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://farfromhomemovie.com/category/blog/classical-physics/ | 1,618,319,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00496.warc.gz | 334,054,788 | 10,046 | Close
Classical Physics
Articles that explain the essential laws of physics
SI system, Path and Relocation
The fundamental models of measurement within the SI system are: unit of measurement of size – meter (1 m), time – a second (1 s), weight – a kilogram (1 kg), quantity of substance – mole (1 mole), temperature – kelvin (1 Ok), the energy of the electrical present is ampere (1 A), For reference: […]
2018-09-27
The theory of optics Part 3
In the glance of gentle by optically denser moderate to optically less compact n 2 < n 1 (e.g. glass to atmosphere) it’s really is likely to discover the occurrence of total internal manifestation, i.e. that the disappearance of this refracted beam. This happening can be seen at angles of prevalence surpassing the crucial angle […]
2018-04-18
The theory of optics Part 2
Diffraction. Diffraction grating Diffraction of lighting is still that the occurrence of lots of lighting by rectilinear propagation management when dying nearby the barrier, whose measurements are equal to the wavelength of lighting (diffraction lighting of these barriers). As experience demonstrates the lighting below certain states will input into the spot of the darkened shadow […]
2018-04-13
The theory of optics Part 1
Lighting is electromagnetic waves that the wavelengths that lie into your common eye is amongst 400 to 760 nm. In those constraints, the lighting is named observable. Lighting with all the lightest wavelength looks reddish to people also with smallest purple. Bear in mind the shifting hues of this spectrum readily state Each Hunter wants […]
2018-04-11
Basics: Kinematics Part 3
The Scope and Complete time of flight of an individual human anatomy pitched angularly into the horizon (presuming the flight endings at an Identical elevation by that we started, i.e. the entire body has been thrown, for Instance, in the ground into the ground): The formulation of Assortment of this entire body pitched in a angle […]
2017-10-18
Basics: Kinematics Part 2
By the Prior formulation, if yet another more prevalent Formulation describing the Reversal of pace Eventually at accelerated Movement. The formulation of Dependence of rate at Time to Time in accelerated Movement. Relocating (although maybe not route) at accelerated motion Is Figured by this System. Formula Proceed into accelerated rectilinear movement. From the latter formulation utilized one particular […]
2017-10-13
Basics: Kinematics Part 1
When executing calculations in SI models angles are measured in radians. In case the physics dilemma isn’t given the components will need to respond, it needs to really be granted in SI units, or based amounts corresponding for the physiological amount, and it is required in the issue. As an instance, in the event the […]
2017-10-10
How to learn the basics of physics Part 2
Also possible to study the thermodynamics of the law, the processes of direct and reverse Carnot cycle. To study, for example, why the internal combustion engine (and any heat engine) is the ultimate real efficiency and equal it is, of course, not 99% or even 70%, and about 40%. In the study of the working […]
2017-08-15
How to learn the basics of physics Part 1
Studying physics at school is built correctly in terms of “unfolding events”, another thing is that, as in the case of mathematics, it is not always clear why do we need to know. Since physics is purely practical, if we are not talking about some special physic mathematical schools, and it needs to be taught […]
2017-08-09 | 785 | 3,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.896201 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=NLR | 1,521,561,299,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647498.68/warc/CC-MAIN-20180320150533-20180320170533-00064.warc.gz | 374,213,019 | 73,535 | Back to list of Stocks See Also: Seasonal Analysis of NLRGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of NLR (VanEck Vectors Uranium+Nuclear Engy ETF)
NLR (VanEck Vectors Uranium+Nuclear Engy ETF) appears to have interesting cyclic behaviour every 46 weeks (1.5585*cosine), 43 weeks (1.4624*sine), and 55 weeks (.9571*sine).
NLR (VanEck Vectors Uranium+Nuclear Engy ETF) has an average price of 46.11 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
Right click on the graph above to see the menu of operations (download, full screen, etc.)
## Fourier Analysis
Using data from 8/15/2007 to 3/12/2018 for NLR (VanEck Vectors Uranium+Nuclear Engy ETF), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
046.11029 0
17.41409 1.931 (1*2π)/553553 weeks
21.79145 .93505 (2*2π)/553277 weeks
36.23562 1.45258 (3*2π)/553184 weeks
41.28872 3.17001 (4*2π)/553138 weeks
5.92882 2.4538 (5*2π)/553111 weeks
6.80005 3.22511 (6*2π)/55392 weeks
7-1.83733 2.62256 (7*2π)/55379 weeks
8-2.01567 -.54331 (8*2π)/55369 weeks
9.14934 -.47829 (9*2π)/55361 weeks
10.06238 .95706 (10*2π)/55355 weeks
11.43298 -.09416 (11*2π)/55350 weeks
121.55852 .80126 (12*2π)/55346 weeks
13.66664 1.46242 (13*2π)/55343 weeks
14-.00117 1.2751 (14*2π)/55340 weeks
15-.38169 1.21722 (15*2π)/55337 weeks
16.00183 1.17147 (16*2π)/55335 weeks
17-.66503 1.33071 (17*2π)/55333 weeks
18-.8601 .42539 (18*2π)/55331 weeks
19-.21364 .64643 (19*2π)/55329 weeks
20-.65582 .09412 (20*2π)/55328 weeks
21.11572 -.06341 (21*2π)/55326 weeks
22-.06085 .3734 (22*2π)/55325 weeks
23.3703 .11744 (23*2π)/55324 weeks
24.11459 .60262 (24*2π)/55323 weeks
25.18485 .91185 (25*2π)/55322 weeks
26-.47815 .86417 (26*2π)/55321 weeks
27-.26735 -.18287 (27*2π)/55320 weeks
28-.18814 .11774 (28*2π)/55320 weeks
29-.01517 -.03129 (29*2π)/55319 weeks
30.26576 .15589 (30*2π)/55318 weeks
31.20827 .62454 (31*2π)/55318 weeks
32.00207 .47584 (32*2π)/55317 weeks
33-.17061 .60885 (33*2π)/55317 weeks
34-.65457 .18684 (34*2π)/55316 weeks
35-.14155 -.24866 (35*2π)/55316 weeks
36.12184 .08628 (36*2π)/55315 weeks
37.18376 -.07429 (37*2π)/55315 weeks
38-.05718 .05693 (38*2π)/55315 weeks
39.13135 -.01529 (39*2π)/55314 weeks
40.25668 .14221 (40*2π)/55314 weeks
41.10198 .35687 (41*2π)/55313 weeks
42-.09375 .10063 (42*2π)/55313 weeks
43.19245 .31996 (43*2π)/55313 weeks
44-.10782 .18784 (44*2π)/55313 weeks
45.01513 -.04643 (45*2π)/55312 weeks
46-.32182 .24103 (46*2π)/55312 weeks
47-.33913 -.05477 (47*2π)/55312 weeks
48.19622 -.01937 (48*2π)/55312 weeks
49.03215 .12839 (49*2π)/55311 weeks
50-.0368 .15131 (50*2π)/55311 weeks
51.1124 .04635 (51*2π)/55311 weeks
52.21187 .20664 (52*2π)/55311 weeks
53.10979 .06132 (53*2π)/55310 weeks
54.09904 .07003 (54*2π)/55310 weeks
55-.03623 .18305 (55*2π)/55310 weeks
56-.08471 -.00578 (56*2π)/55310 weeks
57.37449 -.115 (57*2π)/55310 weeks
58.14653 -.04924 (58*2π)/55310 weeks
59.2397 .16515 (59*2π)/5539 weeks
60.30454 .23588 (60*2π)/5539 weeks
61.13905 .39676 (61*2π)/5539 weeks
62-.06169 .22383 (62*2π)/5539 weeks
63.06202 .25575 (63*2π)/5539 weeks
64-.05748 .19625 (64*2π)/5539 weeks
65.00253 .19995 (65*2π)/5539 weeks
66.062 .13381 (66*2π)/5538 weeks
67-.06349 .0986 (67*2π)/5538 weeks
68.13957 .30428 (68*2π)/5538 weeks
69-.06638 .10768 (69*2π)/5538 weeks
70.10362 .14473 (70*2π)/5538 weeks
71-.03278 .16445 (71*2π)/5538 weeks
72-.15772 .21817 (72*2π)/5538 weeks
73-.18715 .10515 (73*2π)/5538 weeks
74-.20102 -.00092 (74*2π)/5537 weeks
75-.01834 -.1211 (75*2π)/5537 weeks
76.10578 .02416 (76*2π)/5537 weeks
77-.11807 -.01783 (77*2π)/5537 weeks
78.20685 -.23076 (78*2π)/5537 weeks
79.25722 .06974 (79*2π)/5537 weeks
80-.03249 .03604 (80*2π)/5537 weeks
81.22622 .10219 (81*2π)/5537 weeks
82.0091 .0525 (82*2π)/5537 weeks
83.11675 -.03915 (83*2π)/5537 weeks
84.17136 .17938 (84*2π)/5537 weeks
85-.05732 .00752 (85*2π)/5537 weeks
86.16752 .05653 (86*2π)/5536 weeks
87.01528 .15702 (87*2π)/5536 weeks
88-.03932 .20977 (88*2π)/5536 weeks
89-.07488 .17648 (89*2π)/5536 weeks
90-.16717 .06011 (90*2π)/5536 weeks
91-.05619 -.0804 (91*2π)/5536 weeks
92.03529 -.09231 (92*2π)/5536 weeks
93.10166 .01387 (93*2π)/5536 weeks
94.11761 .09443 (94*2π)/5536 weeks
95-.00778 .01558 (95*2π)/5536 weeks
96.07469 .01795 (96*2π)/5536 weeks
97.13962 .08099 (97*2π)/5536 weeks
98.13488 .14942 (98*2π)/5536 weeks
99.07879 .13573 (99*2π)/5536 weeks
100.09433 .09218 (100*2π)/5536 weeks
101.05006 .00169 (101*2π)/5535 weeks
102.00315 .01717 (102*2π)/5535 weeks
103-.03136 -.0145 (103*2π)/5535 weeks
104.16852 .0433 (104*2π)/5535 weeks
105.1409 .05412 (105*2π)/5535 weeks
106.21633 .19127 (106*2π)/5535 weeks
107.09385 .12031 (107*2π)/5535 weeks
108.11641 .1954 (108*2π)/5535 weeks
109.00339 .17031 (109*2π)/5535 weeks
110.08418 .02133 (110*2π)/5535 weeks
111.0348 .05921 (111*2π)/5535 weeks
112.05828 .07679 (112*2π)/5535 weeks
113.19569 .20638 (113*2π)/5535 weeks
114.0902 .12152 (114*2π)/5535 weeks
115.15492 .19955 (115*2π)/5535 weeks
116-.07786 .18378 (116*2π)/5535 weeks
117-.03617 .2437 (117*2π)/5535 weeks
118.03801 .13727 (118*2π)/5535 weeks
119-.04485 .1143 (119*2π)/5535 weeks
120-.03638 .03705 (120*2π)/5535 weeks
121-.01262 .01304 (121*2π)/5535 weeks
122.10715 .0941 (122*2π)/5535 weeks
123.0892 .14608 (123*2π)/5534 weeks
124.02886 .21325 (124*2π)/5534 weeks
125-.07887 .12836 (125*2π)/5534 weeks
126-.08225 .11584 (126*2π)/5534 weeks
127-.05346 .16732 (127*2π)/5534 weeks
128-.03401 -.00346 (128*2π)/5534 weeks
129-.06578 .06438 (129*2π)/5534 weeks
130-.1523 -.00168 (130*2π)/5534 weeks
131.11727 -.17718 (131*2π)/5534 weeks
132.05109 -.00372 (132*2π)/5534 weeks
133-.0401 .08759 (133*2π)/5534 weeks
134.04341 .00931 (134*2π)/5534 weeks
135.05415 .02054 (135*2π)/5534 weeks
136.02424 .0012 (136*2π)/5534 weeks
137.16006 .03955 (137*2π)/5534 weeks
138.0968 .1749 (138*2π)/5534 weeks
139-.11405 .05992 (139*2π)/5534 weeks
140-.01245 -.00899 (140*2π)/5534 weeks
141.06122 .01289 (141*2π)/5534 weeks
142.17042 -.06665 (142*2π)/5534 weeks
143.18532 .07991 (143*2π)/5534 weeks
144.05901 .10467 (144*2π)/5534 weeks
145.06175 .00399 (145*2π)/5534 weeks
146.0658 .11886 (146*2π)/5534 weeks
147.02951 .126 (147*2π)/5534 weeks
148.06837 .06833 (148*2π)/5534 weeks
149-.02555 .11418 (149*2π)/5534 weeks
150-.00258 -.01754 (150*2π)/5534 weeks
151-.0785 -.00716 (151*2π)/5534 weeks
152.04366 -.06827 (152*2π)/5534 weeks
153.09118 .07765 (153*2π)/5534 weeks
154.01675 .05396 (154*2π)/5534 weeks
155.02514 .07462 (155*2π)/5534 weeks
156-.0457 .06485 (156*2π)/5534 weeks
157-.019 .0316 (157*2π)/5534 weeks
158.01369 .04123 (158*2π)/5534 weeks
159-.08482 .05284 (159*2π)/5533 weeks
160.00577 -.1216 (160*2π)/5533 weeks
161.10833 -.02932 (161*2π)/5533 weeks
162.02974 .06251 (162*2π)/5533 weeks
163-.01338 -.0332 (163*2π)/5533 weeks
164.04255 .03117 (164*2π)/5533 weeks
165.1059 .07526 (165*2π)/5533 weeks
166-.02416 -.00895 (166*2π)/5533 weeks
167.05265 .04032 (167*2π)/5533 weeks
168.01518 .07729 (168*2π)/5533 weeks
169-.00068 .03517 (169*2π)/5533 weeks
170.03685 .0144 (170*2π)/5533 weeks
171-.001 .00809 (171*2π)/5533 weeks
172.02109 .019 (172*2π)/5533 weeks
173.06417 .01173 (173*2π)/5533 weeks
174-.01861 .01369 (174*2π)/5533 weeks
175-.02779 .10887 (175*2π)/5533 weeks
176-.04423 .0366 (176*2π)/5533 weeks
177-.0394 .05298 (177*2π)/5533 weeks
178-.05307 .04276 (178*2π)/5533 weeks
179-.0333 -.06885 (179*2π)/5533 weeks
180.0345 -.02881 (180*2π)/5533 weeks
181.00359 -.00296 (181*2π)/5533 weeks
182.04439 .00338 (182*2π)/5533 weeks
183.03747 .04514 (183*2π)/5533 weeks
184.03401 .02078 (184*2π)/5533 weeks
185.07063 .03024 (185*2π)/5533 weeks
186.05791 .02428 (186*2π)/5533 weeks
187.02241 .06146 (187*2π)/5533 weeks
188-.03177 .05027 (188*2π)/5533 weeks
189.05559 .06302 (189*2π)/5533 weeks
190.05273 .07485 (190*2π)/5533 weeks
191.00299 .05294 (191*2π)/5533 weeks
192.04192 .04658 (192*2π)/5533 weeks
193.01817 .06868 (193*2π)/5533 weeks
194.01806 .06736 (194*2π)/5533 weeks
195-.1268 .08816 (195*2π)/5533 weeks
196-.15307 -.06268 (196*2π)/5533 weeks
197.09617 -.05193 (197*2π)/5533 weeks
198.04707 -.00199 (198*2π)/5533 weeks
199.07896 .00377 (199*2π)/5533 weeks
200.0392 .03294 (200*2π)/5533 weeks
201-.03329 -.02709 (201*2π)/5533 weeks
202.06744 .00076 (202*2π)/5533 weeks
203-.01436 .03388 (203*2π)/5533 weeks
204.05342 -.08357 (204*2π)/5533 weeks
205.08284 .0551 (205*2π)/5533 weeks
206.02881 -.01284 (206*2π)/5533 weeks
207.11747 .00949 (207*2π)/5533 weeks
208.16372 .03206 (208*2π)/5533 weeks
209.11593 .11364 (209*2π)/5533 weeks
210.02464 .13516 (210*2π)/5533 weeks
211-.02183 .10991 (211*2π)/5533 weeks
212-.0112 .0412 (212*2π)/5533 weeks
213-.01626 .01718 (213*2π)/5533 weeks
214-.00507 .0183 (214*2π)/5533 weeks
215.07546 -.00412 (215*2π)/5533 weeks
216.03772 .05106 (216*2π)/5533 weeks
217.05795 .08555 (217*2π)/5533 weeks
218-.00723 .08227 (218*2π)/5533 weeks
219-.05339 .00716 (219*2π)/5533 weeks
220.03189 -.01252 (220*2π)/5533 weeks
221.07301 .09771 (221*2π)/5533 weeks
222-.0225 .05262 (222*2π)/5532 weeks
223-.01276 -.01528 (223*2π)/5532 weeks
224-.00886 .02191 (224*2π)/5532 weeks
225-.04292 .03019 (225*2π)/5532 weeks
226-.00998 -.05525 (226*2π)/5532 weeks
227.0332 .0104 (227*2π)/5532 weeks
228-.02927 -.02151 (228*2π)/5532 weeks
229.08586 -.00408 (229*2π)/5532 weeks
230-.04071 .02508 (230*2π)/5532 weeks
231-.01711 -.03599 (231*2π)/5532 weeks
232.09587 -.00951 (232*2π)/5532 weeks
233.07373 .0344 (233*2π)/5532 weeks
234.00894 -.02181 (234*2π)/5532 weeks
235.06988 .00711 (235*2π)/5532 weeks
236.06948 .00445 (236*2π)/5532 weeks
237.05632 .0161 (237*2π)/5532 weeks
238.04297 .02014 (238*2π)/5532 weeks
239.03627 .02683 (239*2π)/5532 weeks
240.01177 .03203 (240*2π)/5532 weeks
241-.05942 -.05584 (241*2π)/5532 weeks
242.01465 -.01307 (242*2π)/5532 weeks
243-.05272 -.03085 (243*2π)/5532 weeks
244.03462 -.0587 (244*2π)/5532 weeks
245.00871 .03163 (245*2π)/5532 weeks
246.01189 -.0479 (246*2π)/5532 weeks
247.07243 .0066 (247*2π)/5532 weeks
248.02623 .06464 (248*2π)/5532 weeks
249-.0285 -.06613 (249*2π)/5532 weeks
250-.02118 -.00705 (250*2π)/5532 weeks
251.00558 -.05173 (251*2π)/5532 weeks
252.04629 -.03974 (252*2π)/5532 weeks
253.06091 -.02895 (253*2π)/5532 weeks
254.01816 -.03215 (254*2π)/5532 weeks
255.05643 -.02514 (255*2π)/5532 weeks
256.06542 .02224 (256*2π)/5532 weeks
257.08879 .00363 (257*2π)/5532 weeks
258.0363 .03236 (258*2π)/5532 weeks
259-.01317 -.02047 (259*2π)/5532 weeks
260.01488 .03127 (260*2π)/5532 weeks
261-.01044 -.00571 (261*2π)/5532 weeks
262.04454 -.03086 (262*2π)/5532 weeks
263.08097 .03997 (263*2π)/5532 weeks
264.08288 .0455 (264*2π)/5532 weeks
265.01055 .01936 (265*2π)/5532 weeks
266-.00315 .00405 (266*2π)/5532 weeks
267-.01118 .03066 (267*2π)/5532 weeks
268.0153 .06009 (268*2π)/5532 weeks
269-.04713 -.07461 (269*2π)/5532 weeks
270.02826 -.13709 (270*2π)/5532 weeks
271.08112 -.0004 (271*2π)/5532 weeks
272.06022 .01127 (272*2π)/5532 weeks
273.06289 -.00948 (273*2π)/5532 weeks
274.04665 .00444 (274*2π)/5532 weeks
275.06802 -.00722 (275*2π)/5532 weeks
276.0148 .04935 (276*2π)/5532 weeks
277.0148 -.04935 (277*2π)/5532 weeks
278.06802 .00722 (278*2π)/5532 weeks
279.04665 -.00444 (279*2π)/5532 weeks
280.06289 .00948 (280*2π)/5532 weeks
281.06022 -.01127 (281*2π)/5532 weeks
282.08112 .0004 (282*2π)/5532 weeks
283.02826 .13709 (283*2π)/5532 weeks
284-.04713 .07461 (284*2π)/5532 weeks
285.0153 -.06009 (285*2π)/5532 weeks
286-.01118 -.03066 (286*2π)/5532 weeks
287-.00315 -.00405 (287*2π)/5532 weeks
288.01055 -.01936 (288*2π)/5532 weeks
289.08288 -.0455 (289*2π)/5532 weeks
290.08097 -.03997 (290*2π)/5532 weeks
291.04454 .03086 (291*2π)/5532 weeks
292-.01044 .00571 (292*2π)/5532 weeks
293.01488 -.03127 (293*2π)/5532 weeks
294-.01317 .02047 (294*2π)/5532 weeks
295.0363 -.03236 (295*2π)/5532 weeks
296.08879 -.00363 (296*2π)/5532 weeks
297.06542 -.02224 (297*2π)/5532 weeks
298.05643 .02514 (298*2π)/5532 weeks
299.01816 .03215 (299*2π)/5532 weeks
300.06091 .02895 (300*2π)/5532 weeks
301.04629 .03974 (301*2π)/5532 weeks
302.00558 .05173 (302*2π)/5532 weeks
303-.02118 .00705 (303*2π)/5532 weeks
304-.0285 .06613 (304*2π)/5532 weeks
305.02623 -.06464 (305*2π)/5532 weeks
306.07243 -.0066 (306*2π)/5532 weeks
307.01189 .0479 (307*2π)/5532 weeks
308.00871 -.03163 (308*2π)/5532 weeks
309.03462 .0587 (309*2π)/5532 weeks
310-.05272 .03085 (310*2π)/5532 weeks
311.01465 .01307 (311*2π)/5532 weeks
312-.05942 .05584 (312*2π)/5532 weeks
313.01177 -.03203 (313*2π)/5532 weeks
314.03627 -.02683 (314*2π)/5532 weeks
315.04297 -.02014 (315*2π)/5532 weeks
316.05632 -.0161 (316*2π)/5532 weeks
317.06948 -.00445 (317*2π)/5532 weeks
318.06988 -.00711 (318*2π)/5532 weeks
319.00894 .02181 (319*2π)/5532 weeks
320.07373 -.0344 (320*2π)/5532 weeks
321.09587 .00951 (321*2π)/5532 weeks
322-.01711 .03599 (322*2π)/5532 weeks
323-.04071 -.02508 (323*2π)/5532 weeks
324.08586 .00408 (324*2π)/5532 weeks
325-.02927 .02151 (325*2π)/5532 weeks
326.0332 -.0104 (326*2π)/5532 weeks
327-.00998 .05525 (327*2π)/5532 weeks
328-.04292 -.03019 (328*2π)/5532 weeks
329-.00886 -.02191 (329*2π)/5532 weeks
330-.01276 .01528 (330*2π)/5532 weeks
331-.0225 -.05262 (331*2π)/5532 weeks
332.07301 -.09771 (332*2π)/5532 weeks
333.03189 .01252 (333*2π)/5532 weeks
334-.05339 -.00716 (334*2π)/5532 weeks
335-.00723 -.08227 (335*2π)/5532 weeks
336.05795 -.08555 (336*2π)/5532 weeks
337.03772 -.05106 (337*2π)/5532 weeks
338.07546 .00412 (338*2π)/5532 weeks
339-.00507 -.0183 (339*2π)/5532 weeks
340-.01626 -.01718 (340*2π)/5532 weeks
341-.0112 -.0412 (341*2π)/5532 weeks
342-.02183 -.10991 (342*2π)/5532 weeks
343.02464 -.13516 (343*2π)/5532 weeks
344.11593 -.11364 (344*2π)/5532 weeks
345.16372 -.03206 (345*2π)/5532 weeks
346.11747 -.00949 (346*2π)/5532 weeks
347.02881 .01284 (347*2π)/5532 weeks
348.08284 -.0551 (348*2π)/5532 weeks
349.05342 .08357 (349*2π)/5532 weeks
350-.01436 -.03388 (350*2π)/5532 weeks
351.06744 -.00076 (351*2π)/5532 weeks
352-.03329 .02709 (352*2π)/5532 weeks
353.0392 -.03294 (353*2π)/5532 weeks
354.07896 -.00377 (354*2π)/5532 weeks
355.04707 .00199 (355*2π)/5532 weeks
356.09617 .05193 (356*2π)/5532 weeks
357-.15307 .06268 (357*2π)/5532 weeks
358-.1268 -.08816 (358*2π)/5532 weeks
359.01806 -.06736 (359*2π)/5532 weeks
360.01817 -.06868 (360*2π)/5532 weeks
361.04192 -.04658 (361*2π)/5532 weeks
362.00299 -.05294 (362*2π)/5532 weeks
363.05273 -.07485 (363*2π)/5532 weeks
364.05559 -.06302 (364*2π)/5532 weeks
365-.03177 -.05027 (365*2π)/5532 weeks
366.02241 -.06146 (366*2π)/5532 weeks
367.05791 -.02428 (367*2π)/5532 weeks
368.07063 -.03024 (368*2π)/5532 weeks
369.03401 -.02078 (369*2π)/5531 weeks
370.03747 -.04514 (370*2π)/5531 weeks
371.04439 -.00338 (371*2π)/5531 weeks
372.00359 .00296 (372*2π)/5531 weeks
373.0345 .02881 (373*2π)/5531 weeks
374-.0333 .06885 (374*2π)/5531 weeks
375-.05307 -.04276 (375*2π)/5531 weeks
376-.0394 -.05298 (376*2π)/5531 weeks
377-.04423 -.0366 (377*2π)/5531 weeks
378-.02779 -.10887 (378*2π)/5531 weeks
379-.01861 -.01369 (379*2π)/5531 weeks
380.06417 -.01173 (380*2π)/5531 weeks
381.02109 -.019 (381*2π)/5531 weeks
382-.001 -.00809 (382*2π)/5531 weeks
383.03685 -.0144 (383*2π)/5531 weeks
384-.00068 -.03517 (384*2π)/5531 weeks
385.01518 -.07729 (385*2π)/5531 weeks
386.05265 -.04032 (386*2π)/5531 weeks
387-.02416 .00895 (387*2π)/5531 weeks
388.1059 -.07526 (388*2π)/5531 weeks
389.04255 -.03117 (389*2π)/5531 weeks
390-.01338 .0332 (390*2π)/5531 weeks
391.02974 -.06251 (391*2π)/5531 weeks
392.10833 .02932 (392*2π)/5531 weeks
393.00577 .1216 (393*2π)/5531 weeks
394-.08482 -.05284 (394*2π)/5531 weeks
395.01369 -.04123 (395*2π)/5531 weeks
396-.019 -.0316 (396*2π)/5531 weeks
397-.0457 -.06485 (397*2π)/5531 weeks
398.02514 -.07462 (398*2π)/5531 weeks
399.01675 -.05396 (399*2π)/5531 weeks
400.09118 -.07765 (400*2π)/5531 weeks
401.04366 .06827 (401*2π)/5531 weeks
402-.0785 .00716 (402*2π)/5531 weeks
403-.00258 .01754 (403*2π)/5531 weeks
404-.02555 -.11418 (404*2π)/5531 weeks
405.06837 -.06833 (405*2π)/5531 weeks
406.02951 -.126 (406*2π)/5531 weeks
407.0658 -.11886 (407*2π)/5531 weeks
408.06175 -.00399 (408*2π)/5531 weeks
409.05901 -.10467 (409*2π)/5531 weeks
410.18532 -.07991 (410*2π)/5531 weeks
411.17042 .06665 (411*2π)/5531 weeks
412.06122 -.01289 (412*2π)/5531 weeks
413-.01245 .00899 (413*2π)/5531 weeks
414-.11405 -.05992 (414*2π)/5531 weeks
415.0968 -.1749 (415*2π)/5531 weeks
416.16006 -.03955 (416*2π)/5531 weeks
417.02424 -.0012 (417*2π)/5531 weeks
418.05415 -.02054 (418*2π)/5531 weeks
419.04341 -.00931 (419*2π)/5531 weeks
420-.0401 -.08759 (420*2π)/5531 weeks
421.05109 .00372 (421*2π)/5531 weeks
422.11727 .17718 (422*2π)/5531 weeks
423-.1523 .00168 (423*2π)/5531 weeks
424-.06578 -.06438 (424*2π)/5531 weeks
425-.03401 .00346 (425*2π)/5531 weeks
426-.05346 -.16732 (426*2π)/5531 weeks
427-.08225 -.11584 (427*2π)/5531 weeks
428-.07887 -.12836 (428*2π)/5531 weeks
429.02886 -.21325 (429*2π)/5531 weeks
430.0892 -.14608 (430*2π)/5531 weeks
431.10715 -.0941 (431*2π)/5531 weeks
432-.01262 -.01304 (432*2π)/5531 weeks
433-.03638 -.03705 (433*2π)/5531 weeks
434-.04485 -.1143 (434*2π)/5531 weeks
435.03801 -.13727 (435*2π)/5531 weeks
436-.03617 -.2437 (436*2π)/5531 weeks
437-.07786 -.18378 (437*2π)/5531 weeks
438.15492 -.19955 (438*2π)/5531 weeks
439.0902 -.12152 (439*2π)/5531 weeks
440.19569 -.20638 (440*2π)/5531 weeks
441.05828 -.07679 (441*2π)/5531 weeks
442.0348 -.05921 (442*2π)/5531 weeks
443.08418 -.02133 (443*2π)/5531 weeks
444.00339 -.17031 (444*2π)/5531 weeks
445.11641 -.1954 (445*2π)/5531 weeks
446.09385 -.12031 (446*2π)/5531 weeks
447.21633 -.19127 (447*2π)/5531 weeks
448.1409 -.05412 (448*2π)/5531 weeks
449.16852 -.0433 (449*2π)/5531 weeks
450-.03136 .0145 (450*2π)/5531 weeks
451.00315 -.01717 (451*2π)/5531 weeks
452.05006 -.00169 (452*2π)/5531 weeks
453.09433 -.09218 (453*2π)/5531 weeks
454.07879 -.13573 (454*2π)/5531 weeks
455.13488 -.14942 (455*2π)/5531 weeks
456.13962 -.08099 (456*2π)/5531 weeks
457.07469 -.01795 (457*2π)/5531 weeks
458-.00778 -.01558 (458*2π)/5531 weeks
459.11761 -.09443 (459*2π)/5531 weeks
460.10166 -.01387 (460*2π)/5531 weeks
461.03529 .09231 (461*2π)/5531 weeks
462-.05619 .0804 (462*2π)/5531 weeks
463-.16717 -.06011 (463*2π)/5531 weeks
464-.07488 -.17648 (464*2π)/5531 weeks
465-.03932 -.20977 (465*2π)/5531 weeks
466.01528 -.15702 (466*2π)/5531 weeks
467.16752 -.05653 (467*2π)/5531 weeks
468-.05732 -.00752 (468*2π)/5531 weeks
469.17136 -.17938 (469*2π)/5531 weeks
470.11675 .03915 (470*2π)/5531 weeks
471.0091 -.0525 (471*2π)/5531 weeks
472.22622 -.10219 (472*2π)/5531 weeks
473-.03249 -.03604 (473*2π)/5531 weeks
474.25722 -.06974 (474*2π)/5531 weeks
475.20685 .23076 (475*2π)/5531 weeks
476-.11807 .01783 (476*2π)/5531 weeks
477.10578 -.02416 (477*2π)/5531 weeks
478-.01834 .1211 (478*2π)/5531 weeks
479-.20102 .00092 (479*2π)/5531 weeks
480-.18715 -.10515 (480*2π)/5531 weeks
481-.15772 -.21817 (481*2π)/5531 weeks
482-.03278 -.16445 (482*2π)/5531 weeks
483.10362 -.14473 (483*2π)/5531 weeks
484-.06638 -.10768 (484*2π)/5531 weeks
485.13957 -.30428 (485*2π)/5531 weeks
486-.06349 -.0986 (486*2π)/5531 weeks
487.062 -.13381 (487*2π)/5531 weeks
488.00253 -.19995 (488*2π)/5531 weeks
489-.05748 -.19625 (489*2π)/5531 weeks
490.06202 -.25575 (490*2π)/5531 weeks
491-.06169 -.22383 (491*2π)/5531 weeks
492.13905 -.39676 (492*2π)/5531 weeks
493.30454 -.23588 (493*2π)/5531 weeks
494.2397 -.16515 (494*2π)/5531 weeks
495.14653 .04924 (495*2π)/5531 weeks
496.37449 .115 (496*2π)/5531 weeks
497-.08471 .00578 (497*2π)/5531 weeks
498-.03623 -.18305 (498*2π)/5531 weeks
499.09904 -.07003 (499*2π)/5531 weeks
500.10979 -.06132 (500*2π)/5531 weeks
501.21187 -.20664 (501*2π)/5531 weeks
502.1124 -.04635 (502*2π)/5531 weeks
503-.0368 -.15131 (503*2π)/5531 weeks
504.03215 -.12839 (504*2π)/5531 weeks
505.19622 .01937 (505*2π)/5531 weeks
506-.33913 .05477 (506*2π)/5531 weeks
507-.32182 -.24103 (507*2π)/5531 weeks
508.01513 .04643 (508*2π)/5531 weeks
509-.10782 -.18784 (509*2π)/5531 weeks
510.19245 -.31996 (510*2π)/5531 weeks
511-.09375 -.10063 (511*2π)/5531 weeks
512.10198 -.35687 (512*2π)/5531 weeks
513.25668 -.14221 (513*2π)/5531 weeks
514.13135 .01529 (514*2π)/5531 weeks
515-.05718 -.05693 (515*2π)/5531 weeks
516.18376 .07429 (516*2π)/5531 weeks
517.12184 -.08628 (517*2π)/5531 weeks
518-.14155 .24866 (518*2π)/5531 weeks
519-.65457 -.18684 (519*2π)/5531 weeks
520-.17061 -.60885 (520*2π)/5531 weeks
521.00207 -.47584 (521*2π)/5531 weeks
522.20827 -.62454 (522*2π)/5531 weeks
523.26576 -.15589 (523*2π)/5531 weeks
524-.01517 .03129 (524*2π)/5531 weeks
525-.18814 -.11774 (525*2π)/5531 weeks
526-.26735 .18287 (526*2π)/5531 weeks
527-.47815 -.86417 (527*2π)/5531 weeks
528.18485 -.91185 (528*2π)/5531 weeks
529.11459 -.60262 (529*2π)/5531 weeks
530.3703 -.11744 (530*2π)/5531 weeks
531-.06085 -.3734 (531*2π)/5531 weeks
532.11572 .06341 (532*2π)/5531 weeks
533-.65582 -.09412 (533*2π)/5531 weeks
534-.21364 -.64643 (534*2π)/5531 weeks
535-.8601 -.42539 (535*2π)/5531 weeks
536-.66503 -1.33071 (536*2π)/5531 weeks
537.00183 -1.17147 (537*2π)/5531 weeks
538-.38169 -1.21722 (538*2π)/5531 weeks
539-.00117 -1.2751 (539*2π)/5531 weeks
540.66664 -1.46242 (540*2π)/5531 weeks
5411.55852 -.80126 (541*2π)/5531 weeks
542.43298 .09416 (542*2π)/5531 weeks
543.06238 -.95706 (543*2π)/5531 weeks
544.14934 .47829 (544*2π)/5531 weeks
545-2.01567 .54331 (545*2π)/5531 weeks
546-1.83733 -2.62256 (546*2π)/5531 weeks
547.80005 -3.22511 (547*2π)/5531 weeks
548.92882 -2.4538 (548*2π)/5531 weeks
5491.28872 -3.17001 (549*2π)/5531 weeks
5506.23562 -1.45258 (550*2π)/5531 weeks
5511.79145 -.93505 (551*2π)/5531 weeks | 9,728 | 21,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.789357 |
https://www.fresherwave.com/ncert-solutions-for-class-3-mathematics-chapter-1-where-to-look-from/ | 1,675,853,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500758.20/warc/CC-MAIN-20230208092053-20230208122053-00224.warc.gz | 780,656,445 | 23,970 | # NCERT Solutions for class 3 Mathematics Chapter-1 Where To Look From?
NCERT Solutions for Class 3 Maths Chapter 1 Where To Look From provided here is extremely helpful in revising complete syllabus and getting a strong base on it. NCERT 3rd Class Maths Where To Look From all questions are solved with detailed explanation available for students. In this article we had given of NCERT solutions for Where To Look From Class 3 Maths step by step solution for each and every question of the chapter. These solutions will also help you with your homework. Best teachers across the India created NCERT solutions for Class 3 Maths Chapter 1 Where To Look From according to curriculum and pattern of syllabus as per guidelines of NCERT (CBSE) Books.
## 3rd Class NCERT Maths Chapter 1 Where To Look From
Name of Organization NCERT Name of Class 3rd Class Name of Subject Maths Name of Chapter Chapter 1 Name of Content Where To Look From Name of Category NCERT Solutions Official site http://ncert.nic.in/
### Chapter 1 Where To Look From covers numerous Questions and answers from all topics and sub-topics which are given below
1.Have you looked at things from different sides? Do they look the sdme or different?
Ans. Yes, I looked at things from different sides. They look different from different sides.
2.Look at the pictures drawn here. How does the table look from the side? Which picture is from the top?
Ans.The table looks like the first picture when seen from the side. It looks like the second picture when seen from the top.
3.Some pictures are drawn below. Imagine how these things will look if seen from the top.
Ans.Yes, they will look like this, when seen from the top.
Practice Time
(A)A cat is peeping into a classroom. Can you help her find out where the teacher is?
Ans. The teacher is sitting on the chair. She is opposite to students.
(B)Here are some pictures. Find out from where you have to look to see the things this way.
Ans. (i) From side (ii) From top (iii) From top
(iv)From side (v) From top (vi) From side
(C)Draw top views of a few things and ask your friends to guess what they Eire.
Ans. Top views of some of the things are given below:
Rangoli
1.Have you ever made a rangoli?
Ans. Yes, I have made simple rEuigoli during diwali.
2.On the dot grid given below, draw the following:
(a)A kite (b) A leaf (c) A flower (d)A boat’ (e)A star (f) A pot.
Ans. Some of the designs are given below:
Tit for Tat
1.The painter had made many such pictures in which he drew only one half of the things. Draw the other half of these pictures and find out what these things are. Try doing it with a mirror.
Ans.On completing the drawing, these will look like following: butterfly, lamp (diya), cat, star and fish.
2.Can we repeat the painter’s trick, while drawing pictures of the following?
Ans. No, for these drawings we cannot repeat the painter’s trick.
3.If you ask the painter to draw things which cannot be divided into two similar mirror halves, then he cannot play the trick. Draw three more such things which do not have similar mirror halves.
Ans. Some of such drawings are given below:
Mirror Halves
1.Look at the pictures given below. Does the dotted line divide each picture into two similar mirror halves?
Ans. 1.Yes 2.Yes 3.No 4.No
5.Yes 6.No 7.Yes 8. Yes
9.No 10.No 11.Yes 12.Yes
13.Yes 14.Yes 15.No 16.Yes
2.Give some more examples:
Ans.
3.Can you guess these letters from their halves?
Ans. The letters are as follows;
A,C, H, U, B and K
4.Guess the words by looking at their halves.
Ans.We can make following words from their halves: COOKED RICE
Scroll to Top | 882 | 3,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-06 | longest | en | 0.925022 |
https://shapeofdata.wordpress.com/2016/01/20/continuous-bayes-theorem/ | 1,500,633,252,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423769.10/warc/CC-MAIN-20170721102310-20170721122310-00246.warc.gz | 720,832,558 | 38,906 | ## Continuous Bayes’ Theorem
Bayes’ Rule is one of the fundamental Theorems of statistics, but up until recently, I have to admit, I was never very impressed with it. Bayes’ gives you a way of determining the probability that a given event will occur, or that a given condition is true, given your knowledge of another related event or condition. All the examples that I’ve read or heard about seemed somewhat contrived and unrelated to the sorts of data analysis I was interested in. But it turns out there’s also an interpretation of Bayes’ Theorem that’s not only much more geometric than the standard formulation, but also fits quite naturally into the types of things that I’ve been discussing on this blog. So in today’s post, I want to explain how I came to truly appreciate Bayes’ Theorem.
But rather than start with the statement of Bayes’ Theorem, I want to use an old math teacher trick (which I realize many students hate) of trying to derive it from scratch, without stating what we’re trying to derive. Rather, we’ll start by modifying a problem that I described in an earlier post on probability distributions.
Lets pretend we have a robot arm with two joints: The first is fixed to the center of a table and spins horizontally. There’s a bar from this first joint to the second joint, which also spins horizontally and is attached to a second bar. The two bars are the same length, as shown on the left in the Figure below. The game is to randomly pick a pair of angles for the two joints, then try to guess the x and y coordinates of the hand at the end of the arm. As I described in the earlier post, we can think of the (density function of the) probability distribution of all the possible (x, y) coordinates, and it would look something like what’s shown on the right of the Figure. Here, darker colors indicate larger values of the function.
In the earlier post, we determined that the best place to predict that the hand would be is near the center, since that’s where the probability density is highest. But now we’re going to modify the game a bit: What if each time after spinning the wheel, we are told something about the x-coordinate; either its exact value or a small range. Then how would that change our prediction? Note that if we’re given the exact x-value and asked to predict y, then this is essentially the regression problem except that the probability distribution doesn’t look anything like the one we used in the post on regression.
But lets start with the case where we’re given a range, for example if we knew that the x-value was between 1/4 and 1/3. Then the probability of getting any point with an x-value outside this range would be zero, so the probability density that we would use to guess the y-value would be equal to zero for all points with x-values outside this range. So it would look something like the Figure on the right.
But dropping those values to zero isn’t enough to get the new distribution; the problem is that when we add this restriction on x, the probability of any point with an x-value in the correct range will increase. The question is: By how much will they increase?
In order to answer this question, we need to look more closely at what the probability density function really means. The first thing to note is that the value of the probability density function at a point is not the probability of choosing that point. In fact, the probability of picking any one point is zero, since there are infinitely many possible x and y values.
In order to understand the meaning of the probability density function, we need to use integrals, but (as usual) we can avoid much of the technical details by describing things in terms of the geometry that underlies those integrals. In particular, we’re going to think of our probability density function as describing the elevations of a mountain whose base is the square in which our robot arm rotates. But it won’t be a normal looking mountain – because of the way the density function looks, it’ll have a high peak in the middle, surrounded by a deep moat, then a high circular ridge (shorter than the central peak) around the outside.
I’ve attempted to draw this on the left, but you’re probably better off using your imagination to picture it. Once we’ve transformed our density function into this mountain, we can replace the word “integral” with “volume” and we’ll be able to calculate some probabilities.
Now, as noted above, if we pick one specific point, the probability that the hand will end up there is zero. However, if we pick a particular region of the square, such as a rectangle defined by a range of x-values and a range of y-values, then there may be a non-zero probability that the hand will stop within A. (Though everything I will say below also holds true for more complex shapes A, as well as for shapes in higher-dimensional probability spaces.) In particular, the density function is defined specifically so that the probability will be equal to the volume of the part of the mountain above the shape A.
In other words, if we were to take a band saw to the mountain, following the outline of A, then the volume of the piece that we cut out would be equal to the probability of the robot’s hand stopping within that region. We’ll call this volume/probability P(A).
So, not only is the value of the probability density function at a point not the probability of getting that point (since it’s always zero), the value of the density function at a point doesn’t even need to be less than one. In particular, if there is a region A with a small area but a very high (though still less than 1) probability, the values of the density function would need to be very high in order to get the appropriate volume.
If we choose region A to be the entire square then P(A) = 1 because the arm is constrained to stay within that region. So the volume of the entire mountain is 1. If we choose a smaller region A, and an even smaller region B contained in A, then we’ll get a piece with smaller volume P(B) < P(A), and thus lower probability as we would expect.
But now lets return to the original question of how to modify our density function after we’ve narrowed down the set of possible outcomes to a smaller range of x-values. This function will define a different mountain that is low and flat everywhere outside of A, but has the same elevations as the original within A, as on the left in the Figure below. We want to modify this function to give us a new probability density defining a volume function which we’ll write P(*|A) where * can be any region of the square. Since we know the robot hand landed in A, the overall probability, i.e. the volume P(A|A) of the new mountain, should be 1. However, since all we did was flatten the parts of the mountain outside the region, its volume is initially quite a bit less than 1.
In order to get the correct volume, we’ll need to scale the function up, i.e. multiply each value of the function by a constant k. The resulting function will define a mountain more like the one shown on the right above. For any region B contained in A, we’ll have P(B|A) = kP(B). Since we want P(A|A) = 1 = P(A)/P(A), the only possible value for k is 1/P(A) and we get P(B|A) = P(B)/P(A).
This is the case when the region B is contained in A. But what if it isn’t? For example, if we want to predict a range of y-values once we know the robot hand is in a certain range of x-values, then A will be a vertical strip of the square, and B will be a horizontal strip of the square, with the two intersecting in a smaller rectangle. (But as I noted above, we could just as easily let A and B be arbitrary blobs in the square or even blobs in a higher-dimensional space, but lets not get too complicated…)
So, if we want to calculate P(B|A) in this case, we need to consider two parts of B separately: The density function above the part of B that is outside of A will all get flattened to zero, so P(B|A) is completely determined by the part of B inside of A, i.e. the intersection A ∩ B. In other words, we have P(B|A) = P(A ∩ B)/P(A). Note that there’s no difference between A and B in this formulation, so we also have P(A|B) = P(A ∩ B)/P(B). We can solve both equations for P(A ∩ B) to get P(B|A)P(B) = P(A ∩ B) = P(A|B)P(A). Finally, if we divide both sides by P(B), we get Bayes’ Theorem:
P(B|A) = P(A|B)P(A)/P(B)
Of course, for the problem we started out with, the original equation P(B|A) = P(A ∩ B)/P(B) may sometimes be more useful. But in the standard setting of Bayes’ Theorem, P(A ∩ B) is the probability that both events happen (or both statements are true) so it might be harder to calculate.
For extra credit, take a minute to think about how you might calculate the probabilities of different y-values if we knew the exact value of x rather than a range. I’ll give you two hints: First, note that the probability density function over the vertical line defined by a single x-value defines a single-variable function like you might find in Calculus I and II, and there is some area (rather than a volume) below this function. Second, note that you can take A to be a small rectangular strip around the line defined by the x-value, calculate its volume, then make the strip smaller and smaller and take a limit.
But this post is already long enough, and I expect that most of my readers either don’t want to read about limits, or would rather work it out themselves. (For me, it’s both.) So I’ll leave it there. | 2,210 | 9,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.950654 |
https://www.exactlywhatistime.com/days-from-date/may-29/112-days | 1,723,669,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00630.warc.gz | 566,799,527 | 6,338 | # Thursday September 18, 2025
Adding 112 days from Thursday May 29, 2025 is Thursday September 18, 2025 which is day number 261 of 2024. This page is designed to help you the steps to count 112, but understand how to convert and add time correctly.
• Specific Date: Thursday May 29, 2025
• Days from Thursday May 29, 2025: Thursday September 18, 2025
• Day of the year: 261
• Day of the week: Thursday
• Month: September
• Year: 2024
## Calculating 112 days from Thursday May 29, 2025 by hand
Attempting to add 112 days from Thursday May 29, 2025 by hand can be quite difficult and time-consuming. A more convenient method is to use a calendar, whether it's a physical one or a digital application, to count the days from the given date. However, our days from specific date calculatoris the easiest and most efficient way to solve this problem.
If you want to modify the question on this page, you have two options: you can either change the URL in your browser's address bar or go to our days from specific date calculator to enter a new query. Keep in mind that doing these types of calculations in your head can be quite challenging, so our calculator was developed to assist you in this task and make it much simpler.
## Thursday September 18, 2025 Stats
• Day of the week: Thursday
• Month: September
• Day of the year: 261
## Counting 112 days forward from Thursday May 29, 2025
Counting forward from today, Thursday September 18, 2025 is 112 from now using our current calendar. 112 days is equivalent to:
112 days is also 2688 hours. Thursday September 18, 2025 is 71% of the year completed.
## Within 112 days there are 2688 hours, 161280 minutes, or 9676800 seconds
Thursday Thursday September 18, 2025 is the 261 day of the year. At that time, we will be 71% through 2025.
## In 112 days, the Average Person Spent...
• 24057.6 hours Sleeping
• 3198.72 hours Eating and drinking
• 5241.6 hours Household activities
• 1559.04 hours Housework
• 1720.32 hours Food preparation and cleanup
• 537.6 hours Lawn and garden care
• 9408.0 hours Working and work-related activities
• 8655.36 hours Working
• 14165.76 hours Leisure and sports
• 7687.68 hours Watching television
## Famous Sporting and Music Events on September 18
• 1922 Nurse Margaret Sanger (43) weds James Noah Henry Slee in Bloomsbury, London
• 1965 Mickey Mantle Day at Yankee Stadium: Mantle play his 2,000th game | 640 | 2,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.918981 |
https://www.crazy-numbers.com/en/27881 | 1,713,195,501,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00299.warc.gz | 677,996,229 | 3,405 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 27881: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 27881: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 27881
Is 27881 a prime number? No
Is 27881 a perfect number? No
Number of divisors 6
List of dividers
Warning: Undefined variable \$comma in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 59
1, 7, 49, 569, 3983, 27881
Sum of divisors 32490
Prime factorization 72 x 569
Prime factors
Warning: Undefined variable \$comma in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 59
7, 569
## How to write / spell 27881 in letters?
In letters, the number 27881 is written as: Twenty-seven thousand eight hundred and eighty-one. And in other languages? how does it spell?
27881 in other languages
Write 27881 in english Twenty-seven thousand eight hundred and eighty-one
Write 27881 in french Vingt-sept mille huit cent quatre-vingt-un
Write 27881 in spanish Veintisiete mil ochocientos ochenta y uno
Write 27881 in portuguese Vinte e sete mil oitocentos oitenta e um
## Decomposition of the number 27881
The number 27881 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
2 iterations of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1 | 579 | 2,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.723648 |
https://discusstest.codechef.com/t/mkwrk-editorial/13445 | 1,627,806,545,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00488.warc.gz | 218,230,947 | 4,212 | # MKWRK - Editorial
https://www.codechef.com/CDSO2016/problems/MKWRK
Medium
Sorting
# Problem:
There is a company having E employees working in town T that you work for. The employees live in N towns in that
area. Some of the employees drive P passengers. When P==1 then it means the driver can only transport themselves to work. You want to ensure that everyone will be able to make it to work and you would like to minimize the number of cars on the road.
You want to calculate the number of cars on the road, with these requirements:
• Every employee can get to town T.
• The only way an employee may travel between towns is in a car belonging to an employee.
• Employees can only take rides from other employees that live in the same town.
• The minimum number of cars is used.
Find whether it is possible for everyone to make it to work, and if it is, how many cars will end up driving to the office.
# Explanation:
From the problem statement we can interpret the following rules to find the number of cars from a given town:
• If there are no employees in a town, we don’t need any cars
• If the town is the location of the office, we don’t need any cars
• If there are employees, and this is not the location of the office, we need cars
• Employees who are drivers can only carry passengers to work from their hometown
The exact number of cars we are told should be the minimum number capable of carrying all the employees from the town. To find this number, sort the employees based on their passenger capacity and sum the capacities until you reach a number greater than or equal to the number of employees in the town. If summing all of the passenger capacities results in a smaller number than the employee headcount, then a solution is impossible. In pseudocode:
int[] employees; // The passenger capacity of each employee.
reverseSort(employees); // sort from high to low
int capacity = 0;
for (int i = 0; i < employees.length; i++) {
capacity += employees[i];
if (capacity >= employees.length)
`````` return i;
``````
}
return “IMPOSSIBLE”
// | 468 | 2,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-31 | latest | en | 0.939021 |
http://www.jiskha.com/display.cgi?id=1363191799 | 1,495,798,548,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00325.warc.gz | 675,922,186 | 3,820 | # math
posted by on .
4x-y-3z=8
2x+2y-z=3
6x+y-3z=2
help me solve matrices
• math - ,
I hope your other one didn't need matrices because..I did it the other way.
