino accepts a spread bet, it gives the player the odds of 10 to 11, or -110. That means that for every 11 dollars the player wagers, the player will win 10, slightly lower than an even money bet. If team A is playing team B, the casino is not concerned with who wins the game; they are only concerned with taking an equal amount of money of both sides. For example, if one player takes team A and the other takes team B and each wager \$110 to win \$100, it doesn’t matter what team wins; the casino makes money. They take \$100 of the \$110 from the losing bet and pay the winner, keeping the extra \$10 for themselves. This is the house edge. The goal of the casino is to set a line that encourages an equal amount of action on both sides, thereby guaranteeing a profit. This also explains how money can be made by the astute gambler. If casinos set lines to encourage an equal amount of money on both sides, it sets them based on the public perception of the team, not necessarily the real strength of the teams. Many things can affect public perception, which moves the line away from what the real line should be. This gap between the Vegas line, the real line, and differences between other sports books betting lines and spreads is where value can be found.
# Since more people wager on football than any other sports it’s usually the easiest game to figure out how to bet. If you don’t know what to do, you can probably ask anyone in the sports book for instructions. The people visiting the casino might not know everything but they can give you the basics. Here’s a fairly thorough look at the different ways you can wager on football in Las Vegas.
```You’ll need to know how the rotation number, point spread, moneyline and over/under are used and what each means. Each of these terms may be called by another name. The rotation number if often referred to as the number or the rotation mark, moneyline becomes line, point spread is called the spread, and over/under becomes the total. When you come down to it, these are the major terms that can be lumped under the heading odds.
```
If you place wagers on US sports, then chances are high that you've heard of point spreads. Here's how they work; if a game has Patriots -9.0 and Vikings +9.0, the Patriots are 9.0 point favorites and the Vikings are 9.0 point underdogs. Unless otherwise stated, no matter which team you bet on, you'll be required to risk \$1.10 for each \$1.00 you want to win. For Patriots bettors to prevail, they need their team to win by 10 or more points. A 9-point Patriot victory would be a push (a tie). For Vikings bettors to take home the victory, they need to either win the game or lose by less than 9 points.
Once you have found the right betting provider and deposited money, there is usually a bonus on the deposited amount. Then you can already place your first bets. But before it comes to that, you should have a look at the website. The offer of the betting providers does not only refer to bets, but also to sports news as well as tips and tricks for successful bets. This information should definitely be read in peace, then you can make the first bet targeted. This step reveals how extensive the bookmakers' platforms are. On average, betting providers manage around forty different sports. The offer ranges from football to Formula 1 and surfing. For each sport several countries, leagues and competitions have to be selected. In football, for example, there are European national leagues such as the German Bundesliga, the DFB Cup or the English Premier League, international competitions such as the Champions League and national championships, i.e. the European Championship and the World Cup. This exists for women and men as well as for almost all countries in the world. So there are numerous possibilities at games, races or competitions on which you can place bets.
### The simplest way to think about a moneyline is to consider a base bet of \$100. A moneyline is a number larger than 100, and it is either positive or negative. A line with a positive number means that the team is the underdog. If the line, for example, was +160 then you would make a profit of \$160 if you were to bet \$100. Obviously, then, the team is a bigger underdog the bigger the number is - a +260 team is perceived to be less likely to win than a +160 team.
You'll also usually be given the option to wager on the game with the money line in football and basketball. In this case, all you have to do is pick the winner of the contest, but there's one drawback. If you bet on the team that's expected to win, you can find yourself risking much more money than you stand to win. Each team is assigned odds, much like in a horse race where a 2-1 favorite will pay out much less than a 15-1 longshot. This method also evens the playing field for bookies, sportsbooks and other gambling institutions.
Making money from betting on any sport requires an understanding of how the betting markets work, and an ability to use that knowledge to gain an advantage. No-one makes a consistent profit from simply following hunches or relying purely on statistics and trends. Sports handicapping is something that people often assume is very complicated, but the basics are actually relatively easy to grasp.
Even if you’re brand-new to betting on football, you’ve probably seen NFL spreads published in the paper or talked about on TV. The point spread levels the playing field, making betting on either the favorite or the underdog equally attractive. Either the favorite has to win by the margin listed, or the underdog has a buffer to lose the game by that margin, or win it outright.
If you'd rather not deal with point spreads, you can do a "Money Line" wager, which is whether the team/player you bet on wins or loses. You will see "Hawaii Tech +150 or Alaska Tech -140". This means if you bet on Hawaii Tech, if you bet \$100, you would profit \$150 (returns \$250), while to win \$100 on Alaska Tech, you must put up \$140. Those ratios work whether you're betting \$100, \$10, or any other multiple of money.
Betting on sporting events has long been the most popular form of spread betting. Whilst most bets the casino offers to players have a built in house edge, betting on the spread offers an opportunity for the astute gambler. When a casino accepts a spread bet, it gives the player the odds of 10 to 11, or -110. That means that for every 11 dollars the player wagers, the player will win 10, slightly lower than an even money bet. If team A is playing team B, the casino is not concerned with who wins the game; they are only concerned with taking an equal amount of money of both sides. For example, if one player takes team A and the other takes team B and each wager \$110 to win \$100, it doesn’t matter what team wins; the casino makes money. They take \$100 of the \$110 from the losing bet and pay the winner, keeping the extra \$10 for themselves. This is the house edge. The goal of the casino is to set a line that encourages an equal amount of action on both sides, thereby guaranteeing a profit. This also explains how money can be made by the astute gambler. If casinos set lines to encourage an equal amount of money on both sides, it sets them based on the public perception of the team, not necessarily the real strength of the teams. Many things can affect public perception, which moves the line away from what the real line should be. This gap between the Vegas line, the real line, and differences between other sports books betting lines and spreads is where value can be found.
For example, a negative value (-11.5) next to a team indicates that they are favored by that many points. So you must deduct 11.5 points from their score to determine if they won the game or not. On the other hand, a positive value on the same game (+11.5) means the underdog starts with an 11.5-point lead before the game even begins. In NFL betting, the favorite must win by 12 points or more to cover the NFL spread. The underdog is able to lose by 11 points and still cover the spread. | 2,761 | 12,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-51 | latest | en | 0.952302 |
https://drawabox.com/community/sketchbook/manasseh/replies | 1,726,850,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00745.warc.gz | 183,363,274 | 11,339 | The Autumn Promptathon is Coming
2024 • 09 • 24 - 2024 • 09 • 30
The Autumn Promptathon is Coming
2024 • 09 • 24 - 2024 • 09 • 30
# manasseh
## manasseh's Sketchbook
##### 10:12 AM, Thursday June 6th 2024
Such a quick response wow thank you so much!! And yeah the wackier shapes appeared are mostly out of boredom haha.
Thank you! On to lesson 3 it is ^^
##### 8:29 AM, Tuesday April 30th 2024
oohh I can see some things are clicking but there's a lot of the same mistakes. If you want, I can do a little draw-over later today? I feel like I can explain better like that than if I try to explain in words haha
0 users agree
##### 10:40 AM, Friday April 26th 2024
Your ghosted lines look amazing! The superimposed ones are also very promising, but it'd be great to see the beautiful straight lines from your ghosted lines exercise here as well. To step it up some more I think it'll be very helpful to work on trying to hitting a little closer to the end point for both these exercises (as well as any line, plane, box following), mostly to help with the fraying (for me it helps to put my pen down after ghosting, look at the end point and then execute the mark!). All in all they're solid though gg.
Ellipses in planes -- Your ellipses look smoothly executed and confident! They do overshoot the edges of the plane, so maybe planning a little more carefully would work, but here too, it's solid and with obvious progress within 2 pages.
Tables of ellipses -- a little more wobbly here! Make sure the ellipses are snug and evenly shaped. Not working too small helps in this exercise.
The same goes for the funnels. Make sure theyre touching the edges. I think you could try the optional version mentioned in the lesson here to start narrow near the center and increase the degree as you move outwards.
Plotted perspective -- only nitpick is to also do hatching lines with care!
Rough perspective -- Make sure you're lining your boxes with the same techniques as all other lines: planned well, ghosted, then a single confident stroke. Here most are redrawn multiple times and on the wobbly side. Reread about guessing, ghosting, and maybe this lesson. Same for organic perspective exercise, and rotated boxes too.
You're underestimating most of the converging lines so apply these tecniques accordingly.
Next Steps:
I was unsure whether or not to mark complete, but I chose to request revisions because of this;
Your lines and ellipses exercises start out showing you have a good grasp of the lesson material. However, later with your boxes, it seems to miss the careful planning of each dot and each line that the drawabox lessons drill in. Plot each of your points and lines with care, and then execute with confidence as you did with your superimposed and ghosted lines and your ellipses in planes which you did very well.
I suggest reading through the lessons on planning, plotting, ghosting and execution, and then redoing the rough perspective and organic perspective exercises while keeping this in mind.
2 users agree
##### 10:11 AM, Friday April 26th 2024
For your ghosted lines, most look nice and straight and confident with minimal fraying.
Your funnels are also mostly fine. Make sure the ellipses are touching eachother.
I think you could redo the 1 point rough perspective boxes exercise, since I see you're capable of much cleaner lines!
Make sure you properly plan and use the ghosting method for each of your lines and draw each once, confidently instead of multiple times as taught here
The horizontal lines should be parallel to the horizon line. Maybe focus on that, tackling the converging lines second. It seems you underestimate how steep some of these lines can get, by how far they end up from the vanishing point. check here again
For the plotted perspective exercise, make sure, again, that your vertical lines are parallel to eachother and at a right angle from the horizon line. Also try to keep any lines, even hatching for shading, clean and intentional.
You may need to show the rest of the exercises before the homework is complete, - rotated boxes, organic perspective, planes and ellipses are leaving a very big gap.
Next Steps:
Finish the rest of the assigned exercises, and try to get in more mileage with your boxes. Above all try to make clean confident lines using the ghosting method in all of your work, especially your boxes, and make sure parallel lines/planes are parallel. Read through the relevant lessons again to make sure you've got everything down.
Good work and good luck!!
##### 9:19 AM, Wednesday March 13th 2024
Thank you so much for tthe crit!! I definitely struggle with ellipses a lot, although curves are just fine somehow. Usually when drawing I just do two halves I suppose. I will warm up with them more.
How do you schedule(?) your warm ups?
I want to fill this sketchbook, so there will be bleeding lines here and there, but I'll clean up my pictures next hand-in :)
Thanks again! Have a good week ~
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### How to Draw by Scott Robertson
When it comes to technical drawing, there's no one better than Scott Robertson. I regularly use this book as a reference when eyeballing my perspective just won't cut it anymore. Need to figure out exactly how to rotate an object in 3D space? How to project a shape in perspective? Look no further. | 1,291 | 5,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.951321 |
https://www.physicsforums.com/threads/motion-in-2d-snowball-problem.175871/ | 1,532,163,393,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592420.72/warc/CC-MAIN-20180721071046-20180721091046-00268.warc.gz | 939,027,809 | 14,689 | # Homework Help: Motion in 2D snowball problem
1. Jul 3, 2007
### heartofaragorn
1. The problem statement, all variables and given/known data
A snowball picks up speed and rolls of a barn roof angled at 40 degrees from horizontal. The edge of the barn is 15.0 m above the ground, and the snowball has a speed of 7.00 m/s as it leaves the roof. Ignore air effects. How far from the edge of the barn does the snowball land? The trail passing near the barn is 5.0 m horizontally from the barn. How tall would a person have to be to be hit by a snowball when on this trail?
2. Relevant equations
Range = velocity squared / g X sin 2 theta
3. The attempt at a solution
I think I solved the first question by solving for the range, and the answer I received was 4.92 m. Is this right? I was looking at the second question and got confused because I thought that regardless of how tall the person was, the snowball would fall short. I tried looking at some kinematic equations and had too many variables left. Just a little help going in the right direction would be much appreciated!!!
