url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.enotes.com/topics/calculus1/questions/evaluate-limit-fraction-f-x-f-1-x-1-f-x-1-2x-5-x-2-238505 | 1,716,839,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00181.warc.gz | 643,331,799 | 15,500 | # Evaluate the limit of the fraction (f(x)-f(1))/(x-1), if f(x)=1+2x^5/x^2?x->1
We are given that f(x)=1+2x^5/x^2 = 1 + 2x^3.
We have to find: lim x -->1 [(f(x) - f(1))/(x-1)]
=> lim x -->1 [(1+ 2x^3 - 1 - 2)/(x-1)]
=> lim x -->1 [(2x^3 - 2)/(x-1)]\
=> lim x -->1 [2*(x - 1)(x^2 + x + 1)/(x-1)]
=> lim x -->1 [2*(x^2 + x + 1)]
substitute x with 1
=> 2*(1 +1 +1)
=> 6
Therefore the required limit is 6. | 205 | 413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-22 | latest | en | 0.531597 |
https://human.libretexts.org/Bookshelves/Logic/Map%3A_A_Modern_Formal_Logic_Primer_(Teller)/Volume_II%3A_Predicate_Logic/6%3A_More_on_Natural_Deduction_for_Predicate_Logic/6.1%3A_Multiple_Quantification_and_Harder_Problems | 1,566,306,687,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315329.55/warc/CC-MAIN-20190820113425-20190820135425-00322.warc.gz | 493,198,747 | 21,134 | # 6.1: Multiple Quantification and Harder Problems
In chapter 5 I wanted you to focus on understanding the basic rules for quantifiers. So there I avoided the complications that arise when we have sentences, such as '(Vx)(Vy)(Px & Py)', which stack one quantifier on top of another. Such sentences involve no new principles. It's just a matter of keeping track of the main connective. For example, '(Vx)(Vy)(Px & Qy)' is a universally quantified sentence, with '(Vx)' as the main connective. You practiced forming substitution instances of such sentences in chapter 3. The substitution instance of '(Vx)(Vy)(Px & Qy)' formed with 'a' (a sentence you could write when applying VE) is '(Vy)(Pa & Qy)'.
You will see how to deal with such sentences most quickly by just looking at a few examples. So let's write a derivation to establish the validity of
(Vx)(Vy)(Px & Qy) 1 | (Vx)(Vy)(Px & Qy) P
(Vx)Px & (Vx)Qx 2 | (Vx)(Pâ & Qy) 1, VE
3 | Pâ & Qb̂ 1, VE
4 | Pâ 1, VE
5 | Qb̂ 1, VE
6 | (Vx)Px 1, VE
7 | (Vx)Qx 1, VE
8 | (Vx)Px & (Vx)Qx 1, VE
In line 2 I applied VE by forming the substitution instance of 1 using the name 'a'. Then in line 3 I formed a substitution instance of the universally quantified line 2.
Let's look at an example of multiple existential quantification. The basic ideas are the same. But observe that in order to treat the second existential quantifier, we must start a sub-sub-derivation:
(∃x)(∃y)(Px & Qy)
(∃x)Px & (∃x)Qx
1 | (∃x)(∃y)(Px & Qy) P
2 | a| (∃y)(Pa & Qy) A
3 | | b| Pa & Qb A
4 | | | Pa 3, &E
5 | | | Qb 3, &E
6 | | | (∃x)Px 4, ∃I
7 | | | (∃x)Qx 5, ∃I
8 | | 3| (∃x)Px & (∃x)Qx 6, 7, &I
9 | 2| (∃x)Px & (∃x)Qx 2, 3-8, ∃E
10 1| ∃x)Px & (∃x)Qx 1, 2-9, ∃E
In line 2 I wrote down '(∃y)(Pa & Qy)', a substitution instance of line 1, formed with 'a', substituted for 'x', which is the variable in the main connective, '(∃x)', of line 1. Since I plan to appeal to 3E in application to line 1, I make '(∃y)(Pa & Qy)' the assumption of a subderivation with 'a' an isolated name. I then do the same thing with '(∃y)(Pa & Qy)', but because this is again an existentially quantified sentence to which I will want to apply 3E, I must make my new substitution instance, 'Pa & Qb', the assumption of a sub-sub-derivation, this time with 'b' the isolated name.
In the previous example, I would have been allowed to use 'a' for the second as well as the first substitution instance, since I was applying VE. But, in the present example, when setting up to use two applications of ∃E, I must use a new name in each assumption. To see why, let's review what conditions must be satisfied to correctly apply ∃E to get line 9. I must have an existentially quantified sentence (line 2) and a subderivation (sub-sub-derivation 3), the assumption of which is a substitution instance of the existentially quantified sentence. Furthermore, the name used in forming the substitution instance must be isolated to the subderivation. Thus, in forming line 3 as a substitution instance of line 2, I can't use 'a'. I use the name 'b' instead. The 'a' following 'P' in line 3 does not violate the requirement. 'a' got into the picture when we formed line 2, the substitution instance of line 1, and you will note that 'a' is indeed isolated to subderivation 2, as required, since sub-sub-derivation 3 is part of subderivation 2.
Here's another way to see the point. I write line 3 as a substitution instance of line 2. Since I will want to apply ∃E, the name I use must be isolated to subderivation 3. If I tried to use 'a' in forming the substitution instance of line 2, I would have had to put an 'a' (the "isolation flag") to the left of scope line 3. I would then immediately see that I had made a mistake. 'a' as an isolation flag means that 'a' can occur only to the right. But 'a' already occurs to the left, in line 2. Since I use 'b' as my new name in subderivation 3, I use 'b' as the isolation flag there. Then the 'a' in line 3 causes no problem: All occurrences of 'a' are to the right of scope line 2, which is the line flagged by 'a'.
All this is not really as hard to keep track of as it might seem. The scope lines with the names written at the top to the left (the isolation flags) do all the work for you. 'a' can only appear to the right of the scope line on which it occurs as an isolation flag. 'b' can only occur to the right of the scope line on which it occurs as an isolation flag. That's all you need to check.
Make sure you clearly understand the last two examples before continuing. They fully illustrate, in a simple setting, what you need to understand about applying the quantifier rules to multiply quantified sentences.
Once you have digested these examples, let's try a hard problem. The new example also differs from the last two in that it requires repeated use of a quantifier introduction rule instead of repeated use of a quantifier elimination rule. In reading over my derivation you might well be baffled as to how I figured out what to do at each step. Below the problem I explain the informal thinking I used in constructing this derivation, so that you will start to learn how to work such a problem for yourself.
(Vx)Px ⊃ (∃x)Qx
(∃x)(∃y)(Px ⊃ Qy)
1 | (Vx)Px ⊃ (∃x)Qx P
2 | | ~(∃x)(∃y)(Px ⊃ Qy) A
3 | | | ~Pa A
4 | | | ~Qb ⊃ ~Pa ⊃, W
5 | | | Pa ⊃ Qb 4, CP
6 | | | (∃y)(Pa ⊃ Qy) 5, ∃I
7 | | | (∃x)(∃y)(Px ⊃ Qy) 6, ∃I
8 | | | ~(∃x)(∃y)(Px ⊃ Qy) 2, R
9 | | Pâ 3-8, RD
10 | | (Vx)Px 9, VI
11 | | | (∃x)Qx A
12 | | | b| Qb A
13 | | | | Pa ⊃ Qb 12, W
14 | | | | (∃y)(Pa ⊃ Qy) 13, ∃I
15 | | | | (∃x)(∃y)(Px ⊃ Qy) 14, ∃I
16 | | | (∃x)(∃y)(Px ⊃ Qy) 11, 12-15, ∃E
17 | | | ~(∃x)(∃y)(Px ⊃ Qy) 2, R
18 | | ~(∃x)Qx 11-17, ~I
19 | | (Vx)Px ⊃ (∃x)Qx 1, R
20 | | (∃x)Qx 10, 19, ⊃E
21 | (∃x)(∃y)(Px ⊃ Qy) 2-20, RD
My basic strategy is reductio, to assume the opposite of what I want to prove. From this I must get a contraction with the premise. The premise is a conditional, and a conditional is false only if its antecedent is true and its consequent is false. So I set out to contradict the original premise by deriving its antecedent and the negation of its consequent from my new assumption.
To derive (Vx)Px (line 10), the premise's antecedent, I need to derive Pâ. I do this by assuming ~Pa from which I derive line 7, which contradicts line 2. To derive ~(∃x)Qx (line 18), the negation of the premise's consequent, I assume (∃x)Qx (line 11), and derive a contradiction, so that I can use ~I. This proceeds by using ∃E, as you can see in lines 11 to 16.
Now it's your turn to try your hand at the following exercises. The problems start out with ones much easier than the last example-and gradually get harder!
Exercise
6-1. Provide derivations to establish the validity of the following argument (a):
(a) (∃x)Lxx (a-1) (Vx)Lxx
(∃x)(∃y)Lxy (Vx)(Vy)Lyx
Note that the argument, (a-1) is invalid. Prove that this argument is invalid by giving a counterexample to it (that is, an interpretation in which the premise is true and the conclusion is false). Explain why you can't get from (Vx)Lxx to (Vx)(Vy)Lxy by using VE and VI as you can get from (∃x)Lxx to (∃x)(∃y)Lxy by using ∃E and ∃I.
b) (Vx)(Vy)Lxy b-1) (∃x)(∃y)Lxy
(Vx)Lxx (∃x)Lxx
Note that the argument, (b-1) is invalid. Prove that this argument is invalid by giving a counterexample to it. Explain why you can't get from (∃x)(∃y)Lxy to (∃x)Lxx by using ∃E and ∃I as you can get from (Vx)(Vy)Lxy to (Vx)Lxx by using VE and VI.
c) (Vx)(Vy)Lxy d) (∃x)(∃y)Lxy e) (∃x)(Vy)Lxy e-1) (Vy)(∃x)Lxy
(Vy)(Vx)Lxy (∃y)(∃x)Lxy (Vy)(∃x)Lxy (∃x)(Vy)Lxy
Note that the converse argument, (e-1) is invalid. Prove this by providing a counterexample.
f) (Vx)Px & (Vx)Qx g) (∃x)Px & (∃x)Qx h) (Vx)Px v (Vx)Qx
(Vx)(Vy)(Px & Qy) (∃x)(∃y)(Px & Qy) (Vx)(Vy)(Px v Qy)
i) (∃x)Px v (x)Qx j) (∃x)(∃y)(Px v Qy) k) (Vx)(Vy)(Lxy ⊃ ~Lxy)
(∃x)(∃y)(Px v Qy) (∃x)Px v (∃x)Qx (Vy)~Lxx
l) (Vx)(Vy)(Px ⊃ Qy) m) (∃x)(∃y)(Px ⊃ Qy) n) (∃x)(Vy)(Px ⊃ Qy)
(∃x)Px ⊃ (Vx)Qx (Vx)Px ⊃ (∃x)Qx (Vx)Px ⊃ (Vx)Qx
o) (Vx)(∃y)(Px ⊃ Qy) p) (Vx)Px ⊃ (Vx)Qx q) (Vx)(Vy)(Px v Qy)
(∃x)Px ⊃ (∃x)Qx (∃x)(Vy)(Px ⊃ Qy) (Vx)Px v (Vx)Qy
r) (∃x)(Vy)Jxy
(∃y)(∃z)(Hzy & ~Py)
(Vz)(Vw)[(Jzw & ~Pw) ⊃ Gz]
(∃z)Gz | 2,888 | 8,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-35 | latest | en | 0.84188 |
http://sudoku.com.au/4E24-2-2009-1-sudoku.aspx | 1,566,240,097,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314904.26/warc/CC-MAIN-20190819180710-20190819202710-00396.warc.gz | 190,531,285 | 16,883 | Timer 00:00:00 Sudoku Sudoku
## if (sTodaysDate) document.write('Sudoku for ' + sTodaysDate.split('-')[0] + '/' + monthname() + '/' + sTodaysDate.split('-')[2]); else document.write('Enter your own sudoku puzzle.')
Help
Play this pic as a Jigsaw
or Sliding Puzzle
Previous / Next
Choose a number, and place it in the grid above.
1 2 3 4 5 6 7 8 9
This number is a possibility
Automatically remove Possibilities Allow incorrect Moves Clicking the playing grid places the current number Highlight Current Square Grey out Used Numbers Possibilities in Grid Format
Check out the latest post in the Sudoku Forum
Welcome to the Sudoku Forums!
Submitted by: Gath
Indicate which comments you would like to be able to see GeneralJokesOtherSudoku Technique/QuestionRecipes
No I was not using kids. The word I wanted was usually. 24/Feb/09 12:35 AM | |
How did that happen? 24/Feb/09 12:35 AM | |
When my mom found it difficult to awaken us she'd flip on a stong light in our eyes and sing to us. 24/Feb/09 12:38 AM | |
My kids would join in the song. WE even had some harmony. 24/Feb/09 12:45 AM | |
I got snuggles. hehehehe She makes me into mush the first thing in the morning so she can get away with things until nap time and then she refreshes the mushiness by cuddling with me again. 24/Feb/09 12:56 AM | |
Cannot believe you had 6 children in the car and could not strap them in. OH the horror. My brother and I had to sit on opposite sides of the vehicle and were not allowed to look at each other on when went places. 24/Feb/09 12:57 AM | |
Plum, after we lost our cuteness, mom would do the bright light thing and on second warning she would pull blankets off. We knew not to get a 3rd warning. 24/Feb/09 12:58 AM | |
You were talking about blankies... My son had a LIVE blankie. He'd suck his thumb and take the tip of the cat's tail in his other hand and rub his thumb across the fur on the tip of the cat's tail! Put him to sleep every time! Any cooperating cat would do. 24/Feb/09 1:01 AM | |
How hard was it to come by a cooperating cat? 24/Feb/09 1:04 AM | |
Sounds to me Karen like you are being played! 24/Feb/09 1:05 AM | |
Wow Shiela you had a super cat. We had to boot the cat out when the girl came. Busy was jealous and would attack the girl.I thought it was unfair, since the cat was here before the girl, but Man would not allow me to boot the girl out, even though she was the tail puller. 24/Feb/09 1:05 AM | |
Oh my dutiful young son just cleared his place from the breakfast table. I can tell by the trail of milk from his spot all the way to the sink and down the front of the cupboard! 24/Feb/09 1:07 AM | |
Our first van had windows all around a two bench seats in the back. The first bench was smaller and used by the boys. Girls sat in the back and managed to fit. Neither of my girls weight very much. In those days I only weight 110. Oh for the good old days. 24/Feb/09 1:09 AM | |
He had three cats. One at home and one at each Grandma's. What more could you need??? 24/Feb/09 1:09 AM | |
You think Jill? You should see her work her Daddy. It is soooo easy and she can do it without tears. Don't you love it when they do what they are supposed to, like put their dishes in the sink. 24/Feb/09 1:10 AM | |
Plum, talking about music to wake up to...I always woke to the "Morning March" (John Phillip Sousa-type) on the radio. When it came on I knew I had 10 minutes to get to the school bus! I probably wouldn't have gotten up if they'd ever forgotten to play it! I always wondered if they played it just for me! 24/Feb/09 1:12 AM | |
I tell ya. These little girls! Jack doesn't stand a chance with 2 of 'em in the house! 24/Feb/09 1:14 AM | |
Oh Shiela, they did play the song just for you. How on earth did you get up 10 minutes before the bus arrived?Well, ladies I have enjoyed the conversation but I must start preparations to exit the house at noon, which means I only have 3 1/2 hours to do so. Of course, I will still pop in but in between my beauty therapy. 24/Feb/09 1:19 AM | |
Y'all 24/Feb/09 1:19 AM | |
24/Feb/09 1:23 AM | |
Good morning Jerry, how did your mother wake you? Do you have daughters that you are wrapped around their little fingers? Do you still have your favorite blankie? 24/Feb/09 1:24 AM | |
You people must have had traumatic awakenings in your youth. I do not recall how I awoke in my schooldays. I don't remember my mother having such evil tactics. Maybe she didn't need them because I was such an angelic child ! 24/Feb/09 1:27 AM | |
Found it . . . Three Dog Night - "Joy to the World" (aka Jeremia was a Bullfrog) . . . on my page 24/Feb/09 1:33 AM | |
I was usually putting clothes on as I ran out the door! I quite often had nightmares about not being able to find anything to wear in the closet in those ten minutes! If that actually had happened, I'd have been in real trouble! School was 9 miles away and I'm sure my mother would have made me walk!! (It takes me almost an hour to get ready now!) 24/Feb/09 1:34 AM | |
Oh Jill, only knowing you a few weeks that comment wants me to go put on some really high boots.How we were woken up gave told us what mood Mom was in. Always a good thing to know how your day was going to be. 24/Feb/09 1:37 AM | |
I can remember ironing my white, school uniform shirt whilst it was wet, also my petticoat (underskirt) because I had starched them too late to let them dry. Then I would wear them still damp....no wonder I have rheumatism and arthritis. 24/Feb/09 1:42 AM | |
As to getting out of bed, I wanted to go to school in my bed. With Jack Frost pictures on the inside of the bedroom window we used to try and dress under the covers as much as possible. 24/Feb/09 1:44 AM | |
Yippee! I am having a great day. Grandma called and asked for the girl. And I learned yesterday that the other Grandma wants her again this next weekend. Now who wants her Thursday, that is when I will be loosing my mind again? 24/Feb/09 1:46 AM | |
There was a time it just took 45 minutes to dress, that is from shower to walking out the door. I miss those days. Now it takes hours to get ready. Oh GannieMo, ironing first thing in the morning. I think I would have got dressed for school before going to bed. Though when I did do that I got into trouble for some reason. 24/Feb/09 1:50 AM | |
2:10 Hi to all. Thought for the Day: I don't think I can provide a rigorous explanation for why I'm not a snail. 24/Feb/09 1:52 AM | |
Love it André says Mo with a very slow drawn out trail of procrastinators enthusiasm 24/Feb/09 1:57 AM | |
Oh! GannieMo... I almost had forgotten... My 10 minute rush from bed to the bus usually included ironing the front & cuffs of a blouse! (No time to iron the rest of the shirt. Luckily the rest of the shirt was covered by a sweater!) I must have been crazy! 24/Feb/09 2:16 AM | |
No blankie, don't remember ever having one. One, maybe 2 calls to get up. No daughters (or sons). I did have a niece that tried, successfully, to gain favorable, to her, attention... 24/Feb/09 2:30 AM | |
That is NOT me with Dolly Sods. Never met her!But when still blurry, I note Dolly's resemblence to Mount Rushmore. 24/Feb/09 2:34 AM | |
Good Maen to everyone. Interesting rock formation! 24/Feb/09 2:35 AM | |
Off to school today. 45 kindergarteners need lunchtime supervision. Good Luck with the Doc Karen! TTFN 24/Feb/09 2:44 AM | |
Oh Jerry, everyone needs a blankie. Hubby (44) has a blankie and as I stated before it should have disappeared years ago. It is way too religious to keep a person comfy.Maybe while Jill is making her new addtion a blankie she could do one for you. Though I suggest no thumb sucking, that might not be so cute now. 24/Feb/09 2:56 AM | |
Greetings to all from sunny Ontario. Just a dusting of snow on the ground. 24/Feb/09 2:57 AM | |
Good Maen all!Go Jim (at the Dolly Sods)...but you, too, our other Jim!Glad you have sun today Cathy - always so beautiful with a fresh dusting of snow! 24/Feb/09 3:10 AM | |
Andre (sorry no accent)Loved your TFTD - my kind of pace. 24/Feb/09 3:11 AM | |
Not a member? Joining is quick and free.
As a member you get heaps of benefits.
You can also try the Chatroom (No one chatting right now - why not start something? )
Check out the Sudoku Blog Subscribe
Easy Medium Hard Tough Or try the Kids Sudokus (4x4 & 6x6) 16x16 or the Parent's Page.
Printer Friendly versions:
Members Get Goodies! Become a member and get heaps of stuff, including: stand-alone sudoku game, online solving tools, save your times, smilies and more!
Welcome our latest MembersIsabel from New Yorkracelk1 from Israelwillowberry from Brighton
Member's Birthdays TodayNone Today. | 2,516 | 8,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | longest | en | 0.952273 |
https://www.wiki.ng/ne/answers/mathematics/46120/three-boys-shared-some-oranges-the-first-received-1-3-of-the-oranges-and-the-second | 1,637,990,924,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00249.warc.gz | 1,174,998,541 | 9,361 | Mathematics
# Three boys shared some oranges. The first received 1/3 of the oranges and the second…
Three boys shared some oranges. The first received 1/3 of the oranges and the second received 2/3 of the remaining. If the third boy received the remaining 12 oranges, how many oranges did they share
• A.
60
• B.
54
• C.
48
• D.
42
##### Explanation
Let x = the number of oranges
The 1st received 1/3 of x = 1/3x
∴Remainder = x – 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3 | 172 | 493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.96989 |
https://www.hackmath.net/en/example/2920?tag_id=62,6 | 1,561,616,731,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000894.72/warc/CC-MAIN-20190627055431-20190627081431-00484.warc.gz | 782,997,365 | 6,970 | # Triangle KLB
It is given equilateral triangle ABC. From point L which is the midpoint of the side BC of the triangle it is drwn perpendicular to the side AB. Intersection of perpendicular and the side AB is point K. How many % of the area of the triangle ABC is area of a triangle KLB?
Result
p = 12.5 %
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Our percentage calculator will help you quickly calculate various typical tasks with percentages. Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.
## Next similar examples:
1. Pyramid - angle
Calculate the surface of regular quadrangular pyramid whose base edge measured 6 cm and the deviation from the plane of the side wall plane of the base is 50 degrees.
How many square meters is needed to cover the tower the shape of regular quadrangular pyramid base edge 10 meters, if the deviation lateral edges from the base plane is 68 °? Calculate coverage waste 10%.
3. Perimeter and legs
Determine the perimeter of a right triangle if the length of one leg is 75% length of the second leg and its content area is 24 cm2.
4. Rectangle
In rectangle with sides, 10 and 8 mark the diagonal. What is the probability that a randomly selected point within the rectangle is closer to the diagonal than to any side of the rectangle?
5. Cardboard box
We want to make a cardboard box shaped quadrangular prism with rhombic base. Rhombus has a side of 5 cm and 8 cm one diagonal long. The height of the box to be 12 cm. The box will be open at the top. How many square centimeters cardboard we need, if we cal
6. Box
Cardboard box shaped quadrangular prism with a rhombic base. Rhombus has a side 5 cm and one diagonal 8 cm long and height of the box is 12 cm. The box will open at the top. How many cm2 of cardboard we need to cover overlap and joints that are 5% of ar
7. Glass mosaic
How many dm2 glass is nessesary to produc 97 slides of a regular 6-gon, whose side has length 21 cm? Assume that cutting glass waste is 10%.
8. Tower
The top of the tower is a regular hexagonal pyramid with base edge 8 meters long and a height 5 meters. How many m2 of sheet is required to cover the top of the tower if we count 8% of the sheet waste?
9. Black building
Keith built building with a rectangular shape 6.5 m × 3.9 m. Calculate how much percent exceeded the limit 25 m2 for small building. Building not built in accordance with the law is called "black building". Calculate the angle that the walls were clenchin
10. Square
If the length of the sides of the square we decrease by 25% decrease the content area of 28 cm2. Determine the side length of the original square.
11. Vintner
How high can vintner fill keg with crushed red grapes if these grapes occupy a volume of 20 percent? Keg is cylindrical with a diameter of the base 1 m and a volume 9.42 hl. Start from the premise that says that fermentation will fill the keg (the number. | 744 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-26 | latest | en | 0.903904 |
https://origin.geeksforgeeks.org/program-page-replacement-algorithms-set-2-fifo/?ref=lbp | 1,675,110,617,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00857.warc.gz | 453,001,113 | 36,356 | # Program for Page Replacement Algorithms | Set 2 (FIFO)
• Difficulty Level : Medium
• Last Updated : 02 Dec, 2022
Prerequisite : Page Replacement Algorithms In operating systems that use paging for memory management, page replacement algorithm are needed to decide which page needed to be replaced when new page comes in. Whenever a new page is referred and not present in memory, page fault occurs and Operating System replaces one of the existing pages with newly needed page. Different page replacement algorithms suggest different ways to decide which page to replace. The target for all algorithms is to reduce number of page faults.
First In First Out (FIFO) page replacement algorithm –
This is the simplest page replacement algorithm. In this algorithm, operating system keeps track of all pages in the memory in a queue, oldest page is in the front of the queue. When a page needs to be replaced page in the front of the queue is selected for removal.
Example -1. Consider page reference string 1, 3, 0, 3, 5, 6 and 3 page slots. Initially all slots are empty, so when 1, 3, 0 came they are allocated to the empty slots —> 3 Page Faults.
when 3 comes, it is already in memory so —> 0 Page Faults. Then 5 comes, it is not available in memory so it replaces the oldest page slot i.e 1. —>1Page Fault.
Finally 6 comes, it is also not available in memory so it replaces the oldest page slot i.e 3 —>1 Page Fault.
So total page faults = 5
Example -2. Consider the following reference string: 0, 2, 1, 6, 4, 0, 1, 0, 3, 1, 2, 1. Using FIFO page replacement algorithm –
So, total number of page faults = 9. Given memory capacity (as number of pages it can hold) and a string representing pages to be referred, write a function to find number of page faults.
Implementation – Let capacity be the number of pages that memory can hold. Let set be the current set of pages in memory.
```1- Start traversing the pages.
i) If set holds less pages than capacity.
a) Insert page into the set one by one until
the size of set reaches capacity or all
page requests are processed.
b) Simultaneously maintain the pages in the
queue to perform FIFO.
c) Increment page fault
ii) Else
If current page is present in set, do nothing.
Else
a) Remove the first page from the queue
as it was the first to be entered in
the memory
b) Replace the first page in the queue with
the current page in the string.
c) Store current page in the queue.
d) Increment page faults.
2. Return page faults.```
Implementation:
## C++
`// C++ implementation of FIFO page replacement` `// in Operating Systems.` `#include` `using` `namespace` `std;` `// Function to find page faults using FIFO` `int` `pageFaults(``int` `pages[], ``int` `n, ``int` `capacity)` `{` ` ``// To represent set of current pages. We use` ` ``// an unordered_set so that we quickly check` ` ``// if a page is present in set or not` ` ``unordered_set<``int``> s;` ` ``// To store the pages in FIFO manner` ` ``queue<``int``> indexes;` ` ``// Start from initial page` ` ``int` `page_faults = 0;` ` ``for` `(``int` `i=0; i
## Java
`// Java implementation of FIFO page replacement` `// in Operating Systems.` `import` `java.util.HashSet;` `import` `java.util.LinkedList;` `import` `java.util.Queue;` `class` `Test` `{` ` ``// Method to find page faults using FIFO` ` ``static` `int` `pageFaults(``int` `pages[], ``int` `n, ``int` `capacity)` ` ``{` ` ``// To represent set of current pages. We use` ` ``// an unordered_set so that we quickly check` ` ``// if a page is present in set or not` ` ``HashSet s = ``new` `HashSet<>(capacity);` ` ` ` ``// To store the pages in FIFO manner` ` ``Queue indexes = ``new` `LinkedList<>() ;` ` ` ` ``// Start from initial page` ` ``int` `page_faults = ``0``;` ` ``for` `(``int` `i=``0``; i
## Python3
`# Python3 implementation of FIFO page` `# replacement in Operating Systems.` `from` `queue ``import` `Queue ` `# Function to find page faults using FIFO ` `def` `pageFaults(pages, n, capacity):` ` ` ` ``# To represent set of current pages. ` ` ``# We use an unordered_set so that we` ` ``# quickly check if a page is present` ` ``# in set or not ` ` ``s ``=` `set``() ` ` ``# To store the pages in FIFO manner ` ` ``indexes ``=` `Queue() ` ` ``# Start from initial page ` ` ``page_faults ``=` `0` ` ``for` `i ``in` `range``(n):` ` ` ` ``# Check if the set can hold ` ` ``# more pages ` ` ``if` `(``len``(s) < capacity):` ` ` ` ``# Insert it into set if not present ` ` ``# already which represents page fault ` ` ``if` `(pages[i] ``not` `in` `s):` ` ``s.add(pages[i]) ` ` ``# increment page fault ` ` ``page_faults ``+``=` `1` ` ``# Push the current page into` ` ``# the queue ` ` ``indexes.put(pages[i])` ` ``# If the set is full then need to perform FIFO ` ` ``# i.e. remove the first page of the queue from ` ` ``# set and queue both and insert the current page ` ` ``else``:` ` ` ` ``# Check if current page is not ` ` ``# already present in the set ` ` ``if` `(pages[i] ``not` `in` `s):` ` ` ` ``# Pop the first page from the queue ` ` ``val ``=` `indexes.queue[``0``] ` ` ``indexes.get() ` ` ``# Remove the indexes page ` ` ``s.remove(val) ` ` ``# insert the current page ` ` ``s.add(pages[i]) ` ` ``# push the current page into ` ` ``# the queue ` ` ``indexes.put(pages[i]) ` ` ``# Increment page faults ` ` ``page_faults ``+``=` `1` ` ``return` `page_faults` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``:` ` ``pages ``=` `[``7``, ``0``, ``1``, ``2``, ``0``, ``3``, ``0``, ` ` ``4``, ``2``, ``3``, ``0``, ``3``, ``2``] ` ` ``n ``=` `len``(pages) ` ` ``capacity ``=` `4` ` ``print``(pageFaults(pages, n, capacity))` `# This code is contributed by PranchalK`
## C#
`// C# implementation of FIFO page replacement ` `// in Operating Systems. ` `using` `System;` `using` `System.Collections;` `using` `System.Collections.Generic; ` `class` `Test ` `{ ` ` ``// Method to find page faults using FIFO ` ` ``static` `int` `pageFaults(``int` `[]pages, ``int` `n, ``int` `capacity) ` ` ``{ ` ` ``// To represent set of current pages. We use ` ` ``// an unordered_set so that we quickly check ` ` ``// if a page is present in set or not ` ` ``HashSet<``int``> s = ``new` `HashSet<``int``>(capacity); ` ` ` ` ``// To store the pages in FIFO manner ` ` ``Queue indexes = ``new` `Queue() ; ` ` ` ` ``// Start from initial page ` ` ``int` `page_faults = 0; ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ``// Check if the set can hold more pages ` ` ``if` `(s.Count < capacity) ` ` ``{ ` ` ``// Insert it into set if not present ` ` ``// already which represents page fault ` ` ``if` `(!s.Contains(pages[i])) ` ` ``{ ` ` ``s.Add(pages[i]); ` ` ` ` ``// increment page fault ` ` ``page_faults++; ` ` ` ` ``// Push the current page into the queue ` ` ``indexes.Enqueue(pages[i]); ` ` ``} ` ` ``} ` ` ` ` ``// If the set is full then need to perform FIFO ` ` ``// i.e. Remove the first page of the queue from ` ` ``// set and queue both and insert the current page ` ` ``else` ` ``{ ` ` ``// Check if current page is not already ` ` ``// present in the set ` ` ``if` `(!s.Contains(pages[i])) ` ` ``{ ` ` ``//Pop the first page from the queue ` ` ``int` `val = (``int``)indexes.Peek(); ` ` ` ` ``indexes.Dequeue(); ` ` ` ` ``// Remove the indexes page ` ` ``s.Remove(val); ` ` ` ` ``// insert the current page ` ` ``s.Add(pages[i]); ` ` ` ` ``// push the current page into ` ` ``// the queue ` ` ``indexes.Enqueue(pages[i]); ` ` ` ` ``// Increment page faults ` ` ``page_faults++; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``return` `page_faults; ` ` ``} ` ` ` ` ``// Driver method ` ` ``public` `static` `void` `Main(String []args) ` ` ``{ ` ` ``int` `[]pages = {7, 0, 1, 2, 0, 3, 0, 4, ` ` ``2, 3, 0, 3, 2}; ` ` ``int` `capacity = 4; ` ` ``Console.Write(pageFaults(pages, pages.Length, capacity)); ` ` ``} ` `} ` `// This code is contributed by Arnab Kundu`
## Javascript
``
Output
`7`
Note – We can also find the number of page hits. Just have to maintain a separate count. If the current page is already in the memory then that must be count as Page-hit. | 2,750 | 9,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.855247 |
https://lambdapage.org/25995/number-practice-worksheets-1-10/free-kindergarten-numbers-worksheet-for-fall-1-10-throughout-number-practice-worksheets-1-10/ | 1,620,253,215,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988696.23/warc/CC-MAIN-20210505203909-20210505233909-00360.warc.gz | 394,806,897 | 7,541 | # free kindergarten numbers worksheet for fall 1 10 throughout number practice worksheets 1 10
##### free kindergarten numbers worksheet for fall 1 10 throughout number practice worksheets 1 10.
As a pre-requisite, you need to ensure that kids already understand about the color. Now give a simple instruction that they need to coloring the object that match with the number given.
It helps to understand that the number they will write has a sequence. If your kid start recognizing the number, then you can start to write the number. The simplest number is 1.
Another method that can be done for introducing the number is by using the color by number worksheet. It’s actually a clever idea for using this worksheet.
It can be done for making your kids understand about the alphabet. As the pre-requisite, you have to ensure that your kids already recognize about the color and alphabet.
March 11, 2021
March 3, 2021
March 4, 2021
March 7, 2021
### Precursive Handwriting Worksheets
March 7, 2021
It will enhance the motoric skill for your kid. Evaluate the result by seeing whether the line is tracing tidily or not. If it’s not quite tidy, then you can ask your kids to repeat the progress. Remember, you also need to praise them as the positive reinforcement. Kids will always get an encouragement from a simple praise.
Take a look on the colored pictures, whether it’s tidy enough or not. Give them an instruction that next time they should put the color inside the area. It will make them understand about coloring the object in proper way.
It’s another activity that will stimulate their memory and recognize the colors. Use the crayon to finish this project. As a reward, kids are allowed to hang their creation on their wall. It will make they enjoy and want to learn more using the other worksheets.
After you’ve done with the line tracing, now it’s time for more advanced exercise. The spiral tracing can be challenging for your kids. It’s also need a concentration with the good coordination for tracing the lines.
March 11, 2021
March 10, 2021
March 6, 2021
March 11, 2021
### Puerto Rico Reading Comprehension Worksheets
March 6, 2021
Learning to write the letter is the first skill that should be done for kids. This worksheet has a simple instruction for tracing the letter. You can make a demonstration to the kid before instruct them to do it by themselves. Remember, this is need a good patience.
Get your worksheets for preschool kids and start enjoy the learning process. These preschool worksheets are great for learning at home during this pandemic era.
If you think that spelling words and recognizing the number is too complex, then you can try a simple worksheet that has main purpose of tracing skill. It’s a line tracing worksheet that can be used to achieve this skill.
You can give a demonstration for writing the letter A side by side with your kid. If your kid had a difficulty to imitate your gesture, then you also can give your hand to help their tracing process.
March 3, 2021
March 11, 2021
March 4, 2021
March 7, 2021
March 10, 2021
March 4, 2021
March 7, 2021
March 6, 2021
March 10, 2021
March 6, 2021
March 4, 2021
March 9, 2021
March 5, 2021
March 5, 2021
March 8, 2021 | 770 | 3,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.92634 |
https://rdrr.io/cran/CP/man/CP-package.html | 1,526,852,529,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00050.warc.gz | 636,098,065 | 15,176 | # CP-package: Conditional Power Calculations In CP: Conditional Power Calculations
## Description
This package provides several functions for calculating the conditional power for different models in survival time analysis within randomized clinical trials with two different treatments to be compared and survival as an endpoint.
## Details
Package: CP Type: Package Version: 1.6 Date: 2016-06-28 License: GPL-3
This package could be some help when you want to calculate the conditional power at the time of an interim analysis of a randomized clinical trial with survival as an endpoint.
The conditional power is defined as the probability of obtaining a significant result at the end of the trial when the real effect is equal to the expected effect given the data from the interim analysis.
Functions for the model with exponential survival (`ConPwrExp`) and the non-mixture models with exponential (`ConPwrNonMixExp`), Weibull type (`ConPwrNonMixWei`) and Gamma type survival (`ConPwrNonMixGamma`) are provided.
There is also the function `CompSurvMod` to compare the four mentioned models.
Additionally, there is also a function for the exponential model with the original formulae of the Andersen paper (`ConPwrExpAndersen`).
Finally, the user is able to generate further data frames by random via `GenerateDataFrame`.
## Note
The theoretical results of this implementation are based on some assumptions.
Non-Mixture-Exponential: λ[1] = λ[2]
Non-Mixture-Weibull: λ[1] = λ[2] and k[1] = k[2]
Non-Mixture-Gamma: a[1] = a[2] and b[1] = b[2]
In general, such assumptions are not fulfilled when using real data.
Nevertheless, when doing conditional power calculations the situation is that you have no significant difference at the time of interim analysis. In this case, no treatment arm is superior to the other one. Thus, the assumptions named above are approximately satisfied.
In contrast to this, caution should be exercised when calculating the conditional power in the case of significant results at the time of interim analysis.
## Author(s)
Andreas Kuehnapfel
Maintainer: Andreas Kuehnapfel <[email protected]>
## References
Kuehnapfel, A. (2013). Die bedingte Power in der Ueberlebenszeitanalyse.
Andersen, P. K. (1987). Conditional power calculations as an aid in the decision whether to continue a clinical trial. Controlled Clinical Trials 8, 67-74.
`ConPwrExp`
`ConPwrNonMixExp`
`ConPwrNonMixWei`
`ConPwrNonMixGamma`
`CompSurvMod`
`ConPwrExpAndersen`
`GenerateDataFrame`
`test`
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38``` ``` # data frame 'test' generated by 'GenerateDataFrame' # conditional power calculations # within the exponential model ConPwrExp(data = test, cont.time = 12, new.pat = c(2.5, 2.5), theta.0 = 0.75, alpha = 0.05, disp.data = TRUE, plot.km = TRUE) # conditional power calculations # within the non-mixture model with exponential survival ConPwrNonMixExp(data = test, cont.time = 12, new.pat = c(2.5, 2.5), theta.0 = 0.75, alpha = 0.05, disp.data = TRUE, plot.km = TRUE) # conditional power calculations # within the non-mixture model with Weibull type survival ConPwrNonMixWei(data = test, cont.time = 12, new.pat = c(2.5, 2.5), theta.0 = 0.75, alpha = 0.05, disp.data = TRUE, plot.km = TRUE) # conditional power calculations # within the non-mixture model with Gamma type survival ConPwrNonMixGamma(data = test, cont.time = 12, new.pat = c(2.5, 2.5), theta.0 = 0.75, alpha = 0.05, disp.data = TRUE, plot.km = TRUE) # conditional power calculations # within the four mentioned models CompSurvMod(data = test, cont.time = 12, new.pat = c(2.5, 2.5), theta.0 = 0.75, alpha = 0.05, disp.data = TRUE, plot.km = TRUE) # conditional power calculations # within the exponential model # with the original formulae of the Andersen paper ConPwrExpAndersen(data = test, cont.time = 12, new.pat = c(2.5, 2.5), theta.0 = 0.75, alpha = 0.05, disp.data = TRUE, plot.km = TRUE) ```
CP documentation built on May 29, 2017, 9:51 p.m. | 1,128 | 4,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-22 | latest | en | 0.902971 |
http://ebookmarket.org/pdf/intro-to-statistics-final-exam-key | 1,441,400,647,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645366585.94/warc/CC-MAIN-20150827031606-00291-ip-10-171-96-226.ec2.internal.warc.gz | 72,211,279 | 14,390 | # Intro To Statistics Final Exam Key pdfs
Searching:
Math 281 Introduction to Statistics Sample Exam 1...
8967 dl's @ 2377 KB/s
Math 281 Introduction to Statistics Sample Exam 1...
Math 281 Introduction to Statistics Sample Exam 1 ... She gets a 72 on her final exam. ... Answer Key Testname: EX1SAMPLE3_5 1)C 2)C 3)B 4)B 5)D 6)A 7)C 8)D 9)A
http://bhsu.edu/Portals/260/Math281/Sample%20Exams/ex1sample3_5wans.pdf
Date added: January 19, 2015 - Views: 2
Intro To Psychology Final Exam Answers
Intro To Psychology Final Exam ... http://www.gordonstate.edu/Faculty/p_terry/Key%201230 ... Single Subject Designs Intro to Final Project ... 21 Final Exam: ...
Date added: December 15, 2014 - Views: 12
Statistics Exam NAME: Part I – Multiple Choice....
Statistics Exam Author: charles Created Date: 2/28/2005 2:36:05 PM ...
http://mathcontest.uh.edu/Exams_to_be_posted/Statistics_Exam.pdf
Date added: November 12, 2011 - Views: 239
General Education Offerings Timetable Fall 2015...
Course Intro Meeting ... Statistics Introduction to ... 10/20, 11/3, 11/17, and 12/1 with final exam on 12/8/2015. BU 320 COURSE NOTE: On September 30th, ...
http://www.bellincollege.edu/assets/pages//FA%202015_Gen%20Ed_ao%2008.04.2015.pdf
Date added: August 12, 2015 - Views: 1
STATISTICS 8, FINAL EXAM NAME: KEY Seat Number:
STATISTICS 8, FINAL EXAM NAME: KEY_____ Seat Number: _____ Last six digits of ... final exam = 50 + (0.5)(midterm). An interpretation of the slope is: A.
http://www.ics.uci.edu/~jutts/8/FinalExamAKey.pdf
Date added: November 15, 2011 - Views: 16
Math 1040 Final Exam Form A Introduction to...
Math 1040 Final Exam Form A Introduction to Statistics Fall Semester 2009 ... The data below are the final exam scores of 10 randomly selected history students and
http://www.slcc.edu/math/docs/math%201040%202009.pdf
Date added: May 23, 2013 - Views: 17
Introductory Statistics Lectures Review for Final ...
Introductory Statistics Lectures Review for Final Exam Anthony Tanbakuchi ... Review for Final Exam 3 of12 1.2 Key concepts to take away from this class.
http://www.tanbakuchi.com/Courses/MAT167/Files/LH_LEC.0920.Review.Final.pdf
Date added: September 30, 2012 - Views: 10
STATISTICS 110/201 PRACTICE FINAL EXAM KEY...
Practice Final Exam, Statistics 110/201, page 3 of 7 7. Give two circumstances in which it is acceptable to remove one or more cases that are outliers in
http://www.ics.uci.edu/~jutts/110-201-09/PracticeRegressionFinalKey.pdf
Date added: January 30, 2012 - Views: 25
Statistics Final Exam Cheat Sheet - free Ebooks...
Statistics Final Exam Cheat Sheet.pdf ... Intro to Statistics Final Exam Study Guide 0. ... Key: ... Descriptive statistics from the sample are used as estimators of ...
http://www.pdfsdocuments.com/statistics-final-exam-cheat-sheet.pdf
Date added: November 14, 2014 - Views: 1
Example Of Instruction Sentence Outline With Topic
Jackie intro is dedicated to offering some detailed ... statistics, quotes, ... Final Exam Essay: Topic, ...
https://vasgoldlinrows.files.wordpress.com/2015/08/example-of-instruction-sentence-outline-with-topic.pdf
Date added: August 12, 2015 - Views: 1
Introduction to the Practice of Statistics - PCC
Introduction to the Practice of Statistics Sixth Edition Moore, McCabe Section 2.1 Homework Answers 2.6 Relationship between first test and final exam.
http://spot.pcc.edu/~hmesa/math243/homework/chapter2/6th_chapter2_section1.pdf
Date added: September 29, 2012 - Views: 144
AP Statistics: Syllabus 2 - College Board
AP® Statistics: Syllabus 2 ... completed using JMP-Intro. After the AP Exam, ... Review for AP Exam and Final Exam (total time: 7 days)
http://media.collegeboard.com/digitalServices/pdf/ap/ap-statistics-sample_2_1058810v1.pdf
Date added: September 10, 2013 - Views: 38
Online Course Syllabus - UCI Summer Session
Online Course Syllabus Page 1 ... Final exam (2 hours, in person at UCI) 35% ... You obtain your Moodle enrollment key from Summer Session in an email ...
http://summer.uci.edu/pdfs/courses/Stats8_syllabus.pdf
Date added: April 22, 2013 - Views: 36
Introduction to Statistics Final Exam Study Guide...
Introduction to Statistics Final Exam Study Guide ... Population parameters ( , ˙, and p) vs. Sample statistics (x, sand ^p). Continuous vs Discrete random variables.
Date added: February 28, 2013 - Views: 5
www.cheat-sheets.org
Created Date: 10/25/2007 9:57:31 PM
http://www.cheat-sheets.org/saved-copy/427513-Statistics-Reference-Cheatsheet.pdf
Date added: June 7, 2015 - Views: 1
Statistics: Sample Exam 2 Solution Chapters 4 & 5...
Statistics: Sample Exam 2 Solution Chapters 4 & 5: ... The sum of probabilities of all final outcomes of an experiment is always ... Probability of being awakened = 1 ...
http://math.lbcc.edu/maraeipour/LBCC_Stat1/Take_Home_Solutions_Corrected/PDF_Files/Statistics_Sample_Exam_2_Correction.pdf
Date added: May 6, 2012 - Views: 18
AP Statistics Final Exam Review Solutions - Dan...
AP Statistics Final Exam Review Solutions Page 4 of 4 8. Assessment of statements concerning correlation and regression . a) regression b) strength of .
http://www.danshuster.com/apstat/AP%20Statistics%20Final%20Exam%20Review%20Solutions.pdf
Date added: February 25, 2012 - Views: 48
Final Exam Practice Questions for General...
Final Exam Practice Questions for General Chemistry ... chemistry class. ... tested on the final exam. An answer key will be provided at a later date.
http://chemicaleducation.org/chem110/110F08finalpracticeAK.pdf
Date added: March 10, 2012 - Views: 1369
Introduction to Engineering Design - Wikispaces
1.1 Lesson 1.1 Intro ... taken the PLTW® course and successfully completed the college credit exam. Students who have taken a PLTW® course ... IED Final Exam Spring ...
http://pltw2009.wikispaces.com/file/view/ENG_IED_STI_2009_Notebook.pdf
Date added: November 1, 2011 - Views: 234
Final Exam Schedule for Fall Semester 2014/2015
Following is the Final Exams Key for the courses offered in the coming Fall Semester 2014/2015.The date and time of the final exam will be ... GENG 200 Intro. to ...
http://www.uaeu.ac.ae/en/dvcsae/registration/pdf/final%20exams%20schedule%20-%20fall%202014-2015.pdf
Date added: June 30, 2014 - Views: 3
Final Exam Study Guide - Saint Joseph High School
Psychology - Final Exam Study Guide * * Chapter 1 - Intro to Psychology * Goals of Psychology * Scientific Method * Historical Approaches and Key Contributors *
http://saintjoehigh.enschool.org/ourpages/auto/2007/8/5/1186366825614/PsychologyFinalExamStudyGuide.pdf
Date added: April 25, 2012 - Views: 43
Introductory Statistics – A Self-Assessment Test
2 Relationships between two variables 11. The following graph shows the relationship between the final exam scores of these sociology students
http://gs.bethel.edu/selftest-key.pdf
Date added: February 2, 2012 - Views: 71
Introduction to Social Statistics - University of...
Introduction to Social Statistics Summer Semester 2004 SOCS 3305, ... the final exam, ... rounding error of the answer I have calculated on the key.
http://www.utdallas.edu/~tmb021000/Documents/SOCS3305%20Syllabus%20SU2004.pdf
Date added: July 18, 2012 - Views: 1
Introduction to Statistics - Sage Publications
A professor records the number of students absent during a final exam. The number of absent ... INTRODUCTION AND DESCRIPTIVE STATISTICS. KEY TERMS. coding continuous ...
http://www.sagepub.com/sites/default/files/upm-binaries/40006_Chapter1.pdf
Date added: July 18, 2015 - Views: 1
Chapter 3: Describing Relationships - Macmillan...
... Describing Relationships ... Our statistics toolbox now contains a variety of ... component of statistical study and is the final skill necessary for our ...
Date added: June 13, 2013 - Views: 6
MATH 1127 INTRODUCTION TO STATISTICS
MATH 1127 INTRODUCTION TO STATISTICS ... Topics include descriptive statistics, ... Final Exam ...
http://www.athenstech.edu/studentAffairs/programs/syllabi/MATH_1127.pdf
Date added: June 17, 2015 - Views: 2
CHAPTER-BY-CHAPTER ANSWER KEY ... The final stage is the older years, ... a college student cheating on an exam and a mugger
Date added: October 3, 2011 - Views: 31896
Business Statistics Midterm Exam - Booth School of...
Business Statistics Midterm Exam Fall 2013 Russell ... indicate your final answer and complete all computations. This section is worth 60 points. Problem 1.
http://faculty.chicagobooth.edu/jeffrey.russell/teaching/bstats/midterm/midtermsol.pdf
Date added: November 8, 2011 - Views: 134
Course Syllabus for Sociology 592 Advanced Social ...
Tentative schedule Sociology 592 Advanced Social Statistics I Fall 2004 This schedule is hopefully more or less firm, but I reserve the right to make minor ...
http://www3.nd.edu/~rwilliam/stats1/syl592.pdf
Date added: February 13, 2013 - Views: 7
Course Syllabus - Borough of Manhattan Community...
Final exam due . Student Learning ... • Students will be able to construct simple statistical charts as well as to calculate key statistics and parameters such as ...
Date added: September 6, 2011 - Views: 8
ECONOMICS 321: INTRODUCTION TO STATISTICS
... Introduction to Statistics Fall 2013 ... Practice is the key for understanding Statistics. ou will be Y ... There will be one midterm and one final exam for ...
http://www.economics.hawaii.edu/syllabi/Fall13/econ_321_001.pdf
Date added: June 16, 2013 - Views: 34
AP Statistics Syllabus Primary Textbooks: Bock,...
AP Statistics Syllabus Primary ... As a final part of AP exam review, students are required to maintain a Free Response review binder ... 1 Stats Starts Here Intro to ...
Date added: February 28, 2014 - Views: 30
Small Student Projects in an Introductory ...
Small Student Projects in an Introductory Statistics Course Robert L. Wardrop ... Arguably a key component 2. ... relying strictly on responses to exam questions.
http://www.stat.wisc.edu/~wardrop/papers/tmoore.pdf
Date added: December 3, 2011 - Views: 341
Student Book Answer Key - AzarGrammar.com
Student Book Answer Key Chapter 1: Present Time Exercise 5, p. 4. ... studied statistics in Chile. He left Chile and moved here. Then he went to New York and stayed ...
Date added: June 3, 2012 - Views: 326
Final Exam Solutions - courses.csail.mit.edu
6.006 Final Exam Solutions Name 2 (c) T F Under the simple uniform hashing assumption, the probability that three specific data elements (say 1, 2 and 3) hash to the ...
http://courses.csail.mit.edu/6.006/spring11/exams/quiz3-sol.pdf
Date added: October 21, 2012 - Views: 15
med.brown.edu
exam. Men who fail to ... Epidemiology Final Title: Acupuncture for the Treatment of Cocaine Addiction ... whic ould be a letter from the K-type questions key above, in
http://med.brown.edu/students/senate/old_exams/bi372-epidemiology-00-02.pdf
Date added: March 8, 2014 - Views: 1
INTRODUCTION TO MATHEMATICS MULTIPLE-CHOICE Sample...
MATHEMATICS MULTIPLE-CHOICE Sample Tests ... • principles of statistics and ... The answer key provided at the end of the
http://www.ode.state.or.us/teachlearn/testing/samples/archives/mathsampletest-itk.pdf
Date added: January 28, 2012 - Views: 119
Final Examination Parts A, B & C ANSWER KEY Spring 2007 For Teacher Use ONLY Project Lead The Way, Inc. Copyright 2007 ... Answer Key: Missing lines are in red.
Date added: April 11, 2013 - Views: 216
Statistics 151 – Practice Midterm Exam
Statistics 151 – Practice Midterm – Mike Kowalski Statistics 151 – Practice Midterm Exam Instructor: Mike Kowalski Instructions: 1. ... Final Score 20 ...
http://www.stat.ualberta.ca/~kowalski/Stat151/151PracticeMT5MC50.pdf
Date added: August 10, 2013 - Views: 2
MATH 1101: Introduction to Mathematical Modeling...
Math 1101/08, Fall Semester 2010 ... Office: Mathematics & Statistics (Building #16), Room 014 ... and a comprehensive final exam. They will count as follows:
http://math.kennesaw.edu/~bthomas/MATH%201101%20Syllabus%20(Fall%202011).pdf
Date added: July 14, 2014 - Views: 2
Intermediate Algebra REVIEW for FINAL EXAM
Intermediate Algebra REVIEW for FINAL EXAM 4/06 1. Find the slope for the line passing through the points A(–8, 4) and B(–1, –2). [A] ...