4 -1 -3 8
2 2 -1 3
6 1 -3 2
You want to manipulate the rows so that in the end, you end up with
1 0 0 a where a = the value for x
0 1 0 b where b = the value for y
0 0 1 c where c = the value for z. | 145 | 367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-22 | latest | en | 0.87242 |
http://mathhelpforum.com/statistics/216652-can-you-check-over-my-work-please-normal-distribution-z-score-word-problem-print.html | 1,529,580,382,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00179.warc.gz | 202,192,173 | 3,458 | # Can you Check Over MY Work Please - Normal Distribution (Z-SCORE) Word Problem
• Apr 4th 2013, 07:37 AM
tdotodot
Can you Check Over MY Work Please - Normal Distribution (Z-SCORE) Word Problem
The lifespan of lightbulbs in a photographic machine is normally distributed with a mean of 210 hours and a standard deviation of 50 hours.
1) Determine the z-score of a light bulb with a lifespan of exactly 124hours.
z= x-mean/standard deviation
z= (124-210)/50
x= -1.72
Z-Score = -1.72
2) What is the probability that a randomly chosen light bulb would have a lifespan of less than 180 hours?
z= x-mean/standard deviation
z= (180-210)/50
z= -0.60
p(x<180) = p(z<-0.60) = 0.2258
Probability= 0.2258 / 22.58%
3) What is the probability that a randomly chosen light bulb would have a lifespan of between 200 and 250 hours?
z= x-mean/standard deviation FOR BOTH 200 and 250:
For200:
z= x-mean/standard deviation
z= (200-210)/50
z= -0.2
For250:
z= x-mean/standard deviation
z= (250-210)/50
z= 0.8
p(200<z<250) = p(-0.2<z<0.8) = p(z<0.8) - p(z<-0.2) = 0.7881 - 0.4207 = 0.3674 **Not sure if those italic/underlined numbers are correct on this line**
Probability = 0.3674 / 36.74%
• Apr 4th 2013, 09:02 PM
chiro
Re: Can you Check Over MY Work Please - Normal Distribution (Z-SCORE) Word Problem
Hey tdotodot.
1) is correct
2) is not. Using R we get
> pnorm(-0.6,0,1)
[1] 0.2742531
3) Looks good. Again using R:
pnorm(0.8,0,1) - pnorm(-0.2,0,1)
[1] 0.3674043
• Apr 5th 2013, 06:39 AM
tdotodot
Re: Can you Check Over MY Work Please - Normal Distribution (Z-SCORE) Word Problem
Quote:
Originally Posted by chiro
Hey tdotodot.
1) is correct
2) is not. Using R we get
> pnorm(-0.6,0,1)
[1] 0.2742531
3) Looks good. Again using R:
pnorm(0.8,0,1) - pnorm(-0.2,0,1)
[1] 0.3674043
Thank you, really appreciate your help. Could you tell me how you got 0.2752531 for question #2? And what is "R"?
I am using the Z-Score chart, row 0.6 / column 0.00 shows the number 0.2258 and column 0.01 shows .2291. Other than that, looks good. :)
• Apr 5th 2013, 03:34 PM
chiro
Re: Can you Check Over MY Work Please - Normal Distribution (Z-SCORE) Word Problem
pnorm(z,0,1) calculates the probability P(Z < z) where Z ~ N(0,1). I used this command to verify your answer.
R is an open source statistical package that is becoming one of the most popular platforms used in statistics.
The R Project for Statistical Computing | 830 | 2,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-26 | latest | en | 0.807358 |
https://www.beatthegmat.com/search.php?author_id=59450&sr=posts | 1,590,858,927,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410284.51/warc/CC-MAIN-20200530165307-20200530195307-00524.warc.gz | 628,015,941 | 9,604 | ## Search found 170 matches
#### Geometry sum
###### Problem Solving
I second that. Its C due to the same reasons although I have to tell you that this is not likely in GMAT.
by yogami
Wed Aug 19, 2009 8:54 am
Forum: Problem Solving
Topic: Geometry sum
Replies: 4
Views: 879
#### verbal bursted my bubble
##### verbal bursted my bubble
Very painful - 680 (Q 50 V 31) Woke up in the morning at 10 am and simply went for a short run and a swim. Had a light breakfast and a shot of redbull and i was set for the battle at 4 pm. Went with an air of confidence feeling good I walk into the test center at 3 pm. Flirted with the test center s...
by yogami
Wed Aug 12, 2009 7:15 pm
Forum: I just Beat The GMAT!
Topic: verbal bursted my bubble
Replies: 7
Views: 1882
#### I got a 460 :(
###### I just Beat The GMAT!
Yeah dude hear ya!!. You've got one choice from now on. That is to fight. GMAT or job you have to make it on your own. Just picture yourself trapped in a warzone and you have to survive. Yuo will do whatever it takes to live. This is what it is for you. And you can do it.
by yogami
Tue Aug 11, 2009 10:25 am
Forum: I just Beat The GMAT!
Topic: I got a 460 :(
Replies: 5
Views: 2095
#### Spray
###### Sentence Correction
In B the "the" threw me off. Is that legal there?
by yogami
Mon Aug 10, 2009 8:37 pm
Forum: Sentence Correction
Topic: Spray
Replies: 6
Views: 1312
#### Tough Algebra Number Properties Question
###### Problem Solving
Hmm. Obviously x and y can be non integers since nothing has been specified. So I just tried a bunch of combinations and found that C fits because x^2 + y^2 can be written as 50 + 50 which is (7.something)^2 + (7.something)^2 and so x = 7.smthng and y is 7.smthng and their sum is 14.smthng which is ...
by yogami
Mon Aug 10, 2009 6:54 pm
Forum: Problem Solving
Topic: Tough Algebra Number Properties Question
Replies: 3
Views: 997
#### what on this planet is an addition table?
###### Problem Solving
I just wanted to squeeze through all the questions i got wrong in my last gmat prep so i thought id just post em' here. Yeah this was more of a puzzle then a math questions and i got lost in creating simultaneous equations instead. I have seen gmat prep putting a strong emphasis on logic in math too...
by yogami
Mon Aug 10, 2009 6:30 pm
Forum: Problem Solving
Topic: what on this planet is an addition table?
Replies: 3
Views: 1132
#### gmat prep sc 5
##### gmat prep sc 5
And I thought but also cannot exist without not only? So how come GMAT Prep's OA looks weird here??
by yogami
Mon Aug 10, 2009 6:24 pm
Forum: Sentence Correction
Topic: gmat prep sc 5
Replies: 3
Views: 781
#### gmat prep sc 3
##### gmat prep sc 3
I am very uncomfortable with all the participle past/c ontinuous/present saga. Can someone throw some light on this one? Tks guys!
by yogami
Mon Aug 10, 2009 6:18 pm
Forum: Sentence Correction
Topic: gmat prep sc 3
Replies: 3
Views: 894
#### gmat prep sc 2
##### gmat prep sc 2
what is wrong with lower than? Why is less than better there?
by yogami
Mon Aug 10, 2009 6:12 pm
Forum: Sentence Correction
Topic: gmat prep sc 2
Replies: 2
Views: 905
#### gmat prep sc 1
##### gmat prep sc 1
OA l8r after some discussion
by yogami
Mon Aug 10, 2009 6:09 pm
Forum: Sentence Correction
Topic: gmat prep sc 1
Replies: 5
Views: 1506
#### what on this planet is an addition table?
##### what on this planet is an addition table?
What is an additional table? This is the first time I am posting an attachment so pls pardon me. This is gmat prep question and i had to guess this one.
by yogami
Mon Aug 10, 2009 6:02 pm
Forum: Problem Solving
Topic: what on this planet is an addition table?
Replies: 3
Views: 1132
#### Well that did not go according to plan
###### I just Beat The GMAT!
Dude if you can get me a V score like that I will personally coach you in Q
Your Q can just go up from where it is. Q can be easily developed compared to V so you will be in a good shape next time for it
by yogami
Mon Aug 10, 2009 8:18 am
Forum: I just Beat The GMAT!
Topic: Well that did not go according to plan
Replies: 3
Views: 1288
#### day after tommorrow
##### day after tommorrow
680,690,700,700,750,720,710 with 46 - 51 in Quant and 35 - 42 in verbal The last two scores are gmat prep 1 and gmat prep 2 which i took with AWAs just today finishing up my journey of CATs for this Gmat attempt. I realised GMAT eventually comes down to your mental clarity and calm disposition more ...
by yogami
Sun Aug 09, 2009 7:00 pm
Forum: GMAT Strategy
Topic: day after tommorrow
Replies: 2
Views: 803
#### 780+ CR's
###### Critical Reasoning
To be honest none of the options MUST be true. But B has fewer variables compared to the rest so I would go with B.
by yogami
Fri Aug 07, 2009 11:39 am
Forum: Critical Reasoning
Topic: 780+ CR's
Replies: 160
Views: 29041
#### Line!!
###### Data Sufficiency
imo it is C The data tells us that both lines have same slopes and gives y intercept of a (1) tells us x intercept of a. So we know two points of line a. So we can plot line a. However line b can be anywhere, although parallel to a!!. So insuff to find out anything about line b (2) tells us the poin...
by yogami
Fri Aug 07, 2009 10:03 am
Forum: Data Sufficiency
Topic: Line!!
Replies: 5
Views: 1684 | 1,577 | 5,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.938897 |
https://news.ycombinator.com/item?id=23514079 | 1,620,375,691,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00259.warc.gz | 424,309,533 | 4,314 | Got it! I'm working to add a "Show solution" button, but I just published another hint for that question. Specifically, the syntax is the same as for summations "m.sum()" but with "prod" instead. Additionally, you need to to multiply each element of "m" by 3 and use parentheses appropriately — (3*m).
I also got stuck here with no way forward. (until reading here)The explanatory text "Note that the matrix first element-wise multiplied by 3." is very unclear.Also, looking at it from a Python novice, it is not clear we were supposed to know that the .prod() operator can act on something that is not purely a named variable, but can work on an already manipulated quantity within parentheses.Also, and this is not your fault but of Python's, but it is not clear (or merits some explanation) why some operators take the form "operator(x)" while others behave like "x.operator" or even "x.operator()" with no argument.
Thanks for the feedback. I think I was being overly clever by trying to make the product example more complicated than just `m.prod()` similar to the `m.sum()` example, but as you mentioned I was fitting two concepts in one question — operating on an already manipulated variable in addition to the new operator.I've simplified question 9 to just `m.prod()` from the previous solution of `(m * 3).prod()` for now.Re: Methods vs. functions — I completely agree! I actually originally implemented Math to Code using Tensorflow.js (which was simpler than using a Python interpreter) where the syntax was:`````` m.pow(2).sum().sqrt() `````` Vs.`````` np.sqrt((m ** 2).sum()) `````` I decided to go with NumPy + Python because I felt it would be more immediately applicable vs. just learning the Tensorflow.js syntax. Also Python allows operator overloading so you can just do `2 * m` vs. `m.mul(2)`.PyTorch has a similarly clean syntax to Tensorflow.js and includes operator overloading, but there doesn't seem to be a PyTorch shim for Skulpt, but that would be a fun project.
also, there are functions which need to be prefixed with np - the module name, but this isn't described. So for someone who isn't quite familiar with python, this doesn't make sense.
Yeah, I had the same issue as the fellow above. I think it would be helpful if the ui displayed the result of the calculation. Debugging the std-dev example was annoying. I ended up opening a python shell so that I could see results of intermediate steps.Also, my difficulty with numpy isn't applying functions to arrays, it's seeing __repr__ of a multi-dimensional arrays and grokking what it's showing me.Site worked well on both my iphone and pc btw.
Agreed. Particularly because NumPy has a mix of functions and methods where you need to balance parentheses.I'd like to more clearly show the answers / I originally didn't show what the test inputs (e.g. if square root is tested with 25, 9, and 4) so you couldn't just return 5 if 25, 3 if 9, etc... but I could display some of the test inputs and not all of them.It definitely would be great to show the intermediate results of the calculations and for things like matrix multiplication `a.dot(b)` isn't very complicated, but it doesn't give you intuition on what's happening under the hood.Thanks for the iPhone feedback — I tried my best to make sure it worked well on mobile. The few examples I could find of interactive Python tutorials usually involved spinning up a VM with multi-second lag for each question and textareas which were not at all mobile-optimized so I wanted to try and do something better :-).
Search: | 802 | 3,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.953792 |
https://tutorpro.us/what-was-the-correlation-between-the-ces-d-scores-in-the-two-waves-of-data-collection/ | 1,718,344,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00828.warc.gz | 521,210,560 | 14,218 | # What was the correlation between the CES-D scores in the two waves of data collection?
Using output from one of the previous four exercises (B3 through B6), create a table to summarize key results of the analyses. Then write a paragraph summarizing the findings.
Exercise b4
In this exercise, run a stepwise multiple regression to predict depression scores, using the same predictors as in Exercise B3. In the first SPSS Linear Regression dialog box, enter cesd as the Dependent variable and the list of predictors in the Independent slot, as in the previous exercise. For Method, select “Stepwise.” For the Statistics options, you can omit Descriptives and Collinearity Diagnostics because you have already examined these, but this time you should select R squared change. Run the analysis and then answer the following questions: (a) How many predictors were entered before the regression stopped? (b) Which predictor variables made it into the regression—and which did not? Was the order of entry of predictors consistent with the values of the zero-order correlations? (c) Looking at the Model Summary panel, what was the progression of the value of R2 from one step to the next—and were these changes significant? What happens to the standard error of the estimate with each progressive step? (d) Were all models statistically significant—that is, was the value of R2 greater than zero at each step? (e) Looking at the “Excluded Variables” panel—and focusing on Model 4, were any of the remaining predictors statistically significant?
Exercise B3
In this exercise, you will run a simultaneous multiple regression analysis to predict the women’s level of depression (scores on the CESD depression scale, cesd) based on several demographic characteristics, socioeconomic characteristics, health status, and self-reported incidence of abuse in the prior year. The list of predictors is as follows: the two race/ethnicity dummy variables you created in exercise B2; age; educatn (educational attainment); worknow (a dummy variable for currently employed); nabuse (number of different types of abuse experienced in the past year, including verbal abuse, efforts to control, threats of harm, and physical abuse); and poorhlth (a dummy coded variable indicating self-reported poor health at the time of the interview. Bring up the regression dialog box by selecting Analyze ➞ Regression ➞ Linear. Insert the variable cesd in the box labeled Dependent. Insert the 7 predictor variables that we just mentioned into the box for Independent(s). Make sure that Method is set to “Enter,” the command for entering all predictors simultaneously. Click the Statistics pushbutton and then select the following options: Estimates (under “Regression Coefficients”); Model Fit; Descriptives; and Collinearity Diagnostics. Then click Continue, and OK to run the analysis. Answer the following questions: (a) How large is the sample on which the regression analysis was run? (b) Interpret the mean value for poor health self-rating. (c) Which predictor has the highest zero-order correlation with cesd? (d) What were the values of R2 and adjusted R2 ? (e) Which predictors in the analysis were significantly predictive of the women’s depression scores, once other predictors were included? Which were not significantly predictive? (f) For this sample of women, which predictor variable appeared to be the most powerful in predicting depression? (g) Did any of the tolerance levels suggest a problem with multicollinearity?
Exercise B2
Now you can create the new dummy-coded variables. Select Transform ➞ Recode ➞ Into Different Variables. Find racethn in the variable list and move it into the slot for “Numeric Variable Output Variable.” On the right, under Output Variable, type in a name for your first new variable (e.g., black, afamer). In the slot for Label, you can type in a longer label, such as “African American” if you so choose. Then click Change, which will confirm the new variable as the output variable. Next, click the “Old and New Values” button, which will bring up a new dialog box. Under “Old Value,” enter 1, which is the code for African Americans in the original (“old”) variable racethn. Then, on the right under New Value, click “Copy old value,” and then click the Add button. Women who were coded 1 on racethn will also be coded 1 on the new African-American variable. Next, under Old Value, click Range and enter 2 and then “through” 3 (the two codes for Hispanic and White/other women on racethn). On the right under New Value, enter 0, then click Add, which will code all non–African-American women as 0. Finally, under Old Value, click “System- or user-missing,” and then under New Value click “System missing,” then click Add. Women who have missing data for racethn will now have a missing values code for the new variable. Click Continue to go back to the original dialog box, then click OK to run the command. If you look at the last variable in your file in Data View, you should see the new variable with values of .00 and 1.00 (which can be changed to 0 and 1 by going into Variable View and changing the number of decimals to 0). Do the analogous procedure for the next new race/ethnic group— except remember to change the values coded 1 and 0. Now, run frequencies on your new variables, and compare the results to those from Exercise B1, making sure that your new variables accurately reflect the original racial/ethnic distribution.
Exercise B1
For these exercises, you will be using the SPSS dataset Polit2SetC to do multiple regression analyses to predict level of depression in the sample of low-income urban women. You will need to begin by dummy coding the variable race/ethnicity. First, create a frequency distribution for racethn (Analyze ➞ Descriptive Statistics ➞ Frequencies), then answer these questions: (a) What percentage of women in this sample was African American, Hispanic, and White or other? (b) Are there any women whose information for racethn is missing? (c) How many new variables need to be created so that race/ethnicity can be used in regression analyses? (d) Which category do you think should be omitted?
Exercise b6
. It is tempting to think of the results obtained in the previous three analyses as suggesting a causal link between the women’s abuse experiences and their level of depression— that is, inferring that being abused caused higher levels of depression. However, the opposite might be the case. For example, women who are depressed, lethargic, or absorbed with personal problems might incite others to yell at them, threaten them, or hit them. In the next analysis, we explore the issue of direction of influence, though we caution against firm causal conclusions. In this hierarchical regression, we will statistically control the women’s level of depression 2 years earlier and then see if recent abuse experiences affected current depression, with earlier depression held constant. This is analogous to asking whether recent abuse was related to changes in depression. In the first SPSS Linear Regression dialog box, enter cesd as the Dependent variable, and cesdwav1 as the predictor in the Independent slot. Then, click the pushbutton Next in the area labeled “Block.” Now enter nabuse (number of types of abuse) as the predictor in the second block. The Method box should say “Enter.” For statistical options select Estimates for the Regression Coefficients, Model Fit, and R squared change. Then run the analysis and answer these questions based on the output: (a) What was the correlation between the CES-D scores in the two waves of data collection? (b) Was R2 statistically significant at both steps of the analysis? (c) What was the change to R2 when abuse was added to the regression? Was this significant—and, if so, what does this suggest? (d) If you wanted to predict a woman’s current CES-D score based on this analysis, what would the unstandardized regression equation be?
Pages (550 words)
Approximate price: -
High Quality Papers
We always make sure that writers follow all your instructions precisely. You can choose your academic level or professional level, and we will assign a writer who has a respective degree.
Experienced Writers
We have a team of professional writers with experience in academic and business writing. Many are native speakers and able to perform any task for which you need help.
Free Revisions
If you think we missed something, send your order for a free revision. You have 10 days to submit the order for review after you have received the final document.
Timely Delivery
All papers are always delivered on time. In case we need more time to master your paper, we may contact you regarding the deadline extension.A 100% refund is guaranteed.
100% Confidential
We use several writing tools checks to ensure that all documents you receive are free from plagiarism. Our editors carefully review all quotations in the text.
Tutorpro support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.
Try it now!
## Calculate the price of your order
Total price:
\$0.00
How it works?
Fill in the order form and provide all details of your assignment.
Proceed with the payment
Choose the payment system that suits you most.
Tutorpro Homework Help Services
At Tutorpro, we have top rated masters and PhD writers who will help you tacke that homework and score A+ grade. Tutorpro services covers all levels of education : high school, college, university undergraduate, masters and PhD academic level.
## Essay Writing
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 2,087 | 9,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.931465 |
https://www.mbatious.com/topic/919/cat-question-bank-quant-gaurav-sharma-set-2/19 | 1,555,892,277,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532948.2/warc/CC-MAIN-20190421235818-20190422021818-00388.warc.gz | 742,655,583 | 18,048 | # CAT Question Bank (Quant) - Gaurav Sharma - Set 2
• Q9) If P Q R are three points on a circle, What is the maximum area of the triangle formed by them where radius of circle is 'a' units?
• Q10) Find the no of positive divisor of N^2 such that the positive divisor are less than N and do not divide N completely. where N = 2^17 * 3^9 * 5^3.
• Q11) The highest power of 2 in 10! + 11! + 12! + 13! + ...+ 1000! is
(a) 8
(b) 9
(c) 10
(d) 11
• Q12) Two points are chosen randomly on the circumference of a circle. What is the probability that the distance between the 2 points is at least 'r', the radius of the circle?
• Q13) How many 10-digit positive integers with distinct digits are multiples of 11111?
a) 1234
b) 2345
c) 3456
d) 4567
e) None of these
• Q14) For how many natural numbers less than 10^5 the sum of their digits equal to 10?
• Q15) f is a real function such that f(x + y) = f(xy) for all real values of x and y. If f(– 5) = 5, then the value of f(– 25) + f(25) is
a. 5
b. 10
c. 0
d. 25
• Q16)
• Q17)
• Q18) The sum of 2 five digit numbers AMC10 and AMC12 is 123422. What is A + M + C ?
a) 10
b) 11
c) 12
d) 13
e) 14
• Q19) A positive integer n has exactly 4 positive divisors that are perfect fifth powers, exactly 6 positive divisors that are perfect cubes, and exactly 12 positive divisors that are perfect squares. Find the least possible number of possible integers that are divisors of n
• Q20) Find the remainder when 1 × 2 + 2 × 3 + 3 × 4 + … + 98 × 99 + 99 ×100 is divided by 101.
• Q21) What are the last two digits of (86789)^41?
• Q22) All the divisors of 360, including 1 and the number itself, are summed up. The sum is 1170. What is the sum of the reciprocals of all the divisors of 360?
a. 3.25
b. 2.75
c. 2.5
d. 1.75
• Q23) Last 2 digits of the number (299)^33
• Q24) Find the remainder when 1^39 + 2^39 + 3^39 + 4^39 + ... + 12^39 is divided by 39.
a. 0
b. 1
c. 12
d. 38
• Q25) Last digit of the LCM of(3^2003-1) and (3^2003+1) :
a. 8
b. 2
c. 4
d. 6
• Q26) What is the remainder when x + x^9 + x^25 + x^49 + x^81 is divided by (x^3 - x)?
• Q27) A number when divided by 8 leaves remainder 3 and quotient Q. The number when divided by 5 leaves remainder 2 and quotient Q + 8. What is the number?
• Q28)
Looks like your connection to MBAtious was lost, please wait while we try to reconnect. | 836 | 2,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-18 | latest | en | 0.78945 |
https://oeis.org/A000240/internal | 1,590,842,667,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409171.27/warc/CC-MAIN-20200530102741-20200530132741-00374.warc.gz | 471,498,273 | 5,460 | 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!)
A000240 Rencontres numbers: number of permutations of [n] with exactly one fixed point. (Formerly M2763 N1111) 36
%I M2763 N1111
%S 1,0,3,8,45,264,1855,14832,133497,1334960,14684571,176214840,
%T 2290792933,32071101048,481066515735,7697064251744,130850092279665,
%U 2355301661033952,44750731559645107,895014631192902120,18795307255050944541,413496759611120779880
%N Rencontres numbers: number of permutations of [n] with exactly one fixed point.
%C a(n) is also the number of permutations of [n] having no circular succession. A circular succession in a permutation p of [n] is either a pair p(i), p(i+1), where p(i+1)=p(i)+1 or the pair p(n), p(1) if p(1)=p(n)+1. a(4)=8 because we have 1324, 1432, 4132, 2143, 2413, 3214, 3241, and 4321. - _Emeric Deutsch_, Sep 06 2010
%C a(n) is also the number of permutations of [n] having no substring in {12, 23, ..., (n-1)n, n1}. For example, a(4) = 8 since we have 1324, 1432, 4213, 2143, 2431, 3214, 3142, 4321 (different from permutations having no circular succession). - _Enrique Navarrete_, Oct 07 2016
%C a(n-1) is also the number of permutations of [n] that allow the substring n1 in the set of permutations of [n] having no substring in {12, 23, ..., (n-1)n}. For example, for n=5 the 8 permutations in S5 having no substring in {12,23,34,45} that allow the substring 51 are {51324,51432,25143,24351,35142,32514,42513,43251} (see link). - _Enrique Navarrete_, Jan 11 2017
%C From _Enrique Navarrete_, Mar 25 2017: (Start)
%C Let D(n,k) be the set of permutations on [n] that for fixed k, 0 < k < n, avoid substrings j(j+k) for 1 <= j <= n - k, and avoid substrings j(j+k) (mod n) for n-k < j <= n. Then the number of permutations in D(n,k) with k relative prime to n, n>=2, is given by a(n). For example, the forbidden substrings in D(4,3) are {14;21,32,43} (the forbidden substrings (mod 4) are written after the semicolon and lie below the diagonal in the chessboard below):
%C 1 2 3 4
%C 1 |_|_|_|x|
%C 2 |x|_|_|_|
%C 3 |_|x|_|_|
%C 4 |_|_|x|_|
%C _
%C Then since 4 and 3 are relatively prime, a(4)=8, and the permutations in D(4,3) are 1234, 1342, 2341, 2413, 3124, 3412, 4123, 4231.
%C For another example, the forbidden substrings in D(8,5) are {16,27,38;41, 52,63,74,85} and the number of permutations in D(8,5) is a(8)=14832 (see the "K-Shift Forbidden Substrings" link).
%C (End)
%D Kaufmann, Arnold. "Introduction à la combinatorique en vue des applications." Dunod, Paris, 1968. See p. 92.
%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A000240/b000240.txt">Table of n, a(n) for n=1..100</a>
%H Bhadrachalam Chitturi and Krishnaveni K S, <a href="https://arxiv.org/abs/1601.04469">Adjacencies in Permutations</a>, arXiv preprint arXiv:1601.04469 [cs.DM], 2016.
%H S. K. Das and N. Deo, <a href="http://www.fq.math.ca/Scanned/25-3/das.pdf">Rencontres graphs: a family of bipartite graphs</a>, Fib. Quart., Vol. 25, No. 3, August 1987, 250-262.
%H FindStat - Combinatorial Statistic Finder, <a href="http://www.findstat.org/St000022">The number of fixed points of a permutation</a>
%H I. Kaplansky, <a href="http://dx.doi.org/10.1090/S0002-9904-1944-08261-X">Symbolic solution of certain problems in permutations</a>, Bull. Amer. Math. Soc., 50 (1944), 906-914.
%H Enrique Navarrete, <a href="https://arxiv.org/abs/1610.01987">Forbidden Patterns and the Alternating Derangement Sequence</a>, arXiv:1610.01987 [math.CO], 2016.
%H Enrique Navarrete, <a href="http://arxiv.org/abs/1610.06217">Generalized K-Shift Forbidden Substrings in Permutations</a>, arXiv:1610.06217 [math.CO], 2016.
%H S. M. Tanny, <a href="http://dx.doi.org/10.1016/0097-3165(76)90063-7">Permutations and successions</a>, J. Combinatorial Theory, Series A, 21 (1976), 196-202.
%F E.g.f.: x*exp(-x)/(1-x). [Corrected by _Vaclav Kotesovec_, Sep 26 2012]
%F a(n) = Sum_{k=0..n-1} (-1)^k*n!/k!.
%F a(n) = A180188(n,0). - _Emeric Deutsch_, Sep 06 2010
%F E.g.f.: x*A(x) where A(x) is the e.g.f. for A000166. - _Geoffrey Critzer_, Jan 14 2012
%F a(n) = n*a(n-1) - (-1)^n*n = A000166(n) - (-1)^n = n*A000166(n-1) = A000387(n+1)*2/(n+1) = A000449(n+2)*6/((n+1)*(n+2)).
%F a(n) = n*floor(((n-1)!+1)/e), n > 1. - _Gary Detlefs_, Jul 13 2010
%F lim_{n->infinity} n!/a(n) = e = 2.71828...
%F a(n) = (n-1)*(a(n-1)+a(n-2)) + (-1)^(n-1), n>=2. - _Enrique Navarrete_, Oct 07 2016
%F O.g.f.: Sum_{k>=1} k!*x^k/(1 + x)^(k+1). - _Ilya Gutkovskiy_, Apr 13 2017
%F a(n) = (-1)^(n-1)*n*hypergeom([1,1-n], [], 1). - _Peter Luschny_, May 09 2017
%e a(3) = 3 because the permutations of {1,2,3} with exactly one fixed point are the transpositions (1 2), (1 3) and (2 3).
%e a(4) = 8 because for each element x of {1,2,3,4} there are exactly two permutations which leave only x invariant, namely the two circular permutations of the three remaining numbers, one being the inverse (and the square) of the other. - _M. F. Hasler_, Jan 16 2017
%p G(x):=exp(-x)/(1-x)*x: f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=1..22);# _Zerinvary Lajos_, Apr 03 2009
%p A000240 := proc(n)
%p end proc: # _R. J. Mathar_, Jul 09 2012
%p a := n -> (-1)^(n-1)*n*hypergeom([1,1-n], [], 1):
%p seq(simplify(a(n)), n=1..22); # _Peter Luschny_, May 09 2017
%t Table[Subfactorial[n]-(-1)^n, {n, 1, 25}] (* _Zerinvary Lajos_, Jul 10 2009, updated for offset 1 by _Jean-François Alcover_, Jan 10 2014 *)
%t Table[n!*Sum[(-1)^k/k!,{k,0,n-1}],{n,1,25}] (* _Vaclav Kotesovec_, Sep 26 2012 *)
%t Table[n!*SeriesCoefficient[x*E^(-x)/(1-x),{x,0,n}],{n,1,25}] (* _Vaclav Kotesovec_, Sep 26 2012 *)
%o (Python)
%o a = 0
%o for i in range(1, 51):
%o a = (a - (-1)**i)*i
%o print(a, end=',') # _Alex Ratushnyak_, Apr 20 2012
%o (PARI) x='x+O('x^66); Vec( serlaplace(x*exp(-x)/(1-x)) ) \\ _Joerg Arndt_, Feb 19 2014
%o (PARI) a(n,p=vector(n,i,i),s=x->!x)=sum(k=1,n!,#select(s,numtoperm(n,k)-p)==1) \\ For illustrative purpose. #select(...) is almost twice as fast as {p=numtoperm(n,k);sum(i=1,n,p[i]==i)}. - _M. F. Hasler_, Jan 16 2017
%Y Cf. A008290, A000166, A000387, A000449, A000475, A129135, etc.
%Y A diagonal of A008291.
%Y Cf. A180188, A170942.
%K nonn,easy,nice
%O 1,3
%A _N. J. A. Sloane_, _Simon Plouffe_
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 30 08:43 EDT 2020. Contains 334712 sequences. (Running on oeis4.) | 2,661 | 6,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-24 | latest | en | 0.783237 |
https://hpmuseum.org/forum/printthread.php?tid=9887 | 1,638,771,110,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00263.warc.gz | 359,412,023 | 4,559 | Triangular number AND sum of first m factorials - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: Not HP Calculators (/forum-7.html) +--- Forum: Not remotely HP Calculators (/forum-9.html) +--- Thread: Triangular number AND sum of first m factorials (/thread-9887.html) Triangular number AND sum of first m factorials - Joe Horn - 01-09-2018 04:31 PM 153 (my favorite number) is both a triangular number (the sum of the integers 1 through $$n$$; in this case $$n=17$$) as well as the sum of the factorials $$1!$$ through $$m!$$ (in this case $$m=5$$). The first three natural numbers which have both of those properties are 1, 3 (both trivial) and 153. Find the next number in this sequence. For extra credit, find the mathematical relationship between $$n$$ and $$m$$ for all members of this sequence (which apparently is not yet in OEIS). RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-09-2018 08:53 PM 001 LBL A 002 DEC X 003 ENTER 004 INC X 005 LBL 00 006 INC X 007 RCL* Y 008 DSE Y 009 GTO 00 010 PSE 10 011 STO+ X 012 +/- 013 1 014 ENTER 015 ENTER 016 R/\ 017 SLVQ 018 X<>Y 019 FP 020 PSE 10 021 RCL L 022 RTN 2 A -> 3 -> 0 -> 2 5 A -> 153 -> 0 -> 17 The next m should be greater than 31, but I can’t find it using the wp34s, at least not with help of this simple program… Edited to fix a typo pointed out by Dieter below. RE: Triangular number AND sum of first m factorials - Valentin Albillo - 01-09-2018 10:16 PM (01-09-2018 04:31 PM)Joe Horn Wrote: 153 (my favorite number) is both a triangular number (the sum of the integers 1 through $$n$$; in this case $$n=17$$) as well as the sum of the factorials $$1!$$ through $$m!$$ (in this case $$m=5$$) It's also a narcissistic number: 153 = 1^3 + 5^3 + 3^3 V. . RE: Triangular number AND sum of first m factorials - Dieter - 01-09-2018 10:19 PM (01-09-2018 08:53 PM)Gerson W. Barbosa Wrote: 001 LBL A 002 DEC X 003 ENTER 004 INC C ?!? – increment register C ? Is this possibly supposed to mean INC X ? (01-09-2018 08:53 PM)Gerson W. Barbosa Wrote: 005 LBL 00 006 INX X Ah, here is the missing X from line 004, and the "C" from there goes here. ;-) Dieter RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-09-2018 11:00 PM (01-09-2018 10:19 PM)Dieter Wrote: Ah, here is the missing X from line 004, and the "C" from there goes here. ;-) :-) Duly swapped. Thanks! Gerson. RE: Triangular number AND sum of first m factorials - John Keith - 01-09-2018 11:10 PM A quick test of sums of factorials up to 69! finds no triangle numbers larger than 153. If one does exist, it has to have well over 100 digits. I may try later on the emulator. John RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-10-2018 04:03 AM Tn&Sf: « { } 3 ROT FOR n n Sfac 8 * 1 + ISPF? { √ 1 - 2 / + n I→R + } { DROP } IFTE NEXT » Sfac: « DUP 1 - 2 FOR m 1 + m * -1 STEP 1 + » ISPF? « 1 » 7 Tn&Sf -> { '(√73-1)/2' 3. '(√265-1)/2' 4. 17 5. '(√6985-1)/2' 6. '(√47305-1)/2' 7. } 49G or 50g in exact mode ISPF? (IsPerfectSquare?) has yet to be implemented. It should return 1 when the argument is a perfect square and 0 otherwise. Then the output would be a list of n and m pairs, separated by dots. Just in case someone wants to try. RE: Triangular number AND sum of first m factorials - Joe Horn - 01-10-2018 04:58 AM (01-10-2018 04:03 AM)Gerson W. Barbosa Wrote: ISPF? (IsPerfectSquare?) has yet to be implemented. It should return 1 when the argument is a perfect square and 0 otherwise. The HP 50g LongFloat library contains a function like the one you're looking for. It's called ZSqrt, and it returns IP(sqrt(x)) to level 2, and a 0 or 1 to level 1, with 1 meaning that x was a perfect square. It works on integers of any length. Gerald H also posted a program HERE which seems to do essentially the same thing. It returns IP(sqrt(x)) to level 2 of the stack, and on level 1 it leaves a SysRPL TRUE if x was a perfect square, otherwise a FALSE. That's the same idea as HP's FPTR2 ^ZSQRT, but Gerald's program is more accurate; see Gerald's posting for evidence of HP's function's inaccuracy. BTW, LongFloat's ZSqrt function gets the same result as Gerald's program when given the example in Gerald's posting, so I surmise that ZSqrt is trustworthy. RE: Triangular number AND sum of first m factorials - Paul Dale - 01-10-2018 06:35 AM I've got a proof that there are only three such numbers. Consider the last pair of digits in $$\sum_1^n i!$$, from n=9 onwards these never change because subsequent factorial terms will always have a factor of 100 present. These digits are '13'. Note that n is triangular iff 8n+1 is a perfect square. For the sum of factorials to be triangular, the last two digits must therefore be '05'. Checking all possibilities shows that there are no square numbers that end '05'. Thus, numbers of the desired form must have n < 9. Checking all cases reveals that only 1, 3 and 153 have the desired properties. Pauli RE: Triangular number AND sum of first m factorials - Joe Horn - 01-11-2018 03:01 AM Paul: Now THAT is beautiful! Thank you! I can stop my futile hunt now. RE: Triangular number AND sum of first m factorials - Paul Dale - 01-11-2018 10:21 AM Thanks Don't let my proof stop your hunt, you'll be able to wile away many hours looking... I hadn't realised that all square numbers that end in '5' actually end in '25'. I must have seen this before but never noticed or remembered it. Pauli RE: Triangular number AND sum of first m factorials - John Cadick - 01-11-2018 02:22 PM A really great thread Joe et. al. Thank you all. John RE: Triangular number AND sum of first m factorials - Gerson W. Barbosa - 01-11-2018 06:29 PM (01-11-2018 10:21 AM)Paul Dale Wrote: Don't let my proof stop your hunt, you'll be able to wile away many hours looking... At least I can do it a little more efficiently now :-) 100 « { } SWAP 0 1 ROT 1 SWAP FOR m m * SWAP OVER + ROT OVER 8 * 1 + ZSqrt { 1 - 2 / + m I→R + } { DROP } IFTE SWAP ROT NEXT DROP2 » EVAL --> { 1 1. 2 2. 17 5. } (about 17 seconds on the real 50g) ZSqrt from the LongFloat Library Yes, that's a consequence of the ever growing number of trailing zeros in factorials and the properties of perfect squares. Gerson. RE: Triangular number AND sum of first m factorials - John Keith - 01-11-2018 10:30 PM (01-11-2018 10:21 AM)Paul Dale Wrote: Thanks Don't let my proof stop your hunt, you'll be able to wile away many hours looking... I hadn't realised that all square numbers that end in '5' actually end in '25'. I must have seen this before but never noticed or remembered it. Pauli I figured it was hopeless since I tested all sums of factorials up to 1000 (over 2500 digits) with no triangle numbers found. Took almost 20 min. on the emulator. An interesting and educational thread indeed! John RE: Triangular number AND sum of first m factorials - John Keith - 01-11-2018 10:43 PM (01-11-2018 06:29 PM)Gerson W. Barbosa Wrote: Yes, that's a consequence of the ever growing number of trailing zeros in factorials and the properties of perfect squares. Gerson. Note, however, that due to the first four numbers, all of the sums of factorials end in 3. | 2,246 | 7,228 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-49 | latest | en | 0.784031 |
https://historyofprivacy.net/dolly-parton/ | 1,680,316,159,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00043.warc.gz | 346,795,395 | 13,906 | # Dolly Parton
Posted on
If you are looking for the answer of Dolly Parton, you’ve got the right page. We have approximately 10 FAQ regarding Dolly Parton. Read it below.
## Dolly has 12 stamps. Mila has twice as many stamps.
Ask: Dolly has 12 stamps. Mila has twice as many stamps. if dolly friend gives 12 more, how many stamps will dolly and Mila have ?
they will have 48 stamps
Step-by-step explanation:
## How did Dolly compare to sheep that were cloned with
Ask: How did Dolly compare to sheep that were cloned with the same DNA?
a. Dolly lived longer
b. Dolly was less healthy
d. Dolly had more genetic material
D
Explanation:
thanks me late 12;\$9\$;20#!*:”0#2;*+!220*!*;\$
## what is saint thomas parton saint of?
Ask: what is saint thomas parton saint of?
Explanation:
He is often reagrded as the patron saint of India, and the name Thomas remains quite popular among Saint Thomas Christians of India.
## S+6=52A.)Dolly is 6 years younger than Alex.The sum of their
A.)Dolly is 6 years younger than Alex.The sum of their ages is 5.how old is dolly?
B.)Dolly is 6 yrs old as Alex. The sum of their ages is 52.how old is dolly?
C.)Dolly is 6 yrs older than
Alex. The difference of their is 52.how old is dolly?
D.)Dolly is 6 yrs older Than Alex. The sum of their ages is 52.how old is dolly?
D. 52
Step-by-step explanation:
A. -1
B. 46
C. 58
D. 52
probably
10) S +6 52
a) Dolly is 6 years younger than Alex. The sum of their ages is 5. How old is
b) Dolly is 6 years old as Alex. The sum of their ages is 52. How old is
Dolly?
c) Dolly is 6 years older than Alex. The difference of their ages is 52. How old
is Dolly?
d) Dolly is 6 years older than Alex. The sum of their ages is 52. How old is
Dolly?
c po tamang sagot
Step-by-step explanation:
i hope this is help
a) 23 + 29 = 52
b) 26
c) 29
dikosure?
## It is a festival which involves different dance ritual in
Ask: It is a festival which involves different dance ritual in honor of the three parton Saint
obando Festival
Explanation:
This festival involves the different dance rituals held for three consecutive days honoring three patron saints: May 17 for St. Paschal (for couples who want a male child), May 18 for St. Claire (for couples who want a female child) and May 19 for the Our Lady of Salambaw (patroness of fishermen).
## frames 29 to 31 are dolly shots? what do you
Ask: frames 29 to 31 are dolly shots? what do you call the dolly shot that moves farther subject?
Yes its Dolly Shot,
and the dolly shot that moves farther from subject is dolly in or dolly out shot.
## what is the meaning of dolly?
Ask: what is the meaning of dolly?
1. A child’s word for a doll.
2. A small platform on wheels used for holding heavy objects, typically film or television cameras.
Explanation:
Hope it help.
## how to picture to dolly
Ask: how to picture to dolly
si tatay jomari kailangan mo
Explanation:
wala na basta malaki batuta
Download stock pictures of Dolly on Depositphotos ✓ Photo stock for commercial use – millions of high-quality, royalty-free photos & images.
Explanation:
sana makatulong:>
## Identification: _____She was an American recording artist. She was best
Ask: Identification: _____She was an American recording artist. She was best known for her hit singles including “How Will I Know”, “I Wanna Dance With Somebody (Who Loves Me)” and her cover of Dolly Parton’s “I Will Always Love You”. | 951 | 3,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.967053 |
https://www.lmfdb.org/ModularForm/GL2/TotallyReal/3.3.1940.1/holomorphic/3.3.1940.1-17.1-b | 1,632,182,351,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00117.warc.gz | 875,589,636 | 5,493 | # Properties
Label 3.3.1940.1-17.1-b Base field 3.3.1940.1 Weight $[2, 2, 2]$ Level norm $17$ Level $[17, 17, -2w^{2} + 15]$ Dimension $6$ CM no Base change no
# Related objects
• L-function not available
## Base field 3.3.1940.1
Generator $$w$$, with minimal polynomial $$x^{3} - 8x - 2$$; narrow class number $$2$$ and class number $$1$$.
## Form
Weight: $[2, 2, 2]$ Level: $[17, 17, -2w^{2} + 15]$ Dimension: $6$ CM: no Base change: no Newspace dimension: $80$
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{6} - 10x^{4} + 9x^{2} - 2$$
Norm Prime Eigenvalue
2 $[2, 2, -w]$ $\phantom{-}1$
3 $[3, 3, w^{2} - 7]$ $\phantom{-}e$
5 $[5, 5, w + 1]$ $-4e^{5} + 38e^{3} - 17e$
5 $[5, 5, -w - 3]$ $-2e^{4} + 19e^{2} - 8$
9 $[9, 3, w^{2} - 2w - 1]$ $-2e^{5} + 19e^{3} - 9e$
17 $[17, 17, -2w^{2} + 15]$ $-1$
17 $[17, 17, 3w + 1]$ $\phantom{-}e^{3} - 9e$
17 $[17, 17, -w^{2} + w + 5]$ $\phantom{-}9e^{5} - 86e^{3} + 41e$
19 $[19, 19, -2w^{2} + w + 15]$ $\phantom{-}3e^{5} - 28e^{3} + 8e$
29 $[29, 29, w^{2} - w - 1]$ $\phantom{-}3e^{5} - 28e^{3} + 9e$
41 $[41, 41, w^{2} - 5]$ $-e^{5} + 9e^{3} - 2e$
43 $[43, 43, w^{2} + w - 3]$ $\phantom{-}4e^{5} - 39e^{3} + 25e$
47 $[47, 47, 2w - 1]$ $\phantom{-}3e^{4} - 28e^{2} + 4$
53 $[53, 53, -2w - 3]$ $-5e^{5} + 46e^{3} - 9e$
59 $[59, 59, w^{2} - w - 11]$ $\phantom{-}13e^{4} - 123e^{2} + 56$
71 $[71, 71, w^{2} - 3]$ $-6e^{5} + 56e^{3} - 21e$
73 $[73, 73, 6w^{2} - 2w - 47]$ $-3e^{5} + 28e^{3} - 6e$
83 $[83, 83, w^{2} - 4w + 1]$ $\phantom{-}10e^{4} - 95e^{2} + 42$
83 $[83, 83, 3w^{2} - 25]$ $\phantom{-}5e^{4} - 46e^{2} + 16$
83 $[83, 83, w - 5]$ $\phantom{-}11e^{4} - 107e^{2} + 52$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$17$ $[17, 17, -2w^{2} + 15]$ $1$ | 983 | 1,831 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-39 | latest | en | 0.224242 |
http://vaporia.com/astro/start/mcmethod.html | 1,606,550,237,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00344.warc.gz | 95,329,870 | 1,590 | Monte Carlo method
(mathematics and simulation using random samples)
A Monte Carlo method is a method of carrying out mathematical calculations and related simulations that uses chance to estimate the answer. Calculations are carried out using numbers chosen at random, yielding an estimate of the solution. It is named for the city of Monte Carlo, which is famous for gambling, something that also depends upon chance.
For example, to find the area of a weird shape for which there is no obvious direct calculation, if there is an easy way to determine whether a specific point is within the weird shape, take random points throughout a rectangle that circumscribes it, then count out what percentage of these points fall within the weird shape. That percentage of the area of the rectangle (an area that is easy to calculate) gives you an estimate of the area of weird shape. Choosing the random points must be done in such a way that they are randomly distributed over the rectangle. The more such points, the better the estimate.
Quite a few challenging math problems are "solvable" by such chance methods. Markov chain Monte Carlo (MCMC) is a more specific type of method that incorporates chance. Quantum Monte Carlo methods are the use of Monte Carlo methods to solve quantum system problems, e.g., using quantum mechanics.
(mathematics,simulations,method)
http://en.wikipedia.org/wiki/Monte_Carlo_method
http://en.wikipedia.org/wiki/Quantum_Monte_Carlo
Referenced by pages:
Hyperion
Markov chain Monte Carlo (MCMC)
N-body problem
quantum Monte Carlo (QMC)
stellar dynamics
Index | 329 | 1,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-50 | latest | en | 0.883639 |
https://www.tutorsglobe.com/homework-help/statistics/basic-definitions-of-statistics-assignment-help-7342.aspx | 1,726,837,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00618.warc.gz | 962,853,222 | 10,142 | #### Basic Definitions of Statistics Assignment Help, Homework Help
Keep your academic writing worries and woes at bay and prefer TutorsGlobe to get the most perfectly crafted paper at the most minimal price range. Don't wait for long and hire our professional Statistics Assignment Help tutors to get the paper that surely scores best grades guaranteed!
Basic Definitions of Statistics:
Definition of Survey:
The collection of information from the elements of a population or sample.
Survey is one of the most important parts in statistics. Without the information, we can not do the estimation or hypothesis testing. The information collected in a survey is in the form of raw data.
Definition of Census:
A survey that includes every element of the target population.
The most accurate survey is when we can include everyone in the target group. Census survey is easier to conduct if the target group is small. However, for a large population, census survey is very costly and time consuming. Moreover it is very difficult to identify each member of a target group.
Definition of Sample Survey:
Survey conducted on a sample to make decisions about the corresponding population.
Sample survey is the alternative method to census survey. Sample survey is easier to conduct and the size of sample depends on needs, budget and availability of the members in the target group.
Definition of Representative Sample:
A sample that represent the characteristics of the populations as closely as possible
For example: to find the average spending of customers in a department store. The selected sample must come from various backgrounds and it in about the same background as the existing population has.
Definition of Random Sample:
A sample drawn in such a way that each element of the population has the same chance of being selected.
Examples of random sample are lucky draw in an annual dinner and draw a representative.
Definition of Element or Member
A specific subject or object about which the information is collected.
Definition of Variable
The characteristics that being studied which assumes some values for each element
Definition of Observation or Measurement
The value of a variable or characteristics for an element.
Definition of Data Set
A collection of observations on one or more variables
Completing your Statistics assignments and coursework won’t be a hassle for you, if you consider TutorsGlobe as your study partner. Our team has been successfully offering flawless solutions to students globally for more than a decade. So, if you would like to secure top-notch grades without putting so much effort, then don’t hesitate to avail our Statistics Assignment Help service right away.
Email based statistics assignment help - homework help at TutorsGlobe
Are you searching statistics tutor for help with Basic Definitions of Statistics questions? Basic Definitions of Statistics topic is not easier to learn without external help? We at www.tutorsglobe.com offer finest service of statistics assignment help and statistics homework help. Live tutors are available for 24x7 hours helping students in their Basic Definitions of Statistics related problems. We provide step by step Basic Definitions of Statistics question's answers with 100% plagiarism free content. We prepare quality content and notes for Basic Definitions of Statistics topic under statistics theory and study material. These are avail for subscribed users and they can get advantages anytime.
Why TutorsGlobe for assignment help
• Higher degree holder and experienced tutors network
• Punctuality and responsibility of work
• Quality solution with 100% plagiarism free answers
• Time on Delivery
• Privacy of information and details
• Excellence in solving accounting questions in excels and word format.
• Best tutoring assistance 24x7 hours
We, TutorsGlobe beneath our Assignment Help service ensure that we help students with almost all the Statistics topics as well. Some of them are: | 752 | 4,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.906939 |
https://deluxeknowledgeexchange.com/helping-741 | 1,675,408,438,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00390.warc.gz | 223,818,886 | 5,530 | # Quick math solution
There is Quick math solution that can make the process much easier. Our website can solve math word problems.
## The Best Quick math solution
This Quick math solution supplies step-by-step instructions for solving all math troubles. Algebra is a fundamental tool in mathematics that allows us to solve equations and understand the relationships between variables. Basic algebra problems usually involve solving for a missing variable in an equation. For example, if we are given the equation "x + 5 = 10", we can solve for "x" by subtracting 5 from both sides of the equation to get "x = 5". Algebra can be used to solve for many different types of variables, making it a very powerful tool.
There’s an app for that. No, really. Photomath is an app that can solve math problems just by taking a photo of them. It’s available for free on iOS and Android, and it’s pretty darn accurate.
There are a number of free online tutor chat services available for math. These services are a great resource for students who need extra help with their math homework or just want to brush up on their skills. The tutors on these chat services are usually math experts who can help students with any questions they may have.
I was never very good at math. I could do the simple arithmetic, but when it came to Story Problems, I was lost. My friends would always laugh at me because I could never solve them. I tried everything to get better, but it just wasn't my thing. One day, I stumbled across a website that claimed to be a math story problem solver. I was skeptical, but I decided to give it a try. I was amazed! The website walked me
Math homework can be a challenge for many students. However, by breaking the task down into smaller chunks, and by seeking assistance when needed, it is possible to complete the homework and improve your math skills.
There is no one app that solves math problems. However, there are many apps that can be used to help with math problems. Some of these apps include Wolfram Alpha, Photomath, and MyScript Calculator. Each of these apps has its own unique set of features and tools that can be used to help solve math problems.
## We will support you with math difficulties
Actually, this this is cool app for any kind of students, teachers or parents. I meant by from any kind "Any level". I'm an AL student, this app helped me a lot solving many hard math problems. Also, I would like to say "Buy the premium version of the app, then you will able to know solution of a question step by step with explanations" it's better than trial version. I recommend you to purchase full version also I believe the app will release a major update. Thanks a lot for developers
Belle Williams
It has been amazingly helpful, when I was using it in college the explanation of problems was very helpful to understand the questions better. Unfortunately, the more complex explanations are a premium feature now but it's still helpful none the less!
Mikayla Cooper
Calculus website solve problems Mathematics online Solving linear equations solver Math solutions App that can do any math problem Mathematical induction problem solver | 656 | 3,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.970382 |
http://www.jiskha.com/display.cgi?id=1291448879 | 1,498,592,850,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00131.warc.gz | 547,828,927 | 3,888 | # Physics
posted by .
An airplane is traveling v = 850 km/h in a direction 38.5° west of north (Fig. 3-34).
Figure 3-34.
(a) Find the components of the velocity vector in the northerly and westerly directions.
1 km/h (north)
2 km/h (west)
(b) How far north and how far west has the plane traveled after 2 h?