2. Jul 3, 2007
### cristo
Staff Emeritus
The range equation will only work if the projectile is shot from the ground and lands back on the ground. You need to use the http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html [Broken]. Are you familiar with these?
Last edited by a moderator: May 3, 2017
3. Jul 3, 2007
### Kurdt
Staff Emeritus
One knows that the snowball is travelling 7 m/s when it leaves the roof at 40 degrees from the horizontal. You can take the components to find its downward speed and its horizontal speed. then knowing the the acceleration due to gravity is 9.81 m/s2 you can solve for the time taken before the snowball hits the ground using kinematic equations (this one isn't easy). Then the distance travelled horizontally will be the horizontal speed multiplied by the time.
4. Jul 4, 2007
### heartofaragorn
I knew exactly what kinematic equation you were speaking of, thank you for the help. I figured the first part out and received an answer of 12.2 m. However, I am still befuddled as to the second question--if someone is on a path that is 5 m from the barn, how tall must they be to be hit with the snowball? I don't understand how someone that's only 5 m away could be hit with an object that lands an extra 7.2 m away from them...do I need to do something with the y-component velocity?
5. Jul 4, 2007
### Kurdt
Staff Emeritus
How far down will the snowball have travelled when its travelled five meters horizontally?
6. Jul 4, 2007
### Astronuc
Staff Emeritus
The problem is really asking "at what elevation (height) does the snowball pass over the trail, which is 5 m from the barn?"
Another good reference on trajectories is
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
With vertical acceleration (due to gravity) and horizontal/vertical velocities, the equations of motion define a parabolic trajectory (if negligible air resistance).
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra6
For the given initial conditions, one can determine the vertical position y, as a function of horizontal position, x, or conversely x as function of y. | 802 | 3,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-30 | latest | en | 0.944569 |
http://stackoverflow.com/questions/14243486/how-to-calculate-table-offsets | 1,406,289,963,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894260.71/warc/CC-MAIN-20140722025814-00083-ip-10-33-131-23.ec2.internal.warc.gz | 280,288,307 | 16,953 | # How to calculate table offsets
I am looking for a cost effective way of accessing records in my 6502 table. The table is 8 bytes long. Using indrect Y, I can offset the fields, but I want a way to get to the next 0th record.
``````lda #id
tax ; now x = x + 8
``````
I have seen others who use ASL while the accumulator has a value that is a power of 2 to do:
``````asl a
tax ; x = x * 2
``````
That would be great if I only had 2 records, but if I ASL again, the result will be x = 32. This would be too far of an offset.
Is there a better way to do this than using ADC?
-
You say the table is 8 bytes long, but it looks like you mean the records are 8 bytes long and the table is several records. How big is the table? – JasonD Jan 9 '13 at 19:06
For a table 8 bytes long, just iterate it. – m0skit0 Jan 9 '13 at 19:07
If the table is actually several 8 bytes records, I would store it as 8 1-byte record tables, which can be efficiently indexed.
i.e struct of arrays, rather than array of structs
This has the advantage of allowing the indexing of 256 records with just register offsets, and also the use of different sized records without changing the address calculation (so if you want 5-byte records, you just use less arrays, rather than changing the calculations).
``````LDA ELEMENT1,Y ; This is an array of the first bytes of our record
STA \$1000
LDA ELEMENT2,Y ; This is the second bytes...
STA \$1001
...etc...
INY ; Just need to inc Y to access next record
``````
As `INX`/`INY` are 2-cycle instructions, you're not going to beat that for updating an index.
This is my primary recommendation as it's fast and simple, though it does mean reformatting your data.
If this isn't an option you could either reformat the data dynamically such that it is (assuming you're going to then access it enough that it's worth the cost, and you have space to do so), or else you could use a variety of techniques, depending on how you actually want to access the data, and how important performance is.
If you really need to add an arbitrary amount to an index, rather than just increment, then the naive way to do it would be:
``````TYA (2)
CLC (2)
TAY (2)
``````
Howver that naive version takes 8 cycles. You can do it in 6 cycles if you omit the clear-carry, which you can do in cases where you know that it's already been cleared (for example, if following some other loop calculation where being in the loop implies that a carry has not been generated). So it can be worth juggling the code around to avoid setting flags. Anything less than incrementing by 4 could be done simply by incrementing multiple times.
If you cannot guarantee that carry is cleared, but you can spare a page in memory for a lookup table, then you could store 0 -> 255 in the bytes in that page and do:
``````LDA table + amt, Y (4)
TAY (2)
``````
The carry flag isn't set by that, but the zero flag is, so if the table wraps around to zero, you can check for that.
If you have to index off an address, then you could do:
``````LDA (zeropagevector), Y
``````
And increment the high-byte of `zeropagevector`. However that takes 5 cycles for the read. If you are only reading from your record in a single instruction, you could simply use normal absolute addressing, and modify the address in the instruction itself, saving a cycle.
Basically, there are loads of ways to optimise this kind of thing in 6502, but it really depends on exactly what your data is and how you want to access it.
-
you need to have 2 bytes (16 bits) for table pointer variable is your data is longer than a 256 bytes block.
`````` tablePointerLo = \$fa ; example zeropage value
tablePointerHi = \$fb ; example zeropage value
tableLength = 100*8 ; example length
sta tablePointerLo
sta tablePointerHi
loop02:
ldy #\$00
loop01:
lda (tablePointerLo),y
sta anywhereYouWant,y
iny
cpy #\$08
bne loop01
lda tablePointerLo
clc
sta tablePointerLo
bcc pass01
inc tablePointerHi
pass01:
; do whatever you want here with the datas
lda tablePointerLo
cmp #<tableEnd
bne loop02
lda tablePointerHi
cmp #>tableEnd
bne loop02
``````
I didn't test the code, it may not work but you should get the main idea.
- | 1,106 | 4,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | latest | en | 0.947945 |
https://amp.doubtnut.com/question-answer/three-angles-of-a-quadrilateral-are-7590-and-75-then-the-fourth-angle-is-26522088 | 1,591,442,629,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513230.90/warc/CC-MAIN-20200606093706-20200606123706-00180.warc.gz | 240,864,286 | 19,982 | IIT-JEE
Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
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Question From class 9 Chapter QUADRILATERALS
# Three angles of a quadrilateral are , then the fourth angle is
D
Solution :
Given, <br> We know that, sum of all the angles of a quadrilateral is . <br> <br> <br> ltBrgt <br> Hence, the fourth angle of a quadrilateral is .
Three angles of a quadrilateral are . Find the measure of the fourth angle.
0:53
Three angles of a quadrilateral are . Its fourth angle is
0:56
Three angles of a quadrilateral are in the ratio If the fourth angle is find the three angles of the quadrilateral.
3:02
Three angles of a quadrilateral are equal. Fourth angle is of measure What is the measure of equal angles.
1:29
Two angles of a quadrilateral are each. The third and fourth angles are in the ratio . Find the measure of third and fourth angles
2:36
The three angles of a quadrilateral are respectively equal to Find its fourth angle.
1:59
Assertion (A) : If three angles of a quadrilateral are and then the fourth angle is <br> Reason(R) : The sum of all the angle of a quadrilateral is <br> The correct answer is : (a) /(b)/(c )/(d).
1:43
The 3 angles of a quadrilateral are The fourth angle is
1:10
Three angles of a quadrilateral measure . Find the measure of the fourth angle.
1:31
Three angles of a quadrilateral measure and .Find the measure of the fourth angle.
1:24
The angle of a quadrilateral are respectively Find the fourth angle.
1:59
A quadrilateral has three acute angles each measures What is the measure of the fourth angle?
1:12
यदि किसी चक्रीय चतुर्भुज के तीन उत्तरोत्तर कोण के अनुपात मे हो, तो चौथे कोण का माप बताए ?
2:03
One angle of a quadrilateral is and the remaining three angles are equal. Find the three equal angles.
1:15
If are three points on a circle with centre such that and then (b) (c) (d)
1:55
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Homework 15 - options = odeset'Events(t,y)hit_wall(t...
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0 5 10 15 20 25 30 35 40 45 50 -2 -1 0 1 2 3 4 Time (s) Position (m) No bounds checking 0 2 4 6 8 10 12 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 Time (s) Position (m) Single zero-crossing handled
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0 5 10 15 20 25 30 35 40 45 50 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 Time (s) Position (m) Final solution function lumped_system % Run entire file, not single cells %% No bounds checking [t,y]=ode45(@(t,a)dydt_solution(t,a),[0 50], [3 0 0]); figure(1), plot(t,y(:,1)); xlabel( 'Time (s)' ) ylabel( 'Position (m)' ) title( 'No bounds checking' ) %% Dealing with zero-crossing for one step options = odeset( 'Events' , @(t,y)hit_wall(t, y), 'Maxstep' , 0.01); [t,y]=ode45(@(t,a)dydt_solution(t,a),[0 50], [3 0 0], options); [t1,y1]=ode45(@(t,a)dydt_solution(t,a),[t(end) 50], [y(end,1) - y(end,2) y(end,3) ], options);
y_total = [y; y1]; t_total = [t; t1]; figure(2), plot(t_total, y_total(:,1)); xlabel( 'Time (s)' ) ylabel( 'Position (m)' ) title( 'Single zero-crossing handled'
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Unformatted text preview: options = odeset( 'Events' , @(t,y)hit_wall(t, y), 'Maxstep' , 0.01); y = [3 0 0]; t = [0]; finalT = 50; while (t(end) < finalT) [t1,y1] = ode45( @(t,a)dydt_solution(t,a),[t(end) finalT], [y(end,1) -y(end,2) y(end,3) ], options ); y = [y; y1]; t = [t; t1]; end figure(3), plot(t,y(:,1)); xlabel( 'Time (s)' ) ylabel( 'Position (m)' ) title( 'Final solution' ) %% Functions used % Returns a column matrix of dy/dt function dydt = dydt_solution(t,a) k1=8; k2=8; F=10; b1=16; b2=16; w=2; m=2; a0=k1/b2; a1=((k1+k2)/k2)+(b1/b2); a2=(b1/k2)+(m/b2); a3=m/k2; dydt = [a(2); ... a(3); ... (F*cos(w.*t) - a0*a(1) - a1*a(2) - a2*a(3)) / a3; ... ]; function [valueToTest, shouldStop, direction] = hit_wall(t, y) valueToTest = y(1); shouldStop = 1; direction = -1;...
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https://popflock.com/learn?s=Electromagnetic_four-potential | 1,660,532,628,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00663.warc.gz | 432,326,969 | 19,567 | Electromagnetic Four-potential
Get Electromagnetic Four-potential essential facts below. View Videos or join the Electromagnetic Four-potential discussion. Add Electromagnetic Four-potential to your PopFlock.com topic list for future reference or share this resource on social media.
Electromagnetic Four-potential
An electromagnetic four-potential is a relativistic vector function from which the electromagnetic field can be derived. It combines both an electric scalar potential and a magnetic vector potential into a single four-vector.[1]
As measured in a given frame of reference, and for a given gauge, the first component of the electromagnetic four-potential is conventionally taken to be the electric scalar potential, and the other three components make up the magnetic vector potential. While both the scalar and vector potential depend upon the frame, the electromagnetic four-potential is Lorentz covariant.
Like other potentials, many different electromagnetic four-potentials correspond to the same electromagnetic field, depending upon the choice of gauge.
This article uses tensor index notation and the Minkowski metric sign convention . See also covariance and contravariance of vectors and raising and lowering indices for more details on notation. Formulae are given in SI units and Gaussian-cgs units.
## Definition
The electromagnetic four-potential can be defined as:[2]
SI units Gaussian units
${\displaystyle A^{\alpha }=\left({\frac {1}{c}}\phi ,\mathbf {A} \right)\,\!}$ ${\displaystyle A^{\alpha }=(\phi ,\mathbf {A} )}$
in which ? is the electric potential, and A is the magnetic potential (a vector potential). The units of A? are V·s·m-1 in SI, and Mx·cm-1 in Gaussian-cgs.