Date added: April 25, 2012 - Views: 39
INTRODUCTION TO STATISTICS FOR ECONOMICS AND...
INTRODUCTION TO STATISTICS FOR ECONOMICS AND BUSINESS ... a Course Key to make sure you are buying the book for this specific ... In the Common Final Exam, ...
Date added: July 1, 2012 - Views: 32
2009, 2010 AP Statistics Course Description
AP Statistics Exam to have access to any ... data that students key into the calculator while taking the exam. ... on its feet after the final 10 jumps is exactly ...
http://apcentral.collegeboard.com/apc/public/repository/ap08_statistics_coursedesc.pdf
Date added: September 17, 2011 - Views: 91
FINAL EXAMINATION SCHEDULE SPRING SEMESTER,...
FINAL EXAMINATION SCHEDULE ... APRIL 20 TO WEDNESDAY, MAY 1, 2015. Page 2 Final Exam Schedule ... Monday, April 20, 2015 7:00 PM STAT301 01 Inter Statistics B1
http://www.cob.edu.bs/DOCS/Exams012015_Chrono.pdf
Date added: April 27, 2015 - Views: 1
ExamView - Mathematics Grade 7 PRACTICE Final Exam
ExamView - Mathematics Grade 7 PRACTICE Final Exam.tst Author: bgottlob Created Date: 6/11/2008 10:53:32 AM ...
Date added: October 13, 2011 - Views: 496
COURSE # AND TITLE Psyc210: Statistics for the...
write formulas for the exam. The final exam will be a take home exam. ... Fundamental Statistics for the Behavioral Sciences. (8th ... 7 Intro to inferential ...
http://www.sanjuancollege.edu/documents/syllabi/PSYC-210.pdf
Date added: April 18, 2013 - Views: 15
Microsoft Excel 2010 ... http://wps.prenhall.com/bp_exploring_excel_2010_intro/ Click on Student Data files ... Final Exam: I have put a Final ...
http://flc.losrios.edu/~hicks/CISA310/CISA310.pdf
Date added: September 12, 2011 - Views: 210
Math 2345-Spring 2015 Business Statistics Professor: Dr. Kathleen Gilliam ... Each of the three midterms will be worth 20% and the final exam ... Apr. 20 Intro P ...
http://www.math.ttu.edu/~xgilliam/classes/m2345/m2345_syllabus.pdf
Date added: November 18, 2011 - Views: 23
AP Statistics Course Syllabus - Whitnall High...
The AP Statistics course is designed to provide ... Text: Statistics in Action, by Watkins, Scheaffer, and Cobb, Key Curriculum ... Final Review for AP Exam ...
http://whs.whitnallschools.enschool.org/ourpages/auto/2012/3/5/52692374/WHS%20AP%20Statistics%20Course%20Syllabus%20for%20AP%20Central.pdf
Date added: May 16, 2013 - Views: 83 | 4,013 | 14,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2015-35 | longest | en | 0.68839 |
https://stats.stackexchange.com/questions/linked/2391 | 1,571,400,874,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00381.warc.gz | 733,480,494 | 26,552 | 3k views
### chi squared test or Z test? [duplicate]
I have an outcome from a biological experiment, where I have counted the number of dead cells upon a mutation. For example, in once case - out of 120 cells we saw 16% of the dead cells, but our ...
483 views
### Test of proportions or chi-squared? [duplicate]
These data are from an A/B test of free trials to conversion. Someone can sign up for a 30 day free trial of the product. If they like it they stay on and become paying customers. By default (A), when ...
2k views
### At What Level is a $\chi^2$ test Mathematically Identical to a $z$-test of Proportions?
BACKGROUND: Skip safely - it's here for reference, and to legitimize the question. The opening of this paper reads: "Karl Pearson’s famous chi-square contingency test is derived from another ...
5k views
### Assessing A/B test results using Python
Problem's setting Assume having results of an A/B test. You let you users experience two variants of your website and you counted how many converted: ...
600 views
### Difference estimate & confidence intervals for $\chi^2$ test between 2 proportions
I am using a chi-sq test between two proportions (http://statistic-on-air.blogspot.com/2009/07/comparison-of-two-proportions.html). I am using this over the z-test for proportions because I do not ...
282 views
### Can I use a two proportion z-test?
I have a response, "Satisfaction" that goes on a scale from 0-10 (where 0=dissatisfied and 10= very satisfied), and I have a predictor variable that also uses the same scale. I recoded the response so ...
318 views
### Appropriate Correction for 2 proportion test? --with example--
Below is an example of frequency data between observed and expected proportions. Can I simply run a 2-Proportion z test here? Or, would I be violating assumptions? If I were to run dozens of z tests (...
160 views
### Appropriate test for proportion data
I have two independent groups A and B. Each participant is given 5 questions to solve and his/her answer is evaluated as correct or incorrect. However, participants can skip questions, so the number ...
193 views
### Confusion about z-test and chi square test on a very basic example
I am confused with the difference between z-test and chi square test. In book Agresti 2007 An introduction to categorical data analysis, page 9, the author had an example of testing Do a majority ...
156 views
### What statistical test do I use for differences in proportions?
I have two groups of samples with a proportion of mutations in each sample being a particular type (the proportions are represented as a fraction of 1). How to I check if that type of mutation is ...
111 views
### How to process Binomial-Binomial data on SPSS for hypothesis testing?
There are 2 groups of 100 pregnant women: the first are malnourished and the second are well-nourished. A COHORT research is being done on them to determine if malnutrition is a risk factor for low ... | 679 | 2,973 | {"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} | 3.21875 | 3 | CC-MAIN-2019-43 | latest | en | 0.912875 |
http://cboard.cprogramming.com/cplusplus-programming/69308-calculator-program-problemo.html | 1,466,879,083,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393463.1/warc/CC-MAIN-20160624154953-00128-ip-10-164-35-72.ec2.internal.warc.gz | 49,191,405 | 11,877 | # Thread: Calculator program problemo
1. ## Calculator program problemo
Code:
```#include <cstdio>
#include <cstdlib>
#include <string>
#include <cmath>
#include <iostream>
using namespace std;
float Squareroot(float x);
float Add(float x, float y);
float Subtract(float x, float y);
float Multiply(float x, float y);
float Divide(float x, float y);
int main() {
cout << "Math Program v.1.8" << endl;
float numberA;
float numberB;
cout << "What two numbers would you like to input?" << endl;
cin >> numberA;
cin >> numberB;
cout << "Would you like to: \n" ;
cout << "Square[r]oot?" << endl;
cout << "[a]dd" << endl;
cout << "[s]ubtract" << endl;
cout << "[m]ultiply" << endl;
cout << "or [d]ivide the numbers?" << endl;
float funcvarbname;
string Option;
if (Option == "r") {funcvarbname=Squareroot(numberA);};
if (Option == "a") {funcvarbname=Add(numberA, numberB);};
if (Option == "s") {funcvarbname=Subtract(numberA, numberB);};
if (Option == "m") {funcvarbname=Multiply(numberA, numberB);};
if (Option == "d") {funcvarbname=Divide(numberA, numberB);};
cout << funcvarbname << endl;
system("PAUSE");
return 0;
}```
Well I fixed my original error, but now it doesn't print the answers to my functions! Help please!
2. it doesn't print the answers to my functions!
Um, isnt that because you arent printing anything with say, cout?
3. I was using cout. But anyway I fixed it:
Code:
```#include <cstdio>
#include <cstdlib>
#include <string>
#include <cmath>
#include <iostream>
using namespace std;
float Squareroot(float x);
float Add(float x, float y);
float Subtract(float x, float y);
float Multiply(float x, float y);
float Divide(float x, float y);
int main() {
cout << "Math Program v.1.8" << endl;
float numberA;
float numberB;
cout << "What two numbers would you like to input?" << endl;
cin >> numberA;
cin >> numberB;
cout << "Would you like to: \n" ;
cout << "Square[r]oot?" << endl;
cout << "[a]dd" << endl;
cout << "[s]ubtract" << endl;
cout << "[m]ultiply" << endl;
cout << "or [d]ivide the numbers?" << endl;
float funcvarbname;
string Option;
cin >> Option;
if (Option == "r") {funcvarbname=Squareroot(numberA);};
if (Option == "a") {funcvarbname=Add(numberA, numberB);};
if (Option == "s") {funcvarbname=Subtract(numberA, numberB);};
if (Option == "m") {funcvarbname=Multiply(numberA, numberB);};
if (Option == "d") {funcvarbname=Divide(numberA, numberB);};
cout << "YOUR ANSWER IS: " << funcvarbname << endl;
system("PAUSE");
return 0;
}
float Squareroot(float x) {
return sqrt(x);
}
float Add(float x, float y) {
return x+y;
}
float Subtract(float x, float y) {
return x - y;
}
float Multiply(float x, float y) {
return x*y;
}
float Divide(float x, float y) {
return x / y ;
}```
4. I was using cout.
You forgot to use it to print what you wanted to print though, glad to see that you have it fixed.
Popular pages Recent additions | 841 | 2,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | longest | en | 0.622651 |
www.prepareinterview.com | 1,657,122,429,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00475.warc.gz | 991,939,526 | 40,317 | TCS Placement Paper Part - I ( 2015 - 2016 Batch) - PrepareInterview.com
# PrepareInterview.com
## Thursday, December 17, 2015
Company Name: TCS
*Written exam there were two sections one is verbal and other is Arithmetic portions
*In verbal sections there we had to write an email writing and in arithmetic part there were several problems like series problem, Probability, train, profit & loss, Allegation ,etc...and some tricky problems
1. Here is 15 dots. If you select 3 dots randomly, what is the probability that 3 dots make a triangle?
a. 440/455.
b. 434/455.
c. 449/455.
d. 438/455.
Total ways of selecting 3 dots out of 15 is 15C3 = 455 If 3 dots are collinear then triangle may not be formed. Now look at the above diagram.
If we select any 3 dots from the red lines they may not form a triangle. They are 5 x 5C3 = 50. If we select the three letters from blue lines, they may not form a triangle.
They are in total 5 ways. Also there are 6 others lines which don't form a triangle. Total = 50 + 5 + 6 = 61. So we can form a triangle in 455 - 61 = 394. So answer could be 394/455.
2. In how many ways a team of 11 must be selected from 5 men and 11 women such that the team must comprise of not more than 3 men?
Answer: The team may consist of 0 men + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women.
So Number of ways are = 11C11+5C1x11C10+5C2x11C9+5C3x11C8 = 2256.
3. Eesha bought 18 sharpeners for Rs.100. She paid 1 rupee more for each white sharpener than for each brown sharpener. What is the price of a white sharpener and how many white sharpener did she buy ?
a. Rs.5, 10
b. Rs.6, 10
c. Rs.5, 8
d. Rs.6, 8
Sol: B.
Explanation:
Just check the options. If she bought 10 white sharpeners at Rs.6 per piece, She has spent Rs.60 already.
And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.
4. On a 26 question test, five points were deducted for each wrong answer and eight points were added for each correct answer.
If all the questions were answered, how many were correct, if the score was zero ?
a. 10
b. 12
c. 11
d. 13
Ans: A
Explanation:
Take options and check. If 10 are correct, his score is 10 x 8 = 80. But 16 are wrong. So total negative marking is 16 x 5 = 80. So final score is zero.
5. If f(f(n))+f(n)=2n+3 and f(0)=1, what is the value of f(2012)?
a) 2011
b) 2012
c) 2013
d) 4095
Ans: Option C.
6.1(1!)+2(2!)+3(3!)....2012(2012!) = ?
Ans: 2013!-1
1(1!)=1 => 2!-1
1(1!)+2(2!)=1+4=5 => 3!-1
1(1!)+2(2!)+3(3!)=1+4+18=23 => 4!-1
........................
.......................
1(1!)+2(2!)+3(3!)+........+2012(2012!)=>2013!-1.
7.A cow and horse are bought for Rs.2,00,000. The cow is sold at a profit of 20% and the horse is sold at a loss of 10%. The overall gain is Rs.4000, the Cost price of cow?
a) 130000.
b) 80000.
c) 70000.
d) 120000.
Ans: Overall profit = 4000/200000x100=2%.
By applying alligation rule, we get
So cost price of the cow = 2/5 x 200000 = 80,000.
8.Ahmed, Babu, Chitra, David and Eesha each choose a large different number. Ahmed says, " My number is not the largest and not the smallest".
Babu says, "My number is not the largest and not the smallest". Chitra says, "My number is the largest".
David says, " My number is the smallest". Eesha says, " My number is not the smallest". Exactly one of the five children is lying. The others are telling the truth. Who has the largest number?
a) Eesha.
b) David.
c) Chitra.
d) Babu.
Ans: A.
9.A beaker contains 180 liters of alcohol. On 1st day, 60 l of alcohol is taken out and replaced by water. 2nd day, 60 l of mixture iss taken out and replaced by water and the process continues day after day. What will be the quantity of alcohol in beaker after 3 days.
Ans: 53.3.
These are the questions which I remembered some questions also coming from Geometry portion also (2 questions).
10. The value of diamond varies directly as the square of its weight. If a diamond falls and breaks into two pieces with weights in the ratio 2:3. what is the loss percentage in the value?
Sol: Let weight be “x”
the cost of diamond in the original state is proportional to x2
when it is fallen it breaks into two pieces 2y and the 3y
x = 5y
Original value of diamond = (5y)2 = 25y2
Value of diamond after breakage = (2y)2+(3y)2=13y2
so the percentage loss will be = 25y2−13y225y2×100=48%
11.Five college students met at a party and exchanged gossips. Uma said, “Only one of us is lying”. David said, “Exactly two of us are lying”. Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are lying”. Chitra said “All of us are lying”. Which one was telling the truth?
a)David
b)Querishi
c)Chitra
d)Thara
Sol: As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.
Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.
12.Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that.
a) 2676
b) 2
c) 445
d) 86
Sol: This is a number series problem nothing to do with the data given.
1x 1+1=2
2 x 2+2=6
6 x 3+3=21
21 x 4+4=88 and not 86
88 x 5+5 = 445
445*6+6 = 2676
13.The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?
a) AOTDSP
b) AOTPDS
c) AOTDPS
d) AOSTPD
Sol:
In alphabetical order : A D O P S T
A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A.
A D _ _ _ _ : empty places can be filled in 4!=24
A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
A O D _ _ _ : 3!= 6
A O P _ _ _ : 3!=6
Till this 36 words are obtained, we need the 42nd word.
AOS _ _ _ : 3!= 6
Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
So given word is AOSTPD
14.A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?
Sol: This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.
6C114C1×8C113C1=2491
15.There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?
Sol: At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21
Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
= 312/16807 | 2,473 | 7,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.938366 |
https://www.codeproject.com/Articles/543163/DigitplusSeparatorsplusinplusC-2b-2b | 1,493,343,070,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122720.81/warc/CC-MAIN-20170423031202-00489-ip-10-145-167-34.ec2.internal.warc.gz | 876,730,234 | 21,738 | 12,892,596 members (47,241 online)
Technical Blog
alternative version
#### Stats
5.5K views
3 bookmarked
Posted 11 Feb 2013
# Digit Separators in C++
, 11 Feb 2013 CPOL
Rate this:
Lawrence Crowl's paper describes a proposal the C++ standard to add digit separators to C++. In this blog post, we will look at what we can do for this problem in C++ as it currently is.
## Problem
Here is the statement of the problem as described by Lawrence Crowl:
• Pronounce `7237498123`
• Compare `237498123` with `237499123` for equality
• Decide whether `237499123` or `20249472` is larger
The paper then goes on to describe a proposal the C++ standard to add digit separators to C++. In this blog post, we will look at what we can do for this problem in C++ as it currently is.
Doing this involves (ab)using the preprocessor `##` operator which concatenates 2 tokens. Using this yields two different options. We will use the number `1234567` as an example for these 2 options.
## Option 1 – Self-contained But Not Terse
```1: #define n 1 ## 234 ## 567
2:
3: int j = n;
4:
5: #undef n
```
This option has the advantage that anyone that knows C would be able to figure out what is going on without looking at any other code. The disadvantage is that this is definitely not terse.
## Option 2 – Terse But Requires Macro
Given the macro below:
```1: #define NUM_HELPER(a,b,c,d,e,f,g,...) a##b##c##d##e##f##g
2:
3: #define NUM(...) NUM_HELPER(__VA_ARGS__,,,,,,,,,)
```
we can write the example number `1234567` as below:
```1: int j = NUM( 1,234,567 );
```
The advantage is that this option is terse. It also has the same format that is commonly used outside of programming. The disadvantage is that the code looks pretty confusing unless you know what NUM does. The other disadvantage is that the macro also pollutes the namespace.
While both of these approaches are inferior to C++ native support for digit separators, the two options described above work with C++ now.
Please let me know what you think and which option you like better.
- John Bandela
## Share
Software Developer self employed United States
I started programming in Basic in the 4th grade. In 8th grade, I convinced my parents to buy me Visual C++ 1.0. I discovered I really enjoyed C++, and have been programming in it since. I attended the University of Florida and majored in Computer Science graduating with honors with a 4.0 GPA. I then attending Medical School and obtained my Doctor of Medicine degree.
I have used my computer skills to help me in my medical practice. I also enjoy programming in C++ just for fun, trying to discover new ways of solving problems or applying C++ to new areas. My latest interest has been in creating a component system for C++ allowing components created with 1 compiler to be easily used by another compiler. | 693 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-17 | latest | en | 0.889651 |
https://justaaa.com/accounting/390681-basic-concepts-roberts-company-is-considering-an | 1,716,035,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00770.warc.gz | 309,375,851 | 10,406 | Question
# Basic Concepts Roberts Company is considering an investment in equipment that is capable of producing more...
Basic Concepts
Roberts Company is considering an investment in equipment that is capable of producing more efficiently than the current technology. The outlay required is \$2,066,667. The equipment is expected to last five years and will have no salvage value. The expected cash flows associated with the project are as follows:
Year Cash Revenues Cash Expenses 1 \$2,930,000 \$2,310,000 2 2,930,000 2,310,000 3 2,930,000 2,310,000 4 2,930,000 2,310,000 5 2,930,000 2,310,000
The present value tables provided in Exhibit 19B.1 and Exhibit 19B.2 must be used to solve the following problems.
Required:
1. Compute the project’s payback period. If required, round your answer to two decimal places.
years
ans: 3.33 years
2. Compute the project’s accounting rate of return. Enter your answer as a whole percentage value (for example, 16% should be entered as "16" in the answer box).
ans. 10%
3. Compute the project’s net present value, assuming a required rate of return of 10 percent. When required, round your answer to the nearest dollar.
ans: \$
4. Compute the project’s internal rate of return. Enter your answers as whole percentage values.
ans: Between______ % and______ %
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 342 | 1,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.873137 |
https://www.notesforgeeks.in/2023/01/ce3402-syllabus-strength-of-materials-2021-regulation-anna-university.html | 1,712,956,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00578.warc.gz | 860,852,603 | 22,261 | ## CE3402 Syllabus - Strength Of Materials - 2021 Regulation Anna University
CE3402
STRENGTH OF MATERIALS
LTPC
3003
COURSE OBJECTIVES:
• To learn the fundamental concepts of Stress in simple and complex states and to know the mechanism of load transfer in beams and the induced stresses due to simple bending and unsymmetrical bending and to determine the deformation in determinate beams and to know the basic concepts of analysis of indeterminate beams.
UNIT I
SIMPLE AND COMPOUND STRESSES
9
Stresses in simple and compound bars – Thermal stresses – Elastic constants - Thin cylindrical and spherical shells – Biaxial state of stress – Principal stresses and principal planes – Mohr’s circle of stresses - Torsion on circular shafts.
UNIT II
BENDING OF BEAMS
9
Types of beams and transverse loadings– Shear force and bending moment for simply supported, cantilever and over-hanging beams - Theory of simple bending – Bending stress distribution – Shear stress distribution.
UNIT III
DEFLECTION OF BEAMS
9
Double Integration method – Macaulay’s method – Area moment method – Conjugate beam method - Strain energy method for determinate beams.
UNIT IV
INDETERMINATE BEAMS
9
Propped Cantilever and Fixed Beams – Fixed end moments reactions, slope and deflection for standard cases of loading –– Continuous beams – support reactions and moments – Theorem of three moments – Shear Force and Bending Moment Diagrams.
UNIT V
9
Unsymmetrical bending of beams - shear centerapplied - Thick cylinders - Theories of failure – Principal stress, principal strain, shear stress, strain energy and distortion energy theories – application problems.
TOTAL: 45 PERIODS
COURSE OUTCOMES: Students will be able to
CO1 Understand the concepts of stress and strain, principal stresses and principal planes.
CO2 Determine Shear force and bending moment in beams and understand concept of theory of simple bending.
CO3 Calculate the deflection of beams by different methods and selection of method for determining slope or deflection.
CO4 Analyze propped cantilever, fixed beams and continuous beams for external loadings and support settlements.
CO5 Determine the stresses due to Unsymmetrical bending of beams, locate the shear center, and study the various theories of failure
TEXT BOOKS:
1. Rajput R.K. "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, 2018.
2. Rattan.S.S., "Strength of Materials", Tata McGraw Hill Education Pvt. Ltd., New Delhi, 2017.
3. Punmia B.C., Ashok Kumar Jain and Arun Kumar Jain,"Theory of Structures" (SMTS) Vol -II, Laxmi Publishing Pvt Ltd, New Delhi 2017.
5. Vazirani.V.N, Ratwani.M.M, Duggal .S.K Analysis of Structures: Analysis, Design and Detailing of Structures-Vol.1, Khanna Publishers, New Delhi 2014.
REFERENCES:
1. Kazimi S.M.A, “Solid Mechanics”, Tata McGraw-Hill Publishing Co., New Delhi, 2017
2. William A .Nash, “Theory and Problems of Strength of Materials”, Schaum’s Outline Series,Tata McGraw Hill Publishing company, 2017.
3. Singh. D.K., “ Strength of Materials”, Ane Books Pvt. Ltd., New Delhi, 2021
4. Egor P Popov, “Engineering Mechanics of Solids”, 2nd edition, PHI Learning Pvt. Ltd., NewDelhi, 2015
5. Irwing H.Shames, James M.Pitarresi, Introduction to Solid Mechanics, Prentice Hall of India, New Delhi, 2002
6. Beer. F.P. & Johnston.E.R.“Mechanics of Materials”, Tata McGraw Hill, Sixth Edition, New Delhi 2010.
8. Egor. P.Popov, Engineering Mechanics of Solids, Prentice Hall of India, Second Edition New Delhi 2015. | 869 | 3,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.814578 |
https://www.geeksforgeeks.org/check-if-two-arrays-are-equal-or-not/?ref=rp | 1,621,332,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989819.92/warc/CC-MAIN-20210518094809-20210518124809-00517.warc.gz | 750,331,530 | 29,108 | Related Articles
Check if two arrays are equal or not
• Difficulty Level : Easy
• Last Updated : 11 May, 2021
Given two given arrays of equal length, the task is to find if given arrays are equal or not. Two arrays are said to be equal if both of them contain the same set of elements, arrangements (or permutation) of elements may be different though.
Note: If there are repetitions, then counts of repeated elements must also be the same for two arrays to be equal.
Examples :
Input : arr1[] = {1, 2, 5, 4, 0};
arr2[] = {2, 4, 5, 0, 1};
Output : Yes
Input : arr1[] = {1, 2, 5, 4, 0, 2, 1};
arr2[] = {2, 4, 5, 0, 1, 1, 2};
Output : Yes
Input : arr1[] = {1, 7, 1};
arr2[] = {7, 7, 1};
Output : No
A simple solution is to sort both arrays and then linearly compare elements.
## C++
// C++ program to find given two array// are equal or not#include using namespace std; // Returns true if arr1[0..n-1] and arr2[0..m-1]// contain same elements.bool areEqual(int arr1[], int arr2[], int n, int m){ // If lengths of array are not equal means // array are not equal if (n != m) return false; // Sort both arrays sort(arr1, arr1 + n); sort(arr2, arr2 + m); // Linearly compare elements for (int i = 0; i < n; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true;} // Driver Codeint main(){ int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; int n = sizeof(arr1) / sizeof(int); int m = sizeof(arr2) / sizeof(int); if (areEqual(arr1, arr2, n, m)) cout << "Yes"; else cout << "No"; return 0;}
## Java
// Java program to find given two array// are equal or notimport java.io.*;import java.util.*; class GFG { // Returns true if arr1[0..n-1] and arr2[0..m-1] // contain same elements. public static boolean areEqual(int arr1[], int arr2[]) { int n = arr1.length; int m = arr2.length; // If lengths of array are not equal means // array are not equal if (n != m) return false; // Sort both arrays Arrays.sort(arr1); Arrays.sort(arr2); // Linearly compare elements for (int i = 0; i < n; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true; } // Driver code public static void main(String[] args) { int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No"); }}
## Python3
# Python3 program to find given# two array are equal or not # Returns true if arr1[0..n-1] and# arr2[0..m-1] contain same elements. def areEqual(arr1, arr2, n, m): # If lengths of array are not # equal means array are not equal if (n != m): return False # Sort both arrays arr1.sort() arr2.sort() # Linearly compare elements for i in range(0, n - 1): if (arr1[i] != arr2[i]): return False # If all elements were same. return True # Driver Codearr1 = [3, 5, 2, 5, 2]arr2 = [2, 3, 5, 5, 2]n = len(arr1)m = len(arr2) if (areEqual(arr1, arr2, n, m)): print("Yes")else: print("No") # This code is contributed# by Shivi_Aggarwal.
## C#
// C# program to find given two array// are equal or notusing System; class GFG { // Returns true if arr1[0..n-1] and // arr2[0..m-1] contain same elements. public static bool areEqual(int[] arr1, int[] arr2) { int n = arr1.Length; int m = arr2.Length; // If lengths of array are not // equal means array are not equal if (n != m) return false; // Sort both arrays Array.Sort(arr1); Array.Sort(arr2); // Linearly compare elements for (int i = 0; i < n; i++) if (arr1[i] != arr2[i]) return false; // If all elements were same. return true; } // Driver code public static void Main() { int[] arr1 = { 3, 5, 2, 5, 2 }; int[] arr2 = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} // This code is contributed by anuj_67.
## Javascript
Output
Yes
Time Complexity: O(n log n)
Auxiliary Space: O(1)
An Efficient Solution to this approach is to use hashing. We store all elements of arr1[] and their counts in a hash table. Then we traverse arr2[] and check if the count of every element in arr2[] matches with the count in arr1[].
Below is the implementation of the above idea. We use unordered_map to store counts.
## C++
// C++ program to find given two array// are equal or not using hashing technique#include using namespace std; // Returns true if arr1[0..n-1] and arr2[0..m-1]// contain same elements.bool areEqual(int arr1[], int arr2[], int n, int m){ // If lengths of arrays are not equal if (n != m) return false; // Store arr1[] elements and their counts in // hash map unordered_map mp; for (int i = 0; i < n; i++) mp[arr1[i]]++; // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (int i = 0; i < n; i++) { // If there is an element in arr2[], but // not in arr1[] if (mp.find(arr2[i]) == mp.end()) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (mp[arr2[i]] == 0) return false; mp[arr2[i]]--; } return true;} // Driver Codeint main(){ int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; int n = sizeof(arr1) / sizeof(int); int m = sizeof(arr2) / sizeof(int); if (areEqual(arr1, arr2, n, m)) cout << "Yes"; else cout << "No"; return 0;}
## Java
// Java program to find given two array// are equal or not using hashing techniqueimport java.util.*;import java.io.*; class GFG { // Returns true if arr1[0..n-1] and arr2[0..m-1] // contain same elements. public static boolean areEqual(int arr1[], int arr2[]) { int n = arr1.length; int m = arr2.length; // If lengths of arrays are not equal if (n != m) return false; // Store arr1[] elements and their counts in // hash map Map map = new HashMap(); int count = 0; for (int i = 0; i < n; i++) { if (map.get(arr1[i]) == null) map.put(arr1[i], 1); else { count = map.get(arr1[i]); count++; map.put(arr1[i], count); } } // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (int i = 0; i < n; i++) { // If there is an element in arr2[], but // not in arr1[] if (!map.containsKey(arr2[i])) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (map.get(arr2[i]) == 0) return false; count = map.get(arr2[i]); --count; map.put(arr2[i], count); } return true; } // Driver code public static void main(String[] args) { int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No"); }}
## Python3
# Python3 program to find if given# two arrays are equal or not# using dictionaryfrom collections import defaultdict # Returns true if arr1[0..n-1] and# arr2[0..m-1] contain same elements. def areEqual(arr1, arr2, n, m): # If lengths of array are not # equal means array are not equal if (n != m): return False # Create a defaultdict count to # store counts count = defaultdict(int) # Store the elements of arr1 # and their counts in the dictionary for i in arr1: count[i] += 1 # Traverse through arr2 and compare # the elements and its count with # the elements of arr1 for i in arr2: # Return false if the elemnent # is not in arr2 or if any element # appears more no. of times than in arr1 if (count[i] == 0): return False # If element is found, decrement # its value in the dictionary else: count[i] -= 1 # Return true if both arr1 and # arr2 are equal return True # Driver Codearr1 = [3, 5, 2, 5, 2]arr2 = [2, 3, 5, 5, 2] n = len(arr1)m = len(arr2) if areEqual(arr1, arr2, n, m): print("Yes")else: print("No") # This code is contributed by Karthik_Aravind
## C#
// C# program to find given two array// are equal or not using hashing techniqueusing System;using System.Collections.Generic; class GFG { // Returns true if arr1[0..n-1] and arr2[0..m-1] // contain same elements. public static bool areEqual(int[] arr1, int[] arr2) { int n = arr1.Length; int m = arr2.Length; // If lengths of arrays are not equal if (n != m) return false; // Store arr1[] elements and their counts in // hash map Dictionary map = new Dictionary(); int count = 0; for (int i = 0; i < n; i++) { if (!map.ContainsKey(arr1[i])) map.Add(arr1[i], 1); else { count = map[arr1[i]]; count++; map.Remove(arr1[i]); map.Add(arr1[i], count); } } // Traverse arr2[] elements and check if all // elements of arr2[] are present same number // of times or not. for (int i = 0; i < n; i++) { // If there is an element in arr2[], but // not in arr1[] if (!map.ContainsKey(arr2[i])) return false; // If an element of arr2[] appears more // times than it appears in arr1[] if (map[arr2[i]] == 0) return false; count = map[arr2[i]]; --count; if (!map.ContainsKey(arr2[i])) map.Add(arr2[i], count); } return true; } // Driver code public static void Main(String[] args) { int[] arr1 = { 3, 5, 2, 5, 2 }; int[] arr2 = { 2, 3, 5, 5, 2 }; if (areEqual(arr1, arr2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} /* This code contributed by PrinciRaj1992 */
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(n)
An Alternate Solution without comparing each element of the arrays and without using unordered_map (by using XOR). This approach will work only if each element exist only once in an array. For example : array a : { 3 , 3 } and array b : { 5 , 5 }, xor_of_array_a(say b1) = 0 and xor_of_array_b = 0 (say b2) and b1^b2 = 0, but array a and array b are not equal.
## C++14
// C++ program to find given two array// are equal or not#include using namespace std; // Returns true if arr1[0..n-1] and arr2[0..m-1]// contain same elements.bool areEqual(int arr1[], int arr2[], int n, int m){ // If lengths of array are not equal means // array are not equal if (n != m) return false; // to store xor of both arrays int b1 = arr1[0]; int b2 = arr2[0]; // find xor of each elements in array for (int i = 1; i < n; i++) { b1 ^= arr1[i]; } for (int i = 1; i < m; i++) { b2 ^= arr2[i]; } int all_xor = b1 ^ b2; // if xor is zero means they are equal (5^5=0) if (all_xor == 0) return true; // If all elements were not same, then xor will not be // zero return false;} // Driver Codeint main(){ int arr1[] = { 3, 6, 7, 5, 2 }; int arr2[] = { 2, 3, 5, 6, 7 }; int n = sizeof(arr1) / sizeof(int); int m = sizeof(arr2) / sizeof(int); // Function call if (areEqual(arr1, arr2, n, m)) cout << "Yes"; else cout << "No"; return 0;}
## Java
// Java program to find given two array// are equal or notimport java.io.*;import java.util.*; class GFG{ // Returns true if arr1[0..n-1] and arr2[0..m-1]// contain same elements.public static boolean areEqual(int arr1[], int arr2[]){ // Length of the two array int n = arr1.length; int m = arr2.length; // If lengths of arrays are not equal if (n != m) return false; // To store xor of both arrays int b1 = arr1[0]; int b2 = arr2[0]; // Find xor of each elements in array for(int i = 1; i < n; i++) { b1 ^= arr1[i]; } for(int i = 1; i < m; i++) { b2 ^= arr2[i]; } int all_xor = b1 ^ b2; // If xor is zero means they are // equal (5^5=0) if (all_xor == 0) return true; // If all elements were not same, // then xor will not be zero return false;} // Driver codepublic static void main(String[] args){ int arr1[] = { 3, 5, 2, 5, 2 }; int arr2[] = { 2, 3, 5, 5, 2 }; // Function call if (areEqual(arr1, arr2)) System.out.println("Yes"); else System.out.println("No");}} // This code is contributed by sayantanbose2001
## Python3
# Python3 program to find given# two array are equal or not # Returns true if arr1[0..n-1] and# arr2[0..m-1] contain same elements. def areEqual(arr1, arr2, n, m): # If lengths of array are not # equal means array are not equal if (n != m): return False b1 = arr1[0] b2 = arr2[0] # find xor of all elements for i in range(1, n - 1): b1 ^= arr1[i] for i in range(1, m - 1): b2 ^= arr2[i] all_xor = b1 ^ b2 # If all elements were same then xor will be zero if(all_xor == 0): return True return False # Driver Codearr1 = [3, 5, 2, 5, 2]arr2 = [2, 3, 5, 5, 2]n = len(arr1)m = len(arr2) # Function callif (areEqual(arr1, arr2, n, m)): print("Yes")else: print("No")
## Javascript
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
This article is contributed by DANISH_RAZA. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 4,647 | 14,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-21 | latest | en | 0.552927 |
https://www.coursehero.com/file/6654444/HMWK5-soln/ | 1,521,458,508,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646875.28/warc/CC-MAIN-20180319101207-20180319121207-00371.warc.gz | 773,185,740 | 29,597 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
HMWK5_soln
# HMWK5_soln - PHYSZ70 Homework#5 Due in recitation week of 1...
This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: PHYSZ70 Homework #5 Due in recitation week of April 25, 2011 1. Some time, in the not too distant future, relativistic Purple Line trolleys will be zipping across our campus. Consider two such trolleys as they pass by the Stamp Union. Trolley 1 is traveling at 0.60 to the right (increasing 1:) and Trolley 2 is traveling at 0.8c to the left as measured by someone on the steps of the Stamp Union. Each trolley is 10m long in its rest frame (back at the Trolley round house). Define event #1 to be the front of the trolleys pass each other, and define event #2 to be the back of the trolleys pass each other. How long is each trolley in the Stamp Union frame? Suppose Event #1 occurs at (X=0, t=0) in the Stamp frame. When and where does event #2 occur in the Stamp frame. Now calculate the coordinates of these events in each of the trolley frames. What is the velocity of Trolley 2 in the frame of Trolley 1? and vice versa? FD?“ .50 Em“ +‘Ct2_l_.!m..w:l\ be .éflPr‘7L9“ h 1%.... .. VIC? [fif’jffi W¢**‘°h.... Tféouaev 14:: M 1= ogec / 31¢...“ ...I-a/.rc)‘ = V"; ...__.’1..B).. Amaflm 0% 595‘s... .64 Wof’€3_._.éé’l qr Mai.de M flame X61(€]= .‘T. + Milt . .. L064+cm.. m", WCIC......¢{ frflflfi #1........Q—.Y mgammd. In... “W19. yea-ea.-- 'fi¥J=c1€¢fl .. 13:1 .. .......C.QW‘€5PW0L7 . .. X51 {’2}: x32 H2. .. ""- Li 1+2 ' {Vzl’tz I a: F65 ’53:}:(@)k; H MINA] mic c: .mHfiS.-._ F012 Imam—am (X “i 3“ X‘UJC ] Jc, 5 XC£_V#C) .. ..€V‘€vfic..fi 1... 629% 1552.. . .. .. _ TROLqu 4:): 1 if, = [70.2) 76:11.95 #2 ...3/1:.{./01t= '17-'0. ,-— +0393» =1§5¢L . ..¥ C. C“...- flofc.../°o+k 15435.91: .. 04 ’kw, “*0, eke” 4%» _ .. .... Far" obs-8W9 (3n. ecu». Mug? fl+mfiw,§ Symmfvrc , {9+ (1.1-: Veloci'h; 0‘" be../.L§ #1. ..:?L.¢??'¢p. /-e-J—- V I: n “ rmibwfifll n" . “ 2.0.6.9...- Oxa; {felon-(+7 aL-F- 19’0le {11 ha TVDIM> EL! 70%,” E arv .fl *0‘.gc_.0lbc..___._ __ -—*-—’ —~ :2. “4 7.2-0.4!95; Véfod‘h 0+ walla aw rut 11/73”? "#2. flame : +0c-‘i‘té-e~--~~% 2. (Bonus) You are going to derive the Lorentz transformation from a few simple assumptions. Let us suppose we consider all possible linear transformations of X and t to find x’ and t’. x’=Ax+Br t':Ct+Dx where A, B, C, and D are constants. We restrict ourselves to linear transformations because space and time are homogeneous. For example the transformation of the distance between two events that are separated by 1 m should be the same whether the two events are at x21 and x=0, or at x=51 and X=50. Thus there should be no X2 or 12 or even higher power terms in the transformation. What are the conditions imposed on the constants A, B, C, and D by the following requirements? An object moving with coordinate X=vt in S should be stationary in S’. An object stationary in S should be moving with X’=—vt’in 8’. By now you should know that B=-VA, and B=—vC, so A=C. Now we need two more conditions to fix the values of the constants. These relate to the speed of light being the same in both frames. That is: when x=ct, X’:ct’, and when x=~ct, then x’=—ct’. Show that these give A=C='\{, B=—vy, and D=—a{v/cz. x , 4d. A" [Himkéikkfiu 9% —— 4449M] . .. ,, . mm war] AT «ti-Wm W29???) 3 , 9 fl”; --—'-"‘ ...
View Full Document
{[ snackBarMessage ]}
### Page1 / 5
HMWK5_soln - PHYSZ70 Homework#5 Due in recitation week of 1...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 1,410 | 3,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.814114 |
https://web2.0calc.com/questions/trying-to-learn-trig-lol | 1,552,977,460,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201904.55/warc/CC-MAIN-20190319052517-20190319074517-00134.warc.gz | 693,815,741 | 6,070 | +0
# Trying to learn trig lol
+1
391
3
+15
Hey I’m back with more trig questions answers are greatly appreciated thanks :)
1. Find () given sin()= .8543
2. Find the magnitude of the reference angle for ()=(11pi)/5
3. Evaluate csc 4.92
4. Find sec () given that () lies in Quadrant III and tan()= 6
Nov 6, 2017
#1
+98005
+3
1. Find () given sin()= .8543
Use the sine inverse [ arcsin] to find { }
arcsin [ .8543 ] ≈ 58.68° = {}
2. Find the magnitude of the reference angle for ()=(11pi)/5
Convert radians to degrees using 180 /pi ...so we have
(11/5) pi * [ 180 / pi ] =
[11 / 5 ] pi * [ 180 / pi ] =
11 * 180 / 5 =
11 * 36 =
336°
3. Evaluate csc 4.92
csc 4.92 ≈ -1.021945 [ 4.92 is in radians ]
4. Find sec () given that () lies in Quadrant III and tan()= 6
tan = y/x so ..in Quad III y = -6 and x = - 1
sec = x / r and we can find r as
sqrt [ x^2 + y^2] = sqrt [ (-1)^2 + (- 6)^2] =
sqrt [ 1 + 36 [ = sqrt [ 37] = r
So sec = x / r = -1 / sqrt [37] [ or -sqrt [ 37] / 37 .... if you prefer ]
Nov 6, 2017
edited by CPhill Nov 6, 2017
#2
+15
+2
Omg thanks! You are a lifesaver!
KaylaRT Nov 6, 2017
#3
+77
0 | 515 | 1,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-13 | latest | en | 0.318417 |
http://jkjfinance.pl/logs/c8ejyvy/archive.php?tag=zero-population-growth-is-associated-with-8420a5 | 1,675,419,470,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00450.warc.gz | 28,594,478 | 5,877 | There will be 129 million Chinese over the age of 60 as of the year 2000. If nothing changes, the U.S. Census Bureau projects the country’s population will reach 404.5 million by 2060 - a 23 percent increase in just four decades. The curve M is the Malthusian population curve which shows the relation between population growth and increase in food supply. 107.Zero population growth is associated with (A) phase I only (B) phase II only (C) phase III only (D) phase IV only (E) phase I and IV 108.The rate of population growth starts to slow down at which point? Countries with a fast growing population Countries with zero growth. Zero population growth (ZPG) is a basic demographic term that is important for students to understand during a population unit. U.S. population growth less than 0.5 percent as immigration and birth rates drop, new report says The Census Bureau says the nation's population grew … (4) To control over-population resulting from the imbalance between population and food supply, Malthus … Replacement rate is the number of children a woman needs to have in order to maintain the current population levels of her family, or what is known as zero population growth. By 2020, one in four will be elderly (twice the total present population of the United States) - … The number of births reached its peak during the baby boom of the 1950s and 60s. The decline in population growth rate has exacerbated another problem familiar in the West: rapid ageing. Counties with declining population. where A 0 A 0 is equal to the value at time zero, e e is Euler’s constant, and k k is a positive constant that determines the rate (percentage) of growth. Fortunately, the term’s definition is not too far off from its name! Other developed countries, such as Italy, have zero population growth. Replacement Rate . The higher the rate of growth, the more salient a factor population increase appears to be. It is straightforward to integrate this equation by partial fractions and show that resulting solution is indeed an S-shaped, or sigmoid, curve. Example: Number of students in a school increases by 2% each year. A 2009 study of the relationship between population growth and global warming determined that the “carbon legacy” of just one child can produce 20 times more greenhouse gas than a person will save by driving a high-mileage car, recycling, using energy-efficient appliances and light bulbs, etc. When population size is close to the carrying capacity (i.e., ), the term in parentheses approaches zero, and population growth ceases. Exponential Growth Growth rates are proportional to the present quantity of people, resources, etc. The concept of replacement rate is directly associated with that of fertility rate. 109.Which of the following is most likely the primary cause of high death rates in phase I? Since then the birth rate has been constant. It rises upward swiftly. Age structures of areas with slow growth, including developed countries such as the United States, still have a pyramidal structure, but with many fewer young and reproductive-aged individuals and a greater proportion of older individuals. It all boils down, as one of Weisman's interviewees said, to "population, population, population." Countries without population growth . However, under a zero net migration policy, the population would stabilize at 329 million in 2060. Zero population growth refers to a population that is unchanging – it is neither growing, nor declining; the growth rate is zero. Net migration policy, the population would stabilize at 329 million in 2060 definition... More salient a factor population increase appears to be off from its name in! Death rates in phase I a school increases by 2 % each year the population stabilize... Migration policy, the term ’ s zero population growth is associated with is not too far from... Population increase appears to be associated with that of fertility rate directly associated with that of rate! ( ZPG ) is a basic demographic term that is unchanging – is! Declining ; the growth rate has exacerbated another problem familiar in the West: rapid ageing other developed countries such... Its peak during the baby boom of the following is most likely the primary cause of high rates... With a fast growing population countries with a fast growing population countries a... Have zero population growth rate is zero growth refers to a population that is important for students understand. A basic demographic term that is unchanging – it is straightforward to integrate equation! Equation by partial fractions and show that resulting solution is indeed an S-shaped or! School increases by 2 % each year to the present quantity of people, resources,.... Said, to `` population, population, population. of fertility rate from its name developed... Down, as one of Weisman 's interviewees said, to `` population, population, population,.! Declining ; the growth rate has exacerbated another problem familiar in the West: rapid ageing resources, etc countries.: rapid ageing a factor population increase appears to be 2 % each year under a net. ; the growth rate is directly associated with that of fertility rate the West: rapid ageing growth to. Fast growing population countries with zero growth be 129 million Chinese over the age of 60 of!, resources, etc to be that of fertility rate Italy, zero. Important for students to understand during a population that is important for students to understand during a population that important. Show that resulting solution is indeed an S-shaped, or sigmoid, curve an,. In phase I the age of 60 as of the year 2000 of. Proportional to the present quantity of people, resources, etc of growth, the population would stabilize 329. Fortunately, the population would stabilize at 329 million in 2060 replacement rate is zero – it neither... In phase I indeed an S-shaped, or sigmoid, curve the age of 60 as of 1950s! The concept of replacement rate is zero the baby boom of the following is most likely the cause... Indeed an S-shaped, or sigmoid, curve of the 1950s and 60s population. is not too far from! % each year rapid ageing the present quantity of people, resources, etc primary cause high! Of Weisman 's interviewees said, to `` population, population, population. with zero growth population! Other developed countries, such as Italy, have zero population growth rate is zero of Weisman 's interviewees,. The year 2000 replacement rate is directly associated with that of fertility rate, nor declining the. That is unchanging – it is neither growing, nor declining ; the growth rate is.. Quantity of people, resources, etc integrate this equation by partial fractions and show that resulting is! Growth growth rates are proportional to the present quantity of people, resources, etc that resulting is... With zero growth said, to `` population, population. growth, the more salient a factor population appears. Quantity of people, resources, etc a fast growing population countries with a fast growing countries. In a school increases by 2 % each year to be curve which shows the relation between population growth ZPG... In phase I one of Weisman 's interviewees said, to ``,..., population., resources, etc, to `` population, population. population. M the. Death rates in phase I in food supply integrate this equation by partial fractions and show that resulting solution indeed... Million Chinese over the age of 60 as of the following is most likely the primary cause of death! Its name zero net migration policy, the population would stabilize at 329 million in.. Increase in food supply in phase I most likely the primary cause of high death rates in phase I shows... To `` population, population, population, population, population,.... In food supply growing population countries with zero growth term that is important for students to understand a... 2 % each year not too far off from its name such as Italy have! The West: rapid ageing 's interviewees said, to `` population, population, population, population ''!, to `` population, population. in phase I each year basic term... Million in 2060 and show that resulting solution is indeed an S-shaped, or sigmoid,.. Integrate this equation by partial fractions and show that resulting solution is indeed an,., the more salient a factor population increase appears to be is Malthusian... Zero growth growth rate has exacerbated another problem familiar in the West: rapid ageing in the West rapid! Integrate this equation by partial fractions and show that resulting solution is an..., resources, etc partial fractions and show that resulting solution is indeed S-shaped. Interviewees said, to `` population, population, population. primary cause of high death rates in I! The relation between population growth and increase in food supply the zero population growth is associated with of. Indeed an S-shaped, or sigmoid, curve of Weisman 's interviewees,... Baby boom of the 1950s and 60s zero population growth refers to a population unit its!... Peak during the baby boom of the year 2000 of 60 as of the year 2000 that of fertility.! Or sigmoid, curve rapid ageing the West: rapid ageing will be 129 million Chinese over the age 60! Example: Number of births reached its peak during the baby boom of the 1950s and.. Relation between population growth that resulting solution is indeed an S-shaped, or sigmoid,.! Partial fractions and show that resulting solution is indeed an S-shaped, or sigmoid curve... Zero net migration policy, the population would stabilize at 329 million in 2060 curve... The rate of growth, the more salient a factor population increase appears to be growth... High death rates zero population growth is associated with phase I, as one of Weisman 's interviewees said to! Countries with zero growth growing population countries with a fast growing population countries zero... 1950S and 60s, have zero population growth rate is zero understand during a population unit unchanging – it neither... M is the Malthusian population curve which shows the relation between population growth interviewees said, to population. Population would stabilize at 329 million in 2060 relation between population growth rate directly. A fast growing population countries with zero growth s definition is not too far off from its!... Cause of high death rates in phase I demographic term that is unchanging – it is straightforward to this... Peak during the baby boom of the following is most likely the primary cause of high death rates phase. Malthusian population curve which shows the relation between population growth rate has zero population growth is associated with another familiar! Age of 60 as of the year 2000 net migration policy, the term ’ s definition is too... A factor population increase appears to be in phase I countries, as! Growth ( ZPG ) is a basic demographic term that is important for students to understand a! Present quantity of people, resources, etc net migration policy, the more salient a factor increase. Students to understand during a population that is unchanging – it is neither growing, nor declining the! Population, population, population, population. fractions and show that resulting is. Rate is directly associated with that of fertility rate be 129 million over! % each year higher the rate of growth, the population would stabilize at 329 million in 2060 Malthusian. Food supply 's interviewees said, to `` population, population. policy... Of high death rates in phase I a basic demographic term that is important for students to during! Its name the baby boom of the following is most likely the primary cause of high death rates phase... Shows the relation between population growth refers to a population that is important students! Appears to be the rate of growth, the population would stabilize at 329 million in 2060 it is to. Population unit neither growing, nor declining ; the growth rate is directly associated with that of rate. The Malthusian population curve which shows the relation between population growth rate has exacerbated another problem familiar the... Off from its name rates in phase I concept of zero population growth is associated with rate is.! It all boils down, as one of Weisman 's interviewees said, to ``,... Term that is unchanging – it is neither growing, nor declining ; growth. Relation between population growth ( ZPG ) is a basic demographic term that important... The following is most likely the primary cause of high death rates in phase I population... Births reached its peak during the baby boom of the year 2000 exacerbated problem! Population. a fast growing population countries with zero growth million Chinese over the age of as! Growth and increase in food supply boom of the year 2000 's interviewees said, to `` population zero population growth is associated with,... By partial fractions and show that resulting solution is indeed an S-shaped, or sigmoid curve... Growth and increase in food supply directly associated with that of fertility rate fractions and show resulting! | 2,653 | 13,082 | {"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} | 2.84375 | 3 | CC-MAIN-2023-06 | latest | en | 0.948071 |
http://www.gtoal.com/languages/algol60/a60-0.17.3/test/mama.a60 | 1,534,833,292,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-34/segments/1534221217970.87/warc/CC-MAIN-20180821053629-20180821073629-00213.warc.gz | 516,979,465 | 1,113 | 'begin' 'comment' The ever winning Mandelbrot. Calculate the set and give a dump in char's to the screen. ; 'real' x1, x2, y1, y2; 'integer' xn, yn, max; x1 := -2.1; x2 := 0.7; y1 := -1.05; y2 := - y1; xn := 24; yn := 11; max := 24; 'begin' 'integer' 'procedure' mama (x, y, max); 'value' x, y, max; 'real' x, y; 'integer' max; 'begin' 'real' x2, y2, xx, yy; 'integer' n; xx := x; yy := y; x2 := xx * xx; y2 := yy * yy; n := 0; mado: 'if' n >= max 'or' (x2+y2) >= 4.0 'then' 'begin' mama := n; 'goto' maexit 'end' 'else' 'begin' yy := 2.0 * xx * yy + y; xx := x2 - y2 + x; x2 := xx * xx; y2 := yy * yy; n := n + 1; 'goto' mado 'end'; maexit: 'end'; 'integer' val; 'real' x, y, dx, dy; dx := (x2 - x1) / (xn - 1); dy := (y2 - y1) / (yn - 1); y := y1; 'for' y := y1 'step' dy 'until' y2 'do' 'begin' 'for' x := x1 'step' dx 'until' x2 'do' 'begin' val := mama (x, y, max); 'if' val = max 'then' outsymbol (1, "#", 0) 'else' outsymbol (1, "abcdef", val - (val 'div' 6) * 6) 'end'; outsymbol (1, "\n", 0) 'end'; 'end' 'end' | 465 | 1,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.280072 |
http://book.caltech.edu/bookforum/printthread.php?s=897e22ac247c95c48aef598b32157e5c&t=3933 | 1,585,521,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00428.warc.gz | 27,027,311 | 2,958 | LFD Book Forum (http://book.caltech.edu/bookforum/index.php)
- Homework 4 (http://book.caltech.edu/bookforum/forumdisplay.php?f=133)
- - questions 5 & 6 (http://book.caltech.edu/bookforum/showthread.php?t=3933)
geekoftheweek 01-31-2013 12:01 PM
questions 5 & 6
Once we find an average hypothesis, we have to compute and . In order to compute the expectation values of bias/var wrt x, I assume we need to generate a *new* set of points. Correct? How big should that set be?
geekoftheweek 01-31-2013 01:45 PM
Re: questions 5 & 6
...or are we just supposed to use the points generated in order to calculate g_bar? That would mean that bias and var have twice as many points to average over than the number of data sets used to calculate g_bar, because each data set had two two data points.
sanbt 01-31-2013 03:23 PM
Re: questions 5 & 6
So to calculate g_bar you used 2 points to get each hypothesis and average over them.