3 km (north)
4 km (west)
• Physics -
a1
• Physics -
a1: 850 cos 38.5
a2: 850 sin 38.5 | 132 | 401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.862105 |
https://brilliant.org/problems/i-cant-have-anymore-pi-pies/ | 1,490,828,664,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191405.12/warc/CC-MAIN-20170322212951-00268-ip-10-233-31-227.ec2.internal.warc.gz | 780,326,509 | 12,461 | I can't have anymore Pi-Pies...
Calculus Level pending
Suppose a Pie is a perfect circle with a radius of $$\pi$$. Suppose $$I$$ is the area of the pie, and $$P$$ is the circumference of the pie, calculate the infinite series $\frac { P }{ I } \quad +\quad \frac { { P }^{ 2 } }{ { I }^{ 2 } } \quad +\quad \frac { { P }^{ 3 } }{ { I }^{ 3 } } \quad +\quad ...$
× | 123 | 367 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-13 | longest | en | 0.781856 |
https://www.questia.com/read/108669322/empirical-direction-in-design-and-analysis | 1,438,425,281,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988650.53/warc/CC-MAIN-20150728002308-00258-ip-10-236-191-2.ec2.internal.warc.gz | 933,515,935 | 22,005 | Empirical Direction in Design and Analysis
By Norman H. Anderson | Go to book overview
PREFACE
Analysis of covariance (Ancova) can reduce variability due to individual differences, thereby yielding narrower confidence intervals and greater power. Prime opportunities arise with experiments that compare effectiveness of methods to produce some change: change in health, behavior, attitude, or skill. The change score, after–before, may seem the natural measure; this change score seems to adjust for individual differences in the before score. But change scores turn out to be full of pitfalls. Ancova not only avoids these pitfalls, but gives a superior analysis.
Furthermore, Ancova is more general than the change score. Ancova can utilize any correlate of the main response measure, even correlates on a different dimension for which a change score makes no sense.
Ancova rests on a simple idea: Combine Anova and regression. Let X denote an individual difference variable and Y the response measure. Run a regression of Y on X and get Ypred as shown in Chapter 9. Since Ypred is predictable error variability, (YYpred) is less variable than Y alone, for it subtracts out that part of Y that is predictable from X. Take advantage of this lesser variability by (in effect) applying Anova to (YYpred).
This chapter aims at a conceptual understanding of Ancova. Formulas and calculations, accordingly, are largely passed over in order to discuss empirical issues. With randomized subject groups, Ancova has considerable potential that seems to have been relatively neglected. If you can find an individual difference variable that is correlated with your response measure, you can decrease your response variability and increase your power, often at little cost.
With nonrandom subject groups, matters are very different. Nonrandom groups differ systematically on individual difference variables before the experiment begins. These systematic differences are confounded with experimental treatments. Many writers assert that Ancova “controls” or “partials out” these individual differences, thereby removing otherwise deadly confounds. Such assertions have little truth. The second part of this chapter shows why Ancova generally fails to “control” with nonrandom groups.
-382-
If you are trying to select text to create highlights or citations, remember that you must now click or tap on the first word, and then click or tap on the last word.
One moment ...
Default project is now your active project.
Project items
Items saved from this book
This book has been saved
Highlights (0)
Some of your highlights are legacy items.
Citations (0)
Some of your citations are legacy items.
Notes (0)
Bookmarks (0)
You have no saved items from this book
Project items include:
• Saved book/article
• Highlights
• Quotes/citations
• Notes
• Bookmarks
Notes
Cited page
Style
Citations are available only to our active members.
(Einhorn, 1992, p. 25)
(Einhorn 25)
1. Lois J. Einhorn, Abraham Lincoln, the Orator: Penetrating the Lincoln Legend (Westport, CT: Greenwood Press, 1992), 25, http://www.questia.com/read/27419298.
Cited page
Empirical Direction in Design and Analysis
• Title Page iii
• Dedication v
• Foreword vi
• Contents vii
• Preface xvi
• Chapter 1 - Scientific Inference 1
• Preface 30
• Chapter 2 - Statistical Inference 31
• How to Do Exercises 54
• Exercises for Chapter 2 54
• Preface 58
• Chapter 3 - Elements of Analysis of Variance I 59
• Notes 75
• Appendix: How to Randomize 77
• Exercises for Chapter 3 84
• Preface 90
• Chapter 4 - Elements of Analysis of Variance II 91
• Notes 111
• Exercises for Chapter 4 113
• Preface 118
• Chapter 5 - Factorial Design 119
• Notes 145
• Appendix: Hand Calculation for Factorial Design 148
• Exercises for Chapter 5 151
• Preface 158
• Chapter 6 - Repeated Measures Design 159
• Notes 177
• Exercises for Chapter 6 181
• Preface 188
• Chapter 7 - Understanding Interactions 189
• Notes 209
• Exercises for Chapter 7 214
• Preface 218
• Chapter 8 - Confounding 219
• Notes 250
• Preface 258
• Chapter 9 - Regression and Correlation 259
• Notes 280
• Exercises for Chapter 9 282
• Preface 286
• Chapter 10 - Frequency Data and Chi-Square 287
• Notes 300
• Exercises for Chapter 10 302
• Preface 306
• Chapter 11 - Single Subject Design 307
• Notes 338
• Exercises for Chapter 11 345
• Preface 350
• Chapter 12 - Nonnormal Data and Unequal Variance 351
• Notes 373
• Exercises for Chapter 12 378
• Preface 382
• Chapter 13 - Analysis of Covariance 383
• Notes 395
• Exercises for Chapter 13 397
• Preface 400
• Chapter 14 - Design Topics I 401
• Notes 431
• Exercises for Chapter 14 437
• Preface 442
• Chapter 15 - Design Topics II 443
• Notes 475
• Exercises for Chapter 15 481
• Preface 484
• Chapter 16 - Multiple Regression 485
• Notes 514
• Exercises for Chapter 16 520
• Preface 524
• Chapter 17 - Multiple Comparisons 525
• Notes 546
• Exercises for Chapter 17 548
• Preface 550
• Chapter 18 - Sundry Topics 551
• Notes 589
• Exercises for Chapter 18 596
• Preface 602
• Chapter 19 - Foundations of Statistics 603
• Notes 637
• Preface 646
• Chapter 20 - Mathematical Models for Process Analysis 647
• Notes 677
• Exercises for Chapter 20 681
• Preface 688
• Chapter 21 - Toward Unified Theory 689
• Notes 729
• Exercises for Chapter 21 742
• Preface 750
• Chapter 22 - Principles and Tactics of Writing Papers 751
• Notes 761
• Preface 764
• Chapter 23 - Lifelong Learning 765
• Notes 780
• Preface 782
• Chapter 0 - Basic Statistical Concepts 783
• Notes 803
• Exercises for Chapter 0 805
• Statistical Tables 808
• References 820
• Author Index 847
• Subject Index 854
Settings
Settings
Typeface
Text size Reset View mode
Search within
Look up
Look up a word
• Dictionary
• Thesaurus
Please submit a word or phrase above.
Why can't I print more than one page at a time?
Help
Full screen
/ 864
How to highlight and cite specific passages
1. Click or tap the first word you want to select.
2. Click or tap the last word you want to select, and you’ll see everything in between get selected.
3. You’ll then get a menu of options like creating a highlight or a citation from that passage of text.
Cited passage
Style
Citations are available only to our active members.
"Portraying himself as an honest, ordinary person helped Lincoln identify with his audiences." (Einhorn, 1992, p. 25).
"Portraying himself as an honest, ordinary person helped Lincoln identify with his audiences." (Einhorn 25)
"Portraying himself as an honest, ordinary person helped Lincoln identify with his audiences."1
1. Lois J. Einhorn, Abraham Lincoln, the Orator: Penetrating the Lincoln Legend (Westport, CT: Greenwood Press, 1992), 25, http://www.questia.com/read/27419298.
Thanks for trying Questia!
Please continue trying out our research tools, but please note, full functionality is available only to our active members.
Your work will be lost once you leave this Web page. | 1,756 | 6,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | latest | en | 0.920379 |
https://www.teacherspayteachers.com/Browse/Search:inquiry%20experiments | 1,540,263,549,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00370.warc.gz | 1,096,862,395 | 53,035 | showing 1-24 of 2,136 results
Scientific Method: Model science process skills and inspire young scientists with this middle school scientific method unit plan. This 140-page resource is a must have for engaging students in meaningful analysis, inquiry and scientific design! Includes worksheets, lesson plans, activity cards and
Types:
\$12.99
312 Ratings
4.0
ZIP (54.55 MB)
In this introductory inquiry activity, students will see that when a GUMMY BEAR is placed in water, the physical properties of the GUMMY BEAR change (physical change in matter). After leaving it overnight, students should notice that the water changed in color and smell and the gummies grew in size
Subjects:
Types:
\$1.35
121 Ratings
4.0
PDF (564.6 KB)
A little sheet of paper can lead to a ton of scientific inquiry and experiments. How? Paper planes! In this activity, students will experiment with different kinds of paper and see how far each kind of paper plane flies. They will first do some inquiry and then a guided experiment. Experiment: How
Subjects:
Types:
\$1.25
78 Ratings
4.0
PDF (395.98 KB)
This is a bundle of 10 lessons on Newton's laws of Motion! Be sure to check out the wealth of activities included! *Newtons Laws of Motion interactive NOTES with demonstrations and activities * Force and Motion Stations-inquiry activities for Newton's 1st law/ Inertia * Newton's Third Law Statio
Subjects:
Types:
\$12.00
\$3.95
106 Ratings
4.0
ZIP (3.92 MB)
Run run as fast as you can! But the gingerbread man gets into trouble when he comes to a river! Have some fun on the days before winter break and get your students thinking! This is a twist on the boat building density (buoyancy) challenge often done with aluminum foil and pennies. Challenge your
Subjects:
Types:
\$1.50
60 Ratings
4.0
PDF (272.37 KB)
Looking to introduce or review the scientific method in a fun way- try this bubble gum mass lab. Students get a piece of bubble gum and chew it to determine if its mass changes with the chewing (works BEST with sugar gums like bubble yum, etc).“Does gum lose mass when you chew it?" This is an introd
Subjects:
Types:
\$1.50
17 Ratings
4.0
PDF (586.46 KB)
Have a little Halloween fun! Students create Static Electricity and try to get some 'ghosts' to dance. Simple materials are used. Two versions of the worksheet are given, one for students who have had some background with static electricity and one for thoese who have not. This can also be used as a
Subjects:
Types:
\$1.25
49 Ratings
4.0
PDF (460.16 KB)
Weather Unit Plan (8 lessons, demonstrations, experiments, web quest/web journey, Venn diagrams, sorting activities, foldable, flapbook, activity sheets, assessment, flash cards, etc.) Grades 3 - 5 Duration: Approximately 3 weeks Looking for a complete unit on Weather to teach your middle to upper
Subjects:
\$9.95
20 Ratings
4.0
PDF (3.49 MB)
Students put eggs in 3 different kinds of liquids to watch what will happen to the eggs. Topics discussed are permeability of egg shells and membranes. Two activities are included- one to dissolve the shell of an egg (inquiry), then an experiment to determine what will go through the egg membrane on
Subjects:
Types:
\$1.25
22 Ratings
4.0
PDF (421.45 KB)
How can you make a paper clip float??? Students are given materials and have to figure out how to use them to get a paper clip to float. Great introduction or conclusion to a surface tension lesson! This is also part of a Scientific Method Inquiry bundle: Easy Scientific Method Inquiry Experiments
Subjects:
Types:
\$1.25
17 Ratings
4.0
PDF (270.23 KB)
In this inquiry activity on force, you build a balloon rocket (rabbit) and time it as it 'flies' across a room. you then challenge students to create a rabbit using simple materials that will fly in a faster time than yours. This is a fun way to teach Newton's Third Law of Motion, but there are 2 v
Subjects:
Types:
\$1.25
23 Ratings
4.0
ZIP (480.38 KB)
This is a super fun hands-on, inquiry based, lesson plan and activity sheets for students to explore solids, liquids, and gases. Included are detailed lesson plans, a directions page, recording sheet, and "In the Body" application based assessment. This activity is designed to be done after studen
Subjects:
\$2.00
19 Ratings
3.7
PDF (649.78 KB)
Inquiry activity using simple materials (marbles and ruler ramps) -Students are given the challenge to increase kinetic energy of a marble. Students practice using the Scientific Method to design and conduct an experiment and change variables.
Subjects:
Types:
\$1.25
13 Ratings
4.0
PDF (190.75 KB)
How can you change the surface tension of water. Part 1: Students compare surface tension of tap, soapy and salt water. Part 2: Students design and conduct their own inquiry experiment using the Scientific Method to try to change surface tension. This is also part of a Scientific Method Inquiry
Subjects:
Types:
\$1.00
10 Ratings
4.0
PDF (337.16 KB)
This is so much fun for 2nd-6th graders! This activity/handout follows the classic "Naked Egg" experiment of submerging an egg in vinegar for 24 hours and observing how it loses it's shell. I have created a clear experiment handout that follows the scientific method and scientific inquiry process.
Subjects:
Types:
\$2.99
\$2.55
12 Ratings
4.0
DOCX (93.55 KB)
This is a fun and exciting experiment designed for 3rd-6th graders. It is the classic Diet Coke and Mentos "geyser" experiment, but with the added rigor of using the scientific method to answer the question "Which will cause a larger reaction: the sugar in regular Coke or the aspartame in Diet Coke
Subjects:
\$3.99
\$2.49
15 Ratings
4.0
DOCX (52.47 KB)
• In this activity students design an experiment using materials to test how meteoroid size affects the size of Moon craters. Students will be dropping different sized rocks into flour (mimicking lunar soil) to simulate meteoroids hitting the moon and forming craters. • Students need to determine h
Subjects:
Types:
\$1.00
11 Ratings
4.0
PDF (471.51 KB)
Who does not love to play with magnets? Students have common misconceptions about what materials will attract to a magnet. This activity allows them to try out many different materials to determine what will attract. Let them try anything reasonable but direct them to try as many different materials
Subjects:
Types:
\$0.50
5 Ratings
3.8
PDF (63.47 KB)
This 10 page packet includes three separate science experiments that can be completed as one large science investigation or as individual lessons. Each science investigation requires students to use inquiry process skills such as hypothesizing, observing and making conclusions. You will hear "Wow!"
Subjects:
\$3.00
8 Ratings
4.0
PDF (154.1 KB)
This resource was created in response to a student inquiry. An Aboriginal educator showed my students deer and moose leather gloves and we started a conversation about its water-repellent property (and compared it to cotton gloves - which get soggy). This connects to the Ontario Grade 1 Science curr
Subjects:
Types:
\$1.00
5 Ratings
4.0
PDF (2.36 MB)
ECOSYSTEMS Unit Plan (8 lessons, differentiated cloze reading passages, student projects, graphic organizers, writing prompts, ecosystem organization pyramid, online research and activities, flipbooks, foldables, hands-on inquiry experiment, worksheets, end-of-unit research project, unit flash car
Subjects:
\$9.95
4 Ratings
3.9
PDF (13.38 MB)
Circuits Series vs. Parallel Grades 4 - 6 An exciting, engaging resource created for all learning styles! Includes one lesson focusing on specific vocabulary terms with illustrations, diagrams, and a fun hands-on inquiry experiment, “Create a Series and Parallel Circuit”, all to reinforce the less
Subjects:
\$1.95
3 Ratings
4.0
PDF (559.56 KB)
Inquiry: Does the volume (amount) of a liquid change if it is placed in a different container? Liquids can take the shape of their container, but have a definite/fixed/certain volume 1. For this inquiry experiment, set up as many stations with two measuring cups/beakers as you have available. I ha
Subjects:
Types:
\$1.00
1 Rating
4.0
PDF (204.84 KB)
Experimentally explore and determine the environment that's most conducive for hatching brine shrimp eggs. Under \$5 (assuming you have some containers, water, toothpicks, and some deionized salt) for brine shrimp eggs at most pet stores. The Shrimp Lab provide a rich opportunity for your students
Subjects:
Types:
\$4.50
\$3.00
1 Rating
4.0
PDF (2.53 MB)
showing 1-24 of 2,136 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,085 | 8,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.873197 |
https://www.gamedev.net/articles/programming/graphics/an-introduction-to-digital-image-processing-r2007/ | 1,571,271,526,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00524.warc.gz | 924,082,762 | 95,275 | • # An Introduction To Digital Image Processing
Graphics and GPU Programming
# Disclaimer
This document is to be distributed for free and without any modification from its original contents. The author declines all responsibility in the damage this document or any of the things you will do with it might do to anyone or to anything. This document (and any of its contents) is not copyrighted and is free of all rights, you may thus use it, modify it or destroy it without breaking any international law.
However, there are no warranties or such that the content of the document is error proof. If you found an error please check the latest version of this paper on my homepage before mailing me.
# Introduction
Digital image processing remains a challenging domain of programming for several reasons. First the issue of digital image processing appeared relatively late in computer history, it had to wait for the arrival of the first graphical operating systems to become a true matter. Secondly, digital image processing requires the most careful optimisations and especially for real time applications. Comparing image processing and audio processing is a good way to fix ideas. Let us consider the necessary memory bandwidth for examining the pixels of a 320x240, 32 bits bitmap, 30 times a second: 10 Mo/sec. Now with the same quality standard, an audio stereo wave real time processing needs 44100 (samples per second) x 2 (bytes per sample per channel) x 2 (channels) = 176Ko/sec, which is 50 times less.
Obviously we will not be able to use the same signal processing techniques in both audio and image. Finally, digital image processing is by definition a two dimensions domain; this somehow complicates things when elaborating digital filters.
We will explore some of the existing methods used to deal with digital images starting by a very basic approach of colour interpretation. As a more advanced level of interpretation comes the matrix convolution and digital filters. Finally, we will have an overview of some applications of image processing.
# I – A simple approach to image processing
## 1 – The colour data: Vector representation
### a - Bitmaps
The original and basic way of representing a digital colored image in a computer's memory is obviously a bitmap. A bitmap is constituted of rows of pixels, contraction of the words 'Picture Element'. Each pixel has a particular value which determines its appearing color. This value is qualified by three numbers giving the decomposition of the color in the three primary colors Red, Green and Blue. Any color visible to human eye can be represented this way. The decomposition of a color in the three primary colors is quantified by a number between 0 and 255. For example, white will be coded as R = 255, G = 255, B = 255; black will be known as (R,G,B) = (0,0,0); and say, bright pink will be : (255,0,255). In other words, an image is an enormous two-dimensional array of color values, pixels, each of them coded on 3 bytes, representing the three primary colors. This allows the image to contain a total of 256x256x256 = 16.8 million different colors. This technique is also know as RGB encoding, and is specifically adapted to human vision. With cameras or other measure instruments we are capable of 'seeing' thousands of other 'colors', in which cases the RGB encoding is inappropriate.
The range of 0-255 was agreed for two good reasons: The first is that the human eye is not sensible enough to make the difference between more than 256 levels of intensity (1/256 = 0.39%) for a color. That is to say, an image presented to a human observer will not be improved by using more than 256 levels of gray (256 shades of gray between black and white). Therefore 256 seems enough quality. The second reason for the value of 255 is obviously that it is convenient for computer storage. Indeed on a byte, which is the computer's memory unit, can be coded up to 256 values.
As opposed to the audio signal which is coded in the time domain, the image signal is coded in a two dimensional spatial domain. The raw image data is much more straight forward and easy to analyse than the temporal domain data of the audio signal. This is why we wiil be able to do lots of stuff and filters for images without tranforming the source data, this would have been totally impossible for audio signal. This first part deals with the simple effects and filters you can compute without transforming the source data, just by analysing the raw image signal as it is.
The standard dimensions, also called resolution, for a bitmap are about 500 rows by 500 columns. This is the resolution encountered in standard analogical television and standard computer applications. You can easily calculate the memory space a bitmap of this size will require. We have 500 x 500 pixels, each coded on three bytes, this makes 750 Ko. It might not seem enormous compared to the standards of hard drives but if you must deal with an image in real time processing things get tougher. Indeed rendering images fluidly demands a minimum of 30 images per second, the required bandwidth of 10 Mo/sec is enormous. We will see later that the limitation of data access and transfer in RAM has a crucial importance in image processing, and sometimes it happens to be much more important than limitation of CPU computing, which may seem quite different from what one can be used to in optimization issues. Notice that, with modern compression techniques such as JPEG 2000, the total size of the image can be easily divided by 50 without losing a lot of quality, but this is another topic.
### b – Vector representation of colors
As we have seen, in a bitmap, colors are coded on three bytes representing their decomposition on the three primary colours. It sounds obvious to a mathematician to immediately interpret colors as vectors in a three dimension space where each axis stands for one of the primary colors. Therefore we will benefit of most of the geometric mathematical concepts to deal with our colors, such as norms, scalar product, projection, rotation or distance. This will be really interesting for some kind of filters we will see soon. Figure 1, illustrates this new interpretation:
Figure 1 : Vector representation of colors
## 2 – Immediate application to filters
### a – Edge Detection
From what we have said before we can quantify the 'difference' between two colors by computing the geometric distance between the vectors representing those two colors. Lets consider two colors C1 = (R1,G1,B1) and C2 = (R2,B2,G2), the distance between the two colors is given by the formula :
(R1)
This leads us to our first filter: edge detection. The aim of edge detection is to determine the edge of shapes in a picture and to be able to draw a result bitmap where edges are in white on black background (for example). The idea is very simple; we go through the image pixel by pixel and compare the colour of each pixel to its right neighbour, and to its bottom neighbour. If one of these comparison results in a too big difference the pixel studied is part of an edge and should be turned to white, otherwise it is kept in black. The fact that we compare each pixel with its bottom and right neighbour comes from the fact that images are in two dimensions. Indeed if you imagine an image with only alternative horizontal stripes of red and blue, the algorithms wouldn't see the edges of those stripes if it only compared a pixel to its right neighbour. Thus the two comparisons for each pixel are necessary. So here is the algorithm in symbolic language and in C.
Algorithm 1 : Edge Detection
• For every pixel ( i , j ) on the source bitmap
• Extract the (R,G ,B) components of this pixel, its right neighbour (R1,G1,B1), and its bottom neighbour (R2,G2,B2)
• Compute D(C,C1) and D(C,C2) using (R1)
• If D(C,C1) OR D(C,C2) superior to a parameter K, then we have an edge pixel !
Translated in C language using allegro 4.0 (references in the sources) gives:
for(i=0;i<temp->w-1;i++){
for(j=0;j<temp->h-1;j++){
color=getpixel(temp,i,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
color1=getpixel(temp,i+1,j);
r1=getr32(color1);
g1=getg32(color1);
b1=getb32(color1);
color2=getpixel(temp,i,j+1);
r2=getr32(color2);
g2=getg32(color2);
b2=getb32(color2);
if((sqrt((r-r1)*(r-r1)+(g-g1)*(g-g1)+(b-b1)*(b-b1))>=bb)||
(sqrt((r-r2)*(r-r2)+(g-g2)*(g-g2)+(b-b2)*(b-b2))>=bb)){
putpixel(temp1,i,j,makecol(255,255,255));
}
else{
putpixel(temp1,i,j,makecol(0,0,0));
}
}
}
This algorithm was tested on several source images of different types and it gives fairly good results. It is mainly limited in speed because of frequent memory access. The two square roots can be removed easily by squaring the comparison; however, the colour extractions cannot be improved very easily. If we consider that the longest operations are the getpixel function,get*32 and putpixel functions, we obtain a polynomial complexity of 4*N*M, where N is the number of rows and M the number of columns. This is not reasonably fast enough to be computed in real time. For a 300x300x32 image I get about 26 transforms per second on an Athlon XP 1600+. Quite slow indeed.
Here are the results of the algorithm on an example image:
Picture 1 : Edge detection Results
A few words about the results of this algorithm: Notice that the quality of the results depends on the sharpness of the source image. If the source image is very sharp edged, the result will reach perfection. However if you have a very blurry source you might want to make it pass through a sharpness filter first, which we will study later. Another remark, you can also compare each pixel with its second or third nearest neighbours on the right and on the bottom instead of the nearest neighbours. The edges will be thicker but also more exact depending on the source image's sharpness. Finally we will see later on that there is another way to make edge detection with matrix convolution.
### b – Color extraction
The other immediate application of pixel comparison is color extraction. Instead of comparing each pixel with its neighbors, we are going to compare it with a given color C1. This algorithm will try to detect all the objects in the image that are colored with C1. This was quite useful for robotics for example. It enables you to search on streaming images for a particular color. You can then make you robot go get a red ball for example. We will call the reference color, the one we are looking for in the image C0 = (R0,G0,B0).
Algorithm 2 : Color extraction
• For every pixel ( i , j ) on the source bitmap
• Extract the C = (R,G ,B) components of this pixel.
• Compute D(C,C0) using (R1)
• If D(C,C0) inferior to a parameter K, we found a pixel which colour's matches the colour we are looking for. We mark it in white. Otherwise we leave it in black on the output bitmap.
Translated in C language using Allegro 4.0 (references in the sources) gives:
for(i=0;i<temp->w-1;i++){
for(j=0;j<temp->h-1;j++){
color=getpixel(temp,i,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
if(sqrt((r-r0)*(r-r0)+(g-g0)*(g-g0)+(b-b0)*(b-b0))<=aa){
putpixel(temp1,i,j,makecol(255,255,255));
}
else{
putpixel(temp1,i,j,makecol(0,0,0));
}
}
}
Once again, even if the square root can be easily removed it doesn't really affect the speed of the algorithm. What really slows down the whole loop is the NxM getpixel accesses to memory and the get*32 and putpixel. This determines the complexity of this algorithm: 2xNxM, where N and M are respectively the numbers of rows and columns in the bitmap. The effective speed measured on my computer is about 40 transforms per second on a 300x300x32 source bitmap. Here are the results I obtained using this algorithm searching for whites shape in the source bitmap:
Picture 2 : White Extraction Results
### c – Color to grayscale conversion
Regarding the 3D color space, grayscale is symbolized by the straight generated by the (1,1,1) vector. Indeed, the shades of grays have equal components in red, green and blue, thus their decomposition must be (n,n,n), where n is an integer between 0 and 255 (example : (0,0,0) black, (32,32,32) dark gray, (128,128,128) intermediate gray, (192,192,192) bright gray, (255,255,255) white etc…). Now the idea of the algorithm is to find the importance a color has in the direction of the (1,1,1) vector. We will use scalar projection to achieve this. The projection of a color vector C = (R,G,B) on the vector (1,1,1) is computed this way :
(R2)
(R3)
However the projection value can reach up to 441.67, which is the norm of the white color (255,255,255). To avoid having numbers above 255 limit we will multiply our projection value by a factor 255/441,67=1/sqrt(3). Thus the formula simplifies a little giving:
(R4)
So in fact converting a color to a grayscale value is just like taking the average of its red, green and blue components. You can also adapt the (R3) formula to other color-scales. For example you can choose to have a red-scale image, Redscale(C)=R, or a yellow-scale image, Yellowscale(C)=(G+B)/sqrt(6) etc…
Algorithm 3 : Grayscale conversion
• For every pixel ( i , j ) on the source bitmap
• Extract the C = (R,G ,B) components of this pixel.
• Compute Grayscale(C) using (R4)
• Mark pixel ( i , j ) on the output bitmap with color (Grayscale(C), Grayscale(C), Grayscale(C)).
C source code equivalent is :
for(i=0;i<temp->w-1;i++){
for(j=0;j<temp->h-1;j++){
color=getpixel(temp,i,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
h=(r+b+g)/3;
putpixel(temp1,i,j,makecol(h,h,h));
}
}
It's impossible to optimize this algorithm regarding its simplicity but we can still give its computing complexity which is given by the number of pixel studied: NxM, (N,M) is the resolution of the bitmap. The execution time on my computer is the same as in the previous algorithm, still about 35 transforms per second. Here is the result of the algorithm on a picture:
Picture 3 : Grayscale conversion Results
## 3 – Grayscale transforms: light and contrast
### a – Grayscale transforms
Grayscale transforms are integer functions from [0,255] to [0,255], thus to any integer value between 0 and 255 we make another value between 0 and 255 correspond. To obtain the image of a color by a grayscale transform, it must be applied to the three red, green and blue components separately and then reconstitute the final color. The graph of grayscale transform is called an output look-up table, or gamma-curve. In practice you can control the gamma-curve of your video-card to set lightness or contrast, so that each time it sends a pixel to the screen, it makes it pass through the grayscale transform first. Here are the effects of increasing contrast and lightness to the gamma-curve of the corresponding grayscale transform:
Normal grayscale transform Increased brightness by 128 Decreased brightness by 128 Increased contrast by 15° Increased contrast by 30° Increased contrast by 30° at 64 and 150 Increased brightness and contrast
Figure 2 : Basic grayscale transforms
Notice that if you want to compose brightness and contrast transforms you should first apply the contrast transform and then the brightness transform. One can also easily create his own grayscale transform to improve the visibility of an image. For example, if you have a very dark image with a small bright zone in it. You cannot increase a lot brightness because the bright zone will saturate quickly but the idea is to increase the contrast exactly where there are the most pixels, in this case: for low values of colors (dark). This will have the effect of flattening the colors histogram and making the image use a wider range of colors, thus making it more visible. So, first the density histogram must be built, counting the number of pixels for each 0-255 value. Then we built the contrast curve we want, finally we integrate to obtain the final transform. The integration comes from the fact that contrast is a slope value, therefore to come back to the contrast transform function we must integrate. Those figures explain this technique:
Bitmap input color histogram: two large amounts of pixels in thin color bandwidths. The contrast slope must be big in those two areas to flatten the input histogram. The contrast transform: It is obtained by integrating the previous plot. The input bitmap color histogram before and after the contrast transform: Now pixels are averagely distributed on the whole color bandwidth.
Figure 3 : Custom grayscale transforms
### b – Light and contrast
Light and contrast transforms are immediate to implement: once the output look-up table is saved into memory once and for all, each pixel is passed trough this table and marked on the output bitmap. The look-up table will be represented simply as a list. The index of the list will stand for the input color value and the contents of the list will be the output. So calling list[153] will give us the image of grayscale level 153 by our transform.
Algorithm 4 : Light or contrast modification
• Generate the look-up table for the desired transform (Light or contrast or any other). For each index j from 0 to 255 is stored its image by the transform in transform_list[ j ].
• For every pixel ( i , j ) on the source bitmap
• Extract the C = (R,G ,B) components of this pixel.
• Mark pixel ( i , j ) on the output bitmap with color (transform_list[R], transform_list[G], transform_list).
In the C source code implementation for contrast, the float variable contrast quantifies the angle of the slope (in radians) of the contrast transform like in figure 2. If contrast is defined as pi/4 there will be no changes in the image, as this corresponds to the 'normal' transform of figure 2. Significant changes start when contrast > pi/4 + 15°. If contrast reaches 0 all the image will be gray because the transform will be a horizontal straight at 128. If contrast becomes very big, any color value becomes either 0 or 255 depending on its position around 128, so the output bitmap will have only the 8 possible colors only made of 255 or 0: (255,255,255), (255,255,0), (255,0,255), (0,255,255), (255,0,0), (0,255,0), (0,0,255), (0,0,0).
for(i=0;i<256;i++){
if(i<(int)(128.0f+128.0f*tan(contrast))&&i>(int)(128.0f-128.0f*tan(contrast)))
Contrast_transform[i]=(i-128)/tan(contrast)+128;
else if(i>=(int)(128.0f+128.0f*tan(contrast)))
Contrast_transform[i]=255;
else
Contrast_transform[i]=0;
}
for(i=0;i<temp->w-1;i++){
for(j=0;j<temp->h-1;j++){
color=getpixel(temp,i,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
putpixel(temp1,i,j,makecol(Contrast_transform[r],Contrast_transform[g],
Contrast_transform[b]));
}
}
C source code implementation for brightness:
for(i=0;i<256;i++){
Light_transform[i]=i+light;
if(Light_transform[i]>255)
Light_transform[i]=255;
if(Light_transform[i]<0)
Light_transform[i]=0;
}
for(i=0;i<temp->w-1;i++){
for(j=0;j<temp->h-1;j++){
color=getpixel(temp,i,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
putpixel(temp1,i,j,makecol(Contrast_transform[r],Contrast_transform[g],
Contrast_transform[b]));
}
}
The efficiency of the algorithm is limited by the getpixel and putpixel calls and the speed is the same as the last two algorithms: about 40 transforms per second on my computer. Here are some examples of outputs:
Picture 4 : Brightness transform result
Picture 5 : Contrast transform result
## 4 – Resizing and rotating algorithms
### a – Resizing algorithm
The resizing of an image is far from being trivial. A few techniques exist, each giving more or less good results in stretching or squeezing. Linear resizing is what we will be studying here, I haven't experienced other methods and I won't be able to compare linear resizing with other methods. However, the resizing process is rarely something done for real-time matters. The aim is usually to resize the image once, and if the algorithm takes 5 seconds or 10 ms, it doesn't matter much. For this kind of algorithm quality is put ahead of speed. For the following explanation let E(x) be the integer part of x (E(3.14) = 3) and F(x) the fractional part of x (E(3.14) = 0.14).
We will first see how to stretch or squeeze an image along its width; the process is the same for height and you can cumulate both of them to give the bitmap the desired resolution. Let us consider a row of pixels of the output bitmap of width 'wout'. Let us take a pixel of this row and name his column number xi. To find which color this pixel should be, we are going to convert his column xi to its equivalent column 'ci' in the source bitmap (width = 'win') using the formula:
(R5)
Thus pixel in column 'xi' in the output bitmap should be in fact the same color as pixel in column 'ci' in the source bitmap. The problem is that 'ci' is a floating point value and one cannot access floating point pixel numbers (no surprise). However, 'ci' falls in between two integer column numbers E(ci) and E(ci)+1 which can be accessed. You can simply convert 'ci' to its actual integer value: E(ci) and choose the color at pixel E(ci) in the source image. This constitutes the 'nearest neighbour' method. There is another way to solve the problem that has the advantage of giving nicer results at high enlarging ratios: linear interpolation.
In fact, the final color we will choose for pixel 'xi' is not exactly that of pixel 'E(ci)' but a linear interpolation between the color of 'E(ci)' and 'E(ci)'+1 using the fractional part of 'ci'. Thus the final color we will choose for the color of the 'xi' pixel in the output bitmap will be computed by this formula:
(R6)
Therefore, when stretching the image, the algorithm will try to compensate the lack of information due to the increase of pixels by using linear interpolation. The difference between choosing linear interpolation to compute the color or simply pick the color at 'E(ci)' is not flabbergasting, but the second gives a nice smooth effect and when you have enormous stretches ratio it looks quite nicer than the first one. The first method is more commonly called 'nearest neighbour' and second one is know as 'bi-linear'. Photoshop© uses a third method called 'bi-cubic', but I don't know about it. I used linear interpolation for this screenshot:
Picture 6 : Resizing algorithm result
Figure 4: Color interpolation in stretching image
The code in C and the algorithm implementing this resizing methods:
Algorithm 5 : Resizing a bitmap
• First we resize the width of the bitmap:
• For every pixel ( i , j ) on the output bitmap
• Compute the ci using formula (R5)
• Extract the (R,G,B) components of pixel E(ci) and E(ci)+1 in source bitmap using (R6)
• Mark pixel ( i , j ) on the output bitmap with color given by formula (R7)
• Repeat the exact same process to resize the height of the bitmap
bs=(float)temp->w/(float)size;
for(i=1;i<size;i++){
for(j=1;j<temp->h-1;j++){
as=(float)i*bs;
es=(int)as;
fs=(int)as+1;
color=getpixel(temp,es,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
colort=getpixel(temp,fs,j);
rt=getr32(color);
gt=getg32(color);
bt=getb32(color);
cs=modf(as,&gs);
dr=(float)(rt-r);
dg=(float)(gt-g);
db=(float)(bt-b);
putpixel(temp_size,i,j,makecol((int)(r+(float)cs*dr),
(int)(g+(float)cs*dg),(int)(b+(float)cs*db)));
//OR the simple E(ci) method (in which case you can remove lots of stuff above):
//putpixel(temp_size,i,j,makecol((int)r,(int)g,(int)b));
}
}
### b – Rotation algorithm
Once again there are different ways of implementing bitmap rotation, the faster they are the more complex they get. We will first study a very basic way and then a much more efficient method. As always there is a little bit of mathematics to understand first. Say x and y are the coordinates of a pixel on the bitmap we can find its image (x',y') by the rotation of center (x0,y0) and of angle α by the following formula:
(R8)
If we apply this formula directly, taking each pixel of the source image and passing them through those equations we will obtain lots of holes in the output bitmap. Those wholes are due to the fact that x and y are integers. Necessarily, there will be some x' and y' that will never be reached and which will stay black. The trick is to start from the output image pixels and compute there inverse rotation to find their antecedent by the rotation. Therefore we will use this formula to find (x,y) for every (x',y') of the output bitmap:
(R9)
Once the antecedent pixel is found, we set the output pixel at the same color. Here is the algorithm and the C source:
Algorithm 6 : Rotating a bitmap
• For every pixel (i,j)=(x',y') on the output bitmap
• Compute the antecedent pixel (x,y) of the input image using (R9)
• Mark pixel (i,j) on the output bitmap with the same color as pixel (x,y) of the input bitmap.
cos_val=cos(rangle);
sin_val=sin(rangle);
half_w=temp->w>>1;
half_h=temp->h>>1;
for(j=0;j<temp->h;j++){
for(i=0;i<temp->w;i++){
bs=i-half_w;
cs=j-half_h;
as=bs*sin_val+cs*cos_val+half_w;
gs=bs*cos_val-cs*sin_val+half_h;
color=getpixel(temp,as,gs);
putpixel(temp1,i,j,color);
}
}
Picture 7 : Rotation algorithm result
With this algorithm I have an even better speed than any of the precedent algorithms, about 42 frames per second. The complexity of the algorithm is N*M where N and M are the dimensions of the source bitmap.
There is a way to improve this algorithm. Instead of computing for each pixel of the output bitmap its inverse rotation, we are going to deal with lines. For each horizontal line in the output image, we compute, as we did before, the inverse rotation of its extreme points. Thus we know the color of all the points of this horizontal line because we know all the color of the pixels on the inverse rotation line on the source bitmap. This way we replace lots of rotations calculations (the ones we did for each pixel on the bitmap) by simple linear interpolation with no multiplications. The difference is not that visible on my machine, I gained 3 fps with this algorithm; moreover, if we look at the complexity, it hasn't change. Indeed even with this linear interpolation there is still N*M pixel accesses. It is really the fact that we avoided all the multiplication of the rotations that makes us win 3 fps.
Algorithm 6 (bis) : Rotating a bitmap
• For every horizontal line in the output bitmap
• Compute the antecedent pixels (x1,y1) and (x2,y2) of the extreme points of the line. (R9)
• For every pixel of line (x1,y1),(x2,y2) on the source image we get its color and set the corresponding pixel on the horizontal line of the output bitmap
cos_val=cos(rangle);
sin_val=sin(rangle);
half_w=temp->w>>1;
half_h=temp->h>>1;
for(j=-1;j<temp->h;j++){
line_count=0;
n=j;
bs=-half_w;
cs=j-half_h;
as=bs*cos_val-cs*sin_val+(half_w);
gs=bs*sin_val+cs*cos_val+(half_h);
r=as;
g=gs;
bs=temp->w-half_w;
cs=j-half_h;
as=bs*cos_val-cs*sin_val+(half_w);
gs=bs*sin_val+cs*cos_val+(half_h);
do_line(temp1,r,g,(int)as,(int)gs,color,Line_make);
if(j==-1)
line_save=line_count;
}
void Line_make(BITMAP* image,int x,int y,int d){
static int lastx=0;
int current_color;
line_count++;
current_color=getpixel(temp,x,y);
if(line_count*temp->w/line_save-lastx>1){
putpixel(temp1,lastx+1,n,current_color);
putpixel(temp1,lastx+2,n,current_color);
}
else{
putpixel(temp1,line_count*
temp->w/line_save,n,current_color);
}
lastx=line_count*temp->w/line_save;
}
## 4 – Blending
### a – Averages
Blending simply consists in averaging the value of two pixels to get a transparency effect, a mix between the two colors. The average can be more or less equilibrated which gives more or less importance to the first or the second pixel. For example if you want to make a blending between two colors C1 and C2 you can use this formula:
(R10)
Blending is usually done to give transparency effects in 3D games or demos, or transition effects in movies. The blending is often done between two bitmaps, or a bitmap and a color. You can also blend a bitmap with itself, but this results in a sort of blurring which you can achieve much more rapidly with algorithms we will see in the second part of this article. Therefore we will consider we have two bitmaps B1 and B2 we want to blend. They are probably not perfectly edged aligned, and not of the same size. To limit iterations, we copy the biggest bitmap (say B1) on the output bitmap, then we take the smallest bitmap (B2) pixel by pixel, test if the pixel belongs to the other image (B1) at the same time, if it does we average with formula (R10) and set the output pixel, if it doesn't we just set the output pixel directly without changing the color. Anyways once you know that the act of blending is just an averaging, the rest depends on what you want to do.
### b – Other blending effects
Blending gives lots of possibilities for cool effects, you can make gradient blending by making the 'α' or 'β' vary; you can have lighting effects, glow effects, transparency effects, water effects, wind effects, lots of the coolest filters in Photoshop© use a simple blending. Once you know this it's rather creativity then technique which makes the difference.
Picture 8 : Blending example
# II – Matrix convolution filters
## 1 – Definition, properties and speed
### a – Definition and a bit of theory
Let us first recall a basic principle in signal processing theory: for audio signals for example, when you have a filter (or any kind of linear system) you can compute its response to an entering signal x(t) by convolving x(t) and the response of the filter to a delta impulse: h(t), you then have:
Continuous time:
(R11)
Discrete time:
(R12)
If you are not familiar with this technique and don't want to get bored, you can skip this section which is not crucial to understand the following algorithms. In the same way that we can do for one-dimensional convolution, we can also compute image convolutions, and basically this is what matrix convolution is all about. The difference here is obviously that we are dealing with a two dimensions object: the bitmap. Likewise audio signal, a bitmap can be viewed as a summation of impulses, that is to say scaled and shifted delta functions. If you want to know the output image of a filter, you will simply have to convolve the entering bitmap with the bitmap representing the impulse response of the image filter. The two dimensional delta functions is an image composed by all zeros, except for a single pixel, which has value of one. When this delta function is passed through a linear system, we will obtain a pattern, the impulse response of this system. The impulse response is often called the 'Point spread function' (PSF). Thus convolution matrixes, PSF, filter impulse response or filter kernel represent the same thing. Two dimensional convolution works just like one dimensional discrete convolutions, the output image y of an entering bitmap x, through a filter system of bitmap impulse response h (width and height M) is given by formula:
(R13)
Notice that the entering signal (Or the impulse response) alike one dimensional signals must first be flipped left for right and top for bottom before being multiplied and summed: this is explained by the (k-i) index in the sums of (R12); and by the (r-j) and (c-i) indexes in (R13). Most of the time we will deal with symmetrical filters, therefore these flips have no effect and can be ignored; keep in mind that for non symmetrical signals we will need to rotate h before doing anything. By the way two dimensional convolutions is also called matrix convolution.
The presence of a normalising factor in gray in formula (R13), can be explained by the fact that we want y[r,c] to be found between 0 and 255. Therefore we can say that computing y[r,c] is like computing the center of gravity of some x[i,j] color values affected by the weights in the filter matrix. This normalising factor is sometimes included into the filter matrix itself, sometimes you have to add it yourself. Just remember to always check that the numbers possible to obtain with y[r,c] are always between 0 and 255.
Let us have an example of matrix convolution:
Figure 5: Matrix Convolution
There are numerous types of matrix filters, smoothing, high pass, edge detection, etc… The main issue in matrix convolution is that it requires an astronomic number of computations. For example is a 320x200 image, is convolved with a 9x9 PSF (Pulse spread function), we already need almost six millions of multiplications and additions (320x200x9x9). Several strategies are useful to reduce the execution time when computing matrix convolutions:
• Reducing the size of the filter reduces as much the execution time. Most of the time small filters 3x3 can do the job perfectly.
• Decompose the convolution matrix of the filter into a product of an horizontal vector and a vertical vector: x[r,c] = vert[r] x horz[c]. In which case we can convolve first all the horizontal lines with the horizontal vector and then convolve the vertical line with the vertical vector. Vert[r] and horz[c] are the projections of the matrix on the x and y axis of the bitmap. You can generate an infinity of seperable matrixes filters by choosing a vert[r] and a horz[c] and multiplying them together. This is a really efficient improvement because it brings the global complexity from N²M² to N²M (where N is the number of rows and columns of the input bitmap and M that of the matrix filter).
• Last method is FFT convolution. It becomes really useful for big matrix filters that cannot be reduced. Indeed Convolution in time or space domain is equivalent to multiplying in frequency domain. We will study in detail this method in the last part of this chapter.
## 1 – A few common filters
### a – sharpness filter
Let us have some examples of different filter kernels. The first one will be the sharpness filter. Its aim is to make the image look more precise. The typical matrix convolution for a sharpness filter is:
(R14)
Looking at those matrixes you can feel that for each pixel on the source image we are going to compute the differences with its neighbors and make an average between those differences and the original color of the pixel. Therefore this filter will be enhancing the edges. Once you have understood how to compute a matrix convolution the algorithm is easy to implement:
Algorithm 7 : Sharpness filter
• For every pixel ( i , j ) on the output bitmap
• Compute its color using formula (R13)
• Set the pixel
#define sharp_w 3
#define sharp_h 3
sumr=0;
sumg=0;
sumb=0;
int sharpen_filter[sharp_w][sharp_h]={{0,-1,0},{-1,5,-1},{0,-1,0}};
int sharp_sum=1;
for(i=1;i<temp->w-1;i++){
for(j=1;j<temp->h-1;j++){
for(k=0;k<sharp_w;k++){
for(l=0;l<sharp_h;l++){
color=getpixel(temp,i-((sharp_w-1)>>1)+k,j-
((sharp_h-1)>>1)+l);
r=getr32(color);
g=getg32(color);
b=getb32(color);
sumr+=r*sharpen_filter[k][l];
sumg+=g*sharpen_filter[k][l];
sumb+=b*sharpen_filter[k][l];
}
}
sumr/=sharp_sum;
sumb/=sharp_sum;
sumg/=sharp_sum;
putpixel(temp1,i,j,makecol(sumr,sumg,sumb));
}
}
Unfortunately this filter is not separable and we cannot decompose the filter matrix in the product of a vertical vector and a horizontal vector. The global complexity of the algorithm is N²M² where and N and M are the dimensions of the filter bitmap and of the source bitmap. This algorithm much slower than any of the others we have studied up to now. Use it carefully.
Picture 9: Sharpness filter effects
### b – Edge Detection
This algorithm uses matrix convolution to detect edges. I found it much slower than the algorithm seen in the first part for an equivalent result. Here are matrix filters exemples for edge detection:
(R15)
The kernel elements should be balanced and arranged to emphasize differences along the direction of the edge to be detected. You may have noticed that some filters emphasize the diagonal edges like the third one, while others stress horizontal edges the fourth and the second. The first is quite efficient because it takes into account edges in any direction. As always, you must choose the filter the most adapted to your problem.
A second important point: notice that the sum of the filter matrix coefficients we usually used as a normalising factor in (R13) is null. Two things to do: we will avoid dividing by zero, we will add to the double sums the value of 128 to bring the color value y[r,c] back into 0-255. The complexity of the algorithm doesn't change N²M².
Algorithm 8 : Edge Detection Filter
• For every pixel ( i , j ) on the output bitmap
• Compute its color using formula (R13)
• Set the pixel
#define edge_w 3
#define edge_h 3
sumr=0;
sumg=0;
sumb=0;
int edge_filter[edge_w][edge_h]={{-1,-1,-1},{-1,8,-1},{-1,-1,-1}};
int edge_sum=0; // Beware of the naughty zero
for(i=1;i<temp->w-1;i++){
for(j=1;j<temp->h-1;j++){
for(k=0;k<edge_w;k++){
for(l=0;l<edge_h;l++){
color=getpixel(temp,i-((edge_w-1)>>1)+k,j-((edge_h-1)>>1)+l);
r=getr32(color);
g=getg32(color);
b=getb32(color);
sumr+=r*edge_filter[k][l];
sumg+=g*edge_filter[k][l];
sumb+=b*edge_filter[k][l];
}
}
sumr+=128;
sumb+=128;
sumg+=128;
putpixel(temp1,i,j,makecol(sumr,sumg,sumb));
}
}
### c – Embossing filter
The embossing effect gives the optical illusion that some objects of the picture are closer or farther away than the background, making a 3D or embossed effect. This filter is strongly related to the previous one, except that it is not symmetrical, and only along a diagonal. Here are some examples of kernels:
(R16)
Here again we will avoid dividing by the normalising factor 0, and we will add 128 to all the double sums to get correct color values. The implementation of this algorithm is just the same as the prvious one except that you must first convert the image into grayscale.
Algorithm 9 : Embossing effect filter
• Convert the image into grayscale
• For every pixel ( i , j ) on the output bitmap
• Compute its color using formula (R13)
• Set the pixel
#define emboss_w 3
#define emboss_h 3
sumr=0;
sumg=0;
sumb=0;
int emboss_filter[emboss_w][emboss_h]={{2,0,0},{0,-1,0},{0,0,-1}};
int emboss_sum=1;
for(i=1;i<temp->w-1;i++){
for(j=1;j<temp->h-1;j++){
color=getpixel(temp,i,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
h=(r+g+b)/3;
if(h>255)
h=255;
if(h<0)
h=0;
putpixel(temp1,i,j,makecol(h,h,h));
}
}
for(i=1;i<temp->w-1;i++){
for(j=1;j<temp->h-1;j++){
sumr=0;
for(k=0;k<emboss_w;k++){
for(l=0;l<emboss_h;l++){
color=getpixel(temp1,i-((emboss_w-1)>>1)+k,j-((
emboss_h-1)>>1)+l);
r=getr32(color);
sumr+=r*emboss_filter[k][l];
}
}
sumr/=emboss_sum;
sumr+=128;
if(sumr>255)
sumr=255;
if(sumr<0)
sumr=0;
putpixel(temp2,i,j,makecol(sumr,sumr,sumr));
}
}
Here are the effects of this algorithm:
Picture 9: Embossing filter
### d – Gaussian blur filter
As its name suggests, the Gaussian blur filter has a smoothing effect on images. It is in fact a low pass filter. Apart from being circularly symmetric, edges and lines in various directions are treated similarly, the Gaussian blur filters have very advantageous characteristics:
• They are separable into the product of horizontal and vertical vectors.
• Large kernels can be decomposed into the sequential application of small kernels.
• The rows and columns operations can be formulated as finite state machines to produce highly efficient code.
We will only study here the first optimisation, and leave the last two to a further learning (you can find more information on these techniques in the sources of this paper). So, the filter kernel of the Gaussian blur filter is decomposable in the product of a vertical vector and an horizontal vector, here are the possible vectors, multiplied by each other they will produce gaussian blur filters:
Figure 6: Gaussian filters coefficients
(source: An efficient algorithm for gaussian blur filters, by F. Waltz and W. Miller)
Gaussian blur filter kernel example (order 2):
(R17)
What increases greatly the algorithm's speed compared to the previous 'brute force' methods is that we will first convolve rows of the image with the (1 2 1) vector and then the columns of pixels with the (1 2 1)T vector. Thus we will use an intermediary bitmap where will be stored the results of the first convolutions. In the final bitmap will be stored the output of the second stage convolutions. So here is the algorithm with an order 6. The fact that we separated our matrix in a product of vectors, ports the complexity of the algorithm to N²M where M is the size of the filter and N that of the source bitmap. Therefore for a order 6 filter we have 6 times less elementary operations. The result is quite satisfying; on my computer I obtained the equivalent transform rate of the order 3 filters we have seen before. We have thus reached the speed of order 3 filters but with the quality result of an order 6!