The electric and magnetic fields associated with these four-potentials are:[3]
SI units Gaussian units
${\displaystyle \mathbf {E} =-\mathbf {\nabla } \phi -{\frac {\partial \mathbf {A} }{\partial t}}}$ ${\displaystyle \mathbf {E} =-\mathbf {\nabla } \phi -{\frac {1}{c}}{\frac {\partial \mathbf {A} }{\partial t}}}$
${\displaystyle \mathbf {B} =\mathbf {\nabla } \times \mathbf {A} }$ ${\displaystyle \mathbf {B} =\mathbf {\nabla } \times \mathbf {A} }$
In special relativity, the electric and magnetic fields transform under Lorentz transformations. This can be written in the form of a tensor - the electromagnetic tensor. This is written in terms of the electromagnetic four-potential and the four-gradient as:
${\displaystyle F^{\mu \nu }=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }={\begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&-B_{z}&B_{y}\\E_{y}/c&B_{z}&0&-B_{x}\\E_{z}/c&-B_{y}&B_{x}&0\end{bmatrix}}}$
assuming that the signature of the Minkowski metric is (+ - - -). If the said signature is instead (- + + +) then:
${\displaystyle F'\,^{\mu \nu }=\partial '\,^{\mu }A^{\nu }-\partial '\,^{\nu }A^{\mu }={\begin{bmatrix}0&E_{x}/c&E_{y}/c&E_{z}/c\\-E_{x}/c&0&B_{z}&-B_{y}\\-E_{y}/c&-B_{z}&0&B_{x}\\-E_{z}/c&B_{y}&-B_{x}&0\end{bmatrix}}}$
This essentially defines the four-potential in terms of physically observable quantities, as well as reducing to the above definition.
## In the Lorenz gauge
Often, the Lorenz gauge condition ${\displaystyle \partial _{\alpha }A^{\alpha }=0}$ in an inertial frame of reference is employed to simplify Maxwell's equations as:[2]
SI units Gaussian units
${\displaystyle \Box A^{\alpha }=\mu _{0}J^{\alpha }}$ ${\displaystyle \Box A^{\alpha }={\frac {4\pi }{c}}J^{\alpha }}$
where J? are the components of the four-current, and
${\displaystyle \Box ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\nabla ^{2}=\partial ^{\alpha }\partial _{\alpha }}$
is the d'Alembertian operator. In terms of the scalar and vector potentials, this last equation becomes:
SI units Gaussian units
${\displaystyle \Box \phi ={\frac {\rho }{\epsilon _{0}}}}$ ${\displaystyle \Box \phi =4\pi \rho }$
${\displaystyle \Box \mathbf {A} =\mu _{0}\mathbf {j} }$ ${\displaystyle \Box \mathbf {A} ={\frac {4\pi }{c}}\mathbf {j} }$
For a given charge and current distribution, and , the solutions to these equations in SI units are:[3]
{\displaystyle {\begin{aligned}\phi (\mathbf {r} ,t)&={\frac {1}{4\pi \epsilon _{0}}}\int \mathrm {d} ^{3}x^{\prime }{\frac {\rho \left(\mathbf {r} ^{\prime },t_{r}\right)}{\left|\mathbf {r} -\mathbf {r} ^{\prime }\right|}}\\\mathbf {A} (\mathbf {r} ,t)&={\frac {\mu _{0}}{4\pi }}\int \mathrm {d} ^{3}x^{\prime }{\frac {\mathbf {j} \left(\mathbf {r} ^{\prime },t_{r}\right)}{\left|\mathbf {r} -\mathbf {r} ^{\prime }\right|}},\end{aligned}}}
where
${\displaystyle t_{r}=t-{\frac {\left|\mathbf {r} -\mathbf {r} '\right|}{c}}}$
is the retarded time. This is sometimes also expressed with
${\displaystyle \rho \left(\mathbf {r} ',t_{r}\right)=\left[\rho \left(\mathbf {r} ',t\right)\right],}$
where the square brackets are meant to indicate that the time should be evaluated at the retarded time. Of course, since the above equations are simply the solution to an inhomogeneous differential equation, any solution to the homogeneous equation can be added to these to satisfy the boundary conditions. These homogeneous solutions in general represent waves propagating from sources outside the boundary.
When the integrals above are evaluated for typical cases, e.g. of an oscillating current (or charge), they are found to give both a magnetic field component varying according to r-2 (the induction field) and a component decreasing as r-1 (the radiation field).[clarification needed]
## Gauge freedom
When flattened to a one-form, A can be decomposed via the Hodge decomposition theorem as the sum of an exact, a coexact, and a harmonic form,
${\displaystyle A=d\alpha +\delta \beta +\gamma }$.
There is gauge freedom in A in that of the three forms in this decomposition, only the coexact form has any effect on the electromagnetic tensor
${\displaystyle F=dA}$.
Exact forms are closed, as are harmonic forms over an appropriate domain, so ${\displaystyle dd\alpha =0}$ and ${\displaystyle d\gamma =0}$, always. So regardless of what ${\displaystyle \alpha }$ and ${\displaystyle \gamma }$ are, we are left with simply
${\displaystyle F=d\delta \beta }$.
## References
1. ^ Gravitation, J.A. Wheeler, C. Misner, K.S. Thorne, W.H. Freeman & Co, 1973, ISBN 0-7167-0344-0
2. ^ a b D.J. Griffiths (2007). Introduction to Electrodynamics (3rd ed.). Pearson Education, Dorling Kindersley. ISBN 978-81-7758-293-2.
3. ^ a b I.S. Grant, W.R. Phillips (2008). Electromagnetism (2nd ed.). Manchester Physics, John Wiley & Sons. ISBN 978-0-471-92712-9. | 1,930 | 6,564 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.789324 |
http://fullhomework.com/downloads/category/expert-tutorials/experttutorials3/ | 1,611,586,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00299.warc.gz | 39,098,200 | 10,058 | # Telegraph
Q. a.) National Telephone and Telegraph (NTT) Company common stock currently sells for \$60 per share. NTT is expected to pay a \$4 dividend during the coming year, and the price of the stock is expected to increase to \$65 a year from now. Determine the expected (ex ante) percentage holding period return on NTT common stock.
b.) Suppose that one year later, NTT’s common stock is selling for \$75 per share. During the one-year period, NTT paid a \$4 common stock dividend. Determine the realized (ex post) percentage holding period return on NTT common stock.
c.) Repeat Part b given that NTT’s common stock is selling for \$58 one year later.
d.) Repeat Part b given that NTT’s common stock is selling for \$50 one year later.
A customer wants to make a teepee in his backyard for his children. He plans to use lengths of PVC plumbing pipe for the supports on the teepee, and he wants the teepee to be 24 feet across and 5 feet tall (see figure). How long should the pieces of PVC plumbing pipe be? Show all work. (3pts)
# SQL Skills
Details: Some of your team members are a little rusty on their SQL skills but need to create the SQL select statements that would produce running summary files for reports on the following:
Customer Volume by Month
CustomerID
Month
Total shipments
Total weight
Driver Performance by Month
Driver ID
Month
Number of manifests on time
Number of manifests delivered late
The team members sent you an e-mail with the following questions:
From where should they source the information for the design?
What difficulties would they have in writing the SQL statements for each of these tables?
Should any of these tables be created on an as-needed basis?
Do these summary tables violate the concept of table normalization?
Assignment Guidelines
Provide answers to the 4 questions outlined in the assignment description in a 2–3-page Word document.
Provide any examples, tables, or SQL statements that you created to arrive at your solutions.
A 1–2-page Word document that contains your answers to the assignment questions and any material that you created to assist you with the assignment.
# Teachers
Teachers should not impose their personal values on students and should remain value neutral. If you were the instructor for an ethics course, outline three distinct strategies you would practice in order to ensure you remain neutral. APA format, with 2 references | 508 | 2,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.930796 |
https://www.exceldemy.com/excel-graph-equation-of-a-line/ | 1,716,498,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00170.warc.gz | 679,281,628 | 69,354 | # How to Display Equation of a Line in Excel Graph (2 Easy Ways)
Imagine, someone has sent an Excel file with a chart, but the equation is not visible there and then you want to know and display the equation that is used to make the chart. Or, if you always want to display the equation while making a chart then there are easy ways to do it in Excel. Today we’ll show 2 easy ways to display the equation of a line in an Excel graph with easy steps and vivid illustrations.
Here’s the dataset that we’ll use to explore the methods. We used the equation y = 2x + 3 to make the line graph.
## 1. Using Display Equation on Chart Option to Show Equation of a Line in Excel
In our first method, we’ll use the built-in option of Excel Chart- Display Equation on Chart to show the equation of the line graph. We can open it in three ways.
Steps:
• First, click anywhere on the chart and you will get three icons beside the right side of the chart
• Click on the Chart Element icon from the right side of the chart and then choose Trendline > More Options.
• Or, click as follows from the Chart Design ribbon: Add Chart Element > Trendline > More Trendline Options.
• The fastest way is to just double-click on the line of the chart.
After a while, the Format Trendline field will appear on the right side of your Excel window.
• Now just mark the Display Equation on Chart option from the field.
Now see, the equation appeared in the chart.
Read More: How to Graph Two Equations in Excel
## 2. Displaying Equation of a Line in Excel Graph Manually
There is no other built-in option to do that but we can apply some alternative tricks manually to display the equation of a line for the Excel chatbox. Like, we can use the text box and equation command to write our equation on our own and then Just place it on the chart. It will look like we have used the option from the first method.
### 2.1. Inserting Text Box
So first, Let’s try the Text Box command to write and display the equation.
Steps:
• Go to the Insert ribbon and then Click on the Text Box command from the Text
Then your cursor will look like this-
• Just click your mouse, hold and drag over anywhere on the sheet to make a box.
• After that, type the equation in the box.
You can apply the text format here too. You can change text color, font color, and alignment just by following as usual ways of Excel. We applied the center and middle alignment here.
• Finally, just place the box on the chart.
And look, It’s looking like the output of the previous method.
Read More: How to Graph a Linear Equation in Excel
### 2.2. Insert Equation
The same thing we can do using the Equation feature of Excel. And looks exactly like an equation.
Steps:
• Firstly, click on the Insert ribbon.
• Then click Equation from the Symbols
Soon, you will get an equation box like the image below.
• Type the equation on the box.
Here also you can change the format. We applied the center and middle alignment here again.
• Finally, place the equation box on the chart by dragging.
The outlook of the equation box is pretty much better than the text box.
You can download the free Excel workbook from here and practice independently.
## Conclusion
That’s all for the article. I hope the procedures described above will be good enough to display the equation of a line in an Excel graph. Feel free to ask any question in the comment section and please give me feedback.
<< Go Back to Plot an Equation |Â Excel Charts | Learn Excel
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Md. Sourov Hossain Mithun, an Excel and VBA Content Developer at Softeko's ExcelDemy project, joined in October 2021. Holding a Naval Architecture & Marine Engineering degree from BUET, Mithun showcases expertise during his two-year tenure. With over 200 published articles on Excel topics, he earned a promotion to Team Leader, excelling in leading diverse teams. Mithun's passion extends to Advanced Excel, Excel VBA, Data Analysis, and Python programming, contributing significantly to the innovative and dynamic environment of ExcelDemy... Read Full Bio
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Advanced Excel Exercises with Solutions PDF | 935 | 4,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.878379 |
https://plainmath.net/6852/kilometer-distribution-deviation-confidence-confidence-confidence | 1,620,504,961,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00200.warc.gz | 457,898,014 | 8,406 | # The service life in kilometer of Goody tires is assumed to follow a normal distribution with a standard deviation of 5,000 km. A random sample of 25 tires yielded a mean service life of 30,000 km. 1) Find the 95% confidence interval for the true mean service life. 2) 2. Find a 75% confidence interval for the true mean service life. 3) Calculate the widths of the intervals found in 1 and 2. How do these widths change as the confidence level decreases?