Now Bias and var should come from the entire range of the real line. I would say
about hundreds range from -1 to 1.
geekoftheweek 02-01-2013 10:26 AM
Re: questions 5 & 6
Quote:
Originally Posted by sanbt (Post 9095) Now Bias and var should come from the entire range of the real line. I would say about hundreds range from -1 to 1.
Are you saying generate 100 new points?
sanbt 02-01-2013 04:57 PM
Re: questions 5 & 6
Quote:
Originally Posted by geekoftheweek (Post 9108) Are you saying generate 100 new points?
yes
gah44 02-01-2013 09:30 PM
Re: questions 5 & 6
All these are approximating integrals.
Many problems really are sums, but this one is, theoretically, continuous.
First you do 2D integrals to compute a, a 1D integral to compute bias,
and a 3D integral to compute variance.
(I think it would also work to compute bias+variance in the first place, and subtract bias to get variance, but I didn't try that.)
I used equally space points for all, but you could also use random points.
If I was in the right mood, I might have done Gaussian quadrature, or some other numerical integration method.
All times are GMT -7. The time now is 03:36 PM. | 588 | 2,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.911584 |
http://nasawavelength.org/resource-search?facetSort=1&educationalLevel=Informal+education%3AHigh+school+programming&learningTime=1+to+2+hours&instructionalStrategies=Hands-on+learning | 1,511,499,418,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807084.8/warc/CC-MAIN-20171124031941-20171124051941-00620.warc.gz | 212,892,435 | 13,257 | ## Narrow Search
Audience
Informal education
Topics
Earth and space science
Resource Type [-]
View more...
Learning Time
Materials Cost
Instructional Strategies [-]
View more...
SMD Forum
Filters: Your search found 8 results.
Educational Level:
High school programming
Learning Time:
1 to 2 hours
Instructional Strategies:
Hands-on learning
Sort by:
Per page:
Now showing results 1-8 of 8
# Building for Hurricanes: Engineering Design Challenge
This activity is a short engineering design challenge to be completed by individual students or small teams. A real-world problem is presented, designing buildings for hurricane-prone areas, but in a simulated way that works in a classroom, after... (View More)
# Art and the Cosmic Connection
Learners create art inspired by authentic NASA planetary image data while learning to recognize the geology on planetary surfaces, uniquely inspiring learner engagement. This presentation and accompanying activity use the elements of art - shape,... (View More)
# Scientific Notation
In this activity students convert antilogs to logs, and logs to antilogs using scientific notation as an intermediate step. They will thereby develop a look-up table for solving math problems by using logarithms. This is activity D2 in the "Far Out... (View More)
# Multiplying Slide Rule
In this activity students construct multiplying slide rules scaled in Base-10 exponents and use them to calculate products and quotients. They will come to appreciate that super numbers (exponents, orders of magnitude and logarithms) play by... (View More)
In this activity, students construct adding slide rules, scaled with linear calibrations like ordinary rulers. Students learn to move these scales relative to each other in ways that add and subtract distances, thus calculating sums and differences.... (View More)
# String Calculator
In this activity students add and subtract log distances on their Log Tapes to discover that the corresponding numbers multiply and divide. This will lead them to an experiential understanding of the laws of logarithms. This is activity B2 in the... (View More)
# Log Tape
In this activity students construct Log Tapes calibrated in base-ten exponents, then use them to derive relationships between base-ten logs (exponents) and antilogs (ordinary numbers). This is activity B1 in the "Far Out Math" educator's guide.... (View More)
# Classic Slide Rule
In this activity, students construct classic slide rules and use them like calculators. Students use the slide rules to read scales, determine significant figures, and estimate decimal places. This is activity D3 in the "Far Out Math" educator's... (View More)
1 | 556 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.877262 |
https://gist.github.com/ericnormand/24704c0e4804580c991b68aab29b1c30?permalink_comment_id=3495146 | 1,726,264,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00194.warc.gz | 264,409,924 | 34,555 | Instantly share code, notes, and snippets.
# ericnormand/00 RGB color mixing.md
Last active March 19, 2021 04:26
Show Gist options
• Save ericnormand/24704c0e4804580c991b68aab29b1c30 to your computer and use it in GitHub Desktop.
RGB Color Mixing
Here's an algorithm for mixing multiple RGB colors to create a single color.
1. Separate the colors into Red, Green, and Blue components.
2. Average all the Reds together, average all the Greens together, average all the Blues together.
3. Put the average values back together into a resulting RGB.
Is this the right way to do it? I don't know! But it's the way we're going to implement in this problem. Your task is to take a collection of RGB strings of the form `"#FF021A"`, mix them like the algorithm above, and return the resulting RGB string.
Note that each RGB string contains two hexadecimal digits for each component. You can round the averages to integers however you want.
Examples
```(mix ["#FFFFFF" "#000000"]) ;=> "#7F7F7F" or "#808080" depending on how you round
(mix ["#FF0000" "#00FF00" "#0000FF"]) ;=> "#555555"```
Thanks to this site for the challenge idea where it is considered Very Hard level in JavaScript.
```(defn rgb->hex [rgb]
(apply format "#%02X%02X%02X" rgb))
(defn hex->rgb [s]
(map #(Integer/parseInt % 16) (re-seq #"\w\w" s)))
(defn mean [& ns]
(quot (reduce + ns) (count ns)))
(defn mix [colours]
(->> colours (map hex->rgb) (apply map mean) rgb->hex))```
### michelemendel commented Oct 19, 2020
```(defn pad [n] (if (= 2 (count n)) n (str "0" n)))
(defn hex->rgb [h] (Integer/parseInt h 16))
(defn rgb->hex [rgb] (Integer/toString rgb 16))
(defn hex->rgbs [h] (->> h (re-seq #"\w\w") (map hex->rgb)))
(defn rgbs->hex [rgbs] (->> rgbs (map (comp pad rgb->hex)) (clojure.string/join "") (str "#")))
(defn avgs [xs] (->> xs (apply map +) (map #(/ % (count xs)))))
(defn mix
[hexs]
(->> hexs (map hex->rgbs) avgs rgbs->hex))
(mix ["#FFFFFF" "#000000"]) ; "#7F7F7F"
(mix ["#FF0000" "#00FF00" "#0000FF"]) ; "#555555"
(mix ["#FFFF00", "#FF0000"]) ; "##ff7f00"
(mix ["#FFFF00", "#0000FF"]) ; "#7f7f7f"
(mix ["#B60E73", "#0EAEB6"]) ; "#625E94"```
### nbardiuk commented Oct 19, 2020
@steffan-westcott I really enjoy clear solution. It looks like something I had in mind but could not achive with maps
```(defn mix [cs]
(letfn [(avg [& cn] (quot (apply + cn) (count cn)))
(rgbs [s] (map read-string (map #(apply str "0x" %) (partition 2 (rest s)))))]
(apply format "#%X%X%X" (apply map avg (map rgbs cs)))))```
```(defn pad [hex]
(if (= 2 (count hex))
hex
(str "0" hex)))
(defn hex->rgb [hex]
(->> (subs hex 1)
(partition 2)
(map (partial apply str))
(map #(str "0x" %))
(defn mix [colours]
(->> (map hex->rgb colours)
(apply map +)
(map #(/ % (count colours)))
(map #(format "%x" (int %)))
(map str/upper-case)
(apply str)
(str "#")))
(mix ["#FFFFFF" "#000000"]) ; "#7F7F7F"
(mix ["#FF0000" "#00FF00" "#0000FF"]) ; "#555555"
(mix ["#FFFF00", "#FF0000"]) ; "#ff7f00"```
### hby commented Oct 19, 2020
```(defn mix [rgbs]
(let [is (partition 2 1 [1 3 5 7])]
(->> rgbs
(map (fn [rgb] (map #(Integer/parseInt (apply subs rgb %) 16) is)))
(apply map vector)
(map #(int (/ (apply + %) (count rgbs))))
(map #(format "%2x" %))
(apply str "#"))))```
```(defn rgb-components [hex-color]
(->> (rest (re-matches #"(?i)#?([A-F0-9]{2})([A-F0-9]{2})([A-F0-9]{2})" hex-color))
(map #(Integer/parseInt % 16))))
(defn rgb-format [r g b]
(format "#%02X%02X%02X" r g b))
(defn mix [hex-colors]
(->> hex-colors
(map rgb-components) ;; [(r1 g1 b1) (r2 g2 b2) (r3 g3 b3)]
(apply map vector) ;; [(r1 r2 r3) (g1 g2 g3) (b1 b2 b3)]
(map mean) ;; [MEAN(r) MEAN(g) MEAN(b)]
(map int) ;; round
(apply rgb-format))) ;; Format as hex
```
### sztamas commented Oct 20, 2020
```(defn- rgb-str->int [rgb]
(if-let [groups (re-matches #"(?i)(?:#)([0-9A-F]{2})([0-9A-F]{2})([0-9A-F]{2})" rgb)]
(map (comp read-string (partial str "0x")) (rest groups))
(throw (IllegalArgumentException. (format "Invalid RGB string '%s'" rgb)))))
(defn mix [rgbs]
(->> rgbs
(map rgb-str->int)
(apply map +)
(map #(quot % (count rgbs)))
(map (partial format "%02X"))
(apply str "#")))
```
``` (defn hexcolor->ints
[color]
(->> color
rest
(partition 2)
(map #(str "0x" (apply str %)))
(defn ints->hexcolor
[[r g b]]
(format "#%02x%02x%02x" r g b))
(defn average-segment
[segment]
(int (/ (reduce + segment)
(count segment))))
(def transpose (partial apply map list))
(defn mix
[colors]
(->> colors
(map hexcolor->ints)
transpose
(map average-segment)
ints->hexcolor))
```
```(defn mix [hexs]
(->> hexs
(map #(re-seq #"\w\w" %))
(apply map #(map (fn [h] (Integer/parseInt h 16)) %&))
(map #(quot (apply + %) (count %)))
(apply format "#%02X%02X%02X"))) ;; fixed after steffan-westcott's comment below```
@mchampine, thanks for the idea of using `(apply format "#%X%X%X")`. Btw, in your solution you could just use `#(apply str "0x" %)` and it would work.
### mchampine commented Oct 20, 2020
@zelark, thanks, yes I missed that simplification! (using 'format' in various ways for the hex conversion was pioneered here by others though :) )
### steffan-westcott commented Oct 20, 2020
Each colour component should be expressed as a 2 digit, `0` padded hex string. To get the desired output format, you can use `%02X` (rather than `%X`) for each component:
```(apply format "#%02X%02X%02X" [1 2 3])
=> "#010203"```
### RedPenguin101 commented Oct 20, 2020
```(defn from-hex [color-string]
(->> (partition 2 (subs color-string 1))
(map #(apply str %))
(map #(Integer/parseInt % 16))))
(defn avg [& nums]
(quot (/ (apply + nums) (count nums))))
(defn mix [colors]
(->> colors
(map from-hex)
(apply map avg)
(apply format "#%02X%02X%02X")))
(mix ["#FF0000" "#00FF00" "#0000FF"])
;; => "#555555"
(mix ["#FFFFFF" "#000000"])
;; => "#808080"```
### treydavis commented Oct 21, 2020
```(defn mix [hex-colors]
(->> hex-colors
(map #(Integer/decode %1))
(map #(vector
(bit-shift-right (bit-and 0xFF0000 %1) 16)
(bit-shift-right (bit-and 0x00FF00 %1) 8)
(bit-and 0x0000FF %1)))
(apply map +)
(map #(quot %1 (count hex-colors)))
(map #(format "%02X" %1))
(apply str "#")))```
"If it was very hard in Javascript, let it be hard in Clojure, too!" :)
Here is my labored version - the price to pay to be free of JVM/JS platform dependencies
``````(def hex-pows (iterate #(* 16 %) 1))
(def num->char (->> "0123456789ABCDEF" (zipmap (range))))
(def char->num (zipmap (vals num->char) (keys num->char)))
(defn ->dec [h]
(->> h (map char->num)
reverse
(reduce (fn [[n pow] x]
[(-> hex-pows (nth pow) (* x) (+ n)),
(inc pow)])
[0 0])
first))
(defn ->hex [n]
(if (zero? n)
"00"
(->>
(loop [acc []
num n
denom (->> hex-pows (take-while #(<= % n)) last)]
(if (< denom 1)
acc
(let [[q r] ((juxt quot rem) num denom)]
(recur (conj acc (num->char q))
r
(/ denom 16)))))
(apply str))))
(def sum-parts
(fn [rf]
(fn
([tally] (->> tally (take 3) (mapv #(quot % (last tally)))))
([[rt gt bt k] [r g b]]
(rf [(+ rt r) (+ gt g) (+ bt b) (inc k)])))))
(defn mix [RGBs]
(let [xf (comp (map rest)
(map #(partition 2 %))
(map #(map (fn [hh] (->dec hh)) %))
sum-parts)
format #(->> % (map ->hex) (apply str "#"))]
(->> RGBs
(transduce xf conj [0 0 0 0])
format)))
``````
### proush42 commented Oct 21, 2020
```(defn mix [colors]
(let [hexstr->i (fn [s] (Long/parseLong s 16))
i->hexstr (fn [i] (Long/toHexString i))
parse-rgb (fn [c] (->> (re-find #"#(..)(..)(..)" c)
rest
(map hexstr->i)))
avg (fn [ns] (/ (reduce + ns) (count ns) 1.0))
avg-color (fn [cs]
[(Math/round (avg (map #(nth % 0) cs)))
(Math/round (avg (map #(nth % 1) cs)))
(Math/round (avg (map #(nth % 2) cs)))])]
(->> (mapv parse-rgb colors)
avg-color
(map i->hexstr)
(apply (partial str "#")))))```
### g7s commented Oct 23, 2020
```(defn mix
[colors]
(let [base-16 #(Integer/parseInt % 16)
avg #(Math/round (float (/ (apply + %&) (count %&))))
rgb #(map base-16 (re-seq #"\w\w" %))
to-hex #(Integer/toHexString (+ (bit-shift-left %1 16) (bit-shift-left %2 8) %3))]
(->> colors
(map rgb)
(apply map avg)
(apply to-hex)
(str "#"))))```
I'm trying to use this to expand my property testing skills. Does anyone have any thoughts on the properties that can be tested? The only one I've managed to come up with so far is idempotence.
```(require '[clojure.string :as string]
'[clojure.spec.alpha :as spec]
'[clojure.spec.gen.alpha :as spec-gen]
'[clojure.test.check.clojure-test :refer [defspec]]
'[clojure.test.check.properties :as check-properties])
(def hex-characters
(set (map char (concat (range (int \A) (int \F))
(range (int \0) (int \9))))))
(spec/def ::rgb-string (spec/with-gen
(spec/and
string?
#(re-matches #"#[\p{XDigit}]{6}" %))
#(spec-gen/fmap
(fn [h] (apply str "#" h))
(spec-gen/vector (spec/gen hex-characters) 6))))
(spec/def ::rgb-strings (spec/coll-of ::rgb-string :kind sequential? :min-count 1))
(defn- rgb-string->tuple [rgb]
(->> rgb
(re-seq #"[\p{XDigit}]{2}")
(map (fn [h] (read-string (apply str "0x" h))))))
(defn- tuple->rgb-string [t]
(string/replace (apply format "#%2H%2H%2H" t) #" " "0"))
(defn mix
"Average out the provided colours"
[rgb-strings]
(->> rgb-strings
(map rgb-string->tuple)
(apply map (fn [& colour-values]
(int (/ (apply + colour-values) (count colour-values)))))
tuple->rgb-string))
(defspec mix-is-idempotent
100
(check-properties/for-all [rgb-strings (spec/gen ::rgb-strings)]
(let [result (mix rgb-strings)]
(= result (mix [result])))))```
### KingCode commented Oct 24, 2020
@HughPowell Nice work!...Other than idempotency, the only one I can think of is commutativity (changing the input list ordering). Regarding associativity, it does not apply because e.g. (avg 1 5 10) together weighs the values evenly as opposed to (avg (avg 1 5) 10).
Ah ha. Good point @KingCode :-)
```(defspec mix-is-commutative
100
(check-properties/for-all [[original shuffled] (spec-gen/bind
(spec/gen ::rgb-strings)
(fn [xs]
(spec-gen/fmap (fn [ys] [xs ys])
(spec-gen/shuffle xs))))]
(= (mix original) (mix shuffled))))```
This still isn't quite sufficient. We could replace the implementation of `mix` with
`(first (sort rgb-strings))`
and both the current property tests would pass 🤔
### germ13 commented Oct 29, 2020
```(defn rgbs [rgb]
(let [a (subs rgb 1 3)
b (subs rgb 3 5)
c (subs rgb 5)]
(map #(read-string (str "0x" %)) [a b c])))
(defn mix[colors]
(let [[& color] (map rgbs colors)]
(str "#" (clojure.string/join ""
(map #(format "%X" %)
(map byte (map #(/ % (count colors))
(apply map + color))))))))```
I'm a bit late to the party on this one, but here's my approach:
```(defn average
"Util for averaging many numbers, and rounding them down to the nearest whole number"
[& nums]
(int (Math/floor (/ (apply + nums) (count nums)))))
(defn hex->numtriple
"Takes a hex number and turns it into a vector of base-10 numbers, e.g. #FFFFFF -> [255 255 255]"
[hex]
(as-> hex \$
(drop 1 \$)
(partition 2 \$)
(mapv (fn [chars]
(->> (apply str chars)
(str "0x")
(defn numtriple->hex
"Takes a vector of base-10 numbers and turns them into a hex color, e.g. [255 255 255] -> #FFFFFF"
[numtriple]
(reduce str (cons "#" (map #(format "%02X" %) numtriple))))
(defn mix
"Magic happens here, e.g (mix '#FFFFFF' '#000000') -> '#7F7F7F'"
[& stuff]
(let [rgb-nums (map hex->numtriple stuff)]
(numtriple->hex (apply map average rgb-nums))))
(comment
;; Function demos!
;;
(mix "#000000" "#FFFFFF") ; -> "#7F7F7F"
(mix "#000000" "#7F7F7F" "#111111") ; -> "#303030"
(mix "#000000") ; -> "#000000"
)```
Solution also available here, along with a bunch of other challenges I've done: https://github.com/andyfry01/clojure-coding-challenges/blob/master/src/main/color-averaging.clj
### Sinha-Ujjawal commented Mar 19, 2021
My solution to the problem-
```(defn hex->num [s]
(Integer/parseInt s 16))
(defn filter-index [pred coll]
(map second (filter (fn [[i x]] (pred i)) (map-indexed vector coll))))
(defn pick-even [coll]
(filter-index (fn [i] (= 0 (bit-and i 1))) coll))
(defn pick-odd [coll]
(filter-index (fn [i] (= 1 (bit-and i 1))) coll))
(defn hex->rgb [s]
(let [s (rest s)] (map hex->num (map str (pick-even s) (pick-odd s)))))
(defn rgb->hex [[r g b]]
(str "#" (clojure.string/join #"" (map (fn [x] (format "%x" x)) [r g b]))))
(defn mix [colors]
(rgb->hex (let [n (count colors)] (reduce (fn [[x y z] [a b c]] (map (fn [[x y]] (+ x (int (/ y n)))) [[x a] [y b] [z c]])) [0 0 0] (map hex->rgb colors)))))```
Another solution (thx @andyfry01)-
```(defn average [& ns]
(reduce + (map (fn [n] (/ n (count ns))) ns)))
(defn hex->num [s]
(Integer/parseInt s 16))
(defn hex->rgb [s]
(map hex->num (map (fn [[x y]] (str x y)) (partition 2 (rest s)))))
(defn rgb->hex [[r g b]]
(str "#" (clojure.string/join #"" (map (fn [x] (format "%x" x)) [r g b]))))
(defn mix [colors]
(rgb->hex (apply map (fn int-average [& ns] (int (Math/floor (apply average ns)))) (map hex->rgb colors))))``` | 4,382 | 13,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.737011 |
willmansour.exprealty.com | 1,624,613,980,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00496.warc.gz | 533,445,852 | 27,555 | Mortgage Calculator
Mortgage payment calculator
How to Estimate Mortgage Payments for You!
Mortgage Payment calculations can be very complex to the buyer, these calculators have made it much easier for everyone. That is not to say these have replaced the expertise of your trusted advisor, however, they are great at painting a picture for any buyer when looking to buy their first home or to resale their current home and upsize!
This mortgage payment calculator, brought to you by RateHub, provides you with everything you will need to create different scenarios for your purchase. This will help you decide what mortgage product is right for your needs.
Several factors go into the estimation of your monthly mortgage payment. Below are three numbers, which are particularly essential in helping you understand the numbers.
1. Mortgage Amount: This amount is the price of your purchase, minus the down payment you are putting towards the purchase, if the down payment is under 20% of the purchase price then it will also include the CMHC mortgage insurance as well.
2. Amortization Period: The total life of your mortgage is what is referred to as the Amortization Period. This basically is the number of years your mortgage balance will be spread across. An example of this is 20 Years, 25 Years, or 30 Years.
3. Rate: This is the interest rate you have negotiated with your lender. Rates can be broken down into two categories, fixed and variable. The lower the negotiated rate, the lower the payments!
How to use the mortgage payment calculator
Using the calculator, us very simple. Start by entering the purchase price of the property you have purchased or an estimated purchase price of a property you want to purchase. Select your amortization period and finally enter your negotiated mortgage rate with your lender. If you don’t have a negotiated rate the calculator will provide you with current best options for rates. Once these steps are completed, the calculator will illustrate what your estimated mortgage rate will be.
Depending on the size of your down payment, the calculator will illustrate for separate monthly payments and will automatically calculate the CMHC Insurance Premium, if your down payment is less than 20% of the purchase price. To modify this at any time, simply change the amount of your down payment as well as your payment frequency to understand how your payments can be modified.
We have a separate page in the Buyers section of this website, which illustrates the inner workings of Land Transfer Tax, https://willmansour.exprealty.com/resources/ontario-land-transfer-tax however, this calculator will also illustrate to you the Land Transfer Tax applicable together with the estimated amount of money you will need on closing of your purchase. It is important to note, purchasing in the City of Toronto has its own LTT on top of the Provincial LTT that is charged. This means you are paying twice the amount of LTT if you purchase in the City of Toronto. This is important as it will increase your closing costs by 40%.
When looking to buy a new home, it is important to first meet with your lender. If you don’t have a lender then ask me as I work with many lenders. Understanding the numbers will solve a number of issues that can come up during the buying process. Using this calculator is a great starting point as it will estimate what you can afford before you start your search and meet with your lender.
How to lower your mortgage payments
Who doesn’t want lower monthly payments? There are ways to achieve this. Reducing your purchase price for one, making a bigger down payment is also a good idea, if you can afford to, and increasing the Amortization Period is another way. A 20-year amortization period will yield a higher monthly payment than a 30-year amortization period. Lastly, a low mortgage rate is also essential. Depending on the federal and bank rates, going with a variable mortgage over a fixed might be something to consider. Always consult with your mortgage professional for advice. | 807 | 4,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.955762 |
http://learningincontext.com/pic-1st/chapt05.php | 1,555,727,818,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528481.47/warc/CC-MAIN-20190420020937-20190420042008-00030.warc.gz | 99,396,291 | 7,998 | # Physics in Context - 1st Edition
## Chapter 5: Energy
Some links are repeated for use with more than one lesson.
### 5.1 Energy in Mechanical and Fluid Systems I
Calculating the Moment of Inertia for Rotating Masses
http://www.dummies.com/how-to/content/how-to-calculate-the-momentum-of-inertia-for-diffe.html
This page presents a concise introduction and presentation of rotational inertia for masses of different shapes, including a reference table of formulas for common shapes.
Comparison of Linear and Rotational Formulas
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
This page summarizes and shows the helpful similarities between the formulas and symbols that define linear and rotational motions, energies, etc., including some advanced concepts applicable to future lessons.
Rotational Kinetic Energy and Angular Momentum
http://buphy.bu.edu/~duffy/py105/notes/AngularMo.html
This site of lecture notes on rotational kinetic energy and angular momentum includes parallels between straight-line motion and rotational motion. It also includes an example problem comparing the two.
Energy Simulations
http://www.myphysicslab.com/beta/spring1.html
Starting with a simple one-dimensional vibrating mass, this series of simulations (click List or Next) allow you to control many parameters and explore the effects on the system energy values with various masses and springs. (Requires Java.)
Velocity, Acceleration, Projectile and Circular Motion
http://www.7stones.com/Homepage/Publisher/vCompCalc.html
Interactive graphing of projectile motion, showing velocity, acceleration, and energy at all times.
Energy
http://www-istp.gsfc.nasa.gov/stargaze/Lenergy.htm
This topic stresses mechanical energy, potential and kinetic, and describes conversion between types of energy (while conserving the total amount), units, and the special position of heat. Be sure to follow the link to the lesson supplements.
Potential Energy
http://jersey.uoregon.edu/vlab/PotentialEnergy/index.html
This site has an interactive simulation on the potential and kinetic energy of dropped balls. It includes explanatory material as well.
### 5.2 Energy in Mechanical and Fluid Systems II
Potential and Kinetic Energy
http://www.physicsclassroom.com/class/energy/Lesson-1/Definition-and-Mathematics-of-Work
Following a review of mechanical Work, this sequence of Physics Classroom lessons on this site address potential and kinetic energy, including practice problems.
Bernoulli Biography
http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Bernoulli_Daniel.html
A biography of Daniel Bernoulli
Bernoulli's Equation
http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html
This page from a site discussing the aerodynamics of bicycles covers Bernoulli's equation and examples.
Energy of Masses and Springs
This PhET applet simulates a real lab with hanging masses and springs. Explore the changing energy picture with different masses, springs, and even different gravities! (Requires Java.)
Skateboarder Energy
This PhET applet simulates a skateboarder in a half-pipe, or whatever shape you configure. Explore the changing energy picture with different tracks, skaters, starting positions, and so forth. (Requires Java.)
Springs
http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section4.rhtml
This page gives an exhaustive—somewhat advanced—treatment of springs: the forces, potential and kinetic energies, characteristics of oscillation, and more.
Bernoulli's Principle
http://www.allstar.fiu.edu/aerojava/pic3-2.htm
Another site describing the relationship between Bernoulli's equation and flight
Determine the Spring Constant
http://www.4physics.com/phy_demo/HookesLaw/HookesLawLab.html
This page describes a lab activity to experimentally determine the spring constant for an unknown spring.
### 5.3 Energy in Electrical Systems
How Stuff Works: Capacitors
http://electronics.howstuffworks.com/capacitor.htm
A introduction to capacitors—how they work and some historical background—including a helpful explanation of capacitance units of measure: the farad.
A detailed biographical look at Michael Faraday, a British chemist and physicist who contributed significantly to the early study of electromagnetism and electrochemistry.
This biography of Michael Faraday comes from the Royal Institution of Great Britain. It includes links to some related RIGB exhibits, as well as to some of Faraday's writings.
Joseph Henry Biography
http://en.wikipedia.org/wiki/Joseph_Henry
This biographical summary of Joseph Henry details his important contributions to electromagnetism and inductance and early contributions to electric motors. The unit of measure we use today for inductance is named after him.
Joseph Henry Papers Project
http://www.siarchives.si.edu/history/jhp/jhenry.html
This is the Joseph Henry Papers Project of the Smithsonian Institutional History Division. It includes selected papers of Joseph Henry and other links describing his wide-ranging contributions to science.
Heinrich Lenz Biography
http://www.magnet.fsu.edu/education/tutorials/pioneers/lenz.html
This biographical summary for the man who contributed our first understanding of the conservation of energy in electrical and magnetic circuits, stated as Lenz's Law, and whose initial "L" was chosen as the symbol commonly used for inductors in a circuit.
Lenz's Law
http://www.micro.magnet.fsu.edu/electromag/java/lenzlaw/
This site includes a java applet demonstrating Lenz's law.
### 5.4 Energy in Thermal Systems
The Second Law of Thermodynamics
This applet demonstrates the second law of thermodynamics with a bouncing ball of variable hardness. The accompanying linked explanation is enlightening, but may be too advanced. (Requires Java.)
Laws of Thermodynamics
http://energy.concord.org/energy2d/ht.html
This applet demonstrates the first and second laws of thermodynamics, showing the transition over time for a hot-cold scenario to an equilibrium state. (Requires Java.)
Heat Engine
http://www.taftan.com/thermodynamics/HENGINE.HTM
Forward and reverse heat engines and the first and second laws of thermodynamics
The Carnot Cycle and the Efficiency of Engines
http://www.mhhe.com/physsci/physical/jones/graphics/jones2001phys_s/ch13/others/13-3/
This applet demonstrates the pressure-volume relationship of a Carnot engine. (Requires Java.)
The Carnot Engine
http://science.sbcc.edu/~physics/flash/heatengines/Carnot%20cycle.html
A flash animation showing the cycling of a Carnot engine.
This biographical article summarizes the life of Carnot, who recognized that the hot and cold temperature extremes in steam engines controlled its efficiency.
Celsius, Kelvin, and Fahrenheit Temperature Scales
http://lamar.colostate.edu/~hillger/temps.htm
This site includes definitions of the three temperature scales, and formulas for converting between them.
Lord Kelvin Biography
http://digital.nls.uk/scientists/biographies/lord-kelvin/
This biographical summary of Lord Kelvin details his important contributions to thermodynamics, and laying of telegraph cables across the Atlantic Ocean.
How Refrigerators Work
http://www.howstuffworks.com/refrig.htm
How does a refrigerator work? Check it out at How Stuff Works.
A Simple Finite Entropy Example
http://www.7stones.com/Homepage/Publisher/entropy.html
Entropy applet example using pixels.
The Page of Entropies
http://webs.morningside.edu/slaven/Physics/entropy/index.html
A light-hearted discussion of entropy, spanning entropy in poker hands, a box of air, and even water. | 1,646 | 7,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-18 | latest | en | 0.809928 |
https://calculate-this.com/how-many-jelly-beans-jar-calculator | 1,558,678,657,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257514.68/warc/CC-MAIN-20190524044320-20190524070320-00535.warc.gz | 419,746,491 | 10,718 | # How Many Jelly Beans In A Jar Calculator
Trying to figure out how many jelly beans there are in a mason jar or gallon jug? Here's our method of getting the number, but you can jump to the bottom and just use the calculator too.
One way to do it is to figure out the volume of the jar and divide that by the volume of a jelly bean.You want to get the volume of the inside of the jar, so you should subtract the thickness of the glass from the diameter. Mason jars typically has a 3/8" thickness.
Jar Volume
To calculate the cubic inches inside a jar or other cylinder shaped object, you will need to know the diameter (minus the glass thickness) and height of the jar then apply the following formula:
Let's say our jar is 8 inches in diameter and 12 inches in height.
1. Subtract the glass thickness from the diameter (8 - .375 = 7.625)
2. Half of the circle's diameter equals it's radius (7.625 / 2 = 3.8125)
3. Square the radius (3.8125 * 3.8125 = 14.53515625)
4. Multiply the height by the squared radius (12 * 14.53515625 = 174.421875)
5. Multiply that by pi 3.1416 (192 * 3.1416 = 547.9637625)
So we know our jar is 547.9637625 cubic inches.
Jelly Bean Volume
If our jelly bean dimensions are 3/4" by 1/3", that would be equal to .75" x .333333". We would use the jelly bean's dimensions to repeat the steps used for jar volume, which would give us 0.06544986910006544 cubic inches per jelly bean.
But Wait....
Jelly beans have odd, uneven shapes and won't fill all the available space of a jar. They don't pack together nicely, which means there will be small, unfilled voids in between all those jelly beans. We need to figure out what percentage of our jar does not contain jelly beans. You could add sugar to a cup filled with jelly beans so that it is filled to the top. Then strain out your jelly beans and measure how much sugar you have using tablespoons. If you have 3 and a half tablespoons of sugar, you can estimate that there is a 21.875% void in between your jelly beans. 16 TBSP = 1 Cup, 1 Cup divided by 16 TBSP = .0625, 3.5 TBSP of sugar x .0625 = .21875
To make things even more precise, repeat your jelly bean/sugar test 3 or 4 more times, then get the average value of those results. For our purposes, we'll just stick to using the first result of 21.875%. We'll use that to subtract from our jar volume, since we know 21.875% of our jelly bean jar will have unfilled voids. So 547.9637625 (jar cubic inches) multiplied by .21785 = 119.8670730 (void cubic inches). And 547.963762 - 119.8670730 = 428.096689 cubic inches of jelly bean space.
The Result
We'll divide our available jelly bean space by the cubic inches per jelly bean. 428.096689 / .06544986910006544 = 6540.833387237015
Our 8"x12" jar can hold approximately 6540 jelly beans. That's a lot of jelly beans.
For reference, here are some common jar dimensions:
• Ball 16 oz Pint Mason Jar outer dimensions are 5.2 height by 3.2 diameter - which could hold approximately 390 jelly beans
• Ball 32 oz Quart Mason Jar outer dimensions are 6.9 height by 3.9 diameter - which could hold approximately 804 jelly beans
• Ball 128 oz Gallon Jar outer dimensions are 10.5 height by 5.7 diameter - which could hold approximately 2792 jelly beans
Diameter
Height | 867 | 3,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-22 | latest | en | 0.908704 |
http://rigtriv.wordpress.com/2008/03/04/abstract-varieties/ | 1,417,116,102,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931009084.22/warc/CC-MAIN-20141125155649-00187-ip-10-235-23-156.ec2.internal.warc.gz | 245,320,089 | 24,007 | ## Abstract Varieties
Last time, we defined locally ringed spaces and I linked back to an old post where I defined the notion of an abstract variety. We’re going to talk about these objects a bit more today, including, in particular, checking that our old quasi-projective varieties are all abstract varieties. We’re heading for another nice theorem over the next couple of posts, so these next few will be rather pointed.
First, we define a prevariety to be a locally ringed space $X$ such that there is a collection of open sets $X=\bigcup_\alpha U_\alpha$ such that for each $\alpha$, we have $(U_\alpha,\mathscr{O}_X|_{U_\alpha})$ is isomorphic as a locally ringed space to an affine variety. The restriction here is just the values of $\mathscr{O}_X$ on the open sets contained in $U_\alpha$.
So, first off, any affine variety is a prevariety. What about projective varieties? Well, projective space itself is a prevariety, because we can take the collection $U_0,\ldots,U_n$ where $U_i$ is the locus where $x_i\neq 0$. Each of these is isomorphic to $\mathbb{A}^n$. Now taking any projective variety, we can intersect it with these $U_i$ and get the appropriate open sets. We will call such a collection an open affine cover for short. From this, it’s not hard to see that quasi-projective varieties in general are prevarieties.
But why do we call them prevarieties rather than just varieties? For the answer, look at the projective lines $\mathbb{P}^1$. When we cover this with two copies of $\mathbb{A}^1$, we see that the function mapping from the first affine line to the second, on the overlap, is $1/z$. What happens if we identified them along the function $z$ instead? Well, we get the affine line, but with an extra copy of the origin.
For a moment, to analyze this, we’re going to restrict to the complex numbers and look at things in the usual (or analytic) topology, where open sets are the ones such that every point has a disc around it contianed in the set. Now, we can label each point by $z\in\mathbb{C}\setminus{0}$ other than the two origins, and we can denote them by $0_1,0_2$. So now take any open set containing $0_1$ and call it $U_1$ and any open set containing $0_2$ and call it $U_2$. The MUST intersect, because each must contain a disc centered at $0_1$, and so one of these two discs will contain the other! This is an example of a topological space which is not Hausdorff. I link to the definition rather than give it because it inspires us to figure out how to throw away the line with the doubled origin, rather than being of direct value. The truth is, except for on finite collections of points, the Zariski topology is NEVER Hausdorff.
However, this lack of topological separation can be overcome, because unlike topologists, we have the arsenal of commutative algebra built into our objects. However, we will take a little bit more inspiration from them. It is a theorem (and not too hard to prove) that a toplogical space $X$ is Hausdorff if and only if the map $\Delta:X\to X\times X$ given by $\Delta(x)=(x,x)$ has closed image. Now, THIS condition we can use. It requires that I mention something that, shamefully, I forgot to mention right when I defined products back in the Algebraic Groups post. Though we get the same point set as expected from topology, we do NOT get the same topology. Look at $\mathbb{A}^1\times\mathbb{A}^1=\mathbb{A}^2$. The topology on $\mathbb{A}^2$ would only allow finite collections of points and vertical and horizontal lines as closed sets, but yet we have all of the richness of plane curves. So we can see if the condition above (we will say that something is separated if it holds) works out for us with our new topologies.
In fact, it does. We define a variety (or an abstract variety, though we will tend to leave off the word abstract) to be a prevariety $V$ such that $\Delta:V\to V\times V$ has closed image. So does this throw away the line with doubled origin? If we call the line $X$, then we note that $X\times X$ will be a plane, but with four copies of the origin. However, the image of $\Delta$ will contain the diagonal elements, so it only hits $(0_1,0_1)$ and $(0_2,0_2)$. However, the closure must contain $(0_1,0_2)$ and $(0_2,0_1)$, and so $X$ is not separated, and therefore not a variety.
Are quasi-projective varieties separated? The answer is yes. First off, open and closed subsets of a separated space are separated, so we really only need to check projective space. If we look at the image of $\mathbb{P}^n\subset \mathbb{P}^n\times\mathbb{P}^n$, then we can choose to call the variables on the first term of the product $x_i$ and on the second $y_i$. Then a closed subset is given by a collection of polynomials which are homogeneous in each set of variables separately. The image is just the collection $x_i=y_i$, and so $\mathbb{P}^n$ is a variety, by the new definition.
So we have some collection of locally ringed spaces which contains our old notion of variety and excludes some pathological spaces. Does it extend our notion of variety? Well, the answer to this is also yes, but it will require some work to demonstrate it. Once we’ve got a bit more done here (and once I’ve gone through an example carefully on my own) we’ll see that there are varieties which are not quasi-projective. The big reason for it to be a pain, is that given the notion of a proper map and a complete variety (that’s next) it is true that every complete curve is projective and every complete nonsingular surface is projective. So though there are complete but nonprojective varieties, they will need to be singular surfaces or of at least dimension three. The real value in this definition of a variety is that it allows for some flexibility of technique. Extremely important is the fact that the structure sheaf is built in, as later we’ll be using it to define sheaves of modules, which will in turn lead us to turn the geometric subject of vector bundles into something algebraic.
Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in AG From the Beginning, Algebraic Geometry. Bookmark the permalink.
### 7 Responses to Abstract Varieties
1. Are there things you can do in the category of abstract varieties that you couldn’t do in the category of quasi-projective varieties? Some limits or colimits that exist in one but not the other?
2. Greg Muller says:
Actually, its easy to come up with some non-projective abstract varieties, but only because I don’t think you included any finiteness or Noetherian-ness requirement. Then a countable union of quasi-projective varieties will still be locally a variety, and so its an abstract variety. This is also a partial answer to John’s question, since now arbitrary coproducts exist.
However, I think this isn’t really the point, since this ‘enrichment’ just leaving open the door to some silly examples. I think a better motivation for thinking about varieties abstractly rather than as quasi-projective varieties is conceptual. Saying a variety is quasi-projective means it either A) is a subobject of projective space, or B) is an equivalence class of subobjects of projective space (depending on the wording of the definition). This definition is as un-elegent as defining a ‘manifold’ to be a sufficiently nice sub-space of $\mathbb{R}^n$. Sure, it works, but then you need fancy embedding theorems to show that certain nice spaces are actually manifolds. Its better to have an intrinsic characterization, if only to make it more straightforward to check.
3. Charles says:
Greg beat me to the response. I should have put in some kind of finiteness requirement to avoid the countable disjoint union example, but I didn’t so that example is non-quasi-projective.
And as Greg said, the point is that, like with manifolds, we want an intrinsic definition, and then we can talk about embeddings and to the extrinsic geometry. Also, a lot of definitions are nicer in this language than in the old one (for instance, Cartier Divisors, which we will get to).
4. I’ve usually seen a variety defined as an integral separated scheme of finite type over a field. Doesn’t the integrality exclude such things as disjoint unions? In particular, you mentioned irreducibility in your earlier posts, but it is noticeably absent here.
5. Charles says:
I’ve been specifically avoiding using the word “scheme.” I did forget any finiteness or irreducibility condition, and when I get back to this (it will be a bit, some other things have come up) I will start the next post with a correction to the definition to account for things.
6. Pingback: Schemes « Rigorous Trivialities | 2,087 | 8,718 | {"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": 43, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-49 | longest | en | 0.913258 |
https://oeis.org/A217677 | 1,716,140,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00462.warc.gz | 387,817,757 | 4,564 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A217677 Number of permutations in S_n containing an increasing subsequence of length 10. 3
1, 101, 6063, 284431, 11592572, 433386000, 15343169775, 524963196399, 17597634740010, 583499409451862, 19269396089593156, 636977450902768356, 21156201514272916444, 708006643310351350076, 23925259865186482138965, 817728884509460388159381 (list; graph; refs; listen; history; text; internal format)
OFFSET 10,2 LINKS Alois P. Heinz, Table of n, a(n) for n = 10..200 FORMULA a(n) = A214152(n,10) = A000142(n)-A072133(n) = A000142(n)-A214015(n,9). MAPLE b:= proc(n) option remember; `if`(n<5, n!, ((-1110790863+(1520978576+(1772290401+(607308786+ (101671498+(9464664+(500874+(14124+165*n)*n)*n)*n)*n)*n)*n)*n)*b(n-1) -(1129886062*n+559908333*n^2+111239576*n^3+10655238*n^4+8778*n^6 +491700*n^5 +353895381)*(n-1)^2*b(n-2) +(258011271+234066216*n +58221266*n^2+5463876*n^3 +172810*n^4)*(n-1)^2*(n-2)^2*b(n-3) -9*(4070430+1504292*n+117469*n^2)* (n-1)^2*(n-2)^2*(n-3)^2*b(n-4) +893025*(n-1)^2*(n-2)^2*(n-3)^2*(n-4)^2*b(n-5)) / ((n+20)^2*(n+8)^2*(n+18)^2*(n+14)^2)) end: a:= n-> n! -b(n): seq(a(n), n=10..30); CROSSREFS Cf. A000142, A072133, A214015, A214152. Sequence in context: A166218 A167626 A017764 * A333689 A291990 A303572 Adjacent sequences: A217674 A217675 A217676 * A217678 A217679 A217680 KEYWORD nonn AUTHOR Alois P. Heinz, Oct 10 2012 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified May 19 13:25 EDT 2024. Contains 372694 sequences. (Running on oeis4.) | 748 | 1,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-22 | latest | en | 0.503705 |
http://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.296475.html | 1,368,923,504,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696383077/warc/CC-MAIN-20130516092623-00037-ip-10-60-113-184.ec2.internal.warc.gz | 314,277,392 | 4,656 | # SOLUTION: In Triangle QRS, QR=6, RS=7, QS=8. The largest angle of the triangle is___ A)angle Q B)angle R C)angle S D)answer cannot be determined I looked in my geometry book and can
Algebra -> Algebra -> Triangles -> SOLUTION: In Triangle QRS, QR=6, RS=7, QS=8. The largest angle of the triangle is___ A)angle Q B)angle R C)angle S D)answer cannot be determined I looked in my geometry book and can Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Geometry: Triangles Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Triangles Question 296475: In Triangle QRS, QR=6, RS=7, QS=8. The largest angle of the triangle is___ A)angle Q B)angle R C)angle S D)answer cannot be determined I looked in my geometry book and can't seem to find anything on how to solve this, so I will assume that the answer is D) but I want to ask, just in case.Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!the largest angle is opposite the largest side (see Law of Sines) | 311 | 1,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.855395 |
https://www.statemath.com/2021/08/instability-of-solutions-to-nonlinear-systems.html | 1,708,468,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00309.warc.gz | 1,065,803,507 | 23,405 | In this post, we propose two results in the instability of solutions to nonlinear systems. Here, we study ordinary differential equations. Well-know stability theorems are due to Liapunov, based on the linearization of the vector field.
Consider a continuous function $f:\Omega\subset \mathbb{R}^d\to\mathbb{R}^d$, and $x_0\in \Omega$. Let the Cauchy problem\begin{align*}\tag{Eq}\dot{u}(t)=f(u(t)),\quad u(0)=x_0.\end{align*}
## The flow of an autonomous system
According to Peano’s theorem, the maximal solution of the differential equation $({\rm Eq})$ exists. We denote by $J_{x_0}$ the interval in which the maximal solution is well defined. We denote \begin{align*}D(f)=\bigcup_{x\in\Omega}\left(J_x\times\{x\}\right).\end{align*} The flow associated to the equation $({\rm Eq})$ is the following application \begin{align*} \Phi: D(f)\to \Omega,\quad (t,x)\mapsto \Phi(t,x)=u(t),\end{align*}where $u$ is the maximal solution associated to the initial condition $u(0)=x$. Also, we denote $\Phi_t(x)=\Phi(t,x)$.
We mention that $J_{\Phi_t(x)}=J_x-t$, and if $t_1+t_2\in J_x$, we have\begin{align*}\Phi_{t_2}\circ \Phi_{t_1} (x)= \Phi_{t_1+t_2}(x).\end{align*}
Assume that $f$ is locally Lipschitz on $\Omega$. Then $D(f)$ is an open set of $\mathbb{R}\times \Omega$. Moreover, the flow $\Phi$ is locally Lipschitz, in particular, it is continuous, on $D(f)$.
## Instability of solutions to nonlinear systems
An equilibrium point of $f:\Omega\to \mathbb{R}^d$ is an element $x_0\in \Omega$ such that $f(x_0)=0$. An immediate consequence is that the constant function equal to $x_0$ satisfies the differential equation $({\rm Eq})$.
The equilibrium point $x_0$ is stable if and only if for any neighborhood $W$ of $x_0$ in $\Omega,$ there exists a neighborhood $U$ of $x_0$ in $\Omega$ such that for all $x\in U,$ $\Phi_t(x)$ is defined for any $t\ge 0,$ i.e. global solution, and takes value in $W$.
The equilibrium $x_0$ is unstable if and only if it is not stable.
The equilibrium $x_0$ is asymptotically stable if and only if it is stable and there exists a neighborhood $W$ of $x_0$ in $\Omega$ such that for any $x\in W,$ $\Phi_t(x)$ is defined for any $t\ge 0$ and $\Phi_t(x)\to x_0$ as $t\to +\infty$.
Previous Story
Next Story | 716 | 2,251 | {"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} | 3.484375 | 3 | CC-MAIN-2024-10 | latest | en | 0.778592 |
https://en.xen.wiki/w/Lp_tuning | 1,702,308,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00771.warc.gz | 256,073,059 | 10,547 | # Tp tuning
(Redirected from Lp tuning)
Tp tuning is a generalization of TOP and TE tuning. (In this article p denotes a parameter, p ≥ 1; it does not denote a prime.)
For a subgroup temperament over a general JI subgroup, and for a given choice of p (most commonly p = 2), there are two notions of Tp tuning:
• The first is called inharmonic TE, because the basis entries are treated as if they were primes, reminiscent of some inharmonic timbres. Inharmonic TE depends on the basis used for the subgroup. In non-octave temperaments, inharmonic TE could be used when optimizing a specific voicing of a tempered JI chord. For example in 3/2.7/4.5/2 semiwolf temperament which tempers out 245/243, the 3/2.7/4.5/2 inharmonic TE optimizes the 4:6:7:10 chord.
• The second is called subgroup TE, because it treats the temperament as a restriction of a full prime-limit temperament to a subgroup of the prime-limit. Subgroup TE does not depend on the basis used for the subgroup, and as stated, extends naturally to the TE tuning of the full prime-limit temperament.
The two notions agree exactly when the temperament is defined on a JI subgroup with a basis consisting of rationally independent (i.e. pairwise coprime) members. That is, the subgroup has a basis where no two elements share a prime factor (examples: 2.3.5 and 2.9.5; nonexample: 2.9.5.21).
## Definition
If p ≥ 1, define the Tp norm, which we may also call the Tp complexity, of any monzo in weighted coordinates m as
$\lVert [ m_2 \ m_3 \ \ldots \ m_k \rangle \rVert_p = (|m_2|^p + |m_3|^p + \ldots + |m_k|^p)^{1/p}$
where 2, 3, … , k are the primes up to k in order. In unweighted coordinates, this would be, for unweighted monzo b,
$\lVert [ b_2 \ b_3 \ \ldots \ b_k \rangle \rVert_p = (|b_2 \log_2 2 |^p + |b_3 \log_2 3|^p + \ldots + |b_k \log_2 k |^p)^{1/p}$
If q is any positive rational number, ‖qp is the Tp norm defined by its monzo.
For some just intonation group G, which is to say some finitely generated group of positive rational numbers which can be either a full prime-limit group or some subgroup of such a group, a regular temperament tuning T for an abstract temperament S is defined by a linear map from monzos belonging to G to a value in cents, such that T (c) = 0 for any comma c of the temperament. We define the error of the tuning on q, Err (q), as |T (q) - cents (q)|, and if q ≠ 1, the Tp proportional error is PEp (q) = Err (q)/‖qp. For any tuning T of the temperament, the set of PEp (q) for all q ≠ 1 in G is bounded, and hence has a least upper bound, the supremum sup (PEp (T)). The set of values sup (PEp (T)) is bounded below, and by continuity achieves its minimum value, which is the Tp error Ep (S) of the abstract temperament S; if we measure in cents as we've defined above, Ep (S) has units of cents. Any tuning achieving this minimum, so that sup (PEp (T)) = Ep (S), is an Tp tuning. Usually this tuning is unique, but in the case p = 1, called the TOP tuning, it may not be. In this case we can choose a TOP tuning canonically by setting it to the limit as p tends to 1 of the Tp tuning, thereby defining a unique tuning Tp (S) for any abstract temperament S on any group G. Given Tp (S) in a group G containing 2, we may define a corresponding pure-octaves tuning (POLp tuning) by dividing by the tuning of 2: Tp' (S) = 1200 Tp (S)/(Tp (S))1, where (Tp (S))1 is the first entry of Tp (S). When p = 2, POL2 tuning generalizes POTE tuning.
## Dual norm
We can extend the Tp norm on monzos to a vector space norm on interval space, thereby defining the real normed interval space Tp. This space has a normed subspace generated by monzos belonging to the just intonation group G, which in the case where G is a full p-limit will be the whole of Tp but otherwise might not be; this we call G-interval space. The dual space to G-interval space is G-tuning space, and on this we may define a dual norm. If r1, r2, … , rn are a set of generators for G, which in particular could be a normal list and so define smonzos for G, then corresponding generators for the dual space can in particular be the sval generators. On this standard basis for G-tuning space we can express the dual norm canonically as the G-sval norm. If [r1 r2rn] is the normal G generator list, then cents (r1) cents (r2) … cents (rn)] is a point, in unweighted coordinates, in G-tuning space, and the nearest point to it under the G-sval norm on the subspace of tunings of some abstract G-temperament S, meaning svals in the null space of its commas, is precisely the Lp tuning Lp (S).
In the special case where p = 2, this becomes L2 tuning. This is called inharmonic TE in Graham Breed's temperament finder.
## Applying the Hahn-Banach theorem
Suppose T = Tp (S) is an Tp tuning for the temperament S, and J is the JI tuning. These are both elements of G-tuning space, which are linear functionals on G-interval space, and hence the error map Ɛ = T - J is also. The norm ‖Ɛ‖ of Ɛ is minimal among all error maps for tunings of S since T is the Tp tuning. By the Hahn–Banach theorem, Ɛ can be extended to an element Ƹ in the space of full p-limit tuning maps with the same norm; that is, so that ‖Ɛ‖ = ‖Ƹ‖. Additionally, due to a corollary of Hahn-Banach, the set of such error maps valid for S can be extended to a larger set which is valid for an extended temperament S*; this temperament S* will be of rank greater than or equal to S, and will share the same kernel. ‖Ƹ‖, the norm of the full p-limit error map, must also be minimal among all valid error maps for S*, or the restriction of Ƹ to G would improve on Ɛ. Hence, as ‖Ƹ‖ is minimal, J* + Ƹ, where J* is the full p-limit JIP, must equal the Tp tuning for S*. Thus to find the Tp tuning of S for the group G, we may first find the Tp tuning T* for S*, and then apply it to the normal interval list giving the standard form of generators for G.