Algorithm 10 : Gaussian blur implementation
• For every pixel ( i , j ) on the intermediate bitmap
• Compute its color using formula (R13), source bitmap pixels and horizontal vector of the matrix decomposition (eg : (1 2 1))
• Set the pixel on the intermediate bitmap
• For every pixel ( i, j ) on the output bitmap
• Compute its color using formula (R13), intermediate bitmap pixels and vertical vector of the matrix decomposition (eg : (1 2 1)T)
• Set the pixel on the output bitmap
#define gauss_width 7
sumr=0;
sumg=0;
sumb=0;
int gauss_fact[gauss_width]={1,6,15,20,15,6,1};
int gauss_sum=64;
for(i=1;i<temp->w-1;i++){
for(j=1;j<temp->h-1;j++){
sumr=0;
sumg=0;
sumb=0;
for(k=0;k<gauss_width;k++){
color=getpixel(temp,i-((gauss_width-1)>>1)+k,j);
r=getr32(color);
g=getg32(color);
b=getb32(color);
sumr+=r*gauss_fact[k];
sumg+=g*gauss_fact[k];
sumb+=b*gauss_fact[k];
}
putpixel(temp1,i,j,makecol(sumr/gauss_sum,sumg/gauss_sum,
sumb/gauss_sum));
}
}
for(i=1;i<temp->w-1;i++){
for(j=1;j<temp->h-1;j++){
sumr=0;
sumg=0;
sumb=0;
for(k=0;k<gauss_width;k++){
color=getpixel(temp1,i,j-((gauss_width-1)>>1)+k);
r=getr32(color);
g=getg32(color);
b=getb32(color);
sumr+=r*gauss_fact[k];
sumg+=g*gauss_fact[k];
sumb+=b*gauss_fact[k];
}
sumr/=gauss_sum;
sumg/=gauss_sum;
sumb/=gauss_sum;
putpixel(temp2,i,j,makecol(sumr,sumg,sumb));
}
}
Picture 10: Gaussian blur filter effects
## 1 – FFT enhanced convolution
### a – FFT with bitmaps
There is a major difference between the use of Fast Fourier Transform in audio signal and in image data. While in the first one fourier analysis was a way to trasform the confusing time domain data into an easy to understand frequency spectrum; the second one converts the straight forward information in spatial domain to a scrambled form in the frequency domain. Therefore, in audio analysis, FFT was a way to 'see' and 'understand' more clearly the signal: now in image processing, FFT messes up the information in an unreadable coding. We will not be able to use FFT in Digital image processing to design filters or to analyse data. Image filters are normally designed in the spatial domain, where the information is encoded in its simplest form. We will always prefer the appelations 'smotthing' and 'shapening' to 'low pass' or 'high pass filter'.
However, some principles of the FFT are still true and very useful in digital image processing: Convolving two functions is equivalent to multiplying in the frequency domain their FFT and then computing an inverse FFT to obtain the resulting signal. The second useful property of the frequency domain is the Fourier Slice Theorem, this therorem deals with the relationship between an image and its projections (the image viewed from its sides. We won't study this theorem here: it would take us far away from our concerns. Let's come back to image convolution using a FFT.
The FFT of an image is very easy to compute once you know how to do one dimensional FFTs (just like in the audio signal). First, if the dimensions of the image are not a power of two we wiil have to increase its size by adding zero pixels; the image must become NxN sized, where N is a power of 2 (32, 64, 128, 256, 1024 …). The two dimensional array that holds the image will be called the real array, and we will use another array of the same dimensions, an imaginary array. To compute the FFT of the image we will take the one-dimensional FFT of each of the rows, followed by the FFT of each of the columns. More precisely, we wiil take the FFT of the N pixel values of row 0 of the real array (original bitmap), store its real part output back into row 0 of real array, and the imaginary part into row 0 of imaginary array. We will repeat this procedure from row 1 to N-1. We will obtain two intermediate images: the real array and the imaginary array. We will then start the process all over again but with columns FFT and using the intermediate images as source. That is to say, compute the FFT of the N pixel values in column 0 of the real array and of the imaginary array (making complex numbers). We will store the result back into column 0 of real and imaginary arrays. After this procedure is repeated through all the columns, we obtain the frequency spectrum of the image, its real part in the real array and its imaginary part in the imaginary array.
Since row and columns are interchangeable in an image, you can start the algorithm by computing FFT on columns followed by rows. The amplitudes of the low frequencies can be found in the corners of the real and imaginary arrays, whereas the high frequencies are close to the center of the output images. The inverse FFT is obviously calculated by taking the inverse FFT of each row and of each column. Properties of the FFT give us a last property of the output images:
(R18)
The idea of passing to the frequency domain to compute an FFT is the following. We will pass both of the images we want to convolve into the frequency domain, adapting their dimensions so that they have both the same size, NxN power of 2, we will fill empty zones with zeros if necessary. Remember to rotate one of the images to convolve by 180° around its center before starting anything. Indeed convolution, according to its mathematical definition, flips one of the source images. Therefore we will have to compensate it by rotating of of the images first. Once we have the real and imaginary arrays FFT outputs for the two source images, we will multiply each of the real values and imaginary values of one image with each of the real values and imaginary values of the other, this will give us our final image in the frequency domain. Finally we compute the inverse FFT of this final image and we obtain X1 convolved with X2. On the following page you will find in figure 7 the process explained in detail.
The convolution by FFT becomes efficient for big kernels (30x30), and you may notice that its execution time doesn't depend on the kernel's size but on the image's size. Standard convolution's execution time depends on both the kernel size and the source image size. Figure 8 is a graph of the execution time for computing convolutions with different kernel sizes and for two sizes of image (128x128 and 512x512), using FFT, standard floating point convolution (FP) and standard integer convolution (INT).
Figure 7: The FFT way of convolving two images
Figure 8: Computing time for different convolution methods and for different kernel sizes
(source: The scientist and engineer's guide to Digital Signal Processing, by Steven W. Smith))
# III – Examples of application
## 1 – Motion Detection and Tracking
### a – Static and Dynamic backgrounds
The aim of motion detection is to detect in a streaming video source, which we will consider as a very rapid succession of bitmaps, the presence of moving objects and to be able to track their position. Motion detection in surveillance systems is rarely implemented with software means because it is not trustable enough and good systems remain very expensive, companies usualy prefer hiring surveillance employees. However, video motion detection does exist and works pretty well as long as you know exactly what you want to do. Are you going to have a static background? Is the camera going to move around? Do you want to spot exclusively living entities? Do you know in advance the object which going to move in the camera's field? Lots of questions we will have to answer before making our choices for the algorithm used.
There are two major types of motion detection: static background motion detection and dynamic background motion detection. The first will be used for surveillance means: the surveillance camera doesn't move and has a capture (in a bitmap for example) of the static background it is facing. The second is much more complex. Indeed in the entering streaming video the background moves at the same time as moving objects we want to spot, all the issue consists in seperating the object from the background. Therefore we will make a major hypothesis: the object moves much slower than the background.
### b – Static background motion detection
Let us suppose a video camera is facing a static background. The static image of this background, say, when no object is in the field of the camera, is stored in a bitmap. The idea is very simple. When surveillance is turned on, for each frame received (or every n frames, depending on the framerate of the recording system and on the relative speed of the potential target objects) the software compares each pixel of the frame with those of the static background image. If a pixel is very different it is marked as white, if not it is left in black. We will calculate the 'difference' between these two pixels by calculating the distance between them with (R1), and comparing it to a threshold value. Some techniques set this threshold value according to the number of color the arriving frame contains, I haven't tested this technique personally but it seems fairly wise. Therefore after this stage, the suspected moving objects are marked in white pixels, while the background is all set to black. Unfortunately, lights reflections, hardware imperfections, other unpredictable factors, will introduce some parasitical white pixels which don't correspond to real objects moving. We will try to remove those isolated white pixels to purify the obtained image.
Thus we will pass the resulting black and white image through an aliasing filter, which is meant to remove noise. This filter will remove all the isolated noisy white pixels and leave, if there are, the big white zones corresponding to moving objects. Basically we scan the image with a window of NxN pixels (depends on the size of the frames), if the center of the pixel of the NxN square is white, and less than K pixels in the window are set to white then it is set back to black because it is probably just an isolated noise not corresponding to a concrete object moving. Oppositly if more than K pixels are set to white and the center pixel is white we leave it as it is. However, this requires a lot of pixel scanning N²M², where M is the dimension of the frames; therefore we will rather scan the image by squares of NxN, for each of them count the number of white pixels they contain, if it superior to the treshhold value the whole NxN square is set to white on the bitmap else we set it black, we then jumps to the next NxN square. With this technique we limit the pixels access to M² which is a lot better. The output image will seem very squary but this is not a problem.
At that point we should have a black image with perhaps some white zones corresponding to moving objects in the fields of the camera. Thus we can already answer to the basic question: is there a moving object in the camera's field? A simple test on the presence of white pixels on the image will answer. Here is an example stage by stage of the processing:
Picture 10: The static background image of the camera's field
Picture 11: My dog (yes, the weird shape in black) has jumped on the bed to look if there isn't something to eat in the red bag
Picture 12: After comparing each pixel of the current frame with the static background we already have a good idea of the moving shape. Notice all the noise (small isolated pixels in white) due to the poor quality of my webcam. The fact that my dog is searching in the bag makes it move a little, which is also detected on the right of the image.
Picture 13: The aliasing filter has removed all the noise and parasites. We finally isolated the moving shape and we are now capable of giving its exact position, width, height etc…
### c – Unknown background (dynamic) motion detection
The case of unknown background motion detection is pretty similar to the previous situation. This becomes useful when you cannot predict the background of the camera's field and therefore cannot apply the previous algorithm. Notice that we will not study here, the motion detection algorithms in which the camera is moving permanently. Here the camera doesn't actually move, nor does it use a background reference like just before. Instead of comparing each received frame to a reference static background we are going to compare the frames received together. This technique is known as frame differentiating.
This is how the algorithm works on a streaming source of frames coming from a video camera: We will store a received frame in bitmap A, wait for the next n frames to pass without doing anything, capture the n+1 frame, compare it to A, store the n+1 frame in A, wait for n frames to pass, capture the 2n+2 frame, compare to A, store in A, and so on. Therefore we will regularly compare two frames together, seperated by n other frames that we will simply loose. This n value, the comparison rate, must be low enough not to miss some moving objects which would go right through the camera's field, but it also must be high enough to see differences between frames if a very slow object is moving in the camera's field. For the example of my webcam and my dog, I took n=10, with a streaming frame rate of 30 fps, this corresponds to 1/3 secs.
When we compare the two frames together we use the same method as seen above, if a pixel changes a lot of color from one frame to another we set it as white, else we leave it in black. We then make the resulting image pass through an aliasing filter and finally we are able to spot the moving objects.
Obviously this technique is less precise than the previous one because we don't spot directly the moving object in itself but we detect the changing zones in the frames and assimilate those changing zones to a part of background which has been recovered recently by a part of the object. Therefore we will only see edges of moving objects. Here is an example of results:
Picture 14: One of the frames we are going to compare (bitmapA)
Picture 15: The second frame to compare, we have skipped 10 frames ever since we saved bitmapA. Therefore the dog has had the time to move and we will see a difference between the frames.
Picture 15: The algorithm sees the zones which have changed colors: corresponding to the head of the dog walking and to my hand I was agitating.
Picture 16: After the aliasing filter only the main zones are left and enable us to situate roughly the moving objects on the camera's field.
## 1 – Shape recognition
### a – Shape recognition by convolution
The aim of shape recognition is to make the computer recognize particular shapes or patterns in a big source bitmap. Like the audio signal with speech recognition, this is far from beeing a trivial issue. Lots of powerful but also very complex methods exist to recognize shapes in an image. The one we are going to study here is very basic and makes a number of assumptions to work correctly. The shape we are going to try to situate on a source image must not be rotated nor distorted on this actual image.This limits a lot the use of this algorithm because in real situations the pattern you search in an image is always a bit rotated or distorted. In one dimensional signal correlation between two signals quantifies how similar they are; in two dimensions the principle is the same. Computing the correlation of the source image and the pattern we are looking for while give us an output image with a peak of white corresponding to the position of the pattern in the image. In fact we make our source image pass through a filter which kernel is the pattern we are looking for. When computing the output image pixel by pixel at one point the kernel and the position of the searched pattern in the image will correspond perfectly, this will give us a peak of white we will be able to situate.
Therefore the implementation of the algorithm is very simple; we compute the matrix convolution of the source image with the kernel, constituted of the pattern we want to search. In this case you will almost be forced to use the FFT convolution, indeed the searched patterns are often more then 30 pixels large, and for kernels of over 30x30 it's faster to use an FFT rather than straight convolution. I tried this algorithm without the FFT optimization and it is very very slow: about one transform per minute!
However the results themselves are quite remarkable, the position of the pattern is well recognized, sometimes with a bit of noise though. A nice way to enhance the results and make the peaks even more precise is to make the kernel pass through an edge enhancement filter; this will make the kernel more selective and precise. Here are the results of the algorithm:
Picture 17: Our source image: will you find the ball before the computer does?
Picture 18: The pattern the algorithm will be looking for.
Picture 19: The white peak corresponding to the ball's position.
Picture 20: Another source image.
Picture 21: The pattern the algorithm will be looking for.
Picture 22: The white peak corresponds to the positions of the patterns.
# Conclusion
Digital image processing is far from being a simple transpose of audio signal principles to a two dimensions space. Image signal has its particular properties, and therefore we have to deal with it in a specific way. The Fast Fourier Transform, for example, which was such a practical tool in audio processing, becomes useless in image processing. Oppositely, digital filters are easier to create directly, without any signal transforms, in image processing.
Digital image processing has become a vast domain of modern signal technologies. Its applications pass far beyond simple aesthetical considerations, and they include medical imagery, television and multimedia signals, security, portable digital devices, video compression, and even digital movies. We have been flying over some elementary notions in image processing but there is yet a lot more to explore. If you are beginning in this topic, I hope this paper will have given you the taste and the motivation to carry on.
# Appendix A : Last minute adds
Figure 9: Plot of the Tranforms per second with the color extraction algorithm according to the number of pixels of the source bitmap (/1000). It is fair to conclude that the fps decrease in an hyperbolic way with the number of pixels in the source bitmap. Notice also that it doesn't have anything to do with the complexity of the algorithm which is a straight according to the number of pixels (a*N²).
Report Article
## User Feedback
There are no comments to display.
## Create an account
Register a new account
• ### Game Developer Survey
We are looking for qualified game developers to participate in a 10-minute online survey. Qualified participants will be offered a \$15 incentive for your time and insights. Click here to start!
• 0
• 0
• 4
• 3
• 16
• 11
• 23
• 42
• 75
× | 14,011 | 59,685 | {"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} | 2.59375 | 3 | CC-MAIN-2019-43 | latest | en | 0.898907 |
https://homework.cpm.org/category/ACC/textbook/acc7/chapter/cc39/lesson/cc39.1.1/problem/9-13 | 1,571,025,491,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00113.warc.gz | 622,729,602 | 16,097 | ### Home > ACC7 > Chapter cc39 > Lesson cc39.1.1 > Problem9-13
9-13.
1. Determine whether the graphs below are functions or not functions. Explain your reasoning. Homework Help ✎
2. a. b. c. d. e. f.
A graph is only a function if there is no more than one y-value for any given x-value.
Are there any points on this line segment that are above or below another point on the line segment? If there are, the graph is not a function.
This graph is a function.
See hint for part (a).
This graph is not a function. Some x-values are related to multiple y values. For example, when x is 3, y can be 2 or ≈ 2.
See hint for part (a).
See hint for part (a).
See hint for part (a).
This graph is not a function.
See hint for part (a). | 201 | 737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.775479 |
http://en.boris-haase.de/bh_index.php?id=1016 | 1,508,836,575,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00062.warc.gz | 111,743,632 | 8,539 | ## Transcendental Numbers
The following section uses the notation from the chapter on Set Theory. We shall consider the behaviour of inconcrete m, n ∈ ωℕ, and we define h, k ∈ ℕ.
Remark: From algebra, we know that the sum, difference, product, and quotient of two algebraic numbers of degree h and k are algebraic of degree at most hk, and that the 1/h-th power of an algebraic number of degree k is algebraic of degree at most hk.
Remark: Transcendental numbers can be viewed as the sum of an algebraic part and a transcendental remainder. When investigating whether a number is transcendental, if the remainder may be expressed as the limit value of a zero sequence (ak) (see Nonstandard Analysis), we cannot simply disregard the values of the sequence for large k: They are important. Transcendental numbers are the numbers that lie between algebraic numbers or on either side of them. By the counting theorem for algebraic numbers, if two distinct transcendental numbers (algebraic of degree h) are sufficiently close, there is no algebraic number (of degree < h) between them.
Bounding theorem for ω-transcendental numbers: Every non-zero complex number whose imaginary or real part has absolute value is ≤ 1/⌈ω⌉ or ≥ ⌈ω⌉, is automatically ω-transcendental.
Proof: In a polynomial equation, set am = 1 and ak = -⌊ω⌋ for k < m, then the claim in the real case follows from the geometric series formula after taking the reciprocal. We can find the exact limit value by replacing ⌈ω⌉ by ω(m) = ⌈ω⌉ - ⌊ω⌋/ω(m)m with ω := ⌈ω⌉ - ⌊ω⌋/ω⌊ω⌋. In the complex case, substitutions of the form x = (1+bi)⌈ω⌉ with b ∈ ωℝ give the desired result.⃞
Coefficient theorem for ω-transcendental numbers: Every normalised irreducible polynomial and series such that |ak| ≥ ⌈ω⌉ for at least one ak only has ω-transcendental zeros.
Proof: The zeros of normalised irreducible polynomials and series are pairwise distinct and uniquely determined. Since they are not ω-algebraic, they must be ω-transcendental.⃞
Product and sum theorem for ω-transcendental numbers: Every unsimplifiable product (or sum) of an infinite complex-rational number a and a complex algebraic or infinite algebraic number b, and every sum and product of an infinite complex algebraic number and a complex ω-algebraic (infinite complex algebraic) number with coprime polynomial or series degrees is ω-transcendental.
Proof: Dividing by an infinite complex-rational number leads always to at least one coefficient with modulus ≥ ⌈ω⌉ in the corresponding minimal polynomial or series. Requiring it to be coprime ensures that there is always at least one infinite algebraic number in the result.⃞
Remark: This simplification is unique if it includes the minimum of the modulus of the coefficients of the minimal polynomial or series of b and every numerator and denominator of the infinite complex-rational number a. Together with the rational numbers, the infinite (complex-)rational numbers already numerically make up the entirety of ℝ (ℂ). Therefore, algebraic and transcendental numbers are (numerically) difficult to distinguish, and approximations are of little use when determining whether a number is algebraic (to a certain degree). The merit of distinguishing between them is therefore questionable. Real continued fractions that do not terminate as a rational number are ω-transcendental, since they are infinite rational. They can only be used as an approximation of ω-algebraic numbers.
Remark: In the case of ω-transcendental numbers we must satisfy ourselves with an arbitrarily precise infinitesimal number, unlike in the case of ω-algebraic numbers, where we can argue using the corresponding minimal polynomial or series. We can therefore represent them in the form of a rational quotient with infinite numerator and denominator. However, when investigating the transcendence of a number using Liouville's approximation theorem, care should be taken to ensure that the approximating rational numbers are in fact conventionally rational and not infinitely rational, such as for example the number 10⌈ω⌉. Similarly, confidence with infinite natural numbers is required when considering prime numbers ≥ ⌈ω⌉ in proofs of transcendence. The following proofs can also be used to show ω-transcendence by replacing the set ωℕ by [⌈ω⌉, n]ℕ for some n that is not excessively large.
Theorem: The sum of all at least conventionally natural powers of a non-zero complex-rational number x that is not a root of unity (geometric series) is already ω-transcendental.
Proof: The modulus of either the numerator or denominator of x⌈ω⌉ is ≥ 2⌈ω⌉/2. Subtracting 1 and dividing by 1 - x preserves ω-transcendence.⃞
Theorem: Euler's number e is ω-transcendental.
Proof: If we accept the exponential series as a representation for e, it follows that e = (k⌊ω⌋+1)/⌊ω⌋! for k ≥ ⌈ω⌉. Therefore, the numerator and the denominator of this fraction must be ≥ ⌈ω⌉, since neither ⌊ω⌋ nor a prime divisor of k in the numerator simplifies with ⌊ω⌋!. However, if we accept the representation (1+1/⌊ω⌋)⌊ω⌋ for e, the claim is trivial. Note that these two representations give different numbers.⃞
Theorem: The twin prime constant C2 is ω-transcendental as the product of (1 - 1/(p - 1)2) over all primes p ∈ ℙ>2.
Proof: By the prime number theorem (see Number Theory), neither the largest nor the second-largest prime number in ℕ - both of which are = ⌈ω⌉ - |O(ln ⌈ω⌉)| - divide the denominator of C2.⃞
Theorem: The Landau-Ramanujan constant K is ω-transcendental as the product of (1 - 1/p2) over all p ∈ ωℙ such that p ≡ 3 mod 4 divided by √2.
Proof: By the (Dirichlet) prime number theorem, neither the largest nor the second-largest prime number ≡ 3 mod 4 of ωℙ - both of which are = ⌈ω⌉ - |O(ln ⌈ω⌉)| - divide the denominator of K.⃞
Theorem: The Glaisher-Kinkelin constant A = 11 22 33 ... ⌊ω⌋⌊ω⌋ (1+1/⌊ω⌋)⌊ω⌋³/4/⌊ω⌋(⌊ω⌋²/2+⌊ω⌋/2+1/12) is ω-transcendental.
Proof: After simplifying, the largest prime number of ωℕ remains in the denominator with exponent > 2.⃞
The greatest-prime criterion for ω-transcendental numbers: If a real number r may be represented as an irreducible fraction a/(bp) ± s/t where a, b, s and t are natural numbers, abst ≠ 0 and b + t > 2, and the (second-)greatest prime number p ∈ ωℙ, p ∤ a and p ∤ t, then r is ω-transcendental.
Proof: We have that r = (at ± bps)/(bpt) with denominator ≥ 2p ≥ 2⌈ω⌉ - |O(ln ⌈ω⌉)| > ⌈ω⌉ by the prime number theorem.⃞
Theorem: Pi π is ω-transcendental.
Proof: This follows from its Wallis product representation, or its product representation using the gamma function with value -½, provided that we accept these representations. It should be noted that these two representations yield distinct numbers. Alternatively, we can apply the greatest-prime criterion to the Leibniz series, or the Taylor series of arcsin(x) at x = 1.⃞
Theorem: The trigonometric and hyperbolic functions and their inverse functions, the digamma function ψ, the Lambert-W function, the Ein function, the (hyperbolic) sine integral S(h)i, Euler's Beta function B, and, for positive natural numbers s and u and natural numbers t the generalised error function Et, the hypergeometric functions 0Ft, the Fresnel integrals C and S and the Bessel function It and the the Bessel function of the first kind Jt, the Legendre function χs, the polygamma function ψs, the generalised Mittag-Leffler function Es,t, the Dirichlet series Σ f(u)/us over u with maximally finite rational |f(u)|, the prime zeta function P(s), the polylogarithm Lis, and the Lerch zeta-function Φ(q, s, r) always yield ω-transcendental values for rational arguments and maximal finite rational |q| and |r| at points where their Taylor series converge.
Proof: The claim follows from the greatest-prime criterion, the Dirichlet prime number theorem, and the Wallis product. For the digamma function, the claim follows from the proof of ω-transcendence of Euler's constant below.⃞
Theorem: The gamma function Γ(z) := m!mz/(z(z + 1) … (z + m)), where m = ⌈ω⌉⌈ω⌉² and z ∈ ωℂ \ -ωℕ, is ω-transcendental for z ∈ ωℚ and for suitable supersets of ωℕ resp. ωℚ.
Proof: The values of Γ(z) are zeros of minimal polynomials or series with infinite integer coefficients.⃞
Theorem: For x ∈ ωℝ, let s(x) be the sum of xk/k over all k ∈ ωℕ*. If we define Euler's constant as γ = s(1) - ln ⌊ω⌋ ∈ ]0, 1[ (since dx/⌊x⌋ - dx/x ≥ 0 and dx/⌊1+x⌋ - dx/x ≤ 0 for x ∈ ω≥1 in the integral representation of γ) and accept m s(½) - s(r/2m) as a representation of ln ⌊ω⌋ with ⌊ω⌋ = 2m - r, m ∈ ωℕ and r ∈ [0, 2m-1[ with a precision of O(m/2⌈ω⌉), then γ is infinite rational and therefore ω-transcendental.
Proof: We obtain -ln(1 - x) = s(x) + O(x⌈ω⌉/(1 - x)) + t(x)dx for x ∈ [-1, 1 - 1/κ] and a real function t(x) such that |t(x)| < ⌈ω⌉ by (exact) integration (see Nonstandard Analysis) of the geometric series. The claim follows from the greatest-prime criterion after applying Fermat's little theorem to the denominator of the k-th summand of s, for the largest or second-largest prime in ωℕ, whose product is greater than m 2m-1 + 2m - r by the prime number theorem.⃞
Remark: At all (higher) precisions, the numerators of the summands contain higher powers of two (after cancelling with the denominator where applicable), so the theorem remains valid. This in particular includes t(x)dx at almost any arbitrary precision, by successively reducing γ to a fraction. If we set dx to be the reciprocal of the maximal power of two in this configuration, with an infinite natural exponent for infinite rational t(x), the claim even holds exactly. In any case, the statement is valid for infinitely many levels of infinity of n, and in particular holds from the conventional perspective.
Definition: When two numbers x, y ∈ ωℂ* or their reciprocals do not satisfy any polynomial or series equation p(x, y) = 0, so they are called ω-algebraically independent.
Theorem: By the greatest-prime criterion, with e = (1+1/p)p for maximal p ∈ ωℙ and π as Wallis product, pairwise ω-algebraically independent representations of A, C2, γ, e, K and π do exist.⃞
Theorem: The BBP series Σp(k)/(q(k)bk) over k for b, k ∈ ωℕ and polynomials resp. series p and q ∈ ωℤ with q(k) ≠ 0 and deg(p) < deg(q) only yield ω-transcendental values.
Proof: We can reduce the sum to a smallest common denominator d ≥ bm > ω with d, m ∈ ℕ*.
Remark: Since all real numbers are (approximately) (infinite) rational numbers, we can compute them in real-time.
Approximation theorem for real algebraic numbers: Every real irrational algebraic number of degree k may be approximated by a real ω-algebraic number of degree h < k with an average asymptotic error of π ζ(h+1)/(2 |ωℤ|h ln h).
Proof: On the conventionally real axis, the number of ω-algebraic numbers approximately evenly distributed between fixed limits increases by a factor of approximately |ωℤ| per degree. The error corresponds to the distance between ω-algebraic numbers. The non-real ω-algebraic numbers are less dense.⃞
Conclusion: Two distinct real ω-algebraic numbers have an average distance of at least π/(|ωℤ|⌊ω⌋ ln ⌈ω⌉). Determining this minimum distance exactly requires an infinite non-linear non-convex optimisation problem to be solved. Therefore, the κ-algebraic numbers have an approximate order of O(κ), disproving Roth's theorem, which essentially amounts to proving the (trivial) minimum distance between two rational numbers, and thus disproves the abc conjecture, but not Liouville's result.
Theorem: The maximum distance between two neighbouring real ω-algebraic numbers ≠ 0 amounts 1/⌊ω⌋2 - |O(1/⌊ω⌋3)|.
Proof: The distance between two real ω-algebraic numbers ≠ 0 is largest around the points ±1. A ω-rational number r > 1 (0 < r < 1) may be better approximated by the real ω-algebraic x that satisfies the polynomial or series equation ⌊ω⌋/r xm - ⌊ω⌋xm-1 = 1 (xm - ⌊ω⌋x = -⌊ω⌋r). An analogous result holds for negative r. If we wish to approximate 1 by a larger real ω-algebraic x, it must necessarily satisfy a polynomial or series equation such that am = ⌊ω⌋ - 1 and a0 = -⌊ω⌋. By setting am = ⌊ω⌋ - 1 and a0 = -⌊ω⌋ the claim follows, since the maximum distance can no longer be reduced.⃞
Gelfond-Schneider theorem: For each α ∈ ωA\{0, 1} and β ∈ ωA\ωℚ, αβ is ω-transcendental.
Proof: In order for αβ to be zero of a minimal polynomial or series, since B := ωA is a field, we would have to perform the invalid substitution α := ξηq/β with ξ, η ∈ B* and q ∈ ωℚ*.⃞
The above theorem and the property that B is a field immediately imply the following result.
Conclusion: As before, given any α, there does not exist any β such that e = αβ, which may be seen by comparing the minimal polynomials or series of e (see above) and αβ. If there is precisely one γ ∈ B* for any given arbitrary ω-transcendental τ ∈ ωℂ such that τγ is ω-algebraic, then every τδ such that δ ∈ B\γωℚ is ω-transcendental.⃞
Definition: A rational number ≠ 0 is said to be power-free if it cannot be represented as the power of a rational number with integer exponent ≠ ±1.
Theorem: For any power-free q ∈ Q := ℚ>0, we have that r := qx ∈ Q, if and only if x ∈ ωℤ and |x| is not excessively large.
Proof: Since there is no contradiction when ±x ∈ ωℕ is not excessively large, we can similarly assume wlog that x ∈ Q\ωℕ. Since this implies qxωAR\Q with R := ℝ>0, we can similarly assume that x ∈ ωAR\Q. Then qx is ω-transcendental by the Gelfond-Schneider theorem. We can therefore assume that x := a/b ∈ ωℝ\ωAR where a, b ∈ ℤ are ω-transcendental. This implies that qa/b ∈ Q\ωℕ* and thus a = cb with c ∈ ℤ, since q is power-free. We deduce the contradiction that qc = qx which proves the claim, since x must be real. ⃞
Remark: These arguments can be extended to finite transcendental numbers by replacing κℕ with ωℕ and making the required adjustments. Inconcrete transcendence implies finite transcendence.
Remark: The above theorem proves the Alaoglu and Erdős conjecture, which states that, given distinct p, q ∈ κℙ, px and qx are κ-rational if and only if x is a κ-integer.
Remark: Conventional differentiation and integration loses the ability to distinguish between transcendental and algebraic in the conventional process of taking limits. This is problematic, for example when we wish to exactly determine the roots of a polynomial. Therefore, the chapter on Nonstandard Analysis on this homepage adopts a different (more precise) approach. | 4,047 | 14,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-43 | latest | en | 0.884542 |
https://aviation.stackexchange.com/questions/39761/how-is-the-mass-flow-between-2-streamlines-derived?answertab=votes | 1,563,831,571,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528220.95/warc/CC-MAIN-20190722201122-20190722223122-00110.warc.gz | 316,243,923 | 35,211 | # How is the mass flow between 2 streamlines derived?
I am reading "Fundamental of Aerodynamics " by J.D.Anderson Fifth edition, At pages 177 and 178 (section 2.14 Stream funciton), the book said: given 2D steady flow, from: $$\frac{dy}{dx}=\frac{v}{u}$$ with u and v are known function of x and y, then we obtain f(x,y) = c by intergration.
If we have steamline 1 is f(x,y) = c1 and streamline 2 is f(x,y) = c2, then why is the mass flow between two streamlines equal (c2 - c1)? The book said that but there is no derivation of that.
As far as I can do:
Short answer: Because that's how we've defined $c_1$ and $c_2$.
Long answer: Read 2.14 carefully. As you've said, we have $f(x,y)=c$ via integration. Anderson then chooses $\bar{\psi}$ as $f$ (a convention) to get $\bar{\psi}=c_1$ and $\bar{\psi}=c_2$ for the two streamlines. However, we know that $c_1$ and $c_2$ are just arbitrary constants of integration that we can set to whatever we want. Anderson just happens to know that later down the road it will be helpful to
define the numerical value of $\bar{\psi}$ such that the difference $\Delta\bar{\psi}$ [between the two streamlines] is equal to the mass flow between the two streamlines.
$\Delta\bar{\psi}$ equaling mass flow (per unit depth) between streamlines is natural. For a steady flow, the mass flow inside a given streamtube is constant along the tube; the mass flow across any cross section of the tube is the same. Since $\Delta\bar{\psi}$ is equal to this mass flow, then $\Delta\bar{\psi}$ itself is constant for a given streamtube. | 435 | 1,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-30 | latest | en | 0.939013 |
https://web2.0calc.com/questions/triangle_49 | 1,619,176,799,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00099.warc.gz | 687,898,838 | 5,486 | +0
# triangle
0
112
1
If UV is 4 and UT is 6.25, what is UW? Round to the nearest thousandth.
Feb 21, 2021
#1
+118069
+1
Note that triangle UVW is similar to triangle UWT
Therefore
UV / UW = UW / UT
4/ UW = UW / 6.25
So
UT * UV = UW * UW
6.25 * 4 = UW^2
25 = UW^2 take the positive root
5 = UW
Feb 21, 2021 | 152 | 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} | 3.796875 | 4 | CC-MAIN-2021-17 | latest | en | 0.698315 |
https://acervodigital.unesp.br/handle/11449/91780?mode=full | 1,620,415,696,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00624.warc.gz | 108,478,559 | 9,322 | You are in the accessibility menu
Please use this identifier to cite or link to this item: `http://acervodigital.unesp.br/handle/11449/91780`
DC FieldValueLanguage
dc.contributor.authorMatulovic, Mariana-
dc.date.accessioned2014-06-11T19:25:28Z-
dc.date.accessioned2016-10-25T19:06:58Z-
dc.date.available2014-06-11T19:25:28Z-
dc.date.available2016-10-25T19:06:58Z-
dc.date.issued2008-07-14-
dc.identifier.citationMATULOVIC, Mariana. A lógica do muito em um sistema de tablôs. 2008. 121 f. Dissertação (mestrado) Universidade estadual Paulista. Faculdade de Filosofia e Ciências, 2008.-
dc.identifier.urihttp://hdl.handle.net/11449/91780-
dc.identifier.urihttp://acervodigital.unesp.br/handle/11449/91780-
dc.description.abstractAmong the several non classical logics that complement the classical first-order logic, we detach the Modulated Logics. This class of logics is characterized by extending the classical logic by the introduction of a new generalized quantifier, called modulated quantifier, that has the attribution of interpreting some inductive aspects of quantifiers in any natural language. As a particular case of Modulated Logic, the Logic of Many formalize the intuitive notion of “many”. The quantifier of many is represented by G. Thus, a sentence of the type Gxα(x) must be understood like “many individuals satisfy the property α”. Semantically, the notion of many is associated with a mathematical structure named proper superiorly closed family. Let E be a non empty set. A proper superiorly closed family F in E is such that: (i) F ⊆ P(E); (ii) E ∈ F; (iii) ∅ ∉ F; (iv) A ∈ F e A ⊆ B ⇒ B ∈ F. Intuitively, F characterizes the sets which have “many” elements. The empty set ∅ does not have many elements. And if A has many elements, then any set which contains A, also has many elements. The logic of many has syntactical elements that caracterize linguisticaly these properties of F. We can verify that the Logic of Many is correct and complete for a first order structure extended by a proper superiorly closed family. The Logic of Many was originally introduced in a Hilbertian deductive system, based only on axioms and rules. In this work, we developed another deductive system for the Logic of Many, but in a tableaux system. We proof that this new system is equivalent to the original one.en
dc.format.extent121 f. : il.-
dc.language.isopor-
dc.sourceAleph-
dc.subjectLógicapt
dc.subjectLógica do muitopt
dc.subjectTablôs analíticospt
dc.subjectModulated logicsen
dc.subjectLogic of manyen
dc.subjectTableaux systemen
dc.titleA lógica do muito em um sistema de tablôspt
dc.typeoutro- | 693 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-21 | latest | en | 0.772358 |
https://ch.mathworks.com/matlabcentral/cody/problems/1442-volume-of-a-simplex/solutions/1217493 | 1,571,110,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655864.19/warc/CC-MAIN-20191015032537-20191015060037-00013.warc.gz | 424,086,374 | 15,711 | Cody
# Problem 1442. Volume of a Simplex
Solution 1217493
Submitted on 21 Jun 2017 by LY Cao
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 0; V_correct = 1.0; assert(abs(simplexvolume(n)-V_correct)<4*eps)
2 Pass
n = 1; V_correct = 1.0; assert(abs(simplexvolume(n)-V_correct)<4*eps)
3 Pass
n = 2; V_correct = 0.433012701892219; assert(abs(simplexvolume(n)-V_correct)<4*eps)
4 Pass
n = 3; V_correct = 0.117851130197758; assert(abs(simplexvolume(n)-V_correct)<4*eps)
5 Pass
n = 4; V_correct = 0.023292374765623; assert(abs(simplexvolume(n)-V_correct)<4*eps)
6 Pass
n = 5; V_correct = 0.003608439182435; assert(abs(simplexvolume(n)-V_correct)<4*eps)
7 Pass
n = 6; V_correct = 0.000459331824838; assert(abs(simplexvolume(n)-V_correct)<4*eps)
8 Pass
n = 7; V_correct = 0.000049603174603; assert(abs(simplexvolume(n)-V_correct)<4*eps)
9 Pass
n = 8; V_correct = 0.000004650297619; assert(abs(simplexvolume(n)-V_correct)<4*eps)
10 Pass
n = 9; V_correct = 0.000000385125244; assert(abs(simplexvolume(n)-V_correct)<4*eps)
11 Pass
n = 10; V_correct = 0.000000028561653; assert(abs(simplexvolume(n)-V_correct)<4*eps)
12 Pass
n = 11; V_correct = 0.000000001917653; assert(abs(simplexvolume(n)-V_correct)<4*eps)
13 Pass
n = 12; V_correct = 0.000000000117613; assert(abs(simplexvolume(n)-V_correct)<4*eps)
14 Pass
n = 13; V_correct = 0.000000000006639; assert(abs(simplexvolume(n)-V_correct)<4*eps)
15 Pass
n = 14; V_correct = 0.000000000000347; assert(abs(simplexvolume(n)-V_correct)<4*eps)
16 Pass
n = 15; V_correct = 0.000000000000017; assert(abs(simplexvolume(n)-V_correct)<4*eps) | 636 | 1,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-43 | latest | en | 0.385731 |
http://list.seqfan.eu/anumref/A001359.html | 1,582,410,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145729.69/warc/CC-MAIN-20200222211056-20200223001056-00343.warc.gz | 92,535,826 | 2,066 | ### References to A001359
Lesser of twin primes.
3, 5, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 227, 239, 269, 281
Index of A-numbers in seqfan: by ascending order by month by frequency by keyword
23 seqfan posts
Wed Jan 18 05:51:59 CET 2012 [seqfan] Re: Large gaps between primes
Sun Dec 18 17:22:01 CET 2011 [seqfan] Re: Naming issue
Sun Dec 18 15:28:35 CET 2011 [seqfan] Naming issue
Thu Aug 12 21:07:41 CEST 2010 [seqfan] Re: Is this right? Main diagonal A[n.n] of array A[k, n] = n-th natural number m such that m and m+2 are both divisible by exactly k primes (counted with multiplicity).
Thu Aug 12 20:48:48 CEST 2010 [seqfan] Re: Is this right? Main diagonal A[n.n] of array A[k, n] = n-th natural number m such that m and m+2 are both divisible by exactly k primes (counted with multiplicity).
Thu Aug 12 20:24:17 CEST 2010 [seqfan] Re: Is this right? Main diagonal A[n.n] of array A[k, n] = n-th natural number m such that m and m+2 are both divisible by exactly k primes (counted with multiplicity).
Thu Aug 12 20:05:40 CEST 2010 [seqfan] Is this right? Main diagonal A[n.n] of array A[k, n] = n-th natural number m such that m and m+2 are both divisible by exactly k primes (counted with multiplicity).
Fri Jan 1 23:25:09 CET 2010 [seqfan] Re: Conjectures relating to twin primes and Lucas numbers
Wed Dec 30 23:01:07 CET 2009 [seqfan] Conjectures relating to twin primes and Lucas numbers
Tue Jun 9 08:57:45 CEST 2009 [seqfan] Re: Twin prime conjecture conjecture
Tue Jun 9 02:04:16 CEST 2009 [seqfan] Re: Twin prime conjecture conjecture
Sun Jun 7 18:07:29 CEST 2009 [seqfan] Twin prime conjecture conjecture
Tue Jun 2 00:24:25 CEST 2009 [seqfan] Re: more digits of sum over squared inverse twin primes
Sun May 31 06:59:23 CEST 2009 [seqfan] Re: more digits of sum over squared inverse twin primes
Sat May 30 19:10:30 CEST 2009 [seqfan] more digits of sum over squared inverse twin primes
Fri Dec 19 10:52:37 CET 2008 [seqfan] Re: Q about A152926
Wed Dec 17 16:56:22 CET 2008 [seqfan] Re: Q about A152926
Wed Dec 17 14:15:02 CET 2008 [seqfan] Re: Q about A152926
Wed Dec 17 09:39:43 CET 2008 [seqfan] Re: Q about A152926
Mon Dec 15 19:00:12 CET 2008 [seqfan] Re: Q about A152926
Mon Dec 15 18:33:39 CET 2008 [seqfan] Re: Q about A152926
Mon Dec 15 18:33:39 CET 2008 [seqfan] Re: Q about A152926
Mon Dec 15 17:30:15 CET 2008 [seqfan] Q about A152926
Index of A-numbers in seqfan: by ascending order by month by frequency by keyword
Links to OEIS content are included according to The OEIS End-User License Agreement . | 926 | 2,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.819969 |
https://gateoverflow.in/220855/madeeasy-test-series-databases-natural-join | 1,670,434,402,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00710.warc.gz | 300,014,219 | 23,433 | 3,431 views
Consider a relation r1(ABC), r2(CDE) and r3(FG) with primary keys A, C and F respectively. Assume that r1 has 150 tupples, r2 has 100 tupples and r3 has 75 tupples. The number of resultant tuple in are ________.
The result of natural join = Cartesian pdt when there are no common attributes isn't it ?
yeah you are right
In $r_1\Join r_2$ $C$ is common ,in which C is foreign key ( assuming that foreign key is present) in r1 so in natural join of r1 and r2 all record of r1 will be present so $r_1\Join r_2$ has maximum no. of tupples 150 .let say $r_1\Join r_2$ as S ,now join of S and r3, in which no element is common( assuming all attributes have diffrent quality as in question nothing is mentioned) so $S\Join r_3$ will become $S \textbf{*}r_3$(cross product) .now every tupple of S will relate to every tupple of $r_3$ ,so maximum no. of tupples =150*75=11250.
Yes!
Then the number of resultant tuples will be less than 11250 right?
I think so.
1 vote | 319 | 979 | {"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} | 3.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.869187 |
https://www.xarg.org/puzzle/codingame/vortex/ | 1,701,684,554,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100527.35/warc/CC-MAIN-20231204083733-20231204113733-00688.warc.gz | 1,192,648,501 | 6,184 | # Codingame Solution: Vortex
## Goal
You must crank an WxH matrix counterclockwise by X positions. To crank a matrix, you must shift each matrix element along the matrix rectangle. For example, for a 7x5 matrix with X = 1 the movements are performed this way:
v < < < < < <
v v < < < < ^
v v 0 0 0 ^ ^
v > > > > ^ ^
> > > > > > ^
where the characters < ^ > v represent the displacement of a value in that position; 0 means no displacement.
Input
Line 1: 2 integers W and H representing the width and height (respectively) of a matrix.
Line 2: An integer X representing the number of positions to crank the matrix.
Next H lines: W matrix values v per line.
Output
H lines: The cranked matrix.
Constraints
1 ≤ W,H ≤ 50
1 ≤ X ≤ 100000000
-100 ≤ v ≤ 100
## Solution
I initially wanted to come up with a mapping function, which takes the coordinate of a matrix cell and $$X$$ and calculates the new position. It turned out to be a quite complicated function and I stoppd the work on it, since it wasn't that readable anymore. The actual solution I finally came up with is the most intuitive. I start at position (0, 0) and go downwards as long as I can, then continue to the right, to the up and back to the left. For each iteration, I add the elements to an array per ring. A ring is just one round around the matrix at a certain depth. On the linear arrays I can add $$X$$ positions modulo the length of the ring and finally bring it back to the original matrix format. But let's do it step by step. We start with reading in the needed information:
function mod(n, m) {
return (n % m + m) % m;
}
var inputs = readline().split(' ')
var w = +inputs[0];
var h = +inputs[1];
var k = +readline();
var x = 0
var y = 0
var dx = 0, dy = 1
var ring = 0
var rings = []
var map = []
for (var i = 0; i < h; i++) {
}
What now follows is exactly what I just have described. I go down, right, up left and continue that pattern one ring deeper in a spiral way:
var cur = []
for (var i = 0; i < w * h; i++) {
cur.push(map[y][x])
if (dx === 0 && dy === 1 && y === h - 1 - ring) {
dx = 1
dy = 0
} else if (dx === 1 && dy === 0 && x === w - 1 - ring) {
dx = 0
dy = -1
} else if (dx === 0 && dy === -1 && y === 0 + ring) {
dx = -1
dy = 0
} else if (dx === -1 && dy === 0 && x === 1 + ring) {
dx = 0
dy = 1
rings.push(cur)
cur = []
ring++
}
x+= dx
y+= dy
}
rings.push(cur)
We now have all rings in a linear way, run over all of these rings and subtract $$X$$ (which I call $$k$$ in the source) as long as we are not of the inner ring of an odd height matrix. After that I screw DRY and walk again in a spiral pattern to write back the shifted data:
var x = 0
var y = 0
var dx = 0, dy = 1
for (var ring = 0; ring < rings.length; ring++) {
if (h % 2 == 1 && ring == rings.length - 1) {
// void
} else {
rings[ring] = rings[ring].map((a, i, map) => map[mod(i - k, map.length)])
}
for (var i = 0; i < rings[ring].length; i++) {
map[y][x] = rings[ring][i]
if (dx === 0 && dy === 1 && y === h - 1 - ring) {
dx = 1
dy = 0
} else if (dx === 1 && dy === 0 && x === w - 1 - ring) {
dx = 0
dy = -1
} else if (dx === 0 && dy === -1 && y === 0 + ring) {
dx = -1
dy = 0
} else if (dx === -1 && dy === 0 && x === 1 + ring) {
dx = 0
dy = 1
}
x+= dx
y+= dy
}
}
for (var i = 0; i < map.length; i++) {
print(map[i].join(" "))
}
« Back to problem overview | 1,103 | 3,339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 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} | 3.765625 | 4 | CC-MAIN-2023-50 | latest | en | 0.768272 |
https://forum.math.toronto.edu/index.php?PHPSESSID=5q1sfetirbptiocm7ausggthh4&action=printpage;topic=839.0 | 1,660,543,791,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572161.46/warc/CC-MAIN-20220815054743-20220815084743-00378.warc.gz | 254,486,855 | 3,038 | # Toronto Math Forum
## APM346-2016F => APM346--Tests => TT2 => Topic started by: Victor Ivrii on November 17, 2016, 03:22:16 AM
Title: TT2-P1
Post by: Victor Ivrii on November 17, 2016, 03:22:16 AM
Solve by Fourier method
\begin{align}
\end{align}
Hint: $\lambda_n\ge 0$. Also remember how solution looks like in the case of double eigenvalues.
Title: Re: TT2-P1
Post by: XinYu Zheng on November 17, 2016, 08:05:25 AM
Introduce $u=X(x)T(t)$ and then separation of variables yield $X''+\lambda X=0$ and $T''+\lambda T=0$. Assume that $\lambda=\omega^2$ where $\omega\geq 0$. Then for $\omega\neq 0$, we have solutions $X_1=\sin \omega x$ and $X_2=\cos\omega x$. Applying the boundary conditions to $X_1$ we find
$$0=\sin \omega \pi$$
$$-1=\cos\omega \pi$$
Applying the boundary conditions to $X_2$ yields the same thing. The first equation suggests that $\omega=n$ where $n=1,2,...$. The second equation suggests that $\omega=2n+1$, $n=0,1,2,...$. To satisfy both, we must take the second one. So we have $\omega_n=2n+1$, $n=0,1,2,...$. Now for $\omega=0$ the solution is $Ax+B$. But the second B.C. requires $B=-B$ so $B=0$. Then applying the first B.C. we find $A=-A$ so $A=0$. So we do not have this eigenvalue. Thus our solutions for the spacial part is
$$X_{1,n}=\sin((2n+1)x), X_{2,n}=\cos((2n+1)x)$$
With $\lambda_n=(2n+1)^2$, $n=0,1,2...$. Then the equation for time can be solved immediately: $T=A\sin((2n+1)t)+B\cos((2n+1)t)$. Applying $u_t|_{t=0}=0$ we find $A=0$, and now we may write down the general solution:
$$u(x,t)=\sum_{n\geq 0}[A_n\sin((2n+1)x)+B_n\cos((2n+1)x)]\cos((2n+1)t)$$
Applying the condition $u_{t=0}=1$ we have
$$1=\sum_{n\geq 0}A_n\sin((2n+1)x)+B_n\cos((2n+1)x)$$
At this point the coefficients can be calculated via the standard method using orthogonality:
$$A_n=\frac{2}{\pi}\int_0^\pi \sin((2n+1)x)\,dx=\frac{2}{\pi(2n+1)}\cos((2n+1)x)|_\pi^0=\frac{4}{\pi (2n+1)}$$
$$B_n=\frac{2}{\pi}\int_0^\pi \cos((2n+1)x)\,dx=0$$
Where the second integral is zero because we will be evaluating sine functions at integer values of $\pi$. So we have our solution:
$$u(x,t)=\frac{4}{\pi}\sum_{n\geq 0}\frac{1}{2n+1}\sin((2n+1)x)\cos((2n+1)t)$$ | 840 | 2,165 | {"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} | 4.34375 | 4 | CC-MAIN-2022-33 | latest | en | 0.716117 |
https://www.sanfoundry.com/engineering-physics-questions-answers-fermi-dirac-distribution/ | 1,656,474,840,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00459.warc.gz | 1,022,804,902 | 22,853 | # Engineering Physics Questions and Answers – Fermi-Dirac Distribution
«
»
This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Fermi-Dirac Distribution”.
1. Fermi-Dirac statistics is for the ________
a) Distinguishable particles
b) Symmetrical Particles
c) Particles with half integral spin
d) Particles with integral spin
Explanation: The Maxwell-Boltzmann statistics is for the distinguishable particles, which are basically the classical particles like atoms and molecules.
2. The difference between fermions and bosons is that bosons do not obey ______
a) Aufbau principle
b) Pauli’s Exclusion Principle
c) Hund’s Rule of Maximum Multiplicity
d) Heisenberg’s Uncertainty Principle
Explanation: The particles that follow Pauli’s Exclusion Principle are defined as Fermions while that don’t are called bosons. Bosons have an integral spin number.
3. The Maxwell-Boltzmann law is given by the expression ni = ________
a) $$\frac{g}{e^{\alpha+\beta E}}$$
b) $$\frac{g}{e^{\alpha+\beta E}-1}$$
c) $$\frac{g}{e^{\alpha+\beta E}+1}$$
d) $$\frac{g}{e^{\alpha+\beta E}+k}$$
Explanation: The correct expression for the Maxwell-Boltzmann law is ni = $$\frac{g}{e^{\alpha+\beta E}+1}$$, where α depends on the volume and the temperature of the gas and β is equal to 1/kT.
Note: Join free Sanfoundry classes at Telegram or Youtube
4. The wave function of fermions is not _________
a) Continuous
b) Single Valued
c) Symmetric
d) Differentiable
Explanation: The particles which have antisymmetric wave function having half odd integral spin number and obey Pauli’s principle are called fermions.
5. Fermi-Dirac statistics cannot be applied to ________
a) Electrons
b) Photons
c) Fermions
d) Protons
Explanation: Fermi-Dirac Statistics can be applied to particles having half odd integral spin number and obey Pauli’s principle which are electrons, fermions and protons. Photon has an integral spin number.
6. The distribution function is given by ________
a) $$\frac{1}{Ae\frac{E}{nT}+1}$$
b) $$\frac{1}{Ae\frac{E}{nT}-1}$$
c) $$\frac{1}{e\frac{E}{nT}+A}$$
d) $$\frac{1}{e\frac{E}{nT}-A}$$
Explanation: The distribution function is given by $$\frac{1}{Ae\frac{E}{nT}+1}$$, where A = $$e^{\frac{-E_F}{nT}}$$ The energy EF is called the Fermi energy and is constant for a given system.
7. At T > 0K, the probability of a state with E > EF filled is zero.
a) True
b) False
Explanation: At T > 0 K, the probability that a state with E > EF is filled is 12. Hence, fermi energy is the energy at which the probability of occupation is 12 at any temperature above 0 K.
8. The expression for mean energy is given by ________
a) $$\frac{3}{5}NE_F$$
b) $$\frac{2}{5}NE_F$$
c) $$\frac{3}{5}E_F$$
d) $$\frac{2}{5}E_F$$
Explanation: The expression for the mean energy of an electron at absolute zero is given by $$\frac{3}{5}NE_F$$. The expression $$\frac{3}{5}E_F$$ is called the zero point energy.