Question
Confidence intervals
The service life in kilometer of Goody tires is assumed to follow a normal distribution with a standard deviation of 5,000 km. A random sample of 25 tires yielded a mean service life of 30,000 km. 1) Find the $$\displaystyle{95}\%$$ confidence interval for the true mean service life. 2) 2. Find a $$\displaystyle{75}\%$$ confidence interval for the true mean service life. 3) Calculate the widths of the intervals found in 1 and 2. How do these widths change as the confidence level decreases?
2021-02-10
Step 1 Given information Sample size $$\displaystyle{\left({n}\right)}={25}$$ Sample mean $$\displaystyle\overline{{{x}}}={30},{000}\ {k}{m}$$ Population standard deviation $$\displaystyle={5},{000}\ {k}{m}$$ 3) Significance level $$\displaystyle{\left(\alpha\right)}={1}\ -\ {0.95}={0.05}$$ $$\displaystyle{95}\%$$ Confidence interval is given by $$\displaystyle{C}.{I}=\ \overline{{{x}}}\ \pm\ {Z}_{{{\frac{{\alpha}}{{{2}}}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={30.000}\ \pm\ {1.9600}\ \times\ {\frac{{{5000}}}{{\sqrt{{{25}}}}}}={\left({28040},\ {31390}\right)}$$
$$\displaystyle{Z}_{{{\frac{{{0.05}}}{{{2}}}}}}={Z}_{{{0.025}}}=\ \pm\ {1.9600}\ \text{(From Excel = NORM.S.INV(0.025))}$$ Step 2 2) Significance level $$\displaystyle{\left(\alpha\right)}={1}\ -\ {0.75}={0.25}$$
$$\displaystyle{75}\%$$ Confidence interval is given by $$\displaystyle{C}.{I}=\ \overline{{{x}}}\ \pm\ {Z}_{{{\frac{{\alpha}}{{{2}}}}}}\ \times\ {\frac{{{5000}}}{{\sqrt{{25}}}}}={\left({28849.7},\ {31150.3}\right)}$$
$$\displaystyle{Z}_{{{\frac{{{0.25}}}{{{2}}}}}}={Z}_{{{0.125}}}=\ \pm\ {1.1503}\ \text{(From Excel = NORM. S. INV(0.125))}$$ Step 3 3) Width of interval for $$\displaystyle{95}\%$$ confidence interval $$\displaystyle={31960}\ -\ {28040}={3920}$$ Width of interval for $$\displaystyle{75}\%$$ confidence interval $$\displaystyle={31150.3}\ -\ {28849.7}={2300.6}$$ On decreasing Confidence level. Width of confidence level decreases.
### Relevant Questions
1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?
A random sample of size $$\displaystyle{n}={25}$$ from a normal distribution with mean $$\displaystyle\mu$$ and variance 36 has sample mean $$\displaystyle\overline{{X}}={16.3}$$
a)
Calculate confidence intervals for $$\displaystyle\mu$$ at three levels of confidence: $$80%, 90% \text{and} 99%$$. How to the widths of the confidence intervals change?
b) How would the CI width change if n is increased to 100?
The marks of DMT students results in June 2020 sessional examinations were normally disyributed with a mean pass mark of 9 and a standard deviation pass mark of 0.15. After moderation, a sample of 30 papers was selected to see if the mean pass mark had changed. The mean pass mark of the sample was 8.95. a) Find the $$\displaystyle{95}\%$$ confidence interval of students mean mark. b) Calculate for the critical regions of the $$\displaystyle{95}\%$$ confidence intervals. c) Using your results in "a" and "b" above, is there evidence of a change in the mean pass mark of the DMT students.
(a) The company's production equipment produces metal discs weighing 200 g. It should be noted that the weight of the discs corresponds to the normal distribution. To check machine consistency, 20 discs are randomly selected with an average weight of 205 g and a standard deviation of 7 g. What is the 99% confidence interval for the average weight of the selected discs?
(b) A company launched a new model of golf ball. It claimed that the driving distance is at least 300m. A sample of 20 balls yields a sample mean of 295m and sample standard deviation of 8m. It is assumed that the driving distance is normally distributed.
Conduct an appropriate hypothesis testing at the 0.05 level of significance. Is there any evidence that the average travel distance stated by the company is true?
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 58 dates, the mean record high daily temperature in a certain city has a mean of $$\displaystyle{83.43}^{{\circ}}{F}$$. Assume the population standard deviation is $$\displaystyle{14.02}^{{\circ}}{F}$$. $$\displaystyle{90}\%=$$ $$\displaystyle{95}\%=$$ Which interval is wider?
You are given the sample mean and standard deviation of the population. Use this information to construct the $$\displaystyle{90}\%{\quad\text{and}\quad}{95}\%$$ confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of 66 dates, the mean record high daily temperature in a certain city has a mean of $$\displaystyle{85.69}^{\circ}{F}$$. Assume the population standard deviation is $$\displaystyle{13.60}^{\circ}{F}.$$
The $$90\%$$ confidence interval is
The $$95\%$$ confidence interval is
Which interval is wider?
Interpret the results.
You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
$$9 6 10 15 19 6 23 26 19 16 11 25 16 11$$
a. To compute the confidence interval use a t or z distribution.
b. With 95% confidence the population mean number of visits per physical therapy patient is between ___ and ___ visits.
c. If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About ___ percent of these confidence intervals will contain the true population mean number of visits per patient and about ___ percent will not contain the true population mean number of visits per patient.
A researcher is interested in finding a $$90\%$$ confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 117 students who averaged 40.9 minutes concentrating on their professor during the hour lecture. The standard deviation was 11.8 minutes. Round answers to 3 decimal places where possible.
With $$90\%$$ confidence the population mean minutes of concentration is between ____ and ____ minutes.
Money reports that the average annual cost of the first year of owning and caring for a large dog in 2017 is $1,448. The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with $$\displaystyle\sigma=\{230}.$$ $$\begin{matrix} 1,902 & 2,042 & 1,936 & 1,817 & 1,504 & 1,572 & 1,532 & 1,907 & 1,882 & 2,153 \\ 1,945 & 1,335 & 2,006 & 1,516 & 1,839 & 1,739 & 1,456 & 1,958 & 1,934 & 2,094 \\ 1,739 & 1,434 & 1,667 & 1,679 & 1,736 & 1,670 & 1,770 & 2,052 & 1,379 & 1,939\\ 1,854 & 1,913 & 2,163 & 1,737 & 1,888 & 1,737 & 2,230 & 2,131 & 1,813 & 2,118\\ 1,978 & 2,166 & 1,482 & 1,700 & 1,679 & 2,060 & 1,683 & 1,850 & 2,232 & 2,294 \end{matrix}$$ (a) What is the margin of error for a $$95\%$$ confidence interval of the mean cost in dollars of the first year of owning and caring for this breed? (Round your answer to nearest cent.) (b) The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use this data set to compute the sample mean. Using this sample, what is the $$95\%$$ confidence interval for the mean cost in dollars of the first year of owning and caring for an Irish Red and White Setter? (Round your answers to nearest cent.)$_______ to \$________ | 2,586 | 9,649 | {"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} | 4.46875 | 4 | CC-MAIN-2021-21 | latest | en | 0.646421 |
https://community.rstudio.com/t/when-i-replace-the-time-series-variables-comes-warnings-number-of-items-to-replace-is-not-a-multiple-of-replacement-length/98219 | 1,618,401,222,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077810.20/warc/CC-MAIN-20210414095300-20210414125300-00199.warc.gz | 283,942,661 | 5,875 | When I replace the time series variables comes Warnings [number of items to replace is not a multiple of replacement length]
Hello community! First post, please pardon any inaccuracies.
So I'm trying to develop a function with loops to conduct the quantile-on-quantile (QQ) regression (a type of time series analysis) between two time series. The function is developed as:
coverLLsr=function(y,x,coverNum=3,h=3,kernel=c('Gauss','Laplace','Bessel'),ylag=1,xlag=NULL)
{
call=match.call()
mAll=sapply((1:coverNum)/(coverNum+1),function(p){
sapply((1:coverNum)/(coverNum+1),function(t){
llsr(y=y,x=x,h=h,theta=p,tao=t,kernel=kernel,ylag=ylag,xlag=xlag)
})
})
coefAll=array(dim=c(coverNum,coverNum,length(ylag)+length(xlag)+2,4))
R2All=matrix(NA,coverNum,coverNum)
AICAll=matrix(NA,coverNum,coverNum)
thetaAll=matrix(NA,coverNum,coverNum)
taoAll=matrix(NA,coverNum,coverNum)
for(i in 1:coverNum)
for(j in 1:coverNum)
{
coefAll[i,j,,]=mAll[[6*(i-1)+6coverNum(j-1)+2]]
R2All[i,j]=mAll[[6*(i-1)+6coverNum(j-1)+3]]
AICAll[i,j]=mAll[[6*(i-1)+6coverNum(j-1)+4]]
thetaAll[i,j]=mAll[[6*(i-1)+6coverNum(j-1)+5]]
taoAll[i,j]=mAll[[6*(i-1)+6coverNum(j-1)+6]]
}
callUse=as.character(call)
dimnames(coefAll)[[3]]=c('intercept',callUse[3],paste0(callUse[2],'_lag',seq(1:length(ylag))))
dimnames(coefAll)[[4]]=c('value','stdError','t value','P-value')
dimnames(thetaAll)[[1]]=dimnames(taoAll)[[1]]=dimnames(R2All)[[1]]=dimnames(AICAll)[[1]]=dimnames(coefAll)[[1]]=paste0('tao',round((1:coverNum)/(coverNum+1),3))
dimnames(thetaAll)[[2]]=dimnames(taoAll)[[2]]=dimnames(R2All)[[2]]=dimnames(AICAll)[[2]]=dimnames(coefAll)[[2]]=paste0('theta',round((1:coverNum)/(coverNum+1),3))
mAll=list(mAll=mAll,coefAll=coefAll,R2All=R2All,AICAll=AICAll,thetaAll=thetaAll,taoAll=taoAll)
class(mAll)='llsrs'
return(mAll)
}
And I first use this function to conduct the regression between two time series as:
r1=data2[,1]
r2=data2[,2]
m2=coverLLsr(r1,r2,coverNum=49,h=1/5);
write.xlsx(m2\$coefAll[,,1,1],'coef.xlsx','intercept')
write.xlsx(m2\$coefAll[,,2,1],'coef.xlsx','b1',append = T)
write.xlsx(m2\$coefAll[,,3,1],'coef.xlsx','aerfa(theta)',append = T)
write.xlsx(m2\$AICAll,'coef.xlsx','AIC',append = T)
write.xlsx(m2\$thetaAll,'coef.xlsx','theta',append = T)
write.xlsx(m2\$taoAll,'coef.xlsx','tao',append = T)
r1 and r2 are the two time series with 5774 observations each
And it works out fine. However, when I replace the time series variable with the other two time series with shorter time range with 157 observations each as:
and again run the above function, and it comes the warning
Error in coefAll[i, j, , ] <- mAll[[6 * (i - 1) + 6 * coverNum * (j - : number of items to replace is not a multiple of replacement length
It troubles me a lot! Any help will be deeply appreciated! Thanks in advance!
Do you really need to program it?
The quatreg package may do what you want.
Thanks for the reply! I will check it though I may need to conduct the quantile on quantile regression rather than the quantile regression.
This topic was automatically closed 21 days after the last reply. New replies are no longer allowed.
If you have a query related to it or one of the replies, start a new topic and refer back with a link. | 1,042 | 3,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.371415 |
https://aaronice.gitbook.io/lintcode/stack/daily-temperatures | 1,716,858,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00823.warc.gz | 60,281,964 | 49,364 | # Daily Temperatures
`stack`, `monotonous stack`
Given a list of daily temperatures`T`, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put`0`instead.
For example, given the list of temperatures`T = [73, 74, 75, 71, 69, 72, 76, 73]`, your output should be`[1, 1, 4, 2, 1, 1, 0, 0]`.