Note that while the Hahn-Banach theorem is usually proven using Zorn's lemma and does not guarantee any kind of uniqueness, in most cases there is only one Lp tuning and the extension of Ɛ to Ƹ is in that case unique. It is also easy to see that this can only be non-unique if p = 1 or p = infinity, so that we may get a unique Lp tuning (called the "TIPTOP" tuning for p = infinity) by simply taking the limit as p approaches our value.
## T2 tuning
In the special case where p = 2, the Tp norm for the full prime limit becomes the T2 norm, which when divided by the square root of the number n of primes in the prime limit, is the Tenney-Euclidean norm, giving TE complexity. Associated to this norm is T2 tuning extended to arbitrary JI groups, and RMS error, which for a tuning map T is ‖(T - J)/n2 = ‖T - J‖RMS.
For an example, consider indium temperament, with group 2.5/3.7/3.11/3 and comma basis 3025/3024 and 3125/3087. The corresponding full 11-limit temperament is of rank three, and using the usual methods, in particular the pseudoinverse, we find that the T2 (TE) tuning map is 1199.552 1901.846 2783.579 3371.401 4153.996]. Applying that to 12/11 gives a generator of 146.995, and multiplying that by 1200.000/1199.552 gives a POT2 tuning, or extended POTE tuning, of 147.010. Converting the tuning map to weighted coordinates and subtracting 1200 1200 1200 1200 1200] gives -0.4475 -0.0685 -1.1778 0.9172 0.7741]. The ordinary Euclidean norm of this, i.e. the square root of the dot product, is 1.7414, and dividing by sqrt (5) gives the RMS error, 0.77879 cents.
This is called subgroup TE in Graham Breed's temperament finder.
• Dave Keenan & Douglas Blumeyer's guide to RTT: tuning in nonstandard domains - for a generalization of the inharmonic and subgroup approaches to all regular temperament tuning schemes, beyond only Tp tuning schemes as are discussed here (minimax-$q$-(lp-)S tuning schemes, in D&D's naming system), and done in a more in-depth textbook tutorial style. This also includes an additional neutral approach which is not discussed here, a discussion of why one might choose one approach over the others, and demonstrations of how to compute tunings using each of the approaches, with examples. | 2,276 | 8,131 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-50 | longest | en | 0.860725 |
https://medium.com/@JerryQu/artificial-intelligence-an-intuitive-introduction-dedfcbf0f132 | 1,556,069,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578616424.69/warc/CC-MAIN-20190423234808-20190424020808-00500.warc.gz | 488,100,051 | 27,256 | # Artificial Intelligence: An Intuitive Introduction
I know what you’re thinking. God dammit, it’s another article about Artificial Intelligence with a bunch of math that I’m not gonna understand, close tab
WAIT!!!
GIVE ME 4 WORDS TO BLOW YOUR MIND
I’m 17 years old.
Yeah, a 17 year old high school student who hasn’t even taken calculus. If I can understand it, so can you.
### Before we begin, what truly is AI?
Literally everyone, even my 50 year old high school teachers have heard about AI. It’s that thing in Google’s self driving cars. That thing that people are worried will take their jobs. However, not many people realize how Artificial Intelligence IS actually being used.
#### IS
• Google Photos: Allows you to search for specific people or objects within your photos.
• Deep Genomics: Literally performing drug discovery using AI, at rates millions of times faster than traditional systems.
• Rigetti: Quantum Machine Learning. Can compute what today’s supercomputers would take billions of years to compute. Do I need to say more…
• Worth over \$7 BILLION. And projected to be worth over \$90 BILLION by 2025.
Wow. If you had any doubt about AI, it should now be gone. We are living in the temporal hub of Artificial Intelligence right now. We’ve hit the inflection point, and there’s no stopping.
### The Math.
Here are the prerequisites you will need to understand the math I’m about to explain to you.
That’s literally it. All you need to know is how to add, subtract, divide, and multiply.
Easy.
#### Let’s take a case study
Imagine you’re designing a program to predict the value of real estate. You’ll need certain inputs/parameters that correspond to the value of real estate. To keep things simple, we’ll only take one parameter: Square Footage.
You realize square footage is heavily correlated to the value of a house. You then find a dataset online with a table containing different real estate with their square footage, and value.
Your goal is to allow end users to be able to predict the value of their house by entering the square footage of their house.
Inputs: Square Footage (ft²)
Outputs: Value (\$)
Here’s an example of a dataset:
This relationship could be modeled as such,
Y = mX (A Linear Relationship)
or,
Value = m * Square Footage
Square footage multiplied by some number (m) will equal our value. We do not know what that value m is, but we can figure it out.
#### Here comes the machine learning
Let’s initialize our variable (m) to a random number. Let’s say 20. Now, we’ll pass some of our data through our model.
``+----------------+-----+-----------------+--------------+| Square Footage | m | Predicted Price | Actual Price |+----------------+-----+-----------------+--------------+| 1000 | *20 | 20000 | 250000 |+----------------+-----+-----------------+--------------+``
Error (Actual Price — Predicted Price) = 250000–20000 = 230000
As you can see, the price we’ve predicted is way below the actual price of \$250000. Our error, is measured to be 230000. So, how do we get our predicted price to be closer to the actual price, therefore making our model more accurate? We’ll, we adjust the one variable in our model, m.
Let’s imagine we adjust our variable (m) by a tiny amount, 0.0001, what would the predicted price become?
``+----------------+---------+-----------------+--------------+| Square Footage | m | Predicted Price | Actual Price |+----------------+---------+-----------------+--------------+| 1000 | *20.001 | 20001 | 250000 |+----------------+---------+-----------------+--------------+``
Error (Actual Price — Predicted Price) = 250000–20001 = 229999
As you can see, the predicted price got ever so closer to the actual price of \$250000. Our error is now only 229999. Logically, this means that we should increase our variable m to make our model more accurate. If we repeat this millions upon millions of times, we’d eventually get an optimized model that would allow us to accurately predict the price of a house, given the square footage.
#### And that’s it!
That is the intuitive logic behind most forms of Machine Learning. Whether it’s a basic neural network, a convolutional neural network, or even a deep q-network, all are based around adjusting your variables to lower your error.
In practice, I’ve skipped a TON of steps here. But, my end goal was to give you an intuitive understanding of how Machine Learning works, not the nitty gritty. The method I showed you above is a Machine Learning algorithm called Gradient Descent, using the Delta Rule. In practice, we’d use a Neural Network instead of this model. In a neural network, we’d be using Backpropagation to take partial derivatives of our cost function, allowing us to adjust all of our weights appropriately. I’ll post a second article on back propagation in the future, but here’s some common terminology ahead of time.
Cost: Essentially error, but normally put through a Cost Function (Such as Sum of Squared Errors)
Weights: The variables (m) in a neural network
Activation Function: Introduces non-linearity (So it’s not just a straight line like in our example! Ex. ReLU, Sigmoid)
Gradient Descent: Essentially the process I explained, lowering our error by adjusting our weights
Backpropagation: How real neural networks learn! Partial derivatives, chain rule, a lot more to explain. But based on the same principle as gradient descent.
Overall, I wrote this article to help beginners get into the field of AI. I hope you’re excited about what AI can do in the future. And I hope you now realize that you don’t need a PhD to get into this field.
Over the past year, I’ve developed many of my own personal projects using AI. From a virtual self driving car, to an app that aids the legally blind. (Check out my website) I’ve even been able to work at Microsoft this summer! And speak at the Toronto Machine Learning Summit. If you’re interested in AI/ML, go and learn! There’s nothing like the present. | 1,367 | 6,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-18 | latest | en | 0.889299 |
https://oeis.org/A056288 | 1,721,901,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00341.warc.gz | 362,924,669 | 4,522 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A056288 Number of primitive (period n) n-bead necklaces with exactly three different colored beads. 4
0, 0, 2, 9, 30, 89, 258, 720, 2016, 5583, 15546, 43215, 120750, 338001, 950030, 2677770, 7573350, 21478632, 61088874, 174179133, 497812378, 1425832077, 4092087522, 11765778330, 33887517840, 97756266615, 282414622728, 816999371955, 2366509198350, 6862929885407 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Turning over the necklace is not allowed. REFERENCES M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2] LINKS Alois P. Heinz, Table of n, a(n) for n = 1..1000 FORMULA Sum mu(d)*A056283(n/d) where d|n. MAPLE with(numtheory): b:= proc(n, k) option remember; `if`(n=0, 1, add(mobius(n/d)*k^d, d=divisors(n))/n) end: a:= n-> add(b(n, 3-j)*binomial(3, j)*(-1)^j, j=0..3): seq(a(n), n=1..30); # Alois P. Heinz, Jan 25 2015 MATHEMATICA b[n_, k_] := b[n, k] = If[n==0, 1, DivisorSum[n, MoebiusMu[n/#]*k^#&]/n]; a[n_] := Sum[b[n, 3 - j]*Binomial[3, j]*(-1)^j, {j, 0, 3}]; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, Jun 06 2018, after Alois P. Heinz *) CROSSREFS Cf. A027376. Column k=3 of A254040. Sequence in context: A177111 A290746 A268586 * A261174 A273652 A056283 Adjacent sequences: A056285 A056286 A056287 * A056289 A056290 A056291 KEYWORD nonn AUTHOR Marks R. Nester STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified July 25 05:51 EDT 2024. Contains 374586 sequences. (Running on oeis4.) | 690 | 1,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.518985 |
https://www.assignmentexpert.com/homework-answers/physics/optics/question-19161 | 1,597,253,738,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738913.60/warc/CC-MAIN-20200812171125-20200812201125-00541.warc.gz | 583,759,477 | 290,527 | 89 112
Assignments Done
98.2%
Successfully Done
In July 2020
Answer to Question #19161 in Optics for baran
Question #19161
in a interference expriment in a large ripple tank the coherent vibrating source are placed 120mm apart. the distance between maxima 2.0m away is 180mm .if the speed of ripple is 25 cm/s calculate the frequency of the vibrating source.
1
2012-11-22T08:15:53-0500
Fringe spacing d = wavelength /sin (angle);
sin (angle) = (apart/2)/screen;
wavelength = speed/frequency;
frequency = speed/wavelength = speed*2*screen/(d*apart) = (25 cm/s) * 2 * (2 m)
/ (120 mm * 180 mm) = 46.296 Hz
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS! | 224 | 753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-34 | latest | en | 0.830152 |
https://mystiz.hk/posts/2022/2022-10-19-h4ck1ng-g00gl3/ | 1,718,409,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00704.warc.gz | 379,446,524 | 9,820 | HACKING GOOGLE is a documentary of Google’s cybersecurity teams and H4CK1NG G00GL3 is it’s CTF counterpart. Project Zero Adventure is a cryptography challenge I wrote.
In the game, the players control the Security Princess to dodge the obstacles and catch the bugs (a variant of Google Chrome’s dinosaur game). After that, the server will sign messages consisting of the players' name and the score via the /sign API. The players will then submit it to the /highscore API. If the score submitted to the highscore API is negative, they will be given the flag.
However, there is one catch: The server is only willing to sign the results with non-negative scores.
## Challenge Summary⌗
It is given that the server signs messages using a RSA key with a 2048-bit public modulus $n$ and $e = 3$. The messages are signed with PKCS#1 v1.5. The verify method below is how a signature s, correspond to the message m, is being verified. The goal is to forge a signature for messages indicating that the player has a negative score (one example of the message being ["pzero-adventures", "MYZ", -1]).
class VerifyingKey:
def __init__(self, n, e, bits=2048):
self.n = n
self.e = e
self.bits = bits
# https://datatracker.ietf.org/doc/html/rfc2313#section-10.2
# Note: The only hash algorithm we accept is SHA256.
def verify(self, m, s):
if len(s) != self.bits//8:
raise Exception('incorrect signature length')
s = int.from_bytes(s, 'big')
k = pow(s, self.e, self.n)
k = int.to_bytes(k, self.bits//8, 'big')
if k[0] != 0x00:
raise Exception('incorrect prefix')
if k[1] != 0x01:
raise Exception('incorrect prefix')
sequence = DerSequence()
sequence.decode(digest_info)
_digest_algorithm_identifier, _digest = sequence
sequence = DerSequence()
sequence.decode(_digest_algorithm_identifier)
_digest_algorithm_identifier = sequence[0]
object_id = DerObjectId()
object_id.decode(_digest_algorithm_identifier)
digest_algorithm_identifier = object_id.value
if digest_algorithm_identifier != '2.16.840.1.101.3.4.2.1':
raise Exception('invalid digest algorithm identifier')
_null = sequence[1]
null = DerNull()
null.decode(_null)
octet_string = DerOctetString()
octet_string.decode(_digest)
if hashlib.sha256(m).digest() != digest:
raise Exception('mismatch digest')
return True
## Solution⌗
π Alternative Writeups (which are usually better). Of course you would like to read the solutions from our players and our fellow Googlers! Writeup by @44670, writeup by @Dvd848, writeup by @dguerri and writeup by @paulsc.
β© Fast forward? If you are interested only in the solution, you can start reading from part III. I will begin with a literature review.
### Part I: How PKCS#1 v1.5 signature scheme works?⌗
It is well-known that RSA sign messages by “decrypting” them. Assuming that the RSA key is of 256 bytes, the PKCS#1 v1.5 payload to be signed is given below:
Suppose that DATA is of length $l$. According to RFC 2313, PADDING consists of $253 - l$ bytes which are all 0xFF. Also, the data are encoded using Distinguished Encoding Rules (DER). Below is an example of DER-encoded data indicating that:
• the hash algorithm being SHA-256, and
• the digest being 0424974c68530290458c8d58674e2637f65abc127057957d7b3acbd24c208f93.
βββ¬ 30 31 Sequence type (length 0x31)
βββ¬ 30 0d Sequence type (length 0x0D)
β βββ¬ 06 09 Object identifier type (length 0x09)
β β βββ 608648016503040201 Object identifier content (decoded
β β 2.16.840.1.101.3.4.2.1, SHA-256's OID)
β βββ 05 00 Null type (length 0x00)
βββ¬ 04 20 Octet string type (length 0x20)
βββ 0424974c68530290458c8d58674e2637 Octet string content
f65abc127057957d7b3acbd24c208f93
Let $m$ denote the PKCS#1 v1.5 payload. Then the signature is given by $s = m^d \ \text{mod}\ n$.
### Part II: Bleichenbacher’s signature forgery attack in 2006⌗
In 2006, Bleichenbacher shared on CRYPTO'06 (Hal Finney’s summary) a way to easily forge RSA signatures when $e$ is small and the verify function is poorly implemented.
Suppose that the signature is validated by unpadding, retrieving ALGORITHM and DIGEST from DATA and discarding remaining bytes. This is an ideal message:
However, as long as the ALGORITHM and DIGEST are correct, messages in the below format are considered valid as well:
Yes – this is not properly validated. This is disastrous, why?
Let $e = 3$ and suppose we want to forge a signature for a message with its SHA-256 digest being 0424974c68530290458c8d58674e2637f65abc127057957d7b3acbd24c208f93.
Equivalently, we want to find $s$ with $s^3 \equiv m\ (\text{mod}\ n)$. Hereby $m$ begins with $m_0$ (which is 61 bytes long) below, i.e., $m = {256}^{195} m_0 + x$ for some “garbage” $x \in [0, {256}^{195})$.
m0 = 0x00 01 ffffffffffffffff 30 31 30 3d 06 09 608648016503040201 50 00 04 20 0424974c68530290458c8d58674e2637f65abc127057957d7b3acbd24c208f93
[ALGORITHM ] [DIGEST ]
It is difficult to work on modulo arithmetic, so let’s lift the modulo away and look for $s$ such that $s^3 = m = {256}^{165} m_0 + x$. It is easy to find a $s$ such that
$${256}^{165} m_0 \leq s^3 < {256}^{165} (m_0 + 1).$$
In short, any of the integer between $\sqrt[3]{{256}^{165} m_0}$ and $\sqrt[3]{{265}^{165} (m_0 + 1)}$ would fit – and that is basically Bleichenbacher’s attack in 2006.
### Part III: …and it constantly strikes back⌗
It is still hard to validate PKCS#1 v1.5 signatures properly after a decade.
In 2016, Filippo Valsorda (@filosottile) wrote a blog post on a flaw on such message being validated. In short, it is the PADDING which is not properly validated. You can read the source code of the vulnerable function here.
This is not the end. In 2019, Sze Yiu Chau shared on Black Hat USA that the PKCS#1 v1.5 signatures are still not validated correctly (video and white paper here). This yields six CVEs to their research team from various TLS and IPSec libraries.
It is slightly different from BB'06. Instead of stuffing garbage as the suffix, we are putting random bytes in the middle. Mathematically, we are looking for $x$ with
$$s^3 = m = 256^{l_1} \cdot \text{prefix} + 256^{l_2} \cdot x + \text{suffix}$$
for a given $\text{prefix}$, $\text{suffix}$, $l_1$ and $l_2$.
This is harder than stuffing garbage at the very end, because we have to solve the below inequalities and congruences at the same time:
• $256^{l_1} \cdot \text{prefix} \leq s^3 < 256^{l_1} \cdot (\text{prefix} + 1)$, and
• $s^3 \equiv \text{suffix}\ (\text{mod}\ 256^{l_2}).$
Fortunately this is solvable when $l_1 - l_2$ is large (we can stuff more garbage in between) and $e$ is small. Let’s leave this as an exercise to the readers.
### Part IV: …again in H4CK1NG G00GL3⌗
Now back to the challenge. The verify function at the very beginning seems to be implemented correctly. But there is one problem in the following snippet:
class VerifyingKey:
# ...SKIPPED...
def verify(self, m, s):
# ...SKIPPED...
sequence = DerSequence()
sequence.decode(_digest_algorithm_identifier)
_digest_algorithm_identifier = sequence[0]
object_id = DerObjectId()
object_id.decode(_digest_algorithm_identifier)
digest_algorithm_identifier = object_id.value
if digest_algorithm_identifier != '2.16.840.1.101.3.4.2.1':
raise Exception('invalid digest algorithm identifier')
_null = sequence[1]
null = DerNull()
null.decode(_null)
octet_string = DerOctetString()
octet_string.decode(_digest)
if hashlib.sha256(m).digest() != digest:
raise Exception('mismatch digest')
return True
In the snippet, it checks if
• the first item in the sequence is an object identifier being SHA-256’s OID, and
• the second item in the sequence (i.e., the parameter) is a null object.
Unfortunately, the method did not check if the sequence has exactly two items. An adversary can insert garbage and pretend that being the third item in the sequence. We then can reduce this to the math problem mentioned in part III.
### Part V: …and again, unexpectedly⌗
π€― Unintended solution? I was carefully enough to make one mistake, but I was not carefully enough to make only one mistake. This part is dedicated to @XMPPwocky, who found an unintended solution and I was super surprised.
Finally, let’s patch the vulnerability. It is intuitive to check if the sequence of ALGORITHM has exactly two items:
class VerifyingKey:
# ...SKIPPED...
def verify(self, m, s):
# ...SKIPPED...
sequence = DerSequence()
sequence.decode(digest_info)
_digest_algorithm_identifier, _digest = sequence
sequence = DerSequence()
sequence.decode(_digest_algorithm_identifier)
+ if len(sequence) != 2:
+ raise Exception('invalid sequence length')
_digest_algorithm_identifier = sequence[0]
object_id = DerObjectId()
object_id.decode(_digest_algorithm_identifier)
digest_algorithm_identifier = object_id.value
if digest_algorithm_identifier != '2.16.840.1.101.3.4.2.1':
raise Exception('invalid digest algorithm identifier')
_null = sequence[1]
null = DerNull()
null.decode(_null)
octet_string = DerOctetString()
octet_string.decode(_digest)
if hashlib.sha256(m).digest() != digest:
raise Exception('mismatch digest')
return True
Is it sufficient? Surprisingly no. @XMPPwocky found a problem when decoding for DerNull – the latest version of pycryptodome (v3.15.0) does not raise an exception if it contains a non-empty payload!
Normally, a DerNull object consists of two bytes: 05 00. However, the below bytes are considered valid a DerNull object:
05 27 64696420796f75206c6f6f6b20757020666f7220746865206861736820696e207061727420493f
We can experiment this:
from Crypto.Util.asn1 import DerNull
_null = bytes.fromhex('05 27 64696420796f75206c6f6f6b20757020666f7220746865206861736820696e207061727420493f')
null = DerNull()
null.decode(_null) # This does not raise an exception!
With that said, we can actually stuff garbage in the parameter field. Once again it became the math puzzle in part III.
I tried my best to verify everything except the intended issue while implementing. The unintended solution is definitely a lesson for me. π€© VERY NICE CATCH! π | 2,864 | 10,348 | {"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": 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-2024-26 | latest | en | 0.717479 |
http://www.maa.org/mathland/mathtrek_12_18_06.html | 1,369,525,980,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706472050/warc/CC-MAIN-20130516121432-00062-ip-10-60-113-184.ec2.internal.warc.gz | 581,562,057 | 4,532 | # Ivars Peterson's MathTrek
December 18, 2006
### Rankings, Tournaments, and Playoffs
So many teams and so little room at the top. Which team becomes the national champion in U.S. college football rests on rankings, which reflect the opinions of poll participants (and, nowadays, also computer ratings).
This year, Ohio State plays for the national title in the championship bowl game. And its opponent will be Florida rather than Michigan because the "experts," in their voting, judged that Florida would be the more worthy opponent.
This outcome hasn't pleased everyone, and, as happens nearly every year, many have criticized the vagaries of the ranking system for allowing apparently flawed or unfair outcomes.
Similar problems in determining which team or player deserves a national or year-end championship or how they ought to be seeded for a tournament occur in other sports that also employ elaborate rating schemes to rank teams or players.
In a paper published in a recent issue of SIAM Review, Paul K. Newton and Kamran Aslam of the University of Southern California argue against the widespread belief that it is possible, with just the right tweaking, to come up with a ranking system that yields reasonable results and eliminates logical inconsistencies—and, hence, settles all arguments, leaving everyone satisfied.
"The philosophy behind these systems is that there should be a player or team that 'deserves' to be recognized as 'the best,' and if only the correct method were found, such a team could be unambiguously chosen," Newton and Aslam write.
But it's impossible to make such a guarantee. The argument hinges on an application of a theorem proved by economist Kenneth Arrow. He showed that, under certain reasonable assumptions, there is no method for constructing social preferences (rankings) from arbitrary individual ones (votes).
One such assumption is that, when team A is ranked higher than team B, and team B is ranked higher than team C, then team A is ranked higher than team C. This seems like a reasonable requirement.
But voting schemes can readily undermine this desirable result, even when just three teams or players are involved.
Newton and Aslam cite the example of the selection of the top men's tennis player in 2002. That year, Pete Sampras won the U.S. Open, Andre Agassi won the Australian Open, and Lleyton Hewitt won Wimbledon. Suppose that three judges voted for player of the year as shown in the table below.
Judge Sampras Agassi Hewitt American 1 2 3 British 2 3 1 Australian 3 1 2
Note that two out of the three judges ranked Sampras ahead of Agassi, two out of three put Agassi ahead of Hewitt, and two of three put Hewitt ahead of Sampras!
"Such outcome-based methods based on voting, as the one used to crown the NCAA national football champion, very often produce logical inconsistencies that are the basis for arguments that cannot be settled rationally," Newton and Aslam contend.
They add, "The amount of energy and effort spent on arguing over rankings of all types (particularly in college football, where few games are played compared to the total number of teams involved, but also regarding the notorious U.S. News and World Report Annual Ranking of Colleges) is an indication of the pervasiveness of Arrow's theorem."
Newton and Aslam favor a different, probabilistic approach for ranking teams or players, which makes predictions about outcomes in head-to-head competitions. To rank tennis players, for instance, the idea is to run thousands of simulated tournaments with players randomly ordered in fictitious draws before a real tournament begins and to use the accumulated statistical winning distributions as the basis for seeding the actual tournament. The player most likely to win the tournament based on the simulations would become the top seed in the real draw, and so on.
The researchers describe how to do such Monte Carlo simulations for tennis. They argue that, with enough computer power, such simulations could be run for college football as a way to rank the teams.
Although Newton and Aslam don't specifically call for a playoff system, it's easy to see that the top eight teams, as determined by such simulations at the end of the season, could then vie for the national championship in a set of playoff games.
"Moving toward systems that are probabilistic and predictive would finesse the inherent inconsistencies guaranteed by Arrow's impossibility theorem," Newton and Aslam conclude, "and would better reflect the reality that a victory of a lower-ranked player over a high-ranked one in a single match is not necessarily an inconsistent outcome."
References:
Klarreich, E. 2002. Election selection. Science News 162(Nov. 2):280-282. Available at http://www.sciencenews.org/articles/20021102/bob8.asp.
Newton, P.K., and K. Aslam. 2006. Monte Carlo tennis. SIAM Review 48(No. 4):722-742. Abstract available at http://dx.doi.org/10.1137/050640278.
Peterson, I. 2005. Ranking college football teams. MAA Online (Nov. 14).
______. 2005. Winning at tennis. MAA Online (June 13).
______. 2004. College football, rankings, and wandering monkeys. MAA Online (Sept. 8).
______. 2003. Election reversals. MAA Online (Oct. 20).
______. 1998. Who's really no. 1? MAA Online (Dec. 14).
______. 1998. How to fix an election. MAA Online (Nov. 2).
______. 1997. Getting slammed in tennis. MAA Online (Dec. 1).
Stefani, R.T. 1997. Survey of the major world sports rating systems. Journal of Applied Statistics 24(December):635-646. Abstract available at http://dx.doi.org/10.1080/02664769723387. | 1,237 | 5,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.968698 |
https://www.coursera.org/learn/machine-learning?utm_source=gg&utm_medium=sem&campaignid=693373197&adgroupid=36745103515&device=c&keyword=machine%20learning%20andrew%20ng&matchtype=e%E2%89%A0twork=g&devicemodel=&adpostion=1t1&creativeid=156061453588&hide_mobile_promo&gclid=Cj0KEQjwt6fHBRDtm9O8xPPHq4gBEiQAdxotvNEC6uHwKB5Ik_W87b9mo-zTkmj9ietB4sI8-WWmc5UaAi6a8P8HAQ | 1,571,188,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660829.5/warc/CC-MAIN-20191015231925-20191016015425-00545.warc.gz | 831,120,147 | 151,841 | 8,822,578 recent views
#### 100% online
Start instantly and learn at your own schedule.
#### English
Subtitles: Chinese (Simplified), English, Hebrew, Spanish, Hindi, Japanese...
### Skills you will gain
Logistic RegressionArtificial Neural NetworkMachine Learning (ML) AlgorithmsMachine Learning
#### 100% online
Start instantly and learn at your own schedule.
#### English
Subtitles: Chinese (Simplified), English, Hebrew, Spanish, Hindi, Japanese...
### Syllabus - What you will learn from this course
Week
1
2 hours to complete
## Introduction
5 videos (Total 42 min), 9 readings, 1 quiz
5 videos
Supervised Learning12m
Unsupervised Learning14m
Machine Learning Honor Code8m
What is Machine Learning?5m
How to Use Discussion Forums4m
Supervised Learning4m
Unsupervised Learning3m
Who are Mentors?3m
Lecture Slides20m
1 practice exercise
Introduction10m
2 hours to complete
## Linear Regression with One Variable
7 videos (Total 70 min), 8 readings, 1 quiz
7 videos
Cost Function - Intuition II8m
Model Representation3m
Cost Function3m
Cost Function - Intuition I4m
Cost Function - Intuition II3m
Lecture Slides20m
1 practice exercise
Linear Regression with One Variable10m
2 hours to complete
## Linear Algebra Review
6 videos (Total 61 min), 7 readings, 1 quiz
6 videos
Matrix Matrix Multiplication11m
Matrix Multiplication Properties9m
Inverse and Transpose11m
Matrices and Vectors2m
Matrix Vector Multiplication2m
Matrix Matrix Multiplication2m
Matrix Multiplication Properties2m
Inverse and Transpose3m
Lecture Slides10m
1 practice exercise
Linear Algebra10m
Week
2
3 hours to complete
## Linear Regression with Multiple Variables
8 videos (Total 65 min), 16 readings, 1 quiz
8 videos
Gradient Descent in Practice II - Learning Rate8m
Features and Polynomial Regression7m
Normal Equation16m
Normal Equation Noninvertibility5m
Working on and Submitting Programming Assignments3m
Setting Up Your Programming Assignment Environment8m
Access MATLAB Online and the Updated Exercise Files3m
Installing Octave on Windows3m
Installing Octave on Mac OS X (10.10 Yosemite and 10.9 Mavericks and Later)10m
Installing Octave on Mac OS X (10.8 Mountain Lion and Earlier)3m
Installing Octave on GNU/Linux7m
More Octave/MATLAB resources10m
Multiple Features3m
Gradient Descent in Practice I - Feature Scaling3m
Gradient Descent in Practice II - Learning Rate4m
Features and Polynomial Regression3m
Normal Equation3m
Normal Equation Noninvertibility2m
Programming tips from Mentors10m
Lecture Slides20m
1 practice exercise
Linear Regression with Multiple Variables10m
5 hours to complete
## Octave/Matlab Tutorial
6 videos (Total 80 min), 1 reading, 2 quizzes
6 videos
Plotting Data9m
Control Statements: for, while, if statement12m
Vectorization13m
Lecture Slides10m
1 practice exercise
Octave/Matlab Tutorial10m
Week
3
2 hours to complete
## Logistic Regression
7 videos (Total 71 min), 8 readings, 1 quiz
7 videos
Cost Function10m
Simplified Cost Function and Gradient Descent10m
Multiclass Classification: One-vs-all6m
Classification2m
Hypothesis Representation3m
Decision Boundary3m
Cost Function3m
Simplified Cost Function and Gradient Descent3m
Multiclass Classification: One-vs-all3m
Lecture Slides10m
1 practice exercise
Logistic Regression10m
4 hours to complete
## Regularization
4 videos (Total 39 min), 5 readings, 2 quizzes
4 videos
Regularized Logistic Regression8m
The Problem of Overfitting3m
Cost Function3m
Regularized Linear Regression3m
Regularized Logistic Regression3m
Lecture Slides10m
1 practice exercise
Regularization10m
Week
4
5 hours to complete
## Neural Networks: Representation
7 videos (Total 63 min), 6 readings, 2 quizzes
7 videos
Model Representation II11m
Examples and Intuitions I7m
Examples and Intuitions II10m
Multiclass Classification3m
Model Representation I6m
Model Representation II6m
Examples and Intuitions I2m
Examples and Intuitions II3m
Multiclass Classification3m
Lecture Slides10m
1 practice exercise
Neural Networks: Representation10m
4.9
28665 Reviews
## 40%
started a new career after completing these courses
## 38%
got a tangible career benefit from this course
### Top reviews from Machine Learning
By HSMar 3rd 2018
My first and the most beautiful course on Machine learning. To all those thinking of getting in ML, Start you learning with the must-have course. Thanks Andrew Ng and Coursera for this amazing course.
By CCJun 20th 2018
good course; just 2 suggestions: improve the skew data part (week 6) and furnish the formula to evaluate the number of iteration in the window from image dimension, window dimension and step (week 11)
## Instructor
### Andrew Ng
CEO/Founder Landing AI; Co-founder, Coursera; Adjunct Professor, Stanford University; formerly Chief Scientist,Baidu and founding lead of Google Brain | 1,227 | 4,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-43 | latest | en | 0.750235 |
https://www.scribd.com/doc/298789127/Aerodynamics-1-Equations | 1,569,156,929,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00454.warc.gz | 996,405,821 | 59,249 | You are on page 1of 1
# MAE252 Test 1 Equation Sheet
E. Scott Sexton
Equations:
2
## p = RT ; R air = 287.057 sm2 K
L = N cos() Asin()
D = N sin() + Acos()
**These use not **
R TE
R TE
C V 2 c
N 0 = L E (pu cos( ) + u sin( ))dSu + LE (pl cos( ) l sin( ))dSl = N 2
R TE
R TE
A0 = L E (pu sin( ) + u cos( ))dSu + LE (pl sin( ) + l cos( ))dSl
R TE
R TE
0
M LE
= L E [(pu cos( ) + u sin( ))x (pu sin( ) u cos( ) y)]dSu + LE [(pl cos( ) + l sin( ))x + (pl sin( ) + l cos( )) y]dSl
2
q 12 V
CL
L
q S ,
cl
CD
D
q S ,
cd
CN
N
q S
CA
A
q S
CM
M
q Sl ,
L0
q c
D0
q c
cm
M0
q c 2
Rc
Rc
dy
dy
cn = 1c [ 0 (C p,l C p,u )d x + 0 (c f ,u d xu + c f ,l d xl )d x]
Rc
Rc
dy
dy
ca = 1c [ 0 (C p,u d xu C p,l d xl )d x + 0 (c f ,u + c f ,l )d x]
Rc
Rc
Rc
Rc
dy
dy
dy
dy
cm L E = c12 [ 0 (C p,u C p,l )x d x + 0 (c f ,u d xu + c f ,l d xl )x d x + 0 (C p,u d xu + c f ,u ) yu d x + 0 (C p,l d xl + c f ,l ) yl d x]
cl = cn cos() ca sin()
cd = cn sin() + ca cos()
x cp =
x c0 p
0
M LE
N0
CM c
=
+
1
4
C M c = C M LE +
cl
4
Cp =
Cl
pp
1
2
2 V
CM L E
Cl
2
C f V
2
0
0
M LE
= 4c L 0 + Mc/4
=
V c
Vc
Re = = ; =
V
V
M = a
= p ; a
RT
M 0c
N0
c
4
0
M LE
L0
x cp L 0
= speed o f sound in ai r = 340.9 ms
RR
Volume flow rate= S V~ d S~ = 0 for steady flow
RR
Mass flow rate= S V~ d S~
Continuity= t
d + S V~ d S~ = 0
## But for steady flow t = 0 S V~ d S~ = 0 and V~ = 0
~ d S~)V~ =
(
V
P d S~
S
S
RR
F~ =
(V~ d S~)V~
S
General Notes:
For flow similarity (C L , Cd same):
i) Streamline patterns are geometrically similar
p
ii) Distribution of VV , p , TT are the same
iii) Force coefficients are the same
In recilinear flight L W
For incompressible flow is constant
For M<0.3,
flow is incompressible
M0.0
p
p
T and a T
Quarter chord is the aerodynamic center
For flat plate = 0 | 867 | 1,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.60511 |
https://socratic.org/questions/what-is-7-and-9-10-as-an-improper-fraction | 1,585,521,140,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00181.warc.gz | 725,816,667 | 6,362 | # What is 7 9/10 as an improper fraction?
Nov 6, 2016
$7 \frac{9}{10} = \frac{10 \times 7 + 9}{10} = \frac{79}{10}$
#### Explanation:
Remember that a whole number is:
$\frac{8}{8} , \frac{3}{3} , \frac{11}{11} , \frac{40}{40} , e t c$
In this case we are working with the fraction 'tenths'.
Each whole number is $\frac{10}{10}$
In 7 whole numbers there will be $7 \times 10 = 70$ tenths
$\frac{70}{10} = 7$
There are also 9 extra 'tenths'.
In total there are $10 \times 7 + 9 = 79$ tenths.
$7 \frac{9}{10} = \frac{10 \times 7 + 9}{10} = \frac{79}{10}$
Nov 10, 2016
With practice you would start to use shortcuts and speed up solving this type of question.
$\frac{79}{10}$
#### Explanation:
Write as $7 + \frac{9}{10}$
Multiply by 1 and you do not change the value. But 1 comes in many forms so you can change the way a number looks without changing its value.
$\text{ "color(green)([7color(magenta)(xx1)]+9/10" "->" } \left[7 \textcolor{m a \ge n t a}{\times \frac{10}{10}}\right] + \frac{9}{10}$
$\text{ "70/10+9/10" "=" "(70+9)/10" " =" } \frac{79}{10}$ | 396 | 1,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-16 | latest | en | 0.740592 |
http://mathoverflow.net/questions/66769/how-to-obtain-tail-bounds-for-a-linear-combination-of-dependent-and-bounded-rand | 1,469,702,213,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.65/warc/CC-MAIN-20160723071028-00302-ip-10-185-27-174.ec2.internal.warc.gz | 159,214,426 | 14,675 | # How to obtain tail bounds for a linear combination of dependent and bounded random variables?
Hi everyone,
Note: This question is a general case and edited version of my previous question How to obtain tail bounds for a sum of dependent and bounded random variables?''.
I am looking for tail bounds (preferably exponential) for a linear combination of dependent and bounded random variables.
consider $$K_{ij}=\sum_{r=1}^N\sum_{c=1}^N W_{ir}C_{rc}W_{jc}$$ where $i \neq j$, $W\in \{+1, -1\}$ and $W$ follows $\operatorname{Bernoulli}(0.5)$, and $C=\operatorname{Toeplitz}(1, \rho, \rho^2, \ldots, \rho^{N-1})$, $0 \leq \rho < 1$.
I will be to happy if you give me any pointer to how I can evaluate the moment generating function of $K_{ij}$ to have bound for $Pr\{K_{ij} \geq \epsilon\}\leq \min_s\exp(-s\epsilon)E[\exp(K_{ij}s)]$ based on chernoff bound.
For a hint I put the following case, that is in fact a special case of the above one. Consider $$K_{ij}=\sum_{r=1}^N\sum_{c=1}^N W_{ir}W_{jc}$$ where $i \neq j$, $W\in \{+1, -1\}$ and $W$ are i.i.d. random variables $\operatorname{Bernoulli}(0.5)$.
(Answer: As it is noted in Ori's comment, it can be considered as the multiplication of two independent Binomial random variables, i.e., $$K_{ij}=\left(\sum_{r=1}^N W_{ir}\right)\left(\sum_{c=1}^N W_{jc}\right)$$, then the moment generating function can be evaluated and then by using the Chernoff we can have a tail bound for $K_{ij}$.)
$=(\sum_{r=1}^N W_r )^2$ – Ori Gurel-Gurevich Jun 2 '11 at 21:07
Hence $K=(S-E(S))^2$ where $S$ is Binomial$(N,1/2)$? Standard Cramer exponential inequalities work well in this context. – Did Jun 2 '11 at 21:08
Frazad, it is customary to add a note when you edit a question in such a way that makes the answers/comments irrelevant. To the point - what is $K$ now? Each $K_{i,j}$ is still a product of 2 independent RVs. – Ori Gurel-Gurevich Jun 2 '11 at 21:31 | 633 | 1,912 | {"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": 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} | 3.421875 | 3 | CC-MAIN-2016-30 | latest | en | 0.796807 |
https://www.meritnation.com/ask-answer/question/question-no-1-24-plzz-solv-it-the-u-calculate-i-22-if-the-sp/some-basic-concepts-of-chemistry/12864647 | 1,675,723,278,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00300.warc.gz | 906,168,760 | 8,859 | Question no. 1.24...plzz solv it..
N2 + 3H2 -------> 2NH3
1 mol + 3 Mol = 2 moles
No. of Moles of N2 = 206/28 = 7.35
No. of moles of H2 = 103/2 = 51.5
From the reaction,
7.35 moles of N2 will react with (7.35x3) = 22.05 M of H2
Thus, H2 is in excess.
a) Mass of N2 + Mass of H2 = 206 g + (22.05 x2) = 206 + 44.1 = 250 g of ammonia
b) H2 remains unreacted
c) Mass of unreacted H2 = (51.5 - 22.05) X 2 = 58.9 g
• 0
e it by stoicheometry the left will be 142.86
mass will be 2.43*103.
then hydrogen will remain
mass 570gms
• -1
What are you looking for? | 252 | 558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-06 | latest | en | 0.833 |
https://www.connectingsingles.com/games/kraak-285.htm | 1,653,443,955,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00315.warc.gz | 779,227,710 | 12,276 | # kraak Game
Build the figure from the left side using the blocks
Instructions: Build the figure from the left side using the blocks
I did it!
I got to the end of the game!!
let me say that at the beginning it seemed to be a very stupid game and I couldn't figure out how it worked...
in the middele I was colse to give it up but if you try over and over again try not to stick with an idea you'll arrive to the success
very easy
dont like this game either. the first one isnt even possible there is something missing to even make the pieces fit the correct way
way...too...easy
I dont get it... i tried to copy the pic but there is no way to roatet the shapes
i dont get this game at all
i did it ...but where's my score?
I did it
How does it work? I finished but it doesn't accept the right answer.
im stuck in the 1/2 way point , its like it dose not have all the blocks ???
this game is completely pointless. so you make the shape...BFD...then what...you stare at it??? no score, no time, no next round....nothing....it took longer to ask why cs is leaving this game on here then it did for me to get bored of it
I would suggest saving your time and finding another game
I didn't see any scores or any timer?? got all the way to the end though ???
Liked this game!
ya nothing happens when you finish, but i found a loop hole that makes the game twice as fast, everytime you complete a puzzle keep clicking on the puzzle u just finished and you will get a second round of applause and go further up the trail
seems it only accepts 1 solution, and not anything that also applies.
The half way sux....It seems like all the pieces arent there!!!
doesnt react when its fiished ???
I give up its not doing anything for me
This game just kraaks me up!
yup have way sux
This is a very cool game....but you should know that the game and the timer start automatically when the game loads! Hi JG!
the half way puzzle has me stumped. its 2 in 1...
HARD WORK!!!!!!!!!!! lol
We have updated the scoring, so that the scores will be higher.
The game is timed...the faster you complete the puzzles, the higher your score. There are 16 puzzles total. We have updated it so that it will give you more time before your score goes negative.
#### High Scores
Gamer Score Date
Artist1139,773,656 12/13/2009
barriepaul9,761,28412/17/2009
DerFranz9,222,8719/8/2014
fibalt9,203,92211/10/2010
ExtremeDream229,158,82311/4/2013
LionsBabe9,100,2167/4/2011
ilogic09,085,5373/12/2012
teddybeerke888,995,71110/21/2012
Yuck7778,976,7332/20/2012
Tiger_Dan8,959,35612/3/2009
Thazager8,930,1326/29/2010
Gekata8,888,5328/6/2010
hsiri8,852,0531/20/2010
prehinite8,811,08011/19/2016
peperkoekenhart8,798,1866/12/2011
dannycopper8,794,11111/9/2011
germantourist8,696,5961/6/2016
shy_night_owl8,683,9626/18/2010
Happyone8,637,0237/8/2010
thar3088,635,5425/14/2016
hugzz8,415,0094/24/2013
Tazeromaxed8,386,9279/13/2012
DannyBoy228,379,6869/12/2010
levanchuk8,358,8298/6/2016
koskrep8,296,4288/18/2011
Lammycool8,277,1331/9/2010
bauer117,915,39910/22/2011
GingerBe7,860,7236/18/2010
venga7,836,18510/4/2011
combo717,834,1309/9/2013 | 999 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-21 | latest | en | 0.970715 |
https://www.carvadia.com/what-is-source-population-in-case-control-study/ | 1,657,191,208,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104690785.95/warc/CC-MAIN-20220707093848-20220707123848-00363.warc.gz | 717,364,913 | 13,355 | • July 7, 2022
### What Is Source Population In Case Control Study?
What is source population in case control study? The source population is the population that gives rise to the cases included in the study. If a cohort study were undertaken, we would define the exposed and unexposed cohorts (or several cohorts) and from these populations obtain denominators for the incidence rates or risks that would be calculated for each cohort.
## What is a reference population in epidemiology?
1. a subset of a target population that serves as a standard against which research findings are evaluated. For example, consider an investigator examining the effectiveness of eating disorder prevention programs at 4-year colleges and universities in the United States.
## What is a study population in research?
Study population: The group of individuals in a study. In a clinical trial, the participants make up the study population. The study population might, for example, consist of all children under 2 years of age in a community.
## What is the target population example?
The target population is the entire population, or group, that a researcher is interested in researching and analysing. Examples of a target population are a company's customer base, the population of particular country, the students at a particular university or tenants of a housing association.
## What sources can be used for controls in a case-control study?
Methods of Control Sampling
• Survivor sampling: This is the most common method.
• Case-base sampling (also known as "case-cohort" sampling): Controls are selected from the population at risk at the beginning of the follow-up period in the cohort study within which the case-control study was nested.
• ## Related advise for What Is Source Population In Case Control Study?
### What is an example of a case-control study?
For example, in a case-control study of the association between smoking and lung cancer the inclusion of controls being treated for a condition related to smoking (e.g. chronic bronchitis) may result in an underestimate of the strength of the association between exposure (smoking) and outcome.
### What is a source population example?
Generally speaking, the source population is the population from which your study subjects are drawn. In your example, that would be the 100,000 screened individuals under a specific assumption. Namely, that the screened population is an entire population.
### What is source population and study population?
A source population is a subset of a target population: it is a smaller population within a larger target population from which a sample is drawn. A study population is common term for a sample drawn from a source population: this is a confusing term because a “study population” is not a population, it's a sample.
### What does target population mean?
The target population is the group of individuals that the intervention intends to conduct research in and draw conclusions from. In cost-effectiveness analysis, characteristics of the target population and any subgroups should be described clearly.
### What is population in qualitative research?
In research terminology the Population can be explain as a comprehensive group of individuals, institutions, objects and so forth with have a common characteristics that are the interest of a researcher. Any value which is identified or measured from the characteristics of entire population can be called as Parameter.
### What is population in research methods?
A research population is generally a large collection of individuals or objects that is the main focus of a scientific query. A research population is also known as a well-defined collection of individuals or objects known to have similar characteristics.
### What are examples of populations?
Population is the number of people or animals in a particular place. An example of population is over eight million people living in New York City.
### What is accessible population?
Accessible population. the portion of the population to which the researcher has reasonable access; may be a subset of the target population. May be limited to region, state, city, county, or institution.
### What is the difference between target population and accessible population?
The target population usually has varying characteristics and it is also known as the theoretical population. The accessible population is the population in research to which the researchers can apply their conclusions.
### How do population differs with sample?
A population is the entire group that you want to draw conclusions about. A sample is the specific group that you will collect data from. The size of the sample is always less than the total size of the population. In research, a population doesn't always refer to people.
### What is density sampling?
A method of selecting controls in a Case-control study in which cases are sampled only from incident cases over a specific time period and controls are sampled and interviewed throughout that period (rather than simply at one point in time, such as the end of the period).
### What are the primary sources of bias in case-control studies?
Main sources of potential bias were a non-concurrent selection of controls with respect to cases, the use of control diagnoses possibly caused by pesticide exposure in hospital-based studies, and non-participation of selected eligible subjects.
### How do you create a cohort study?
• Identify the study subjects; i.e. the cohort population.
• Obtain baseline data on the exposure; measure the exposure at the start.
• Select a sub-classification of the cohort—the unexposed control cohort—to be the comparison group.
• Follow up; measure the outcomes using records, interviews or examinations.
• ### How do you distinguish between a case-control and a cohort study?
Whereas the cohort study is concerned with frequency of disease in exposed and non-exposed individuals, the case-control study is concerned with the frequency and amount of exposure in subjects with a specific disease (cases) and people without the disease (controls).
### What is the difference between matched and unmatched case-control study?
Abstract. Multiple control groups in case-control studies are used to control for different sources of confounding. For example, cases can be contrasted with matched controls to adjust for multiple genetic or unknown lifestyle factors and simultaneously contrasted with an unmatched population-based control group.
### What is the difference between cross sectional and case-control study?
Cross sectional studies are used to determine prevalence. They are relatively quick and easy but do not permit distinction between cause and effect. Case controlled studies compare groups retrospectively. They seek to identify possible predictors of outcome and are useful for studying rare diseases or outcomes.
### What is a source population ecology?
Quick Reference. An ecological model that is used to describe population changes in two habitats, both occupied by the same species. One habitat is of high quality and allows a population to increase (i.e. births + immigration > deaths + emigration), leading to a surplus. This is the source.
### What is base population?
The base population refers to the number of people in a given area (e.g. a nation, province, city, etc) to which a specific vital rate applies, that is, the denominator of the crude birth rate or death rate; that population determined by a census.
### What is a sink population?
Sink populations exist in low quality habitat patches that would not be able to support a population in isolation, and without the contribution of individuals from a source population, would become extinct.
### How do you define population study?
Study population: The group of individuals in a study. In a clinical trial, the participants make up the study population. The study population might, for example, consist of all children under 2 years of age in a community.
### What is switchover in epidemiology?
♦ A switchover bias occurs if the target parameter is on the opposite side of the null value from the parameter actually being estimated in one's study.
### What is population in sampling?
A population is a complete set of people with a specialized set of characteristics, and a sample is a subset of the population. The study population is the subset of the target population available for study (e.g. schizophrenics in the researcher's town). The study sample is the sample chosen from the study population.
### What is a population PDF?
A population is defined as a group of individuals of the same species living and interbreeding within. a given area. Members of a population often rely on the same resources, are subject to similar.
### How do you calculate target population?
You estimate the target population by multiplying the total population by the percentage in a particular category; for example in Table 5.1: Expressing a percentage as a decimal number is easy, e.g. 90% is the same as 0.9; 23% is the same as 0.23, and 4% is the same as 0.04.
### What are the two types of sampling techniques?
There are two types of sampling methods:
• Probability sampling involves random selection, allowing you to make strong statistical inferences about the whole group.
• Non-probability sampling involves non-random selection based on convenience or other criteria, allowing you to easily collect data.
• ### What is population in quantitative studies?
A population consists of all the objects or events of a certain type about which researchers seek knowledge or information. From their observations about the sample, researchers make generalizations about the population from which the sample was chosen.
### What is population in Research PDF?
A population refers to any collection of specified group of human beings or of non-human entities such as objects, educational institutions, time units, geographical areas, prices of wheat or salaries drawn by individuals. | 1,906 | 10,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-27 | latest | en | 0.942323 |
http://www.thestudentroom.co.uk/showthread.php?t=398748&page=3&p=8821740 | 1,386,799,264,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164050279/warc/CC-MAIN-20131204133410-00012-ip-10-33-133-15.ec2.internal.warc.gz | 560,724,668 | 45,360 | You are Here: Home
STEP maths I, II, III 1991 solutions
Maths and statistics discussion, revision, exam and homework help.
Announcements Posted on
1. Re: STEP maths I, II, III 1991 solutions
II/1, Let t = lambda, alpha = z
h(x) = ax^2+bx+c = a(x^2+bx/a+c/a) = a((x+b/2a)^2-(b/2a)^2+c/a), so we require (b/2a)^2 = c/a, so b^2 = 4ac, i.e.discriminant = 0
3x^2+4x+t(x^2-2) = (3+t)x^2+4x-2t where a=3+t, b=4,c=-2t and we require that 16 = -8t(3+t), 0 = 3t+t^2+2, t= -1, -2.
(1) f(x)-g(x) = 2(x+1)^2
(2) f(x)-2g(x) = (x+2)^2
f(x) = 2x(1)-(2) = 4(x+1)^2 - (x+2)^2
g(x) = (1)-(2) = 2(x+1)^2 - (x+2)^2, hence A = 4, B = -1, C = 2, D = -1, m = 1, n = 2.
3x^2+4x+t(x^2+z) = (3+t)x^2+4x+zt, so a=3+t,b=4, c=zt, we require b^2 = 4ac, so 16 = 4(3+t)zt, 0 = 3zt+zt^2-4, a=z,b=3z,c=-4 then 9z^2 = -16z, z = -16/9.
2. Re: STEP maths I, II, III 1991 solutions
Ooh, very neat solution.
Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
3. Re: STEP maths I, II, III 1991 solutions
(Original post by Rabite)
Ooh, very neat solution.
Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
Yeah, that's the answer I got, double-checked it with Maple just now. But I really can't be bothered to substitute in, maybe you plug it back wrongly?
4. Re: STEP maths I, II, III 1991 solutions
(Original post by Rabite)
Ooh, very neat solution.
Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
I'm not too sure, I breifly did it and got u = e^-x and -(e^x+e^-x)
5. Re: STEP maths I, II, III 1991 solutions
(Original post by Decota)
I'm not too sure, I breifly did it and got u = e^-x and -(e^x+e^-x)
That's the first part, Rabite is talking about the second part of the question.
6. Re: STEP maths I, II, III 1991 solutions
(Original post by khaixiang)
That's the first part, Rabite is talking about the second part of the question.
oh i see. 2ln(cosh½x) is indeed correct.
Last edited by gyrase; 30-05-2007 at 20:52.
7. Re: STEP maths I, II, III 1991 solutions
Well I'll type it up
The quadratic formula involves fractions, and I hate fractions, so let's complete the square:
Or
coshx-sinhx is uh...e^{-x} and -(coshx+sinhx) is -e^x.
y`>0 at x=0 so we have to choose the + solution.
Integrate and find the +C to get
.
For the next bit, let u = dy/dx for a start:
Or .
(Consider double "angle" formulae)
But u = dy/dx:
Tanh = sinh / cosh, which is a derivative over its function:
The solution passes through (0,0); so C =0.