9. For all the quantum states with energy greater than Fermi energy to be empty in a Fermi-Dirac system, the temperature should be ______
a) 273 K
b) 373 K
c) 0 K
d) 100 K
Explanation: We know that the Fermi-Dirac distribution is given by:
FFD(E) = $$\frac{1}{e\frac{E-E_F}{nT}+1}$$
For all the quantum states with energy greater than Fermi energy to be empty,
FFD(E) = 0, for E > EF and FFD(E) = 1, for E < EF
Therefore, for E < EF $$\frac{1}{e\frac{E-E_F}{nT}+1}=1$$
$$e\frac{E-E_F}{nT}= 0$$
As, E < EF, E- EF < 0. Therefore, to satisfy the given statement,
T = 0 K
Thus, we can define Fermi energy as the energy of the uppermost occupied level at 0 K.
10. What is the relationship between T1 and T2?
a) T1 > T2
b) T1 < T2
c) T1 = T2
d) Insufficient Information
Explanation: The given figure shows the variation of Fermi-Dirac distribution with energy E. In this case, T2 > T1, according to the expression: FFD(E) =$$\frac{1}{e\frac{E-E_F}{nT}+1}$$.
11. The density of silver is 10.5 g/cm3 and its atomic weight is 108. If each atom contributes one electron for conduction, what is the fermi energy?
a) 2.12 eV
b) 3.31 eV
c) 4.69 eV
d) 5.51 eV
Explanation: Fermi Energy, $$E_F = \frac{h^2}{8m} (\frac{3N}{\pi v})^{2∕3}$$
Here, N /V = 10.5 X 6.02 X 1023/108
= 5.85 X 1028 m-3
Therefore, $$E_F = \frac{6.626 X 6.626}{8 X 9.1} \times 10^{-19} \times \frac{3 \times 5.585}{\pi}^{2∕3}$$
= 8.816 X 10-19J
= 5.51 eV.
12. The Fermi energy of a material is 3.45 eV. What is the zero-point energy of the material?
a) 1.02 eV
b) 2.07 eV
c) 3.45 eV
d) 4.16 eV
Explanation: As we know, the fermi energy of the material is 3.45 eV.
Now, zero-point energy =$$\frac{3}{5}E_F$$
= 3/5 X 3.45 eV
= 2.07 eV.
13. The average energy of one electron silver is 3.306 eV. What is the fermi-energy of Silver at 0 K?
a) 2.32 eV
b) 3.78 eV
c) 4.12 eV
d) 5.51 eV
Explanation: We know, Average Energy, E =$$\frac{3}{5}NE_F$$
Here, E = 3.306 eV, N = 1
Therefore, we get: EF = 5E/3
= 5.51 eV.
14. In Fermi-Dirac Statistics, one energy state can be occupied by more than one particle.
a) True
b) False
Explanation: In Fermi-Dirac statistics, one energy state can be occupied by only one particle. Therefore, when one state is filled, the particle will go on to another state.
15. Which of the following is the curve for Fermi-Dirac statistics?
a) X
b) Y
c) Z
d) None
Explanation: As seen the curve the order of the ration of f(K) with E/kT is in the order X > Y > Z. Here, Y is Maxwell-Boltzmann Statistics, X is Bose-Einstein Statistics and Z is Fermi-Dirac Statistics.
Sanfoundry Global Education & Learning Series – Engineering Physics.
To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers. | 1,866 | 5,670 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-27 | longest | en | 0.72167 |
http://docplayer.net/23034482-9-1-basic-principles-of-hypothesis-testing.html | 1,542,796,789,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00132.warc.gz | 96,951,097 | 26,171 | # 9.1 Basic Principles of Hypothesis Testing
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 9. Basic Principles of Hypothesis Testing Basic Idea Through an Example: On the very first day of class I gave the example of tossing a coin times, and what you might conclude about the fairness of the coin depending on the outcome of this experiment. If you got 55 heads, would you conclude that the coin was not fair? What if you got 95 heads? What if you got 75 heads? The further away our sample is from the expected value, the more suspicious we would be that the coin is not fair. Now having some knowledge about probabilities, you could find out how unlikely it is to get 75 heads or more out of if in fact the coin was fair (the probability is zero). So in a case like that, we could conclude that the coin must not be fair. In chapter 9 we will use hypothesis testing to make decisions about a population parameter ( µ or p ) based on the value of a sample statistics. The idea behind hypothesis testing is the same as the idea behind a criminal trial. The person is presumed to be not guilty until there is sufficient evidence to declare the person guilty. A null hypothesis, denoted H, states that some parameter is equal to a specific value, ex. H : µ = 27 (always use an equal sign for H ). The null hypothesis is a claim about a population parameter that is assumed to be true until declared false. An alternative hypothesis, denoted H, states that the value of the parameter differs from that specified by the null hypothesis, ex. H : µ 27 (a two-tailed test), or H : 27 µ > (a right-tailed test), or H : 27 µ < (a left-tailed test). The alternative hypothesis is a claim about a population parameter that will be true if the null hypothesis is false. Examples: Null hypothesis: H : The coin is fair (.5 Alternative hypothesis: H : The coin is not fair ( p.5 or p.5 or p.5 Null hypothesis: H : The person is not guilty Alternative hypothesis: H : The person is guilty Consider the M&M experiment that we did in class. What were the hypotheses? Null hypothesis: Alternative hypothesis:
2 Conclusions We reject H if the value of our test statistics falls too far away from the assumed population parameter. This is the same as saying that we support the H claim. We usually write this conclusion as "There is sufficient evidence to show that... H... ", where H is stated in words. We fail to reject H, if the difference between the sample statistics and the assumed population parameter is small, since the difference may be due to chance. We usually write this conclusion as "There is not sufficient evidence to show that... H... ", where H is stated in words. Note how we don't say that H is true. But we could say that " H may be true". Errors Our conclusion will be to either reject the null hypothesis or fail to reject it. Such conclusions are sometimes correct and sometimes not. There are two types of errors: Type I error occurs when rejecting the null hypothesis when it is actually true. The probability of making a type I error is α, the significance level. ex. Person is found guilty in court even though he did not commit the crime ex. Coin was found to be unfair even though it is fair. Type II error occurs when failing to reject the null hypothesis, when it is actually false. The probability of making a type II error is β (beta). ex. Person is found not guilty in court even though he did commit the crime. ex. Coin was found to be fair even though it is unfair. Fill each cell in the table with one of the following: Type I Error, Type II Error, or Correct Decision. Decision Reject H The Truth H is true H is false Do not reject H
3 . Test if the mean weight of women who won Miss America titles is still equal to 2 lb, or if it has changed. State hypotheses: Is it a left-tailed, right-tailed, or two-tailed test? If the null hypothesis is rejected, state an appropriate conclusion: What kind of error could we have made? 2. Plain M&M candies supposedly haves a mean weight that is.8535g. Test if the candies' weight is less than that. State hypotheses: Is it a left-tailed, right-tailed, or two-tailed test? If the null hypothesis is not rejected, state an appropriate conclusion: What kind of error could we have made? 3. Test if more than 25% of Internet users pay bills online. State hypotheses: Is it a left-tailed, right-tailed, or two-tailed test? If the null hypothesis is not rejected, state an appropriate conclusion: What kind of error could we have made? 4. Test if the mean hours spent per week on house chores by all housewives is less than 3. State hypotheses: Is it a left-tailed, right-tailed, or two-tailed test? If the null hypothesis is rejected, state an appropriate conclusion: What kind of error could we have made?
4 9.2 - Hypothesis Tests About µ when σ is Known The Test Statistic is a value used in making a decision about the null hypothesis, and it is found by converting 2 the sample statistic ( x, ˆp, or s) to a score (such as z, t, or χ ) with the assumption that the null hypothesis is true. The probability that we use to determine whether an event is unusual is called the significance level of the test, and is denoted by α. If we reject H after choosing a significance level α, we say that the result is statistically significant at the α level (note that statistically significant doesn't always mean practically significant). We also say that H is rejected at the α level. The P-value is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. "At least as extreme" means " as far away from the expected parameter as we got, or even further". If the P-value is very small, it means that the probability of getting what we got in our sample is very small, which may indicate that our original assumption H is not true. H is rejected if the P-value is smaller than the significance level α. In a left tailed test: P-value = area to the left of the test statistic P-value Test Stat In a right tailed test: P-value = area to the right of the test statistic P-value In a two tailed test: P-value = twice the area in the tail beyond the test statistic P-value / 2 Test Stat P-value is the total area of both tails P-value / 2 Test Stat or Test Stat
5 Hypothesis Tests About µ when σ is Known: Requirements. The value of the population standard deviation σ is known. 2. Either or both of these conditions is satisfied: The population is normally distributed or n > 3. If above is satisfied we know that the sampling distribution of x is Normal. Test Statistics: x µ z = where σ x = σ x σ n 8-Step Procedure for Performing a Hypothesis Test: ) State the null and alternative hypotheses of the test. 2) Choose and/or state the significance levelα. 3) State type of test, the standardized sampling distribution that should be used, and check that all of the required assumptions for using that distribution are satisfied. 4) Compute the test statistic. 5) Draw a picture of the standardized sampling distribution you are using. Label the test statistic. 6) Calculate the P-value. 7) Interpret the P-value and make a decision. If P-value < α then we reject the null hypothesis, and we have sufficient evidence for the alternative hypothesis. If P-value > α then we do NOT reject the null hypothesis, and we do NOT have sufficient evidence for the alternative hypothesis. 8) State a conclusion in the form of a detailed sentence that addresses the alternative hypothesis. When we Reject H, we say there is sufficient evidence to show that H, where H is stated in words. When we Fail to Reject H, we say there is not sufficient evidence to show that H, where H is stated in words.
6 When presenting the results of a hypothesis test, one should report the P-value or the value of the test statistics. That way the reader can tell exactly how likely or unlikely the test statistics was, and he/she can determine whether H could be rejected at a different level. Relationship between hypothesis tests and confidence intervals: If we test H : µ = µ vs. H: µ µ, then if the (-α )% confidence interval contains µ, then H will not be rejected at the α level if the (-α )% confidence interval does not contain µ, then H will be rejected at the α level P(type I error) = α and P(type II error) = β The smaller the probability of a type I error becomes, the larger the probability of a type II error becomes. Examples:. (a) The print on the package of -watt General Electric soft-white light-bulbs says that these bulbs have an average life of 75 hours. Assume that the lives of all such bulbs have a normal distribution with a standard deviation of 55 hours. The mean life of a simple random sample of 25 such bulbs was 725 hours. We are concerned that the stated average life on the package is exaggerated. Test at 5% significance level if this is true. (b) Would your conclusion be different if the significance level was %?
7 2. A certain type of children's pain reliever states that it contains 325 mg of acetaminophen in each ounce of the drug. If 7 one ounce samples are tested for acetaminophen and it is determined that the sample mean is 39 mg of the drug and a population standard deviation of 26 mg. With a =., test the claim that the population mean is equal to 325 mg. We can check our answers by using programs in our calculators. Press STAT and go to TESTS
8 3. A random sample of 6 second-graders in a certain school district are given a standardized mathematics skills test. The sample mean score is 52. Assume the standard deviation of test scores is 5. The nationwide average score on this test is 5.The school superintendent wants to know whether the second-graders in her school district have greater math skills than the nationwide average. Use a. level of significance to test this. We never accept the null hypothesis Note, that if we Fail to Reject H, we never say we accept H because the sample data is evidence against H (in favor of H ). It is not evidence in favor of H. We cannot prove something is true when we had to assume it was true to get the test started. ex. In the famous O.J. Simpson trial there was not sufficient evidence to show that O.J. Simpson was guilty of murder. That does not mean he was innocent.
9 4. When 4 people used the Atkins diet for one year, their mean weight change was -2. lb (new weight - old weight). Assume that the standard deviation of all such weight changes is σ =4.8 lb and use a.5 significance level to test the claim that the mean weight change is less than. Based on the results, does the diet appear to be effective? Does the mean weight change appear to be substantial enough to justify the special diet?
10 9.3 - Hypothesis Tests About µ when σ is Not Known Requirements. The value of the population standard deviation σ is NOT known. 2. Either or both of these conditions are satisfied: The population is normally distributed or n > 3. If above is satisfied we know that the sampling distribution of x is Normal. x µ Test Statistics: t = where sx s x = s n The 8-step procedure for performing a hypothesis test:. State the null and alternative hypotheses. 2. State significance level α 3. State which test to use and why. 4. Calculate the test statistic. 5. Draw a picture and label the test statistic. 6. Calculate the p-value. 7. Interpret p-value and make a decision. 8. State your conclusion.. A soft-drink manufacturer claims that its 2-ounce cans do not contain, on average, more than 3 calories. A random sample of 64 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the alternative hypothesis that the manufacturer's claim is false? Use a significance level of 5%.
11 2. The insurance Institute for Highway Safety conducted tests with crashes of new cars traveling at 6mi/h. Total cost of the damage was found. Results are listed below for a simple random sample of the tested cars. Assume the population has a distribution that is approximately normal. \$7448 \$49 \$95 \$6374 \$4277 Use a. significance level to test the claim that when tested under the same conditions, the damage costs for the population of cars have a mean of \$5. 3. A simple random sample of 4 recorded speeds (in mph) is obtained from cars traveling on a section of Highway 45 in Los Angeles. The sample has a mean of 68.4 mph and a standard deviation of 5.7 mph. Use a.5 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65 mph.
12 9.4 - Hypothesis Tests About a Population Proportion Requirements. We have a simple random sample 2. The population is at least 2 times as large as the sample. 3. The items in the population are divided into two categories. 4. The sample must contain at least individuals in each category. If above is satisfied we know that the sampling distribution of ˆp is approximately Normal. pˆ p p ( p) Test Statistics: z = where σ pˆ = σ n pˆ The 8-step procedure for performing a hypothesis test:. State the null and alternative hypotheses. 2. State significance level α 3. State which test to use and why. 4. Calculate the test statistic. 5. Draw a picture and label the test statistic. 6. Calculate the p-value. 7. Interpret p-value and make a decision. 8. State your conclusion.. Trials in an experiment with a polygraph (lie detector) has 98 results that include 24 cases of wrong results and 74 cases of correct results. Use a.5 significance level to test the claim that such polygraph results are correct less than 8% of the time. Based on the results, should polygraph test results be prohibited as evidence in trials?
13 2. (a) 9-year-old Emily Rosa chose the topic of touch therapy for a science fair project. She convinced 2 experienced touch therapists to participate in a simple test of their ability to detect human energy field. Emily constructed a cardboard partition with two holes for hands. Each touch therapist would put both hands through the two holes, and Emily would place her hand just above one of the therapist's hands and ask the therapist to identify the hand that Emily had selected. Emily used a coin toss to randomly select which hand to use. This test was repeated 28 times. If the touch therapists really did have the ability to sense a human energy field, they should have identified the correct hand much more than 5% of the time. If they just guessed, they should have been correct about 5% of the time. The touch therapists identified the correct hand 23 times out of the total 28 times, which is a success rate of 44%. Use a.5 significance level to test the claim that touch therapists are correct less than 5% of the time. (b) Would a different significance level change your conclusion in part (a)? (c) Can you argue against the validity of this study?
### Chapter 8. Hypothesis Testing
Chapter 8 Hypothesis Testing Hypothesis In statistics, a hypothesis is a claim or statement about a property of a population. A hypothesis test (or test of significance) is a standard procedure for testing
### Section 7.1. Introduction to Hypothesis Testing. Schrodinger s cat quantum mechanics thought experiment (1935)
Section 7.1 Introduction to Hypothesis Testing Schrodinger s cat quantum mechanics thought experiment (1935) Statistical Hypotheses A statistical hypothesis is a claim about a population. Null hypothesis
### 8-2 Basics of Hypothesis Testing. Definitions. Rare Event Rule for Inferential Statistics. Null Hypothesis
8-2 Basics of Hypothesis Testing Definitions This section presents individual components of a hypothesis test. We should know and understand the following: How to identify the null hypothesis and alternative
### Chapter 8 Hypothesis Testing Chapter 8 Hypothesis Testing 8-1 Overview 8-2 Basics of Hypothesis Testing
Chapter 8 Hypothesis Testing 1 Chapter 8 Hypothesis Testing 8-1 Overview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim About a Proportion 8-5 Testing a Claim About a Mean: s Not Known 8-6 Testing
### MATH 10: Elementary Statistics and Probability Chapter 9: Hypothesis Testing with One Sample
MATH 10: Elementary Statistics and Probability Chapter 9: Hypothesis Testing with One Sample Tony Pourmohamad Department of Mathematics De Anza College Spring 2015 Objectives By the end of this set of
### Example Hypotheses. Chapter 8-2: Basics of Hypothesis Testing. A newspaper headline makes the claim: Most workers get their jobs through networking
Chapter 8-2: Basics of Hypothesis Testing Two main activities in statistical inference are using sample data to: 1. estimate a population parameter forming confidence intervals 2. test a hypothesis or
### Chapter III. Testing Hypotheses
Chapter III Testing Hypotheses R (Introduction) A statistical hypothesis is an assumption about a population parameter This assumption may or may not be true The best way to determine whether a statistical
### Introduction to Hypothesis Testing OPRE 6301
Introduction to Hypothesis Testing OPRE 6301 Motivation... The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief, or hypothesis, about
### Hypothesis Testing --- One Mean
Hypothesis Testing --- One Mean A hypothesis is simply a statement that something is true. Typically, there are two hypotheses in a hypothesis test: the null, and the alternative. Null Hypothesis The hypothesis
### Null Hypothesis H 0. The null hypothesis (denoted by H 0
Hypothesis test In statistics, a hypothesis is a claim or statement about a property of a population. A hypothesis test (or test of significance) is a standard procedure for testing a claim about a property
### Chapter 8 Introduction to Hypothesis Testing
Chapter 8 Student Lecture Notes 8-1 Chapter 8 Introduction to Hypothesis Testing Fall 26 Fundamentals of Business Statistics 1 Chapter Goals After completing this chapter, you should be able to: Formulate
### Introduction to Hypothesis Testing. Hypothesis Testing. Step 1: State the Hypotheses
Introduction to Hypothesis Testing 1 Hypothesis Testing A hypothesis test is a statistical procedure that uses sample data to evaluate a hypothesis about a population Hypothesis is stated in terms of the
### A Trial Analogy for Statistical. Hypothesis Testing. Legal Trial Begin with claim: Statistical Significance Test Hypotheses (statements)
A Trial Analogy for Statistical Slide 1 Hypothesis Testing Legal Trial Begin with claim: Smith is not guilty If this is rejected, we accept Smith is guilty reasonable doubt Present evidence (facts) Evaluate
### 22. HYPOTHESIS TESTING
22. HYPOTHESIS TESTING Often, we need to make decisions based on incomplete information. Do the data support some belief ( hypothesis ) about the value of a population parameter? Is OJ Simpson guilty?
### Chapter Additional: Standard Deviation and Chi- Square
Chapter Additional: Standard Deviation and Chi- Square Chapter Outline: 6.4 Confidence Intervals for the Standard Deviation 7.5 Hypothesis testing for Standard Deviation Section 6.4 Objectives Interpret
### Hypothesis testing allows us to use a sample to decide between two statements made about a Population characteristic.
Hypothesis Testing Hypothesis testing allows us to use a sample to decide between two statements made about a Population characteristic. Population Characteristics are things like The mean of a population
### Hypothesis Testing. Concept of Hypothesis Testing
Quantitative Methods 2013 Hypothesis Testing with One Sample 1 Concept of Hypothesis Testing Testing Hypotheses is another way to deal with the problem of making a statement about an unknown population
### Chapter 7 Part 2. Hypothesis testing Power
Chapter 7 Part 2 Hypothesis testing Power November 6, 2008 All of the normal curves in this handout are sampling distributions Goal: To understand the process of hypothesis testing and the relationship
### Lecture 8 Hypothesis Testing
Lecture 8 Hypothesis Testing Fall 2013 Prof. Yao Xie, yao.xie@isye.gatech.edu H. Milton Stewart School of Industrial Systems & Engineering Georgia Tech Midterm 1 Score 46 students Highest score: 98 Lowest
### Step 1: Set up hypotheses that ask a question about the population by setting up two opposite statements about the possible value of the parameters.
HYPOTHESIS TEST CLASS NOTES Hypothesis Test: Procedure that allows us to ask a question about an unknown population parameter Uses sample data to draw a conclusion about the unknown population parameter.
### Section 12.2, Lesson 3. What Can Go Wrong in Hypothesis Testing: The Two Types of Errors and Their Probabilities
Today: Section 2.2, Lesson 3: What can go wrong with hypothesis testing Section 2.4: Hypothesis tests for difference in two proportions ANNOUNCEMENTS: No discussion today. Check your grades on eee and
### Ch. 8 Hypothesis Testing
Ch. 8 Hypothesis Testing 8.1 Foundations of Hypothesis Testing Definitions In statistics, a hypothesis is a claim about a property of a population. A hypothesis test is a standard procedure for testing
### Sampling and Hypothesis Testing
Population and sample Sampling and Hypothesis Testing Allin Cottrell Population : an entire set of objects or units of observation of one sort or another. Sample : subset of a population. Parameter versus
### Math 251, Review Questions for Test 3 Rough Answers
Math 251, Review Questions for Test 3 Rough Answers 1. (Review of some terminology from Section 7.1) In a state with 459,341 voters, a poll of 2300 voters finds that 45 percent support the Republican candidate,
### Chapter 7. Hypothesis Testing with One Sample
Chapter 7 Hypothesis Testing with One Sample 7.1 Introduction to Hypothesis Testing Hypothesis Tests A hypothesis test is a process that uses sample statistics to test a claim about the value of a population
### Introduction to Hypothesis Testing
I. Terms, Concepts. Introduction to Hypothesis Testing A. In general, we do not know the true value of population parameters - they must be estimated. However, we do have hypotheses about what the true
### How to Conduct a Hypothesis Test
How to Conduct a Hypothesis Test The idea of hypothesis testing is relatively straightforward. In various studies we observe certain events. We must ask, is the event due to chance alone, or is there some
### HypoTesting. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.
Name: Class: Date: HypoTesting Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A Type II error is committed if we make: a. a correct decision when the
### Chapter 9, Part A Hypothesis Tests. Learning objectives
Chapter 9, Part A Hypothesis Tests Slide 1 Learning objectives 1. Understand how to develop Null and Alternative Hypotheses 2. Understand Type I and Type II Errors 3. Able to do hypothesis test about population
### Lecture 13 More on hypothesis testing
Lecture 13 More on hypothesis testing Thais Paiva STA 111 - Summer 2013 Term II July 22, 2013 1 / 27 Thais Paiva STA 111 - Summer 2013 Term II Lecture 13, 07/22/2013 Lecture Plan 1 Type I and type II error
### Confidence Interval: pˆ = E = Indicated decision: < p <
Hypothesis (Significance) Tests About a Proportion Example 1 The standard treatment for a disease works in 0.675 of all patients. A new treatment is proposed. Is it better? (The scientists who created
PROBLEM SET 1 For the first three answer true or false and explain your answer. A picture is often helpful. 1. Suppose the significance level of a hypothesis test is α=0.05. If the p-value of the test
### Hypothesis Testing I
ypothesis Testing I The testing process:. Assumption about population(s) parameter(s) is made, called null hypothesis, denoted. 2. Then the alternative is chosen (often just a negation of the null hypothesis),
### Homework #3 is due Friday by 5pm. Homework #4 will be posted to the class website later this week. It will be due Friday, March 7 th, at 5pm.
Homework #3 is due Friday by 5pm. Homework #4 will be posted to the class website later this week. It will be due Friday, March 7 th, at 5pm. Political Science 15 Lecture 12: Hypothesis Testing Sampling
### Chapter 7. Section Introduction to Hypothesis Testing
Section 7.1 - Introduction to Hypothesis Testing Chapter 7 Objectives: State a null hypothesis and an alternative hypothesis Identify type I and type II errors and interpret the level of significance Determine
### HYPOTHESIS TESTING: POWER OF THE TEST
HYPOTHESIS TESTING: POWER OF THE TEST The first 6 steps of the 9-step test of hypothesis are called "the test". These steps are not dependent on the observed data values. When planning a research project,
### Statistical inference provides methods for drawing conclusions about a population from sample data.
Chapter 15 Tests of Significance: The Basics Statistical inference provides methods for drawing conclusions about a population from sample data. Two of the most common types of statistical inference: 1)
### 15.0 More Hypothesis Testing
15.0 More Hypothesis Testing 1 Answer Questions Type I and Type II Error Power Calculation Bayesian Hypothesis Testing 15.1 Type I and Type II Error In the philosophy of hypothesis testing, the null hypothesis
### Chapter 8. Professor Tim Busken. April 20, Chapter 8. Tim Busken. 8.2 Basics of. Hypothesis Testing. Works Cited
Chapter 8 Professor April 20, 2014 In Chapter 8, we continue our study of inferential statistics. Concept: Inferential Statistics The two major activities of inferential statistics are 1 to use sample
### Hypothesis Testing: p-value
STAT 101 Dr. Kari Lock Morgan Paul the Octopus Hypothesis Testing: SECTION 4.2 andomization distribution http://www.youtube.com/watch?v=3esgpumj9e Hypotheses In 2008, Paul the Octopus predicted 8 World
### The alternative hypothesis,, is the statement that the parameter value somehow differs from that claimed by the null hypothesis. : 0.5 :>0.5 :<0.
Section 8.2-8.5 Null and Alternative Hypotheses... The null hypothesis,, is a statement that the value of a population parameter is equal to some claimed value. :=0.5 The alternative hypothesis,, is the
### Hypothesis testing. c 2014, Jeffrey S. Simonoff 1
Hypothesis testing So far, we ve talked about inference from the point of estimation. We ve tried to answer questions like What is a good estimate for a typical value? or How much variability is there
### Module 7: Hypothesis Testing I Statistics (OA3102)
Module 7: Hypothesis Testing I Statistics (OA3102) Professor Ron Fricker Naval Postgraduate School Monterey, California Reading assignment: WM&S chapter 10.1-10.5 Revision: 2-12 1 Goals for this Module
### Hypothesis testing: Examples. AMS7, Spring 2012
Hypothesis testing: Examples AMS7, Spring 2012 Example 1: Testing a Claim about a Proportion Sect. 7.3, # 2: Survey of Drinking: In a Gallup survey, 1087 randomly selected adults were asked whether they
### Module 5 Hypotheses Tests: Comparing Two Groups
Module 5 Hypotheses Tests: Comparing Two Groups Objective: In medical research, we often compare the outcomes between two groups of patients, namely exposed and unexposed groups. At the completion of this
### Homework 5 Solutions
Math 130 Assignment Chapter 18: 6, 10, 38 Chapter 19: 4, 6, 8, 10, 14, 16, 40 Chapter 20: 2, 4, 9 Chapter 18 Homework 5 Solutions 18.6] M&M s. The candy company claims that 10% of the M&M s it produces
### Hypothesis testing for µ:
University of California, Los Angeles Department of Statistics Statistics 13 Elements of a hypothesis test: Hypothesis testing Instructor: Nicolas Christou 1. Null hypothesis, H 0 (always =). 2. Alternative
### CHAPTER 11 SECTION 2: INTRODUCTION TO HYPOTHESIS TESTING
CHAPTER 11 SECTION 2: INTRODUCTION TO HYPOTHESIS TESTING MULTIPLE CHOICE 56. In testing the hypotheses H 0 : µ = 50 vs. H 1 : µ 50, the following information is known: n = 64, = 53.5, and σ = 10. The standardized
### Name: (b) Find the minimum sample size you should use in order for your estimate to be within 0.03 of p when the confidence level is 95%.
Chapter 7-8 Exam Name: Answer the questions in the spaces provided. If you run out of room, show your work on a separate paper clearly numbered and attached to this exam. Please indicate which program
### reductio ad absurdum null hypothesis, alternate hypothesis
Chapter 10 s Using a Single Sample 10.1: Hypotheses & Test Procedures Basics: In statistics, a hypothesis is a statement about a population characteristic. s are based on an reductio ad absurdum form of
### Wording of Final Conclusion. Slide 1
Wording of Final Conclusion Slide 1 8.3: Assumptions for Testing Slide 2 Claims About Population Means 1) The sample is a simple random sample. 2) The value of the population standard deviation σ is known
### Correlational Research
Correlational Research Chapter Fifteen Correlational Research Chapter Fifteen Bring folder of readings The Nature of Correlational Research Correlational Research is also known as Associational Research.
### Chapter 21. More About Tests and Intervals. Copyright 2012, 2008, 2005 Pearson Education, Inc.
Chapter 21 More About Tests and Intervals Copyright 2012, 2008, 2005 Pearson Education, Inc. Zero In on the Null Null hypotheses have special requirements. To perform a hypothesis test, the null must be
### HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1. used confidence intervals to answer questions such as...
HYPOTHESIS TESTING (ONE SAMPLE) - CHAPTER 7 1 PREVIOUSLY used confidence intervals to answer questions such as... You know that 0.25% of women have red/green color blindness. You conduct a study of men
### BA 275 Review Problems - Week 6 (10/30/06-11/3/06) CD Lessons: 53, 54, 55, 56 Textbook: pp. 394-398, 404-408, 410-420
BA 275 Review Problems - Week 6 (10/30/06-11/3/06) CD Lessons: 53, 54, 55, 56 Textbook: pp. 394-398, 404-408, 410-420 1. Which of the following will increase the value of the power in a statistical test
### Basic Statistics Self Assessment Test
Basic Statistics Self Assessment Test Professor Douglas H. Jones PAGE 1 A soda-dispensing machine fills 12-ounce cans of soda using a normal distribution with a mean of 12.1 ounces and a standard deviation
### 4) The role of the sample mean in a confidence interval estimate for the population mean is to: 4)
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Assume that the change in daily closing prices for stocks on the New York Stock Exchange is a random
### Chapter 1 Hypothesis Testing
Chapter 1 Hypothesis Testing Principles of Hypothesis Testing tests for one sample case 1 Statistical Hypotheses They are defined as assertion or conjecture about the parameter or parameters of a population,
### Reasoning with Uncertainty More about Hypothesis Testing. P-values, types of errors, power of a test
Reasoning with Uncertainty More about Hypothesis Testing P-values, types of errors, power of a test P-Values and Decisions Your conclusion about any null hypothesis should be accompanied by the P-value
### Hypothesis Testing with One Sample. Introduction to Hypothesis Testing 7.1. Hypothesis Tests. Chapter 7
Chapter 7 Hypothesis Testing with One Sample 71 Introduction to Hypothesis Testing Hypothesis Tests A hypothesis test is a process that uses sample statistics to test a claim about the value of a population
### Hypothesis testing - Steps
Hypothesis testing - Steps Steps to do a two-tailed test of the hypothesis that β 1 0: 1. Set up the hypotheses: H 0 : β 1 = 0 H a : β 1 0. 2. Compute the test statistic: t = b 1 0 Std. error of b 1 =
### Chapter 08. Introduction
Chapter 08 Introduction Hypothesis testing may best be summarized as a decision making process in which one attempts to arrive at a particular conclusion based upon "statistical" evidence. A typical hypothesis
### Probability, Binomial Distributions and Hypothesis Testing Vartanian, SW 540
Probability, Binomial Distributions and Hypothesis Testing Vartanian, SW 540 1. Assume you are tossing a coin 11 times. The following distribution gives the likelihoods of getting a particular number of
### The Goodness-of-Fit Test
on the Lecture 49 Section 14.3 Hampden-Sydney College Tue, Apr 21, 2009 Outline 1 on the 2 3 on the 4 5 Hypotheses on the (Steps 1 and 2) (1) H 0 : H 1 : H 0 is false. (2) α = 0.05. p 1 = 0.24 p 2 = 0.20
### 8-1 8-2 8-3 8-4 8-5 8-6
8-1 Review and Preview 8-2 Basics of Hypothesis Testing 8-3 Testing a Claim About a Proportion 8-4 Testing a Claim About a Mean: s Known 8-5 Testing a Claim About a Mean: s Not Known 8-6 Testing a Claim
### Chapter 8 Hypothesis Testing
Chapter 8 Hypothesis Testing Chapter problem: Does the MicroSort method of gender selection increase the likelihood that a baby will be girl? MicroSort: a gender-selection method developed by Genetics
### Hypothesis Tests for a Population Proportion
Hypothesis Tests for a Population Proportion MATH 130, Elements of Statistics I J. Robert Buchanan Department of Mathematics Fall 2015 Review: Steps of Hypothesis Testing 1. A statement is made regarding
### Extending Hypothesis Testing. p-values & confidence intervals
Extending Hypothesis Testing p-values & confidence intervals So far: how to state a question in the form of two hypotheses (null and alternative), how to assess the data, how to answer the question by
### Chapter 2. Hypothesis testing in one population
Chapter 2. Hypothesis testing in one population Contents Introduction, the null and alternative hypotheses Hypothesis testing process Type I and Type II errors, power Test statistic, level of significance
### Online 12 - Sections 9.1 and 9.2-Doug Ensley
Student: Date: Instructor: Doug Ensley Course: MAT117 01 Applied Statistics - Ensley Assignment: Online 12 - Sections 9.1 and 9.2 1. Does a P-value of 0.001 give strong evidence or not especially strong
### Hypothesis Testing - II
-3σ -2σ +σ +2σ +3σ Hypothesis Testing - II Lecture 9 0909.400.01 / 0909.400.02 Dr. P. s Clinic Consultant Module in Probability & Statistics in Engineering Today in P&S -3σ -2σ +σ +2σ +3σ Review: Hypothesis
### Third Midterm Exam (MATH1070 Spring 2012)
Third Midterm Exam (MATH1070 Spring 2012) Instructions: This is a one hour exam. You can use a notesheet. Calculators are allowed, but other electronics are prohibited. 1. [40pts] Multiple Choice Problems
### Testing: is my coin fair?
Testing: is my coin fair? Formally: we want to make some inference about P(head) Try it: toss coin several times (say 7 times) Assume that it is fair ( P(head)= ), and see if this assumption is compatible
### Notes 8: Hypothesis Testing
Notes 8: Hypothesis Testing Julio Garín Department of Economics Statistics for Economics Spring 2012 (Stats for Econ) Hypothesis Testing Spring 2012 1 / 44 Introduction Why we conduct surveys? We want
### C. The null hypothesis is not rejected when the alternative hypothesis is true. A. population parameters.
Sample Multiple Choice Questions for the material since Midterm 2. Sample questions from Midterms and 2 are also representative of questions that may appear on the final exam.. A randomly selected sample
### Purpose of Hypothesis Testing
Large sample Tests of Hypotheses Chapter 9 1 Purpose of Hypothesis Testing In the last chapter, we studied methods of estimating a parameter (μ, p or p 1 p 2 ) based on sample data: point estimation confidence
### Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 9 Introduction to Hypothesis Testing
Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 9 Introduction to Hypothesis Testing 1) Hypothesis testing and confidence interval estimation are essentially two totally different statistical procedures
### 5/31/2013. Chapter 8 Hypothesis Testing. Hypothesis Testing. Hypothesis Testing. Outline. Objectives. Objectives
C H 8A P T E R Outline 8 1 Steps in Traditional Method 8 2 z Test for a Mean 8 3 t Test for a Mean 8 4 z Test for a Proportion 8 6 Confidence Intervals and Copyright 2013 The McGraw Hill Companies, Inc.
### The Basics of a Hypothesis Test
Overview The Basics of a Test Dr Tom Ilvento Department of Food and Resource Economics Alternative way to make inferences from a sample to the Population is via a Test A hypothesis test is based upon A
### Confidence Intervals (Review)
Intro to Hypothesis Tests Solutions STAT-UB.0103 Statistics for Business Control and Regression Models Confidence Intervals (Review) 1. Each year, construction contractors and equipment distributors from
### Homework 6 Solutions
Math 17, Section 2 Spring 2011 Assignment Chapter 20: 12, 14, 20, 24, 34 Chapter 21: 2, 8, 14, 16, 18 Chapter 20 20.12] Got Milk? The student made a number of mistakes here: Homework 6 Solutions 1. Null
### Introduction to. Hypothesis Testing CHAPTER LEARNING OBJECTIVES. 1 Identify the four steps of hypothesis testing.
Introduction to Hypothesis Testing CHAPTER 8 LEARNING OBJECTIVES After reading this chapter, you should be able to: 1 Identify the four steps of hypothesis testing. 2 Define null hypothesis, alternative
### Hypothesis Testing: General Framework 1 1
Hypothesis Testing: General Framework Lecture 2 K. Zuev February 22, 26 In previous lectures we learned how to estimate parameters in parametric and nonparametric settings. Quite often, however, researchers
### Chapter Five: Paired Samples Methods 1/38
Chapter Five: Paired Samples Methods 1/38 5.1 Introduction 2/38 Introduction Paired data arise with some frequency in a variety of research contexts. Patients might have a particular type of laser surgery
Chapter 11 Testing Hypotheses About Proportions Hypothesis testing method: uses data from a sample to judge whether or not a statement about a population may be true. Steps in Any Hypothesis Test 1. Determine
### Mind on Statistics. Chapter 12
Mind on Statistics Chapter 12 Sections 12.1 Questions 1 to 6: For each statement, determine if the statement is a typical null hypothesis (H 0 ) or alternative hypothesis (H a ). 1. There is no difference
### The Purpose of Hypothesis Testing
Section 8 1A: An Introduction to Hypothesis Testing The Purpose of Hypothesis Testing Seeʼs Candy states that a box of itʼs candy weighs 16 oz. They do not mean that every single box weights exactly 16
### An Introduction to Statistics Course (ECOE 1302) Spring Semester 2011 Chapter 10- TWO-SAMPLE TESTS
The Islamic University of Gaza Faculty of Commerce Department of Economics and Political Sciences An Introduction to Statistics Course (ECOE 130) Spring Semester 011 Chapter 10- TWO-SAMPLE TESTS Practice
### 7 Hypothesis testing - one sample tests
7 Hypothesis testing - one sample tests 7.1 Introduction Definition 7.1 A hypothesis is a statement about a population parameter. Example A hypothesis might be that the mean age of students taking MAS113X
### Hypothesis Testing. Bluman Chapter 8
CHAPTER 8 Learning Objectives C H A P T E R E I G H T Hypothesis Testing 1 Outline 8-1 Steps in Traditional Method 8-2 z Test for a Mean 8-3 t Test for a Mean 8-4 z Test for a Proportion 8-5 2 Test for
### Calculating P-Values. Parkland College. Isela Guerra Parkland College. Recommended Citation
Parkland College A with Honors Projects Honors Program 2014 Calculating P-Values Isela Guerra Parkland College Recommended Citation Guerra, Isela, "Calculating P-Values" (2014). A with Honors Projects.
### 3.4 Statistical inference for 2 populations based on two samples
3.4 Statistical inference for 2 populations based on two samples Tests for a difference between two population means The first sample will be denoted as X 1, X 2,..., X m. The second sample will be denoted
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
STA2023 Module 10 Test Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. A hypothesis test is to be performed. Determine the null and alternative
### 6: Introduction to Hypothesis Testing
6: Introduction to Hypothesis Testing Significance testing is used to help make a judgment about a claim by addressing the question, Can the observed difference be attributed to chance? We break up significance
### IQ of deaf children example: Are the deaf children lower in IQ? Or are they average? If µ100 and σ 2 225, is the 88.07 from the sample of N59 deaf chi
PSY 511: Advanced Statistics for Psychological and Behavioral Research 1 All inferential statistics have the following in common: Use of some descriptive statistic Use of probability Potential for estimation
### Introduction to Hypothesis Testing. Point estimation and confidence intervals are useful statistical inference procedures.
Introduction to Hypothesis Testing Point estimation and confidence intervals are useful statistical inference procedures. Another type of inference is used frequently used concerns tests of hypotheses.
### NUMB3RS Activity: Candy Pieces. Episode: End of Watch
Teacher Page 1 NUMB3RS Activity: Candy Pieces Topic: Chi-square test for goodness-of-fit Grade Level: 11-12 Objective: Use a chi-square test to determine if there is a significant difference between the
### Single sample hypothesis testing, II 9.07 3/02/2004
Single sample hypothesis testing, II 9.07 3/02/2004 Outline Very brief review One-tailed vs. two-tailed tests Small sample testing Significance & multiple tests II: Data snooping What do our results mean? | 9,757 | 41,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-47 | latest | en | 0.910439 |
https://www.stat.math.ethz.ch/pipermail/r-help/2011-November/296209.html | 1,652,790,342,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00304.warc.gz | 1,214,735,072 | 2,351 | # [R] Function gives numeric(0) for every input
Peter Langfelder peter.langfelder at gmail.com
Sun Nov 20 05:31:53 CET 2011
```Well, you assign numeric(0) or numeric(0)+1, which is still
numeric(0). No wonder the return value is always numeric(0).
You probably need to replace numeric(0) simply by 0. Numeric(0) does
not mean 0, it means a numeric vector of length zero (i.e., empty).
HTH,
Peter
On Sat, Nov 19, 2011 at 6:52 PM, alex_janssen <janssena at uoguelph.ca> wrote:
> Hi,
>
> I am trying to code buffons needle in R for a class
>
> This is my code w/ output from R, if anyone could tell me why this is
> happening it would be amazing,
>
> I can generate correct results without putting the steps into a function but
> alas that is the assignment.
>> buffon = function(n){
> + x = NULL
> + theta = NULL
> + a = NULL
> + phat = numeric(0)
> + i = 1
> +
> + while ( i <= n){
> + x[i] <- runif(1,0,1/2)
> + theta[i] <- runif(1,0,pi/2)
> + a[i] <- 1/2 * (sin(theta[i])) - x[i]
> +
> +
> + if (a[i] <= 0)phat <- phat + 1
> + else phat<-phat
> + i <- i + 1
> +
> + }
> + return(phat)
> + }
>> buffon(10)
> numeric(0)
>>
> Thanks Again
>
> the response should be the number of times a[i] is less than or equal to
> zero, which i am calling phat
>
> --
> View this message in context: http://r.789695.n4.nabble.com/Function-gives-numeric-0-for-every-input-tp4087834p4087834.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
```
More information about the R-help mailing list | 565 | 1,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.708564 |
http://mathoverflow.net/questions/tagged/pr.probability+it.information-theory | 1,410,918,197,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657120446.62/warc/CC-MAIN-20140914011200-00321-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 178,226,623 | 23,373 | Tagged Questions
67 views
Entropy on a draw from a random distribution.
Suppose I am attempting to calculate the entropy of a continuous, normally distributed random variable $X$, from the distribution $\mathcal{N}(\mu, \sigma)$. This is easy to to do - I just calculate ...
76 views
Mutual information staying constant under composition of channels
Consider the following scenario: one has 2 communication channels $C_1$ and $C_2$. Denote by $p(x)$ the input probability distribution. The mutual information between the input and the output of ...
679 views
A conjecture about the entropy of matrix vector products
Consider a random $m$ by $n$ partial circulant matrix $M$ whose entries are chosen independently and uniformly from $\{0,1\}$ and let $m < n$. Now consider a random $n$ dimensional vector $v$ ...
76 views
Lower convex envelope of a function involving entropy
Suppose two discrete random variables $X$ and $Y$ defined on finite sets $\mathcal{X}$ and $\mathcal{Y}$ are given and also suppose the conditional distribution $P_{Y|X}$ (i.e, channel) is fixed. We ...
532 views
conjectures regarding a new Renyi information quantity
In a recent paper http://arxiv.org/abs/1403.6102, we defined a quantity that we called the "Renyi conditional mutual information" and investigated several of its properties. We have some open ...
161 views
145 views
Moments of random matrices - when are they finite
I need to evaluate the moment $$\mathbb{E} (AX)^n,$$ where A is an NxN Hermitian square matrix, and X is $$X=ZZ^{\ast},$$ where $Z=\mu+Y$, where $\mu$ is mean of $Z$ and $Y$ is a zero-mean complex ... | 404 | 1,623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-41 | longest | en | 0.872025 |
http://math.stackexchange.com/questions/tagged/integral-transforms+calculus | 1,406,134,470,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997880800.37/warc/CC-MAIN-20140722025800-00044-ip-10-33-131-23.ec2.internal.warc.gz | 205,808,805 | 13,584 | # Tagged Questions
25 views
### Fourier Transform-1
I am trying to solve a Fourier transform problem and I am stuck. The problem is: $$f(t)= \frac{\sin(2t)}{e^{|t|}}.$$ I have used integration, but the answer that I come up with is different than ...
15 views
80 views
### Change of variables in double integral - what's wrong?
I have a homework problem, as follows: Evaluate the double integral by making an appropriate change of variables. $\iint_R 9\sin(49x^2+16y^2)\,dA$, where $R$ is the region in the first ...
91 views
### How to find the inverse mellin transform?
On the wikipedia page http://en.wikipedia.org/wiki/Mellin_transform the second formula is an integral transformation for the inverse Mellin transform. Being new to integral transforms, I wonder how ...
105 views
### Laplace Transformation Applications
In one of our Mathematics lecture our Prof told us that similar to Logarithmic Transformations we can use Laplace Transformations to solve difficult equations. What kind of equations do Laplace ...
86 views
### Continuity of the inverse Laplace Transform
If I know $Y(s)$, can I predict when $\mathscr{L}^{-1}[Y(s)]=y(t)$ will be continuous or continuously differentiable or even stronger conditions? For example; I'm solving an ODE with the Laplace ...
Recently I was studying the mellin convolution technique to solve definite integrals. I am just wondering is the technique valid only for any definite integrals (with any range $\int^b_a$)? or only ... | 351 | 1,491 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2014-23 | latest | en | 0.869548 |
https://math.stackexchange.com/questions/2821321/find-all-functions-such-that-fx2y-xfxfy | 1,607,221,171,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00192.warc.gz | 380,113,056 | 35,675 | # Find all functions such that $f(x^2+y)=xf(x)+f(y)$
Find all functions (over real numbers) such that $$f(x^2+y)=xf(x)+f(y).$$
My idea: Put $x=1$. Therefore the given equation becomes $f(y+1)=f(y)+f(1)$. It is in the Cauchy's first form, so we get $f(x)=cx$. It also satisfies that given functional equation.
Is my answer correct?? If not then what is the complete solution?? Please help me!
• That's not in Cauchy's form. – Saad Jun 16 '18 at 6:21
• @Alex Francisco $f(x+y)=f(x)+f(y)$ is a Cauchy form! And its solution is $f(x)=cx$ – Sufaid Saleel Jun 16 '18 at 6:23
• What you derived is $f(y+1)=f(y)+f(1)$. – Saad Jun 16 '18 at 6:24
• No, you are not right. If $g$ is a function of period $1$ with $g(0)=0,$ then $f(x)=g(x) +cx$ satisfies $f(x+1)=f(x)+f(1).$ – Thomas Andrews Jun 16 '18 at 6:24
• Is there any statement of continuity? – DanielV Jun 16 '18 at 6:25
Here is a proof that $f$ does satisfy Cauchy's functional equation. For now, I do not have a complete answer, unless some sort of continuity conditions is given.
Let $P(x,y)$ denote the condition that $f\left(x^2+y\right)=x\,f(x)+f(y)$. Then, $P(1,0)$ implies that $f(0)=0$. Now, $P(x,0)$ implies that $f\left(x^2\right)=x\,f(x)$ for all $x\in\mathbb{R}$. This shows that $f\left(x^2+y\right)=f\left(x^2\right)+f(y)$ for all $x,y\in\mathbb{R}$. Next, $P\left(x,-x^2\right)$ leads to $f\left(-x^2\right)=-f\left(x^2\right)$, and consequently, $f(-x)=-f(x)$ for every $x\in\mathbb{R}$. From this, it can be easily shown that $f$ satisfies Cauchy's functional equation.
• $$( x + 1 ) f ( x ) + ( x + 1 ) f ( 1 ) = ( x + 1 ) f ( x + 1 ) = f \left( ( x + 1 ) ^ 2 \right) \\ = f \left( x ^ 2 + 2 x + 1 \right) = f \left( x ^ 2 \right) + f ( 2 x ) + f ( 1 ) = ( x + 2 ) f ( x ) + f ( 1 )$$ and thus $f ( x ) = f ( 1 ) x$. – Mohsen Shahriari Nov 9 at 19:35
Assuming $f \in C$ and arranging as
$$\frac{f(y+x^2)-f(y)}{x^2}=\frac{f(x)}{x} = \phi(x)$$
So
$$\lim_{x\to 0}\frac{f(y+x^2)-f(y)}{x^2}=\phi(0)$$
which is independent of $y$ hence $f(x) = C_0 x$
• It's enough to assume $f$ is continuous. Then your argument shows that the derivative exists and is constant. – Wojowu Jun 16 '18 at 8:26
• Yes. I left an exponent in excess at $C$. Thanks. – Cesareo Jun 16 '18 at 9:01
• Why do you know that $\lim_{x\rightarrow 0} \phi(x)$ exists? – Severin Schraven Jun 16 '18 at 11:46
• Because this implies on that the first limit does not exists for any $y$. – Cesareo Jun 16 '18 at 11:48 | 961 | 2,452 | {"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-2020-50 | latest | en | 0.71738 |
http://survey.constantcontact.com/survey/a07ee50n3zbj2bjpnxu/results | 1,568,839,857,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573331.86/warc/CC-MAIN-20190918193432-20190918215432-00336.warc.gz | 182,093,455 | 15,262 | ## Overall Survey Results:
Text Block:
Please give your honest and fair assessment of the performance of all the players who took part in this game, including playing substitutes. I would prefer that those who complete this function have watched 90 minutes of the game in question.Don't forget to grade those players mentioned in the second section below.
1Here are the first ten team members:-
1 = 1 , 2 = 2 , 3 = 3 , 4 = 4 , 5 = 5 , 6 = 6 , 7 = 7 , 8 = 8 , 9 = 9 , 10 = 10
1 2 3 4 5 6 7 8 9 10
Number of
Responses
Rating
Score*
Lloris 142 6.2
Walker 143 4.6
Alderweireld 143 5.3
Vertonghen 143 5.8
Davies 143 5.7
Dier 142 5.4
Wanyama 141 6.1
Eriksen 142 5.6
Alli 142 5.2
Son 143 5.2
*The Rating Score is the weighted average calculated by dividing the sum of all weighted ratings by the number of total responses.
2Don't forget to grade these players too!
1 = 1 , 2 = 2 , 3 = 3 , 4 = 4 , 5 = 5 , 6 = 6 , 7 = 7 , 8 = 8 , 9 = 9 , 10 = 10
1 2 3 4 5 6 7 8 9 10
Number of
Responses
Rating
Score*
Kane 143 5.2
Dembele 143 5.4
Janssen 141 5.1
Trippier 138 5.6
*The Rating Score is the weighted average calculated by dividing the sum of all weighted ratings by the number of total responses.
3I would also like you to give marks out of 10 in the following skill areas:-
1 = 1 , 2 = 2 , 3 = 3 , 4 = 4 , 5 = 5 , 6 = 6 , 7 = 7 , 8 = 8 , 9 = 9 , 10 = 10
1 2 3 4 5 6 7 8 9 10
Number of
Responses
Rating
Score*
SKILL 134 5.2
PLUCK (EFFORT) 134 5.5
UNFORCED ERRORS (The less errors, the better the mark) 133 4.7
RESOLVE (Defensive ability) 134 4.9
SCORE RATIO (Goals scored in relation to shots) 131 3.7
*The Rating Score is the weighted average calculated by dividing the sum of all weighted ratings by the number of total responses.
4
Now please grade Mauricio Pochettino on three areas - Team Selection, Tactics, and Substitutions
1 = 1 , 2 = 2 , 3 = 3 , 4 = 4 , 5 = 5 , 6 = 6 , 7 = 7 , 8 = 8 , 9 = 9 , 10 = 10
1 2 3 4 5 6 7 8 9 10
Number of
Responses
Rating
Score*
Team Selection 138 6.2
Tactics 136 5.2
Substitutions 136 5.2
*The Rating Score is the weighted average calculated by dividing the sum of all weighted ratings by the number of total responses.
5Finally, here's a chance for you to give a single mark out of ten for the performance of the officials. Today's referee was Anthony Taylor..