Note:The length of`temperatures`will be in the range`[1, 30000]`. Each temperature will be an integer in the range`[30, 100]`.
## Analysis
### Brute-force
Brute-force的方法就略去不表，时间复杂度O(n^2)。
### Stack
Brute-force有很多是重复搜索，更有效率的是利用Stack，一个单调递增的stack。逆序的方式进行扫描。
`T = [73, 74, 75, 71, 69, 72, 76, 73]`
``````When i = 7, stack = [7 (73)]. ans[i] = 0.
When i = 6, stack = [6 (76)]. ans[i] = 0.
When i = 5, stack = [5 (72), 6 (76)]. ans[i] = 1.
When i = 4, stack = [4 (69), 5 (72), 6 (76)]. ans[i] = 1.
When i = 3, stack = [3 (71), 5 (72), 6 (76)]. ans[i] = 2.
When i = 2, stack = [2 (75), 6 (76)]. ans[i] = 4.
When i = 1, stack = [1 (74), 2 (75), 6 (76)]. ans[i] = 1.
When i = 0, stack = [0 (73), 1 (74), 2 (75), 6 (76)]. ans[i] = 1.``````
`ans[i] = stack.isEmpty() ? 0 : stack.peek() - i;`
`T[i] >= T[stack.peek()]`时候要不断`pop()`，然后再放入当前下标i。这里要允许`"="`的条件，因为需要寻找第一个出现的位置，因此重复位置要`pop()`出来。
When I saw the question in this competition, I firstly think about Monotonous stack inspired by Largest Rectangle in Histogram. Because Monotonous stack can help us find first largest element in O(n) time complexity.
Time Complexity - O(n)
Space Complexity - O(n)
## Solution
Stack with reverse order iteration T.length - 1 to 0 - Time: O(n), Space: O(n) - (52 ms)
``````class Solution {
public int[] dailyTemperatures(int[] T) {
int[] ans = new int[T.length];
Deque<Integer> stack = new ArrayDeque<>();
for (int i = T.length - 1; i >= 0; i--) {
while (!stack.isEmpty() && T[i] >= T[stack.peek()]) {
stack.pop();
}
ans[i] = stack.isEmpty() ? 0 : stack.peek() - i;
stack.push(i);
}
return ans;
}
}``````
Stack with 0 to T.length - 1 order iteration (17 ms, faster than 92.98%)
Using ArrayDeque() will significantly increase performance than Stack()
``````class Solution {
public int[] dailyTemperatures(int[] T) {
Deque <Integer> stack = new ArrayDeque<>();
int[] ret = new int[T.length];
for (int i = 0; i < T.length; i++) {
while (!stack.isEmpty() && T[i] > T[stack.peek()]) {
int idx = stack.pop();
ret[idx] = i - idx;
}
stack.push(i);
}
return ret;
}
}``````
Brute-force O(n^2) - (348 ms)
``````class Solution {
public int[] dailyTemperatures(int[] T) {
int n = T.length;
int[] ans = new int[n];
int days = 0;
for (int i = 0; i < n; i++) {
days = 0;
for (int j = i; j < n; j++) {
if (T[j] > T[i]) {
ans[i] = days;
break;
} else {
days++;
}
}
}
return ans;
}
}``````
## Reference
https://leetcode.com/problems/daily-temperatures/solution/
https://leetcode.com/problems/daily-temperatures/discuss/109832/Java-Easy-AC-Solution-with-Stack
Last updated | 1,056 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.538421 |
https://www.datamagazine.co.uk/understanding-jackknife-statistics-assessing-the-stability-and-reliability-of-statistical-estimators/ | 1,713,346,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00254.warc.gz | 650,947,894 | 26,158 | # Understanding Jackknife Statistics: Assessing the Stability and Reliability of Statistical Estimators
Navigate the Article
## Key Takeaways
– Jackknife statistics is a powerful resampling technique used in statistical analysis.
– It helps estimate the bias and variance of a statistical estimator.
– The jackknife method is based on systematically leaving out one or more observations from a dataset.
– It can be used to assess the stability and reliability of statistical results.
– Jackknife statistics is widely used in various fields, including finance, biology, and social sciences.
## Introduction
In the world of statistics, accuracy and reliability are of utmost importance. Researchers and analysts often rely on various techniques to assess the quality of their statistical estimators. One such technique is jackknife statistics, a resampling method that provides valuable insights into the stability and reliability of statistical results. In this article, we will explore the concept of jackknife statistics, its applications, and its significance in statistical analysis.
## The Concept of Jackknife Statistics
Jackknife statistics, also known as the jackknife method or the delete-one jackknife, is a resampling technique used to estimate the bias and variance of a statistical estimator. The method was first introduced by Maurice Quenouille in 1949 and later refined by John W. Tukey in 1958. The name “jackknife” refers to the idea of systematically leaving out one or more observations from a dataset, similar to how a jackknife is used to remove a blade from a pocketknife.
#### How Does Jackknife Statistics Work?
The jackknife method works by repeatedly estimating a statistic of interest after systematically removing one observation at a time from the dataset. This process is repeated for each observation, resulting in a set of “jackknife replicates.” These replicates are then used to calculate the bias and variance of the estimator.
#### Applications of Jackknife Statistics
Jackknife statistics finds applications in various fields, including finance, biology, and social sciences. In finance, it can be used to assess the stability of portfolio returns and estimate the risk associated with different investment strategies. In biology, jackknife resampling is often used to assess the accuracy of phylogenetic trees and estimate species diversity. In social sciences, it can help evaluate the impact of influential observations on regression models and estimate the reliability of survey data.
The jackknife method offers several advantages over other resampling techniques. Firstly, it is computationally efficient, as it does not require generating multiple bootstrap samples. Instead, it systematically removes one observation at a time, making it less computationally intensive. Secondly, the jackknife method provides an unbiased estimate of the variance, unlike the bootstrap method, which can be biased in certain situations. Lastly, the jackknife method allows for the assessment of the stability and reliability of statistical results, providing valuable insights into the robustness of estimators.
#### Limitations of Jackknife Statistics
While jackknife statistics is a powerful resampling technique, it does have some limitations. Firstly, it assumes that the observations in the dataset are independent and identically distributed, which may not always be the case in real-world scenarios. Secondly, the jackknife method may not be suitable for small sample sizes, as removing one observation at a time can lead to a significant reduction in the available data. Lastly, the jackknife method may not be appropriate for certain types of estimators, such as those based on maximum likelihood estimation.
## Conclusion
Jackknife statistics is a valuable tool in the field of statistical analysis. It provides insights into the stability and reliability of statistical estimators, allowing researchers and analysts to make informed decisions based on their results. By systematically leaving out one or more observations from a dataset, the jackknife method helps estimate the bias and variance of estimators, making it a powerful resampling technique. Despite its limitations, jackknife statistics continues to be widely used in various fields, contributing to the advancement of statistical analysis and decision-making processes. | 792 | 4,388 | {"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} | 3.140625 | 3 | CC-MAIN-2024-18 | latest | en | 0.904308 |
https://terrorsofjoy.com/doing-homework-800 | 1,675,062,758,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00335.warc.gz | 573,797,013 | 5,030 | # Website that does math for you
Apps can be a great way to help learners with their math. Let's try the best Website that does math for you. We can solving math problem.
## The Best Website that does math for you
Website that does math for you can help students to understand the material and improve their grades. Whether you're a math whiz or just need a little help to understand a problem, there's a lot of good online resources out there. There are lots of free online calculators like the ones that come pre-installed on phones or tablets, but they don't always have all the features you'll need. You can also search for "math calculator" on Google or other search engines to find more options. Some sites also have video tutorials with step-by-step explanations that might be helpful if you're having trouble understanding something. You can also go to math help sessions at your local community college or library if you don't have access to high schools or teachers near your home. They usually offer tutoring for free, and some even have programs that help students prepare for the SATs and ACTs.
Natural logarithm (ln) can be easily solved by equation. There is no need to guess values and there are no complex calculations required. The basic formula for solving ln is as follows: math>ln(x) = frac{ln(y)}{1 + y}/math> Therefore, if math>y = 35/math>, then math>ln(35)/math> will be calculated as follows: math>frac{34}{1 + 35}/math> This value can then be used in any calculations to get results that are relative to the original value, such as math>frac{2}{1 + 3}/math>. If math>y = 10)/math>, then math>ln(10)/math> will be calculated as follows: math>frac{9}{1 + 10}/math>. Finally, math>frac{1}{0.5 + 1} = frac{1}{4} = 0.25/math>. Therefore, the natural logarithm of 10 is 25. The calculation process goes like this: 1. Input x and calculate y based on the formula given above 2. Then calculate ln(x). 3. Repeat step 2 with y = x to verify that the answer is correct Note that the l
This is a great way to get your child used to doing math in a fun way. Another option is to have your child use his or her phone or tablet to do simple addition and subtraction math problems. Once they know the answer, they can simply tap the screen and it will tell them the correct answer. This can be very helpful for kids who don’t like to write their own numbers on paper. Another way you can help your child learn math is by explaining how each number works. For example, when talking about the number 100, explain that this is ten times 10, which means 100. They may not understand this at first, but if you keep trying and explaining it over and over again, they will eventually start to get it.
Solve each proportion of the equation by breaking down the fraction into two terms: If one side is a whole number, the other term can be simplified. If both sides are whole numbers, the equation is true. If one side is a fraction, the other side must be a whole number. To solve proportions when one side has a variable, simply divide both sides by the variable. To solve proportions when both sides have variables, simply multiply both sides by the variable. Example: If 17/20 = 0.8 and 9/10 = 1, what is 9 ÷ 10? The answer is 9 ÷ (10 × 0.8) = 9 / 10 = 0.9 or 9 out of 10
Life saver, it's a must have! I need to see the full problem worked down to understand it. This app helps me not only with the answers it helps me understand what I am doing right or wrong. I get that aha! Feeling with this app.
Reina Jenkins
What's that? You want to get this math homework done so you can go to your friend’s house to watch a movie but then leave and play games with them instead half way through because you don’t want to waste time because the last time you saw this friend was in 2012? This app got your back. Oh, what's that? You think cheating is bad? This app shows you how to solve the equations that you take a photo of. One last thing, the photo system is dummy smart. 10/10 would smash if it was a human.
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Math-solving Value Write Website that gives you answers to math problems How to solve for x and y intercepts Solve algebra equations Differnetial equation solver Algebra problems to solve | 1,010 | 4,223 | {"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} | 4.3125 | 4 | CC-MAIN-2023-06 | latest | en | 0.929331 |
https://qa.auth.gr/en/class/1/600099248 | 1,718,335,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00513.warc.gz | 450,122,172 | 9,378 | Sampling
Title ΔΕΙΓΜΑΤΟΛΗΨΙΑ / Sampling Code 0566 Faculty Sciences School Mathematics Cycle / Level 1st / Undergraduate Teaching Period Spring Common Yes Status Active Course ID 40000532
Programme of Study: UPS of School of Mathematics (2014-today)
Registered students: 110
OrientationAttendance TypeSemesterYearECTS
CoreElective CoursesSpring-5
Academic Year 2017 – 2018 Class Period Spring Instructors from Other Categories Vasileios Karagiannis 39hrs Weekly Hours 3 Class ID 600099248
Course Type 2016-2020
• Scientific Area
• Skills Development
Course Type 2011-2015
Specific Foundation / Core
Mode of Delivery
• Face to face
Digital Course Content
Erasmus
The course is also offered to exchange programme students.
Language of Instruction
• Greek (Instruction, Examination)
Prerequisites
General Prerequisites
Simple knowledge of basic concepts of Statistics (Mean, Spread, Standard Deviation, Coefficient of Variation).