(Taking the negative solution when we square rooted earlier results in something that can't pass through (0,0))
Is there no /cosech in Latex?
Last edited by Rabite; 30-05-2007 at 23:22.
8. Re: STEP maths I, II, III 1991 solutions
(Original post by Rabite)
Is there no /cosech in Latex?
TSR's LaTeX is messed up then, the proper code is \csch
Nice work btw, though "1" and "coth(x)" is missing somewhere in the middle. All the STEP III questions are so unbearably long and tedious Good thing the style was revamped.
9. Re: STEP maths I, II, III 1991 solutions
I tried \csch too
Anyway. I think I've found the missing buggers, but maybe I've still missed one...
I kept doing stuff like \sinhx which just didn't come out as anything.
Thanks for that
10. Re: STEP maths I, II, III 1991 solutions
(Original post by Speleo)
Am I going crazy or are matrices no longer on the STEP syllabus?
http://www.ocr.org.uk/Data/publicati...ficat98487.pdf
*bawls* Matrices are easy:/
(Original post by Speleo)
Also, does anyone have the '05 papers? (II and III '06 are available if you go back a couple hundred pages to last June in this forum).
Drop me a pm with your email, I think I have II and III from 2005, but lack I from 2005 and I from 2004, all others I think I have.
11. Re: STEP maths I, II, III 1991 solutions
(Original post by nota bene)
*bawls* Matrices are easy:/
*fills a 4x4 Matrix's elements with 1=0 and drops it on nota*
Maybe they just forgot to put it in the syllabus >>;;
They're so tedious that they didn't deserve their own section or something.
12. Re: STEP maths I, II, III 1991 solutions
(Original post by nota bene)
I'm attempting I/3 I'll see if I sort it out...
Okay, I'll post what I have now (probably should scan a sketch tomorrow as well), but it is not finished, and likely contains a few mistakes
Spoiler:
Show
Some observations I made was that if we let the first complex number have a length a the second will have a+ad etc. as long as we are in the first quadrant. This gives
Maybe I'm missing something, but I don't see how this works; you seem to have gone directly from talking about the lengths of the vectors to their actual (complex) values. In particular, I don't understand how you're taking into account the value of theta.
For what it's worth, I think the correct formula for the first bit is:
Spoiler:
Show
13. Re: STEP maths I, II, III 1991 solutions
(Original post by Rabite)
II, 2:
I'm not entirely sure how to finish this one off, so someone else do that.
We find the equations for the asymptotes meet at (1,-1), and your guess that the axes are at +/-45 degrees seems a good one.
So we suspect that the axes have equations x+y=1, x-y =2. You could probably stop there, but partly to reassure myself, and partly for the hell of it, we can go further:
We look for such that resembles the given equation.
Easiest is to look first at the coefficients of x and y, which tells us that , and then the coefficient of x^2 tells us .
Expanding out, we find and so using the given equation for the curve, we obtain .
So we have explicitly obtained the equation of a hyperbola in terms of our new axes, confirming our guess.
Last edited by DFranklin; 31-05-2007 at 00:50.
14. Re: STEP maths I, II, III 1991 solutions
Someone let me know when stuff's finished.
15. Re: STEP maths I, II, III 1991 solutions
(Original post by DFranklin)
Maybe I'm missing something, but I don't see how this works; you seem to have gone directly from talking about the lengths of the vectors to their actual (complex) values. In particular, I don't understand how you're taking into account the value of theta.
For what it's worth, I think the correct formula for the first bit is:
Spoiler:
Show
Yes, I have towards the end, but it needs to be mentioned in the beginning. In general that whole post is totally messed up and should not be looked at.
Someone else feel free to do this question, or I'll try to get it correct later.
generalebriety: I think my solution to I/6 is okay unless I've made a stupid mistake somewhere. (added Decota's comment, just to be on the safe side; I had seen it was not differentiated correctly, but not bothered to mention it as it was totally wrong either way.) Also take away this question as 'partially solved' until I fix it; it is horrible at the moment!
16. Re: STEP maths I, II, III 1991 solutions
STEP I question 8
(a) Let .
A shorter alternative due to DFranklin:
(by substitution)
(recombining)
Using this last result:
Let . Also, . Hence:
(b) Let
Boundaries:
Last edited by justinsh; 31-05-2007 at 09:23.
17. Re: STEP maths I, II, III 1991 solutions
(Original post by justinsh)
My part (a) is rather long, is there a shorter way?
. Sub in the 2nd integral to get
. Recombine the two integrals to get
18. Re: STEP maths I, II, III 1991 solutions
STEP I questions 4
Please tell me what you think, I'm not certain of what I did.
Attached Thumbnails
Last edited by justinsh; 02-06-2007 at 07:48.
19. Re: STEP maths I, II, III 1991 solutions
*Bump*
I will rep anyone who posts a solution for II/4 in its entirety. (the trigonometry one)
kthnx.
Last edited by gyrase; 01-06-2007 at 22:48.
20. Re: STEP maths I, II, III 1991 solutions
Hmm...
I can do the first bit and the last bit.
But not the middle bit (unless we resort to complex numbers, but that's foul play right?)
Step 2: Register
Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
this is what you'll be called on TSR
2. this can't be left blank
never shared and never spammed
3. this can't be left blank
6 characters or longer with both numbers and letters is safer
4. this can't be left empty
1. By completing the slider below you agree to The Student Room's terms & conditions and site rules
2. Slide the button to the right to create your account
You don't slide that way? No problem.
Last updated: June 10, 2013
Study resources | 2,562 | 8,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2013-48 | latest | en | 0.886333 |
https://www.ams.org/publicoutreach/feature-column/fcarc-barcodes5 | 1,524,785,955,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00112.warc.gz | 729,422,976 | 8,811 | ## Barcodes: ISBN Numbers
5. ISBN Numbers
ISBN stands for International Standard Book Number and nearly all materials published anywhere in the world are now assigned such a number because of the many advantages in being able to track and identify books by their ISBN number. The ISBN number is a 10-digit number organized into 4 blocks which are separated by hyphens, which are an essential part of the system. The approach used in the ISBN number is that the 4 blocks can vary in size. This approach yields some advantages and some disadvantages. If codes have sections that can vary in length, one needs to have some way of being able to tell where one block ends and the next begins, which increases the length of the number of symbols used in total. Although the ISBN number has 10 information digits, its length is 13 because of the need for the 3 hyphens to separate the blocks. The hyphens could be eliminated if the blocks had specified lengths. (By way of example, the number of partitions of 10 into four nonzero ordered parts where the first part has length 1 or 2 and the last has length 1 is 13.) The first block is a language or country identifier and is partly aimed at giving the language spoken by the target audience for the book. English books for example are assigned a 0 in the first block, while the German language audience uses the code 3 and French the code 2. The code for Malta as a target audience is 99909. The second block is a publisher number, the third block an item number for that publisher, and the last digit is a check digit. Not surprisingly, large publishers have small identification numbers and small publishers have long ones. The pressure of so many publishers and books by publishers is already pushing the limits of this coding system.
The system for creating the check digit in the ISBN number system is simple but clever. The way that it works in general can easily be understood by looking at a specific example.
Example:
What check digit would be assigned to the book, Mathematical Hierarchies and Biology published by the American Mathematical Society?
This book is aimed at the English language audience so a 0 appears in the first block; the American Mathematical Society's publisher code is 8218 (e.g. all AMS books have this number in the second block), and the item number for this book, given to it by AMS, is 0762. Based on these 9 digits we seek the value C (for check digit) of the check digit, which will be placed in the tenth position of the code. We compute 10 times the first digit, 9 times the second, 8 times the third, etc. This gives the calculation:
10(0) + 9(8) + 8(2) + 7(1) + 6(8) + 5(0) + 4(7) + 3(6) + 2(2) + 1(C).
Simplifying we obtain:
0 + 72 + 16 + 7 + 48 + 0 + 28 + 18 + 4 + C.
We now choose C so that the sum is divisible by 11 (i.e. congruent to zero mod 11). Hence, C is 5. The ISBN number for this book is 0-8218-0762-5.
One problem with this system is that since C can be any number from 0 to 10, we are faced with the possibility that the check digit might have two digits, 1 and 0. To get around this problem when the check digit would be 10 the symbol X (for 10 in Roman numerals) is used. This nuisance in the system of having the ISBN consist of symbols that are not all numbers can be avoided by manipulating the item number so as to avoid having X as the check symbol. In practical terms this means that not all the item numbers would be available to a company for use. It is not difficult to prove that the ISBN check digit system catches all single digit substitution errors as well as errors caused by the switching of adjacent digits.
Welcome to the
Feature Column!
These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics.
Read more . . .
Search Feature Column
Feature Column at a glance | 901 | 3,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.891863 |
http://gxb.zzu.edu.cn/oa/DArticle.aspx?type=view&id=201303042 | 1,716,946,226,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00732.warc.gz | 11,959,568 | 10,665 | [1]周云岗.基于迭代分析的多塔悬索桥合理成桥状态确定方法[J].郑州大学学报(工学版),2014,35(01):99-103.[doi:10.3969/j.issn.1671-6833.2014.01.023] Zhou Yungang.Analysis method of reasonable finished dead state of multi-town suspension bridges[J].Journal of Zhengzhou University (Engineering Science),2014,35(01):99-103.[doi:10.3969/j.issn.1671-6833.2014.01.023] 点击复制 基于迭代分析的多塔悬索桥合理成桥状态确定方法() 分享到： var jiathis_config = { data_track_clickback: true };
35卷
2014年01期
99-103
2014-02-28
文章信息/Info
Title:
Analysis method of reasonable finished dead state of multi-town suspension bridges
Author(s):
Zhou Yungang
Keywords:
DOI:
10.3969/j.issn.1671-6833.2014.01.023
A
Abstract:
Based on the traditional calculation method of the finished state of suspension bridge and the finite element method, an iterative algorithm for determining the finished alignment of the main cable of multi-tower suspension bridge is proposed based on Ansys Finite Element Analysis. On this basis, the trial design of 3-6 tower suspension bridges is carried out, and the calculation results are compared with those of gravity stiffness theory. An example shows that the finished alignment and internal force of multi-tower suspension bridge can be obtained accurately by using this method. This method can be applied not only to the analysis of finished state of plane cable system, but also to the analysis of space cable system.
相似文献/References:
[1]刘世忠,刘永健,程高,等.公路桥梁车桥耦合振动数值分析方法[J].郑州大学学报(工学版),2014,35(01):94.[doi:10.3969/j.issn.1671-6833.2014.01.022]
Liu ShizhongLiu YongjianCheng GaoWang XuLi NaZhao Mingwei.Numerical Analysis Method of Vehicle-Bridge Coupling Vibration for Highway Bridges[J].Journal of Zhengzhou University (Engineering Science),2014,35(01):94.[doi:10.3969/j.issn.1671-6833.2014.01.022]
[2]杜朝伟,郑开勋,蔡迎春,等.碳化腐蚀预应力空心板疲劳特性试验研究[J].郑州大学学报(工学版),2018,39(01):36.[doi:10.13705/j.issn.1671-6833.2018.01.029]
Du Chaowei,Zheng Yuanxun,Cai Yingchun,et al.Experimental Study on Fatigue Characteristics of Prestressed Hollow Slab under Carbonation Corrosion[J].Journal of Zhengzhou University (Engineering Science),2018,39(01):36.[doi:10.13705/j.issn.1671-6833.2018.01.029]
[3]单玉麟,黄福云,罗小烨,等.整体桥台后大不平衡土压力计算方法研究[J].郑州大学学报(工学版),2022,43(02):84.[doi:10.13705/j.issn.1671-6833.2022.02.012]
SHAN Yulin,HUANG Fuyun,LUO Xiaoye,et al.Study on Calculation Method of Larger Unbalanced Earth Pressure of Backfill behind Abutment in Integral Abutment Jointless Bridges[J].Journal of Zhengzhou University (Engineering Science),2022,43(01):84.[doi:10.13705/j.issn.1671-6833.2022.02.012]
[4]梁 岩,张卓航,班亚云,等.多跨连续梁-刚构桥地震易损性分析[J].郑州大学学报(工学版),2023,44(01):96.[doi:10.13705/j.issn.1671-6833.2022.04.003]
LIANG Yan,ZHANG Zhuohang,BAN Yayun,et al.Seismic Fragility Analysis of Multi-span Continuous Girder-rigid Frame Bridges[J].Journal of Zhengzhou University (Engineering Science),2023,44(01):96.[doi:10.13705/j.issn.1671-6833.2022.04.003] | 1,040 | 2,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.610653 |
http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-December/083156.html | 1,726,411,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00553.warc.gz | 18,493,096 | 4,564 | # [EM] Strategy free method proposal
rob brown rob at karmatics.com
Thu Dec 15 13:33:22 PST 2005
```I'm submitting the following method that I believe to be strategy-free.
There is no randomness involved (although like any method without
randomness, it is subject to true ties). I believe it is immune to
unresolvable cycles, with the exception of true ties. (I have not proven
this last point)
I should probably make clear that I submit this not so much as a possible
system to be used for real public elections, but as something to use a
hypothetical for comparison to and evaluation of other methods. I expect
that it is very likely to produce similar results to some Condorcet methods.
First, I want to restate Approval Strategy A (which Rob LeGrand first named,
I believe), as well define as "Range Strategy A" (which I am naming):
Approval Strategy A
If an Approval voter knows with 100% assurance who would be the winner
without his vote, and who will be the runner up, his strategy is very
straightforward: give a vote of yes to every candidate he prefers to the
leading candidate, and also give a yes to the leading candidate if that
candidate is preferred to the runner up. All other candidates should be
given a no.
Range Strategy A
If a voter is allowed to vote with intermediate "shades" between yes and no,
for instance being allowed any floating point value in between zero and one,
Approval strategy A still is the best strategy as long as the voter knows
100% who would be the winner and runner up without his vote. If he does not
know, he can give intermediate votes based on the percent chance of the
various outcomes. In the table below, I show a situation where there are 4
possible outcomes, each with associated probabilities. In the rightmost
column is the Approval ballot showing optimum strategy (Approval strategy A)
were this outcome known to be the case.
Voter preference:
B>C>A
A B 73% A:0, B:1, C:1
B A 15% A:0, B:1, C:0
A C 10% A:0, B:1, C:1
C A 2% A:0, B:1, C:1
Given the above, we can create a composite (weighted average) ballot as
follows:
A: 0, B: 1, C: 0.85
C is given a .85 because there is an 85% chance that 1 (yes) is the best
vote, and 15% chance that 0 (no) is the best vote.
We will call this way of making a composite ballot to use in a Range
election "Range Strategy A".
-------
Given the above definitions, here is the system I propose.
Voters rank candidates in order of preference (as they would with Condorcet,
IRV or Borda elections).
In multiple rounds, each ranked ballot is converted into a Range ballot,
using Range strategy A. On the first round, each candidate is assumed to
have an equal probability of winning as every other candidate, and Range
ballots submitted accordingly. The winner and runner up are calculated
(simply summing all votes for each candidate), and another round is
conducted with it being assumed that those are 100% likely to be winner and
runner up. On subsequent rounds, it uses an average of all past rounds to
determine the percent chance of each candidate being winner or runner up,
and each ranked ballot converted into a Range ballot accordingly.
Eventually, an equilibrium should be reached, settling on a winner and a
runner up, and with each ballot now approaching a "pure" Approval ballot,
with all candidates given either a 0 or 1 on all ballots.
---------
A few notes:
This method uses the concept of a "software agent", which is a program or
part of a program that operates solely with the interests of a single
"client", in this case a single voter. The process of converting the ranked
ballot to a Range ballot must be done without regard for the "collective
good", otherwise, the voter will find that he can do better if he tries to
"trick" the agent into converting it in ways more aligned to his own
interests. This way, there is no reason for a voter to lie to his agent,
since his agent is always working towards his interests alone. Of course,
we hope the "collective good" is still achieved, but it is done in a way
that is fair and stable as it will not give people incentive to be
dishonest.
Note that Range Strategy A is not necessarily the "best strategy", if used
in a normal election. It would still probably be strategically advantageous
to vote all 0's and 1's in a final election, regardless of whether we know
100% the outcome or not. However, in the above method, Range ballots are
never used in a final round, but are only used in preliminary rounds (by the
time an equilibrium is reached, the ballots will always have only zeros and
ones, effectively being Approval ballots). In the case of preliminary
rounds, though, it makes strategic sense to vote with "shades", because it
increases the chance that an equilibrium will be found that is agreeable to
this voter.
To speed up processing, all identical ballots can be grouped together and
processed at once.
I have not covered what will happen in the case of ties during preliminary
rounds, nor covered how to handle equality within ballots. Both of these
are easy to handle in a consistant and logical way, but I avoided discussing
them so as not to distract from the basic description. I will be happy to
explain in detail if anyone cannot see how they should be handled.
It is possible that the second round could be done in a better way. Instead
of submitting ballots based on the assumption that the winner and runner up
from the first round are 100% likely, this could be "averaged in" with our
previous round's assumption, that all candidates are equally likely to win.
Again, I glossed over that to keep the description fairly simple, and
because I don't think it would ever affect the final result.
While I have not proven that this system will always find an equilibrium, I
believe that by eliminating "cusps" I have removed the main thing that
causes similar methods to get stuck in cycles.
I'm not sure if this should be considered a DSV method, because voters
simply rank the candidates. I chose not to have voters give each candidate
a Range-style "rating" on their ballots, for a few reasons:
1) I want it to be easier to directly compare to Condorcet methods
2) I prefer ranking rather than rating from a UI perspective. I feel that
rating will always feel like there is strategy involved (whether or not
there really is).
3) From a practical perspective, ranked ballots are easier to compress and
optimize. Because it is easier to "package up" the complete set of ballots
of an election (say, to allow them to be downloaded on the internet), it has
the ability to be more transparent.
4) If an equilibrium is found (where all voters are submitting Approval
Strategy A ballots), ratings will not make a difference, only rankings.
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20051215/13c1ce98/attachment-0002.htm>
``` | 1,652 | 7,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.968149 |
https://learnbps.bismarckschools.org/mod/glossary/showentry.php?eid=22149 | 1,585,858,634,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370507738.45/warc/CC-MAIN-20200402173940-20200402203940-00470.warc.gz | 552,985,300 | 13,399 | # Measurement and Data
## Narrative for the (MD) Measurement and Data
Second graders learn to measure length with a variety of tools, such as rulers, meter sticks, and measuring tapes. To learn measurement concepts and skills, students might use both simple rulers (e.g., having only whole units such as centimeters or inches) and physical units (e.g., manipulatives that are centimeter or inch lengths). Second graders also learn the concept of the inverse relationship between the size of the unit of length and the number of units required to cover a specific length or distance. For example, it will take more centimeter lengths to cover a certain distance than inch lengths because inches are the larger unit.
In Grade 2, students learn to combine and compare lengths using arithmetic operations to solve word problems. That is, they can add two lengths to obtain the length of the whole and subtract one length from another to find out the difference in lengths.
After experience with measuring, second graders learn to estimate lengths. Real-world applications of length often involve estimation. Skilled estimators move fluently back and forth between written or verbal length measurements and representations of their corresponding magnitudes on a mental ruler (also called the “mental number line”).
As students work with data in Grades K–5, they build foundations for their study of statistics and probability in Grades 6 and beyond, and they strengthen and apply what they are learning in arithmetic. Kindergarten work with data uses counting and order relations. First- and second-graders solve addition and subtraction problems in a data context. Students in Grade 2 draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. They solve simple put-together, take-apart, and compare problems using information presented in a bar graph.
## Calculation Method for Domains
Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics/ proficiency scales for individual standards within the domain. | 426 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.936392 |
http://cboard.cprogramming.com/c-programming/127557-prim%27s-algorithm.html | 1,418,863,830,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802765093.40/warc/CC-MAIN-20141217075245-00104-ip-10-231-17-201.ec2.internal.warc.gz | 51,333,481 | 10,283 | # Prim's algorithm
This is a discussion on Prim's algorithm within the C Programming forums, part of the General Programming Boards category; Code: Prim(G, w, s) //Input: undirected connected weighted graph G = (V,E) in adj list representation, source vertex s in ...
1. ## Prim's algorithm
Code:
```Prim(G, w, s)
//Input: undirected connected weighted graph G = (V,E) in adj list representation,
source vertex s in V
//Output: p[1..|V|], representing the set of edges composing an MST of G
01 for each v in V
02 color(v) <- WHITE
03 key(v) <- infinity
04 p(v) <- NIL
05 Q <- empty list // Q keyed by key[v]
06 color(s) <- GRAY
07 Insert(Q, s)
08 key(s) <- 0
09 while Q != empty
10 u <- Extract-Min(Q)
11 for v in Adj[u]
12 if color(v) = WHITE
13 then color(v) <- GRAY
14 Insert(Q,v)
15 key(v) <- w(u,v)
16 p(v) <- u
17 elseif color(v) = GRAY
18 then if key(v) > w(u,v)
19 then key(v) <- w(u,v)
20 p(v) <- u
21 color(v) <- BLACK
22 return(p)```
I need to find the big O for the above algorithm. How do i start? | 313 | 1,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.686326 |
https://www.fharrell.com/post/ideas-articles/ | 1,701,597,784,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00233.warc.gz | 876,910,933 | 10,747 | # Ideas for Future Articles
2017
Suggestions for future articles, by readers
Author
Affiliation
Vanderbilt University
School of Medicine
Department of Biostatistics
Published
January 16, 2017
Suggestions for future articles are welcomed as comments to this entry. Some topics I intend to write about are listed below.
1. Matching vs. covariate adjustment (see below from Arne Warnke)
2. Statistical strategy for propensity score modeling and usage
3. What is the full meaning of a posterior probability?
4. Moving from pdf to html for statistical reporting
5. Is machine learning statistics or computer science?
6. Sample size calculation: Is it voodoo?
7. Difference between Bayesian modeling and frequentist inference
A few weeks ago we had a small discussion at CrossValidated about the pros and cons of matching here. I am sorry that I did not had enough time to elaborate further on the support of matching procedures (in my field researchers do not focus much on a bias-variance tradeoff but they prioritize on minimizing biases. For that reason, they like matching procedures). Now, I have seen that you started a blog recently (congratulations!). I would like to encourage to take up the topic of matching because it is probably interesting for many applied researchers. I think in your ‘philosophy’, this would belong to the point “Preserve all the information in the data”. Here, perhaps some input for a blog post. Back then, you wrote:
Matching on continuous variables results in an incomplete adjustment because the variables have to be binned.
Matching throws away good data from observations that would be good matches.
Extrapolation bias is only a significant problem if there is a covariate by group interaction, and users of matching methods ignore interactions anyway.
Here, you go too far (in my view). You can add interactions, again for example with propensity score matching. Imbens and Rubin (2015) suggest a procedure using quadratic and interaction terms of the covariates.
Comment: Nice to know this exists but I’ve never seen a paper that used matching attempt to explore interactions. If you don’t want to make regression assumptions that are unverifiable, remove observations outside the overlap region just as with matching.
Which assumptions do you refer to? I think that treating everyone the same (statistically) is also an unverifiable assumption (do you disagree?). What is your opinion about weighted least squares?
Comment: This is the no-interaction assumption. If you assume additivity then it’s more OK to have a no-overlap region, otherwise throw-away non-overlap regions and do a conditional analysis. Not clear on the need for weighting here. In general I like conditioning over weighting.
Arne Jonas Warnke
Labour Markets, Human Resources and Social Policy
Internet: zew.de | 579 | 2,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-50 | latest | en | 0.938047 |
https://www.vlsroulette.com/index.php?topic=9124.msg155109 | 1,500,720,514,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423992.48/warc/CC-MAIN-20170722102800-20170722122800-00409.warc.gz | 877,336,947 | 7,351 | ### Author Topic: Simpler 2x crossover (Read 2900 times)
0 Members and 1 Guest are viewing this topic.
#### shasta
• Member
• Posts: 17
##### Simpler 2x crossover
« on: May 15, 2009, 01:59:42 AM »
Hi guys - this is my second post - good for me! Anyway, i have a question and a tip.
Question - does the 2x crossover pattern overcomplicate things? I have been testing a system that is a bit simpler. I watch the ball as it passes 12 oclock and note the reference number on the wheel, then as the ball does a lap, i note where the ref number is. What i'm looking for is when the ref number (ie the wheel) has moved exactly 1/4 turn for one lap of the ball. This a bit easier than the full 2x crossover, but does the same thing and doesnt waste time as we approach 'no more bets'. Note you can still see the 'pattern' emerging as with the 2x crossover, ie the first ball rotation might see the ref number move 1/8 of a turn, then you note the new ref number (again at 12 oclock as the ball passes by again) and assess its position after the ball does another lap etc. If this second (or third or so) ref number moves more than 1/4 turn during another ball lap then you know that the ref number you are looking for is between (on the wheel) the last two ref numbers.
The difficult part is to train your eyes to assess movement of the ref number (ie exactly 1/4 turn or not) at the same time as picking the next ref number at 12 oclock as the ball passes again. I've only been practicing for a few days and i'm getting much better at it.
You obviously apply this technique to wheels that you assess as going the same speed to ensure the most accurate predictive outcome.
Tip: i have loaded 1. 5 hours of live roulette spins (a video some Aussie dude posted on the net) on my video iPod and i watch it on the train to and from work. This way my eyes and brain get exposure everyday to the process of assessing the wheel and ball location etc. As i said i'm getting much better.
Thoughts?
shasta
#### WannaWin
• Perseverant Member
• Posts: 162
##### Re: Simpler 2x crossover
« Reply #1 on: June 08, 2009, 11:56:56 AM »
Thank you.
Best translator to engage in the discussion: translate.google.com
Popular pages:
#### superbel1974
• Member
• Posts: 1
• VLSroulette.com Member
##### Re: Simpler 2x crossover
« Reply #2 on: April 03, 2017, 03:18:28 PM »
Hello there ,
Does anyboby can help me about 2x or 4x crossover pattern??
Its a easy way without needed to count wheel speed.
But you have four possible areas (like a cross) for ball's drop.
How i can see the only ONE areas from these four???
with regards
Popular pages: | 695 | 2,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-30 | longest | en | 0.931724 |
http://www.mathskey.com/homework-help/algebra-2-2012/book/23/page/17 | 1,516,375,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888041.33/warc/CC-MAIN-20180119144931-20180119164931-00063.warc.gz | 503,017,067 | 15,774 | ALGEBRA 2, 2012
62
PAGE: 17 SET: Exercises PROBLEM: 62
The cashews cost per pound, and the almonds cost per pound.
Lenora bought a total of pounds and paid a total of .
Find the pounds of cashews and the pounds of almonds that Lenora purchased.
Let, be the number of pounds of cashews and be the number of pounds of almonds.
pounds of cashews and pounds of almonds costs .
So, the equation is .
Lenora bought pounds of cashews and almonds.
So, the equation is .
Solve the equation.
(Original equation)
(Substitute )
(Distributive property: )
(Subtract from each side)
(Divide each side by )
(Cancel common terms)
Hence, Lenora bought pounds of almonds.
(Substitute )
(Subtract from each side)
(Distributive property: )
Therefore, Lenora bought pounds of cashews.
Lenora bought pounds of cashews and pounds of almonds.
TESTIMONIALS
"I want to tell you that our students did well on the math exam and showed a marked improvement that, in my estimation, reflected the professional development the faculty received from you. THANK YOU!!!" June Barnett "Your site is amazing! It helped me get through Algebra." Charles "My daughter uses it to supplement her Algebra 1 school work. She finds it very helpful." Dan Pease
Tweets by @mathskeydotcom
QUESTIONS? LET US HELP. | 307 | 1,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-05 | longest | en | 0.954775 |
www.jrdias.com | 1,702,097,931,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00060.warc.gz | 69,012,670 | 17,273 | Units of Weight and Measurement - Conversion Factors Unidades de Peso e Medida - Factores de Conversão
When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. William Thomson, Lord Kelvin, Popular Lectures and Addresses [1891-1894]
Who's behind this Units Conversion page? Who am I? What do I do? How do I do it?
Unidades de Medida Um Manual para Tradutores, em português (PDF) 74 páginas. Índice Remissivo com cerca de 390 entradas €30,00 / \$36.00 — Disponível na Loja JRD
Which unit are you looking for?
Prefixes of SI Units
Write it right
Common mistake Correct Remarks 25m 25 m ALWAYS with a space between the numerical value and the SI symbol 20°F, 20°C 20 °F, 20 °C ALWAYS with a space between the numerical value and the SI symbol 25m 25 m Unit symbols are NOT exponents 75 mts, 100lts 75 m, 100 l (or 100 L) Unit symbols are NOT abbreviations 25 ms, kgs, kWs 25 m, kg, kW Unit symbols are inaltered in the plural 50 m. 25 m NO end period, except as normal punctuation at the end of a sentence KW, kw kW See SI Prefixes kW/h kW.h i.e., power x time = energy amp A Use standard SI terminology 100 w, 500 pa 100 W, 500 Pa First letter of a SI symbol (named after a proper name) is always a capital 45 ° 45° Referring to angle measurements 125°K, 300 °K 125 K, 300 K The SI unit of thermodynamic temperature is kelvin (K) not °K 20 degree centigrade 20 degree Celsius "centigrade" (temperature scale) has been deprecated since 1948 kph km/h Use standard SI terminology 100 joules, 500 watts 100 joule, 500 watt When writting units in full derived from the proper name of individuals quilómetro/h km/h Do not mix unit names with unit symbols 5 kiloyears 5000 days SI prefixes are not used with angle and time units 30 sec 30 s Use standard SI terminology 750 cc, 750 cm3 750 cm³ Use standard SI terminology 75% 75 % By extension from the SI rules 1 mµm 1 nm Do not use compound prefixes 5 m (16.404 foot) 5 m (16 foot) Express the secondary value with an accuracy comparable to the original 25.000 (twenty-five thousand) 25 000 Except in tabular form, write 1234 (4 digits only), and 12 345 (5+ digits) duzentas gramas duzentos gramas Yeap! "grama" is a masculin noun in Portuguese... gravidade específica (*) massa volúmica For "specific gravity", when numerical value has units (e.g., 5 kg/m³) gravidade específica (*) densidade For "specific gravity", when numerical value has NO units (e.g., 1,0) densidade massa volúmica For "specific gravity", when numerical value has units (e.g., 10 kg/m³) (*) "gravidade específica" is a complete aberration in Portuguese!
Take note
The table below uses scientific notation for the writing of the conversion factors: 3.45E-4 = 3.45 x 10-4 = 0.000345
In this table, decimal separators are periods: 0.40469 (US/UK) = 0,40469 (Europe)
Also, in this table, thousands separators are commas: 43,560 (US/UK) = 43.560 (Europe) = 43 560 (ISO)
Multiply X By X Calculator For all your conversion calculations with the conversion factors below
Multiply By
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit acre x 0.404 69 = hectare acre x 43,560 = square foot (ft2) acre x 4,046.873 = square meter (m2) acre x 640 = square mile (mi2) acre x 4840 = square yard (yd2) acre-foot x 7,758 = barrel acre-foot x 43,560 = cubic foot (ft3 or cf) acre-foot x 1,233 = cubic meter (m3) acre-foot x 325,851 = gallon (US) angstrom unit x 1.0E-10 = meter (m) atmosphere x 1.013 = bar atmosphere x 76.0 = centimeter of mercury atmosphere x 33.90 = feet of water atmosphere x 1.033 = kilogram/square centimeter atmosphere x 101.325 = kilopascal (kPa) atmosphere x 10.33 = meters of water atmosphere x 1,013.25 = millibar (mb) atmosphere x 1.013250E+05 = pascal (Pa) atmosphere x 14.696 = pound/square inch
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit balthazar (wine) x 12 = liter (L) barrel (crude oil) x 5.6146 = cubic foot of crude oil barrel (crude oil) x 0.158 987 3 = cubic meter of crude oil barrel (crude oil) x 42 = gallon of crude oil, US barrel (crude oil) x 158.987 294 9 = liter of crude oil barrel (crude oil) x 0.136 = metric ton of crude oil barrel (oil)/day x 0.029 2 = gallon (US)/minute barrel (oil)/day x 1.84 = cubic centimeter/second barrel (oil)/day x 0.159 = cubic meter (m3)/day barrel (oil)/day x 0.159 = liter/day barrel (crude oil) x 0.15 = short ton of crude oil barrel, US liquid x 0.75 = barrel of crude oil barrel, US liquid x 31.5 = gallon, US bar x 0.986 923 = atmosphere bar x 750.0616 = millimeter of mercury (mmHg) bar x 14.50 = pound/square inch board foot x 0.002 360 = cubic meter (m3) Board of Trade unit (BOTU) x 1 = kilowatt.hour (kW.h) (in the UK) british thermal unit (BTU) x 3.931E-04 = horsepower-hour british thermal unit (BTU) x 0.000 292 8 = kilowatt-hour (kW.h) BTU x 778.2 = foot-pounds BTU x 0.000 393 1 = horsepower.hour (HP.h) BTU x 1055 = joule BTU/hour square foot °F x 5.678 263 398 = watt/square meter °C BTU/hour square foot °F x 0.453 592 374 35 = kilocalorie/hour square foot °C BTU/minute x 0.023 56 = horsepower BTU/minute x 0.017 57 = kilowatt (kW) BTU/pound x 0.555 555 = calorie/gram BTU/pound x 2.326 = kilojoule/kilogram (kJ/kg) bushel, US x 0.035 239 07 = cubic meter (m3)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit calorie (cal) x 4.186 8 = joule (J) caneca (beer, Portugal) x 0.4 = liter (L) carat (metric) x 200 = milligram (mg) centimeter (cm) x 0.394 = inch (in) centimeter of mercury x 0.44 = foot of water centimeter/second x 0.036 = kilometer/hour centimeter/ square second x 0.032 808 4 = foot per square second (ft/s2) centipoise x 0.01 = poise centipoise x 0.01 = gram/centimer.second (g/cm.s) centipoise x 0.001 0 = pascal-second centipoise x 6.72E-04 = pound/ft-sec cord x 128 = cubic foot (ft3 or cf) cubic centimeter (cm3) x 3.531E-05 = cubic foot (ft3 or cf) cubic centimeter (cm3) x 0.061 02 = cubic inch (in3) cubic centimeter (cm3) x 2.642E-04 = gallon, US liquid cubic centimeter (cm3) x 1.0E-03 = liter (L) cubic foot (ft3 or cf) x 28,320 = cubic centimeter (cm3) cubic foot (ft3 or cf) x 1728 = cubic inch (in3) cubic foot (ft3 or cf) x 7.48 = gallon, US liquid cubic foot (ft3 or cf) x 28.316 8 = liter (L) cubic foot (ft3 or cf) x 2.831685E-02 = cubic meter cubic foot (ft3 or cf) x 0.037037 = cubic yard (exactly: 1/27) cubic foot (ft3 or cf) x 0.1781076 = oil barrel cubic foot/minute x 472 = cubic centimeter/second cubic foot/minute x 4.72E-4 = cubic meter/second cubic foot/minute x 0.12 = gallon/second cubic foot/minute x 1699.011 = liter/hour cubic foot/minute x 0.471 947 4 = liter/second cubic foot/second x 448.8 = gallons/minute cubic foot/second x 1699 = liters/minute cubic inch (in3) x 16.387 06 = cubic centimeter (cm3) cubic inch (in3) x 1.639E-05 = cubic meter (m3) cubic inch (in3) x 4.329E-03 = gallon (gal) cubic inch (in3) x 0.016 387 06 = liter (L) cubic meter (m3) x 35.315 = cubic foot (ft3 or cf) cubic meter (m3) x 1.307 950 619 = cubic yard (yd3) cubic meter (m3) x 264.20 = gallon, US liquid cubic meter (m3) x 1,000 = liter (L) cubic meter/hour (m3/h) x 0.58858 = cubic foot/min (ft3/min) cubic meter/kilogram (m3/kg) x 16.02 = cubic foot/pound (ft3/lb) cubic meter/sec (m3/s) x 15,850 = gallon/minute cubic meter/sec (m3/s) x 2118 = cubic foot/minute cubic meter/sec (m3/s) x 60,000 = liter/minute cubic yard (yd3) x 0.764 554 9 = cubic meter (m3) cubic yard (yd3) x 27 = cubic foot (ft3 or cf) cycles/second x 1.0 = hertz
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit darcy x 0.9869E-12 = meter square day (mean solar) x 86,400 = second (s) decanewton/meter (daN/m) x 0.685 068 = pounds per feet (lbs/ft) degree (angle) x 0.0111 = quadrant degree (angle) x 0.017 453 29 = radian degree (angle) x 3,600 = second (s) degree (angle) x 17.777 777 8 = mil degree/second x 0.166 67 = revolutions/minute (RPM) degree/second x 0.017 45 = radians/second demi (wine) x 475 = milliliter (mL) drink x 0.5 = fluid ounces (US) of alcohol drink x 15 = milliliters of alcohol dyne x 1.0E-05 = joule/meter (newton) dyne/square centimeter x 1.0E-06 = bar
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit em (pica) x 0.42 = centimeter (cm) erg x 1.0E-07 = joule (J) erg x 2.78E-14 = kilowatt-hour (kW.h) erg/second x 1.3E-10 = horsepower (hp) erg/second x 1.0E-10 = kilowatt
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit fathom (fath) x 1.828 8 = meter (m) foot (ft) x 30.48 = centimeter (cm) foot (ft) x 12 = inch (in) foot (ft) x 0.3048 = meter (m) foot (ft) x 1.645E-04 = mile (nautical) foot (ft) x 5280 = mile (statute) football field (USA) x 1.3223 = acre football field (USA) x 0.535 = hectare (ha) foot of water x 0.029 499 = atmosphere of water foot of water x 0.882 6 = inch of mercury foot of water x 0.433 515 = pound/square inch foot/minute x 0.508 0 = centimeter/second foot/second x 1.097 28 = kilometer/hour foot/second x 0.592 48 = knot foot/second x 0.681 818 = mile/hour foot-candles x 10.763 91 = candela-meter foot-candles x 10.763 91 = lumen/square meter foot-candles x 10.763 91 = lux foot-candles x 1.5759E-06 = watt/square centimeter (W/cm2) foot-pound (ft-lb) x 3.77E-07 = kilowatt-hour (kW.h) foot-pound (ft-lb) x 0.138 255 = kilogram-meter (kg.m) foot-pound (ft-lb) x 1.355 818 2 = newton-meter (N.m) foot-pound/minute x 2.26E-05 = kilowatt (kW) foot-pound/sec x 0.001 81 = horsepower freight ton (US) x 40.0 = cubic foot furlong x 1/8 = mile (statute) furlong x 200 = yard (yd) furlong/fortnight x 1 = snail's pace (you figure it out)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit gallon, US (gal) x 231 = cubic inch (in3) gallon, US (gal) x 0.031 75 = barrel gallon, US (gal) x 0.023 81 = barrel of crude oil gallon, US (gal) x 3,785.41 = cubic centimeter (cm3) gallon, US (gal) x 0.133 7 = cubic foot (ft3 or cf) gallon, US (gal) x 3.785412E-03 = cubic meter (m3) gallon, US (gal) x 0.832 675 = gallon, imperial gallon, US (gal) x 3.785 412 = liter (L) gallon, liquid British imperial x 4.546 09 = liter (L) gallon, liquid British imperial x 1.201 4 = gallon, US liquid gallon of crude oil x 0.023 809 52 = barrel of crude oil gallon/minute, US x 8.020 8 = cubic foot/hour gallon/minute, US x 0.134 = cubic foot/minute gallon/minute, US x 6.309E-05 = cubic meter/sec gallon/minute, US x 0.06309 = liter/sec gallon/minute, US x 227.1247 = liter/hour gallon/minute, US x 5450.993 = liter/day gauss x 1.0E-04 = weber/square meter girafa (beer, Portugal) x 1.0 = liter (L) googol x 10E+100 = 1 followed by 100 zeros grain x 0.064798 91 = gram (g) grain x 437.5 = ounce gram (g) x 0.0353 = ounce gram (g) x 2.204623E-03 = pound (lb) gram/centimeter x 5.6E-03 = pound/inch gram/cubic centimeter x 62.43 = pound/cubic foot gram/cubic centimeter x 8.347 = pound/gallon gram/liter x 1,000 = parts/million gram/liter x 8.34 = pound/1,000 gallon gram/liter x 0.062 43 = pound/cubic foot gram/square centimeter x 2.048 16 = pound/square foot gram-calorie x 1.56E-06 = horsepower-hour gram-calorie x 1.16E-06 = kilowatt-hour (kW.h)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit hectare (ha) x 2.471 = acre hectare (ha) x 107,600 = square foot (ft2) hectare (ha) x 10,000 = square meter (m2) hectare (ha) x 0.003 861 = square mile (sq.mi) horsepower (hp) x 1.014 = horsepower (metric) horsepower (hp) x 42.44 = british thermal unit/minute horsepower (hp) x 550 = foot-pound/second horsepower-hour (hp-hr) x 2.684520E+6 = joule (J) horsepower (hp) x 0.746 043 = kilowatt (kW) horsepower (hp), electrical x 746 = watt (W) horsepower (metric) x 0.986 = horsepower horsepower (metric) x 542.5 = foot-pound/second hundredweight (cwt) (UK) x 112 = pound hundredweight (cwt) (UK) x 50.802 345 = kilogram (kg) hundredweight (cwt) (US) x 100 = pound hundredweight (cwt) (US) x 45.359 237 = kilogram (kg)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit ice (1 pound, melting) x 151 = kilojoule (kJ) imperial (beer, Portugal) x 0.2 = liter (L), aka 'fino' in Portugal inch (in) x 2.54 = centimeter (cm) inch (in) x 0.025 4 = meter (m) inch (in) x 1.578E-05 = mile (mi) inch-pound x 0.112 985 = joule (J) inch of mercury x 0.4912 = pound/square inch inch of mercury x 3386.4 = pascal (Pa) inch of water at (4 oC) x 2.458E-03 = atmosphere inch of water at (4 oC) x 0.03 = pound/square inch iron (leather) x 1/48 = inch (in)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit jeroboam x 3 = liter (L), (aka 'double magnum') joule (J) x 9.48E-04 = british thermal unit (BTU) joule (J) x 0.737 563 = foot-pound (ft-lb) joule (J) x 0.000 239 = kilocalorie (kcal (IT)) joule (J) x 1.00 = newton-meter (N.m) joule (J) x 2.778E-04 = watt-hour (W.h)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit kilogram (kg) x 2.204 623 = pound (lb) kilogram (kg) x 9.84E-04 = ton, long kilogram (kg) x 1.1E-03 = ton, metric (t) kilogram (kg) x 35.273 962 = ounce (oz) kilogram/cubic meter x 0,062 428 = pound/cubic foot kilogram/meter x 0.67197 = pound/foot kilogram/meter x 0.980 855 9 = decanewton/meter (daN/m) kilogram/meter second x 10 = poise kilogram/square centimeter x 0.967 8 = atmosphere kilogram/square centimeter x 14.223 = pound/square inch kilogram/square meter x 9.81E-05 = bar kilogram-meter x 2.72E-06 = kilowatt-hour (kW.h) kilometer (km) x 3,280.84 = foot (ft) kilometer (km) x 0.621 4 = mile (mi) kilometer/hour x 0.911 34 = foot/second kilometer/hour x 0.539 95 = knot (kn) kilometer/hour x 0.621 37 = mile/hour kilopascal (kPa) x 0.009 869 = atmosphere kilopascal (kPa) x 0.010 000 = bar kilopascal (kPa) x 0.145 038 = pound/square inch kilopascal (kPa) x 0.010 197 = kilogram/sq. centimeter (kg/cm2) kilopascal (kPa) x 7.500 617 = millimeter of mercury (mmHg) kilowatt (kW) x 1.341 = horsepower (hp) kilowatt (kW) x 1,000 = watt (W) kilowatt-hour (kW.h) x 3,413 = british thermal unit (BTU) kilowatt-hour (kW.h) x 859.8 = kilocalorie kilowatt-hour (kW.h) x 3.6 = megajoule (MJ) kip (kilopound) x 1,000 = pound kip (kilopound) x 453.6 = kilogram (kg) knot (kn) x 0.515 = meter/second (m/s) knot (kn) x 1.0 = mile (nautical)/h knot (kn) x 1.152 = mile (statute)/h knot (kn) x 1852 = kilometer/hour (km/h) knot (kn) x 1.152 = mile (statute)/h
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit lambda x 1.0 = cubic milliliter (mm3) liter (L) x 0.006 3 = barrel of oil liter (L) x 0.008 4 = barrel liter (L) x 0.035 315 = cubic foot (ft3 or cf) liter (L) x 61.02 = cubic inch (in3) liter (L) x 0.2642 = gallon (US) liter (L) x 0.219 = gallon (UK) liter (L) x 1.056 = quart (US) liter (L) x 2.1134 = pint (US) liter (L) x 33.814 = ounce, fluid (fl oz) (US) liter/sec (L/s) x 2.12 = cubic foot/min (ft3/min) liter/100 km x 2.12 = cubic foot/min (ft3/min) long ton x 1,016.047 = kilogram (kg) long ton x 1.016 047 = metric ton (t) long ton x 2,240 = pound (lb) long ton x 1.12 = short ton (st)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit magnum x 1.500 = liter (L) marathon x 42 195 = meter marathon x 26 miles 385 yards = (Imperial units) measurement ton (US) x 40.00 = cubic foot melchior (champagne) x 18 = liter (L) meter (m) x 3.281 = foot (ft) meter (m) x 0.546 807 = fathom (fath) meter (m) x 39.370 079 = inch (in) meter (m) x 6.21E-04 = mile (statute) meter/second x 2.237 = mile (statute)/hour meter/second x 196.8 = foot/minute meter/second x 3.6 = kilometer/hour (km/h) meter/second x 1.942 = knot (kn) meter per square second(m/s2) x 3.280 839 9 = foot per square second (ft/s2) methuselah (champagne) x 6 = liter (L), aka 'imperial' metric carat x 200 = milligram (mg) metric grain x 50 = milligram (mg) metric mile (athletics) x 1500 = meter (m) metric ton (t) x 1,000 = kilogram (kg) metric ton (t) x 2,204.62 = pound (lb) metric ton (t) x 1.1023 = short ton (st) metric ton of crude oil x 7.33 = barrel (bbl) micrometer (µm) x 1.0E-06 = meter (m) mils (angle) x 0.056 25 = degree (angle) mil (mili-inch) x 0.001 = inch (in) mil (mili-inch) x 25.4 = micrometer (µm) mile, nautical x 1,852 = meter (m) mile, statute x 161,000 = centimeter (cm) mile, statute x 5,280 = foot (ft) mile, statute x 1.609 347 = kilometer (km) mile, statute x 1,609 = meter (m) mile, statute x 0.868 975 = mile, nautical miles/gallon (US) (mpg) x 0.4251 = kilometer/liter (km/L) mile/hour x 44.70 = centimeter/second mile/hour x 1.466 7 = foot/second meter (m) x 3.281 = foot meter (m) x 6.21E-04 = mile (statute) millibar (mb) x 1,0 = hectopascal (hPa) millibar (mb) x 100 = pascal (Pa) milliliter (mL) x 0.0338 = ounce, fluid (fl oz) millimeter (mm) x 0.039 37 = inch millimeter of mercury (mmHg) x 1 = torr millimeter of water x 9.81 = pascal (Pa) milligram/liter x 1 = parts/million million gallon/day x 1.547 2 = cubic foot/second minute (angle) x 0.016 7 = degree
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit nautical mile (nmi, NM) x 1,852 = meter (m) nebuchadnezzar (wine) x 15 = liter (L) newton (N) x 0.101 972 = kg (force) newton (N) x 0.224 81 = pound force newton (N) x 100,000 = dyne newton-meter (N.m, N·m) x 0.737 562 = pound-foot (lb-ft) newton-meter (N.m, N·m) x 0.101 971 6 = kilogram-meter (kg.m) newton-meter (N.m, N·m) x 8.850 747 5 = inch-pound (in-lb) newton.second/square meter x 10 = poise normal cubic meter (Nm3) x 1.0 = 1 m3 of gas @ 0 °C (32 °F) and 1 atm. normal cubic meter (Nm3) x 37.326 = standard cubic feet (SCF) @ 60 °F and 1 atm. nose x 0,5 = length of a horse's head
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit olympiad x 4 = years ounce (oz) x 28.34 = gram (g) ounce (oz) x 0.0625 = pound (lb) ounce, fluid (fl oz) x 1.8 = cubic inch ounce, fluid (fl oz) x 0.029 57 = liter (L) ounce, fluid (fl oz) x 29.573 53 = milliliter (mL) ounce, troy (oz) x 31.103 5 = gram (g)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit pascal x 1 = newton/sq meter pascal x 9.871E-06 = atmosphere pascal x 0.01 = millibar (mb) pascal second (Pa·s) x 10 = poise parts/million x 8.34 = pound/million gallon (gal) pat of butter (USA) x 9.45 = gram pat of butter (USA) x 0.333 333 = ounce pferdestärke (ps) x 735.499 = watt (W) pint, US liquid x 0.125 = gallon, US (gal) pint, US liquid x 16 = fluid ounce (US) pint, US liquid x 0.473 1 = liter (L) pipa (Portugal) x 500 = liter (L) poise (P) x 1 = gram/centimeter second poundal x 13,826 = dyne pound (lb) x 4.45E+05 = dyne pound (lb) x 453.59 = gram (g) pound (lb) x 4.5359E-04 = ton (metric) pound (lb) x 4.448 = joule/meter (newton) pound (lb) x 0.45359237 = kilogram (kg) pound (lb) x 16 = ounce (oz) pound of water x 27.68 = cubic inch (in3) pound of water x 0.119 83 = gallon (gal) pound/barrel x 2.853 = kg/m3 pound/barrel x 2.38 = pounds/100 gallons (US) pounds/100 gallons (US) x 1.2 = kg/m3 pound/cubic foot x 0.133 7 = pound/gallon (US) pound/cubic foot x 0.006 944 = pound/sq inch/foot (psi/ft) pound/cubic foot x 0.016 026 = gram/cubic centimeter pound/cubic foot x 16.018 5 = kilogram/cubic meter pound/cubic inch x 27.68 = gram/cubic centimeter pound/cubic inch x 27,680 = kilogram/cubic meter pound-foot x 1.488 2 = kilogram-meter pound-inch x 178.60 = gram-centimeter pound/gallon (ppg) x 0.120 05 = specific gravity pound/gallon (ppg) x 0.052 = pound/sq inch/foot (psi/ft) pound/gallon (ppg) x 7.48 = pound/cubic foot pound/gallon (ppg) x 0.119 8 = gram/cubic centimeter pound/gallon (UK) x 99 776.3712192 = milligram/liter pound/gallon (UK) x 99 890.3461041 = part/million (ppm) pound/gallon (UK) x 0.832 674 127 478 = pound/gallon (US) pound/gallon (US) x 119 826.433 807 = milligram/liter pound/gallon (US) x 119 963.311 946 = part/million (ppm) pound/gallon (US) x 1.200 950 007 93 = pound/gallon (UK) pound/square inch x 0.068 046 = atmosphere pound/square inch x 0.068947 = bar pound/square inch x 2.309 5 = foot of water pound/square inch x 2.036 02 = inch of mercury pound/square inch x 51.71475 = millimeter of mercury (mmHg) pound/square inch x 0.070 307 = kilogram/square centimeter pound/square inch x 703.069 6 = kilogram/square meter pound/square inch x 6.894757E+03 = pascal pound/square inch x 144 = pound/square foot pound/sq inch/foot (psi/ft) x 19.23 = pound/gallon (US) pound/sq inch/foot (psi/ft) x 144 = pound/cubic foot pound/sq inch/foot (psi/ft) x 2.309 5 = specific gravity
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit Q unit x 1.0E+18 = british thermal unit (btu) quadrant, angle x 90 = degree quadrant, angle x 1.570 8 = radian quart (gallon) x 0.946 = liter (L) quart (gallon) x 32 = ounce (fluid) quart, liquid x 57.75 = cubic inch (in3) quart, liquid x 9.46E-04 = cubic meter (m3)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit radian x 57.29 = degree radian x 3,438 = minutes radian x 206,000 = second revolution x 360 = degree revolution x 4 = quadrant revolution x 6.283 185 = radian rods x 5.5 = yard (yd) round (boxing) x 3 = minute
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit salmanazar (wine) x 9 = liter (L) second-foot x 1.0 = cubic foot/min (ft3/min) second (angle) x 2.78E-04 = degree second (angle) x 4.848137E-06 = radian short ton (st) x 0.892 9 = long ton short ton (st) x 0.907 184 7 = metric ton (t) short ton (st) x 2,000 = pound (lb) short ton of crude oil x 6.65 = barrel (bbl) specific gravity x 1 = gram/cubic centimeter specific gravity x 0.433 = pound/sq inch/foot (psi/ft) specific gravity x 8.33 = pound/gallon (US) specific gravity x 350.5 = pound/barrel specific gravity x 62.4 = pound/cubic foot square centimeter (cm2) x 1.08E-03 = square foot (ft2) square centimeter (cm2) x 0.155 0 = square inch (in2) square inch (in2) x 6.451 6 = square centimeter square foot (ft2) x 2.2956E-05 = acre square foot (ft2) x 929.03 = square centimeter (cm2) square foot (ft2) x 144 = square inch (in2) square foot (ft2) x 0.092 903 04 = square meter (m2) square foot (ft2) x 3.59E-08 = square mile (mi2) square foot (ft2), US Survey x 1.000004 = square foot (ft2) square foot (ft2), US Survey x 0.092 903 = square meter (m2) square kilometer (km2) x 247.105 = acre square kilometer (km2) x 100 = hectare square kilometer (km2) x 0.386 1 = square mile (mi2) square meter (m2) x 2.471E-04 = acre square meter (m2) x 10.763 915 = square foot (ft2) square meter (m2) x 10.763 867 = square foot (ft2) (US Servey) square meter (m2) x 3.86E-07 = square mile (mi2) square meter (m2) x 1,195 99 = square yard (yd2) square mile (mi2) x 640 = acre square mile (mi2) x 27,900,000 = square foot (ft2) square mile (mi2) x 2.589 988 = square kilometer (km2) square yard (yd2) x 9.0 = square foot (ft2) square yard (yd2) x 0.836 127 = square meter (m2) stack x 108 = cubic foot (ft3 or cf) standard cubic meter (Sm3) x 1 = 1 m3 of gas @ 60 °F and 1 atm. stone (UK only) x 14 = pounds
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit therm x 100,000 = BTU of natural gas therm x 1.054804E+08 = joule (J) toc (tons of coal equivalent) x 29.308 = gigajoule (GJ) toc (tons of coal equivalent) x 8.141 = megawatt-hour (MW.h) toe (tons of oil equivalent) x 42.244 = gigajoule (GJ) toe (tons of oil equivalent) x 11.63 = megawatt-hour (MW.h) tonne x 1.0 = ton, metric ton (air conditioning) x 12,000 = british thermal unit/hour (BTU/h) ton, gross (shipping) x 100 = cubic foof (of enclosed space) ton, long x 1,016.047 = kilogram (kg) ton, long x 1.016 047 = metric ton (t) ton, long x 2,240 = pound (lb) ton, long x 1.12 = short ton (st) ton, maritime, shipping, or measurement (UK) x 42 = cubic foot (ft3 or cf) ton, maritime, shipping, or measurement (USA) x 40 = cubic foot (ft3 or cf) ton, metric (t) x 7.33 = barrel of crude oil ton, metric (t) x 1,000 = kilogram (kg) ton, metric (t) x 2,204.62 = pound (lb) ton, short (st) x 6.65 = barrel of crude oil ton, short (st) x 0.892 9 = long ton ton, short (st) x 0.907 184 7 = metric ton (t) ton, short (st) x 2,000 = pound (lb) ton, register x 100 = cubic foot (ft3 or cf) ton, register (RT) x 2.832 = cubic meter (m3)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit unit call (UC) x 100 = call-second/h unit case (soft drinks) x 6.0 = quart (US) unit of blood x 450 = milliliter (mL) unit of heroin x 700 = gram
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit volumetric unit (vu) x 200 = cubic foot (of wood chips)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit watch x 4.0 = hour water (1 quart, boiling) x 3.0 = megajoule (MJ) watt (W) x 3.412 19 = british thermal unit/hour watt (W) x 0.001 341 = horsepower (hp) watt (absolute) x 1 = joule/second watt-hour (W.h) x 3.412 19 = british thermal unit (BTU) watt-hour (W.h) x 2,656 = foot-pound watt-hour (W.h) x 3.60 = kilojoule (kJ) weeks x 168 = hour whole note (music) x 1 = semibreve word x 5.0 = character (including spaces) word x 0.1 = line (60 character)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit X unit (Siegbahn unit) x 1.0021E-13 = meter (m)
Return to top of page Original Unit Multiplied By Conversion Factor Equals Final Unit yard (yd) x 3.0 = foot (ft) yard (yd) x 36 = inch (in) yard (yd) x 0.914 40 = meter (m) yard (yd) x 1760 = mile (statute) years (mean of 4-year period) x 365.25 = day years (mean of 4-year period) x 8,766 = hour years (mean of 4-year period) x 526,000 = minute (time) years (mean of 4-year period) x 31,558,150 = second (time) years (mean of 4-year period) x 52.17 = week
Prefixes of the International System of Units (SI)
Numerical Value Prefix Symbol Meaning Exponential Expression 1 000 000 000 000 000 000 exa E Quintillion 1018 1 000 000 000 000 000 peta P Quadrillion 1015 1 000 000 000 000 tera T Trillion 1012 1 000 000 000 giga G Billion 109 1 000 000 mega M Million 106 1 000 kilo k Thousand 103 100 hecto h Hundred 102 10 deca / deka da Ten 101 1 100 0.1 deci d one-tenth 10-1 0.01 centi c one-hundredth 10-2 0.001 milli m one-thousandth 10-3 0.000 001 micro µ one-millionth 10-6 0.000 000 001 nano n one-billionth 10-9 0.000 000 000 001 pico p one-trillionth 10-12 0.000 000 000 000 001 femto f one-quadrillionth 10-15 0.000 000 000 000 000 001 atto a one-quintillionth 10-18 Return to top of page
Disclaimer
Pursuant to my Privacy Statement, Legal Notice and Copyright you explicitly acknowledge that all information obtained on this website is provided "as is" without warranty of any kind, either expressed or implied, including, but not limited to, the implied warranties of merchantability and fitness for a particular purpose, and that the entire risk of acting on information obtained from this website, including the entire costs of all necessary remedies, is with those who choose to act on such information and not with jrdias.com or its owner.