1 = 1 , 2 = 2 , 3 = 3 , 4 = 4 , 5 = 5 , 6 = 6 , 7 = 7 , 8 = 8 , 9 = 9 , 10 = 10
1 2 3 4 5 6 7 8 9 10
Number of
Responses
Rating
Score*
138 5.1
*The Rating Score is the weighted average calculated by dividing the sum of all weighted ratings by the number of total responses.
Online Surveys by | 981 | 2,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-39 | latest | en | 0.817294 |
https://www.geeksforgeeks.org/c-c-program-for-longest-increasing-subsequence/ | 1,611,698,817,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803737.78/warc/CC-MAIN-20210126202017-20210126232017-00141.warc.gz | 804,311,780 | 23,576 | Related Articles
C/C++ Program for Longest Increasing Subsequence
• Difficulty Level : Medium
• Last Updated : 04 Dec, 2018
The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80}.
More Examples:
```Input : arr[] = {3, 10, 2, 1, 20}
Output : Length of LIS = 3
The longest increasing subsequence is 3, 10, 20
Input : arr[] = {3, 2}
Output : Length of LIS = 1
The longest increasing subsequences are {3} and {2}
Input : arr[] = {50, 3, 10, 7, 40, 80}
Output : Length of LIS = 4
The longest increasing subsequence is {3, 7, 40, 80}
```
Optimal Substructure:
Let arr[0..n-1] be the input array and L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS.
Then, L(i) can be recursively written as:
L(i) = 1 + max( L(j) ) where 0 < j < i and arr[j] < arr[i]; or
L(i) = 1, if no such j exists.
To find the LIS for a given array, we need to return max(L(i)) where 0 < i < n.
Thus, we see the LIS problem satisfies the optimal substructure property as the main problem can be solved using solutions to subproblems.
Following is a simple recursive implementation of the LIS problem. It follows the recursive structure discussed above.
`/* A Naive C/C++ recursive implementation of LIS problem */` `#include ` `#include ` ` ` `/* To make use of recursive calls, this function must return ` ` ``two things: ` ` ``1) Length of LIS ending with element arr[n-1]. We use ` ` ``max_ending_here for this purpose ` ` ``2) Overall maximum as the LIS may end with an element ` ` ``before arr[n-1] max_ref is used this purpose. ` ` ``The value of LIS of full array of size n is stored in ` ` ``*max_ref which is our final result */` `int` `_lis(``int` `arr[], ``int` `n, ``int``* max_ref) ` `{ ` ` ``/* Base case */` ` ``if` `(n == 1) ` ` ``return` `1; ` ` ` ` ``// 'max_ending_here' is length of LIS ending with arr[n-1] ` ` ``int` `res, max_ending_here = 1; ` ` ` ` ``/* Recursively get all LIS ending with arr[0], arr[1] ... ` ` ``arr[n-2]. If arr[i-1] is smaller than arr[n-1], and ` ` ``max ending with arr[n-1] needs to be updated, then ` ` ``update it */` ` ``for` `(``int` `i = 1; i < n; i++) { ` ` ``res = _lis(arr, i, max_ref); ` ` ``if` `(arr[i - 1] < arr[n - 1] && res + 1 > max_ending_here) ` ` ``max_ending_here = res + 1; ` ` ``} ` ` ` ` ``// Compare max_ending_here with the overall max. And ` ` ``// update the overall max if needed ` ` ``if` `(*max_ref < max_ending_here) ` ` ``*max_ref = max_ending_here; ` ` ` ` ``// Return length of LIS ending with arr[n-1] ` ` ``return` `max_ending_here; ` `} ` ` ` `// The wrapper function for _lis() ` `int` `lis(``int` `arr[], ``int` `n) ` `{ ` ` ``// The max variable holds the result ` ` ``int` `max = 1; ` ` ` ` ``// The function _lis() stores its result in max ` ` ``_lis(arr, n, &max); ` ` ` ` ``// returns max ` ` ``return` `max; ` `} ` ` ` `/* Driver program to test above function */` `int` `main() ` `{ ` ` ``int` `arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 }; ` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` ``printf``(``"Length of lis is %d\n"``, ` ` ``lis(arr, n)); ` ` ``return` `0; ` `} `
Output:
```Length of lis is 5
```
Please refer complete article on Dynamic Programming | Set 3 (Longest Increasing Subsequence) for more details!
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes arrow_drop_up
Recommended Articles
Page : | 1,292 | 3,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-04 | latest | en | 0.830358 |
http://mathforum.org/kb/message.jspa?messageID=9396444 | 1,474,895,394,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660760.0/warc/CC-MAIN-20160924173740-00204-ip-10-143-35-109.ec2.internal.warc.gz | 159,668,185 | 11,224 | Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Topic: Infinity: The Story So Far
Replies: 298 Last Post: Apr 7, 2014 7:30 AM
Messages: [ Previous | Next ]
thenewcalculus@gmail.com Posts: 1,361 Registered: 11/1/13
Re: Infinity: The Story So Far
Posted: Feb 27, 2014 1:37 PM
The sum (or addition) of two numbers (2 and 3) is that number (6) whose difference with either of the two numbers (2 and 3) is either of the two numbers (3 and 4).
Evidently 6 cannot be a sum, but Martin Shobe thinks it can. Tsk, tsk.
6 - 4 = 2 Not Okay. But 4 is NOT one of the numbers.
6 - 3 = 3 Okay.
Yet in this individual's errant brain cells, he reckons that my axioms produce the fallacy 2+3=6.
Sigh, ... :-)
Date Subject Author
2/25/14 Dan Christensen
2/25/14 Brian Q. Hutchings
2/25/14 ross.finlayson@gmail.com
2/25/14 ross.finlayson@gmail.com
2/26/14 Leo Sgouros
2/26/14 ross.finlayson@gmail.com
2/26/14 William Elliot
3/1/14 Dan Christensen
3/2/14 Brian Q. Hutchings
2/26/14 thenewcalculus@gmail.com
2/26/14 Dan Christensen
2/26/14 thenewcalculus@gmail.com
2/26/14 Dan Christensen
2/26/14 Virgil
2/26/14 Dan Christensen
2/26/14 Martin Shobe
2/26/14 ross.finlayson@gmail.com
2/27/14 Dan Christensen
2/27/14 Martin Shobe
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/26/14 thenewcalculus@gmail.com
2/26/14 Dan Christensen
2/26/14 thenewcalculus@gmail.com
2/26/14 Dan Christensen
2/26/14 thenewcalculus@gmail.com
2/26/14 Dan Christensen
2/26/14 Brian Q. Hutchings
2/26/14 mueckenh@rz.fh-augsburg.de
2/26/14 Dan Christensen
2/26/14 mueckenh@rz.fh-augsburg.de
2/26/14 Dan Christensen
2/26/14 Brian Q. Hutchings
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Dan Christensen
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Dan Christensen
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 fom
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Tanu R.
2/27/14 Tanu R.
2/28/14 Virgil
2/27/14 Virgil
2/27/14 Tanu R.
2/27/14 fom
2/27/14 thenewcalculus@gmail.com
2/28/14 Virgil
2/28/14 mueckenh@rz.fh-augsburg.de
2/28/14 Virgil
2/28/14 mueckenh@rz.fh-augsburg.de
2/28/14 Virgil
2/28/14 fom
2/27/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Tanu R.
2/27/14 thenewcalculus@gmail.com
2/27/14 Virgil
2/26/14 Virgil
2/26/14 mueckenh@rz.fh-augsburg.de
2/26/14 Virgil
2/26/14 Dan Christensen
2/26/14 fom
2/26/14 mueckenh@rz.fh-augsburg.de
2/26/14 fom
2/26/14 Dan Christensen
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/27/14 Dan Christensen
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 fom
2/27/14 Peter Percival
2/27/14 fom
2/27/14 Peter Percival
2/27/14 fom
2/27/14 thenewcalculus@gmail.com
4/7/14 Aatu Koskensilta
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Virgil
2/26/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/26/14 thenewcalculus@gmail.com
2/26/14 Brian Q. Hutchings
2/26/14 Virgil
2/26/14 mueckenh@rz.fh-augsburg.de
2/26/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Virgil
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Tanu R.
2/26/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Martin Shobe
2/27/14 thenewcalculus@gmail.com
2/27/14 Martin Shobe
2/27/14 thenewcalculus@gmail.com
2/27/14 thenewcalculus@gmail.com
2/27/14 thenewcalculus@gmail.com
2/27/14 Martin Shobe
2/27/14 thenewcalculus@gmail.com
2/27/14 Martin Shobe
2/27/14 thenewcalculus@gmail.com
2/27/14 Martin Shobe
2/27/14 thenewcalculus@gmail.com
2/28/14 Martin Shobe
2/28/14 thenewcalculus@gmail.com
2/28/14 Martin Shobe
2/28/14 John Gabriel
2/28/14 Martin Shobe
2/28/14 John Gabriel
2/28/14 Sal Honda
2/27/14 Wizard-Of-Oz
2/27/14 fom
2/27/14 thenewcalculus@gmail.com
2/27/14 fom
2/27/14 mueckenh@rz.fh-augsburg.de
2/27/14 Wizard-Of-Oz
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/28/14 Wizard-Of-Oz
2/28/14 thenewcalculus@gmail.com
2/28/14 Wizard-Of-Oz
2/28/14 Dan Christensen
2/28/14 Wizard-Of-Oz
2/28/14 Wizard-Of-Oz
2/28/14 Wizard-Of-Oz
2/28/14 thenewcalculus@gmail.com
2/28/14 Wizard-Of-Oz
2/27/14 thenewcalculus@gmail.com
2/27/14 thenewcalculus@gmail.com
2/27/14 thenewcalculus@gmail.com
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/27/14 Dan Christensen
2/28/14 thenewcalculus@gmail.com
2/28/14 Dan Christensen
2/27/14 thenewcalculus@gmail.com
2/28/14 Dan Christensen
2/28/14 thenewcalculus@gmail.com
2/28/14 Dan Christensen
2/28/14 John Gabriel
2/28/14 Dan Christensen
2/28/14 John Gabriel
2/28/14 fom
2/28/14 John Gabriel
2/28/14 fom
2/28/14 John Gabriel
2/28/14 Dan Christensen
2/28/14 Wizard-Of-Oz
2/28/14 John Gabriel
2/28/14 Harman Kardan
2/28/14 John Gabriel
2/28/14 Wizard-Of-Oz
2/28/14 Johnny Oiler
3/1/14 thenewcalculus@gmail.com
3/1/14 Dan Christensen
3/1/14 thenewcalculus@gmail.com
3/1/14 Dan Christensen
3/1/14 thenewcalculus@gmail.com
3/1/14 Dan Christensen
3/1/14 thenewcalculus@gmail.com
3/1/14 Dan Christensen
3/1/14 thenewcalculus@gmail.com
3/2/14 Wizard-Of-Oz
3/1/14 Dan Christensen
3/2/14 Dan Christensen
3/1/14 Wizard-Of-Oz
3/1/14 thenewcalculus@gmail.com
3/2/14 Wizard-Of-Oz
3/1/14 thenewcalculus@gmail.com
3/1/14 thenewcalculus@gmail.com
3/2/14 Dan Christensen
3/2/14 thenewcalculus@gmail.com
3/2/14 Dan Christensen
3/2/14 John Gabriel
3/2/14 Wizard-Of-Oz
3/2/14 Dan Christensen
3/2/14 thenewcalculus@gmail.com
3/2/14 Dan Christensen
3/2/14 thenewcalculus@gmail.com
3/2/14 Dan Christensen
3/2/14 thenewcalculus@gmail.com
3/2/14 Dan Christensen
3/2/14 John Gabriel
3/2/14 Dan Christensen
3/3/14 John Gabriel
3/3/14 Dan Christensen
3/3/14 Wizard-Of-Oz
3/3/14 thenewcalculus@gmail.com
3/3/14 Virgil
3/2/14 Wizard-Of-Oz
3/2/14 thenewcalculus@gmail.com
3/2/14 Jürgen R.
3/2/14 thenewcalculus@gmail.com
3/3/14 New Balance
3/2/14 Wizard-Of-Oz
3/2/14 New Balance
3/4/14 thenewcalculus@gmail.com
3/4/14 thenewcalculus@gmail.com
3/4/14 Dan Christensen
3/4/14 thenewcalculus@gmail.com
3/4/14 Dan Christensen
3/4/14 Dan Christensen
3/4/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/4/14 ross.finlayson@gmail.com
3/5/14 ross.finlayson@gmail.com
3/5/14 Dan Christensen
3/5/14 ross.finlayson@gmail.com
3/5/14 Brian Q. Hutchings
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/6/14 thenewcalculus@gmail.com
3/6/14 Dan Christensen
3/6/14 Dan Christensen
3/6/14 thenewcalculus@gmail.com
3/5/14 YBM
3/6/14 New Balance
3/6/14 thenewcalculus@gmail.com
3/6/14 Brian Q. Hutchings
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 thenewcalculus@gmail.com
3/5/14 Dan Christensen
3/5/14 Peter Percival
3/5/14 thenewcalculus@gmail.com
3/5/14 Martin Shobe
3/5/14 Wizard-Of-Oz
3/5/14 Wizard-Of-Oz
3/5/14 Wizard-Of-Oz
3/5/14 thenewcalculus@gmail.com
3/6/14 Dan Christensen
3/6/14 thenewcalculus@gmail.com
3/6/14 thenewcalculus@gmail.com
3/11/14 thenewcalculus@gmail.com
3/11/14 Wizard-Of-Oz
3/11/14 Martin Shobe
3/11/14 thenewcalculus@gmail.com
3/11/14 Martin Shobe
3/11/14 thenewcalculus@gmail.com
3/11/14 thenewcalculus@gmail.com
3/11/14 Martin Shobe
3/11/14 thenewcalculus@gmail.com
3/11/14 thenewcalculus@gmail.com
3/11/14 Martin Shobe
3/12/14 thenewcalculus@gmail.com
3/12/14 Martin Shobe
3/12/14 ross.finlayson@gmail.com | 3,399 | 8,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-40 | longest | en | 0.590062 |
https://www.jiskha.com/questions/1511985/If-x-2y-2-x-3y-4-3xy-2-then-xy-2-a-1-5-b-4-c-8-d-16-e-64 | 1,558,953,660,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00012.warc.gz | 840,587,500 | 5,556 | # Algebra
If x^2y^2+x^3y^4=(3xy)^2,then xy^2=
a. 1.5
b.4
c.8
d.16
e.64
1. 👍 0
2. 👎 0
3. 👁 37
1. x ^ 2 * y ^ 2 + x ^ 3 * y ^ 4 = ( 3 x y ) ^ 2
x ^ 2 * y ^ 2 + x ^ 2 * y ^ 2 ( x * y ^ 2 ) = 9 x ^ 2 * y ^ 2
x ^ 2 * y ^ 2 * ( 1 + x * y ^ 2 ) = 9 * x ^ 2 * y ^ 2 Divide both sides by x ^ 2 * y ^ 2
1 + x * y ^ 2 = 9 Subtract 1 to both sides
1 + x * y ^ 2 - 1 = 9 - 1
x * y ^ 2 = 8
x y ^ 2 = 8
1. 👍 0
2. 👎 0
posted by Bosnian
## Similar Questions
Posted by Jessie on Monday, October 18, 2010 at 3:48pm. I think I figured it out-please check I'm really stuck on these problems- Simplify 6xy(3xy)^-2? I think this is 2/3xy choices were 2/xy 3xy 2xy 2/3xy Simplify 12xyz^-2/8x^-2z
asked by Jessie on October 18, 2010
Posted by Jessie on Monday, October 18, 2010 at 3:48pm. I think I figured it out-please check I'm really stuck on these problems- Simplify 6xy(3xy)^-2? I think this is 2/3xy choices were 2/xy 3xy 2xy 2/3xy Simplify 12xyz^-2/8x^-2z
asked by Jessie on October 18, 2010
3. ### Simultaneous equation
X+y+z=1......(1) x^2+y^2+z^2=35........(2) x^3+y^3+z^3=97.........(3) solution but from 3 (x^3+y^3)=(x+y)^3-3xy(x+y)......(4) puting 4 into 3 bck (x+y)^3-3xy(x+y)+z^3=97.......(5) now from 1 x+y=1-z......(6) putin 6 into 7 we have
asked by Collins on October 7, 2015
4. ### polynomials
I have to simplify (-x^5y^4)^4(3x^2y)^2(1/3xy) I did (x^20y^16)(9x^4y^2)(1/3xy) (9x^24y^18)(1/3xy) 3x^25y^19 Is this correct? Thank you for your help!
asked by sara on September 25, 2016
5. ### Derive
Find the derivative of sin(2x + 3y)= 3xy + 5y-2. I have this so far 1) cos(2x+3y) * d/dx (2x+3y)= 3xy' + 3y + 5y' 2) cos (2x+3y) * [2 + 3y']= 3xy' + 3y + 5y' 3) 3y' cos (2x+3y) + 2cos(2x+3y)= 3xy' + 3y + 5y' 4) 3xy' + 5y' -
asked by Mary on March 7, 2013
6. ### Math help polynomial
multiply the following polynomial -3x(y^2+2y^2) answer choices 3xy^2-6xy^2' -3xy^2-gxy^2 -3xy^2+5y^2 -9xy^2 The last one was the answer. Can you tell me how the answer was solved? Thanks!
asked by Alex on February 15, 2014
7. ### differential equations
the directions are to determine whether or not the equation is exact if so then solve it. the question is (x-y^3+ysinx)dx = (3xy^2+2ycosx)dy i solve and i got that it is exact because My is -3y^2+2ysinx and Nx is -3y^2+2ysinx
asked by allison on October 8, 2008
8. ### Math
I have no clue how to do this as no values were given and im confused. can someone explain plz? if x^3-y^3=m and x-y=N, express xy in terms of M and N This is one of those pure manipulation problems. you have to know that x^3-y^3
asked by luke on May 3, 2007
9. ### Math!
1) What is the value of 19 + 5a^4 + 3b^2 when a = 2 and b = -4? (this is not a multiple choice question) but my answer is 147. 2)Combine the like terms to simplify the expression: 4 + 7xy^2 + 5y^3 + 10x -4xy^2 - 8x - 4y^3 + 9 -
asked by Anonymous on September 14, 2014
10. ### bobpursly,
Is this correct now after i revised it Multiply. 3xy^2(5xy-4x+6y) 3xy^2(5xy)=15x^2y^3 3xy^2(-4x)=-12x^2y^2 3xy^2(6y)=18xy^3 so i get from this 15x^2y^3-12x^2y^2+18xy^3 correct awsome i just thought that the 15x^2y^3-12x^2y^2
asked by jas20 on February 28, 2007
11. ### algebra 2
(w + 2t)(w^2 - 2wt + 4t^2) (x + y)(x^2 -3xy + 2y^2) =x^3 - 3x^2y + 2xy^2 + x^2y -3xy^2 + 2y^3 = x^3-2x^2y-xy^2+2y^3
asked by Reina Bonilla on January 21, 2015
More Similar Questions | 1,532 | 3,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.863369 |
https://oeis.org/A215511 | 1,708,500,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00027.warc.gz | 447,389,670 | 4,126 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A215511 A sequence of prime numbers expressed as minimum bases using only digits 0 and 1. 1
10, 10, 11, 101, 111, 1011, 10001, 10011, 11101, 11111, 11101, 10011, 10111, 101111 (list; graph; refs; listen; history; text; internal format)
OFFSET 3,1 COMMENTS 3 = 10 base 3 = 3 + 0 5 = 10 base 5 = 5 + 0 7 = 11 base 6 = 6 + 1 37 = 101 base 6 = 36 + 0 + 1 43 = 111 base 6 = 36 + 6 + 1 223 = 1011 base 6 = 216 + 0 + 6 + 1 1297 = 10001 base 6 = 1296 + 0 + 0 + 0 + 1 1303 = 10011 base 6 = 1296 + 0 + 0 + 6 + 1 1549 = 11101 base 6 = 1296 + 216 + 36 + 0 + 1 2801 = 11111 base 7 = 2401 + 343 + 49 + 7 + 1 4673 = 11101 base 8 = 4096 + 512 + 64 + 0 + 1 6571 = 10011 base 9 = 6561 + 0 + 0 + 9 + 1 10111 = 10111 base 10 = 10000 + 0 + 100 + 10 + 1 101111 = 101111 base 10 = 100000 + 0 + 1000 + 100 + 10 + 1 LINKS Table of n, a(n) for n=3..16. FORMULA Step 1: Starting at the first prime number (3), convert to the minimum base (3, as all primes may be expressed in binary). Step 2: If the next prime number can be converted into the same base using only 0 and 1 without exceeding the value of the next prime number in the next base, this is the next item in the sequence. Step 3: If the next prime number cannot be expressed in this base before exceeding the value of the next prime number in the next base, skip this prime number and move on to the next prime number and repeat Step 2. Step 4: If the next prime number cannot be expressed in this base before exceeding the value of the next prime number in the next base, but can be expressed in the next base, this is the next item in the sequence. EXAMPLE The first term is 3 in base 3. The next prime in that base is 13, which is greater than the value of the prime in the next base, which is 5 in base 4, so the second term is 5 in base 4. CROSSREFS A126359 Sequence in context: A004209 A162720 A112867 * A280105 A173198 A111381 Adjacent sequences: A215508 A215509 A215510 * A215512 A215513 A215514 KEYWORD nonn,base AUTHOR Jason Betts, Aug 14 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 February 21 02:19 EST 2024. Contains 370219 sequences. (Running on oeis4.) | 844 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.794164 |
https://www.wedugo.com/category/area/partB/ | 1,600,984,242,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400220495.39/warc/CC-MAIN-20200924194925-20200924224925-00415.warc.gz | 1,040,484,108 | 7,246 | # Area | Multiple Choice Questions (MCQ)
## Curved surface area of a cylinder is 528 sq cm. If circumference of its base is 44 cm, find the height of the cylinder?
12 cm
24 cm
36 cm
6 cm
Area
1 year ago | 08/03/2019 16:02:57 | gknowledge
ID - 480dab6b4f-sec2
## In a triangle the length of the side opposite the right angle is 9√3 cm, what is the length of the side opposite to the angle which measures 30 degree?
9 cm
3√3 cm
6 cm
(9√3)/2 cm
Area
1 year ago | 08/03/2019 16:00:56 | gknowledge
ID - 830f6e12f1-sec2
## In ΔDEF, G and H are points on side DE and DF respectively. GH is parallel to EF. If G divides DE in the ratio 1:3 and HF is 7.2 cm, find length of DF
2.4 cm
4.8 cm
3.6 cm
9.6 cm
Area
1 year ago | 08/03/2019 15:59:35 | gknowledge
ID - 0a664913d5-sec2
## If the radius of a circle is increased by 15% its area increases by _________
30 percent
32.25 percent
15 percent
16.125 percent
Area
1 year ago | 06/03/2019 18:40:44 | gknowledge
Next Page
Previous Page | 355 | 984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.801337 |
https://hockey-graphs.com/2015/09/10/rate-metrics-matter/ | 1,501,080,067,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426169.17/warc/CC-MAIN-20170726142211-20170726162211-00129.warc.gz | 671,556,437 | 37,225 | # Rate Metrics Matter
The other day, @Moneypuck_ and @SteveBurtch had a conversation about the Prospect Cohort Success Model:
While the PCS model is interesting in its own right, I found the discussion about the methods we use to analyze players to be interesting as well.
The above conversation shows that the same player can appear differently depending on which metric is used. Burtch seems to argue that Orr’s accumulation of games played is misleading, since he apparently did not play much in any individual game.
Justin Bourne touched off a Hockey Twitter maelstrom with his tweets regarding rate metrics. I don’t want to unfairly castigate Justin here, but it’s clear that there is still some disagreement in the hockey community on how to measure player production.
Rate metrics are one of the many innovations of modern hockey analysis. Analysts have begun to substitute them for traditional point totals. Point totals ignore one of the most important drivers of production: opportunity. Skaters play different amounts of games in a season, and skaters can be allotted different amounts of time within those games.
Doing the math quickly, if Skater A scores 60 points in 82 games, that’s a .73 points per game (PPG) rate. If Skater B scores 55 points in 70 games, that results in a .78 PPG rate. Skater B clearly produced at a higher rate, but you wouldn’t see that using point totals alone.
We can further investigate the amount of opportunity that players receive using time on ice (TOI). Assume that Skater A played 20 minutes a night and scored 60 points in 82 games. That would be a 2.2 Points per 60 minutes pace ((60 points/1640 minutes)*60 minutes). If Skater B played 15 minutes a night and scored 55 points in 70 games, he would perform at a rate of 3.14 Points per 60 minutes pace ((55 points/1050 minutes)*60 minutes).
Skater A scored more total points in more total games, but Skater B performed at a higher rate with less opportunity. Perhaps Skater B could surpass Skater A’s point totals with more TOI per game (TOI/Gm).
The variance between TOI and GP increases as you go up and to the right. The clear takeaway from this graph is that TOI and GP do not increase in lockstep with each other. If there was no variation in the distribution of TOI within games, TOI would increase solely as a function of GP. If that were the case, we would see a straight line. Instead, we see a curved line with increasing variance as GP increases.
We can see that there are players whose TOI lags behind where it should be if you assume a uniform distribution of minutes. Grinders, defensive specialists, and enforcers (Chris Neil, Craig Adams, and Colton Orr), and defensive defensemen (Corey Sarich, Tim Gleason, and Anton Volchenkov) are examples of players whose TOI lag behind their GP.
This graph shows that Time on Ice Per Game (TOI/Gm) is not tightly distributed around the mean (though the distribution does not vary much across the two positions). This means that there are wide differences in how many minutes skaters can play in a game.
Looking at how TOI interacts with Points and Points per 60 (P60), it’s clear that how many Points a player accumulates is heavily influenced by how much TOI they are allotted. The direction of the slope holds for each position, but the effect is stronger for forwards. The higher R^2 values indicate how much the metrics are correlated with each other. The R^2 value of .92 (0-1 range) indicates that TOI and Points (forwards) are deeply connected, while the relationship between P60 and TOI is much weaker.
TOI/Gm also affects both Points and Points/Gm, though less so than TOI. The same pattern holds here as well.
If you argue that TOI is not worth adjusting for, you are arguing that there are not significant differences in player TOI/Gm. This is clearly not the case. Furthermore, that position would need to hold that TOI allocation does not affect point production. That is also not supported by the evidence.
I think it is clear that TOI is the superior metric to standardize player performance against because it is sensitive to the varying levels of opportunity that players experience within individual games. While it does not control for other factors (zone starts, quality of teammate, score situation deployment), it is an incremental improvement well worth using. Point totals do not control for any of these variables. It’s time to move on.
I created an interactive viz in which you can explore the data further.
Data from War-On-Ice | 995 | 4,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-30 | longest | en | 0.949663 |
https://goopennc.oercommons.org/authoring/694-t4t-ducks-on-the-pond/view | 1,723,709,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00079.warc.gz | 217,477,797 | 13,262 | # T4T Ducks on the Pond
Lesson Excerpt:
Launch:
1. Ducks on the Pond (3-5 minutes)
During whole group time, show Picture Card K to the entire class and model the activity.
The teacher asks students, “What do you see on the card?”
Possible responses: I see ducks. I see water.
The teacher says:
Let’s see if I can come up with a story problem for this picture for you to solve. There are 2 dark ducks and 1 white duck. How many ducks are on the water?
As students give their solutions, it is acceptable to allow many students to share their solution, even if they all say “3.”
The teacher may ask: How many ducks are in the water? Does anyone have a different solution?
Regardless of the solution, the teacher should not lead students to think whether their solution is correct or incorrect.
The teacher then asks students: How can we find out how many ducks are in the water?
Possible responses:
We can count the number of ducks.
I know that 1 more than 2 is 3.
I know that 2 and 1 is 3.
I counted them all: 1, 2, 3. | 253 | 1,048 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.967972 |
https://mechasco.wordpress.com/2015/02/19/imaginary-powers/ | 1,500,611,301,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423716.66/warc/CC-MAIN-20170721042214-20170721062214-00492.warc.gz | 688,998,702 | 37,287 | Imaginary Powers
Imaginary numbers ($i = \sqrt{-1}$) are used to relate Cartesian coordinates to polar coordinates as shown in the figure below. The argument $\varphi$ is the angle from the real axis along the unit circle, and can be expressed as two terms: a real and an imaginary, giving the Cartesian (x,y) position.
$e^{i\varphi} = \text{cos}(\varphi) + i \text{sin}(\varphi) = x + i y$
(If you want to expand beyond the unit circle just multiply by the desired radius: $e^{i\varphi} \rightarrow re^{i\varphi}$)
This is known as the Euler relation, and it gives us some neat results, like:
$e^{i\pi} = -1$
$e^{i \pi /2} = i$
But what happens when you exponentiate $i$ with itself? $i^i$.
From the Euler relation we know that $i=e^{i \pi /2}$, so it’s really a matter of substitution:
$i^i = e^{i \pi / 2 i} = e^{-\pi / 2} = 0.20788...$
So $i^i$ is a real number!
But wait a second, this isn’t the whole answer. We can add or subtract $2\pi$ from any angle and wind up in the same place on the circle. So $i = e^{i \pi /2 + i 2 \pi n}$, where $n$ is any integer. So we end up with a set of values:
$i^i = e^{- \pi /2 - 2 \pi n}$
which are all real!
The curve above goes through the set of values for $i^i$, where $n$ is an integer.
$i^i = \{...,111.32,0.20788,0.00038820,...\}$
But 0.20788… will be looked at as the primary value for now since it is the result when $n=0$, and having one value keeps things simple.
So what happens when we repeat this exponentiation? $i^{i^i}$
First, this can be written in a simpler way:
$^3i = i^{i^i}$
This operation is called “tetration” or “power tower”.
You start from the top and work your way down. So $^3i = i^{i^i} = i^{0.20788}$
Now use the Euler relation again (ignoring $2 \pi n$ terms in the exponential):
$i^{0.20788} = e^{i \pi/2 \times 0.20788} = e^{i 0.32653}$
So the result is complex:
$^3i = 0.94715+0.32076i$
We can keep going using results from the last iteration:
$^4i = i^{0.94715+0.32076 i} = 0.05009+0.60211 i$
Eventually you find that with $^mi$ as $m$ increases, the results start to converge on one number. This implies there’s a value for $^\infty i$. After plugging and chugging we can eventually conclude that:
$^\infty i = 0.43828 + 0.36059 i$
Convergence can be seen in both the real and imaginary components, around $m \approx 70$ if we require the difference between iterations to be less than 0.1%. (See figures below).
Each result has a different radius associated with it, since the imaginary component of the previous result puts a real term in the exponential of the next. This real term can be expressed as a factor $r$:
$e^{i\varphi} = a + bi \rightarrow i^{a+bi} = e^{i \pi/2 (a + bi)} = e^{-b\pi/2}e^{i a \pi/2} = r e^{i a \pi/2}$
The radius can be found by multiplying the result by its complex conjugate and then find the square root (i.e. the Pythagorean theorem).
We can then plot radius of $^mi$ as a function of $m$ (see plot below).
The python code for making these plots is:
import os
import sys
import math
import numpy
import pylab
REAL = []
IMAG = []
ITERA = []
pp = 1j
for n in range(100):
ppL = pp #save previous value for cut off
pp = (1j)**pp #interate power tower
RA = 0.001 #ratio for cut off
if (abs(pp.real - ppL.real)/(ppL.real + 0.00001) < RA) and (abs(pp.imag - ppL.imag)/(ppL.imag + 0.00001) < RA): #establish a cut off
print n
print pp
break
else:
REAL.append(pp.real) #save real component
IMAG.append(pp.imag) #save imaginary component
ITERA.append(n+2) #save iteration "m"
ITERA = [1]+ITERA
REAL = [0]+REAL
IMAG = [1]+IMAG
pylab.figure(1)
pylab.xlabel("m")
# pylab.show()
pylab.figure(2)
pylab.xlabel("m")
pylab.ylabel("imaginary component")
pylab.plot(ITERA,IMAG,'k')
pylab.savefig('Imfig.png')
pylab.figure(3)
pylab.xlabel("m")
pylab.ylabel("real component")
pylab.plot(ITERA,REAL,'k')
pylab.savefig('Refig.png') | 1,228 | 3,843 | {"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": 35, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.830171 |
https://math.stackexchange.com/questions/1524297/how-to-compute-lim-n-to-inftyey-ny-n-1/1524386 | 1,716,470,870,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00592.warc.gz | 333,677,666 | 37,359 | # How to compute $\lim_{n\to +\infty}E[Y_{n}Y_{n-1}]$? [closed]
Let $X_{1}, X_{2},\ldots, X_{n},\ldots$ be independent and identically distributed random variables with distribution
$P(X_{n}=0)=P(X_{n}=1)=P(X_{n}=2)=P(X_{n}=3)=\frac{1}{4}$
Assume that the sequence $\{Y_n\}$ is defined as: $Y_{0}=0$ and for all $n\in\mathbb{N}$ we have
$Y_n=\begin{cases} 3 &\text{if } X_n=3,\\\min{\{Y_{n-1},X_n\}} &\text{if } X_n<3. \end{cases}$
Compute $\displaystyle \lim_{n\to +\infty}E[Y_{n}Y_{n-1}]$?
I don't know how to start.
Behavior of $Y_i$ is described by an aperiodic irreducible Markov chain on $4$ states which has a unique limiting distribution $\pi$ (a row vector). Let $f=(0,1,2,3)^T$ (column vector - values of $Y_i$ on $4$ states). Then $$\lim E(Y_{n-1}Y_n)=\pi(f\cdot Pf)$$ where $\cdot$ presents coordinate-wise product. You are left to explicitely contruct $P$ and find $\pi$ s.t. $\pi P=\pi$.
• Thank you for your post but i am not familiar with Markov chain theory, so this problem seems to be too hard for me. Why I recived -1 point for my question? What is wrong with it?
– Leon
Nov 12, 2015 at 6:02
• @Leon Ignore the haters. I don't see a way to approach this problem without Markov Chain theory.
– A.S.
Nov 12, 2015 at 15:56
$$E(Y_{n}Y_{n+1})=E(E(Y_{n}Y_{n+1}|X_{n+1}))\\=\frac{1}{4}[E(Y_{n}\min\{Y_{n},0\})+E(Y_{n}\min\{Y_{n},1\})+E(Y_{n}\min\{Y_{n},2\})+3E(Y_n)]$$ Now, note that for all $n\ge 1,\ Y_n\in \{0,1,2,3\}$. Then, \begin{eqnarray*} E(Y_{n}\min\{Y_{n},0\})&=&0\\ E(Y_{n}\min\{Y_{n},1\})&=&P(Y_{n}=1)+2P(Y_{n}=2)+3P(Y_{n}=3)\\ E(Y_{n}\min\{Y_{n},2\})&=&P(Y_{n}=1)+4P(Y_{n}=2)+6P(Y_{n}=3) \end{eqnarray*} Let $$p_n:=P(Y_n=1),\ q_n:=P(Y_n=2),\ r_n=P(Y_n=3)$$ Then, we have the recursion, $$E(Y_{n+1}Y_{n})=\frac{5p_n+12q_n+18r_n}{4},\ n\ge 1$$ Now, we note that $$p_{n+1}=P(Y_{n+1}=1)=P(Y_n= 1,1\le X_{n+1}<3)+P(1< Y_n, X_{n+1}=1)\\=\frac{2p_n+q_n+r_n}{4}\quad(\because X_{n+1}\perp Y_n)\\ q_{n+1}=\frac{(q_n+r_n)}{4}\\ r_{n+1}=\frac{1}{4}$$ Thus, defining $$v_{n+1}=[p_{n+1}\quad q_{n+1}]^T$$ we get the recursive equation, $$v_{n+1}=Av_n+a$$ where $$A=\begin{bmatrix} 2/4 & 1/4\\ 0 & 1/4 \end{bmatrix}\\ a=[1/16\quad 1/16]^T$$ Also, note the initial conditions $$p_1=0,\ q_1=0$$ Thus, noting that the maximum eigenvalue of matrix $A$ is $1/2$, the sequence $v_n$ converges, and we can find the limit as $$\lim_{n\to \infty}v_n=(I-A)^{-1}a$$ Thus $$\lim_{n\to \infty}E(Y_nY_{n-1})=[5/4\quad 3](I-A)^{-1}a+\frac{9}{8}.$$
• $P(Y_{n+1}=1)=P(Y_n\ge 1,1\le X_{n+1}<3)$ is not quite correct. If $Y_n=X_{n+1}=2$ then $Y_{n+1}=2$, not $Y_{n+1}=1$.
– user940
Nov 11, 2015 at 18:35
• In the first line should be $E(Y_{n}Y_{n-1})=E[E(Y_{n}Y_{n-1}|X_{n})]$?
– Leon
Nov 12, 2015 at 5:58
• @ByronSchmuland, sorry, that was a mistake, I am editing it. Nov 12, 2015 at 6:08
• @ByronSchmuland, I have edited my answer. Nov 12, 2015 at 6:45
• Your equations for $p_{n+1}$ and $q_{n+1}$ are still incorrect...
– user940
Nov 12, 2015 at 16:57 | 1,337 | 2,957 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.680833 |
https://www.breathinglabs.com/monitoring-feed/genetics/how-many-cells-are-represented-in-a-sample-going-through-ngs/ | 1,624,601,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00381.warc.gz | 605,258,014 | 16,469 | # How many cells are represented in a sample going through NGS
3 hours ago by
Netherlands, Rotterdam, ErasmusMC
It is easy to compute. You just need the ng of DNA input in the NGS experiment. For example, assume we input 250 ng of DNA (that is what is required in a standard WGS experiment*). Each cell contains 6 pg of DNA, hence, the number of cells is:
``````250 ng * (1000 pg / 1 ng) * (1 cell / 6 pg**) = 41,667.67 cells
``````
*The amount of 6 pg per cell can be estimated from: human genome is around 6200 Mbp, the average nucleotide molecular weight is about 660 g/bp mol and Avogadro's number is 6.022 10^23. Hence:
``````6200 Mbp * (10^6 bp/ 1 Mbp) * (1 bp mol/ 6.022 * 10^23 bp) * (660 g/ 1 bp mol) * (10^12 pg/ 1 g)= 6.795 pg
``````
Of course, as stated by ATpoint in the comment, the amount of input DNA highly depends on the NGS experiment. | 276 | 861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-25 | longest | en | 0.919156 |
https://www.interviewcake.com/question/csharp/simulate-7-sided-die | 1,670,423,181,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00074.warc.gz | 866,798,850 | 43,043 | ### You only have free questions left (including this one).
But it doesn't have to end here! Sign up for the 7-day coding interview crash course and you'll get a free Interview Cake problem every week.
You have a method Rand5 that generates a random integer from 1 to 5. Use it to write a method Rand7 that generates a random integer from 1 to 7.
Rand5 returns each integer with equal probability. Rand7 must also return each integer with equal probability.
Simply running Rand5 twice, adding the results, and taking a modulus won't give us an equal probability for each possible result.
Not convinced? Count the number of ways to get each possible result from 1..7.
Your method will have worst-case infinite runtime, because sometimes it will need to "try again."
However, at each "try" you only need to make two calls to Rand5. If you're making 3 calls, you can do better.
We can get away with worst-case space. Does your answer have a non-constant space cost? If you're using recursion (and your language doesn't have tail-call optimization), you're potentially incurring a worst-case infinite space cost in the call stack.
Because we need a random integer between 1 and 7, we need at least 7 possible outcomes of our calls to Rand5. One call to Rand5 only has 5 possible outcomes. So we must call Rand5 at least twice.
Can we get away with calling Rand5 exactly twice?
Our first thought might be to simply add two calls to Rand5, then take a modulus to convert it to an integer in the range 1..7:
public static int Rand7Mod() { return (Rand5() + Rand5()) % 7 + 1; }
However, this won't give us an equal probability of getting each integer in the range 1..7. Can you see why?
There are at least two ways to show that different results of Rand7Mod have different probabilities of occurring:
1. Count the number of outcomes of our two Rand5 calls which give each possible result of Rand7Mod
2. Notice something about the total number of outcomes of two calls to Rand5
If we count the number of ways to get each result of Rand7Mod:
result of Rand7Mod # pairs of Rand5 results that give that result
1 4
2 3
3 3
4 3
5 3
6 4
7 5
So we see that, for example, there are five outcomes that give us 7 but only three outcomes that give us 5. We're almost twice as likely to get a 7 as we are to get a 5.
But even without counting the number of ways to get each possible result, we could have noticed something about the total number of outcomes of two calls to Rand5 , which is 25 (5*5). If each of our 7 results of Rand7Mod were equally probable, we'd need to have the same number of outcomes for each of the 7 integers in the range 1..7. That means our total number of outcomes would have to be divisible by 7, and 25 is not.
Okay, so Rand7Mod won't work. How do we get equal probabilities for each integer from 1 to 7?
Is there some number of calls we can make to Rand5 to get a number of outcomes that is divisible by 7?
When we roll our die n times, we get 5^n possible outcomes. Is there an integer n that will give us a 5^n that's evenly divisible by 7?
No, there isn't.
That might not be obvious to you unless you've studied some number theory. It turns out every integer can be expressed as a product of prime numbers (its prime factorization). It also turns out that every integer has only one prime factorization.
Since 5 is already prime, any number that can be expressed as 5^n (where n is a positive integer) will have a prime factorization that is all 5s. For example, here are the prime factorizations for 5^2, 5^3, 5^4:
5^2 = 25 = 5 * 5 5^3 = 125 = 5 * 5 * 5 5^4 = 625 = 5 * 5 * 5 * 5
7 is also prime, so if any power of 5 were divisible by 7, 7 would be in its prime factorization. But 7 can't be in its prime factorization because its prime factorization is all 5s (and it has only one prime factorization). So no power of 5 is divisible by 7. BAM MATHPROOF.
So no matter how many times we run Rand5 we won't get a number of outcomes that's evenly divisible by 7. What do we dooooo!?!?
Let's ignore for a second the fact that 25 isn't evenly divisible by 7. We can think of our 25 possible outcomes from 2 calls to rand5 as a set of 25 "slots" in an array:
int[] results = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };
Which we could then try to evenly distribute our 7 integers across:
int[] results = { 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4 };
It almost works. We could randomly pick an integer from the array, and the chances of getting any integer in 1..7 are pretty evenly distributed. Only problem is that extra 1, 2, 3, 4 in the last row.
Any way we can sidestep this issue?
What if we just "throw out" those extraneous results in the last row?
21 is divisible by 7. So if we just "throw out" our last 4 possible outcomes, we have a number of outcomes that are evenly divisible by 7.
But what should we do if we get one of those 4 "throwaway" outcomes?
We can just try the whole process again!
Okay, this'll work. But how do we translate our two calls to Rand5 into the right result from our array?
What if we made it a 2-dimensional array?
int[][] results = { new[] { 1, 2, 3, 4, 5 }, new[] { 6, 7, 1, 2, 3 }, new[] { 4, 5, 6, 7, 1 }, new[] { 2, 3, 4, 5, 6 }, new[] { 7, 1, 2, 3, 4} };
Then we can simply treat our first roll as the row and our second roll as the column. We have an equal chance of choosing any column and any row, and there are never two ways to choose the same cell!
public static int Rand7Table() { int[][] results = { new[] { 1, 2, 3, 4, 5 }, new[] { 6, 7, 1, 2, 3 }, new[] { 4, 5, 6, 7, 1 }, new[] { 2, 3, 4, 5, 6 }, new[] { 7, 0, 0, 0, 0 } }; // Do our die rolls int row = Rand5() - 1; int column = Rand5() - 1; // Case: we hit an extraneous outcome // so we need to re-roll if (row == 4 && column > 0) { return Rand7Table(); } // Our outcome was fine. Return it! return results[row][column]; }
This'll work. But we can clean things up a bit.
By using recursion we're incurring a space cost in the call stack, and risking stack overflow. This is especially scary because our method could keep rerolling indefinitely (though it's unlikely).
How can we rewrite this iteratively?
Just wrap it in a while loop:
public static int Rand7Table() { int[][] results = { new[] { 1, 2, 3, 4, 5}, new[] { 6, 7, 1, 2, 3 }, new[] { 4, 5, 6, 7, 1 }, new[] { 2, 3, 4, 5, 6 }, new[] { 7, 0, 0, 0, 0 } }; while (true) { // Do our die rolls int row = Rand5() - 1; int column = Rand5() - 1; // Case: we hit an extraneous outcome // so we need to re-roll if (row == 4 && column > 0) { continue; } // Our outcome was fine. Return it! return results[row][column]; } }
One more thing: we don't have to put our whole 2-d results array in memory. Can you replace it with some arithmetic?
We could start by coming up with a way to translate each possible outcome (of our two Rand5 calls) into a different integer in the range 1..25. Then we simply mod the result by 7 (or throw it out and try again, if it's one of the last 4 "extraneous" outcomes).
How can we use math to turn our two calls to Rand5 into a unique integer in the range 1..25?
What did we do when we went from a 1-dimensional array to a 2-dimensional one above? We cut our set of outcomes into sequential slices of 5.
How can we use math to make our first roll select which slice of 5 and our second roll select which item within that slice?
We could take something like:
int outcomeNumber = roll1 * 5 + roll2;
But since each roll gives us an integer in the range 1..5 our lowest possible outcome is two 1s, which gives us 5 + 1 = 6, and our highest possible outcome is two 5s, which gives us 25 + 5 = 30. So we need to do some adjusting to ensure our outcome numbers are in the range 1..25:
int outcomeNumber = ((roll1 - 1) * 5 + (roll2 - 1)) + 1;
(If you're a math-minded person, you might notice that we're essentially treating each result of Rand5 as a digit in a two-digit base-5 integer. The first roll is the fives digit, and the second roll is the ones digit.)
Can you adapt our method to use this math-based approach instead of the results array?
Because Rand5 has only 5 possible outcomes, and we need 7 possible results for Rand7, we need to call Rand5 at least twice.
When we call Rand5 twice, we have 5*5=25 possible outcomes. If each of our 7 possible results has an equal chance of occurring, we'll need each outcome to occur in our set of possible outcomes the same number of times. So our total number of possible outcomes must be divisible by 7.
25 isn't evenly divisible by 7, but 21 is. So when we get one of the last 4 possible outcomes, we throw it out and roll again.
So we roll twice and translate the result into a unique outcomeNumber in the range 1..25. If the outcomeNumber is greater than 21, we throw it out and re-roll. If not, we mod by 7 (and add 1).
public static int Rand7() { while (true) { // Do our die rolls int roll1 = Rand5(); int roll2 = Rand5(); int outcomeNumber = (roll1 - 1) * 5 + (roll2 - 1) + 1; // If we hit an extraneous // outcome we just re-roll if (outcomeNumber > 21) { continue; } // Our outcome was fine. Return it! return outcomeNumber % 7 + 1; } }
Worst-case time (we might keep re-rolling forever) and space.
As with the previous question about writing a Rand5 method, the requirement to "return each integer with equal probability" is a real sticking point.
Lots of candidates come up with clever -time solutions that they are so sure about. But their solutions aren't actually uniform (in other words, they're not truly random).
In fact, it's impossible to have true randomness and non-infinite worst-case runtime.
If you don't understand why, go back over our proof using "prime factorizations," a little ways down in the breakdown section.
Reset editor | 2,794 | 9,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.926117 |
https://web2.0calc.com/questions/inverted-cone-problem | 1,553,547,778,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00228.warc.gz | 666,599,911 | 5,880 | +0
# Inverted Cone Problem
+5
338
2
+637
A tank that is in the form of an inverted cone contains liquid. The height h, in meters, of the space above the liquid is given by the formula h = 21 - 7/2r, where r is the radius of the liquid surface, in meters. The circumference of the top of the tank, in meters is...
(a) 9 pi
(b) 12 pi
(c) 15 pi
(d) 18 pi
(e) 21 pi
The answer says it is 12 pi but there is no solution, can some please explain?
Thanks,
Oct 8, 2017
#1
+7352
+4
When the height of the space above the liquid equals 0, the radius of the liquid's surface will be the same as the radius of the top of the tank.
So, plug in 0 for h and solve for r .
0 = 21 - $$\frac72$$ r Add $$\frac72$$r to both sides of the equation.
$$\frac72$$r = 21 Multiply both sides of the equation by $$\frac27$$ .
r = 6
This is the radius (in meters) of the liquid surface when there is no space above it, and so this is the radius of the top of the tank. So we have a circle with a radius of 6 meters.
circumference = 2pi * 6
circumference = 12pi
Oct 8, 2017
#2
+637
+1
Thank you so much!!!
supermanaccz Oct 9, 2017 | 399 | 1,162 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-13 | latest | en | 0.795786 |
http://bioinfo.iric.ca/implementing-a-siamese-neural-network-with-mariana-1-0/ | 1,505,919,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687324.6/warc/CC-MAIN-20170920142244-20170920162244-00364.warc.gz | 40,284,217 | 34,209 | # Implementing a “Siamese” Neural Network with Mariana 1.0
## Implementing a “Siamese” Neural Network with Mariana 1.0
Mariana was previously introduced in this blog by Geneviève in her May post Machine learning in life science.
The Mariana codebase is currently standing on github at the third release candidate before the launch of the stable 1.0 release. This new version incorporates a large refactorization effort as well as many new features (a complete list of the changes found in the 1.0 version can be found in the changelog). I am taking this opportunity to present here a small tutorial on extending the functionalities of Mariana 1.0.
### Implementing a “Siamese” Neural Network
The Mariana documentation is ripe with use cases for the various networks already implemented within Mariana, but what about functionalities that aren’t yet implemented? Extending Mariana can be done with minimal effort. Let’s take the “Siamese” network described in Signature Verification using a “Siamese” Time Delay Neural Network as example [1].
In this example, the network attempts to identify forged signatures. When faced with a pair of signatures, the network attempts to identify if both signatures are genuine or if one of the signatures is a forgery. To do so, the network is composed of two parallel feed-forward networks with tied weights. The cosine similarity measure of the output of both network is used as the distance measure between both inputs. A threshold is selected to determine whether the pair is composed of both genuine or genuine and forged signatures.
$\text{Cosine Similarity} = \cos(\theta) = {\mathbf{A} \cdot \mathbf{B} \over \|\mathbf{A}\| \|\mathbf{B}\|} = \frac{ \sum\limits_{i=1}^{n}{A_i B_i} }{ \sqrt{\sum\limits_{i=1}^{n}{A_i^2}} \sqrt{\sum\limits_{i=1}^{n}{B_i^2}} }$
In order to implement such a network in Mariana, we must add a new output layer. In order to do so, two modifications must be done to the base Mariana output class : 1) manage connections between the layers of the two branches of the network and 3) define the cosine similarity measure at the output.
Let’s get started with our new layer. This layer, aptly named Siamese, will need to abstract the base class for all output layers in Mariana, Output_ABC.
import theano.tensor as tt
from Mariana.layers import Output_ABC
class Siamese(Output_ABC):
def __init__(self, **kwargs):
Output_ABC.__init__(self, size=1, **kwargs)
self.targets = tt.ivector(name="targets_" + self.name)
self.inpLayers = []
Our Siamese layer must also implement some restrictions on its connections. Here, the function _femaleConnect will enforce a maximum of 2 input connections and will ensure that the outputs of these two input layers are of the same size.
def _femaleConnect(self, layer):
if self.nbInputs is None:
self.nbInputs = layer.nbOutputs
elif self.nbInputs != layer.nbOutputs:
raise ValueError("All inputs to layer %s must have the same size, \
got: %s previous: %s" % (self.name, layer.nbOutputs, self.nbInputs))
if len(self.inpLayers) > 2:
raise ValueError("%s cannot have more than 2 input layers." % (self.name))
self.inpLayers.append(layer)
Manipulation of the outputs of our layer is described in the function _setOutputs. This is where the cosine similarity will be implemented.
def _setOutputs(self):
"""Defines self.outputs and self.testOutputs"""
if len(self.inpLayers) != 2:
raise ValueError("%s must have exactly 2 input layers." % (self.name))
num = tt.sum(self.inpLayers[0].outputs * self.inpLayers[1].outputs, axis=1)
d0 = tt.sqrt(tt.sum(self.inpLayers[0].outputs**2, axis=1))
d1 = tt.sqrt(tt.sum(self.inpLayers[1].outputs**2, axis=1))
num_test = tt.sum(self.inpLayers[0].testOutputs * self.inpLayers[1].testOutputs, axis=1)
d0_test = tt.sqrt(tt.sum(self.inpLayers[0].testOutputs**2, axis=1))
d1_test = tt.sqrt(tt.sum(self.inpLayers[1].testOutputs**2, axis=1))
self.outputs = num / (d0 * d1)
self.testOutputs = num_test / (d0_test * d1_test)
Nearly done! Now that we have our Siamese output layer, let’s move on to constructing our network.