Learning Outcomes
!) They can design and participate in conducting sampling. 2) They may budget the size and structure of the sample. 3) Ηandle Tables of Random Numbers (TRN) 4) Manage irresolute voters
General Competences
• Apply knowledge in practice
• Retrieve, analyse and synthesise data and information, with the use of necessary technologies
• Make decisions
• Work autonomously
• Work in teams
• Respect natural environment
• Demonstrate social, professional and ethical commitment and sensitivity to gender issues
Course Content (Syllabus)
Population and Sample. Definition of Sampling and usefulness of Probability Sampling. Eστιματορσ: Basic Properties and τειρ role in the Planning of sampling. Main types of Sampling techniques: A) Simple Random Sampling (SRS), B) Stratified Sampling (StS) (various versions of), C) Systematic Sampling (SyS) (Introductory, Cyclic Law, SyS in 2-dimensional populations, Best Selection of samples) D) Cluster Sampling (CluS) (equal size clusters and introduction to techniques with non equal size). Comparison of sampling methods. Applications of Sampling in Economy, Ecology and Politics. Indicators, Indices. Classic application examples from the literature and daily practice. Management of Irresolute voters electronic or not.
Educational Material Types
• Notes
• Book
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P: n/a [a+b for a,b in zip(xrange(1,51), xrange(50,0,-1))] [51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51] i want to add all the elemtns a s well. can i do this all in a listcomprehension? i can do this ofc: reduce(lambda x,y:x+y,[a+b for a,b in zip(xrange(1,51), xrange(50,0,-1))]) but reduce is a functional way of doing it, what is the more pythonic way of doing this? Jun 27 '08 #1
7 Replies
P: n/a On Jun 22, 6:32*pm, cirfu
P: n/a On Jun 23, 10:32*am, cirfu
P: n/a On Jun 23, 9:23 am, Paul Hankin
P: n/a John Machin wrote: > Instead of sum(a + b for a, b in zip(foo, bar)) why not use sum(foo) + sum(bar) ? or even sum(foo+bar) as may apply. Cheers, BB Jun 27 '08 #5
P: n/a Le Monday 23 June 2008 11:39:44 Boris Borcic, vous avez écrit*: John Machin wrote: Instead of sum(a + b for a, b in zip(foo, bar)) why not use sum(foo) + sum(bar) ? or even sum(foo+bar) as may apply. Because some are better than others : sum(foo+bar) is the worst, it create a superfluous list of len(foo) + len(bar) elements. sum(a + b for a, b in zip(foo, bar)), creates a list of max(len(foo), len(bar)) elements, in most cases it is the same as the former. This could have been corrected using itertools.izip. So the winner is sum(foo) + sum(bar), which does not create anything not needed. But if the question is "how to build the list and sum up all elements in a efficient way for sequences of arbitrary length ", it's important to make it in the same iteration, so the most effective/clear, and so "pythonic", way to do this is (untested) : res, sum = [], 0 for s in (a + b for a, b in zip(itertools.izip(xrange(1, 51), xrange(50, 0, -1)))): sum += s res.append(sum) Which is "pythonic" in first place is to provide sequences of datas as iterators when they can grow in size. -- _____________ Maric Michaud Jun 27 '08 #6
P: n/a On Jun 23, 9:16 pm, Maric Michaud
P: n/a Le Monday 23 June 2008 13:51:34 John Machin, vous avez écrit*: On Jun 23, 9:16 pm, Maric Michaud or even sum(foo+bar) as may apply. Because some are better than others : sum(foo+bar) is the worst, it create a superfluous list of len(foo) + len(bar) elements. sum(a + b for a, b in zip(foo, bar)), creates a list of max(len(foo), len(bar)) elements, in most cases it is the same as the former. This could have been corrected using itertools.izip. So the winner is sum(foo) + sum(bar), which does not create anything not needed. But if the question is "how to build the list and sum up all elements in a efficient way for sequences of arbitrary length ", it's important to make it in the same iteration, so the most effective/clear, and so "pythonic", way to do this is (untested) : res, sum = [], 0 Please use tot or total, not sum! for s in (a + b for a, b in zip(itertools.izip(xrange(1, 51), Perhaps you should not have left zip() in there ... xrange(50, 0, -1)))): sum += s res.append(sum) Do you mean res.append(s) ? Yes, my mistakes, all remarks are indeed true. Sorry for this out of thoughts sketch. I would have thought that it would have been better to create the list and then sum it: res = [a + b for a, b in itertools.izip(foo_iter, bar_iter)] total = sum(res) This is good enough if you accept the overhead of iterating twice over the whole list. But this may be a bit more complicated, for efficiency, there is a better way of allocating memory than successively appending new item. I guess this should be a far better solution, but you'll need to know the required number of iteration (the size of your iterators, the xrange in this sample) : res, total = [None] * n, 0 for i, s in enumerate(a + b for a, b in izip(xrange(1, n+1), xrange(n, 0, -1)): total += s res[i] =s -- _____________ Maric Michaud Jun 27 '08 #8 | 1,275 | 4,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.869809 |
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# meaning of standard error in multiple regression Daingerfield, Texas
Changing the value of the constant in the model changes the mean of the errors but doesn't affect the variance. The graph below presents X1, X4, and Y2. Conversely, the unit-less R-squared doesn’t provide an intuitive feel for how close the predicted values are to the observed values. The distribution of residuals for the example data is presented below.
The value of R can be found in the "Model Summary" table of the SPSS/WIN output. INTERPRET ANOVA TABLE An ANOVA table is given. It will prove instructional to explore three such relationships. I would really appreciate your thoughts and insights.
In this case the variance in X1 that does not account for variance in Y2 is cancelled or suppressed by knowledge of X4. The P value is the probability of seeing a result as extreme as the one you are getting (a t value as large as yours) in a collection of random data The interpretation of the "Sig." level for the "Coefficients" is now apparent. Additional analysis recommendations include histograms of all variables with a view for outliers, or scores that fall outside the range of the majority of scores.
Interpreting the variables using the suggested meanings, success in graduate school could be predicted individually with measures of intellectual ability, spatial ability, and work ethic. Example data. Entering X1 first and X3 second results in the following R square change table. If the correlation between X1 and X2 had been 0.0 instead of .255, the R square change values would have been identical.
If some of the variables have highly skewed distributions (e.g., runs of small positive values with occasional large positive spikes), it may be difficult to fit them into a linear model In a scatterplot in which the S.E.est is small, one would therefore expect to see that most of the observed values cluster fairly closely to the regression line. In fact, if we did this over and over, continuing to sample and estimate forever, we would find that the relative frequency of the different estimate values followed a probability distribution. It can be thought of as a measure of the precision with which the regression coefficient is measured.
Therefore, the standard error of the estimate is There is a version of the formula for the standard error in terms of Pearson's correlation: where ρ is the population value of The standard error here refers to the estimated standard deviation of the error term u. The standard error is an estimate of the standard deviation of the coefficient, the amount it varies across cases. But outliers can spell trouble for models fitted to small data sets: since the sum of squares of the residuals is the basis for estimating parameters and calculating error statistics and
X Y Y' Y-Y' (Y-Y')2 1.00 1.00 1.210 -0.210 0.044 2.00 2.00 1.635 0.365 0.133 3.00 1.30 2.060 -0.760 0.578 4.00 3.75 2.485 1.265 1.600 5.00 If you are regressing the first difference of Y on the first difference of X, you are directly predicting changes in Y as a linear function of changes in X, without The adjustment in the "Adjusted R Square" value in the output tables is a correction for the number of X variables included in the prediction model. Use of the standard error statistic presupposes the user is familiar with the central limit theorem and the assumptions of the data set with which the researcher is working.
In "classical" statistical methods such as linear regression, information about the precision of point estimates is usually expressed in the form of confidence intervals. Accessed: October 3, 2007 Related Articles The role of statistical reviewer in biomedical scientific journal Risk reduction statistics Selecting and interpreting diagnostic tests Clinical evaluation of medical tests: still a long It is particularly important to use the standard error to estimate an interval about the population parameter when an effect size statistic is not available. However, S must be <= 2.5 to produce a sufficiently narrow 95% prediction interval.
In the case of the example data, the following means and standard deviations were computed using SPSS/WIN by clicking of "Statistics", "Summarize", and then "Descriptives." THE CORRELATION MATRIX The second step Predicting y given values of regressors. The estimated coefficients for the two dummy variables would exactly equal the difference between the offending observations and the predictions generated for them by the model. For example: R2 = 1 - Residual SS / Total SS (general formula for R2) = 1 - 0.3950 / 1.6050 (from data in the ANOVA table) =
Statistical Methods in Education and Psychology. 3rd ed. EXAMPLE DATA The data used to illustrate the inner workings of multiple regression will be generated from the "Example Student." The data are presented below: Homework Assignment 21 Example Student This surface can be found by computing Y' for three arbitrarily (X1, X2) pairs of data, plotting these points in a three-dimensional space, and then fitting a plane through the points The table of coefficients also presents some interesting relationships.
In this case it indicates a possibility that the model could be simplified, perhaps by deleting variables or perhaps by redefining them in a way that better separates their contributions. A minimal model, predicting Y1 from the mean of Y1 results in the following. Therefore, it is essential for them to be able to determine the probability that their sample measures are a reliable representation of the full population, so that they can make predictions This is a model-fitting option in the regression procedure in any software package, and it is sometimes referred to as regression through the origin, or RTO for short.
Allison PD. In regression analysis terms, X2 in combination with X1 predicts unique variance in Y1, while X3 in combination with X1 predicts shared variance. In this case the value of b0 is always 0 and not included in the regression equation. The estimated CONSTANT term will represent the logarithm of the multiplicative constant b0 in the original multiplicative model. | 1,385 | 6,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-39 | latest | en | 0.910686 |
https://www.plus2net.com/python/numpy-exercise-a.php | 1,685,352,347,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00535.warc.gz | 1,045,307,730 | 6,504 | # Numpy Exercise : Bascis
Numpy
1. Create an array ( x1 ) with sequence of number starting from 9 to 17 ( inclusive )
2. Create one more array x2 in which each element I = 2 (element of x1) + 8
3. Create an array with 4 numbers starting from 2 to 20
4. Create an array starting from 2 to 20 and each element is increased by 3
5. Create another array of same shape of x1
6. Create another array of same shape of x1 but with different random values.
## Create an array ( x1 ) with sequence of number starting from 9 to 17 ( inclusive )
how to create an array with start, stop and step by using arange()
``````import numpy as np
x1=np.arange(9,18)
print(x1) # [ 9 10 11 12 13 14 15 16 17] ``````
## Create one more array x2 in which each element I = 2 (element of x1) + 8
``````x2=(2*x1) +8
print(x2) # [26 28 30 32 34 36 38 40 42]``````
## Create an array with 4 numbers starting from 2 to 20
How to create an array with start stop and number of steps by using linspace()
``````x3=np.linspace(2,20,4)
print(x3) # [ 2. 8. 14. 20.]``````
## Create an array starting from 2 to 20 and each element is increased by 3
``````x4=np.arange(2,21,3)
print(x4) # [ 2 5 8 11 14 17 20]``````
## Create another array of same shape of x1
``````import numpy as np
x1=np.arange(9,18)
print(x1)
s1=x1.shape
y1=np.ones(s1)
print(y1)``````
Output
``````[ 9 10 11 12 13 14 15 16 17]
[1. 1. 1. 1. 1. 1. 1. 1. 1.]``````
## Create another array of same shape of x1 but with different random values.
``````import numpy as np
x1=np.arange(9,18)
print(x1)
s1=x1.shape
y1=np.ones(s1)
y1=np.random.randint(4,high=10,size=s1)
print(y1)``````
Output
``````[ 9 10 11 12 13 14 15 16 17]
[7 4 5 7 7 6 4 4 4]``````
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Question
In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position when they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with 41% of the participants saying they start in this position. Use a normal distribution to find a 95% confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016.
in progress 0
2 weeks 2021-11-19T18:47:12+00:00 1 Answer 0 views 0
1. Answer: = ( 0.411, 0.409)
Therefore at 95% confidence interval (a,b) = (0.411, 0.409)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean gain x = 0.41
Standard deviation r = 0.016
Number of samples n = 1000
Confidence interval = 95%
z(at 95% confidence) = 1.96
Substituting the values we have;
0.41+/-1.96(0.016/√1000)
0.41+/-1.96(0.000506)
0.41+/-0.00099
0.41+/-0.001
= ( 0.411, 0.409)
Therefore at 95% confidence interval (a,b) = (0.411, 0.409) | 405 | 1,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-49 | latest | en | 0.865808 |
https://www.chegg.com/homework-help/algebra-and-trigonometry-mymathlab-and-student-solutions-manual-9th-edition-chapter-3.6-solutions-9780321786463 | 1,550,472,391,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484689.3/warc/CC-MAIN-20190218053920-20190218075920-00059.warc.gz | 811,132,240 | 23,153 | Algebra and Trigonometry, MyMathLab, and Student Solutions Manual (9th Edition) View more editions Solutions for Chapter 3.6
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Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 13
We have the following graph for .