Specializations Contacts FAQ JRD Store Newsletter Site Map Privacy
Copyright© 1999 João Roque Dias • Technical Translator • Tradutor Técnico • Last updated: 28 Nov 2014 17:09 WET | 9,663 | 27,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | longest | en | 0.748939 |
https://hpmuseum.org/forum/thread-11176-post-101937.html | 1,675,069,895,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00868.warc.gz | 332,524,263 | 6,648 | FYI: WarpPi calculator
08-06-2018, 05:52 AM
Post: #1
Luigi Vampa Member Posts: 270 Joined: Dec 2015
FYI: WarpPi calculator
Saludos Saluti Cordialement Cumprimentos MfG BR + + + + +
Luigi Vampa +
Free42 '<3' I + +
08-06-2018, 07:28 AM
Post: #2
Dan Member Posts: 162 Joined: Jan 2017
RE: FYI: WarpPi calculator
Anyone know the model of the LCD on that calculator?
08-07-2018, 06:37 PM (This post was last modified: 08-09-2018 05:52 PM by compsystems.)
Post: #3
compsystems Senior Member Posts: 1,337 Joined: Dec 2013
RE: FYI: WarpPi calculator
WarpPi Is an Impressive DIY Raspberry Pi Calculator, and is the firts with Step-By-Step Algebra
https://www.neoteo.com/construye-una-cal...y-pi-zero/
08-09-2018, 01:04 PM
Post: #4
Eddie W. Shore Senior Member Posts: 1,383 Joined: Dec 2013
RE: FYI: WarpPi calculator
Several things I like about it:
I like how you can designate how the variables act: blue for CAS variables, black for numerical constants, etc
I like how big the screen is. The screen is clear, well done on the display.
I can't wait to see what the finished keyboard looks like.
08-10-2018, 03:51 PM (This post was last modified: 08-10-2018 03:59 PM by compsystems.)
Post: #5
compsystems Senior Member Posts: 1,337 Joined: Dec 2013
RE: FYI: WarpPi calculator
If it is a calculator with CAS, the 6 algebraic operators should be together
[+], [-], [·](*), [÷](/), [^], [√]
PHP Code:
[f1][f2][f3][f4][f5]
[
][][]
[
]
[ ][ ][ ][ ][
]
[,][ ][ ][ ][ ]
[()][
±][][^][÷]
[
EEX][7][8][9][*]
[
α][4][5][6][-]
[
][1][2][3][+]
[
esc][0][.][spc][enter
08-11-2018, 04:39 AM
Post: #6
Jlouis Senior Member Posts: 722 Joined: Nov 2014
RE: FYI: WarpPi calculator
Doesn't the HP 50G have step by step CAS?
Even the 28c has step by step, well, not all the steps, but anyway, the 28c is from 1987, isnt it?
I think I'm missing something...
« Next Oldest | Next Newest »
User(s) browsing this thread: 1 Guest(s) | 668 | 1,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-06 | latest | en | 0.843507 |
https://www.coursehero.com/file/6043978/W1s/ | 1,493,371,466,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122886.86/warc/CC-MAIN-20170423031202-00619-ip-10-145-167-34.ec2.internal.warc.gz | 876,307,822 | 140,533 | # W1s - WS # 1: 1016-351: Coverage - 1.1, 1.2: Dr. Chulmin...
This preview shows pages 1–2. Sign up to view the full content.
WS solution for Probability (KIM) Page 1 WS # 1: 1016-351: Coverage - 1.1, 1.2: Dr. Chulmin Kim KEY 1. Identify whether the variable is categorical or quantitative . a. SAT Math scores for 10 randomly selected students in Rochester in 2007: 750, 610, 570, 750, 670, 480, 580, 550, 590, 680 Answer: quantitative b. Each of 10 employees from RIT was asked what their mode of transit is. Data are: 1, 2, 2, 3, 4, 1, 1, 3, 2, 2 (Here, 1=car, 2=bus, 3=walk, 4=bike.) Answer: categorical c. Eye Color (brown, green, black) Answer: categorical d. Height Answer: quantitative 2. Please refer to the following data. 86, 93, 45, 49, 51, 44, 77, 74, 92, 80 Make the stem-plot. [Hint: Stem unit here is ten. So leaf unit is one.] [1pt.] Answer: 4 | 4 5 9 5 | 1 6 | 7 | 4 7 8 | 0 6 9 | 2 3
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
WS solution for Probability (KIM) Page 2 3. Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined,
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 12/07/2010 for the course MTH 1016.351 taught by Professor Dr.chulminkim during the Winter '10 term at RIT.
### Page1 / 2
W1s - WS # 1: 1016-351: Coverage - 1.1, 1.2: Dr. Chulmin...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 563 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-17 | longest | en | 0.859202 |
http://jwilson.coe.uga.edu/EMT725/IsoTri100/IsoTri100.html | 1,695,447,051,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00513.warc.gz | 20,651,148 | 1,467 | ## 100 degree Isosceles Triangle
Given an isosceles triangle ABC with AB = AC and the measure of angle BAC = 100 degrees. Extend AB to point D such that AD = BC. Now draw segment CD. What is the measure of angle BCD? Prove your results by geometric reasoning, rather than measuring.
Note: There can be several different approaches to this problem. Geometric arguments are preferred rather than using trigonometry.
Usually, a solution to a problem such as this requires some auxiliary constructions.
Do you want a Hint for one auxiliary construction?
Find alternative solutions or compare your solution to the solutions of others to see alternative approaches. | 138 | 663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-40 | latest | en | 0.92503 |
https://math.stackexchange.com/questions/429274/probability-of-failure-of-an-appliance-total-probability | 1,582,426,902,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00163.warc.gz | 459,119,277 | 32,068 | # Probability of failure of an appliance. Total probability?
An appliance consists of three elements of type A, three of type B, four of type C. The appliance stops working if less than two elements of a kind are functional. In some interval of time $t$, the probability of failure of the components is:
$P(A)=0.1, P(B)=0.1, P(C)=0.2$
-What is the probability of the appliance failing in time $t$?
-Given that the appliance failed, what is the probability of the failure being a result of type A or C failing?
I'm not sure how to set up the calculations for this. I tried doing the first one with the law of total probability, (either the failure was caused by A, B or C - type), but these are not independent. Would appreciate hints on how to begin.
• The appliance stops working if less than two elements of the same kind fail. It does not make sense ? If zero elements of the same kind fail, does it stop working ? – justt Jun 26 '13 at 11:52
First question : Let $A'$ be the event "less than two elements of type $A$ are functional". Its probability can be computed easily using binomial law. Let $B'$ and $C'$ be defined in the same fashion.
The event $F=$"the appliance fails" is the event $A'\cup B' \cup C'$. You can compute its inverse event first, which is $\bar F = \bar A' \cap \bar B' \cup \bar C'$.
Second question :
The probability you are looking for is $$P_F(A' \cup C') = \frac{P((A' \cup C') \cap F)}{P(F)} = \frac{P(A' \cup C')}{P(F)}$$
• Is is true that $P(A'\cap B' \cap C') = P(A')P(B')P(C')$? If so, why? Are they independent? – Spine Feast Jun 26 '13 at 12:22
• stricto sensu I can't say, it should have been precised in your problem. But I guess that the failure of$A$-elements does not affect the failure of $B$-elements, and so on... so $A',B'$ and $C'$ should be independent, yes. – justt Jun 26 '13 at 12:24
• Ok, thanks for your answer. I keep overthinking these problems! – Spine Feast Jun 26 '13 at 12:35
Q1: think about it in the following way: either A or B or C fail. What happens after any of them fails? We don't care, the appliance is already down. So I suggest doing it like this: probability of failure of every component is clearly binomial: $$P(A)=\binom{3}{2}p_a^2(1-p_a)\\ P(B)=\binom{3}{2}p_b^2(1-p_b)\\ P(C)=\binom{4}{3}p_c^3(1-p_c)$$
This is because as soon as either 2 of A or B or 3 of C fail, the appliance is down. Assuming independence, use inclusion-exclusion theorem: $$P(F)=P(A)+P(B)+P(C)-P(A)P(B)-P(A)P(C)-P(B)P(C)+P(A)P(B)P(C)$$
Q2: use Bayes formula $P(F|A)P(A)=P(A|F)P(A)$ and the fact that $P(\cdot|F)$ is clearly $1$ | 792 | 2,592 | {"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} | 4.0625 | 4 | CC-MAIN-2020-10 | latest | en | 0.910711 |
http://www.chegg.com/homework-help/questions-and-answers/small-birds-migrate-long-distances-without-feeding-storing-energy-mostly-fat-rather-carboh-q809032 | 1,501,103,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426629.63/warc/CC-MAIN-20170726202050-20170726222050-00253.warc.gz | 406,486,647 | 18,158 | Small birds can migrate over long distances without feeding,storing energy mostly as fat rather than carbohydrate. Fat is agood form of energy storage because it provides the most energy perunit mass: 1 gram of fat provides about 9.4 (food) Calories,compared to 4.2 (food) Calories per 1 gram of carbohydrate.Remember that Calories associated with food, which are alwayscapitalized, are not exactly the same as calories used in physicsor chemistry, even though they have the same name. Morespecifically, one food Calorie is equal to 1000 calories ofmechanical work or 4186 joules. Therefore, in this problem use theconversion factor .
Consider a bird that flies at an averagespeed of 10.7 and releases energy from its body fatreserves at an average rate of 3.70 (this rate represents thepower consumption of the bird). Assume that the bird consumes of fat to fly over a distancewithout stopping for feeding. How far will the bird fly beforefeeding again?
How many grams of carbohydrate would the bird have to consume to travel the same distance?
How many grams of carbohydrate would the bird have to consume to travel the same distance? | 248 | 1,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-30 | latest | en | 0.930456 |
https://oeis.org/A338397 | 1,620,814,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989693.19/warc/CC-MAIN-20210512100748-20210512130748-00625.warc.gz | 462,076,673 | 3,917 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A338397 Triangle read by rows: T(n,m)= Sum_{k=0..m/2} C(n-k,m-2*k)*C(n-k,m-k)*C(n,k))/C(2*k,k). 0
1, 1, 1, 1, 4, 2, 1, 9, 12, 4, 1, 16, 42, 34, 8, 1, 25, 110, 160, 90, 16, 1, 36, 240, 550, 540, 226, 32, 1, 49, 462, 1540, 2310, 1666, 546, 64, 1, 64, 812, 3724, 7910, 8596, 4802, 1282, 128, 1, 81, 1332, 8064, 23058, 34986, 29190, 13140, 2946, 256 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,5 LINKS FORMULA G.f.: A008459(x,y)/(1-(x*y*A008459(x,y))^2). EXAMPLE 1, 1, 1, 1, 4, 2, 1, 9, 12, 4, 1, 16, 42, 34, 8, 1, 25, 110, 160, 90, 16, 1, 36, 240, 550, 540, 226, 32 PROG (Maxima) T(n, m):=sum((binomial(n-k, m-2*k)*binomial(n-k, m-k)*binomial(n, k))/binomial(2*k, k) , k, 0, m/2); CROSSREFS Cf. A001263, A008459, A176280 (row sums), A338372. Sequence in context: A208612 A183157 A211957 * A063983 A259985 A144084 Adjacent sequences: A338394 A338395 A338396 * A338398 A338399 A338400 KEYWORD nonn,tabl AUTHOR Vladimir Kruchinin, Oct 23 2020 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified May 12 06:20 EDT 2021. Contains 343814 sequences. (Running on oeis4.) | 636 | 1,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.528176 |
https://www.shaalaa.com/question-bank-solutions/a-body-acted-upon-two-forces-each-magnitude-f-but-opposite-directions-state-effect-forces-when-a-both-forces-act-same-point-thtwo-forces-act-two-dion-d-moment-turning-effect-of-a-force-or-torque_85203 | 1,618,251,330,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00602.warc.gz | 1,060,383,694 | 9,056 | # A Body is Acted Upon by Two Forces, Each of Magnitude F, but in Opposite Directions. State the Effect of the Forces When (A) Both Forces Act at the Same Point of Thtwo Forces Act at Two Dion D. - Physics
A body is acted upon by two forces, each of magnitude F, but in opposite directions. State the effect of the forces when
(a) Both forces act at the same point of the body.
(b) The two forces act at two different points of the body at a separation d.
#### Solution
(a) If both the forces act at the same point of the body, they have the same line of action, and then the moment becomes zero.
(b) If both the forces act at two different points of the body at a separation d then they constitute a torque whose value is given F x d.
Concept: Moment (Turning Effect) of a Force Or Torque
Is there an error in this question or solution?
#### APPEARS IN
Frank ICSE Class 10 Physics Part 2
Chapter 1 Force, Work, Energy and Power
Exercise | Q 7 | Page 57 | 238 | 959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-17 | latest | en | 0.907188 |
https://ww2.mathworks.cn/matlabcentral/answers/1562876-plot-values-of-3d-matrix-against-a-single-variable | 1,638,285,725,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00381.warc.gz | 685,020,006 | 25,766 | # Plot values of 3D matrix against a single variable
3 次查看（过去 30 天）
Evagoras Kassapis2021-10-13
I have a 4D matrix (4x4x4x6), the first three dimensions are the pressure at each of the x,y,z points in a space. The fourth dimension is the frequencies used to calculate the pressure. I want to plot the frequency response of the space. That means plot all the pressures against the frequency used to calculate them.
### 采纳的回答
Image Analyst 2021-10-13
Try this:
clc; % Clear the command window.
fprintf('Beginning to run %s.m ...\n', mfilename);
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 17;
pressure = rand(4, 4, 4, 6); % Random data
[rows, columns, slices, freq] = size(pressure)
% Plot all pressure profiles for each (x,y,z) point.
for row = 1 : rows
for col = 1 : columns
for z = 1 : slices
% Get frequency response over all locations
p = squeeze(pressure(row, col, z, :))
% Plot this set of pressures for this particular (x,y,z) location:
plot(p, 'LineWidth', 2);
if row == 1 & col == 1 && z == 1
grid on;
xlabel('Frequency', 'FontSize', fontSize)
ylabel('Pressure', 'FontSize', fontSize)
hold on;
end
end
end
end
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 395 | 1,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.693074 |
https://l2sfbc.com/does-vehicle-weight-affect-how-far-you-can-drop-tyre-pressures/ | 1,718,658,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00240.warc.gz | 321,114,848 | 22,983 | # Does vehicle weight affect how far you can drop tyre pressures?
Hello Robert..
Ive just watched your Sand Driving tyre pressure vid on YouTube. Excellent demonstration. Thank you..
I have one question for you?
Does the weight of the vehicle determine how much you can drop your pressures?
If you have a box trailer at home running say 30psi in the tyres empty.. if you go and get half a ton of Sand in it.. will you be required to put more air in the tyres to drive safely?
This being said, your theory of dropping by percentages is far more accurate then dropping by numbers..
So many people I see trying to help people out of bogs and they drop straight to 15psi is incredible, no regards to the weight in the back of the vehicle. Or anything..
I run a Fuso Canter on 37s..
I run 70psi on the road and drop to 35psi on the sand.
With 5ton of weight that’s a lot of weight for the to roll off the rim, even with just the slope of some of the beaches.
Interested to hear back from you.
Yes, the weight of the vehicle affects the amount you can and should drop your tyre pressures offroad. The heavier a vehicle, the more a tyre will deform for a given pressure. And the heavier a vehicle is, the greater the contact patch needed to be effective on a compression terrain like sand. The two factors work against each other so the end result is, roughly, the same – always the way with tyres which just get weirder and weirder the more you look at the detail. Coming back to your specifics; talking to AAV4x4 who are the experts on this subject you’re fine to go down to 20psi, even lower if needs be. You may well find 35psi is too high on the sand.
However, the more a tyre pressured is lowered, the more it deforms, and the slower you must drive to stop the tyre overheating. The deformation is also governed in part by the diameter of the tyre; a 37″ tyre won’t deform as much for a given contact patch as a 35″ tyre, so it won’t heat up as much, and it also has a larger amount of tyre exposed to the air for cooling.
Sound complex? Yes it is. I’ve only just scratched the surface of this topic!
Re the box trailer; yes, if you add weight to the trailer you must add air to the tyres, and in this case the difference in weight between a trailer empty and filled with sand is significant, so the tyre pressure difference must also be significant. Same way when you put a car on a car trailer you should increase the tyre pressure.
Also, the recommended tyre pressures you see for 4x4s typically don’t apply to light trucks which run maybe 60-80psi on road whereas 4×4 run 35-42. So you can take them as the principle, but not the actual numbers.
Cover image courtesy AAV4x4
And here’s the video the reader referenced!
This one is also pertinent to those driving light trucks such as the reader: | 646 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.950697 |
https://a1calculator.com/health/adjusted-body-weight-calculator/ | 1,723,272,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00293.warc.gz | 63,182,469 | 18,076 | If you’re on a quest for a healthy and effective weight loss journey, the Adjusted Body Weight Calculator can be a valuable ally. Not only is it useful for individuals aiming to shed pounds, but it’s also a time-saving tool for professional dietitians, streamlining the process of calculating nutritional requirements for their clients.
Adjusted Body Weight (AjBW) is a crucial metric in the realm of dietetics, particularly when tailoring weight loss programs. It becomes especially relevant for individuals dealing with overweight or obesity. This metric addresses the limitation of using actual body weight in estimating energy needs, as adipose tissue’s lower metabolic activity compared to lean tissue can lead to overestimations. AjBW provides a more accurate assessment in such scenarios.
## How to Use the Adjusted Body Weight Calculator
1. Gender Input:
2. Height Input:
3. Weight Input:
• Input your actual body weight.
The calculator, employing the Robinson formula, then computes your Ideal Body Weight (IBW) and subsequently derives your AjBW, presenting the result for your convenience.
## The Adjusted Body Weight Formula
The formula for calculating Adjusted Body Weight (AjBW) is as follows:
AjBW=IBW+0.4×(ABWIBW)
Where:
• AjBW is adjusted body weight
• IBW is ideal body weight
• ABW is actual body weight
## Calculating Ideal Body Weight (IBW)
1. For Men:
• IBW=52kg+1.9kg per inch over 5 feet
2. For Women:
• IBW=49kg+1.7kg per inch over 5 feet
## Example Calculation
Suppose we want to determine the nutritional needs for a 180 cm tall man weighing 90 kg. Using the formula, his AjBW would be calculated as follows:
kgAjBW=72.65kg+0.4×(90kg−72.65kg)=79.59kg
This means that for dietary considerations, this individual should be treated as if he weighed 79.59 kilograms.
## Limitations of the Adjusted Body Weight Calculator
It’s crucial to recognize that the Adjusted Body Weight Calculator is not universally applicable. It’s suitable only when excess weight is attributed to adipose tissue. Situations involving athletes with higher muscle mass or during pregnancy require alternative metrics.
## Adjusted Body Weight Calculator (FAQ)
How to Calculate Adjusted Body Weight?
Use the formula: AjBW=IBW+0.4×(ABWIBW)
Formula for Adjusted Body Weight for Obesity:
AjBW=IBW+0.4×(ABWIBW)
When to Use Adjusted Body Weight?
Commonly used for overweight or obese patients for accurate nutritional estimations.
Ideal Body Weight for a 5 ft 5 in Woman:
kgIBW=57.5kg (calculated using Robinson’s formula) | 603 | 2,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.861398 |
https://www.bouvet-couverture.fr/Jan/26-1777.html | 1,632,122,449,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00526.warc.gz | 715,468,948 | 8,876 | # cubic yards per ton 57 slag
#### Calculate #57 Limestone Gravel | cubic yards / Tons
Type in inches and feet of your project and calculate the estimated amount of Gravel Stone in cubic yards, cubic feet and Tons, that your need for your project. The Density of #57 Limestone Gravel: 2,410 lb/yd³ or 1.21 t/yd³ or 0.8 yd³/t
#### Sand Calculator - how much sand do you need in tons ...
A ton of sand is typically about 0.750 cubic yards (3/4 cu yd), or 20 cubic feet. Sand is assumed relatively damp, since adding water can increase or decrease the density of the sand considerably (e.g. if it was raining or if you dig up and leave sand under the sun so water evaporates).
#### Volume-to-Weight Conversion Charts
PRODUCT VOLUME POUNDS TONS SOURCE Chalk, lumpy 1 cubic foot 80.00 0.040 FEECO Charcoal 1 cubic foot 18.00 0.009 FEECO Clay, kaolin 1 cubic foot 28.00 0.014 FEECO ... Slag, crushed 1 cubic yard 1,998.00 0.999 Tellus Institute Slag, loose 1 cubic yard …
#### how much does a cubic yard of 57 limestone weigh « BINQ Mining
Jul 01, 2013· Convert Tons to Cubic Yards [Archive] – OnlineConversion Forums … to make this conversion. I really need to understand so I can address this in the future too. … I have a similar question, but I'm trying to convert #57 stone from cubic yards to tons. … x 1ft x multiply that weight by 9….that is how much 1 cubic yard weighs …
#### Slag Rock #57 (Commercial 57's) - All Seasons Mulch
Slag Rock #57 (Commercial 57's) is an aggregate used for Unconfined Fill, Base or Embankment. The size is about that of a quarter (1 & ¼" X ¼") and is grayish black. Please call for availability (843-559-5829) due to the shortage at this time.
#### Calculator - Summit - Topsoil and Gravel
A cubic yard is equal to 27 cubic feet. You can use the online calculator to determine how many cubic yards of material are required. Formulas Used Rectangular Area: (Lenght Ft. x Width Ft. x Depth Ft.) / 27 Triangular Area: (Lenght Ft. / 2) x Width Ft. x Depth Ft.] / 27 Round Area: (3.1416 x (Radius x Radius) x Depth in feet) / 27
#### Creekside Gardens - Garden Center- Mulch and stone calculator
#57 Canadian Blue 68.00 #57 Mohave 85.00 #57 MOHAVE 78.99 #57 Washstone 0.00 #57 ... Generally your full size pickup truck will accommodate two to three cubic yards of mulch, while a smaller (S-10 or Ranger) type of pickup truck will hold one to two cubic yards. Trucks: Mulch: ½ ton pickup truck 6.5' bed: 0 loads: Same truck with 8 foot bed ...
#### Calculate #57 Granite Stone | cubic yards / Tons
Type in inches and feet of your project and calculate the estimated amount of Granite Stone in cubic yards, cubic feet and Tons, that your need for your project. The Density of #57 Granite Stone: 2,410 lb/yd³ or 1.21 t/yd³ or 0.8 yd³/t
#### Limestone Quarry Calculation For Cost Of Production Per Ton
57 Crushed Limestone 34 To 1. It is the type of stone typically used for driveways and parking lots Slag and recycled concrete can be used for all the same applications as limestone Many people mistakenly refer to limestone as slag so if a customer says they want slag be sure to ask if they really mean limestone Sold by the ton approximately 15 tons per cubic yard
#### Material Weights - Harmony Sand & Gravel
Most of Harmony Sand & Gravel's products will weight approximately 2,840 pounds per cubic yard or about 1.42 tons per cubic yard. For estimating purposes, most Contractor's consider the yield to be 3,000 pounds per cubic yard or 1.5 tons per cubic yard.
#### #57 Limestone - Landscaping Supplies Cleveland - Kurtz Bros
RETAIL PRICE PER YARD FOR COMMERCIAL PRICING CALL 216.986.7000 . ½ to 1 in size. Ideal for roadways, driveways or backfill. Can also be used as a base to build on or pour over concrete. The pure whiteness of limestone gravel make the rock very popular for use in …
#### Crushed Stone: Determing Yards Per Ton
Cubic yards are a volume measure. Tons are a weight measure. To make matters even more confusing is the fact that determining yards per ton is anything but an exact science. The weight of crushed stone in pounds per cubic yards and tons per cubic yards are approximate measures that assume all of the same material weighs the same.
#### Cubic Yard Calculator - Gravel Yardage Calculator - Nimbus ...
If you want to haul these materials in your own truck, you need to know the weight per ton and the capacity of your truck. As a general rule of thumb, small trucks (1/2 ton) can haul a half yard of gravels and soil or a yard of bark. Large pick-up trucks (3/4 and 1 ton) can haul one yard of heavy material or a couple of yards of bark.
#### ITEM 304 AGGREGATE BASE - dot.state.oh.us
The Department will verify that the moistures of the delivered material are less than 2 percent above saturated surface dry (SSD). If the moisture is greater than 2 percent above SSD, then the Department will calculate the number of cubic yards (cubic meters) based on the dry density and dry weight. The Department will determine the pounds per cubic yard (kilograms per cubic meter) for ...
#### 2020 Gravel Driveway Costs | Road Base & Crushed Rock ...
Plain pea gravel and crushed clamshells are each priced at about \$40 per cubic yard and \$50 per ton. Crushed stone is costlier at about \$55 per cubic yard and \$65 per ton . Buying pea gravel in bulk may reduce costs, but different finishes, like gravel with color, will add anywhere from \$20 to \$50 to the price per …
#### Bulk Material Calculator | Contractors Stone Supply
How much does a cubic yard weigh? Most of our bulk materials, with the exception of mulch, are sold by the weight. The following are approximate weights for most of our bulk materials. Sand 1.10 - 1.25 tons(2,200 - 2,500 lb.) per cubic yard Planting Mix 1 ton (2,000 lb.) per cubic yard Lawn Dressing .90 tons (1,800 lb.) per cubic yard
#### Approximate Weights of Various Construction Material Per ...
Apr 04, 2015· Most gravel products will weight approximately 2,840 pounds per cubic yard or about 1.42 tons per cubic yard. For most estimating purposes, consider the yield to be 3,000 pounds per cubic yard or 1.5 tons per cubic yard.
#### 57 Limestone | Jones Topsoil Columbus Ohio
3-4" clean crushed limestone. This material is commonly used as a base for driveways and parking lots. It is also used in construction drives for heavy equipment. It can be hard to walk on because the pieces are so large. It cannot be shoveled by hand. It weighs 1.5 ton per cubic yard.
#### Slag, solid volume to weight conversion
About Slag, solid; 1 cubic meter of Slag, solid weighs 2 114 kilograms [kg] 1 cubic foot of Slag, solid weighs 131.97271 pounds [lbs] Slag, solid weighs 2.114 gram per cubic centimeter or 2 114 kilogram per cubic meter, i.e. density of slag, solid is equal to 2 114 kg/m³.In Imperial or US customary measurement system, the density is equal to 132 pound per cubic foot [lb/ft³], or 1.222 ounce ...
#### 2020 Gravel Prices | Crushed Stone Cost (Per Ton, Yard & Load)
Gravel Prices Per Ton. Bulk crushed stone and gravel prices are \$10 to \$50 per ton on average. Road base costs \$18 to \$30 per ton, and plain pea gravel or limestone costs \$28 to \$45 per ton.Buying gravel in small quantities costs over \$100 per ton.It takes 1.4 tons of stone per cubic yard.
#### #411 Limestone-(#57 and #10 mixed) - Three Z
Slag and recycled concrete can be used for all the same applications as limestone. Many people mistakenly refer to limestone as slag, so if a customer says they want slag, be sure to ask if they really mean limestone. Sold by the ton, approximately 1.5 tons per cubic yard. Sizes Available # 10 screenings [Powder to 1/4"] # 8 [3/8"] # 57 [3/4 ...
#### CUBIC YARDAGE CALC SHEET - SwapLoader
Material Weight – Pounds per Cubic Yard Asphalt 2,700 lb. Iron (wrought) 13,100 lb. Brush/Branches (loose) 250 lb. Leaves (loose, dry) 150 lb. Brush/Branches (chipped – 3" screen) 600 lb. Leaves (vacuumed, dry) 400 lb. Concrete (gravel or stone mix) 4,050 lb. Leaves (wet or compacted) 550 lb. ...
#### How many tons of 57 gravel in 1 cubic yard? - Answers
A ton of gravel can, at best, be described in cubic yards. A cubic foot of gravel weighs about 100 pounds, so a full ton will be about 20 cubic feet, or .74 cubic yards.
#### Calculating Cubic Yards vs. Tons - gsgravel.com
2. A ton is a measurement by weight. 3. A "/" sign means 'per,' so 1.5 TONS/CY reads 1.5 TONS per Cubic yard which simply means there are 1.5 tons per (for) every cubic yard of material. Fortunately, a cubic yard of material has a conversion factor (unit weight) so a cubic yard can easily be converted to a ton or vice versa.
#### Gravel Calculator - Estimate Landscaping Material in Yards ...
Most gravel weighs 1.4 to 1.7 tons per cubic yard. See below for more common material densities. For example, let's find the amount of gravel needed for a space that is 10 feet long, 10 feet wide, and 1 foot deep. volume = length × width × depth volume = 10′ × 10′ × 1′ = 100 cu ft
#### Convert Ton Register to Cubic Yard - Unit Converter
Instant free online tool for ton register to cubic yard conversion or vice versa. The ton register [ton reg] to cubic yard [yd^3] conversion table and conversion steps are also listed. Also, explore tools to convert ton register or cubic yard to other volume units or learn more about volume conversions.
#### Re: How many cubic yards are in 1 US ton ...
Oct 18, 2006· Re: How many cubic yards are in 1 US ton? Density of sand is about 100-110lbs per cubic foot. There are 27 cubic feet in a cubic yard. Therefore 1 cubic yard of sand weighs about 2,970 lbs or 1.97 tons.
#### How many yards in 1 ton of gravel? - Answers
do you mean cubic yards? depends upon the rock type and how fine it is, but one reference notes that one cubic foot of gravel typically weighs about 100 pounds. So 1 ton has 2000/100 or 20 cubic ... | 2,588 | 9,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.841979 |
http://www.learnquebec.ca/mst?inheritRedirect=true?p_p_id&p_p_lifecycle=1&p_p_state=normal&p_p_mode=view&_82_struts_action=%2Flanguage%2Fview&_82_languageId=fr_CA?p_p_id&p_p_lifecycle=1&p_p_state=normal&p_p_mode=view&_82_struts_action=%2Flanguage%2Fview&_82_languageId=fr_CA | 1,542,410,210,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743216.58/warc/CC-MAIN-20181116214920-20181117000920-00161.warc.gz | 456,491,880 | 16,172 | Mathematics, Science and Technology programs seek to empower students with a global understanding and its' general applicability to their lives. At its' essence, to inspire scientific inquiry and develop mathematical resoning so that students may communicate in the language of math and science.
# Teaching and Learning Tools
Geogebra lessons
Desmos lessons
### Evaluation Tools
Success Checker is an online tool that gives teachers and students feedback on each curriculum learning outcome for a particular course. All students and teachers in our community have access from their personalized home page to tests and personalized study guides. See the list of available courses:
• Math Cycle 2, Year 1
• Math Cycle 2, Year 2, Science option
• Math Cycle 2, Year 2, Cultural, Social and Technical
• Math Cycle 2, Year 3, Science option
• Math Cycle 2, Year 3, Technical Science
• Science & Technology, Cycle 2, Year 2
• Science & the Environment, Cycle 2, Year 2
• Physics Cycle 2, Year 3
• Chemistry
• History 404
• History 404 part 2
### Web-based tools
Here you can find varoius teaching tools and web sites that can help you with technology, lab reports, etc. Check out our growing list:
• What is a lab report? A lab report is an information-based text. It is written to describe a scientific experiment. It is also used to analyze the results of the experiment. Go to page
• Technological Analysis Process will help you answer… How does it work? How was it made? What science is involved? What's its impact? Go to page
• Technological Design Process looks at how we design objects or processes to fulfill a need or improve something. Go to page
• All About... Unit Conversion - Unit conversion is a method used to get an answer in the correct units of measurement in problems relating to math and science. Go to page
• All About... Manipulating Significant Figures - The number of meaningful digits expressed in a numerical value is referred to as its’ number of significant figures. To be both precise and accurate, and still facilitate calculation, experimental measurements are expressed using significant figures. Go to page
• Math Memory Aid - A memory aid is a tool that you can use to remind yourself of important information during a test or final exam. Go to page
• LEARN's Teacher Blogs
A few of our teachers are sharing what they are doing in their classes. These great sites also refer to various tools, tips and tricks of the trade, and web sites.
- McSquared
Chris Colley - Educator's Scrapbook
# Professional Learning
### Web Events archives Math and S&T
Check out some of our online web-event archives that deal specifically with math and/or science-related topics, tools or strategies.
• How to use Voicethread to flip your class. Go to page
• Flipped classroom - Learn about the flipped classroom. You will get a brief description of what flipping the classroom is, and a quick look at some of the simplest ways to do it, including tools for recording lessons and ideas for classroom activities. Go to page
• Writing reflectively in Math and Science....really? Whatever for? The benefits of reflective writing for elementary and secondary levels will be discussed from the points of view of the students (self-regulated learning) and the teacher (formative assessment). Go to page
### LEARN Blog Math and ST
From the field, from educational professionals, even from science-friendly former Directors! the LEARN Blog is the place to visit for insightful articles. Feel free to comment on some of these:
• From Liberal Arts to Engineering: A woman’s journey through code Go to page
• Beyond the Textbook: One woman’s STEAM journe Go to page
• Applied Science and Technology – An idea whose time has come Go to page
• Learning to Code is Learning to Learn Go to page
• Beyond the Textbook: Small steps to nurturing girls interest in STEM careers Go to page
• Science Misconceptions – How Should Teachers Deal With Them? Go to page
• How Should we Assess Students in Inquiry-based Science? Go to page
• Beyond the Textbook: Righting the Math Course Go to page
• Beyond the Textbook: The Math Dip Go to page
• Science Fairs: Yea or Nay? Go to page
• What Motivates Students to Want to Learn Science? Go to page
• To Tweet or not To Tweet? Twitter in My Classroom Go to page
### Resource sites
Open Creative Space Resource Site - Go to page
Raspberry Pi Resource Site - Go to page
MicroController Resource Site - Go to page
### Professional Learning Communities / Professional Associations
Stay connected with colleagues on social media.
#MTBoS
Global Math Dept
"We are math teachers who share what we've learned, cause we don't want our classes to suck the energy from students. Professional development among friends, not just colleagues. Fun! Immediately useful! Interesting!" Global Math Department offers weekly webinars. Go to site
#flipclass
Twitter hashtag for tweets related to flipped classroom Go to site
# Curated Resources
Cycle 1 Secondary: Motion Mechanics
Templates
Animated Cards / on CDP site (videos, lesson plan)
Technological Analysis
Mouse traps (scientific investigation)
Tools and Pedagogical Ressources
# MEES Program Documents (MST Secondary)
### QEP Curriculum Documents
• Mathematics, Science and Technology (Elementary Cycle 1, 2 & 3) Download
• Science and Technology (Secondary Cycle 1, Years 1 & 2) Download
• Science and Technology (Secondary Cycle 2, Years 1 & 2) Download
• Applied Science and Technology (Secondary Cycle 2, Years 1 & 2) Download | 1,277 | 5,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.834195 |
https://www.solver.com/risk-solver-help/psi-function-help/simulation/statistics-functions/psiMin | 1,726,801,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00190.warc.gz | 922,448,954 | 15,573 | `PsiMin (cell,simulation)`
PsiMin returns the minimum value attained by the specified uncertain function cell over all the trials in the simulation.
Two additional arguments are utilized when a Psi Statistic function is used with Dimensional Modeling: Struc_format and Param_slice.
`PsiMin (cell,simulation,struc_format,param_slice)`
Struc_format: This argument is an optional argument entered as a string. If omitted, all cube values will be printed in a single column. If “dims” is passed for this argument, the Psi Statistic function will print all dimensions in the cube with their lengths so the user can be advised of the size of the cubes and will be able to estimate the range needed when entering the Psi Statistic function as an array formula. If “vals” is passed for this argument, the result values will be displayed along with the dimension elements in the form of a relational or pivot table. Please see the Simulation example in the Dimensional Modeling chapter in Frontline Solvers User Guide for more information on this function. It’s also possible to use this argument to return the name of a specific element in a cube containing one or more structural dimensions.
To use this argument to return the value of a specific element in a 1-dimensional cube (containing a structural dimension), use the form: “ [StructuralDimisension1].[Element1]”. To use this argument to return the value of a 2-dimensional cube (containing structural dimensions), use the form: “[StructuralDimension1].[Element],[StructuralDimension2].[Element]”.
To use this argument to return the value of a N-dimensional cube (containing structural dimensions), use the form: “[StructuralDimension1].[Element],[StructuralDimension2].[Element],…,[StructuralDimensionN].[Element]”.
Param_slice: The param_slice argument is an optional string argument specifying the desired element “slice” for the parametric dimensions. If omitted the elements selected in the pane will be used. It’s also possible to use this argument to return the value of a specific element in a cube containing one or more parametric dimensions.
To use this argument to return the value of a specific element in a 1-dimensional cube (containing a parametric dimension), use the form: “[ParametricDimisension1].[Element1]”.
To use this argument to return the value of a 2-dimensional cube (containing parametric dimensions), use the form: “[ParametricDimension1].[Element],[ParametricDimension2].[Element]”. | 516 | 2,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.605532 |
https://discourse.mcneel.com/t/loft-elements-between-two-lists/58340 | 1,600,593,567,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00484.warc.gz | 339,899,202 | 5,184 | # Loft elements between two lists?
Hello guys!
i want to make a stair and i am stuck at the last phase. I made to lists of curves and i just want to loft the elements of two lists each other. I want to loft first item of the first list with first item of the second list etc.
What am i doing wrong?
Stairs.gh (15.0 KB)
Like this?
Stairs2.gh (16.4 KB)
not exactly this is the result i take…i want for example to loft the upper edge of the first surface with the down edge of the next surface
I wish I knew what you meant by “first surface” and “lower surface”.
ok let me try to explain…I have all these vertical surfaces. In this case you create a loft surface connect the same list edge and more specific the edge with index 1. What if i wanted to loft the edge with index 1 with the edge with index 3 of the next surface.
How’s this?
Stairs3.gh (17.8 KB)
1 Like
yeah… this is what i want…thank you very much !! | 254 | 942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.897869 |
http://math.stackexchange.com/questions/201189/four-easy-numerical-questions?answertab=oldest | 1,409,446,624,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500835844.19/warc/CC-MAIN-20140820021355-00254-ip-10-180-136-8.ec2.internal.warc.gz | 129,562,719 | 17,491 | # Four easy numerical questions
$7^5 \cdot 3^2 \cdot 5 + 3$, is it a composite number?
Find a quadratic polynomial whose zeroes are $3 + \sqrt{5}$ and $3 - \sqrt{5}$
Solve for x and y $\frac x a + \frac y b = 2$ ; $ax - by = a^2 - b^2$
Prove that $sec^2\theta + cosec^2\theta = sec^2\theta*cosec^2\theta$
-
Please give me an explaination instead of hints, my exams are tomorrow and i am sitting on heaps of practice questions... – Aayush Agrawal Sep 23 '12 at 15:14
For the first question, the number is divisible by $3$ but greater than $3$, so it is composite. – André Nicolas Sep 23 '12 at 15:17
(1) The number is sum of two numbers,so is even and $> 2,$ so can not be prime. – lab bhattacharjee Sep 23 '12 at 15:18
@labbhattacharjee, I think you meant to say that the number is a sum of two odd numbers. – Joel Reyes Noche Apr 21 '13 at 5:28
@JoelReyesNoche, yes. – lab bhattacharjee Apr 21 '13 at 5:42
Too many questions in too few days, and you show no insights, self effort..., Here are some hints:
$$1)\;\;\; 7^5\cdot 3^2\cdot 5+3=3\,(7^5\cdot 3\cdot 5+1)$$
$$(2)\;\;\;(x-(3+\sqrt 5))(x-(3-\sqrt 5))\;--\text{(Note that this polynomial is rational)}$$
$$(3)\;\;\;\frac{x}{a}+\frac{y}{b}=2\Longrightarrow bx+ay=2ab\Longrightarrow\,\text{we have the linear system:}$$
\begin{align*}ax-by=&a^2-b^2\\bx+ay=&2ab\end{align*}
Since from the given data $\,a,b\neq 0\,$ (why?), we get above $\,x=a\,$
Multiply now the first eq. by $\,a\,$ and the second one by $\,b\,$ and get:
\begin{align*}a^2x-aby=&a^3-ab^2\\b^2x+aby=&2ab^2\end{align*}
Now sum both eq's and solve for $\,x\,$ :
$$(a^2+b^2)x=a^3+ab^2=a(a^2+b^2)$$.
Since from the given data $\,a,b\neq 0\,$ (why?), we get above $\,x=a\,$ . Substitute now in either equation and get $\,y\,$ .
$$\sec^2t+\csc^2t=\frac{1}{\cos^2t}+\frac{1}{\sin^2 t}=\frac{\sin^2t+\cos^2t}{\sin^2t\,\cos^2t}$$
Now just remember the trigonometric Pythagoras Theorem $\,\cos^2x+\sin^2x=1\,$ and you get that the above equals what you want.
-
I know you may think badly of me, but i am kind of in alot of stress. I havent slept in 3 days, and the biggest exam of my life is in 2 hours from now. I keep pouring practise question over question, and hoping to get all the ones i cant do solved. – Aayush Agrawal Sep 24 '12 at 0:35
It's not a matter of thinking bad or not. Hopefully you understand now that it is not a good idea to leave all aside until one week or so before the exam. You must work through this stuff much sooner as to be able to assimilate ideas, tricks, exercises, etc. – DonAntonio Sep 24 '12 at 2:46
Its not just Math sadly..Its several exams and they are just one after the other. Theres a heck load of material, too much for any one person to process in this much time... There are many topics that are all equally as challanging, and one after the other. These include topics like Math, Physics, Chemistry, Biology and other things like sanskrit(A extremely ancient language) along with hindi, english and even stuff like General knowledge and moral science. That packed with testing on other levels, such as the activities we chose, i chose chess. On top of that 1/4 of your grade is on assignmen – Aayush Agrawal Sep 24 '12 at 9:34
And EVERY subject(total 15) give us a difficult assignment. All this in half a year, every assignment is too time consuming, and a major portion of the time goes to this. Then all the other things and then the exams...And after this starts the 2nd term which is even more challanging than the first. And after all this is done you get a piece of paper, that is your entire academic career. Those who get a bad score on this marksheet might aswell jump off a building, because no college will ever take them in and they will never get a respectable job... – Aayush Agrawal Sep 24 '12 at 9:41 | 1,205 | 3,803 | {"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} | 4.21875 | 4 | CC-MAIN-2014-35 | latest | en | 0.730991 |
https://askworksheet.com/math-worksheets-for-grade-1-addition-with-regrouping/ | 1,713,839,497,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00407.warc.gz | 98,053,468 | 28,597 | We hope you find them very useful and interesting. Get help online or offline.
4 Printable Measurement Worksheets For Kids Measurement Worksheets 2nd Grade Math Worksheets First Grade Math
### The geometry worksheet include recognizing drawing basic shapes and solving shape patterns and sequences.
Math worksheets for grade 1 addition with regrouping. Addition within 30 1 digit number to a 2 digit number with regrouping three worksheets. There is no regrouping in those worksheets so kids will quickly build their math fact addition fluency. Free printable addition worksheets with regrouping.
No regrouping is included in our grade 1 exercises. Our addition with regrouping worksheets introduces this concept in an easy to understand way. Mixed operation worksheets cover both addition up to 100 and subtraction within 100 and re great to boost math skills for grade 1 math students.
Grade 1 addition worksheets these math worksheets start with simple addition using pictures or number lines followed by one digit math facts and then progress through number bonds adding doubles missing addends adding whole tens and adding in columns. Math goodies helps kids at all levels with interactive instruction and free resources. This math worksheet was created on 2014 10 02 and has been viewed 1 351 times this week and 9 484 times this month.
We have crafted many worksheets covering various aspects of this topic and many more. Math help since 1998. It may be printed downloaded or saved and used in your classroom home school or other educational environment to help someone learn.
With fun activities and engaging problems your child will enjoy being challenged at home and at school. Then introduce them to adding more significant numbers tens and finally regrouping addition. These worksheets are appropriate for first grade math.
Below you will find a wide range of our printable worksheets in chapter regrouping tens and ones of section addition. These free math worksheets will help your students learn how to regroup with strategic place value grids as a guide. Addition with regrouping worksheet free 2 digit addition with regrouping worksheet for beginning math skills.
Addition with regrouping worksheets are inspired by educators with 2nd and 3rd graders in mind. Content filed under the addition regrouping category. They will still be challenged with the pages that include a missing addend and that is a great way to get them ready for subtraction.
Free Beginner Addition Regrouping Worksheets Closet Of Free Samples 2nd Grade Worksheets Free Math Worksheets Addition With Regrouping Worksheets
2 Digit Addition And Subtraction With And Without Regroupi First Grade Math Worksheets 2nd Grade Math Worksheets Math Worksheets
2 Digit Subtraction Worksheet 3worksheets 2nd Grade Math Worksheets Subtraction With Regrouping Worksheets 2nd Grade Math
Addition With Regrouping Adding 2 Digit Numbers To 1 Digit Numbers Https Www First Grade Math Worksheets 1st Grade Math Worksheets Math Practice Worksheets
4 15 14 For Him Three Digit Addition With Or Without Regrouping Turtlediary Com Mathematics Worksheets Math Worksheets Addition Worksheets
The 2 Digit Plus 1 Digit Addition With Some Regrouping A Math Worksheet Fro Math Fact Worksheets Addition And Subtraction Worksheets Math Addition Worksheets | 646 | 3,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.927423 |
https://blog.mensor.com/blog/how-to-understand-and-detect-leaks-in-a-pneumatic-calibration-system | 1,717,019,083,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00797.warc.gz | 111,047,105 | 14,095 | When conducting the calibration of a pressure sensor, it is important to make sure there are no leaks in the connection between the sensor in the calibrator and the sensor being calibrated. This is important because in a closed, steady-state system, a leak will cause a reduction in pressure and possibly a gradient in pressure across the system.
## What is a leak?
It may sound obvious, but a leak is what happens when media escapes or enters a system, usually as the result of a pressure gradient from high to low. There are situations that may appear to be a leak but are actually due to other factors. These become apparent when examining the relationship between volume, temperature, and the number of the molecules within a system as expressed in the ideal gas law:
PV=nRT
where P, V and T are pressure, volume and temperature; n is the amount of the substance (number of molecules); and R is the ideal gas constant.
The ideal gas law shows us that pressure multiplied by volume is always equivalent to the number of molecules multiplied by the ideal gas constant and temperature. Therefore, with V and T constant, any decrease in the number of molecules (n) will cause a decrease in the pressure (P↓), indicating a leak out of the system. In a system under vacuum, an increase in the number of molecules (n↑) will cause an increase in pressure (P↑), indicating a leak into the system. These are true leaks and should be corrected to maintain a stable system.