As it was for previous Mariana versions, describing the structure and connections of the network remains childsplay. Once instantiated, the network components can be strung together using the > operator. This detail allows for rapid network prototyping and the easy creation of many differents network structures such as forks, fusions and loops (a complete implementation of recurrent neural networks are, however, not yet implemented in Mariana).
Our network will be trained by gradient descent with a learning rate of 0.01. A mean squared error is used as the cost function.
ls = MS.GradientDescent(lr=0.01)
cost = MC.MeanSquaredError()
The layers of the network must now be instantiated. Two branches, A and B, with one input layer (_i) and two hidden layers (_h1, _h2), are introduced which use a rectifier (MA.ReLU) as activation function as well as a bit of L2 regularization (MR.L2). The size of each layer is passed as the first argument. Finally, our Siamese output layer is instantiated and linked up with the other layers using the > operator.
import Mariana.activations as MA
import Mariana.layers as ML
import Mariana.costs as MC
import Mariana.regularizations as MR
import Mariana.scenari as MS
A_i = ML.Input(28 * 28, name='inpA')
A_h1 = ML.Hidden(28 * 28, activation=MA.ReLU(), regularizations=[MR.L2(0.01)], name='A_h1')
A_h2 = ML.Hidden(50, activation=MA.ReLU(), regularizations=[MR.L2(0.01)], name='A_h2')
B_i = ML.Input(28 * 28, name='inpB')
B_h1 = ML.Hidden(28 * 28, activation=MA.ReLU(), regularizations=[MR.L2(0.01)], name='B_h1')
B_h2 = ML.Hidden(50, activation=MA.ReLU(), regularizations=[MR.L2(0.01)], name='B_h2')
o = Siamese(learningScenario=ls, name='Siamese', costObject=cost)
network = A_i > A_h1 > A_h2 > o
network = B_i > B_h1 > B_h2 > o
All that remains to do is to tie the weights of the hidden layers of both branches. We must first instantiate the network in order for the weights to be initialized with the same values.
network.init()
A_h1.W = B_h1.W
A_h2.W = B_h2.W
The saveHTML function offers us a peak at the final structure of our network:
Version 1.0 is expected to be released in the coming months. Keep an eye out on the github page for the newest updates!
I’d like to thank Tariq Daouda, author of Mariana, for his help in writing this post.
### References
[1] Bromley, Jane, et al. “Signature verification using a “Siamese” time delay neural network.” International Journal of Pattern Recognition and Artificial Intelligence 7.04 (1993): 669-688.
By | 2017-04-29T16:24:07+00:00 November 7, 2016|Categories: Machine learning, Python||0 Comments | 1,754 | 6,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-39 | longest | en | 0.856887 |
http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-2-in-a-series/ | 1,429,405,405,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246636650.1/warc/CC-MAIN-20150417045716-00027-ip-10-235-10-82.ec2.internal.warc.gz | 27,415,207 | 9,292 | Michi’s blog » Introduction to Algebraic Geometry (2 in a series)
## Introduction to Algebraic Geometry (2 in a series)
• February 21st, 2008
• 10:43 pm
I want to lead this sequence to the point where I am having trouble understanding algebraic geometry. Hence, I won’t take the usual course such an introduction would take, but rather set the stage reasonably quickly to make the transit to the more abstract themes clear.
But that’s all a few posts away. For now, recall that we recognized already that any variety is defined by an ideal, and that intersections and unions of varieties are given by sums and intersections or products of ideals.
This is the first page of what is known as the Algebra-Geometry dictionary. The dictionary is made complete by a pair of reasonably famous theorems. I won’t bother proving them – the proofs are a good chunk of any decent commutative algebra course – but I’ll quote the theorems and discuss why they matter.
We call a ring Noetherian if all ideals are finitely generated. If a ring R is Noetherian, then quotients are Noetherian.
Hilbert’s Basis Theorem: If R is Noetherian, then so is R[x].
We define the radical of an ideal I to be the ideal consisting of all elements a such that some power of a is actually in I. We call an ideal I radical if . This concept is relevant for our considerations since if for a point p the function f^n(p) vanishes, then f(p) also vanishes. Thus, the set of points such that f^n vanishes is the same set as the set of points where f vanishes. The relevancy of this is captured in:
Hilbert’s Nullstellensatz: Let k be an algebraically closed field. For any ideal I in , there is an equality of ideals
Note, for the statement of this theorem that we write V(I) for the variety defined by simultaneous vanishing of all elements in I, and we write I(V) for the ideal of all polynomials in that vanish on all of I(V).
So – and here is the beautiful part – affine algebraic varieties correspond bijectively to radical ideals in polynomial rings. For every ideal, there is a variety and for every variety, there is an ideal. But we can push this further.
## Coordinate rings
Let’s consider polynomial functions from to k. These are precisely the polynomials in . Given a variety V, we can take a polynomial and restrict it to a function .
Two different polynomials give the same restricted function precisely when their difference vanishes on all of V. So polynomial functions on V are precisely the equivalence classes in the quotient ring . We call the resulting ring the coordinate ring and denote it by k[V].
Conversely, if R is a Noetherian k-algebra such that there are no nilpotent elements in R, then R is a quotient of some polynomial ring with some radical ideal. Hence it is the coordinate ring of some variety in some affine space somewhere. We call a ring lacking nilpotents reduced.
We get, out of all this, a bijective correspondence
{ Noetherian reduced k-algebras } { Affine algebraic varieties }
The really beautiful part will come in my next post. We can introduce homomorphisms of varieties in a reasonably natural way so that this bijective correspondence ends up being functorial – i.e. any homomorphisms on one side gives rise to a corresponding homomorphism on the other side. Thus, the categories of Noetherian reduced k-algebras and of affine algebraic varieties are equivalent.
### 2 People had this to say...
[…] 21 – Aaron Roth explores social welfare in Nash equilibrium, but the aspect that catches his attention is the “Price of Malice.” 21 – In Basics of Patch Theory Brent explores the mathematical theory behind version control systems, particularly as it applies to collaborative editing. 21 – Suresh brings a guest blogger in to the Geomblog (Ganesh Gopalakrishnan) to discuss work on model checking by Clarke, Emerson, and Difakis (won the ACM Turing award). 21 – Michi finished his thesis, and found the time to post an introduction to algebraic geometry (in two parts). […]
• lhrrwcc
• March 6th, 2008
• 23:34
Another common and important theorem that is worth to mention are the Weak Nullstelensatz, this theorem says that:
If an ideal I in the polynomial ring k[X_1,…, X_n] over an algebraic closed field k does not contain the idenity (ie I!=(1)) then V(I) != {}.
Also, the corollary of this theorem (in some books this is called the Weak Nullstelensatz) says that:
A maximal ideal of the polynomial ring k[X_1,…, X_n] (k is algebraic closed field) has the following form:
(X_1 -a_1, …, X_n – a_n) with a_i in k.
The beauty of this is that it gaves us a biyective correspondence between
A^n (ie points) and maximal ideals of k[X_1,…,X_n]
This, will be clear in the next post | 1,110 | 4,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | latest | en | 0.942282 |
https://galeon.com/common-maths-problems-students-faced/ | 1,695,730,749,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00322.warc.gz | 299,742,742 | 34,259 | # 13 Common Maths Problems Students Faced
Maths can be an extremely frustrating subject, especially for many students. Here we examine some of the problems that students normally face and experience.
### 1. Remembering the Facts of Mathematics
Source: pexels.com
For many younger students who are less familiar with the rules of mathematics, they struggle with various basic mathematical facts, such as addition, subtraction, multiplication, and division. Without remembering these basic rules, students often cannot move on to solve more complex questions, since a higher order of thinking. They have inconsistency in the mastery of math facts. In secondary school, for example, students are exposed to algebra where numbers are replaced with letters. These letters are often used to represent unknown values. However, to solve the equations, the student has to be able to perform simple manipulation to make the unknown the subject of interest. If the student is unable to recollect the multiplication or division rules, he or she will not be able to find the required solution in a timely fashion. This will certainly have an impact on the final score.
### 2. Familiarity with Computational Methodologies
It is common for students to experience difficulties during calculations. It could be carrying a different figure during the process of multiplication or division, or listing figures under different columns during long divisions, or writing different signs and symbols. As marks are awarded for applying the correct formula, displaying the right workings, and deriving the correct answers, students who commit calculation errors will lose marks heavily on the step-by-step workings and final solutions.
### 3. Learning Disabilities
Students who are suffering from dyslexic, have difficulties in comprehension and understanding logic. They have issues with numbers perception and handling arithmetic. To them, it is a real challenge to match, compare numbers and their relationships. They have trouble retrieving number sequences or producing reasonable estimates. They find it hard to accept math vocabulary and translate word problems into numbers. They are unable to remember schedules, sequences of past and future events. They often appear absentminded or lost in their world and have no means to visualize. They also possess limited strategic planning ability.
### 4. Inattentiveness
Source: hillcenter.org
To do well in mathematical exams, students need to be focused in class and be highly attentive. If they fail to double-check their workings, they tend to arrive at the wrong answer and spend numerous amounts of time correcting for it. For them, memorizing instead of being able to comprehend the mathematical principles leads to frustration for students. This is especially the case when they are unable to recollect the exact steps needed to resolve a problem. Thus, students who practice answering maths problems are better off since they are exposed to more questions.
### 5. Panic Attacks
Some students experience anxiety when they deal with math. For these students, math can bring about determinantal feelings of fear and negatively impact their capability to do well. The intensity and the lack of confidence exhibited by these students create a mental freeze when they do maths, often causing them to forget things they have so well memorized.
### 6. Weak Foundation
Students who struggle with mathematics often do so due to their weak foundation, and may not be due to their learning difficulties. They may be forced to move on to more complex topics before they have the requirements, causing them to always lag behind their peers.
### 7. Math is Abstract
Source: pexels.com
Math tends to be an extremely abstract subject for students to relate to. If they cannot relate to it or find an analogy in real life, they will find maths hard to grasp. As it becomes more complex and advanced, students will find the questions harder and harder to do. They also would have to practice a lot more to understand the more convoluted math concepts.
### 8. Correct or Wrong
There is no room for mistakes in math. It is often they arrive at the correct answer or veer off track from the start. Often students do not even know how to start. Due to this reason, math can very fast degrade to a mentally frustrating process for students.
### 9. Cumulative Knowledge is Required
Being a cumulative subject means that maths needs to know what comes before it. Students certainly need to have a solid foundation before they can move on to newer topics. If a student fails to catch on in a topic, he or she will find it hard to understand advanced topics that are built on the previous ones or you can find help on SageSchool.
### 10. Math is Expected to be Tough
Source: pexels.com
Since students already expect the subject to be difficult, students tend to fall into the case of self-fulfilling prophecy. When they have the mindset that the subject is going to be tough, they will very rapidly throw in the towel. Such negativeness can cause them to have low self-confidence and produce poor results.
### 11. Memorize Instead of Understanding
When students memorize formulas and the workings without understanding them, they are setting up a trap for themselves. What works for a few example questions typically do not work for the more complex cases set in the examinations. Very soon when they realized old tricks do not work, they become visibly frustrated and stop practicing altogether. | 1,067 | 5,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.954644 |
https://www.coursehero.com/file/11192701/Ch-12-CQ/ | 1,600,453,351,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188049.8/warc/CC-MAIN-20200918155203-20200918185203-00239.warc.gz | 802,913,700 | 45,761 | Ch 12 CQ - Question 1 You hear an ambulance with a certain loudness and you are about 100 yards away Another person hears the same ambulance with an
# Ch 12 CQ - Question 1 You hear an ambulance with a certain...
This preview shows page 1 - 7 out of 20 pages.
Question 1 You hear an ambulance with a certain loudness and you are about 100 yards away. Another person hears the same ambulance with an intensity that is about 10 times softer . Roughly, how far is the other person from the ambulance? A) about the same distance B) about 300 yards C) about 1000 yards D) about 3000 yards E) about 10000 yards
Question 2 A CD playing softly has an intensity level of about 40 dB . Light street traffic has a level of about 70 dB . How much greater is the intensity of the street traffic compared to the background music? A) about the same B) about 10 times C) about 100 times D) about 1,000 times
Question 3 You are playing your stereo at a sound level of 90 dB. Your roommate turns on her stereo at a sound level of 90 dB. The new sound level is A) 87 dB B) 90 dB C) 93 dB D) 180 dB
Question 4 How many more stereos must be turned on to increase the sound level to 96 dB ? A) 1 B) 2 C) 3 D) 4
Question 5 You have an empty 2-liter soda bottle and an empty 12-oz soda bottle . Which one has the higher frequency when you blow across the top? A) the 2-liter bottle B) the 12-oz bottle C) both have the same frequency D) depends on how hard you blow
Question 6 You blow into an open pipe and produce a tone.
#### You've reached the end of your free preview.
Want to read all 20 pages?
• Winter '08
• REHSE
• Physics, Eduardo
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 632 | 2,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.906826 |
http://physicshelpforum.com/advanced-electricity-magnetism/13581-how-calculate-current-across-two-points.html | 1,506,417,230,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695375.98/warc/CC-MAIN-20170926085159-20170926105159-00075.warc.gz | 258,025,894 | 10,147 | Physics Help Forum How to calculate the current across two points
Aug 25th 2017, 08:52 AM #1 Junior Member Join Date: Aug 2017 Posts: 5 How to calculate the current across two points How to calculate the current across two points in this : So what i did is i used Ohms law to find the total current across the whole circuit which i computed to be 3 A and i am given the EMF of the battery and asked to neglect the internal resistance. So i am stuck at calculating the current across point A-B. Please help
Aug 30th 2017, 04:07 AM #2 Junior Member Join Date: Apr 2017 Posts: 1 Use electric induction
Aug 30th 2017, 05:28 AM #3 Physics Team Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,261 It would be helpful if you could show us how you calculated the total current is 3A. What are the values of E and the four resistors? Sometimes there are "tricks" than can come into play for solving problems like this depending on the specific values of the resistors. But nevertheless, the objective is to find the current running through each resistor, because once you have that then from Kirchoff's Law the current flowing in the A-B link needs to make up for any differences between current coming into A from R1 and flowing out of A into R2, and similarly balance the current flowing into B from R3 and out of B to R4. The procedure is to start by recognizing the voltage is the same at points A and B (do you see what that is?). From this you know that the voltage drop across R1 is equal to the voltage drop across R2, and hence from Ohm's Law: $\displaystyle I_1R_1 = I_2R_2$ Similarly: $\displaystyle I_3R_3 = I_4R_4$ In addition since you know the total currents is 3 amps, from Kirchoffs Law you have: $\displaystyle I_1 + I_2 = I_3+ I_4 = 3$ Amps. That's four equations in 4 unknowns. Now you can simplify: From the first equation: $\displaystyle I_1R_1 = (3-I_1)R_2$ Solve for $\displaystyle I_1$, and then use that value to determine $\displaystyle I_2 = 3-I_1$. Similarly: $\displaystyle I_3R_3 = (3-I_3)R_4$ So now you have all four currents. Use Kirchoff to determine the flow from B to A (remember that sum of currents flowing into A must equal currents flowing out of A). Hope this helps. Last edited by ChipB; Aug 30th 2017 at 08:37 AM.
Aug 30th 2017, 12:28 PM #4 Senior Member Join Date: Nov 2013 Location: New Zealand Posts: 505 This looks like the exact same question: How to calculate the current across two points ? PLEASE HELP __________________ Burn those raisin muffins. Burn 'em all I say.
Tags calculate, current, points
Thread Tools Display Modes Linear Mode
Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post KingLee Kinematics and Dynamics 8 Aug 25th 2017 05:24 PM rcmango Electricity and Magnetism 2 Oct 9th 2014 02:33 PM ling233 Waves and Sound 0 Jul 13th 2014 08:14 AM HenryW Kinematics and Dynamics 1 Feb 8th 2011 02:33 PM iamET Electricity and Magnetism 3 Oct 28th 2010 11:49 AM | 811 | 2,968 | {"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} | 4.34375 | 4 | CC-MAIN-2017-39 | longest | en | 0.92591 |
https://math.stackexchange.com/questions/3153361/solving-system-of-first-order-differential-equations-using-eigenvalue-eigenvect | 1,718,373,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00754.warc.gz | 355,181,886 | 35,305 | Solving system of first order differential equations using eigenvalue /eigenvector and matrix exponential approaches
I am attempting to solve the following first order linear system of differential equations using the eigenvalue /eigenvector approach and then by the matrix exponential approach. $$\mathbf{x}^{\prime}(t) = \begin{pmatrix} 6 & 1 \\ 4 & 3 \end{pmatrix} \mathbf{x} (t)$$
First, using the eigenvalue, eigenvector approach (and letting the solution be in the form $$\mathbf{X(t)}=\mathbf{K}e^{\lambda t}$$, where $$\lambda$$ is the eigenvalue and $$\mathbf{K}$$ is the corresponding eigenvector), I get solutions $$\mathbf{X_1}= \begin{pmatrix} 1 \\ -4 \end{pmatrix} {e^{2t}} \text{ and } \mathbf{X_2}= \begin{pmatrix} 1 \\ 1 \end{pmatrix} {e^{7t}}$$ which gives the homogeneous solution as \begin{align} \mathbf{X}(t) &= C_{1}\begin{pmatrix} 1 \\ -4 \end{pmatrix} {e^{2t}} + C_{2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} {e^{7t}} \\ &= \begin{bmatrix} e^{2t} & e^{7t} \\ -4e^{2t} & e^{7t} \end{bmatrix} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} \end{align} \nonumber
However, when I try to solve the same equation using the matrix exponential approach, where $$\mathbf{X}(t)=e^{\mathbf{A}t}\mathbf{c}$$: \begin{align} e^{\mathbf{A}t} &= \mathcal{L}^{-1}\{(s\mathbf{I}-\mathbf{A})^{-1}\} \\ &= \mathcal{L}^{-1} \left\{\begin{bmatrix} s-6 & -1 \\ -4 & s-3 \end{bmatrix}^{-1}\right\} \\ &= \mathcal{L}^{-1} \left\{\frac{1}{(s-2)(s-7)}\begin{bmatrix} s-3 & 1 \\ 4 & s-6 \end{bmatrix}\right\} \\ &= \mathcal{L}^{-1} \left\{\begin{bmatrix} \frac{s-3}{(s-2)(s-7)} & (\frac{1}{(s-2)(s-7)} \\ \frac{4}{(s-2)(s-7)} & \frac{s-6}{(s-2)(s-7)} \end{bmatrix}\right\} \end{align} \nonumber
But the inverse Laplace transform is not agreeing with my original solution for $$\mathbf{X}(t)$$; As an example, the term in row 1, column 1 of the above matrix has an inverse Laplace transform of $$\frac{4}{5}e^{7t} + \frac{1}{5}e^{2t}$$.
Can someone please guide me as to where I am going wrong?
The columns of your eigenvector-based matrix $$\begin{pmatrix}e^{2t}\\-4e^{2t}\end{pmatrix}$$ and $$\begin{pmatrix}e^{7t}\\e^{7t}\end{pmatrix}$$ are one such basis.
The columns of the Laplace transform-based matrix $$\begin{pmatrix}\frac45e^{7t}+\frac15e^{2t}\\ \frac45e^{7t}-\frac45e^{2t}\end{pmatrix}$$ and $$\begin{pmatrix}\frac15e^{7t}-\frac15e^{2t}\\ \frac15e^{7t}+\frac45e^{2t}\end{pmatrix}$$ are another. | 920 | 2,398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-26 | latest | en | 0.537276 |
https://www.teacherspayteachers.com/Product/3rd-Grade-Thanksgiving-Activities-3rd-Grade-Thanksgiving-Math-1525804 | 1,492,928,565,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118477.15/warc/CC-MAIN-20170423031158-00595-ip-10-145-167-34.ec2.internal.warc.gz | 973,733,931 | 26,193 | Total:
\$0.00
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
11.34 MB | 14 pages
### PRODUCT DESCRIPTION
This Thanksgiving Math pack is filled with high-interest goodies that will keep your kiddos engaged and practicing their skills during the busy and sometimes unpredictable holiday season! Thanksgiving Math includes ten pages of thoughtfully designed printables covering many Common Core standards and concepts taught or reviewed in the first half of third grade. Thanksgiving Math also provides a review opportunity for fourth graders.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Thanksgiving Math is also featured in this money-saving BUNDLE:
Thanksgiving Math & Christmas Math BUNDLE (TWO 10 Page Packets)
You may also be interested in Thanksgiving Multiplication!
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Thanksgiving Math printables and concepts include:
• A Thanksgiving Riddle: solving 3-digit addition and subtraction, with regrouping
• Gobble, Gobble, Math: identifying place value to 10,000
• Order 'em Up: ordering numbers to 10,000 (with an emphasis on numbers with 0s, 1s and 9s in them)
• Watch the Clock: telling and writing time from analog clocks and solving problems involving intervals of time
• Number Roundup: rounding numbers to the nearest 10 and 100
• Name That Number: interpreting whole number products and solving problems using symbols for unknown quantities in equations
• Thanksgiving Shopping: solving word problems involving equal groups
• Thanksgiving Feast-O: interpreting whole number products
• Follow Along: solving multi-step word problems involving addition, subtraction, and multiplication
• A Thanksgiving Feast: an extended response problem that challenges students to detail their mathematical thinking when planning a Thanksgiving feast, given a price list and budget
Thanksgiving Math blackline cover page for student packet included!
*** Please sneak a peek at the Thanksgiving Math preview to see the variety in this pack! ***
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
If you enjoy using fun, seasonal, CCSS-aligned printables to reinforce math skills, you might also like Christmas Math: Grade 3 Packet!
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
To earn TPT credits to use on future purchases:
• Please go to your My Purchases page (you may need to login). Beside each purchase you'll see a Provide Feedback button. Simply click it and you will be taken to a page where you can give a quick rating and leave a short comment for the product. You can check your credit balance in your TpT Credit Balance page. Every 100 Credits is worth \$5 that you can apply towards future TpT purchases in my store, or any TpT store. I value and appreciate your feedback!
Be the first to know about my new products, discounts, and freebies:
• Look for the green star next to my store logo and click it to become a follower. You’ll receive customized email updates about this store, and be the first to know when new products are added.
Total Pages
14
N/A
Teaching Duration
N/A
### Average Ratings
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
94 ratings
\$3.00
User Rating: 4.0/4.0
(2,829 Followers)
\$3.00 | 1,044 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.759297 |
https://socratic.org/questions/how-do-you-write-the-following-as-an-algebraic-expression-using-x-as-the-variabl-5 | 1,632,350,916,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00439.warc.gz | 525,963,000 | 5,806 | # How do you write the following as an algebraic expression using x as the variable: 6 more than 3 times x?
May 6, 2018
$3 x + 6$ or $3 \left(x + 2\right)$
#### Explanation:
We have to break this down. What do we mean by "3 times x"?
It means x taken 3 times, or
$3 x$ (i.e. $x + x + x = 3 x$)
What is 6 more than this? It would be $3 x + 6$
This can also be written $\left(x + 2\right) + \left(x + 2\right) + \left(x + 2\right) = 3 \left(x + 2\right)$ | 171 | 457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-39 | latest | en | 0.915036 |
https://lauterdar.com/blog/simple-light-activated-switch-circuit/34a-2822bff | 1,653,150,235,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00745.warc.gz | 444,935,889 | 14,581 | Home
# Simple light circuit with switch
### Circuit With a Switch Basic Concepts and Test Equipment
This circuit is most impressive when the wires are long, as it shows how the switch is able to control circuit current no matter how physically large the circuit may be. Measure voltage across the battery, across the switch (measure from one screw terminal to another with the voltmeter), and across the lamp with the switch in both positions In building Simple Bots, we are going to use a series of photo-coupled devices called modules. But, before we can do this, we need a good photo-couple to use as a light-activated switch. To make one of these switches, we will be using a special OSRAM-brand photo transistor and any standard NPN transistor (like a 2N2222, 2N3904, BC546, etc.) A simple lighting circuit is where the light switch is installed between the supply and the light fitting. A common place to find in-line switches can be in an attic, in a garage or a shed. One way lighting circuit using in-line switching Line diagram of a one way lighting circuit using in line method (fig 1) Need help wiring a 3 way switch? With easy to follow diagrams and instructions, you can have that convenience in no time. Read More. Wire a Switched Outlet. Mar 09, 21 09:56 PM. Want to turn a lamp on with a light switch? Sometimes it is handy to have an outlet controlled by a switch. Step by step instructions on how to wire a switched outlet.
This circuit is wired with a 2-wire cable running from the light to the switch location. The neutral from the source is connected directly to the neutral terminal on the light and the source hot is spliced with the white loop wire. The white wire is marked black on both ends to identify it as hot. At SW1 it is connected to one of the terminals Wiring a light switch is very simple. The white (neutral) wire from the power source and the white (neutral) wire that goes to the light fixture get connected to each other. The black (hot) wires are what get connected to the light switch. Again, excuse my paint-covered hands
Run a two-conductor wire to a switch in a desired location and install electrical box with switch. Mark the extension wires with colored tape to identify switch wires (both ends). Connect the marked-white wire to the black lead going to light. Wait to connect the marked-black wire going to switch to hot (black) last Light switch wiring diagram depicted here shows the power from the circuit breaker panel going to a wall switch and then continues to a ceiling light with a three conductor cable. From the ceiling box an electrical receptacle outlet is fed power. The diagram above shows a two conductor cable from the circuit breaker panel going to a wall switch. If the switch is ON, the electric circuit is closed and the lamp glows, and if the switch is OFF, it will disconnect the power supply to the lamp. For safe operation this wiring is placed in a box called a switch box. The switch wire and live wire are a single wire; it is just cut in between to connect the switch I had added more pictures on how the wires connect to the bulb, battery and switch - 13/11batteryinsulated wirelight bulbducktapeon'off switchRead more here:.. This can be done using a simple switch. How to make a simple switch The basic circuit below has a switch made using a small piece of paper and 2 paper clips to break the flow of electricity which allows the bulb to be turned on and off
How To Make A Simple Light Bulb Circuit With Switch Posted by Margaret Byrd Posted on May 18, 2021. Simple electric circuit project kit how to make a electrical 3 ways wikihow switch energizer circuits phoebe s site electricity experiments for kids thunderbolt LDRs or Light Dependent Resistors are very useful especially in light/dark sensor circuits. Normally the resistance of an LDR is very high, sometimes as high as 1000 000 ohms, but when they are illuminated with light resistance drops dramatically. LDR switch circuit diagrams. This is a light switch or light activated relay circuit
### Simple Light Switch : 5 Steps (with Pictures) - Instructable
Circuit 3 of Simple LED Circuits (LEDs in Parallel) The final circuit in the simple LED Circuits tutorial is LEDs in Parallel. In this circuit, we will try to connect three 5mm White LEDs in parallel and light them up using a 12V Supply. The Circuit Diagram for LEDs in Parallel Connection is shown in the following image Connect the source wire to the bottom terminal of the switch and the other black wire to the top terminal. In this picture, you can see the strip gauge on the back of the switch. This gauge is when you want to use the quick-connect method when wiring a light switch. Simply strip the wire to the length specified and push the wire into the hole
A Simple Light activated switch Description. This is the circuit diagram of a light activated switch based on National Semiconductors comparator IC LM 311 and a LDR. The circuit is based on a voltage comparator circuit wired around IC 1.The non inverting in put of IC1 is given with a reference voltage of 6V using resistors R3 and R4 1 switch. Assorted batteries and holders. Assorted Solar Cells. 1 bread board for testing. 1 multi meter. Capacitors; a must for the voltage multipliers. 1.2nF, 100pF, one of each. Inductors. Two 0.47mH. One 22mH. If you make the circuits in the garden light IC datasheet you will need the parts listed in the datasheets By wiring a 2-way switch, The circuit below shows the basic concept of electricity flow to the load. Let's assume the load you are controlling is a light. The electricity flows from the hot wire (black) through the 2-way switch (shown in off position) and then to the light and returns through the neutral wire (white). This is a completed circuit
An open circuit means the path is broken so that the electricity can't flow back to the power source. Your light bulb won't light up if there's an open circuit. When the light is turned on in your kitchen, you have a closed circuit. When you flip the switch to turn it off, it breaks that path and becomes an open circuit The 4 simple light activated day night switch circuits explained here can all be used for controlling a load, normally a 220V lamp, in response to the varying levels of the surrounding ambient light The schematic of the dark-activated switch circuit we will build is shown below: So, for this circuit, 3 volts is powering the circuit. This 3 volts is in parallel to a 100KΩ resistor and the photoresistor. In the middle of these 2 components is connected the base of the NPN 2N3904 transistor. This is how the circuit works: When exposed to.
### Wire a light switch in one way lighting circuits - DIY How T
• al. You can also open the circuit by removing the light bulb from the holder.
• A simple electrical circuit. An opportunity to discover the basic vocabulary. View in 1080p HD.Switch (open or closed), generator, wire, connection, banana c..
• A Simple Electric Circuit is a circuit including a power source (battery), a resistor (light bulb) and a switch connected to each other in series (meaning that wires connect the battery to the switch, the switch to the light bulb and the light bulb back to the other end of the battery)
A simple electrical circuit Once one of the wires is removed from the power source or a break is made in the flow, the circuit is now open and the lamp will no longer light. In practical application, circuits are opened by such devices as switches, fuses, and circuit breakers When you rotate the paperclip and make contact with the second thumbtack you close and complete the circuit, current flows through the circuit and lights up the bulb (Figure 3). The switch, bulb holder, and portable power pack are a complete circuit and arrangement of conductors; they allow the passage of electric current through the wire This is a simple light switch circuit where the wires are treated as transmission lines. When you hit the switch, it takes a while for the light and the source to get the message. Of course, in real life, it takes more than a nanosecond to flip a switch. Next: Memristor. Previous: Low-Pass Filter. Index. Simulator Hom Simple Light Activated Switch Circuit. The light activated switch circuit works on a very basic principle, that of amplification of the current produced by a photo-transistor when light impinges upon it. The current is amplified by op-amp 741 that drives T2, energising a relay. The circuit shown can be used to activate a relay with a. Simple Light-activated relay circuit with PCB. So, I select a light switch controller Because there are fewer devices and easy build. Which is the circuit diagram below: In today's post, I'm going to share a simple light activated relay circuit. Why should you learn it? Your daily life is busy
This is Simple light controller circuit that is controlled light-activate. Which use the inverter gate IC-40106 as main components in compare and controlled switches, and common phototransistors as sensors. The Operation of switch has several forms. The most basic way is to press the switch directly. If modern must be controlling by Infrared. Simple Light Sensor Circuit using LDR. The light sensor circuit is one of the coolest circuits in basic electronics projects. It detects the quantity of light present in the environment and the results can be detected by the brightness of the LED. This circuit may be useful for knowing the working of LDR (Light Dependent Resistor) or. Automatic Bathroom Light Switch Circuit Diagram. The circuit diagram of this device consists of two main components: Operational Amplifier IC LM741 and Decade Counter IC CD4017. One end of the reed switch is connected to +5V power supply while the other is connected to the base of BC558 PNP transistor. The collector is connected to the +5 volts. Light activating and deactivating switches are simple to build when you are provided with a LDR (sensing element) and 555 (switch activator). These kind of sensor circuits are widely employed in many applications and may come in handy when you intend to build a simple light sensing circuit Lighting Circuits. Key to wiring diagrams. Key to types of switch. 2 way light switch with power feed via switch. 2 way light switch with power feed via light. 2 way light switch with power feed via switch (two lights) 3-way light switch with power feed via the light switch. Three way light switch with power feed via the light
### Light Switch Wiring Diagram (Single Pole
1. The circuit is disrupted at the second switch, so the light is turned off. When the second switch is flipped ON, the power will run into the switch and the light, as well as into the neutral (white) cable while turning the light on. If the first switch is turned (or flipped OFF), the electricity still makes it to the second switch.
2. A basic circuit of a programmed night/light depicted right now in this tutorial. This automatic circuit switch ON a load when darkness falls and switch it OFF when dawn breaks. In this DIY project tutorial, we will depict a basic automatic day and night switch. The circuit can perform both day and night/light and dark detecting errands
3. On Friday, we talked about simple circuits using everyday materials. Your kids now know that a circuit is a circular path through which electricity flows unbroken. Now, we're going to take that simple circuit and add a switch. When your kids experiment with batteries, wires, and bulbs, they learn that a current of electricity needs a power.
4. Light Activated Switch is a simple electrical project circuit by which we can switch on and off the electrical load appliances like lights, fans, coolers, air conditioners, street lights, etc., automatically based on the day-light intensity instead of manually operating the switches. By using this method, manpower can be reduced to a great extent
How to Wire a Light Switch: Wire New Switch. Wrap your black wire clockwise around one of the brass screws. Repeat the process with the white wire, wrapping it clockwise around the other brass screw and tightening it with a screwdriver. Photo By: Jalynn Baker In contrast to a simple light switch, which is a single pole, single throw (SPST) switch, multiway switching uses switches with one or more additional contacts and two or more wires are run between the switches. When the load is controlled from only two points, single pole, double throw (SPDT) switches are used By using this type of resistor to bias a transistor, we can make a transistor switch power to a load depending upon the light conditions. Simple transistor circuits with LDR are very common and utilised in night lighting and emergency lighting systems. The transistor may directly drive a low-current LED or bulb, or a relay coil, which then acts. The circuit of light detector is very simple and easy to build with very few components. As you can see in the LDR circuit diagram , it can be a distinguished as two smaller circuits; a) Voltage divider made using LDR (LDR1) and a Potentiometer (RV1) b) Output (LED D1) in our switching circuit made using a transistor BC547 Q1
### Light Switch Wiring Diagrams - Do-it-yourself-help
This switch is super easy to make and is great for controlling projects with multiple LEDS. In order to help you get started, we've included (4) FREE paper circuit templates that use this type of paper clip switch. Once you learn the basics, you can create your own custom circuit very easily. Get creative and have fun ! Project Time: 15 Minute Relay Switch Circuit Diagram Working of the Basic 5V Relay Circuit. In the above circuit, 5V relay is powered by a 9V battery. An ON/OFF switch is added for the switching purpose of the relay. At the initial condition when switch is open, no current flow through coil, hence Common Port of relay is connected to NO (Normally Open) Pin, so the. You made a very simple circuit. There's a push from a power source (the battery), a path for the electrons to follow (wires, lights, and a switch) and a final path back to the power source. The electron movement begins on the negative (-) pole of the battery. They get pushed through the circuit to the positive (+) pole by an electric field Schematic of the Dimmer for Light Bulbs. So as you can see it is a very simple circuit and I think everybody can build it. Mount the BTB12 triac on a little heatsink as you can see in the pictures above. You only need 2 terminals from the 680K potentiometer and when adjusted you'll notice that the highest brightness will be when its.
If the light is connected to a live supply (Line and Neutral wires) it will light up. In order to make the switching possible we need to introduce a switch in the Line side of the circuit, effectively breaking or making the circuit whenever we need the light on or off As the light level decreases and LDR meets the maximum threshold resistance, the circuit automatically switches on the LED D1. Dark sensor with variable resistor: A dark detector can be made using a variable resistor. The sensitivity of the circuit can be adjusted with a variable resistor. High resistance-> more darkness to switch on the LED A few light sensor circuit based projects can be listed as a solar highway lighting system with auto turn off in daytime, security alarm system by photo electric sensor, sunset to sunrise lighting switch, Arduino managed high sensitive LDR based power saver for street light control system, etc,
### Electrical Basics - Wiring A Basic Single-Pole Light Switc
1. Table 2-3: Light switches and their light circuits Item Switch Application (classical light circuits) Chap ter 1 Off switch The light is switched from one location in the room. 2.2.1 2 Series connection (Typically with chandeliers) 3 Changeover switch The light is switched from two locations. 2.2.2 4 Intermediate switch
2. Simple Soft Latch Switch using Push-Button. Latching Toggle switches are one of the most popular kinds of switches, from their use by kids in basic electricity classes to their use in prototypes and more advanced products, they provide a familiar and reliable way to close or open a circuit. This makes them the go to switch for designers in most.
3. DC-based Light Switch. When the connection is made the lamp will glow. Connect a switch in between anyone wire that will cut off our supply DC voltage to the LED bulb. We discussed a few simple electrical circuits, let's continue a few simple electrical devices. Also, see the circuit functioning and uses of these devices. Thermocouple Circuit
4. Wiring Parallel Circuit Lights With Switch on Other End Be careful cause when there's a series of lights between the switch(es) and the power supply the wiring is a little bit trickier. Study this diagram to understand what I mean. This type of wiring is also accounted for when wiring 3-way switches with a series of lights between the power box and the switches
5. The circuit diagram presented above is an classic illustration of a dimmer switch, where a triac continues to be employed for managing the depth of light. When AC mains is provided to the above circuit, as per the setting of the pot, C2 charges fully after a specific delay supplying the necessary firing voltage to the diac
6. One of these transistors drives a heavier transistor which controls a lamp. This circuit can switch on and off a light, a fan or a radio etc by the sound of a clap. This circuit is constructed using basic electronic components like resistors, transistors, relay, transformer, capacitors. This circuit turns 'ON' light for the first clap
7. i-project, made with the help of the basic components. Clap Switch has the ability to turn ON/OFF any electrical component or circuit by the clap sound. We will use two basic clap switch circuit diagrams i.e. (With IC 555 Timers and without 555 Timer ). It is known as Clap Switch because the condenser mic.
How good it will be if the lights turn on or off at the moment when u open or close a door. Automatic Washroom Light. In this project, I will tell you the best way to plan and manufacture a straightforward Automatic Washroom Light Switch Circuit, which will consequently turn on lights when you enter the washroom and turns it off when you leave Jul 2, 2021 - Explore student's board Simple circuit, followed by 1008 people on Pinterest. See more ideas about simple circuit, circuit, circuit diagram This is the simplest Light Dimmer Circuit Diagram or Fan Regulator Circuit Diagram. The circuit is based on the principle of power control using a Triac.The circuit, and design are the same For Fan or Light the only difference is the output load to be varied, that is, fan or light. Assemble the circuit on a good quality PCB or common board A very simple series circuit is shown below. Notice the space left in the aluminum foil for both the switches and the battery: Depending on the grade level of the students, you can have them build a very simple circuit or a very complex circuit with both series and parallel circuits combined
### How to Wire a Simple 120v Electrical Circuit (with Pictures
1. Simply light circuit. Description: Connection of one or more luminaire points (Lights) controlled by a simple switch. This kind of connection is used in almost all interior electrical installations. General diagrams. Single line diagra There are different types of sensors such as light sensors, temperature sensor, humidity sensor, pressure sensor, fire sensor, ultrasonic sensors, IR sensor, touch sensor, and so on.. What is a Light Sensor Circuit? The light sensor circuit is a simple electrical circuit, which can be used to control the (switch on and off) electrical load appliances like lights, fans, coolers, air. The circuit itself is a simple one, controlled by a switch operated either directly by the brake pedal or hydraulically by the increase in pressure in the brake pipes. Simple checks Start by making a visual check on the lights
The circuit idea of this dimmer switch for incandescent lights is pretty simple in design. The given components can be easily procured and assembled over a piece of general PCB. The whole assembled unit may be fixed and wired inside your existing domestic switchboard (as shown in the adjoining diagram), allowing only the shaft or the knob of. Electric circuits have switches that allow people to control the flow of the electric current through the circuit. When someone flips a light switch in a room or pushes a button on a flashlight that person is helping to complete the circuit. The current can then flow to the light or the bulb Here we are using a simple PLC to control this lamp using two switches, one switch at ground floor and second switch at first floor to control one lamp as shown in below figure. Note : we can also build the circuit using simple relays/switches also. This article only for understanding the basic concept of 2 way switch using a PLC Ladder Logic
Some light units incorporate their own switch, for these fittings, the power circuit is then connected directly to the fitting. Light switches Most room lights are controlled by wall mounted toggle switches (although alternatively touch sensitive or rotary light dimmers can be fitted), The cable normally runs down the wall within conduit within. The basic circuit of light dimmer using TRIAC is shown below and this circuit depends on phase control. The variable resistor VR1 plays as a main controller in the light dimmer circuit. The capacitor 'C2' in the circuit below gets charge from the main supply. In the circuit, variable resistor VR1 and D1 DIAC are used to control the. To construct and analyze the behavior of four circuits: a simple switch circuit, a simple rheostat (variable resistance) circuit, a series circuit, and a parallel circuit. Equipment Leads, alligator clips, two 6.3-V light bulbs and sockets, an electric motor, two SPDT switches, 9-V battery and cap, multimeter. Procedure Part A. Simple Circuit 1 Nov 12, 2019 - We bring to you the best simple circuits which will gives output without any boring. 1. Light Activated Relay This circuit reacts as a light activated relay, we can use this circuit as a switch to activate any work when light is appeared. For an example we can connect garden lamp into the relay (in normally close pin) whenever the sunlight appears the relay contact moves to (N/o.
### Light Switch Wiring Diagrams for Your Residenc
1. Simple Light-operated switch Circuit Diagram. It is ideal to use as switch less night lamps. This circuit uses a flip-flop arrangement of Ql and Q2. Normally Ql is conducting heavily. Light on CDS photocell causes Ql bias to decrease, cutting it off, turning on Q2, removing the remaining bias from Q1. Reset is accomplished by depressing S1
2. The most basic diagrams are for simple circuits involving one switch that turns one light on or off. This type of switch is commonly called a single pole single throw (SPST) switch. Current flows through a switch to the load, in this case, a light fixture. When you flip the switch off, the circuit is broken, the flow of electricity is.
3. A series circuit room: two lights in series with a switch in one room 2. A parallel circuit room: three lights in parallel with a switch in one room 3. A single outside light with a separate switch 2. Approval of Building Plans (Circuit Diagram) You must provide one diagram for the complete circuit in your building (all rooms). It must
4. g from the top.
### 10 Simple Electric Circuits with Diagrams - Bright Hub
The simplest and most common light switch is actually referred to by hardware dealers and electricians as a single-pole light switch. With a single-pole light switch, flipping the toggle or paddle up completes the circuit, turning lights or appliances on, and flipping it down breaks the circuit, turning lights or receptacles off Construct a simple working electrical circuit using a flashlight battery, a switch, and a light. Make drawings of three kinds of bridges and explain their differences. Construct a model bridge of your choice. Make a simple crane using a block and tackle and explain how the block and tackle is used in everyday life Circuit Description :-. Today we are going to make a simple clap switch circuit.This is a simple clap switch circuit with high sensitivity. It switches ON/OFF electrical appliances through claps.In the circuit, I will control a single relay using clap switch. The circuit can sense the sound of claps from a distance of 1-2 meters A continuity tester is a simple electrical device with a metal probe, a tester light, and a wire with a clip at one end. All it does it test for continuity or a complete electrical circuit. A circuit that is open is broken and has no continuity. A circuit that is closed has continuity. A light switch opens and closes a lighting circuit Paper Circuit with Switch. Now that you've seen how to create the basic circuit, we're going to step it up a notch and add a switch to control the LED. Step 1 - Print Circuit Template. Download our Simple Circuit w/ Switch template and print it out on standard printer paper. Card stock is not needed for this particular paper circuit project
This diagram provides an example to control high power loads such as motors, lamps, and heater. In this circuit, we want to control 12 volts load from a digital logic AND gate. But the output of the AND gate is only 5 volts. By using a transistor as a switch, we can drive 12v or even high voltage loads with a 5-volt digital logic signal Figures 3 to 5 show some practical relay-output light-activated switch circuits based on the LDR. Figure 3 shows a simple non-latching circuit, designed to activate when light enters a normally-dark area, such as the inside of a safe or cabinet, etc. FIGURE 3. Simple non-latching light-activated relay switch
The circuit shown in figure 1 is a very simple light sensor circuit that will activate an LED when the LDR in the circuit receives light. The circuit shown in figure 2 is using a relay at its output and when light falls on the LDR it will activate the relay. switch and hence any AC and DC appliance connected with the relay will become activated The toggle switch is a switch that can play crucial switching roles in circuits. A SPST toggle switch can act as a simple ON-OFF switch in a circuit. While SPDT and DPDT toggle switches can flip different devices on or off in a cirucit. Toggle switches are common components in many different types of electronic circuits home:: categories:: sensor:: light-sensor:: simple optical switch. How to build Simple optical switch November 30, 2010 - category: Light sensor. Circuit diagram. Introduction. The 555 is proved to be the most versatile and ubiquitous IC all over the world.This is a possible use: simple inverting schmitt trigger. Circuit explanation This is the description: Simulate a circuit for controlling a hallway light that has switches at both ends of the hallway. Each switch can be up or down, and the light can be on or off. Toggling either switch turns the light on or off. I have these methods but unsure about how to use them. public int getFirstSwitchState () (0 for down off, 1. 4. Tape down a simple switch. We want our light up cards to work again and again. If we have the paper circuit in an always ON mode the battery will run out after a day or two. To allow your circuit to be ON or OFF at a given time you need a switch. I like simple push button switches that have the circuit ON only when the button is compressed Simple Deluxe 1-Pack 20 Feet Extension Hanging Lantern Pendant Light Lamp Cord Cable E26/E27 Socket (no Bulb Included) On/Off Switch, 1 Pack, White 4.7 out of 5 stars 1,616 \$8.99 \$ 8 . 9 | 5,949 | 27,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-21 | latest | en | 0.923795 |
https://cracku.in/25-61-72-73-59-74-58-x-ibps-rrb-clerk-2017 | 1,716,053,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00790.warc.gz | 174,728,250 | 22,603 | Question 25
61, 72, ?, 73, 59, 74, 58
Solution
For first, third, fifth and seventh number:
61-1 = 60
60-1 = 59
59-1 = 58
For fourth and sixth number:
72+1 = 73
73+1 = 74 | 75 | 178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-22 | latest | en | 0.792157 |
http://www.mathworks.fr/fr/help/matlab/ref/hess.html?s_tid=gn_loc_drop&nocookie=true | 1,405,132,797,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776430645.99/warc/CC-MAIN-20140707234030-00033-ip-10-180-212-248.ec2.internal.warc.gz | 360,993,413 | 7,905 | Accelerating the pace of engineering and science
# Documentation Center
• Trials
• Product Updates
# hess
Hessenberg form of matrix
## Syntax
H = hess(A)
[P,H] = hess(A)
[AA,BB,Q,Z] = hess(A,B)
## Description
H = hess(A) finds H, the Hessenberg form of matrix A.
[P,H] = hess(A) produces a Hessenberg matrix H and a unitary matrix P so that A = P*H*P' and P'*P = eye(size(A)) .
[AA,BB,Q,Z] = hess(A,B) for square matrices A and B, produces an upper Hessenberg matrix AA, an upper triangular matrix BB, and unitary matrices Q and Z such that Q*A*Z = AA and Q*B*Z = BB.
## Definitions
A Hessenberg matrix is zero below the first subdiagonal. If the matrix is symmetric or Hermitian, the form is tridiagonal. This matrix has the same eigenvalues as the original, but less computation is needed to reveal them.
## Examples
H is a 3-by-3 eigenvalue test matrix:
```H =
-149 -50 -154
537 180 546
-27 -9 -25```
Its Hessenberg form introduces a single zero in the (3,1) position:
```hess(H) =
-149.0000 42.2037 -156.3165
-537.6783 152.5511 -554.9272
0 0.0728 2.4489```
## See Also
Was this topic helpful? | 366 | 1,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2014-23 | latest | en | 0.779086 |
https://www.vedantu.com/question-answer/with-the-help-of-a-circuit-diagram-explain-how-a-class-12-physics-cbse-5f60f8dc7d7dc34d3b8bc585 | 1,601,137,850,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244231.61/warc/CC-MAIN-20200926134026-20200926164026-00361.warc.gz | 1,111,462,389 | 80,780 | Question
# With the help of a circuit diagram, explain how a meter bridge can be used to find the unknown resistance of a given wire.
Hint: First draw the circuit diagram of the meter bridge. Then, try to figure out the null point of the Galvanometer and continue this with variable resistance. Now find the unknown resistance with the help of the Wheatstone bridge principle. The Wheatstone bridge works on the principle of null deflection, that is in normal condition the ratio of their resistance are unbalanced and current passes through it but in the other case when the galvanometer does not show any deflection the ratio of resistances is same and they get nullified and thus no current flows through it.
Above is the diagram of a meter bridge in which, R and S are two resistance, where R is a resistance box with variable resistance and S is a fixed resistance(this resistance is made up of the wire whose resistance is unknown). B, A, and C are metallic plates G is a galvanometer which is connected at point B with which a jockey D is connected, the meter scale is a 1 m scale divided into 100 divisions, K$_{1}$ is the key, and $\xi$ is a battery that is inside the circuit.
Now let us consider that ${{l}_{1}}$ is the length where the galvanometer shows zero deflections,
Therefore ${{l}_{2}}=100-{{l}_{1}}$ .
At first, the key K$_{1}$ is closed to pass a current and we are also choosing a suitable resistance R in the resistance box.
Then the jockey is tapped along the wire to locate the null point that is the point where the galvanometer shows zero deflection. The bridge is then said to be balanced.
And after that by the principle of Wheatstone bridge.
$\dfrac{R}{{{l}_{1}}}=\dfrac{S}{{{l}_{2}}}$ ,
$S=\dfrac{R\times {{l}_{2}}}{{{l}_{1}}}$,
Therefore, by this formula, we can find the unknown resistance of the wire.
As R is a known value so unknown resistance S can be calculated.
Note: We have to locate the exact zero point to apply the Wheatstone bridge principle, sometimes terminal error occurs that is the terminal of the devices get exchanged, check all the connections properly, and after doing the observation of one resistance switch of the circuit because if the wire gets heated up too much resistance will increase. | 517 | 2,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-40 | longest | en | 0.930031 |
https://www.northernarchitecture.us/sound-transmission/t5-ykf.html | 1,542,511,654,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743963.32/warc/CC-MAIN-20181118031826-20181118053826-00477.warc.gz | 968,114,752 | 7,696 | ## T5
IOO bL
io so no bO
THICKNESS, cm
Figure 12.16 Forces and Velocities for Soft Floor Covering (Ver, 1971)
Figure 12.16 Forces and Velocities for Soft Floor Covering (Ver, 1971)
and arad = 1. If we assume a 15 cm (6 in) dense concrete slab and use the figure, we get about 78 dB, and for a lightweight concrete structure of the same thickness we get about 92 dB with no surface covering. Human heel drops are not this loud, but the numbers are close to the measured levels for a tapping machine test (Fig. 12.18). Since it is usually impractical to increase the slab thickness and density enough to make a significant change, we turn to the floor surface covering for improvement.
### Improvement Due to Soft Surfaces
Ver (1971) and Cremer (1973) have analyzed the impact of a carpet or other similar elastic surfaces on tapping machine noise transmitted through a floor. An illustration of Ver's model is given in Fig. 12.16.
The falling weight strikes a surface, whose stiffness is the elasticity of the carpet. In this model damping is ignored and the weight is assumed to strike the surface and recoil elastically once without multiple bounces. The equation of motion of the spring mass system is and d2x m-^ - kx = 0 (12.34)
dt j m0
where the natural frequency ofthe spring mass system is
The spring constant is given by
where Ed = dynamic Young's modulus of elasticity (N/m2) which is about twice the static modulus Sh = area of the striking surface of the hammer = 0.0007 m2 h = thickness of the elastic layer (m) m = mass of the hammer = 0.5 kg When the hammer is dropped it strikes the elastic layer with a velocity u0 at time t = 0 and its subsequent motion can be calculated from Eq. 12.34 to be u (t) = u0 cos (&>nt) for 0 < t <n/rnn (12.38)
and u (t) = 0 for t < 0 and t > (12.39) Figure 12.16 also shows this velocity function. The force is given by
F(t) = m — = uQ œn m sin (œn t) for 0 < t < n/œn (12.40)
Now the Fourier amplitude of the tapping machine pulse train as given in Eq. 12.22
where n = 1, 2, 3,... This yields the force coefficients of the Fourier series
n n4 V a fi in terms of the coefficients given in Eq. 12.43 and
The improvement due to the elastic surface in the impact noise isolation is given in terms of a level
which is shown graphically in Fig. 12.17.