• Step 2 of 13
(a)
Let be a point on the graph .
• Step 3 of 13
We know that the distance between two points and is
• Step 4 of 13
Now distance from to the origin is
Since is a point on the graph of , we substitute for . Then
• Step 5 of 13
The distance from to the origin as a function of is
.
• Step 6 of 13
(b)
For , we have
• Step 7 of 13
Thus for , .
• Step 8 of 13
(c)
For , we have
• Step 9 of 13
Thus for , .
• Step 10 of 13
(d)
We have
• Step 11 of 13
• Step 12 of 13
The graph of is
• Step 13 of 13
(e)
Using a graphing utility, we find that when , the value of is smallest.
Corresponding Textbook
Algebra and Trigonometry, MyMathLab, and Student Solutions Manual | 9th Edition
9780321786463ISBN-13: 0321786467ISBN: Michael SullivanAuthors:
This is an alternate ISBN. View the primary ISBN for: Algebra and Trigonometry 9th Edition Textbook Solutions | 358 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-09 | latest | en | 0.824357 |
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# Erlang: elegant tuple_to_list/1
I'm getting myself introduced to Erlang by Armstrongs "Programming Erlang". One Exercise is to write a reeimplementation of the tuple_to_list/1 BIF. My solution seems rather inelegant to me, especially because of the helper function I use. Is there a more Erlang-ish way of doing this?
``````tup2lis({}) -> [];
tup2lis(T) -> tup2list_help(T,1,tuple_size(T)).
tup2list_help(T,Size,Size) -> [element(Size,T)];
tup2list_help(T,Pos,Size) -> [element(Pos,T)|tup2list_help(T,Pos+1,Size)].
``````
Thank you very much for your ideas. :)
-
Make yourself aware which of the implementations below is tail-recursive. – Tilman Apr 25 '13 at 22:23
There is nothing wrong with having helper functions, often you need them and they are the best way of doing things. And don't be over concerned with tail-recursion, see erlang.org/doc/efficiency_guide/listHandling.html#id64720 – rvirding Apr 26 '13 at 10:50
I think your function is ok, and more if your goal is to learn the language. As a matter of style, usually the base case when constructing lists is just the empty list []. So I'd write
``````tup2list(Tuple) -> tup2list(Tuple, 1, tuple_size(Tuple)).
tup2list(Tuple, Pos, Size) when Pos =< Size ->
[element(Pos,Tuple) | tup2list(Tuple, Pos+1, Size)];
tup2list(_Tuple,_Pos,_Size) -> [].
``````
you can write pretty much the same with list comprehension
``````[element(I,Tuple) || I <- lists:seq(1,tuple_size(Tuple))].
``````
it will work as expected when the tuple has no elements, as lists:seq(1,0) gives an empty list.
-
+1 for list comprehension. I like the usage of lists:seq. Feels a bit pythonian to me. ;) – Zakum Apr 25 '13 at 19:16
Your code is good and also idiomatic way how to make this sort of stuff. You can also build this list backward which in this case will be a little bit faster because of tail call but not significant.
``````tup2list(T) -> tup2list(T, size(T), []).
tup2list(T, 0, Acc) -> Acc;
tup2list(T, N, Acc) -> tup2list(T, N-1, [element(N,T)|Acc]).
``````
-
Better to use `tuple_size/1` instead of `size/1`. See here: erlang.org/doc/efficiency_guide/commoncaveats.html – coffeMug May 24 '15 at 10:08
In Erlang R16B you can also use erlang:delete_element/2 function like this:
``````tuple2list({}) -> [];
tuple2list(T) when is_tuple(T) ->
[element(1, T) | tuple2list(erlang:delete_element(1, T))].
``````
-
+1 for getting rid of the helper function and the need for counter and accumulator. I read that the speedup of tail-recursion vs. head-recursion in current versions is more a matter of processor-architecture* so I quit like this solution! *erlang.org/doc/efficiency_guide/myths.html – Zakum Apr 26 '13 at 10:31
-1 for a solution which creates a new tuple when an element is added to the list. Remember there is no mutable data in erlang and `erlang:delete_element/2` creates a new tuple! – rvirding Apr 26 '13 at 10:54
@Zakum: You should read this myth demystification more carefully. body-recursive list function and tail-recursive function that calls `lists:reverse/1` at the end will use exactly the same amount of memory Note the `lists:reverse/1` which mine version doesn't do. Mine tail recursive version definitely will be faster than body recursive ppolv's solution, no doubt. But when most tuples are small ti doesn't matter. But know the difference. – Hynek -Pichi- Vychodil Apr 27 '13 at 12:33
@rvirding If I understand the question correctly it was about elegant `tuple_to_list/1` not about most optimal one, because you can't beat erlang:tuple_to_list/1 anyway. – hdima Apr 27 '13 at 17:49
@hdima True, but an elegant solution is not needlessly inefficient. IMAO. – rvirding Apr 28 '13 at 3:18
I am attempting exercises from Joe Armstrong book, Here is what I came up with
``````my_tuple_to_list(Tuple) -> [element(T, Tuple) || T <- lists:seq(1, tuple_size(Tuple))].
``````
-
Erlang 17.0, you should build list in natural order, solutions above is incorrect from the point of efficiency. Always add elements to the head of an existing list:
``````%% ====================================================================
%% API functions
%% ====================================================================
my_tuple_to_list({}) ->
[];
my_tuple_to_list(Tuple) ->
tuple_to_list_iter(1, size(Tuple), Tuple, [])
.
%% ====================================================================
%% Internal functions
%% ====================================================================
tuple_to_list_iter(N, N, Tuple, List) ->
lists:reverse([element(N, Tuple)|List]);
tuple_to_list_iter(N, Tuplesize, Tuple, List) ->
L = [element(N, Tuple)|List],
tuple_to_list_iter(N + 1, Tuplesize, Tuple, L)
.
``````
-
```mytuple_to_list(T) when tuple_size(T) =:= 0 -> []; mytuple_to_list(T) -> [element(1, T)|mytuple_to_list(erlang:delete_element(1, T))].```
- | 1,370 | 5,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-30 | latest | en | 0.864085 |
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# WhatsApp Image 2019-03-29 at 10.19.07 PM.jpeg - \5 To Find...
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Unformatted text preview: \5\\ To Find_ the Points where the_Curve has a horizontal_ tangent line _ _dys dyl do_ dx /do = 0 dy Fle) Sin (A ) + F 10 ) copp) d x F ( A ) COs ( D ) _ F (@ ) sin ( b ) < =\ FQ ) = 5 CSC(DL = _ Sun( D ) FIA ) - 5 Cos(D ) sin = ( D ) 5_ Cos(D )_ siAter_ Cos(D) dy since + Sin?(@) O 5 -- Sin ( A ) X P - 5 cos ( A )_ Cosle)_ sin? ( D) Sin CAL 5 COS ( D ) + 5 COS ( Q) - O sin A sing - 5 Cos (0 ) sing + 5 Cos ( 4 ) Sin ( D) = Q Sins ( Q ) - 5 Cos ( Q ) Sin ( 6 ) + 5 Cos ( D ) = Q - 5 toslot- (1 - 5ing ) = 0 _5 {es( 4) = 6 L_ since) _ Of # = It 3 TT T 2) D = [ 7 So the Points are Scanned_ with Cam Scanner...
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Jill Tulane University ‘16, Course Hero Intern | 628 | 2,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-22 | latest | en | 0.842995 |
https://www.ozproblems.com/walkabout/walkabout2017/feb2 | 1,726,668,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00043.warc.gz | 858,549,014 | 41,285 | # Two selfmates by Laimons Mangalis
### 2 Feb. 2017 | by Peter Wong
In December’s column regarding a new e-book on Laimons Mangalis, I mentioned that he was a proficient composer of selfmates. Since this major problem genre is rarely discussed on this site, it seems a good opportunity to delve into some examples discovered in the book. In selfmates, White plays first and compels Black to mate in the specified number of moves, while Black does not cooperative and resists giving mate. The two selections below are both fine illustrations of the type, presenting appealing and accessible ideas.
Laimons Mangalis
The Problemist 1977
Selfmate in 2
In the first problem, Black’s B + K battery pointed at the white king suggests that it will be forced to open and deliver mate at some stage. Hence if Black were to play 1…Sf3 guarding e5 and d4, then 2.Qg7+ Kxg7 mate. This is the only set variation, however, as no selfmates-in-one are prepared for the other black moves. Consider the checks 1…Kf8+ and 1…e6+; each by attacking d5 means that 2.Ke5 could be followed by 2…Sf3 mate, but Black is not obliged to move the knight. The key 1.Qh8!, a waiting move, deals with these checks because: (1) 1…Kf8+ now self-pins the black bishop (and also the e7-pawn), so that 2.Ke5 does force 2…Sf3 by zugzwang, and (2) after 1…e6+ 2.Ke5, the black king is confined by the queen from its corner position, leaving 2…Sf3 again as Black’s only legal move. The black royal battery fires twice as the mating move: 1…Sf3 2.Qg7+ Kxg7 as in the set play, and 1…e5 2.Qf6+ Kxf6, when White makes a different queen sacrifice to ensure that the black king covers the e5-flight.
Laimons Mangalis
The Problemist 1977
Selfmate in 2
The second position contains a R + B battery on the first rank, and any bishop move would mate immediately. A set line utilises the battery thus: 1…Ke3 – attacking f2 – 2.Rxe4+ Bxe4. The surprising key 1.Kh1! (waiting) unpins the queen, sparking a full-length variation when Black moves the bishop: 1…B~+ 2.Qg1+ Rxg1. The black king has two flights on e3 and c5. Taking the first gives 1…Ke3 2.Rxe4+ Bxe4 – unchanged from the set play, though here the queen is pinned on the diagonal instead. If Black takes the second flight, White exploits the unpin of the black knight with 1…Kxc5 2.Qf2+ Sxf2. The final defence 1…c2 admits 2.Qxc2, immobilising the black king, and Black is forced by zugzwang to open the R + B battery once more: 2…Bxc2. | 705 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.923843 |
https://forum.piedao.org/answers/2557429-this-runner-completed-a-100-meter-race-with-a-time-of-13-75-seconds-what-was-her-average-speed | 1,709,328,513,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00568.warc.gz | 248,828,133 | 6,757 | # This runner completed a 100 meter race with a time of 13.75 seconds. what was her average speed
Taking into account the definition of speed, you get that the average speed of the runner is 7.27 .
It is necessary to know that speed is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it.
In other words, speed can be defined as the amount of space traveled per unit of time with which a body moves and can be calculated using the expression:
In this case:
• distance traveled= 100 meters
• time= 13.75 seconds
Replacing these values in the speed definition:
Solving:
Speed= 7.27
In summary, the average speed of the runner is 7.27 .
Explanation: The average speed of an object in an interval of time is the distance traveled by the object divided by the duration of the interval.