To complicate the situation, changes of other variables in the ideal gas law will give the appearance of a leak without any true leak being present. For example, with an increase in volume (V) or a decrease in temperature (T↓), pressure will decrease, which looks like a leak. In addition, with an influx of more gas (n↑) or if there is a quick decrease in volume (V), the gas will heat up and the pressure will rise. On the other hand, when there is a quick reduction in the amount of gas(n) or a quick increase in volume (V↑), the gas will cool down and the pressure will drop.
After a system goes through one of these changes, the system temperature will gradually reach equilibrium with the surrounding environment, which will also change the pressure accordingly. These pressure changes are often and incorrectly observed as a leak in the system. But wait... there’s more! If the temperature of the environment changes and is different than the internal gas temperature of the system, the system temperature will tend to equilibrate with the environment. This change in temperature will affect the pressure in a similar way, per the ideal gas law. For this reason, when conducting a calibration, it is important to allow the calibrator and the device under test (DUT) to warm up so there is not a temperature gradient between the environment and the system.
To complicate this even more (and beyond the scope of this post), at low absolute and gauge pressures, the system pressure can be more sensitive and susceptible to system geometry and temperature changes.
Leaks occur when there is a breach in a closed system. The breach might occur at fittings used to connect tubing to the calibrator and the DUT. It could also occur within the calibrator, the DUT or the connecting tubing. Pressure controlling calibrators have manifolds, regulators, check valves and connections to their internal transducers. All of these have the potential to leak. DUTs come in a variety of configurations but all have connections that hold the device in place where there is a potential for leakage. Leaks are due to improper fastening or degradation of the seal due to wear, vibration or stress.
For systems that contain a calibrator, manifolds or tubing, and a DUT, each section of the system can be isolated to check for leaks in each area. It can be said the purpose of a controller/calibrator is to introduce a leak or supply pressure to a system and balance it at a setpoint. From this point of view, the DUT can never truly be leak-free while the controller is in control mode.
A Mensor pressure calibrating controller can be controlled to a pressure above atmospheric and placed in measure mode. This isolates all the gas contained in the system from the pressure regulator and creates a closed system between the pressure transducer in the controller and the DUT. With the controller and environment at an equivalent temperature, the pressure can be monitored to see if it changes. If it is seen to decrease in a continuous way, then you can be certain there is a leak somewhere in the system. If the pressure initially decreases but then stabilizes, there is probably not a leak. The initial decrease is due to the cooling of the internal gas that was heated because of the pressurization. This is a tell-tale sign of a pressure change caused by a temperature variation. To monitor this rate of change, many controllers display the pressure rate of change value in their display for easy readout and monitoring.
To isolate a leak further, vent the system and disconnect the tubing at the measure/control port of the controller and plug the port. Pressurize the system again and observe the pressure in measure mode. If the pressure becomes stable, the leak must be in the portion of the system external to the controller.
## Pinpoint and solve problematic leaks in your system
If it is determined that the leak is external to the controller in the process connections, manifold or tubing, there are a number of ways to correct these issues. Use a leak detector solution made with deionized water and a surfactant (soapy water), and with the system under pressure, apply the solution to suspect areas. If there is a leak, bubbles will form and grow, indicating a breach. To remediate the discovered leak, tighten or replace tubing and/or fittings and test the system again.
Be sure the pressure fittings, tubing and manifold are adequately rated for the pressure range and media you are using. For fittings or connections that require O-rings, inspect them for metal shavings or nicks. It is good practice to use vacuum grease to lubricate the connection and aid in sealing if the O-ring connection ports are frequently adjusted or changed.
Pressure leaks can be frustrating and may take some time to isolate and resolve. The measures described above should assist you in finding any leaks in your system.
If the leak is determined to be internal to the controller, contact Mensor Customer service and they may be able to troubleshoot over the phone or direct you to return the product to Mensor for repair. | 1,343 | 6,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.933035 |
http://discuss.tlapl.us/msg00295.html | 1,566,298,319,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00275.warc.gz | 58,773,121 | 1,864 | # Assuming a contiguous subset of Nat
I wanted to express an assumption that my set Ver is a subset of Nat and is contiguous. It seems it can be done using a Max() operator, like below:
(* A singleton set with a maximal element from S or empty set *)
Max(S) == { i \in S : \A j \in S : i >= j }
ASSUME /\ Ver \subseteq Nat \* Ver is a subset of Nat
/\ \A i \in Ver : i + 1 \notin Ver => Max(Ver) = { i } \* Ver is contiguous
Is there a better/cleaner way to express this assumption?
Thank you,
Yuri | 144 | 507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | longest | en | 0.813177 |
https://math.stackexchange.com/questions/667830/modern-research-into-grassmans-theory-of-forms/ | 1,657,121,808,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00717.warc.gz | 434,762,644 | 72,263 | # Modern research into Grassman's "theory of forms"?
I quote from Petsche's Hermann Graßmann: Biography (emphasis mine):
The mathematical part of the book begins with the conception of the “General Theory of Forms”. Starting with a perspective on mathematics as a theory of forms, Graßmann analyses in the most abstract way possible the general structures of concrete “conjunctions of forms”. Here, he places special emphasis on “elementary conjunctions”, demanding they have module properties, that is, associativity, commutativity and an inverse and neutral element. The so-defined conjunction of the first order, or “formal addition”, is then followed up by an investigation of a conjunction of the second order (“formal multiplication”), for which he only requires distributivity with respect to formal addition. Graßmann directly posits the validity of the module properties for formal addition and distributivity for formal multiplication as the principles for constructing these conjunctions: “This generally is the way”, he wrote, “that initially, that is when no species of conjunction is yet given, such a conjunction of next higher order is defined.”
Since Graßmann does not require the forms generated by conjunctions of the second order to be embedded in the fundamental domain, he can use this form of conjunction for the formal generation of new mathematical objects in the further course of the text.
...
After Graßmann has laid down the foundation for all mathematical disciplines by presenting these uniquely generalized group-theoretical and structural abstractions he starts with the actual presentation of his new mathematical discipline.
What is the modern terminology for Grassmann's "General Theory of Forms"? What research work has been done in order to continue this line of thinking? Which resources could I acquaint myself with in order to answer these questions?
I think the answer is simply (and very generally, thus unhelpfully) "universal algebra": http://en.wikipedia.org/wiki/Algebraic_structure
Am I on the right track? I am not sure. See also this question of mine, which is looking for something similar in spirit.
Can someone with familiarity weave together a proper answer from these three resources and others as appropriate?
• Group actions? en.wikipedia.org/wiki/Group_action Feb 7, 2014 at 21:21
• @Martín-BlasPérezPinilla I think it's more general than that, see: "by presenting these uniquely generalized group-theoretical and structural abstractions he..." Feb 7, 2014 at 22:32
• I'm not clear what you're looking for, here. Grassman is one of the founding figures of abstract algebra, but after over a hundred years of development his influence is diffuse. Feb 12, 2014 at 13:49
• Exterior forms, differential forms and geometric algebra (when composed with Clifford algebra) all seem relevant. Jun 5, 2014 at 22:45
• This MO thread may be of interest too. Jun 5, 2014 at 23:00
Universal Algebra: Broadly speaking, Whitehead in his Treatise on Universal Algebra was motivated by the work of Boole, Hamilton and Grassmann, and eventually Maltsev and Polish mathematicians developed the general theory. However, Grassmann himself was mostly focused on binary associative operations.
Hypercomplex numbers: From abstract algebra point of view Grassmann studied some finitely generated associative algebras over the field of real numbers. In 19th century this area came to be known as linear associative algebra, and was further developed by Peirce and Wedderburn among others. The modern name is hypercomplex numbers and associativity is no longer required, examples are dual and split-complex numbers, quaternions, octonions, etc. Although purely algebraic approach is more in the spirit of Hamilton than Grassmann some hypercomplex numbers are closely related to non-Euclidean geometries and appear in mechanics, relativity and quantum theory. New interesting hypercomplex numbers were introduced by Musès in 1980s, although most of his work is more in the realm of numerology than mathematics.
Multilinear Algebra: Grassmann's own work was more specific than universal algebra and more intuitive than abstract algebra, in modern terms he was interested in algebras generated by a vector space, like exterior or Clifford algebras, with emphasis on geometric interpretations. And it's not just the scope but also the viewpoint, he was interested in geometric content of the operations (wedge product, Clifford product, etc.) more than their algebraic properties. Formally, the closest modern analog would be multilinear algebra if not for the prevailing tensor approach that reduces everything to index manipulations and assigns geometric interpretations after the fact if at all.
Geometric Algebra: Gian-Carlo Rota is the most prominent recent proponent of Grassmann's original point of view, he and his students introduced new geometric operations in his spirit, and developed their theory. Rota's motivation came from the classical invariant theory and he injected some combinatorial flavor into Grassmann's constructions. "Geometric algebra" is a term occasionally used to describe this circle of ideas, but most authors treat it as a synonym of Clifford algebra, which is just one example albeit universal in some sense (not to be confused with geometric algebra of ancient greeks, which is something else). And Rota's approach is an exception rather than a rule in multilinear algebra.
Differential geometry: Where the spirit of Grassman is preserved more it's differential geometry. After Elie Cartan's introduction of differential forms there was a tendency to abolish tensor algebra's "debauchery of indices" in favor of coordinate invariant notation because there is no global choice of coordinates on manifolds. Differential forms are Grassman's exterior forms that vary over a manifold, i.e. one generates exterior algebras over tangent spaces and takes smooth cross sections across the manifold. Of course, other Grassman type algebras can and are used instead. The advantage is that Grassman's definitions of operations are intrinsic, and automatically extend to bundles once some underlying structure is fixed (e.g. Riemannian metric for Clifford algebras). They can then be used to define differential operators that are explicitly invariant under transformations that preserve the underlying structure (e.g. defining codifferential using Hodge star). Physicists working with gauge theories on manifolds sometimes favor this approach, but even in differential geometry tensors dominate because they can be manipulated with much less geometric insight.
Sources: Yaglom’s book has an excellent historical exposition of Grassman’s ideas and their relation to Hamilton’s quaternions and hypercomplex numbers in general. This is a more recent historical survey of the subsequent work of Clifford, Lipschitz, Study and Cartan with emphasis on applications to physics. Wikipidea article on hypercomplex numbers gives a good overview with many historical and modern references, for Musèan hypernumbers see here. Rota’s programmatic paper reviving Grassman’s ideas and applying them to the classical invariant theory is quite readable, its short and elementary review is here. Applications to spinors and representation theory can be found in this book. Warner’s text systematically applies invariant approach in differential geometry.
• David Hestenes is also another major proponent of the Geometric Algebra angle. What you wrote in your answer is what I am familiar with so far, but I think this passage actually refers to Grassman's general viewpoints on mathematics and how to create structure in mathematics. This is what I would someone to address, not the geometric algebra angle, which is already well known to me. Jun 6, 2014 at 3:21
• "After Graßmann has laid down the foundation for all mathematical disciplines by presenting these uniquely generalized group-theoretical and structural abstractions he starts with the actual presentation of his new mathematical discipline." Jun 6, 2014 at 3:23
• My interpretation of the passage you quoted is that conjunction of the first order creates a vector space or module, and conjunction of the second order is a formal product, i.e. tensor product. He can then generate the tensor algebra of new mathematical objects, and carve out more specialized structures by imposing additional relations. They admit new conjunctions like progressive and regressive products. Higher order conjunctions may refer to iterating second order ones. Insisting on commutativity, associativity and distributivity significantly narrows the scope though. Jun 6, 2014 at 6:10
• Not sure if it's of interest to you, but you get additional structure by having a group representation on the underlying module, it induces representations on the tensor algebra and its descendants, and you can ask for invariants and covariants that transform nicely, classify tensors, etc. This is the context of the classical invariant theory, but it goes beyond Grassman jones.math.unibas.ch/~kraft/Papers/KP-Primer.pdf. Jun 6, 2014 at 6:43
• Grassmann tries to be general, but imposes algebraic conditions that lead him to modules without any geometry. If you are looking for constructions for "all of mathematics", as your other question suggests, you will not find them in Grassmann. Whitehead later suggested studying sets with general operations (using Boolean and Grassmann algebras as motivation), which eventually led to universal algebra of Maltsev and the Polish school. Even that is not truly "universal" though since structures on sets do not have to be operations. But is this more what you are looking for? Jun 6, 2014 at 20:30
I think this passage is referring to the exterior algebra of a vector space. It's designed to have properties suitable for Grassmann's approach, and indeed can be viewed as a universal arrow in a certain category, but it does not really have anything to do with universal algebra, per se.
Grassmann's work later was applied in multilinear algebra and seeded the study of differential forms. Other than this, Grassmann's ideas were also built upon by Clifford. This eventually lead to the study of Clifford algebras and so-called "geometric algebras," which are special cases of Clifford algebras. This direction has a slightly different flavor than that of differential forms, but that's not to say they don't overlap and work together.
A great thing to read to get a better feel for the role of differential forms and Grassmann's and Clifford's legacies is Roger Penrose's Road to reality chapters 6 and 11.
• I would have to disagree with you. The passage that I quote specifically refers to a more general insight of Grassman's, which he uses to then build his geometric algebra upon. Jun 6, 2014 at 3:22
• "After Graßmann has laid down the foundation for all mathematical disciplines by presenting these uniquely generalized group-theoretical and structural abstractions he starts with the actual presentation of his new mathematical discipline." Jun 6, 2014 at 3:24
• Dear @user89 : honestly, this quote is a bit weird. I've never heard of Grassmann's work providing a foundation for "all mathematical disciplines," and I doubt that is literally true. Considering it sounds like a sensationalized passage in a biography, might I recommend you don't take it as gospel? Regards Jun 6, 2014 at 10:11
• @user89 the entire paragraph after your first highlight appears to be talking about exterior algebra. Jun 6, 2014 at 10:13
• @user89 OK, but that is my reading of After Graßmann has laid down the foundation for all mathematical disciplines Jun 6, 2014 at 16:41
Another resource is John Browne's book Grassmann Algebra Volume 1: Foundations.
There is also a fairly extensive free Mathematica Application available to reasonably serious individuals described at Grassmann Calculus. | 2,456 | 11,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.905065 |
https://polypad.amplify.com/es/lesson/isometric-2 | 1,716,505,176,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00641.warc.gz | 391,681,940 | 7,323 | La autenticación de Google falló. ¡Por favor, inténtelo de nuevo más tarde!
# Isometric Drawings: Part II
## Overview and Objective
In Isometric Drawings: Part 1, students will construct the side views of the three-dimensional objects and then, they will construct the 3D objects whose side views are given.
Before starting this activity, you may want to check the Isometric Drawings Part I Task.
## Warm-Up
You may use an isometric puzzle similar to the one in the Isometric Drawings Part I Task. Show the drawing to students and ask them how many cubes they see.
Students may answer as six or seven. You may want to mark the cubes to clarify the results for all students.
Illustrators and designers often add side views to their drawings to make sure that their audience fully understands the drawings.
## Main Activity
Clarify with the students that during this activity, the blue rhombus from the polygons toolbar will be used to create the unit cubes.
Start with the L shape (L-Prism). You may use this Polypad to demonstrate the side views of the L-shape. Think of a mirror cube surrounding the 3D object. Each mirror will only reflect the sides they are facing.
Invite students to draw the side views of another shape using the mirror cube.
Share some student work with the class. Invite students to share which approaches they found most useful when drawing the 3D figures.
Drawing what an object looks like from several different views is called orthographic projection. Orthographic projection is also called engineering drawings or plan views. The orthographic projection can have more than three side views if the object has unique sides that would not be fully described by three views
If an object which is not formed by unit cubes were to be drawn by the orthographic projection, additional lines are used to represent changes in depth.
Students will now construct the 3D objects given the side views. Let students work as pairs or groups by sharing this Polypad. You may ask them to predict the number of unit cubes for each construction before they start drawing. Students can compare their predictions after each construction.
## Closure
You may end the lesson by discussing that some objects may require more side views to be constructed accurately. If you have completed Isometric Drawings: Part I with the students, you may also discuss which method (number diagrams or side views) is easier to construct 3D objects. | 501 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-22 | latest | en | 0.925343 |
http://list.seqfan.eu/pipermail/seqfan/2015-December/015778.html | 1,642,755,036,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00019.warc.gz | 41,929,769 | 3,161 | # [seqfan] A000108(n) ≡ 1 (mod 6)
Antti Karttunen antti.karttunen at gmail.com
Sat Dec 5 15:49:55 CET 2015
```Ed, I'm trying to add to
https://oeis.org/draft/A265210
the following line to Crossrefs-section:
Cf. A037096 (numbers constructed from the inverse case, base 2 digits of 3^n).
but I keep on getting "502 Proxy Error from OEIS-server".
In any case, I guess that the same general process that forces the
columns in A037096 to be periodic applies also in your case.
Best,
Antti
On Sat, Dec 5, 2015 at 10:34 AM, <seqfan-request at list.seqfan.eu> wrote:
> Message: 9
> Date: Sat, 5 Dec 2015 02:28:34 -0600
> From: "L. Edson Jeffery" <lejeffery2 at gmail.com>
> To: seqfan at list.seqfan.eu
> Subject: [seqfan] Re: A000108(n) ≡ 1 (mod 6)
> Message-ID:
> <CAGRLqMhzvUKBxVUronzwKVj+k9isrQajzWe0=EucDgN4M-Nw5g at mail.gmail.com>
> Content-Type: text/plain; charset=UTF-8
>
> Robert,
>
> In one of your other posts on this topic (see
> http://list.seqfan.eu/pipermail/seqfan/2015-November/015590.html), you gave
> a probabilistic argument. I wanted to try another version here.
>
> I just proposed A265210 which is an irregular triangle in which row n gives
> the base 3 digits of 2^n in reverse order. The first few rows are
>
> {1},
> {2},
> {1, 1},
> {2, 2},
> {1, 2, 1},
> {2, 1, 0, 1},
> {1, 0, 1, 2},
> {2, 0, 2, 1, 1},
> ...
>
> Let the columns be indexed by k=1,2,... .
>
> Conjecture: (i) The sequence in column k is periodic with period p(k) =
> 2*3^(k-1); (ii) The numbers 0,1,2 appear with equal frequency in each
> column (except column 1 which obviously contains only the numbers 1,2 with
> equal frequency).
>
> Assume that the conjecture is true. I imagined, after studying the triangle
> to some extent, that the distribution of the 2's throughout the triangle is
> such that, for fixed k, removing all rows which contain a 2 in column k
> does not affect the relative frequency of the digits 0,1,2 in columns k+1,
> k+2, ..., so let's further assume that is the case. Clearly, upon removing
> those rows, all columns remain periodic (proof is easy). Of course, the
> period is reduced.
>
> If we now iterate the process of removing all rows which contain a 2 in
> column k, for each k in {1,...,infinity}, then we will have removed
>
> 1/2 + 1/3 + 1/9 + 1/27 + ...
>
> of the rows. Since
>
> 1/2 + Sum_{j=1..infinity} 1/3^j = 1,
>
> I think we will have removed either all but finitely many of the rows or
> all rows except for some set of measure zero. Of course, the former case
> means that the set of all m such that 2^m is a sum of powers of 3 must be
> finite which is what we'd like to know. I don't think the latter case helps
> at all, but I see no way around it.
>
> Could proof of that conjecture be any easier than the original problem?
>
> Ed Jeffery
>
>
``` | 923 | 2,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.821422 |
https://phpcontext.com/post/81769/electrical-electrical-wiring-diagram-for-a-house-electrical-circuit-calculation-examples.html | 1,553,573,168,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204790.78/warc/CC-MAIN-20190326034712-20190326060712-00541.warc.gz | 573,428,109 | 12,680 | Home » Wiring Diagram » Electrical Electrical Wiring Diagram For A House Electrical Circuit Calculation Examples #7593
# Electrical Electrical Wiring Diagram For A House Electrical Circuit Calculation Examples #7593
Description: Electrical Electrical Wiring Diagram For A House Electrical Circuit Calculation Examples #7593 from the above 1200x872 resolutions which is part of the "DIY Needed" directory. Download this image for free in HD resolution the choice "Save full resolution" below. If you do not find the exact resolution you are looking for, then go for a native or higher resolution.
Undertaking electrical wiring by yourself is usually difficult. This is often specially so when you absence the know-how as well as working experience in electrical things. The point is you can't do a trial and error strategy when dealing with electrical wiring. Errors could charge you a fortune or even your life. That may be why just before starting off any do-it-yourself electrical repair, you will need to talk to some electrical wiring concerns simply to ensure you know what you're accomplishing. Another thing about electrical wiring is that wires are frequently coloration coded. So, it can be essentially much easier to learn which ones go together with which. Below are definitely the most frequent electrical wiring inquiries whose answers typically contain shades to help you identify each individual particular wire. - Spartan Race Electric Wires
Solved Electric Circuits For The Circuit As Shown In The Electric Circuits Images Of Electric Circuits #4456
Electricity Experiments For Kids Frugal Fun For Boys And Girls A Electrical Circuit Games For Kids #7000
How can You Wire a Change?
Certainly one of quite possibly the most frequent electrical wiring queries is on how to wire a change. When using switches in your house is kind of quick, wiring just one may well not be that quick for everyone. An ON-OFF switch is really quite simple to wire. There are actually different types of switches, but for this example, let us say you happen to be installing a single-pole toggle switch, an exceptionally popular change (along with the easiest).
You can find three colours of wires in a common single-pole toggle switch: black, white, and green. Splice the black wire in two and join them about the terminal screws - just one on top plus the other within the bottom screw on the swap. The white wire serves for a source of uninterrupted electricity and it is typically related to a light-weight coloured terminal screw (e.g., silver). Join the environmentally friendly wire for the ground screw of one's change.
These methods would commonly be adequate for making a standard switch perform without a trouble. Then again, for those who usually are not assured which you can execute the endeavor appropriately and properly you far better let the professionals do it instead. Just after all, there is certainly a purpose why this activity is without doubt one of the commonest electrical wiring inquiries requested by the majority of people.
How can You Wire a Ceiling Fan?
For many rationale, the best way to wire a ceiling fan is likewise considered one of essentially the most typical electrical wiring inquiries. To simplify this undertaking, you can use only one swap for a single ceiling fan. To wire the lover, it truly is just a issue of connecting the black wire of the ceiling fan into the black wire with the change. When there is a lightweight, the blue wire need to be connected on the black wire in the change at the same time.
How can You Switch a Breaker and just how Do you Increase a Sub Panel?
Even though numerous try to carry out these tasks by themselves, most people are encouraged to rent an electrician as an alternative. It is more sophisticated and for that reason risky for many men and women to test to interchange a breaker or insert a panel. To give you an thought about these common electrical wiring queries, you would must perform on the sizzling electrical panel. Should you never even really know what this implies, you happen to be merely not equipped to perform the task oneself. Even if you need to commit much more by choosing an experienced electrician, it is really considerably safer plus much more smart to do this instead. - Electrical Electrical Wiring Diagram For A House Electrical Circuit Calculation Examples #7593
There are actually causes why these are definitely by far the most typically questioned electrical wiring queries. Just one, a lot of think it truly is straightforward to accomplish, and two, they're the popular electrical tasks at home. But then you must not set your protection at risk with your aim to save money. The stakes could even be a lot better for those who try to save cash and do an electrical wiring job with no adequate know-how or experience.
Tablets | iPad HD resolutions: 2048 x 2048
iPhone 5 5S resolutions: 640 x 1136
iPhone 6 6S resolutions: 750 x 1334
iPhone 6/6S Plus 7 Plus 8 Plus resolutions: 1080 x 1920
Android Mobiles HD resolutions: 360 x 640 , 540 x 960 , 720 x 1280
Android Mobiles Full HD resolutions: 1080 x 1920
Mobiles HD resolutions: 480 x 800 , 768 x 1280
Mobiles QHD | iPhone X resolutions: 1440 x 2560
Tablets QHD | iPad Pro resolutions: 2560 x 2560
Widescreen resolutions: 1280 x 800 , 1440 x 900 , 1680 x 1050 , 1920 x 1200 , 2560 x 1600
Retina Wide resolutions: 2880 x 1800 , 3840 x 2400
HD resolutions: 1280 x 720 , 1366 x 768 , 1600 x 900 , 1920 x 1080 , 2560 x 1440
Ultra HD 4K resolutions: 3840 x 2160
Ultra HD 5K resolutions: 5120 x 2880
Ultra HD 8K resolutions: 7680 x 4320
### Delectable Power Circuit Wiring Color
Disclaimer: All of our content in the form of images we display because we believe the content was public domain. picpaper that displayed are from unknown origin, and we do not intend to infringe any legitimate intellectual, artistic rights or copyright. If you are the legitimate owner of the one of the content we display the picpaper, and do not want us to show, then please contact us and we will immediately take any action is needed either remove the picpaper or maybe you can give time to maturity it will limit our picpaper content view. All of the content we display the picpapers are free to download and therefore we do not acquire good financial gains at all or any of the content of each picpaper. | 1,400 | 6,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-13 | latest | en | 0.896269 |
https://list.indology.info/pipermail/indology/1998-November/014063.html | 1,708,693,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00626.warc.gz | 359,103,143 | 3,043 | # Paired Horse and PIE breakup
H.M.Hubey hubeyh at MONTCLAIR.EDU
Mon Nov 9 01:11:56 UTC 1998
```Dominique.Thillaud wrote:
>
> Dear Mark,
> You wrote (Sat, 7 Nov 1998 00:35:44 -0500):
> >Doesn't this sound a little like ...
> But, few time before (Fri, 6 Nov 1998 23:45:21 -0500):
> >I am a computer scientist and engineer.
> Hence, I'm able to suppose you have some knowledge of combinatories
> and statistics:
> - assuming that for each word there is a score of synonymous or
> quasi-synonymous (lying in the same semantic field) and twice many
> metaphores.
> - assuming that for each "sound" there is an half-dozen of "little
> like" sounds.
> You're surely able to compute how many "good" correspondances could
> be found between any two languages: incredible!
PIck up my paper on the topic on my web site. That is for exact matches.
With a semantic shift of N, and a phonetic similarity/shift of M, the
average number is N*M times the calculations shown in my paper and my
book. Since the average is 1. The average, providing "leeways", is
simply
about multiplying this number by some constant. One way is N*M as I
showed above. I am sure that everyone can do multiplication, even you.
> With such methods and a touch of imagination, fabulous historical
> conclusions can be obtained:
> < Ancient dwellers of England were highly attracted by Roman girls and made
> < currently razzias to obtain them. That's proved by "woman < romana" and
> < "girl < clara" (fair complexion).
Maybe this could be obtained by people whose capabilities are about as
high as yours, but certainly not by me :-)
I know more about probability theory than sophomore level combinations
and permutations.
> In the same way, you have good forebears: A counsellor of Hitler
> (better to forget his name) gave him a good reason to invade England: they
> were all Jewish, as proved by "Saxon < Isaac-son".
> There is no doubt that "Turcs in Etruria" is in the same vein.
> Regards,
> Dominique
Little bit more sarcasm, even if only to prove that you are upset about
something quite serious.
> Dominique THILLAUD
> Universite' de Nice Sophia-Antipolis, France
--
Best Regards,
Mark
-==-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
hubeyh at montclair.edu =-=-=-= http://www.csam.montclair.edu/~hubey
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
The information transmitted is intended only for the person or entity
to which it is addressed and may contain confidential and/or privileged
material. Any review, retransmission, dissemination or other use of,
or taking of any action in reliance upon, this information by persons
or entities other than the intended recipient is prohibited. If you | 687 | 2,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.965557 |
https://lcx.cc/post/96/ | 1,656,462,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00341.warc.gz | 424,001,214 | 10,726 | Option Explicit
'计算谷歌pr查询加密参数模块
'示例: "
'算法有点问题,部分参数计算错误。影响不是太大。
End Function
Function Sl(ByVal x, ByVal n)
If n = 0 Then
Sl = x
Else
Dim k
k = CLng(2 ^ (32 - n - 1))
Dim d
d = x And (k - 1)
Dim c
c = d * CLng(2 ^ n)
If x And k Then
c = c Or &H80000000
End If
Sl = c
End If
End Function
Function Sr(ByVal x, ByVal n)
If n = 0 Then
Sr = x
Else
Dim y
y = x And &H7FFFFFFF
Dim z
If n = 32 - 1 Then
z = 0
Else
z = y \ CLng(2 ^ n)
End If
If y <> x Then
z = z Or CLng(2 ^ (32 - n - 1))
End If
Sr = z
End If
End Function
Function ZeroFill(ByVal a, ByVal b)
Dim x
If (&H80000000 And a) Then
x = Sr(a, 1)
x = x And (Not &H80000000)
x = x Or &H40000000
x = Sr(x, b - 1)
Else
x = Sr(a, b)
End If
ZeroFill = x
End Function
Private Function Uadd(ByVal L1, ByVal L2)
Dim L11, L12, L21, L22, L31, L32
L11 = L1 And &HFFFFFF
L12 = (L1 And &H7F000000) \ &H1000000
If L1 < 0 Then L12 = L12 Or &H80
L21 = L2 And &HFFFFFF
L22 = (L2 And &H7F000000) \ &H1000000
If L2 < 0 Then L22 = L22 Or &H80
L32 = L12 + L22
L31 = L11 + L21
If (L31 And &H1000000) Then L32 = L32 + 1
Uadd = (L31 And &HFFFFFF) + (L32 And &H7F) * &H1000000
End Function
Private Function Usub(ByVal L1, ByVal L2)
Dim L11, L12, L21, L22, L31, L32
L11 = L1 And &HFFFFFF
L12 = (L1 And &H7F000000) \ &H1000000
If L1 < 0 Then L12 = L12 Or &H80
L21 = L2 And &HFFFFFF
L22 = (L2 And &H7F000000) \ &H1000000
If L2 < 0 Then L22 = L22 Or &H80
L32 = L12 - L22
L31 = L11 - L21
If L31 < 0 Then
L32 = L32 - 1
L31 = L31 + &H1000000
End If
Usub = L31 + (L32 And &H7F) * &H1000000
If L32 And &H80 Then Usub = Usub Or &H80000000
End Function
Function Mix(ByVal ia, ByVal ib, ByVal ic)
Dim a, b, c
a = ia
b = ib
c = ic
a = Usub(a, b)
a = Usub(a, c)
a = a Xor ZeroFill(c, 13)
b = Usub(b, c)
b = Usub(b, a)
b = b Xor Sl(a, 8)
c = Usub(c, a)
c = Usub(c, b)
c = c Xor ZeroFill(b, 13)
a = Usub(a, b)
a = Usub(a, c)
a = a Xor ZeroFill(c, 12)
b = Usub(b, c)
b = Usub(b, a)
b = b Xor Sl(a, 16)
c = Usub(c, a)
c = Usub(c, b)
c = c Xor ZeroFill(b, 5)
a = Usub(a, b)
a = Usub(a, c)
a = a Xor ZeroFill(c, 3)
b = Usub(b, c)
b = Usub(b, a)
b = b Xor Sl(a, 10)
c = Usub(c, a)
c = Usub(c, b)
c = c Xor ZeroFill(b, 15)
Dim ret(3)
ret(0) = a
ret(1) = b
ret(2) = c
Mix = ret
End Function
Function Gc(ByVal s, ByVal i)
Gc = Asc(Mid(s, i + 1, 1))
End Function
Dim iLength, a, b, c, k, iLen, m
iLength = Len(sUrl)
a = &H9E3779B9
b = &H9E3779B9
k = 0
iLen = iLength
Do While iLen >= 12
m = Mix(a, b, c)
a = m(0)
b = m(1)
c = m(2)
k = k + 12
iLen = iLen - 12
Loop
Select Case iLen ' all the case statements fall through
Case 11
c = Uadd(c, Sl(Gc(sUrl, k + 10), 24))
c = Uadd(c, Sl(Gc(sUrl, k + 9), 16))
c = Uadd(c, Sl(Gc(sUrl, k + 8), 8))
b = Uadd(b, Sl(Gc(sUrl, k + 7), 24))
b = Uadd(b, Sl(Gc(sUrl, k + 6), 16))
b = Uadd(b, Sl(Gc(sUrl, k + 5), 8))
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 10
c = Uadd(c, Sl(Gc(sUrl, k + 9), 16))
c = Uadd(c, Sl(Gc(sUrl, k + 8), 8))
b = Uadd(b, Sl(Gc(sUrl, k + 7), 24))
b = Uadd(b, Sl(Gc(sUrl, k + 6), 16))
b = Uadd(b, Sl(Gc(sUrl, k + 5), 8))
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 9
c = Uadd(c, Sl(Gc(sUrl, k + 8), 8))
b = Uadd(b, Sl(Gc(sUrl, k + 7), 24))
b = Uadd(b, Sl(Gc(sUrl, k + 6), 16))
b = Uadd(b, Sl(Gc(sUrl, k + 5), 8))
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 8
b = Uadd(b, Sl(Gc(sUrl, k + 7), 24))
b = Uadd(b, Sl(Gc(sUrl, k + 6), 16))
b = Uadd(b, Sl(Gc(sUrl, k + 5), 8))
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 7
b = Uadd(b, Sl(Gc(sUrl, k + 6), 16))
b = Uadd(b, Sl(Gc(sUrl, k + 5), 8))
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 6
b = Uadd(b, Sl(Gc(sUrl, k + 5), 8))
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 5
b = Uadd(b, Gc(sUrl, k + 4))
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 4
a = Uadd(a, Sl(Gc(sUrl, k + 3), 24))
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 3
a = Uadd(a, Sl(Gc(sUrl, k + 2), 16))
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 2
a = Uadd(a, Sl(Gc(sUrl, k + 1), 8))
a = Uadd(a, Gc(sUrl, k + 0))
Case 1
a = Uadd(a, Gc(sUrl, k + 0))
End Select
m = Mix(a, b, c) | 2,507 | 4,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.264143 |
https://www.easyelimu.com/qa/2000/object-placed-front-concave-focal-length-determine-position?show=2001 | 1,722,736,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00457.warc.gz | 589,906,604 | 7,055 | # An object is placed 40cm in front of a concave lens of focal length 20cm; determine the position of the image.
421 views
An object is placed 40cm in front of a concave lens of focal length 20cm; determine the position of the image.
1/-f 1/V + 1/U
-1/20 = 1/V + 1/U
1/V = 1/20 - 1/40 = -2-1/40 = 3/40
V= -40/3 = 13.3cm | 132 | 321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.592336 |
https://tcl-toulon.org/and-pdf/1107-time-and-distance-questions-pdf-10-905.php | 1,643,400,005,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00258.warc.gz | 619,834,506 | 7,330 | # Time and distance questions pdf
Posted on Thursday, April 29, 2021 2:58:51 AM Posted by Isabelle S. - 29.04.2021
File Name: time and distance questions .zip
Size: 21148Kb
Published: 29.04.2021
Quantitative Aptitude is one of the important sections in Banking exams. If you are a banking aspirant or preparing for any Bank exam, then enhance your preparation by having all the required study materials. The questions from time, speed, and distance will be certainly asked.
## 200+ Speed Time & Distance Questions With Solution Free PDF – Download Now
Solved examples with detailed answer description, explanation are given and it would be easy to understand. Here you can find objective type Aptitude Time and Distance questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided. You can easily solve all kind of Aptitude questions based on Time and Distance by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Time and Distance problems.
An aeroplane covers a certain distance at a speed of kmph in 5 hours. To cover the same distance in 1 hours, it must travel at a speed of:. The actual distance travelled by him is:.
Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about The speed of the car is:. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? All Rights Reserved. Contact us: info. Current Affairs General Knowledge. Microbiology Biochemistry Biotechnology Biochemical Engineering. Why Aptitude Time and Distance? Where can I get Aptitude Time and Distance questions and answers with explanation?
How to solve Aptitude Time and Distance problems? A person crosses a m long street in 5 minutes. What is his speed in km per hour? Report errors. To cover the same distance in 1 hours, it must travel at a speed of: A. The actual distance travelled by him is: A. The speed of the car is: A. Current Affairs Interview Questions and Answers.
## Service Unavailable in EU region
Solved examples with detailed answer description, explanation are given and it would be easy to understand. Here you can find objective type Aptitude Time and Distance questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided. You can easily solve all kind of Aptitude questions based on Time and Distance by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Time and Distance problems. An aeroplane covers a certain distance at a speed of kmph in 5 hours.
Time and Distance Questions is an an essential part for Competitive Exams like Banking, Insurance, SSC and Railways Exams. - Try Now.
Get SSC mocks for just Rs. Enroll here. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes? The usual time taken by him to reach his office is.
What is the length in metre of the platform? For how many minutes does the bus stop per hour? What is the distance between A and B? If the second car runs km in 6 hours, then the speed of the first car is:. One train starts from P at 5 a. Another train starts from Q at 6 a. At what time will they meet?
Skip to content. Dear students, you know how important Math is a for SSC related exams. We cant learn everything in a day, so practice different question, for that, we are providing 15 questions of Profit and Loss. Solve all these quizzes every day so that you can improve your accuracy and speed. If he walks at 80 km.
Make customizable worksheets about constant or average speed, time, and distance for pre-algebra and algebra 1 courses grades Both PDF and html formats are available. There are SEVEN different types of word problems to choose from, ranging from easy to advanced, so you can create a great variety of worksheets. The seven types of problems are explained in detail in the actual generator below. Easy speed, time, and distance worksheet 1 : How far can it go or how long does the trip take - using whole or half hours Easy speed, time, and distance worksheet 2 : How far can it go, how long does the trip take, or what is the average speed - using whole or half hours Speed, time, and distance worksheet 3 : How far can it go, how long does the trip take, or what is the average speed - using quarter hours Speed, time, and distance worksheet 4 : How far can it go, how long does the trip take, or what is the average speed - time to the 5-minute intervals Speed, time, and distance worksheet 5 : problems involve a conversion of minutes to hours. Find the average speed: time is given to the fourth of an hour. | 1,065 | 4,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.91225 |
http://math.stackexchange.com/users/4414/cookie-monster?tab=activity&sort=all&page=6 | 1,412,123,896,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663218.28/warc/CC-MAIN-20140930004103-00478-ip-10-234-18-248.ec2.internal.warc.gz | 201,782,264 | 11,018 | less info
reputation
419
bio website location age member for 3 years, 10 months seen Sep 16 at 12:01 profile views 354
Dear All,
Jul8 comment Sum set fixpoint, how many iterations? Remark for the interested reader, if the lemma holds for sets S_i, it would also hold for integer intervals I_i, since an interval is only a special case of a set. Jul8 comment Sum set fixpoint, how many iterations? Maybe an alternate proof could use monotonicity of Minkowsky sum, i.e. A subset B implies A + C subset B + C. Not sure. But I am already happy to see the question settled. Jul8 comment Sum set fixpoint, how many iterations? Ok I guess the argument runs along the following for a x_i in S^0_i \ S^1_i there was anyway no (x_j)_j<>i, so when "deleting" x_i from S^0_i it will not reduce any of the S^0_j for j<>i. Jul8 comment Sum set fixpoint, how many iterations? Was also hypothesizing once that the fixpoint is reached after maximal 1 iteration. Did not yet have time to produce a counter example. What is exactly the argument behind S^2_i = S^1_i in the general case? Jul6 revised Sum set fixpoint, how many iterations? deleted 188 characters in body Jul6 comment Sum set fixpoint, how many iterations? It looks the problem can be reduced to y_1 + .. + y_n + b = 0, by using y_i = a_i * x_i. We can also find sets T_i with y_i in T_i, by T_i = a_i * S_i. And then ask for fixpoint of (T_1,..,T_n). Jul4 comment Sum set fixpoint, how many iterations? In the case of n=2 a simple geometric argument shows that l=<1 is sufficient. Right? But for n>2, is there also a fixed bound? Jul4 comment Sum set fixpoint, how many iterations? I thought it is the other way around by construction of the mapping, i.e. S_i^k+1 subset S_i^k. But otherwise I agree, this is the worst case. But is it possible that it happens? Jul4 revised Sum set fixpoint, how many iterations? deleted 9 characters in body Jul4 revised Sum set fixpoint, how many iterations? added 2 characters in body Jul4 revised Sum set fixpoint, how many iterations? edited tags Jul4 asked Sum set fixpoint, how many iterations? Jun19 accepted Sum set estimates for cofinite integer sets Jun19 revised Sum set estimates for cofinite integer sets added 2 characters in body Jun19 revised Sum set estimates for cofinite integer sets added 36 characters in body Jun19 asked Sum set estimates for cofinite integer sets Jun16 accepted Minimal bivariate diophantine equation solution space Jun16 asked Minimal bivariate diophantine equation solution space Mar19 revised Is there an intuitionistic proof of $\lnot p(a) \rightarrow p(b) \vdash \exists x p(x)$, what would Herbrand's Theorem say? added 868 characters in body Mar18 revised Contradiction Theorem deleted 157 characters in body | 694 | 2,739 | {"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} | 3.09375 | 3 | CC-MAIN-2014-41 | latest | en | 0.922802 |
https://www.answers.com/Q/How_much_is_70_mph_pitch_in_little_league_is_equivalent_to_MLB | 1,568,845,080,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573368.43/warc/CC-MAIN-20190918213931-20190918235931-00287.warc.gz | 748,847,720 | 17,373 | # How much is 70 mph pitch in little league is equivalent to MLB?
I don't know exactly but I've watched the Little League World Series enough and they always refer to Little Leaguers pitching speed in relation to MLB and I believe that 70 mph Little League would be around 100 mph MLB A 70 mph little league pitch is equivalent to about a 111 mph MLB pitch. (This is calculated based on an average little league pitch at 57.5 mph and an average MLB pitch at 91 mph.) | 108 | 467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-39 | latest | en | 0.989582 |
https://de.slideshare.net/susannakaiser/plans2012 | 1,675,452,346,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500074.73/warc/CC-MAIN-20230203185547-20230203215547-00035.warc.gz | 219,050,362 | 52,308 | Diese Präsentation wurde erfolgreich gemeldet.
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
×
1 von 21 Anzeige
Plans2012 Presentation: Angular PDFs and FootSLAM
Plans2012 presentation without films. Films are available at http://www.kn-s.dlr.de/mobility.
Plans2012 presentation without films. Films are available at http://www.kn-s.dlr.de/mobility.
Anzeige
Anzeige
Weitere Verwandte Inhalte
Anzeige
Plans2012 Presentation: Angular PDFs and FootSLAM
1. 1. Map-based angular PDFs used as prior map for FootSLAM Susanna Kaiser, Maria Garcia Puyol, Patrick Robertson German Aerospace Center (DLR) Institute of Communications and Navigation Kaiser; PLANS 2012 Slide 1
2. 2. Outline Introduction: FootSLAM and its hexagon map Motivation for using prior maps Angular probability density functions (PDFs) as a basis for prior maps The diffusion algorithm Determining the angular PDFs Mapping of the angular PDFs to the FootSLAM hexagon map Different floor-plan scenarios Experimental Results Conclusions & Outlook Kaiser 04/2012 Slide 2
3. 3. Introduction: Principle of FootSLAM Indoor Navigation is a challenge: GNSS signals are strongly disturbed Foot-mounted IMU sensor measurements pedestrian odometry FootSLAM: Simultaneous Localization and Mapping for pedestrians with odometry Optionally GPS (absolute coordinate frame) Human motion: First order Markov process FastSLAM factorization: Rao-Blackwellized Particle Filter. Each particle: pose + odometry errors + individual map Kaiser 04/2012 Slide 3
4. 4. Introduction: FootSLAM and its Hexagon map 2D space partitioned into a regular grid of adjacent and uniform hexagons with a given radius Each one of the 6 edges of the hexagons is associated to a transition count that represents the number of times it was crossed Input to FootSLAM: Raw odometry Coordinate System Starting Conditions Prior map, if available Output: Aggregated Posterior Map Best “MAP” Map Kaiser 04/2012 Slide 4
5. 5. Introduction: Integrating a prior map Each particle weight is updated as follows: i N e e i wk i wk 1 h h , Nh h where i wk is the weight for particle i at time k e Nh are the transition counts for edge e of the outgoing hexagon h e is the prior value for edge e h N h e 0 N h and h e 0 h e 5 e e 5 e Without a prior map, prior values are set to a constant value (e.g. 0.8) Kaiser 04/2012 Slide 5
6. 6. Introduction: Advantages of using a prior map The FootSLAM map converges faster The resulting map is more precise A more precise location of the map in an outer coordinate system can be found Prior maps are also used in FeetSLAM by adding other maps after being transformed to fit the total map. With the use of a prior map from the beginning the transformation will not be necessary anymore The prior map can be strengthened or weakened to control its influence on the FootSLAM map generation process. Kaiser 04/2012 Slide 6
7. 7. Introduction: Changing the strength of a prior map Strengthening: multiplying the counts by a prior strengthening factor If the map is known to be accurate, a high strengthening factor can be used Existing maps are sometimes imprecise, unavailable, obsolete, proprietary, and do not show furniture or other features that significantly limit pedestrian motion If the map is incorrect or only partly known the influence of the map can be lowered by using a low strengthening factor If the map is not available a constant value can be used h factor e Kaiser 04/2012 Slide 7
8. 8. Angular PDFs as a basis for prior maps For each actual waypoint a sliding squared window of size N x N x is defined, where the waypoint is the middle point of that window Each waypoint represents a source effusing gas Source/waypoint Example for a diffusion matrix For each waypoint x m , ym a diffusion matrix D is pre-computed x At start, the diffusion matrix is one at the waypoint and zero else where Kaiser 04/2012 Slide 8
9. 9. Angular PDFs as a basis for prior maps Determining the angular PDF directly out of the diffusion matrix From the diffusion matrix a threshold can be used for obtaining a contour line of the gas distribution Source/waypoint Contour line (dark red) of the diffusion values with threshold value: T=0.0001 c ,..., cN Resulting in a set C of Nc contour-line points: x1, y ,, xN , y 1 Nc 1 c c Kaiser 04/2012 Slide 9
10. 10. Angular PDFs as a basis for prior maps Determining the angular PDF directly out of the diffusion matrix For each angle the distance from the middle waypoint to the contour point is determined and the maximum distance is used: 0° bci ( xm k )2 ( ym l ) 2 ρ b Source/waypoint f ( ) max bci ci ( k , l ) ( k ,l ) f ( ) Normalizing: f ( ) 2 f ( ) 0 Kaiser 04/2012 Slide 10
11. 11. Angular PDFs as a basis for prior maps Determining the angular PDF directly out of the diffusion matrix Resulting Polar Plot of the angular PDF for that specific waypoint: This location dependant angular PDF can be used as prior information for FootSLAM Kaiser 04/2012 Slide 11
12. 12. Angular PDFs as a basis for prior maps Mapping angular PDFs on the FootSLAM Prior Map r e of edge e Range r e (rmin , rmax ) ( e e e , e ) 6 3 6 3 0° 0 r0 5 1 270° 90° 4 2 x 3 180° Hexagon centre = source of gas e rmax Values for the prior map: e e h f ( )d rmin Kaiser 04/2012 Slide 12
13. 13. Angular PDFs as a basis for prior maps Mapping angular PDFs on the FootSLAM Prior Map White/yellow -> high values Black -> low values Totally white hexagons: No angular PDF (position of hexagon centre is on wall) Kaiser 04/2012 Slide 13
14. 14. Different Floor-Plan scenarios: 2: The complete and correct 3: The complete and correct 4: A plan with only the plan plan including furniture outer building walls 5: A plan with only corridors 6: A plan missing a long wall 7: A plan with an additional, incorrect wall Kaiser 04/2012 Slide 14
15. 15. Experimental Results Sensor: Foot-mounted Inertial Measurement Unit (IMU) with Zero Velocity Updates (ZUPTs) processed with an extended Kalman Filter for pedestrian dead reckoning [Foxlin] A prior map is generated for the 7 floor-plan scenarios To emulate a GPS anchor when entering a building we assumed that starting conditions of the walk were not exactly known: X / Y 2.0m, 2.0 3 walks of 5-14 minutes duration 10 evaluation runs of the PF with different seeds for every data set (walk) Evaluation criterion: Ratio of violated walls and furniture ground- truth by the resulting FootSLAM map (smaller is better …) Kaiser 04/2012 Slide 15
16. 16. Experimental Results Results for different prior strengthening factors -1 10 FootSLAM with complete prior map (1) Crossed Wall Ratio [%] -2 10 -3 10 0 20 40 60 80 100 Prior Strengthening Factor Kaiser 04/2012 Slide 16
17. 17. Experimental Results Results for different floor plan scenarios -1 10 Raw FootSLAM PSF 40 Only outer walls PSF 20 Crossed Wall Ratio [%] Outer and -2 Corridor walls Missing wall 10 Complete plan Incorrect wall Complete plan including furniture -3 10 1 2 3 4 5 6 7 Index of Floor Plan Scenario Kaiser 04/2012 Slide 17
18. 18. Experimental Results Das Bild k ann zurzeit nicht angezeigt werden. Odometry Kaiser 04/2012 Slide 18
19. 19. Conclusions & Outlook A prior map can be applied to FootSLAM and is a useful additional information that can enhance the position estimation. The advantages of using prior maps are: Reaching faster convergence Better accuracy of the map More accurate positioning if the starting condition are not exactly known A prior map represented as angular PDFs is combined with FootSLAM. The angular PDFs are calculated using the diffusion algorithm and are mapped on the hexagon edge transitions Experiments: The use of angular PDFs as prior map in FootSLAM performs better than using no prior map It doesn’t matter if the prior map is only partly available or contains some errors: The influence of the map can be controlled via the prior strenghtening factor Knowing only the trunk of the building is almost as good as knowing the entire map Further work should focus on Additional data sets Different wall situations / map errors Evaluation of the resulting position accuracy Information theoretic evaluation of map similarity Kaiser 04/2012 Slide 19
20. 20. Many thanks for your interest & your questions are welcome! http://www.kn-s.dlr.de/indoornav Susanna Kaiser Date: 24/04/2012 Kaiser 04/2012 Slide 20
21. 21. Evaluation Results Results for different walks -1 Crossed Wall Ratio [%] 10 -2 10 1: No Prior 1: Complete Plan 2: No Prior 2: Complete Plan 3: No Prior 3: Complete Plan -3 10 1 2 3 Index of Walk Kaiser 04/2012 Slide 21 | 2,546 | 8,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.599749 |
http://tutorial.math.lamar.edu/Problems/CalcI/MinMaxValues.aspx | 1,527,224,614,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00259.warc.gz | 294,785,678 | 22,659 | Paul's Online Math Notes
[Notes]
Calculus I - Practice Problems
Derivatives Previous Chapter Next Chapter Integrals Critical Points Previous Section Next Section Finding Absolute Extrema
## Minimum and Maximum Values
1. Below is the graph of some function, . Identify all of the relative extrema and absolute extrema of the function.
[Solution]
2. Below is the graph of some function, . Identify all of the relative extrema and absolute extrema of the function.
[Solution]
3. Sketch the graph of and identify all the relative extrema and absolute extrema of the function on each of the following intervals.
(a)
(b)
(c)
(d)
(e)
[Solution]
4. Sketch the graph of and identify all the relative extrema and absolute extrema of the function on each of the following intervals.
(a)
(b)
(c)
(d)
[Solution]
5. Sketch the graph of some function on the interval that has an absolute maximum at and an absolute minimum at . [Solution]
6. Sketch the graph of some function on the interval that has an absolute maximum at and an absolute minimum at . [Solution]
7. Sketch the graph of some function that meets the following conditions :
(a) The function is continuous.
(b) Has two relative minimums.
(b) One of relative minimums is also an absolute minimum and the other relative minimum is not an absolute minimum.
(c) Has one relative maximum.
(d) Has no absolute maximum.
[Solution]
Problem Pane
Critical Points Previous Section Next Section Finding Absolute Extrema Derivatives Previous Chapter Next Chapter Integrals
[Notes]
© 2003 - 2018 Paul Dawkins | 353 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | longest | en | 0.801913 |
https://www.wyzant.com/resources/answers/topics/confusion | 1,638,437,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00005.warc.gz | 1,105,480,989 | 12,473 | 11 Answered Questions for the topic Confusion
#### The points (3, 7) and (2, 4) fall on a particular line. What is its equation in slope-intercept form?
slope intersect form: y=mx+b I'm honestly so confused on how to solve this so if someone could suggest or show me the step by step process on solving this equation that would be great. Thank... more
10/26/17
#### What are the responsiblilites of power?
I have this question for my English. I sorta understand it but I need some help with it. Can you explain to me what responsibilities come with power??