Distance traveled= 100 m
Time taken = 13.75 s
Average speed=
Average speed=
## Related Questions
The coefficients in the balanced chemical equation are used to determine mole ratios.Write the mole ratios found in the reaction "Making water
H2: _____0
H2;
H20
___02: _
H20
1 mole of HCl or NaOH gives you 1 mole of H2O , then the number of moles in H2O is: [ 1÷1×1 ] = 1 mole.
Explanation:
Out of sodium,rubidium,aluminum,and indium which is the most reactive metal
Explanation:
Among the metals Sodium, Rubidium, Aluminum, and Indium, Rubidium is the most reactive due to its position in the alkali metals group in the periodic table. The reactivity of metals increases as you go down in any group.
### Explanation:
The question asked was: 'Out of Sodium, Rubidium, Aluminum, and Indium, which is the most reactive metal?' In the periodic table of elements, reactivity of metals increases as we move down in any group. Sodium and Rubidium are alkali metals present in Group 1. Out of these two elements, Rubidium is further down in the group and is thus more reactive compared to Sodium. However, Aluminum and Indium belong to different groups and are relatively less reactive when compared to alkali metals.
Passivation, or the formation of nonreactive film of oxide on metals such as Aluminum, makes them less reactive compared to alkali metals, which readily react with water and oxygen. In conclusion, among Sodium, Rubidium, Aluminum, and Indium, Rubidium is the most reactive metal.
brainly.com/question/20570629
#SPJ3
How do Crocodiles See, Hear, and Smell?
They are reptiles.. and reptiles are born from good habitats.. and they have ears so they can smell.. they also have keen eyes so they could see good... and they have tiny noses so they could smell well.
How many grams of CO2 are in 2.1 mol of the compound?a. 21.0 g
b. 44.0 g
c. 66.0 g
d. 92.4 g
Molar mass CO₂ = 44.0 g/mol
1 mole ------- 44.0 g
2.1 mole ----- > ?
g = 44.0 * 2.1 / 1
g = 92.4 / 1
= 92.4 g
hope this helps!
Answer quickly plzzzz 100 pointsWhat are the two types of genetic molecules, contained in living things, that the bases formed from formamide make?
Nucleobases and Nitrogen
Hydrogen and Oxygen
DNA and ATGC
DNA and RNA
hydrogen and oxygen
Explanation:
DNA AND ATGC
Explanation:
SORRY IF ITS WRONG
When and acid and base mix together it makes | 824 | 3,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-10 | latest | en | 0.924024 |
https://cooking.stackexchange.com/questions/13357/why-does-a-microwave-rotate-the-food-being-cooked-and-is-there-ever-a-time-when | 1,653,601,551,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00566.warc.gz | 230,514,087 | 67,995 | # Why does a microwave rotate the food being cooked? And is there ever a time when it shouldn't rotate?
Most microwaves I've used have a plate at the bottom that rotates when the microwave is running. I presume that rotating helps more evenly heat the food.
Interestingly, my microwave rotates by default but has a button you can press to disable the rotation. Under what circumstances or scenarios would you want to stop the microwave from rotating the food?
• If you press stop and start again some modles rotate the opposite direction. Because microwaves cook from the insjde out moving plate helps distrbute the heat Apr 28, 2017 at 3:19
• Some models restart in a pseudo random direction. Jul 4, 2018 at 23:50
Microwave ovens have a turntable because the microwaves themselves form what are known as 'standing waves'. This means there are essentially static columns of microwave energy inside the cabinet itself.
You can see this for yourself - spread a tray or plate with grated cheese, take the turntable out (or press that button), and zap it for a couple of minutes. You should see bands of melted cheese interleaved with unmelted cheese. Thus, the food is rotated to ensure even heating.
• By measuring the distance between the melted spots of cheese, you can work out the speed of light! Mar 22, 2011 at 10:54
• Yes you can. Or you can just remember that it's 299,792,458m/s, using 'We guarantee certainty, clearly referring to this light mnemonic' (the number of letters in the words match the figures) :) This of course the speed of light in a vacuum. Halve it for diamonds! Mar 22, 2011 at 12:03
• @Skizz Actually, since the size of a meter is now defined in terms of the speed of light (that is, the speed of light is defined to be precisely 299,792,458m/s, and the length of a second is defined in terms of the speed of a certain excitation change in Cesium, which means the size of a meter is a function of those two definitions), you can really only use this to either check how much slower light travels in air, or alternatively (if you suck all the air out of your microwave before performing the test), you could use it to check the accuracy of your ruler. Feb 19, 2013 at 19:06
• ...and obviously, the time when you want the microwave to stop rotating your food, is when you want to measure the speed of light!
– SF.
Apr 27, 2017 at 8:18
You stop the turntable when there's not room for your dishes to rotate - think large rectangular dish, or two plates fitting into the corners. At best, the turntable will strain and do nothing, and at worst, it'll make a mess.
Otherwise, it does help to get your food heated evenly. It can't fix everything, of course. The center point is always in the center, and even if your food is small enough to avoid the center, not all distances from the center will get quite the same heating even as they sweep out a circle. And no matter where your food is, the middle is always going to take a while to heat up. But it's a lot better than fixed hot and cold spots without a turntable.
• About the last paragraph; that's why you put your food as far away from the center of the platter as possible. Jun 8, 2015 at 19:40
• [center] Why is you put your food on one or both sides of the turntable. Dec 16, 2015 at 19:26
• If the turntable is straining, you can often replace the heavy glass with a smaller, lighter plate.I use an 8" Corel, and it runs on the wheels better than the original spinner, which met a dire fate upon hitting the floor at a couple meters per second. The solution was far cheaper than buying a new microwave, or digging up an old plate somewhere. Jul 4, 2018 at 23:55
There are at least 2 types of units, one with the turntable and one without. The turntable rotates to more evenly distribute the RF waves from the magnitron. The units without a turntable use a rotating reflector just under the magnitron to distribute the RF waves more evenly. Neither are perfect, as there are always hot spots. Microwaves have come down in price a lot since their introduction because they now make a magnitron not used in the radar industry. The magnitron use to be half of the price along with their power supply. Now they are just more throw-away crap like most electronic devices.
• Some models have both the turntable and the rotating reflector. That adds an extra little bit of randomness to the heating process. Jul 4, 2018 at 23:52
• +1 for mentioning flatbeds. May 4, 2019 at 14:23 | 1,068 | 4,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.93734 |
https://www.iykoongchallenge.com/how-to-find-slope | 1,725,932,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00594.warc.gz | 762,922,330 | 15,155 | # How to Find Slope: A Detailed Explanation
Baca Cepat
## Welcome, Challenger!
Before we dive into the nitty-gritty of finding slope, let’s take a moment to discuss why it matters. Understanding slope is crucial in many fields, including mathematics, engineering, physics, and even geography. Slope is a measure of the steepness of a line or curve and can help us solve problems ranging from finding the optimal path for a hiking trail to designing a roller coaster. By the end of this article, you’ll have a complete understanding of how to find slope and how to apply it to real-life scenarios. So, let’s get started!
## Introduction: What is Slope?
Slope is defined as the ratio of the vertical change to the horizontal change between two points on a line or curve. In other words, it’s how much the line rises or falls compared to how far it moves left or right. The slope of a line can be positive, negative, zero, or undefined, depending on its characteristics.
Positive Slope: A line with a positive slope moves upward as it moves to the right. This means that the y-coordinate is increasing as the x-coordinate increases.
Negative Slope: A line with a negative slope moves downward as it moves to the right. This means that the y-coordinate is decreasing as the x-coordinate increases.
Zero Slope: A line with a slope of zero is horizontal. This means that the y-coordinate does not change as the x-coordinate increases.
Undefined Slope: A line with an undefined slope is vertical. This means that the x-coordinate does not change as the y-coordinate increases.
Now that we have an understanding of what slope is, let’s move on to the different methods of finding it.
## How to Find Slope: The Different Methods
### Method 1: Using the Slope Formula
The slope formula is a popular method of finding slope in both mathematics and physics. This method involves using the coordinates of two points on the line to calculate the slope.
x1 y1 x2 y2
a b c d
Using the table above, we can plug in the values for x1, y1, x2, and y2 to find the slope:
Slope = ( y2 – y1 ) / ( x2 – x1 )
It’s important to note that if the line is vertical, the slope is undefined. This is because we cannot divide by zero.
### Method 2: Graphing the Points
Another method of finding slope is by graphing the points on a coordinate plane. The slope is then determined by counting the rise and run between the two points. The rise is the vertical distance between the two points, and the run is the horizontal distance.
For example, if we have two points: (2,4) and (5,9), we can draw a line connecting them and count the rise and run. The rise is 5 (9 – 4), and the run is 3 (5 – 2). So, the slope is 5/3.
### Method 3: Calculating the Slope of a Curve
If you’re dealing with a curve rather than a straight line, the mathematical method of finding slope won’t work. Instead, you need to use calculus. To find the slope of a curve at a specific point, you need to take the derivative of the function at that point.
The derivative is defined as the rate of change of a function at a given point. By taking the derivative of a function, we can find the slope of the tangent line at that point, which is the slope of the curve at that point.
These are just a few of the methods used to find slope. It’s important to understand each method and choose the one that best fits the problem you’re trying to solve.
### Q1: What is slope used for?
Slope is used to quantify the steepness of a line or curve. This can be helpful in many fields, including mathematics, engineering, physics, and geography. It can help us solve problems ranging from finding the optimal path for a hiking trail to designing a roller coaster.
### Q2: Can a line have a negative slope and a positive y-intercept?
Yes, a line can have a negative slope and a positive y-intercept. The slope determines the direction and steepness of the line, while the y-intercept determines where the line intersects the y-axis.
### Q3: What does the slope of a horizontal line represent?
A horizontal line has a slope of zero. This means that the y-coordinate does not change as the x-coordinate increases.
### Q4: What does an undefined slope mean?
An undefined slope means that the line is vertical. This means that the x-coordinate does not change as the y-coordinate increases.
### Q5: How do you find slope if you only have one point?
If you only have one point, you cannot find the slope. The slope requires two points on a line or curve.
### Q6: Is the slope of a circle constant?
No, the slope of a circle is not constant. The slope changes at every point on the circle.
### Q7: How do you find the slope of a curve?
To find the slope of a curve, you need to take the derivative of the function at the point of interest. This will give you the slope of the tangent line at that point, which is the slope of the curve at that point.
### Q8: What is the slope of a vertical line?
A vertical line has an undefined slope. This is because we cannot divide by zero.
### Q9: How do you find the slope of a line in standard form?
To find the slope of a line in standard form (Ax + By = C), we need to rearrange the equation to get y = mx + b format. Once we have the equation in slope-intercept form, we can easily identify the slope (m).
### Q10: Can the slope of a line be greater than 1?
Yes, the slope of a line can be greater than 1. This means that the line is moving upward at a steep angle.
### Q11: How do you find the slope of a line that passes through two given points?
To find the slope of a line that passes through two given points, we can use the slope formula. The slope formula is: Slope = ( y2 – y1 ) / ( x2 – x1 )
### Q12: Can the slope of a line be negative and undefined?
No, the slope of a line cannot be both negative and undefined. If the line is vertical, the slope is undefined. Otherwise, it can be positive or negative.
### Q13: Can the slope of a line be zero?
Yes, the slope of a line can be zero. This means that the line is horizontal.
## Conclusion: Apply What You’ve Learned!
Congratulations, Challenger! You now have a complete understanding of how to find slope and how to apply it to real-life scenarios. Now it’s time to put your skills to the test. Try finding the slope of different lines and curves, and see if you can solve related problems. As you continue to practice and improve, you’ll be able to solve even more complex problems using slope. So, keep on challenging yourself!
### Disclimer
This article is provided for informational purposes only, and the author makes no representations as to the accuracy, completeness, or suitability of any information contained herein. The information should not be construed as professional, financial, or legal advice. You should consult with a professional before making any decisions based on the information provided in this article. | 1,567 | 6,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-38 | latest | en | 0.925991 |
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