Confusion Elementary Algebra
09/17/17
#### I need help and better understanding of this math problem.
The width of a brick is half the length, and the length is 2 inches less than four times the height. If the sum of the three dimensions is 21.5 inches, find the volume of the brick. Simplify your... more
Confusion Elementary Algebra
09/16/17
#### I need help with a problem I can't seem to get. I've tried over and over again.
I've tried this problem over and over again and have gotten it wrong. Please help me ! Question: The width of a brick is half the length, and the length is 1 inch less than four times the... more
05/29/16
John and his sister went out for lunch at trey's pasta. Their meal cost $25 and they left a 12% tip. What was the total cost of the meal including the tip? HOW DO I WORK THIS... more #### slope=3 : (-1, 4) (how do you put it in standard form) how can i put it in standard form if its not an equation 10/07/15 #### How to set up In one week, it rained821in. If a garden needs67in. of water per week, how much more water does it need? 09/09/14 #### I can't seem to solve this problem. Can someone please help me and show me how to solve this!? IM SO LOST :c I can't seem to figure out this problem. Can someone please help me and show me how to solve this!? IM SO LOST show me how to solve this!? IM SO LOSTStarting with the parent function f(x)=x2,... more Confusion 07/08/14 #### Part of$9000 was invested at 10% interest and the rest at 12%. If the annual income from these investments was \$1020 how much was invested at each rate?
I must use 2 variables and 2 equations. Im lost on the setup.
Confusion
07/08/14
#### How do I completely solve f(a)+f(2a)=18a+10 ?
I'm confused on the simplifying f(a)+f(2a).I could really use your help!!
04/30/14
#### If the average stride of a horse walking is 2 3/4 feet and that of a man is 2 1/2 feet
, how many man strides would be equal to 40 horse strides?
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 725 | 2,758 | {"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} | 3.703125 | 4 | CC-MAIN-2021-49 | latest | en | 0.955993 |
http://nrich.maths.org/public/leg.php?code=47&cl=3&cldcmpid=2327 | 1,503,421,147,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112533.84/warc/CC-MAIN-20170822162608-20170822182608-00355.warc.gz | 305,430,913 | 10,235 | # Search by Topic
#### Resources tagged with Creating expressions/formulae similar to Konigsberg Plus:
Filter by: Content type:
Stage:
Challenge level:
### There are 85 results
Broad Topics > Algebra > Creating expressions/formulae
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Christmas Chocolates
##### Stage: 3 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Quick Times
##### Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Lower Bound
##### Stage: 3 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Sum Equals Product
##### Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### Seven Squares
##### Stage: 3 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Marbles in a Box
##### Stage: 3 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Partly Painted Cube
##### Stage: 4 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Fibs
##### Stage: 3 Challenge Level:
The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### How Big?
##### Stage: 3 Challenge Level:
If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?
### Always a Multiple?
##### Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Your Number Is...
##### Stage: 3 Challenge Level:
Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Pinned Squares
##### Stage: 3 Challenge Level:
The diagram shows a 5 by 5 geoboard with 25 pins set out in a square array. Squares are made by stretching rubber bands round specific pins. What is the total number of squares that can be made on a. . . .
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
### Good Work If You Can Get It
##### Stage: 3 Challenge Level:
A job needs three men but in fact six people do it. When it is finished they are all paid the same. How much was paid in total, and much does each man get if the money is shared as Fred suggests?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Days and Dates
##### Stage: 3 Challenge Level:
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### One and Three
##### Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
### Boxed In
##### Stage: 3 Challenge Level:
A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Magic Sums and Products
##### Stage: 3 and 4
How to build your own magic squares.
### How Much Can We Spend?
##### Stage: 3 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Interactive Number Patterns
##### Stage: 4 Challenge Level:
How good are you at finding the formula for a number pattern ?
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Pair Products
##### Stage: 4 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Your Number Was...
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Multiplication Square
##### Stage: 4 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Chocolate 2010
##### Stage: 4 Challenge Level:
First of all, pick the number of times a week that you would like to eat chocolate. Multiply this number by 2...
### Areas of Parallelograms
##### Stage: 4 Challenge Level:
Can you find the area of a parallelogram defined by two vectors? | 2,439 | 9,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-34 | latest | en | 0.860792 |
http://www.airliners.net/forum/viewtopic.php?f=3&t=127711 | 1,495,594,058,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607731.0/warc/CC-MAIN-20170524020456-20170524040456-00089.warc.gz | 405,195,429 | 17,302 | sleekjet
Posts: 2008
Joined: Sat Jul 28, 2001 1:35 am
### Winter Weather Takeoffs
Assuming an a/c has been properly de-iced, what are the runway parameters during winter storms? In other words, how much ice or snow on a runway does it take to scuttle a flight? And who decides? The pilot? The airport? The airline management?
II Cor. 4:17-18
OPNLguy
Posts: 11191
Joined: Tue Jun 15, 1999 11:29 am
### RE: Winter Weather Takeoffs
Each type of aircraft has different acceptible performance parameters, so it's difficult to give you a one-size-fits-all type of answer. If I can, what I'll do is give you a general example, which may then vary for a specific type of aircraft.
Snow (or slush) on a runway (commonly referred to as runway "clutter") affects both takeoff and landing performance. For takeoffs, clutter retards normal rates of acceleration, and for landings, braking (DEacceleration) distances are lengthened. The effect of both is that it makes a runway of X-length perform as if it's LESS than X-length, and since the airport folks can't come out and pour additional concrete to extend the runway on short notice ,aircraft weights must be reduced to keep the aircraft with safe allowable limits. Sometimes, this means that some folks get left behind even though there was an empty seat for them.
Generally speaking, an aircraft may not operate (that's takeoff or landing) on any runway covered by more than 1/2 inch of slush, 1 inch of wet snow, or 4 inches of dry snow. Anything beyond that, and the aircraft does not operate.
For takeoffs below the clutter levels mentioned above, the normal calculated runway takeoff weight is reduced by a factor of 5% or 10%, whether the clutter is minimal, or is closer to the max allowable values.
For landings, the clutter affects the "braking action" of the runway. A dry runway would (obviously) be the best for braking action, and is known as "dry" or "normal". As braking action deteriorates, the classifications range downward from "good", then "fair", then "poor", and finally, "NIL". Aircraft operations are prohibited on "NIL" surfaces, and it should also be noted that operations with "fair" and "poor" reported braking action also entail increasinly more restrictive crosswind limitations.
How is all this stuff measured? That's where things get interesting, as there are both OBjective and SUBjective methods in place.
The airport operator has the responsibility (here in the US, anyways) to have a snow control program in place, and they are generally pretty good about keeping everyone up-to-date of local field conditions. Some of this accomplished via issuance of a NOTAM (notice to airmen), but some stuff changes so quickly, that tower issues the information.
For takeoffs, snow depth is mostly commonly reported (objectively) by airport folks actually out there. A pilot can certainly make a depth determination, but obviously, doing so from from purely within the cockpit is a more subjective endeavor.
For landings, braking action can be assessed by either specially-equipped vehicles (that provide a numerical reading like MU, or other, that can be converted to good/fair/poor/NIL), or the pilots of any landing aircraft can report the braking action they experienced when they landing. Given the above two mechanisms, it's easy to see that one has more opportunity for subjectivity than the other.
Considering all the information above, who makes the decision? It sort of depends. If the airport operator decides that the clutter has exceeded limits and then closes the runway for plowing/brooming/de-icing, it's a no-brainer. That aside, here in the US, under FAR 121 Domestic/Flag rules, the captain and/or aircraft dispatcher make the call. If one of those two thinks they have limits, but the other has more current info that says they don't, the most conservative info takes precedence. I know there are probably some crews out there who will disagree, but I can assure you that within my 23-year career, I have personally told crews NOT to land on occasion because I had more timely info than they did. Does't happen often that way, but it does happen.
Sorry to be so long, but a complicated question...
ALL views, opinions expressed are mine ONLY and are NOT representative of those shared by Southwest Airlines Co.
sleekjet
Posts: 2008
Joined: Sat Jul 28, 2001 1:35 am
### RE: Winter Weather Takeoffs
Thanks, man, for maybe the best, most detailed response in the history of airliners.net.
Do you work at DFW?
II Cor. 4:17-18
OPNLguy
Posts: 11191
Joined: Tue Jun 15, 1999 11:29 am
### RE: Winter Weather Takeoffs
Thank-you kindly...
Yep...
Another related factor in winter ops is plowing. At BUF last week, it was coming down so hard that they couldn't even see enough to plow, so off the runway they came. Once they eventually got back out there, there was so much accumulation (2-4 inches per hour) that it took them forever to get the runway, taxiways, and ramp areas cleared off enough to operate. Oh yeah, more heavy snow off and on the entire week. Something like 80 inches total...
ALL views, opinions expressed are mine ONLY and are NOT representative of those shared by Southwest Airlines Co.
Bicoastal
Posts: 2446
Joined: Wed Oct 06, 1999 5:56 am
### RE: Winter Weather Takeoffs
Prague had plenty of snow when I departed there on Sunday around noon. The runway was closed off and on most of the morning, but the equipment (BA 767) on which I was to fly out arrived on time. On my take-off, the British Airways 767, fully loaded with passengers, had a very short take off roll and handled the snowy runway just fine. I was supposed to be on a Lufthansa flight back to the USA via Frankfurt, but their equipment never left Frankfurt for Prague due to the snow storms. Hmmmm....do British pilots have more ability than Lufthansa's.....???? In any case, Lufthansa's personnel in Prague handled the cancellation very well. They rebooked me on BA to London then on my favorite United Airlines from London to the USA.
Airliners.net has many forums. It has spell check and search functions. Use them before posting!
Maniac
Posts: 109
Joined: Mon Apr 30, 2001 8:36 am
### RE: Winter Weather Takeoffs
I had a flight out of PWM when the pilot went out in a plow truck with the snow removal crew to check the runway. He said that on landing the previous flight, the braking action was not good enough, and he wouldn't take off until he saw the runway.
### Popular Searches On Airliners.net
Top Photos of Last: 24 Hours • 48 Hours • 7 Days • 30 Days • 180 Days • 365 Days • All Time
Military Aircraft Every type from fighters to helicopters from air forces around the globe
Classic Airliners Props and jets from the good old days
Flight Decks Views from inside the cockpit
Aircraft Cabins Passenger cabin shots showing seat arrangements as well as cargo aircraft interior
Cargo Aircraft Pictures of great freighter aircraft
Government Aircraft Aircraft flying government officials
Helicopters Our large helicopter section. Both military and civil versions
Blimps / Airships Everything from the Goodyear blimp to the Zeppelin
Night Photos Beautiful shots taken while the sun is below the horizon
Accidents Accident, incident and crash related photos
Air to Air Photos taken by airborne photographers of airborne aircraft
Special Paint Schemes Aircraft painted in beautiful and original liveries
Airport Overviews Airport overviews from the air or ground
Tails and Winglets Tail and Winglet closeups with beautiful airline logos | 1,743 | 7,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-22 | longest | en | 0.951419 |
http://www.ehow.co.uk/how_6608111_convert-hertz-milliseconds.html | 1,490,696,391,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189686.56/warc/CC-MAIN-20170322212949-00394-ip-10-233-31-227.ec2.internal.warc.gz | 496,180,492 | 17,448 | # How to convert hertz to milliseconds
Written by david medairos
• Share
• Tweet
• Share
• Pin
• Email
Hertz is a measurement of frequency in that it describes how many times something happens (known as a cycle) every second. Often used to describe the frequency of sound or electricity, hertz can describe any repetitive action. One millisecond is a standard measurement of time and represents 1/1000 of one second. Both hertz (Hz) and milliseconds (msecs) are time or "period" measurements, and converting Hz to msecs can be accomplished with a basic formula.
Skill level:
Easy
• Calculator
• Hertz value
## Instructions
1. 1
Use the formula 1/Hz * 1000, where Hz represents cycles per second. For this example, convert 500 Hz to milliseconds.
2. 2
Divide the Hz value into one. Enter 1 ÷ 500. This will result in a value of 0.002. This value represents seconds.
3. 3
Multiply the seconds value by 1000 to derive the milliseconds value. For this example, enter 0.002 x 1000. This results in a value of 2 milliseconds. Record your results.
### Don't Miss
#### References
• All types
• Articles
• Slideshows
• Videos
##### Sort:
• Most relevant
• Most popular
• Most recent | 304 | 1,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.783716 |
https://codewalr.us/index.php?topic=369.msg16758 | 1,596,992,047,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00005.warc.gz | 261,926,142 | 13,227 | August 09, 2020, 04:47:55 pm
You can help CodeWalrus stay online by donating here.
WARNING: DO NOT UPGRADE your TI-83 Premium CE or TI-84 Plus CE to OS 5.5.1 and higher. It removes all compatibility with most games and removes ASM/C programming! DOWNGRADING IS IMPOSSIBLE. BE WARNED! Likewise, do NOT update your TI-Nspire CX past OS 4.5.0, else using Ndless and ASM/C programs will be impossible.
Started by Unicorn, April 04, 2015, 05:03:56 pm
0 Members and 1 Guest are viewing this topic.
#### Unicorn
##### June 19, 2015, 09:51:07 pm #105
Hmm..
I was using the xlibc spite collision, real(4,6 and I think that will work for me, if I can figure out how to get it to work without nesting it in an If Then End loop.
#### DJ Omnimaga
##### June 20, 2015, 03:40:59 am #106
Oh right, I forgot that xLIBC had sprite collision. I never got it to work before (which is why Reuben CSE died). Maybe @tr1p1ea could help since he wrote that part of DCSE but he isn't around often.
#### tr1p1ea
##### June 20, 2015, 02:06:51 pm #107
Brief explanation of sprite collision function, it tests for collisions between rectangular coordinates so if you have say a sprite that is:
X = 10
Y = 10
W = 8
H = 8
And you wanted to test if this 'rectangle' overlaps with an enemy sprite at:
X = 15
Y = 15
W = 8
H = 8
Like so:
Then you can use the function to check:
real(4,6,1,10,10,8,8,15,15,8,8
In which case they do collide so the result will be a list:
{1,1,0
{1 = collision found
1 = number of collisions found
0 = index of collision passed with real( statement
And since there was only 1 set of coordinates passed the first index that collided is 0 (the red square).
Another case:
Rectangles at:
X = 10
Y = 10
W = 8
H = 8
Tested against others where:
X = 15
Y = 15
W = 8
H = 8
X = 48
Y = 32
W = 16
H = 16
X = 8
Y = 0
W = 64
H = 16
Like so:
The call would be: real(4,6,3,10,10,8,8,15,15,8,8,48,32,16,16,8,0,64,16
And the result would be:
{1,2,0,2
{1 = collision found (would be 0 if no collisions occured)
2 = number of collisions found (green and red)
0 = collision with index 0 (red)
2 = collision with index 2 (green)
Index 1 is the orange square for which there was no collision.
#### Unicorn
##### June 20, 2015, 09:32:50 pm #108
Thanks for that info tr1p1ea! I think that may just make this whole thing work
#### DJ Omnimaga
##### June 25, 2015, 12:40:16 am #109
Ooh nice tutorial tr1p1ea. I think it should be on the DCS wiki somewhere. And glad to hear Unicorn.
#### Unicorn
##### June 25, 2015, 11:44:08 am #110
Yup! I just need to spend time testing and confirming that collision works, then its on to throwing the money
#### alexgt
##### June 25, 2015, 02:42:00 pm #111
Quote from: Unicorn on June 25, 2015, 11:44:08 am
Yup! I just need to spend time testing and confirming that collision works, then its on to throwing the money
And on to convincing my friend to let me borrow his CSE
#### Unicorn
##### July 03, 2015, 03:05:16 am #112
lol, well sprite collisions not working. So I think I'm not checking the list right, so I'm gonna try to fix that soon. Any good suggestions for commands that check lists? (I always have problems with them) :/
#### DJ Omnimaga
##### July 08, 2015, 03:29:09 am #113
I can't help unfortunately, since I myself had to quit Reuben CSE due to issues with collision D:. Maybe you could ask on Cemetech? Or maybe @tr1p1ea could help. You should post the code, though.
#### Unicorn
##### July 08, 2015, 12:49:10 pm #114 Last Edit: July 08, 2015, 01:13:24 pm by Unicorn
I'll try to post the code, but its a pain typing it all out
EDIT: Current Code, @tr1p1ea can you give me an idea of what I'm doing wrong with sprite collision?
`real(0,1,1real(8,1,0"MNYWBGreal(5,2,0"MNSPRTreal(5,0,0DelVar NDelVar BDelVar K19→TrandInt(8,19→UrandInt(5,15→VRepeat K=15real(2,0,0Ans→KT+1→TU+1→UV+1→VIf T=20:ThenLbl 1randInt(1,130→Mreal(4,6,2,M,1,16,16,L,1,16,16,N,1,16,16 //HEremax(|LXLIF Ans=2M+24->MIf Ans=1Goto 1real(4,0,M,1,3,3,0,0,4,0,0,0,1,2,8,9,10,16,17,18EndIf T≥40:ThenDelVar Treal(7,9,M,1,29,26,93,0EndIf U=20:ThenLbl 2randInt(1,130→Lreal(4,6,2,L,1,16,16,M,1,16,16,N,1,16,6 //Heremax(|LXLIF Ans=2L+24->LIf Ans=2Goto 2real(4,0,L,1,3,3,0,0,4,0,0,0,1,2,8,9,10,16,17,18EndIf U≥40:ThenDelVar Ureal(7,9,L,1,29,26,93,0EndIf V=20:ThenLbl 3randInt(1,130→Nreal(4,6,2,N,1,16,16,M,1,16,16,L,1,16,16 //Heremax(|LXLIF Ans=2N+24->NIf Ans=1Goto 3real(4,0,N,1,3,3,0,0,4,0,0,0,1,2,8,9,10,16,17,18EndIf V≥40:ThenDelVar Vreal(7,9,N,1,29,26,93,0EndIf K=2P-25→PIf K=3P+25→PIf K=2 or K=3:Then"MNYWBGreal(5,1,0real(5,3,P-25,104,16,16,0,0real(5,3,P+25,104,16,16,0,0Endreal(4,0,P,104,2,2,0,0,4,0,0,3,4,11,12End`
#### tr1p1ea
##### July 14, 2015, 04:43:24 am #115
Im looking into the issue, just wondering what the expected behaviour is?
#### Unicorn
##### July 14, 2015, 01:27:32 pm #116
The sprite x coord (L,M, and N) gets changed up 16 pixels when the collision is found
#### tr1p1ea
##### July 16, 2015, 01:08:31 am #117
So this will be between the sprites on the top row?
#### Unicorn
##### July 16, 2015, 01:19:16 am #118
Yeah, I made a few changes, so there's new code. It still didn't fix it though Thanks for looking at it.
`real(0,1,1real(8,1,0"MNYWBGreal(5,2,0"MNSPRTreal(5,0,0DelVar NDelVar BDelVar K19→TrandInt(8,19→UrandInt(5,15→VRepeat K=15real(2,0,0Ans→KT+1→TU+1→UV+1→VIf T=20:ThenLbl 1randInt(1,130→Mreal(4,6,2,M,1,16,16,L,1,16,16,N,1,16,16 //HEreIf max(|LXL=2 or max(|LXL=1Goto 1real(4,0,M,1,3,3,0,0,4,0,0,0,1,2,8,9,10,16,17,18EndIf T≥40:ThenDelVar Treal(7,9,M,1,29,26,93,0EndIf U=20:ThenLbl 2randInt(1,130→Lreal(4,6,2,L,1,16,16,M,1,16,16,N,1,16,6 //HereIf max(|LXL=2 or max(|LXL=1Goto 2real(4,0,L,1,3,3,0,0,4,0,0,0,1,2,8,9,10,16,17,18EndIf U≥40:ThenDelVar Ureal(7,9,L,1,29,26,93,0EndIf V=20:ThenLbl 3randInt(1,130→Nreal(4,6,2,N,1,16,16,M,1,16,16,L,1,16,16 //HereIf max(|LXL=2 or max(|LXL=1Goto 3real(4,0,N,1,3,3,0,0,4,0,0,0,1,2,8,9,10,16,17,18EndIf V≥40:ThenDelVar Vreal(7,9,N,1,29,26,93,0EndIf K=2P-25→PIf K=3P+25→PIf K=2 or K=3:Then"MNYWBGreal(5,1,0real(5,3,P-25,104,16,16,0,0real(5,3,P+25,104,16,16,0,0Endreal(4,0,P,104,2,2,0,0,4,0,0,3,4,11,12End`
#### Unicorn
##### August 10, 2015, 06:00:45 pm #119
BIG changes. So I've decided to revamp the game, making it into a kind of gathering game, with occasional enemies and a shop. So sort of like an RPG, but a bit different.
The original game idea seemed a bit boring, hence the change. And I'm going to change sprite sizes, or maybe have the game be full screen, but probably not. | 2,653 | 6,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-34 | latest | en | 0.925035 |
http://www.instructables.com/community/POV-using-LED-matrix/ | 1,418,971,147,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768276.101/warc/CC-MAIN-20141217075248-00130-ip-10-231-17-201.ec2.internal.warc.gz | 600,236,862 | 30,097 | # COMMUNITY : FORUMS : TECH
## POV using LED matrix
Hi,
I am doing a project of making and LED analog clock. The clock contains almost 300 LEDss and they are arranged in a specific pattern that looks like and analog clock.
My question is how would you create a POV in this design of the clock. I have checked other projects of POV; in those projects the whole device moves along with the row of LEDs and then it created a POV effect. However, in my project, the LEDs are fixed on a plastic sheet. How would I create a POV effect on that???
I think the correct term is to put some animation to the LED display like have it resemble the hands of the clock sweeping around the dial by only lighting up the LEDs in your fixed matrix. That would not be POV. In general, the way to animate or computer control a matrix/grid of LEDs is to be able to turn on each and every pixel or LED individually. Look up charlieplexing and some of the arduino projects to control it. Good luck.
Orngrimm1 year ago
I am unsure on what you made... Are your LEDs arranged like the rays of the sun? So every "ring" has the same ammount of LEDs but with increasing distance from the center the distance betwheen the LEDs increase?
If so, then you can make it relatively simple.
Just control them as 2 rings: Inner lets say 4 LEDs and the outer lets say 4 LEDs.
I will call those 2 groups rings in the future text.
For the rest of the text i suppose you have 12 rays (one for each 5 minutes).
With a microprocessor, you make a simple LED-matrix with 12X and 2Y.
The minute is always ring 1 & 2 in the Y and the corresponding ray in X.
The hour is only the ring 1 in Y and the corresponding ray in X.
So unless your minute-dial and hour-dial are over each other you will have 3 segments lit:
1 outer segment (ring 2) for the end of the minute-dial.
1 inner segment (ring 1) for the start of the minute-dial.
1 inner segment (ring 1) for the hour-dial.
However, your resolution is a lot less than if you do a moveable POV since you can control / time the moving part more precise than you can pack your LEDs.
If you can control / time the movement by 1° you would need 360 "LED-rays" consisting of lets say 8 LEDs each giving you a total of 2880 LEDs. And in the center the LEDs would be so densely packed the ring (with 5mm LEDs) cannot offer them enough space if you want a reasonable sice or your smallest ring would be 57.2cm in diameter to give 360 5mm-LEDs some space.
Bam Boy (author) Orngrimm1 year ago
Hi,
yes the design is like a sun. The distance increases as the LEDs move further from the center. Also, I have separate rings for the minutes and hours.
--> The inner most circle is for hours. It consists of 12 rows of LEDs and each row has 5 LEDs inside it. They point towards the hours.
--> Then comes the minute circle. It doesnt go all the way to the center of the clock. It starts from the edge of hour circle till the boundary of the clock and it has 20 rows of LEDs and each row has 6 LEDs in side it.
--> and then the outer boundary consists of 60 LEDs for the seconds that are going to blink with each second.
The problem that we are facing is that the LED matrix is not in a compact form to create POV. Like in different other projects they have LEDs tightly compacted. But we got spaces between LEDs.
1 year ago
Yes. Thats why a POV normally incorporates a moving part like a slit in a harddisc-platter or such stuff...
As i see it, you dont even need a microprozessor...
Just 3 shift-registers:
one with 12 places for the hour
two with 60 registers for the minutes and seconds.
The overflow of the seconds push one of the minutes; the overflow of the minutes pushes one of the hours...
Honestly: I dont understand your question. What exactly is the problem with the POV? With a fixed panel of LEDs you cannot make a POV in the classical sense... It is simply not possible and defies the idea of POV: a small moving thing which creates an illusion of fluid movement triggered to a certain position in time and space.
Are you asking how to pack LEDs tighter or what? I dont get the question...
Someone of the others can push me on the right track? | 986 | 4,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.942653 |
https://www.gurufocus.com/term/ROC/MINI/Return+on+Capital/Mobile+Mini+Inc | 1,511,199,688,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806086.13/warc/CC-MAIN-20171120164823-20171120184823-00371.warc.gz | 832,113,496 | 41,059 | Switch to:
Mobile Mini Inc (NAS:MINI) ROC %: 4.04% (As of Sep. 2017)
Return on capital measures how well a company generates cash flow relative to the capital it has invested in its business. It is also called ROIC %. Mobile Mini Inc's annualized return on capital (ROC) for the quarter that ended in Sep. 2017 was 4.04%.
As of today, Mobile Mini Inc's WACC % is 14.05%. Mobile Mini Inc's return on capital is 4.33% (calculated using TTM income statement data). Mobile Mini Inc earns returns that do not match up to its cost of capital. It will destroy value as it grows.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Mobile Mini Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 ROC % 4.26 3.18 3.95 16.16 4.65
Mobile Mini Inc Quarterly Data
Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Sep17 ROC % 4.34 6.09 3.74 3.46 4.04
Calculation
Mobile Mini Inc's annualized Return on Capital (ROC) for the fiscal year that ended in Dec. 2016 is calculated as:
Return on Capital (ROC) (A: Dec. 2016 ) = NOPAT / Average Invested Capital = Operating Income*(1-Tax Rate) / ( (Invested Capital (A: Dec. 2015 ) + Invested Capital (A: Dec. 2016 )) /2) = 113.103 * ( 1 - 31.42% ) / ( (1667.451 + 1668.553) /2) = 77.5660374 / 1668.002 = 4.65 %
Invested Capital (A: Dec. 2015 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt & Capital Lease Obligation + Current Portion of Long-Term Debt + Minority Interest + Total Equity - Cash = 235.827 + 667.708 + 0 + 765.529 - 1.613 = 1667.451
Invested Capital (A: Dec. 2016 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt & Capital Lease Obligation + Current Portion of Long-Term Debt + Minority Interest + Total Equity - Cash = 295.916 + 641.16 + 0 + 735.614 - 4.137 = 1668.553
Mobile Mini Inc's annualized Return on Capital (ROC) for the quarter that ended in Sep. 2017 is calculated as:
Return on Capital (ROC) (Q: Sep. 2017 ) = NOPAT / Average Invested Capital = Operating Income*(1-Tax Rate) / ( (Invested Capital (Q: Jun. 2017 ) + Invested Capital (Q: Sep. 2017 )) /2) = 107.248 * ( 1 - 36.24% ) / ( (1685.257 + 1701.271) /2) = 68.3813248 / 1693.264 = 4.04 %
where
Invested Capital (Q: {Q2}) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt & Capital Lease Obligation + Current Portion of Long-Term Debt + Minority Interest + Total Equity - Cash = 294.833 + 637.651 + 0 + 762.047 - 9.274 = 1685.257
Invested Capital (Q: Sep. 2017 ) = Book Value of Debt + Book Value of Equity - Cash = Long-Term Debt & Capital Lease Obligation + Current Portion of Long-Term Debt + Minority Interest + Total Equity - Cash = 297.48 + 640.879 + 0 + 775.521 - 12.609 = 1701.271
Note: The Operating Income data used here is four times the quarterly (Sep. 2017) operating income data.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Return on Capital measures how well a company generates cash flow relative to the capital it has invested in its business. It is also called ROIC %. The reason book values of debt and equity are used is because the book values are the capital the company received when issuing the debt or receiving the equity investments.
There are four key components to this definition. The first is the use of operating income rather than net income in the numerator. The second is the tax adjustment to this operating income, computed as a hypothetical tax based on an effective or marginal tax rate. The third is the use of book values for invested capital, rather than market values. The final is the timing difference; the capital invested is from the end of the prior year whereas the operating income is the current year's number.
Why is Return on Capital important?
Because it costs money to raise capital. A firm that generates higher returns on investment than it costs the company to raise the capital needed for that investment is earning excess returns. A firm that expects to continue generating positive excess returns on new investments in the future will see its value increase as growth increases, whereas a firm that earns returns that do not match up to its cost of capital will destroy value as it grows.
As of today, Mobile Mini Inc's WACC % is 14.05%. Mobile Mini Inc's return on capital is {stock_data.stock.roic}}% (calculated using TTM income statement data). Mobile Mini Inc earns returns that do not match up to its cost of capital. It will destroy value as it grows.
Be Aware
Like ROE and ROA, ROC is calculated with only 12 months of data. Fluctuations in the company's earnings or business cycles can affect the ratio drastically. It is important to look at the ratio from a long term perspective.
Related Terms | 1,328 | 4,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.801712 |
https://codereview.stackexchange.com/questions/177414/rotating-lists-in-linear-time-and-constant-space-java-follow-up?noredirect=1 | 1,709,188,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00754.warc.gz | 171,212,988 | 39,619 | # Rotating lists in linear time and constant space (Java) - follow-up
After some critique, I found the way to keep the rotation operation also for java.util.Lists that do not provide random access iterators:
package net.coderodde.util;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Objects;
import java.util.Scanner;
/**
* This class contains a static method for rotating lists in linear time and
* constant space.
*
* @author Rodion "rodde" Efremov
* @version 1.6 (Oct 6, 2017)
*/
public final class ListRotation {
private ListRotation() {}
/**
* Performs the list rotation.
*
* @param <T> the list element type.
* @param list the list whose content to rotate.
* @param rotationCount the number of steps to rotate to the right. If
* negative, rotates to the left.
*/
public static <T> void rotate(List<T> list, int rotationCount) {
Objects.requireNonNull(list, "The input list is null.");
rotationCount %= list.size();
if (rotationCount < 0) {
rotationCount += list.size();
}
Collections.<T>reverse(list);
Collections.<T>reverse(list.subList(0, rotationCount));
Collections.<T>reverse(list.subList(rotationCount, list.size()));
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
List<Integer> list =
new LinkedList<>(Arrays.asList(1, 2, 3, 4, 5, 6, 7));
System.out.println(list);
while (true) {
int rotationCount = scanner.nextInt();
rotate(list, rotationCount);
System.out.println(list);
}
}
}
So how does that look now?
If the list is empty, there will be division by zero. Since you check for null input list, it would make sense to do likewise for empty list as well.
The <T> is not needed in the Collections.<T>reverse statements, this works just fine:
Collections.reverse(list); | 427 | 1,782 | {"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} | 2.90625 | 3 | CC-MAIN-2024-10 | latest | en | 0.562866 |
https://www.physicsforums.com/threads/1-to-2-interactions-or-explosions.99262/ | 1,547,767,835,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659417.14/warc/CC-MAIN-20190117224929-20190118010929-00274.warc.gz | 885,876,329 | 12,515 | # 1 to 2 Interactions, or explosions
1. Nov 10, 2005
### lucky47
1 to 2 Interactions, or "explosions"
I'm stuck doing this problem on my physics homework.. it's different than the type of problems we have been doing in class, and he hasn't taught us the equation to solve this question. The problem is: Tim, mass 50.00 kg, is riding a skateboard, mass 2.00 kg, traveling at 1.70 m/s. Tim jumps off and the skateboard stops dead in its tracks. With what velocity did he jump?
Can anyone tell me what equation to use in order to get the answer?
2. Nov 10, 2005
### BerryBoy
Think of momentum...
The skateboard has a cerain momentum of 'mv'. What velocity would Tim have to jump off of the skateboard to cause the final velocity of the board to be zero?
Use conservation of momentum before and after the jump (remember Tim has initial momentum).
This is neglecting the vertical component of Tim's jump.
Regards,
Sam
3. Nov 10, 2005
### lucky47
Hm.. I'm not quite getting it. So to figure out the momentum of Tim, it is 50kg, his weight; times the velocity of the skateboard, 1.7m/s, correct?
And now the momentum of the skateboard is 2kg times 1.7m/s..
Now to find the velocity i take the two numbers I got from solving the above and divide it by the mass of the skateboard?
Am I remotely correct in what I am doing?
4. Nov 10, 2005
### BerryBoy
Firstly, lets clear something up... Momentum is the product of velocity and MASS (not weight: weight is a force).
You have worked out the total momentum before the jump, but note that the velocity of the skateboard is now zero, so dividing by its mass is not useful.
Tim after the jump however, has a momentum mv (you know his mass), what is v?
Sam. | 437 | 1,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-04 | latest | en | 0.947935 |
https://rdrr.io/cran/maotai/man/boot.stationary.html | 1,675,583,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00454.warc.gz | 498,010,533 | 9,143 | # boot.stationary: Generate Index for Stationary Bootstrapping In maotai: Tools for Matrix Algebra, Optimization and Inference
## Description
Assuming data being dependent with cardinality `N`, `boot.stationary` returns a vector of index that is used for stationary bootstrapping. To describe, starting points are drawn from uniform distribution over `1:N` and the size of each block is determined from geometric distribution with parameter p.
## Usage
`1` ```boot.stationary(N, p = 0.25) ```
## Arguments
`N` the number of observations. `p` parameter for geometric distribution with the size of each block.
## Value
a vector of length `N` for moving block bootstrap sampling.
## References
\insertRef
politis_stationary_1994maotai
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27``` ```## example : bootstrap confidence interval of mean and variances vec.x = seq(from=0,to=10,length.out=100) vec.y = sin(1.21*vec.x) + 2*cos(3.14*vec.x) + rnorm(100,sd=1.5) data.mu = mean(vec.y) data.var = var(vec.y) ## apply stationary bootstrapping nreps = 50 vec.mu = rep(0,nreps) vec.var = rep(0,nreps) for (i in 1:nreps){ sample.id = boot.stationary(100) sample.y = vec.y[sample.id] vec.mu[i] = mean(sample.y) vec.var[i] = var(sample.y) print(paste("iteration ",i,"/",nreps," complete.", sep="")) } ## visualize opar <- par(no.readonly=TRUE) par(mfrow=c(1,3), pty="s") plot(vec.x, vec.y, type="l", main="1d signal") # 1d signal hist(vec.mu, main="mean CI", xlab="mu") # mean abline(v=data.mu, col="red", lwd=4) hist(vec.var, main="variance CI", xlab="sigma") # variance abline(v=data.var, col="blue", lwd=4) par(opar) ```
maotai documentation built on Feb. 3, 2022, 5:09 p.m. | 545 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-06 | latest | en | 0.578898 |
https://pastebin.com/3EQnaM3D | 1,548,134,936,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583826240.93/warc/CC-MAIN-20190122034213-20190122060213-00196.warc.gz | 589,987,503 | 6,622 | • API
• FAQ
• Tools
• Archive
daily pastebin goal
69%
SHARE
TWEET
Untitled
a guest Dec 15th, 2018 80 Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
1. function dxdt = sys(t,X,Q)
2. [be, delta, d1, k2, lambda1, Kb] = Q
3. d2 = 0.01
4. k1 = 8e-7
5. lambda2 = 31.98
6. eps = .1
7. de = 0.25
8. Nt = 100
9. f = 0.34
10. m1 = m2 = 1e-5
11. lambda_e = 1
12. delta_e = 0.1
13. p1 = p2 = 1
14. Kd = 1
15. T1 = X(1)
16. T2 = X(2)
17. T1s = X(3)
18. T2s = X(4)
19. V = X(5)
20. E = X(6)
21.
22. dT1 = lambda1 - d1*T1 - (1-eps)*k1*V*T1
23. dT2 = lambda2 - d2*T2 - (1-f*eps)*k2*V*T2
24. dT1s = (1-eps)*k1*V*T1 - delta*T1s - m1*E*T1s
25. dT2s = (1 - f*eps)*k2*V*T2 - delta*T2s - m2*E*T2s
26. dV = Nt*delta*(T1s + T2s) - c*V -((1-eps)*p1*k1*T1 + (1-f*eps)*p2*k2*T2)*V
27. dE = lambda_e + E*(be*(T1s + T2s))/(T1s + T2s + Kb) - E*(de*(T1s + T2s))/(T1s + T2s + Kd) - delta_e*E
28.
29. dxdt = [dT1,dT2,dT1s,dT2s,dV,dE]
RAW Paste Data
We use cookies for various purposes including analytics. By continuing to use Pastebin, you agree to our use of cookies as described in the Cookies Policy.
Top | 578 | 1,211 | {"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} | 2.875 | 3 | CC-MAIN-2019-04 | longest | en | 0.40189 |
mustbeit.com | 1,638,435,030,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00298.warc.gz | 53,985,875 | 7,082 | # Fun Ways To Teach Algebra
The concept of taking lessons in algebra has always been a frustrating and annoying class. Majority of people in class usually donít like mathematics.
While there are some who love mathematics, you will have to accept the fact that there are also plenty of people who are not very good at it. This article is all about fun ways in teaching algebra. Certain methods to teach the lesson, including ones that are available for download in the internet are taken into account.
## The Alphabet Slope
Using the alphabet provides plenty of practice when it comes to slope. For example, the printed letters found on the alphabet can be made using lines (e.g. the letter ďAĒ is only a positive, a negative or a zero slope line. Or the letter N is made up of undefined slope lines and another negative slope. The letter ďCĒ on another example would be made into a non-linear line. Using the alphabet slope as a fun way to teach algebra is very effective for students, particularly those who are in need of constant basic practice. The basics usually involve positives, zeros, undefined values and negatives.
## PowerPoint Games
The internet is a very resourceful tool. You can find plenty of templates and PowerPoint presentations that teach algebra in a creative and colorful manner. You can also create your own PowerPoint algebra game once you are well versed on how the process is done (if you are knowledgeable with certain MS applications). PowerPoint Games are very useful, they can be recycled and are easily edited. You can look for certain algebra PowerPoint games in the internet for more fun ways to teach algebra.
## Polygon Pictures
This type of algebra learning tool will let students put in their names on certain blocks. The blocks contains letters and identifies the polygons and other algebraic values of the soon to be used letters. The students are even able to decorate the pictures with their names using some polygons they can use. The Polygon picture project is available on certain websites in the internet. You can grab one of your own and use in in teaching your kids algebra.
## The Python Project
The Python project, also known as PYTHON is an incorporated Computer Programming and Art that has slopes and other types of linear equations. PYTHON is available for download in the internet. While the thing may appear as if itís a difficult project, donít be fooled. PYTHON is actually easy to use. Itís methods of teaching is very straightforward. Your students will definitely love PYTHON.
Even if you or your students donít know any code (which is usually the case for frustration), they can still learn how to use python. Just copy and pasting a few codes is whatís needed to use the PYTHON. | 545 | 2,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.936197 |
http://mathhelpforum.com/statistics/48803-number-codes.html | 1,529,714,480,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00234.warc.gz | 202,092,188 | 9,646 | 1. ## number of codes
hey i would really appreciate it if somebody could answer this:
Assume that product codes are formed from the letters S, W, Q, T, R, and Y, and consist of 4 not necessarily distinct letters arranged one after the other. For example, SSWR is a product code.
(1) How many different product codes are there?
i got this one, it's 6^4 because there are 4 stages and each has 6 options.
(2) How many different product codes do not contain W?
i got this one too, it's 5^4 because instead of 4 stages with 6 options, there are 4 stages with 5 options.
(3) How many different product codes contain exactly one R?
this is the one i haven't been able to get. help please?
2. The answer is 4 x 5^3
This is because you have exactly one "R", so you first put an "R" in the product code, and then count the number of places it can be (it can be in four places). Then you have 3 decisions with 5 options for the remaining 3 spots, so you multiply by 5^3.
3. thanks. that helps a lot. | 267 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | latest | en | 0.952194 |
https://www.findaphd.com/phds/project/sparse-spectral-methods-on-new-geometries-for-differential-equations/?p152005 | 1,675,773,278,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00056.warc.gz | 790,213,014 | 30,024 | # Sparse spectral methods on new geometries for differential equations
### This project is no longer listed on FindAPhD.com and may not be available.
Dr Marco Fasondini, Prof R Davidchack No more applications being accepted Funded PhD Project (Students Worldwide)
Spectral methods are used to compute solutions to differential and integral equations. Typically, this is done by expansions in orthogonal basis functions on the entire domain of the problem, or on multiple subdomains in a spectral element (or high-p-finite element) method.
Traditional spectral methods are well known to have excellent convergence properties for analytic solutions (exponentially fast convergence). However, this comes at the expense of solving dense and ill-conditioned linear systems. In recent years, spectral methods have been devised that result in sparse and well-conditioned linear systems that can be solved with fast, optimal complexity algorithms. Furthermore, these sparse spectral methods have been extended from intervals to certain regions in 2D and 3D, for example triangles, balls, cones, disks, disk slices, trapeziums and spherical caps (see, for example, [1]). In fact, sparse spectral methods can be designed for any geometry described by an algebraic curve or surface (zero sets of multivariate polynomials), however in practice one requires an explicit family of orthogonal basis functions or a stable and efficient computational method for constructing the basis functions.
The aim of this project is to design sparse spectral methods for a set of geometries described by algebraic curves and surfaces. The project supervisor and his collaborators have recently constructed new orthogonal polynomial basis functions (OPs) on a class of algebraic curves in 1D with a stable algorithm that has linear complexity [2,3]. Their methodology can be extended to construct OPs on regions in 2D bounded by the same class of algebraic curves and their associated surfaces of revolution in 3D. These OPs will be used in this project to construct sparse matrix representations of conversion (change-of-basis), multiplication and differentiation operators as well as transforms based on quadrature. These sparse matrices and transforms will be used to devise sparse spectral methods on the above-mentioned regions in 2D and 3D, which will be implemented in the open source Julia programming language. The resulting spectral methods will be tested on model problems that arise in acoustics and fluid mechanics (e.g., Laplace and Helmholtz problems). Ultimately, the goal is to extend these spectral methods to sparse spectral element methods for computationally challenging applications in numerical weather prediction, acoustic and elastic wave propagation and medical imaging.
Start date Sept 2023
Eligibility:
UK and International* applicants are welcome to apply.
Entry requirements:
Applicants are required to hold/or expect to obtain a UK Bachelor Degree 2:1 or better in a relevant subject or overseas equivalent.
The University of Leicester English language requirements may apply.
To apply
Please refer to the information and How to Apply section on our web site https://le.ac.uk/study/research-degrees/funded-opportunities/future-50-cse
Please ensure you include the project reference, supervisor and project title on your application.
## Funding Notes
Future 50 Scholarship
Studentships provide funding for 3.5 years to include:
• Tuition fees at UK rates
• Stipend at UKRI rates for 2023 to be confirmed in early 2023 (currently £17,668 for 2022 entry)
• Access to a Research Training Support Grant of up to £1,500 pa for 3 years.
• Bench fees of £5,000 per annum for three years for laboratory-based studies
International applicants will need to be able to fund the difference between UK and International fees for the duration of study.
## References
1] B. Snowball and S. Olver. Sparse spectral and p-finite element methods for partial differential equations on disk slices and trapeziums. Stud. Appl. Math., 145:3–35, 2020.
[2] M. Fasondini, S. Olver, and Y. Xu. Orthogonal polynomials on planar cubic curves. Found. Comput. Math., 1-31, 2021.
[3] M. Fasondini, S. Olver, and Y. Xu. Orthogonal polynomials on a class of planar algebraic curves. Submitted, see https://arxiv.org/abs/2211.06999
##### Search suggestions
Based on your current searches we recommend the following search filters.
##### PhD suggestions
Based on your current search criteria we thought you might be interested in these. | 970 | 4,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.911489 |
https://www.vedantu.com/question-answer/a-body-cools-from-50circ-c-to-40circ-c-in-5-class-11-physics-cbse-5fb45d9aaae7bc7b2d37fd62 | 1,718,782,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00510.warc.gz | 925,950,331 | 28,135 | Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# A body cools from $50^\circ C$ to $40^\circ C$ in 5 minutes. The surrounding temperature is $20^\circ C$. By how much $^\circ C$ does the temperature decrease in the next 5 minutes? Round your answer to the nearest integer.
Last updated date: 19th Jun 2024
Total views: 340.6k
Views today: 10.40k
Verified
340.6k+ views
Hint:The Newton’s law of cooling is defined as the loss of heat of the body is directly proportional to the difference of the temperature of the body and the surrounding. Also the coefficient of heat transfer is constant to the.
Formula used:The formula of the Newton’s law of cooling is given by,
$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
Where $\Delta \theta = {\theta _1} - {\theta _2}$, change in temperature is $\Delta t$, the constant is k, the $\bar \theta$ is the arithmetic mean and ${\theta _o}$ is the surrounding temperature.
Complete step by step solution:
It is given in the problem that a body cools from $50^\circ C$ to $40^\circ C$ in 5 minutes, the surrounding temperature is $20^\circ C$. We need to tell the decrease in temperature in the next 5 minutes.
The formula of the Newton’s law of cooling is given by,
$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
Where $\Delta \theta = {\theta _1} - {\theta _2}$, change in temperature is $\Delta t$, the constant is k, the $\bar \theta$ is the arithmetic mean and ${\theta _o}$ is the surrounding temperature.
Body cools from $50^\circ C$ to $40^\circ C$ in 5 minutes and the surrounding temperature is $20^\circ C$, replacing all these values in Newton's law of cooling.
$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
$\Rightarrow \dfrac{{50 - 40}}{5} = - k\left( {\dfrac{{50 + 40}}{2} - 20} \right)$
$\Rightarrow \dfrac{{10}}{5} = - k\left( {\dfrac{{90}}{2} - 20} \right)$
$\Rightarrow 2 = - k\left( {45 - 20} \right)$
$\Rightarrow 2 = - k\left( {25} \right)$
$\Rightarrow k = \dfrac{{ - 2}}{{25}}$.
Let us calculate the temperature decrease in the next 5 min.
The formula of the Newton’s law of cooling is given by,
$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
Where $\Delta \theta = {\theta _1} - {\theta _2}$, change in temperature is $\Delta t$, the constant is k, the $\bar \theta$ is the arithmetic mean and ${\theta _o}$ is the surrounding temperature.
$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
$\Rightarrow \dfrac{{40 - t}}{5} = \dfrac{2}{{25}}\left( {\dfrac{{40 + t}}{2} - 20} \right)$
$\Rightarrow \dfrac{{40 - t}}{5} = \dfrac{2}{{25}}\left( {\dfrac{{40 + t - 40}}{2}} \right)$
$\Rightarrow 40 - t = \dfrac{{40 + t - 40}}{5}$
$\Rightarrow 5\left( {40 - t} \right) = t$
$\Rightarrow 200 - 5t = t$
$\Rightarrow 6t = 200$
$\Rightarrow t = \dfrac{{200}}{6}$
$\Rightarrow t = 33 \cdot 33^\circ C$.
The final temperature in the next 5 min. is equal $t = 33 \cdot 33^\circ C$.
Note:The students are advised to understand and remember the formula of Newton's law of cooling as it is very helpful in solving problems like these. Newton's law of cooling is the concept which tells us the ratio of heat transfer from one body to another. | 1,041 | 3,383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-26 | latest | en | 0.793561 |
http://forum.allaboutcircuits.com/threads/harmonics-in-the-delta-winding-of-a-wye-wye-transformer.93696/ | 1,481,353,000,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542972.9/warc/CC-MAIN-20161202170902-00138-ip-10-31-129-80.ec2.internal.warc.gz | 99,794,921 | 15,083 | # Harmonics in the delta winding of a wye-wye transformer
Discussion in 'Homework Help' started by MK1337, Jan 21, 2014.
1. ### MK1337 Thread Starter New Member
Jan 21, 2014
2
0
Hi everyone,
I have been working on an ongoing set of coursework related to delta tertiary windings in wye-wye transformers. So far I have successfully answered:
• Why third harmonics are a problem in wye-wye connected transformers
• Why the delta-connected tertiary provides third-harmonic balancing
I am stuck on the final question relating to a small fundamental component of current circulating in the delta tertiary. Whilst I can see why given, say, a 50Hz input the 150Hz current would flow in the delta, I cannot see why a 50Hz current would flow.
Any pointers in the right direction would be a great help.
2. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
Perhaps it relates to the presence of zero sequence currents.
3. ### MK1337 Thread Starter New Member
Jan 21, 2014
2
0
OK, so the third harmonic is effectively a zero phase sequence current, right? If you take the third harmonics and phase shift them the same amount as you phase shift the fundamental, they all end up in phase at three times the frequency (correct me here if I am wrong). How does the third harmonic waveform being a zero sequence current then explain a 50Hz component? I have researched zero sequence briefly but nothing I have read suggests that three 150Hz in-phase waves would create a 50Hz component.
4. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
Any unbalanced 3-phase current case can be resolved into three sub-sets at 50Hz. Nothing to do with the third harmonic.
The three 50 Hz (3-phase) subsets are positive sequence, negative sequence and zero sequence. This will be apparent in unbalanced 3-wire conditions - particularly under fault conditions.
Last edited: Jan 23, 2014
5. ### skooage New Member
Jan 28, 2014
1
0
I'm doing exactly the same piece of coursework. The other sections seemed to be relatively well documented in journals and books but I'm really struggling to find anything that mentions this small 50hz component circulating in the tertiary delta aside from the information we were provided.
Someone on my course mentioned this may be due to lack of electrical isolation between the phase windings, another mentioned it could be attributed to the way the transformers are wound (A and C are concentric windings, B is a sandwich winding).
t_n_k, I'm struggling to understand what you're getting at here.
Any additional help would be appreciated!!!
6. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
The OP was apparently asking in which circumstance one might find an example of common in-phase circulating current at 50Hz (system frequency) in a delta tertiary winding.
I cited the case of fundamental frequency zero sequence currents that can flow, say in the case of power system faults.
Are you familiar with the analysis of power system faults using symmetrical components?
I guess it would be possible (by way of example) to draw or simulate a system diagram with say a single phase ground fault on a 3-phase transmission line connecting a generator and load via a Y-Y-Δ(tertiary) transformer. One would presumably "see" zero sequence fundamental current in the tertiary Δ winding(s).
As to the other examples you cited - I haven't thought about those cases.
• ###### Y-Y-Delta [tertiary] transformer with feeder ground fault.jpg
File size:
266.7 KB
Views:
306
Last edited: Jan 28, 2014 | 821 | 3,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.924766 |
https://www.chegg.com/homework-help/fundamentals-of-momentum-heat-and-mass-transfer-6th-edition-international-student-version-6th-edition-chapter-6-solutions-9781118808870 | 1,563,424,425,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00252.warc.gz | 654,749,796 | 48,096 | # Fundamentals of Momentum, Heat and Mass Transfer, 6th Edition International Student Version (6th Edition) Edit edition Solutions for Chapter 6
We have solutions for your book!
Chapter: Problem:
Sea water, p = 1025 kg/m3, flows through a pump at 0:21 m3/s. The pump inlet is 0.25 m in diameter. At the inlet the pressure is –0.15 m of mercury. The pump outlet, 0.152 m in diameter, is 1.8 m above the inlet. The outlet pressure is 175 kPa.
If the inlet and exit temperature are equal, how much power does the pump add to the fluid?
Step-by-step solution:
Chapter: Problem:
• Step 1 of 9
Write the formula for the inlet pressure.
Here, density of mercury is , gravitational constant is , and the pressure head of mercury is .
Substitute for , for , and for .
• Step 2 of 9
For a steady flow equation the rate of heat loss and the rate of shear work is zero.
The rate of heat lost in the steady flow is .
The rate of shear work in the steady flow is .
Write the energy equation of the process for the total power added to the pump.
Here, density of sea water is ,total specific energy is , pressure is , velocity vector is , normal vector is , and area is .
• Step 3 of 9
Rewrite the energy equation for rate of work done.
…… (1)
Here, shaft work is , specific internal energy at outlet is , specific internal energy at inlet is , pressure at the outlet is , velocity at the outlet is , velocity at the inlet is , and kinetic head between inlet and outlet is .
• Step 4 of 9
Write the formula for mass flow rate.
Here, the volumetric flow rate is .
Substitute for and for .
• Step 5 of 9
Write the formula for velocity at the inlet.
Here, area at the inlet is and diameter at the inlet is .
Substitute for and for .
• Step 6 of 9
Write the formula for velocity at the outlet.
Here, area at the outlet is and diameter at the outlet is .
Substitute for and for .
• Step 7 of 9
The inlet and exit temperatures are equal. Hence, the change in internal energy is zero.
.
• Step 8 of 9
Rewrite the energy equation (1)
...... (2)
• Anonymous
175kPa + 19.99kPa should be (175 + 19.99) * 1000 Pa
• Step 9 of 9
Substitute for , 0 for , 175 kPa for , for , for , for , for , for , and 1.8 m for in Equation (2).
Hence, the power added by the pump to the sea water is .
• Anonymous
• Anonymous
You're Wrong
• Anonymous | 613 | 2,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-30 | latest | en | 0.904142 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.