Note that in this model we have ignored the contribution of the floor impedance and have assumed that it is very stiff compared with the elastic covering. At very low frequencies—that is, below the spring mass resonance—the improvement due the covering is zero. Above this frequency the surface covering becomes quite effective, giving a 12 dB per
0 0 | 686 | 2,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.877897 |
https://scribesoftimbuktu.com/evaluate-13-16/ | 1,674,832,780,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00520.warc.gz | 512,576,526 | 12,459 | # Evaluate 13/(16%)
1316%
Convert 16% to a decimal.
130.16
Divide 13 by 0.16.
81.25
Evaluate 13/(16%)
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
Scroll to top | 73 | 218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-06 | latest | en | 0.595202 |
https://ethosspace.com/programmers/c-program-to-identify-even-odd-numbers-from-user-input/ | 1,726,289,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00733.warc.gz | 205,161,400 | 13,408 | # C Program to Identify Even Odd numbers from user input.
Below program will first accept total number which user want to enter. Then user has to enter numbers one by one.
Program will identify even and Odd numbers and store their total count in variable.
``````// Program Name: Accept n numbers from user and then identify even or odd Numbers
#include<stdio.h>
void main()
{
int totalNum, i, ary[100];
int eNoCnt = 0, oNoCnt = 0;
printf("\n\t Enter the Size of an Array (total numbers) : \t");
scanf("%d", &totalNum);
printf("\n\t Enter the Numbers to identify even numbers\t");
for(i = 0; i < totalNum; i++)
{
scanf("%d", &ary[i]);
}
for(i = 0; i < totalNum; i ++)
{
if(ary[i] % 2 == 0)
{
eNoCnt++;
}
else
{
oNoCnt++;
}
}
printf("\n\t Total Number of Even Numbers are = %d ", eNoCnt);
printf("\n\t Total Number of Odd Numbers are = %d ", oNoCnt);
}``````
``````
Enter the Size of an Array (total numbers) : 6
Enter the Numbers to identify even numbers 1
2
3
4
5
6
Total Number of Even Numbers are = 3
Total Number of Odd Numbers are = 3
``````
## Program to print Odd Numbers till user input
``````// Program to print odd numbers from 1 to n
#include <stdio.h>
int main() {
int range;
printf("Enter range of number to print odd numbers");
scanf("%d",&range);
printf("Odd Numbers from 1 to %d are \n",range);
for(int i =1;i<=range;i++)
{
if (i%2 != 0)
printf("\t%d",i);
}
return 0;
}``````
Output
``````Enter range of number to print odd numbers 15
Odd Numbers from 1 to 15 are
1 3 5 7 9 11 13 15``````
## Program to Print Even Numbers till User Input
``````// Program to print even numbers from 1 to n
#include <stdio.h>
int main() {
int range;
printf("Enter range of number to print even numbers ");
scanf("%d",&range);
printf("Even numbers from 1 to %d are \n",range);
for(int i =1;i<=range;i++)
{
if (i%2 == 0)
printf("\t%d",i);
}
return 0;
}``````
Output
``````Enter range of number to print even numbers 15
Even numbers from 1 to 15 are
2 4 6 8 10 12 14`````` | 618 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.706833 |
https://www.studysmarter.us/textbooks/math/linear-algebra-and-its-applications-5th/vector-spaces/q15e-if-a-is-a-bf6-times-bf8-matrix-what-is-the-smallest-pos/ | 1,685,879,348,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00084.warc.gz | 1,101,177,037 | 22,542 | • :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon
Suggested languages for you:
Americas
Europe
Q15E
Expert-verified
Found in: Page 191
Linear Algebra and its Applications
Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384
If A is a $${\bf{6}} \times {\bf{8}}$$ matrix, what is the smallest possible dimension of Null A?
The smallest possible dimension of Null A is 2.
See the step by step solution
Step 1: Describe the given data
From the given $$6 \times 8$$ matrix, the number of pivots cannot exceed 6. That is
$${\rm{rank}}\,A \le 6$$.
Step 2: Use the rank theorem
By the rank theorem, you get
\begin{aligned} n &= {\rm{rank}}\,A + \dim \,{\rm{Null}}\,A\\8 &\le 6 + \dim \,{\rm{Null}}\,A\\8 - 6 &\le \dim \,{\rm{Null}}\,A\\2 &\le \dim \,{\rm{Null}}\,A.\end{aligned}
Step 3: Draw a conclusion
Hence, the smallest possible dimension of Null A is 2. | 330 | 961 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.601755 |
https://community.wolfram.com/groups/-/m/t/2197532 | 1,624,408,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00034.warc.gz | 175,883,499 | 29,111 | # Solving system of equations with Solve takes too long?
Posted 4 months ago
1091 Views
|
12 Replies
|
1 Total Likes
|
Hello, I have been trying to solve the system below but it executes forever and doesn't seem to find an answer. F[x_]:=x/(1+x) lambda=l1+l2+l3 c1=RandomReal[3] c2=RandomReal[3] c3=RandomReal[3] c4=RandomReal[3] pn=(P1^(-2)+P2^(-2)+P3^(-2)+(P4*F[lambda]/2)^(-2))^(-1/2) s1=(P1/pn)^(-1/2) s2=(P2/pn)^(-1/2) s3=(P3/pn)^(-1/2) s4=(P4/pn)^(-1/2) Solve[P1\[Equal] (2*(1-s1))/(3*(1-s1)+s1)*c1&&P2\[Equal] (2*(1-s2))/(3*(1-s2)+s2)*c2&&P3==(2*(1-s3))/(3*(1-s3)+s3)*c3&& P4\[Equal] (2*(1-s4))/(3*(1-s4)+s4)*c4&&y*(1-c1/P1)*2*s1*(s4)\[Equal](1+F[lambda]/2)^3 /(D[F[lambda],l1]/2)&&y*(1-c2/P2)*2*s2*(s4)\[Equal](1+F[lambda]/2)^3 /(D[F[lambda],l2]/2)&&y*(1-c3/P3)*2*s3*(s4)\[Equal](1+F[lambda]/2)^3 /(D[F[lambda],l3]/2),{P1,P2,P3,P4,l1,l2,l3}] I have been trying to solve the equations one after the other in the following way : sol1=Solve[P1 == (2*(1-s1))/(3*(1-s1)+s1)*c1,{P1}] X1=Array[x1,{Length@sol1}] x1[k_] /; MemberQ[Range[Length@sol1],l] :=sol1[[k,1,2]] P1=X1 sol2=Solve[P2 == (2*(1-s2))/(3*(1-s2)+s2)*c2,{P2}] X2=Array[x2,{Length@sol2}] x2[k_] /; MemberQ[Range[Length@sol2],l] :=sol2[[k,1,2]] P2=X2 The first part works and I was planning to continue this way but the second part returns with a serie of errors. And still, even the first part takes a very long time to run...Can you tell please tell me why it doesn't work or/and whether there is an other way to have a result in a faster way?
Answer
12 Replies
Sort By:
Posted 4 months ago
I think, this 2=(P2/pn)^(-1/2) is a typo. Shouldn't it be s2=(P2/pn)^(-1/2)
Answer
Posted 4 months ago
Yes, thank you, I have edited my post
Answer
Posted 4 months ago
Are you sure it is F(lambda) instead of F[lambda]? If you define c1 = RandomReal[3,1] you get a list instead of a number: is that really what you want? IÂ would simply write c1 = RandomReal[3].
Answer
Posted 4 months ago
You are right about F[lambda], thank you. About the RandomReal[3,1] I guess you are right too, even if I am not sure it is problematic to have a list of one element instead of a number. I have corrected my post anyway, for both of your remarks.
Answer
Posted 4 months ago
And what is pPit is nowhere defined
Answer
Posted 4 months ago
Another typo, I am sorry. I have corrected it.
Answer
Posted 4 months ago
So, then be so kind and show the corrected code. But these equations look complicated.
Answer
Posted 4 months ago
I have edited my post to correct for your observations, thank you.
Answer
Posted 4 months ago
I could not solve your problem, but tried a lot which might be helpful. The basic problem is that your equations are far too complicated and must be simplified before use. But even this does not work straight forward.See the attached notebook "210219 Solve Simplify.nb" to understand what I was thinking about.There is pretty strange and incorrect code within your first post: Use a block or module, use semicolons and commas, avoid symbols starting with uppercase letters. The code can be formulated shorter. Solve and everything else should run over Reals.Since you don't solve for y is it true that y is some fixed given general parameter? I have added the symbol eqns for your system of equations and simplified it. I realized that your original system is too complicated. It can't even be simplified, let alone solved, in a reasonable amount of time.Even by assuming some conditions conds the system cannot be simplified. Just by examining your original system I found necessary conditions (that are hopefully correct for your original problem): - The p1, p2, p3, p4 have to be positive reals. - The l1, l2, l3 have to be real. - The sum lambda = l1+l2+l3 has to be real and unequal to -1 and unequal to 0. - The given value y has to be real.Unfortunatly this system cannot be simplified by Simplify or FullSimplify. I try to solve it anyway for the variables p1, p2, p3, p4, l1, l2, l3, lambda, which doesn't succeed. This is expected and the same what you observed.All of this is still too complicated to be simplified and solved. See the following code marked with comments "First Attempt". This is exactly your code, only reformulated.There are several problems with your equations which pre-empt simplification: - The complicated expression for pn appears over and over again. - The expression l1+l2+l3 appears over and over again.That's why in the second attempt (marked with "Second Attempt" and the changes marked with "--#--"): - I didn't assign pn first, but left it unevaluated for now, then simplified everything and inserted it only at the very end. - I didn't assign lambda first, but left it unevaluated for now, then simplified everything and inserted it only at the very end. - Actually I reused fl = f[lambda] and dfls = D[f[lambda],li], i=1,2,3, from the first attempt and replaced the sums l1+l2+l3 by lambda. - I used the fact that all partial derivatives dfls are equal and independend of the li. I name them dfl. This gives the new equations eqns2.The conditions conds2 slighty change: - From the definition of pn one can see that pn must be real positive. - The value of lambda has to be real and unequal to -1. - l1, l2, l3 do no longer occur.This system can now be significantly simplified. I try to solve it then for the variables p1, p2, p3, p4, pn, lambda. But this does not succeed. Hence we are not further on than before.If this succeeded I would back-replace pn and lambda into the solution, possibly solve again a new System and verify that the solution satisfies the original system. This is not yet coded.As an extra step I try to proof that after back-replacement of pn and lambda into eqns2 and a further simplification step the new system eqns2a is equivalent to the original system eqns under conds. Usually such proofs are done with Resolve, but this doesn't succeed either. May be the proof is too complicated for Resolve, or I made some error and eqns2a is indeed not equivalent to eqns.I think you need to significantly simplify your original problem and thus its equations to arrive at a solution. Formal simplification, as I have tried, is not sufficient, but you have to start with the underlying physics (or whatever).The following code within the attached notebook executes both attempts. Set test=True to see what happens. Set maxTime as the maximal time for execution of FullSimplify and Resolve, maxTimeS as the maximal time for execution of Solve. I use them just for forcing continuation and termination of the program within reasonable time. Block[{test = True, maxTime = 5, maxTimeS = 10, maxT, f, lambda, l1, l2, l3, fl, dfls, attempt, c1, c2, c3, c4, pn, s1, s2, s3, s4, p1, p2, p3, p4, y, eqns, conds, eqns0, abort, sol, pn2, lambda2, fl2, dfl2s, dfl2, eqns2, conds2, sol2, eqns2a, proof, proofRes, sol2a}, (* Defining the base function and calculating its partial \ derivatives *) f[x_] := x/(1 + x); (*... See the attached notebook for the complete coding ... *) (* Solving the system *) maxT = If[! abort, maxTimeS, maxTime]; abort = False; sol2a = TimeConstrained[ Solve[Join[eqns2a, conds2], {p1, p2, p3, p4, l1, l2, l3}, Reals] , maxT, abort = True]; If[abort, sol2a = {}; Echo[Row[{"no solution after ", maxT, " sec"}], attempt <> "Solve aborted: "]]; If[test && ! abort, Echo[Column[sol2a], attempt <> "sol2a:\n"]]; (* Return solution *) Print[Style[Row[{attempt, "sol2a = ", Column[sol2a]}], Blue, Bold]]; Print[]; ] Attachments:
Answer
Posted 4 months ago
Strange, Olimpia, that you don't answer. I would like to hear if I could help you a little.
Answer
Posted 4 months ago
Dear Werner, Thank you very much for all your work and great explanations. I have been working on it for the last couple of days, and your help was very useful. First of all thanks a lot for the reformulation of my code in the "first attempt", as you have noticed I am very new to this language, and it helped me a lot to learn.It is true that I consider y as a fixed general parameter and you are totally right, that the first derivatives are the same and this simplifies the code. The conditions that pn must be a real positive is totally fine by assumption (and lambda different from one too). For your second attempt, I think fixing pn and lambda in a first step should be a godd way to proceed even if it not working yet and I think I will continue on this way, 1) simplifying the problem (I will try to remove one equation by imposing p3=l3=0), 2) trying to show that this second system - where we replace lambda and pn afterwards is equivalent to the first one. I will certainly let you know here if this way I manage to solve it.I also wanted to tell that one of the reasons I wanted to solve this problem was to find whether their actually is a solution (and I have the intuition there might not be). Again, thanks a lot!
Answer
Posted 4 months ago
I'm glad that I could help a little.
Answer
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments | 2,561 | 9,051 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-25 | latest | en | 0.764728 |
https://stats.stackexchange.com/questions/47969/metrics-of-correlation-for-3-variables | 1,701,676,439,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00634.warc.gz | 615,083,247 | 41,350 | # Metrics of correlation for 3+ variables
I have a basic question, which hopefully you all can resolve for me. What is the best way to determine correlations between 3 or more variables? I have a dataset in which 5 continuous variables each correlate (Pearson, R ~= 0.7) with another continuous variable (which I'll call Z).
I know I can use partial correlation between groups of 2 variables and Z, but I don't think this is exactly what I'm looking for, if I understand it correctly.
• You may be looking for the coefficient of determination of a regression of $z$ on $x_1, \dots, x_5$: en.wikipedia.org/wiki/Coefficient_of_determination But beware, interpreting it can be misleading. Jan 17, 2013 at 19:30
• If I read that correctly, the coefficient of determination gives an idea about how several measurements correlate together with a single variable (Z in this case). What if I want to get a feel for the relationship between these other variables, without reference to Z? Jan 17, 2013 at 21:01
• @learner, if you want summarize the strengths of intercorrelations among p variables in a single number you could compute, for example, (1) average abs. correlation or (2) geometric mean abs. correlation, or (3) determinant of abs(corr. matrix). These three are very different ways to conceptualize p-variate association by a single value. Jan 18, 2013 at 9:45
• @ttnphns, I have heard that the difference between the arithmetic mean and the geometric mean is that the geometric mean reduces the impact of unusually high and low values in the quantities examined (correlations, here). What insight is the determinant of the correlation matrix providing? Jan 18, 2013 at 16:32
• Geometric mean has also another obvious sense: the product is really big when its factors are big jointly. Determinant is the "volume" of the correlation matrix, - the product of its eigenvalues. When correlations are small, det is high; when either they are big or there is collinearities, det is close to 0. Jan 18, 2013 at 18:06 | 476 | 2,015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-50 | longest | en | 0.924019 |
http://economicsessays.com/analysis-of-borrower-and-lender-households/ | 1,695,541,483,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00093.warc.gz | 13,926,081 | 15,556 | - Analysis of Borrower and Lender Households - Free Economics Essays Analysis of Borrower and Lender Households
HELLO, GUEST
Analysis of Borrower and Lender Households
Macro Comp Question 1 August 2015
Suppose there are two types of households in the economy. “Borrower” households borrow funds from “lender” households to purchase housing. Borrower and lender households differ in two ways: first, only borrowers value housing, while both types value non-durable consumption. Second, borrowers discount future utility flows more heavily than lenders: 0 < βB < βL < 1. Both borrowing and lending are subject to constraints.
If you need assistance with writing your essay, our professional essay writing service is here to help!
First, consider the problem of borrowers. Borrowers maximize the expected discounted value of period utility flows with an infinite horizon, where the period utility flow in period t depends on both housing consumption H and non-housing consumption C, as follows:
∞
(1)Max E0 ∑ (βB)t [u(CBt) + w(Ht)]
t=0
where u(C) and w(H) are increasing flow utility functions. Housing depreciates at the rate δ, between 0 and 1. There are no costs of adjustment for housing. Households enter period t with Ht units of housing, and have (1-δ)Ht units of housing available to retain or sell at the end of the period. In period t they choose their housing for next period Ht+1. The house price in period t is Pt, which may be time varying and stochastic. Non-housing consumption is the numeraire, with constant price 1.
In period t, borrowers may borrow Dt+1 ≥ 0 in funds from lender households, at a risk-free interest rate Rt+1, which may be time varying but is known at t. Borrowing is limited to a fraction of the value of next period’s housing stock by the following inequality constraint:
(2)Dt+1≤ θ Pt Ht+1
Borrower households enter period t with a stock of debt RtDt which they must repay to lenders, and receive an endowment of non-durable consumption goods Yt. The household’s period budget constraint is given by
(3) CBt + Pt Ht+1 = Yt + (1 – δ) Pt Ht – RtDt + Dt+1
(a) [5 points]: Identify the minimum set of exogenous and endogenous state variables and the minimum set of control variables, and write down the Bellman Equation for the borrower’s problem. Don’t forget the inequality constraint (2); let μt denote the multiplier on this constraint.
(b) [20 points]: Take first order and envelope conditions, and state the complementary slackness condition. Derive two Euler equations: one involving only μt and values of u'(C) at t and t+1, and the other involving μt, values of u'(C) at t and t+1, and w'(H) at t+1.
Next, consider the problem of lenders. They also maximize expected discounted utility over an infinite horizon, but their period utility does not include housing:
∞
(4)Max E0 ∑ (βL)t u(CLt)
t=0
In period t, lenders may give loans Lt+1 ≥ 0 to borrowers at a risk-free interest rate Rt+1. Lending is limited by the following inequality constraint, where ≥ 0 is a given parameter:
(5)Lt+1≤ L
Lender households enter period t with a stock of loans RtLt for which they receive payment from borrowers, and receive an endowment of non-durable consumption goods Xt. The lender household’s period budget constraint is given by
(6) CLt = Xt + RtLt – Lt+1
(c) [5 points] Identify the minimum set of exogenous and endogenous state variables and control variables and write down the lender’s Bellman equation. Let λt denote the multiplier on the inequality constraint (5).
(d) [15 points] Take first order and envelope conditions and state the complementary slackness condition. Derive the Euler equation for non-durable consumption.
In equilibrium, housing demand by the borrowers must equal the fixed housing supply in all periods. Also, supply and demand for loans must equal: Dt+1 = Lt+1 in all periods. In equilibrium, at least one of the inequality constraints (2) and (5) will bind; both may bind.
Important: for the rest of the problem, assume that the utility function for nondurable consumption for both borrowers and lenders is linear: u(C) = C.
(e) [15 points] Below is a picture of the borrower’s demand curve for funds for this model. Using the complementary slackness condition and the Euler Equation for nondurable consumption from part (b), explain the shape of this demand curve and express the maximum interest rate RMax and the maximum quantity of funds DMax in terms of model parameters and the housing price P.
(f) [20 points] Using the complementary slackness condition and the Euler Equation from part (d), draw a picture of the lender’s supply curve of funds, expressing any threshold interest rates or quantity of funds in terms of model parameters.
Our academic experts are ready and waiting to assist with any writing project you may have. From simple essay plans, through to full dissertations, you can guarantee we have a service perfectly matched to your needs.
(g) [20 points] Equilibrium in the market for funds can be expressed graphically by combining the supply and demand curves for funds in a single graph. Prove that for a given θ, the equilibrium interest rate will equal (1/βL) when L is sufficiently large, and (1/βB) when L is sufficiently small. (Draw one graph for a high value of L and another for a low value of L).
Most Used Categories
Recommendation
EssayHub’s Community of Professional Tutors & Editors
Tutoring Service, EssayHub
Professional Essay Writers for Hire
Essay Writing Service, EssayPro
Professional Custom
Professional Custom Essay Writing Services
In need of qualified essay help online or professional assistance with your research paper?
Browsing the web for a reliable custom writing service to give you a hand with college assignment?
Out of time and require quick and moreover effective support with your term paper or dissertation? | 1,393 | 5,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.931203 |
http://www.math.kit.edu/ianm4/seite/ma-algebraic-geometry/de | 1,537,292,297,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155634.45/warc/CC-MAIN-20180918170042-20180918190042-00178.warc.gz | 359,545,280 | 7,085 | Arbeitsgruppe: Numerische Simulation, Optimierung und Hochleistungsrechnen
Sekretariat
Kollegiengebäude Mathematik (20.30)
Zimmer 3.039
Karlsruher Institut für Technologie (KIT)
Institut für Angewandte und Numerische Mathematik
Englerstrasse 2
76131 Karlsruhe
Öffnungszeiten:
Tel.: +49 721 608 - 42062
Fax.: +49 721 608 - 44178
# Der Modellansatz: Modell171 - Algebraic Geometry
Bei genauem Hinsehen finden wir die Naturwissenschaft und besonders Mathematik überall in unserem Leben, vom Wasserhahn über die automatischen Temporegelungen an Autobahnen, in der Medizintechnik bis hin zum Mobiltelefon. Woran die Forscher, Absolventen und Lehrenden in Karlsruhe gerade tüfteln, erfahren wir im Modellansatz Podcast aus erster Hand.
Gudrun spent an afternoon at the Max Planck Institute for Mathematics in the Sciences (MPI MiS) in Leipzig. There she met the Colombian mathematician Eliana Maria Duarte Gelvez. Eliana was a PostDoc at the MPI MiS in the Research group in Nonlinear Algebra. Its head is Bernd Sturmfels. Now she works as Postdoc at the University of Magedburg.
They started the conversation with the question: What is algebraic geometry? It is a generalisation of what one learns in linear algebra insofar as it studies properties of polynomials such as its roots. But it considers systems of polynomial equations in several variables so-called multivariate polynomials. There are diverse applications in engineering, biology, statistics and topological data analysis. Among them Eliana is mostly interested in questions from computer graphics and statistics.
In any animated movie or computer game all objects have to be represented by the computer. Often the surface of the geometric objects is parametrized by polynomials. The image of the parametrization can as well be defined by an equation. For calculating interactions it can be necessary to know what is the corresponding equation in the three usual space variables. One example, which comes up in school and in the introductory courses at university is the circle. Its representation in different coordinate systems or as a parametrized curve lends itself to interesting problems to solve for the students.
Even more interesting and often difficult to answer is the simple question after the curve of the intersection of surfaces in the computer representation if these are parametrized objects. Moreover real time graphics for computer games need fast and reliable algorithms for that question. Specialists in computer graphics experience that not all curves and surfaces can be parametrized. It was a puzzling question until they talked to people working in algebraic geometry. They knew that the genus of the curve tells you about the possible vs. impossible parametrization.
For the practical work symbolic algebra packages help. They are based on the concept of the Gröbner basis. Gröbner basis help to translate between representations of surfaces and curves as parametrized objects and graphs of functions. Nevertheless, often very long polynomials with many terms (like 500) are the result and not so straightforward to analyse.
A second research topic of Eliana is algebraic statistics. It is a very recent field and evolved only in the last 20-30 years. In the typical problems one studies discrete or polynomial equations using symbolic computations with combinatorics on top. Often numerical algebraic tools are necessary. It is algebraic in the sense that many popular statistical models are parametrized by polynomials. The points in the image of the parameterization are the probability distributions in the statistical model. The interest of the research is to study properties of statistical models using algebraic geometry, for instance describe the implicit equations of the model.
Eliana already liked mathematics at school but was not always very good in it. When she decided to take a Bachelor course in mathematics she liked the very friendly environment at her faculty in the Universidad de los Andes, Bogotá. She was introduced to her research field through a course in Combinatorial commutative algebra there. She was encouraged to apply for a Master's program in the US and to work on elliptic curves at Binghamton University (State University of New York) After her Master in 2011 she stayed in the US to better understand syzygies within her work on a PhD at the University of Illinois at Urbana-Champaign. Since 2018 she has been a postdoc at the MPI MiS in Leipzig and likes the very applied focus especially on algebraic statistics.
In her experience Mathematics is a good topic to work on in different places and it is important to have role models in your field.
## Podcasts
• P. Schwer: Metrische Geometrie, Gespräch mit G. Thäter im Modellansatz Podcast, Folge 102, Fakultät für Mathematik, Karlsruher Institut für Technologie (KIT), 2016.
Diese Podcast-Episode zitieren | 1,058 | 4,897 | {"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} | 2.875 | 3 | CC-MAIN-2018-39 | longest | en | 0.600128 |
http://cn.metamath.org/mpeuni/ballotlemelo.html | 1,660,379,020,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00570.warc.gz | 8,966,799 | 4,665 | Mathbox for Thierry Arnoux < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > ballotlemelo Structured version Visualization version GIF version
Theorem ballotlemelo 30889
Description: Elementhood in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
Hypotheses
Ref Expression
ballotth.m 𝑀 ∈ ℕ
ballotth.n 𝑁 ∈ ℕ
ballotth.o 𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}
Assertion
Ref Expression
ballotlemelo (𝐶𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀))
Distinct variable groups: 𝑀,𝑐 𝑁,𝑐 𝑂,𝑐
Allowed substitution hint: 𝐶(𝑐)
Proof of Theorem ballotlemelo
Dummy variable 𝑑 is distinct from all other variables.
StepHypRef Expression
1 fveq2 6332 . . . 4 (𝑑 = 𝐶 → (♯‘𝑑) = (♯‘𝐶))
21eqeq1d 2773 . . 3 (𝑑 = 𝐶 → ((♯‘𝑑) = 𝑀 ↔ (♯‘𝐶) = 𝑀))
3 ballotth.o . . . 4 𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}
4 fveq2 6332 . . . . . 6 (𝑐 = 𝑑 → (♯‘𝑐) = (♯‘𝑑))
54eqeq1d 2773 . . . . 5 (𝑐 = 𝑑 → ((♯‘𝑐) = 𝑀 ↔ (♯‘𝑑) = 𝑀))
65cbvrabv 3349 . . . 4 {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀} = {𝑑 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑑) = 𝑀}
73, 6eqtri 2793 . . 3 𝑂 = {𝑑 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑑) = 𝑀}
82, 7elrab2 3518 . 2 (𝐶𝑂 ↔ (𝐶 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀))
9 ovex 6823 . . . 4 (1...(𝑀 + 𝑁)) ∈ V
109elpw2 4959 . . 3 (𝐶 ∈ 𝒫 (1...(𝑀 + 𝑁)) ↔ 𝐶 ⊆ (1...(𝑀 + 𝑁)))
1110anbi1i 610 . 2 ((𝐶 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀) ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀))
128, 11bitri 264 1 (𝐶𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀))
Colors of variables: wff setvar class Syntax hints: ↔ wb 196 ∧ wa 382 = wceq 1631 ∈ wcel 2145 {crab 3065 ⊆ wss 3723 𝒫 cpw 4297 ‘cfv 6031 (class class class)co 6793 1c1 10139 + caddc 10141 ℕcn 11222 ...cfz 12533 ♯chash 13321 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-sep 4915 ax-nul 4923 This theorem depends on definitions: df-bi 197 df-an 383 df-or 835 df-3an 1073 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-eu 2622 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ral 3066 df-rex 3067 df-rab 3070 df-v 3353 df-sbc 3588 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-nul 4064 df-if 4226 df-pw 4299 df-sn 4317 df-pr 4319 df-op 4323 df-uni 4575 df-br 4787 df-iota 5994 df-fv 6039 df-ov 6796 This theorem is referenced by: ballotlemscr 30920 ballotlemro 30924 ballotlemfg 30927 ballotlemfrc 30928 ballotlemfrceq 30930 ballotlemrinv0 30934
Copyright terms: Public domain W3C validator | 1,477 | 2,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.234902 |
http://calculator.academy/blended-rate-calculator/ | 1,591,143,235,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00547.warc.gz | 19,845,376 | 44,237 | # Blended Rate Calculator
Enter up to 5 different loan or mortgage amounts and their current interest rates. The blended rate calculator will display the blended interest rate of all of the amounts.
## Blended Rate Formula
The following formula is used to calculate a blended rate from a series of different loan amounts and interest.
BR = Sum ( r * a) / #
• Where BR is the blended rate (%)
• r is the interest rate (%) of each loan
• a is the amount of each loan
• # is the total number of loans
How to calculate a blended rate
1. First, gather all of the interest rates and loan amounts
Calculate or determine the amount of each loan and the % interest rate.
2. Next, multiply the interest rates by the loan amounts
For each loan, multiply the total amount by the % rate. Keep each calculated number separate.
3. Sum the values from step 2
Add up all of the values calculated in step 2.
4. Calculate the blended rate
Finally, divide the sum of the values calculated in step 3 by the total number of loans.
## FAQ
What is a blended rate?
A blended rate is a term used in finance used to describe the weighted average of a set of different loan rates. It takes into account the total values of the loans, so that the blended rate is the effective rate of all loans together.
Is each rate weighted equal in a blended rate?
No, blended rates are weight by their total loan amounts. For example, if you had two loan amounts and rates, say 10,000 at 5% and 1,000 at 1%, the blended rate would be weighted towards 5% at 10 times the weight of 1%. In short, the blended rate would be more than 4.5%. | 373 | 1,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-24 | latest | en | 0.949965 |
https://gmatclub.com/forum/my-gmat-journey-9-25-is-the-day-100978-20.html?sort_by_oldest=true | 1,498,437,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320595.24/warc/CC-MAIN-20170625235624-20170626015624-00418.warc.gz | 757,344,823 | 59,773 | It is currently 25 Jun 2017, 17:37
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# My GMAT Journey ( 9/25 is the DAY)
Author Message
TAGS:
### Hide Tags
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
17 Sep 2010, 00:00
I m just concerned about the relevance of quant and verbal questions with the real Gmat..
Scoring does not matter
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Senior Manager
Joined: 25 Jun 2009
Posts: 302
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
17 Sep 2010, 00:03
gurpreetsingh wrote:
I m just concerned about the relevance of quant and verbal questions with the real Gmat..
Scoring does not matter
As per my experience, questions are pretty much relevant.
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
17 Sep 2010, 00:06
thanks !!
I will give it tomorrow.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Manager
Joined: 09 Sep 2010
Posts: 76
Schools: Haas-Columbia, Wharton, UCLA, Kellog, Booth
WE 1: 10 yrs in Healthcare IT
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
18 Sep 2010, 11:33
Nitish, you are definetly scoring well. I am sure you will crack GMAT this time. My test is on 9/27. please do post your scores once you complete the exam. Good luck!
Senior Manager
Joined: 25 Jun 2009
Posts: 302
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
25 Sep 2010, 01:39
1
KUDOS
willneverquit wrote:
Nitish, you are definetly scoring well. I am sure you will crack GMAT this time. My test is on 9/27. please do post your scores once you complete the exam. Good luck!
Thanks Mate, I just came from the test centre and I have scored dot 700 this time, I am satisfied but not very happy.
I will write a detailed debrief tomorrow.
and yeah best of luck for your test on 27th.
Cheers,
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
25 Sep 2010, 02:05
nitishmahajan wrote:
willneverquit wrote:
Nitish, you are definetly scoring well. I am sure you will crack GMAT this time. My test is on 9/27. please do post your scores once you complete the exam. Good luck!
Thanks Mate, I just came from the test centre and I have scored dot 700 this time, I am satisfied but not very happy.
I will write a detailed debrief tomorrow.
and yeah best of luck for your test on 27th.
Cheers,
You have improved a lot. Congrats Nitish !!
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Senior Manager
Status: Upset about the verbal score - SC, CR and RC are going to be my friend
Joined: 30 Jun 2010
Posts: 316
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
25 Sep 2010, 09:40
nitishmahajan wrote:
willneverquit wrote:
Nitish, you are definetly scoring well. I am sure you will crack GMAT this time. My test is on 9/27. please do post your scores once you complete the exam. Good luck!
Thanks Mate, I just came from the test centre and I have scored dot 700 this time, I am satisfied but not very happy.
I will write a detailed debrief tomorrow.
and yeah best of luck for your test on 27th.
Cheers,
Cool. Nice score. Waiting for your debrief.
_________________
My gmat story
MGMAT1 - 630 Q44V32
MGMAT2 - 650 Q41V38
MGMAT3 - 680 Q44V37
GMATPrep1 - 660 Q49V31
Knewton1 - 550 Q40V27
Manager
Status: He who asks is a fool for five minutes, but he who does not ask remains a fool forever
Joined: 20 Aug 2010
Posts: 101
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
26 Sep 2010, 06:07
Gr8 score Nitish. Congrats . Waiting for debrief....
Manager
Joined: 09 Sep 2010
Posts: 76
Schools: Haas-Columbia, Wharton, UCLA, Kellog, Booth
WE 1: 10 yrs in Healthcare IT
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
26 Sep 2010, 12:13
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
16 Oct 2010, 06:53
nitishmahajan wrote:
Gave Kaplan Free test today and scored 690( Q46, V41) [ 3 dispersed mistakes in Q and 8 in Verbal]
I found Kaplan test pretty much aligned to the Real Deal but unlike MGMAT I did not find a lot of difficult questions in Kaplan Test.
Moreover, I guess the sectional marking is kind of weird, I scored 46 in Q after 3 mistakes
Have you folks also faced the same situation ?
670 - Q47 V 38 => 2 mistakes in quant and 12 in verbal.
lol how can they score q47 with 2 mistakes. If any Kaplan representative is reading this, pls let me know the possible reason. Even if I miss the first two easiest questions, I won't score less than 50 or 49.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Moderator
Status: battlecruiser, operational...
Joined: 25 Apr 2010
Posts: 982
Schools: Carey '16
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
16 Oct 2010, 07:30
maybe it depends on the question you got wrong? maybe they were really easy questions so they dock more points?
idk i don't think any of these CAT's really are able to gauge the exact score. the best they can do is estimate. if we knew the algorithm we'd be able to know exactly which questions to answer correct to get the score we want
_________________
Moderator
Status: battlecruiser, operational...
Joined: 25 Apr 2010
Posts: 982
Schools: Carey '16
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
16 Oct 2010, 07:54
btw, do you mean 10/25? 9/25 was last month
_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
16 Oct 2010, 08:00
vwjetty wrote:
btw, do you mean 10/25? 9/25 was last month
This thread was started by nitish who scored 700 on 9/25
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Intern
Joined: 15 Oct 2010
Posts: 12
Location: United States
Concentration: Finance
Schools: HBS '15, Stanford '15
GMAT Date: 05-04-2012
GPA: 3
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink]
### Show Tags
19 Oct 2010, 19:31
Was there ever a debrief?
Re: My GMAT Journey ( 9/25 is the DAY) [#permalink] 19 Oct 2010, 19:31
Go to page Previous 1 2 [ 34 posts ]
Similar topics Replies Last post
Similar
Topics:
11 My GMAT journey - 760 25 27 Apr 2011, 20:06
7 My GMAT journey.... 61 30 May 2011, 22:21
My GMAT Journey 1 08 Jan 2011, 09:49
2 My GMAT journey 12 17 Dec 2010, 20:40
My GMAT journey 5 17 Oct 2007, 19:55
Display posts from previous: Sort by | 2,794 | 9,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-26 | longest | en | 0.82188 |
https://mynursingaffiliate.com/youve-taken-your-neighbors-young-child-to-the-carnival-to-ride-the-rides-she-wants-to-ride-the-rocket-eight-rocket-shaped-cars-hang-by-chains/ | 1,628,213,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00494.warc.gz | 411,988,230 | 10,424 | You’ve taken your neighbor’s young child to the carnival to ride the rides. She wants to ride The Rocket. Eight rocket-shaped cars hang by chains…
You’ve taken your neighbor’s young child to the carnival to ride the rides. She wants to ride The Rocket. Eight rocket-shaped cars hang by chains from the outside edge of a large steel disk. A vertical axle through the center of the ride turns the disk, causing the cars to revolve in a circle. You’ve just finished taking physics, so you decide to figure out the speed of the cars while you wait. You estimate that the disk is 5 m in diameter and the chains are 6 m long. The ride takes 10 s to reach full speed, then the cars swing out until the chains are 20 from vertical. | 162 | 724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.924512 |
https://thekidsworksheet.com/writing-equations-of-parallel-and-perpendicular-lines-worksheet-pdf/ | 1,713,196,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00154.warc.gz | 525,731,180 | 22,255 | # Writing Equations Of Parallel And Perpendicular Lines Worksheet Pdf
Write the equation of a line perpendicular to y 3 8. Parallel and perpendicular lines write the slope intercept form of the equation of t he line described.
Slope Parallel Perpendicular Lines Determining Coloring Activity Color Activities Parallel And Perpendicular Lines Activities
### 1 y 4x 7 2 9.
Writing equations of parallel and perpendicular lines worksheet pdf. It is given or plug the slope and a point into y mx b and solve for b. This worksheet is supported by a flash presentation under mausmi s math movies. Write the equation of the line that is parallel to the graph of 6 2 1 y x and whose y intercept is 2.
Equations and slopes. M1 1 3 3 0 2 3. Write the equation of the line that is parallel to the graph of y 4x 9 and whose y intercept is 3.
Write the equation in slope intercept form of the line that is perpendicular to. Write the equation of a line perpendicular to y 5 6 x 3 and whose y intercept is 0 11. Write the equation in slope intercept form of the line parallel and line perpendicular to given line through given point.
Equations parallel and perpendicular lines. With this pdf worksheet students learn to rewrite the given equations to the form y mx b and figure out the slopes. 3 5 write and graph equations of lines.
Ahead of discussing equations of parallel and perpendicular lines worksheet with answers please understand that training will be our own key to an improved next week as well as mastering doesn t just stop as soon as the university bell rings of which staying mentioned most of us provide you with a number of straightforward still educational articles or blog posts as well as web templates. Write the equation of a line that is perpendicular to y 2x 4 that passes through the point 8 8 the next questions are a bit different. If the slopes are equal the lines are parallel and if their product is 1 the lines are perpendicular.
Parallel and perpendicular lines worksheet. To write equations of lines using slope intercept form. Two lines are neither parallel nor perpendicular.
Write the equation of the line by plugging in m and b into y mx b. R da cltlq jrzi mgchat gso ir kebsoe or1vre4dp l n am2a hdze 6 cwaivtzh x si 3nxfbion yitxen bagllgle cbir ea1 z1 o 4 worksheet by kuta software llc. Worksheet 3 6 parallel and perpendicular lines date.
Worksheet by kuta software llc geometry camp writing equations of parallel perpendicular lines name date period t s2x0n1x8l vkyuttyaz rsxovf tqwgatrke slflicl g u kawlelk grriygdhatrsq crbetsveqravoewdq. Writing parallel and perpendicular equations worksheet with answers a ced 1 2 this pdf worksheet contains 8 problems that required the student to calculate an equation that is parallel and perpendicular to the given slope intercept equation and passes through a given set of points. Write the slope intercept form of the equation of the line described.
Equations Of Parallel Perpendicular Lines Silly Math Story Line Math Parallel And Perpendicular Lines Equations
Pin On Worksheets
Geometry Worksheets Parallel And Perpendicular Lines Worksheets Geometry Worksheets Parallel And Perpendicular Lines Writing Equations
Writing Linear Equations Given The Slope And A Point Writing Linear Equations Graphing Linear Equations Graphing Linear Inequalities
Parallel And Perpendicular Lines Equations Foldable Algebra Interactive Notebooks Teaching Algebra Graphing Linear Equations
Parallel Lines Perpendicular Lines Geometry Worksheet 4 G 1 4th Grade Geometry Worksheets Line Geometry Angles Worksheet
Parallel And Perpendicular Lines Parallel And Perpendicular Lines Geometry Lessons Perpendicular
Parallel Perpendicular Or Intersecting Lines Parallel And Perpendicular Lines Geometry Worksheets Angles Worksheet
Pin On Absolute Algebra My Pins For Tpt
Equations Of Parallel And Perpendicular Lines Maze Parallel And Perpendicular Lines Equations Perpendicular
Writing The Equation Of Parallel And Perpendicular Lines Algebra Writing Equations Parallel And Perpendicular Lines Algebra
Inb Activity Mazes Writing Equations Of Parallel Perpendicular Lines Parallel And Perpendicular Lines Writing Equations Equations
Parallel Perpendicular And Intersecting Lines Basic Geometry Worksheet Parallel Perpendicular And Intersecting Geometry Worksheets Basic Geometry Line Math
Geometry Worksheets Parallel And Perpendicular Lines Worksheets Geometry Worksheets Parallel And Perpendicular Lines Writing Equations
Math Worksheets Parallel Perpendicular Intersecting Geometry Worksheets Math Worksheets Math Charts
Parallel Lines And Transversals Worksheet Finding The Unknown Worksheets Student Writing Math Teacher
Equations Of Parallel And Perpendicular Lines Chain Activity Teaching Math Teaching Algebra Middle School Math Lesson Plans
Algebra Writing Equations Of Parallel And Perpendicular Lines Writing Equations Writing Linear Equations Equations
Types Of Lines Parallel Intersecting Or Perpendicular In 2020 Types Of Lines Parallel And Perpendicular Lines Line Math | 1,036 | 5,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-18 | latest | en | 0.914849 |
https://gmatclub.com/forum/two-trains-started-simultaneously-from-opposite-ends-of-a-110263.html | 1,571,734,260,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00327.warc.gz | 522,569,770 | 155,709 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 22 Oct 2019, 01:51
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Two trains started simultaneously from opposite ends of a
Author Message
TAGS:
### Hide Tags
Intern
Joined: 26 Feb 2011
Posts: 29
Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
Updated on: 27 Jun 2013, 02:14
3
38
00:00
Difficulty:
25% (medium)
Question Stats:
78% (02:11) correct 22% (02:26) wrong based on 635 sessions
### HideShow timer Statistics
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?
A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5
Originally posted by Madelaine88 on 02 Mar 2011, 06:18.
Last edited by Bunuel on 27 Jun 2013, 02:14, edited 2 times in total.
Edited the question
Math Expert
Joined: 02 Sep 2009
Posts: 58410
### Show Tags
02 Mar 2011, 06:39
13
14
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?
A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5
As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.
_________________
Senior Manager
Joined: 17 Sep 2013
Posts: 322
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
19 Sep 2013, 05:00
11
Speed of first train: 100/5 = 20 Mph
Speed of second train: 100/3 = 33.33 Mph
Now in 1 hr distance covered by X is 20 & that by Y is 33.33.In the next hr it will be 40 & 66.66 resp.
40+66.66=106.6> the distance between them...So they would have met by now & the answer would be something less than 40..which leaves us with A
_________________
Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..
##### General Discussion
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 653
### Show Tags
02 Mar 2011, 06:30
3
1
6
Train X : 100 m in 5 hrs
speed = 100/5 = 20mph
Train Y : 100 m in 3 hrs
speed = 100/3 mph
The distance shrinking at effective speed (20 + 100/3 ) mph
Time of intersect = 100 / (20+100/3) = 15/8 hrs
Distance travelled by X < Distance travelled by Y
Distance travelled by X = time * speed = 15/8 * 20 = 75/2
Manager
Joined: 15 Apr 2011
Posts: 59
### Show Tags
13 Apr 2012, 00:42
Bunuel, could you please explain this. It is not clear to me. thanks!
_________________
Manager
Joined: 13 Mar 2012
Posts: 202
Concentration: Operations, Strategy
### Show Tags
13 Apr 2012, 02:02
vel x/ vel y=3/5
let they met after t time, distance travelled by X=x
then,
x/vel x=(100-x)/ vel y
=> x/(100-x)= 3/5
=> x= 300/8= 37.5
_________________
Practice Practice and practice...!!
If there's a loophole in my analysis--> suggest measures to make it airtight.
Math Expert
Joined: 02 Sep 2009
Posts: 58410
### Show Tags
13 Apr 2012, 02:12
2
3
Bunuel, could you please explain this. It is not clear to me. thanks!
Since, the ratio of times of X and Y to cover the same distance of 100 miles is is 5:3, then the ratio of their rates is 3:5. Consider this, say the rates of trains X and Y are X and Y respectively, then:
Distance=Rate*Time --> X*5=Y*3 --> ratio of the rates is X:Y=3:5. At the time they meet, so after they travel the same time interval the ratio of distances covered by X and Y would also be in that ratio (for example if X=3 mph and Y=5 mph then they would meet in 100/(3+5)=100/8 hours, hence train X would cover 300/8 miles and train Y would cover 500/8 miles --> ratio of distances covered (300/8):(500/8)=3:5).
Now, since the the ratio of distances covered by X and Y is 3:5 then X covered 3/(3+5)=3/8 of the total distance.
Hope it's clear.
_________________
Manager
Joined: 13 Jul 2013
Posts: 61
GMAT 1: 570 Q46 V24
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
19 Sep 2013, 02:37
Why is that when the meet they would have covered 100 miles? This bit is little confusing.
Math Expert
Joined: 02 Sep 2009
Posts: 58410
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
19 Sep 2013, 04:55
theGame001 wrote:
Why is that when the meet they would have covered 100 miles? This bit is little confusing.
When they meet one train covers some part of 100-mile distance and another covers the remaining part of 100-mile distance, so combined they cover 100 miles.
_________________
Intern
Joined: 04 May 2015
Posts: 2
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
30 Dec 2015, 10:48
6
1
1
Speed of train X= Distance/Time
= 100/5
= 20 mph
Speed of train Y= Distance/Time
= 100/3
= 33.33 mph
Relative Speed (opposite direction) = 20+33.33
= 53.33 mph
Now,
Time when X & Y meet:
Distance = Relative Speed x Time
Time= D/RS
= 100/53.33
= 1.87 hrs
After 1.87 hrs X & Y meet at some distance
In 1.87 hrs, X traveled
Distance= 20 x 1.87
And Y traveled,
Distance= 33.33 x 1.87
= 62.3 (approx.)
I hope it makes your life somewhere easy.
Hit kudos if like.
Manager
Joined: 17 Nov 2013
Posts: 77
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
02 Apr 2016, 18:29
RxT = D. D/R = T, SO WHEN YOU SEE TIME = 100 AND THE CHALLENGE IS THAT BOTH TRAINS MUST MEET SO YOU NEED TO ADD (+) THE INDIVIDUAL RATES. ALSO, TO GET THE INDIVIDUAL RATES USE FOR X = 100/5 = 20 M/H, RATE FOR Y = 100/3 M/H. SO IN YOUR FORMULA YOU HAVE: $$\frac{100}{(20+(100/3))} = T$$ => T = 15/8, MEANING BOTH TRAINS MET AT 15/8 TIME. TO FIND OUT THE DISTANCE AT WHICH X MET Y, YOU DO R x T = D => 20 x (15/8) => D = 37.5 MILES FOR X WHET IT MET Y.
VP
Joined: 09 Mar 2016
Posts: 1230
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
11 Apr 2018, 04:49
Bunuel wrote:
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?
A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5
As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.
Speed X = 100/5= 20
Speed y = 100/3 = 33.33
relative speed 20+33 =53
time when they meet = 100/53 =1.8
Distance done by X when meeting Y --> 20*1.8 = 36 (I got approximate answer , is it correct ? )
thank you!
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3577
Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
11 Apr 2018, 05:19
1
dave13 wrote:
Speed X = 100/5= 20
Speed y = 100/3 = 33.33
relative speed 20+33 =53
time when they meet = 100/53 =1.8
Distance done by X when meeting Y --> 20*1.8 = 36 (I got approximate answer , is it correct ? )
thank you!
Hey dave13 ,
Yes, your approach is 100% correct. If you get the time to two decimal places, you will get more closer answer to 37.5.
Thanks!
_________________
My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place
GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.
New! Best Reply Functionality on GMAT Club!
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free
Check our new About Us Page here.
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 737
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
12 Apr 2018, 10:24
1
1
dave13 wrote:
Bunuel wrote:
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?
A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5
As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.
Speed X = 100/5= 20
Speed y = 100/3 = 33.33
relative speed 20+33 =53
time when they meet = 100/53 =1.8
Distance done by X when meeting Y --> 20*1.8 = 36 (I got approximate answer , is it correct ? )
thank you!
dave13 Perfect approach .Relative speed concept is used nicely with proper simplification of numbers.
However if you keep the relative speed as is: 100 ( 1/5 + 1/3 ) = 100*8/15 (Speed)
time = 100/(100*8/15) = 15/8 hr
distance traveled by 20*15/8 = 75/2 = 37.5
It would help if the options were very close. Supposed we have 36 & 37.5 both in the options...
Best,
_________________
Regards,
“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
Senior Manager
Joined: 15 Jan 2017
Posts: 340
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
30 Jul 2018, 23:25
Solved it this way:
train x: 100/5 = 20 mph
train y: 100/ 3 = 33.4 mph (approx)
relative speed = 100/53.4 = 25/13.1 = approx 2 hours
so time for X --> 20mph*2 = 40 --> but since its not actually 2 hours but close to it, it must closer to 40 not 40. Which is option A) 37.5
Kudos if you liked this solution
Intern
Joined: 16 May 2017
Posts: 48
GPA: 3.8
WE: Medicine and Health (Health Care)
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
20 Oct 2018, 02:29
Train X travelling speed= 100/5 mph
Train Y travelling speed=100/3 mph
Relative speed as they are travelling in opposite direction= 100/5+100/3= 800/15 mph
Time after which X and Y would meet= distance/relative speed= 100/800/15= 100*15/800= 15/8 hr
X travelled distance when it met with train Y= speed*time they met= 100*15/5*8= 37.5 miles.
Ans (A)
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8125
Location: United States (CA)
Re: Two trains started simultaneously from opposite ends of a [#permalink]
### Show Tags
09 Sep 2019, 10:58
1
Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?
A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5
The combined distance traveled by the two trains was 100 miles. Each train traveled for t hours. We can create the distance equation:
100/5 * t + 100/3 * t = 100
Multiplying by 15 to clear the fractions from the equation, we have:
300t + 500t = 1500
800t = 1500
t = 15/8
Thus, train X had traveled 15/8 x 100/5 = 15/8 x 20 = 37.5 miles by the time it reached Y.
_________________
# Scott Woodbury-Stewart
Founder and CEO
Scott@TargetTestPrep.com
122 Reviews
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.
Re: Two trains started simultaneously from opposite ends of a [#permalink] 09 Sep 2019, 10:58
Display posts from previous: Sort by | 3,991 | 13,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-43 | latest | en | 0.919466 |
https://the-algorithms.com/algorithm/determinant?lang=javascript | 1,701,506,998,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00002.warc.gz | 647,524,569 | 20,505 | #### Determinant
N
/**
* Given a square matrix, find its determinant using Laplace Expansion.
* Time Complexity : O(n!)
*
*
* @param {number[[]]} matrix - Two dimensional array of integers.
* @returns {number} - An integer equal to the determinant.
*
* @example
* const squareMatrix = [
* [2,3,4,6],
* [5,8,9,0],
* [7,4,3,9],
* [4,0,2,1]
* ];
*
* const result = determinant(squareMatrix);
* // The function should return 858 as the resultant determinant.
*/
const subMatrix = (matrix, i, j) => {
let matrixSize = matrix[0].length
if (matrixSize === 1) {
return matrix[0][0]
}
let subMatrix = []
for (let x = 0; x < matrixSize; x++) {
if (x === i) {
continue
}
subMatrix.push([])
for (let y = 0; y < matrixSize; y++) {
if (y === j) {
continue
}
subMatrix[subMatrix.length - 1].push(matrix[x][y])
}
}
return subMatrix
}
const isMatrixSquare = (matrix) => {
let numRows = matrix.length
for (let i = 0; i < numRows; i++) {
if (numRows !== matrix[i].length) {
return false
}
}
return true
}
const determinant = (matrix) => {
if (
!Array.isArray(matrix) ||
matrix.length === 0 ||
!Array.isArray(matrix[0])
) {
throw new Error('Input is not a valid 2D matrix.')
}
if (!isMatrixSquare(matrix)) {
throw new Error('Square matrix is required.')
}
let numCols = matrix[0].length
if (numCols === 1) {
return matrix[0][0]
}
let result = 0
let setIndex = 0
for (let i = 0; i < numCols; i++) {
result +=
Math.pow(-1, i) *
matrix[setIndex][i] *
determinant(subMatrix(matrix, setIndex, i))
}
return result
}
export { determinant } | 475 | 1,637 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-50 | latest | en | 0.263912 |
https://math.stackexchange.com/questions/3104918/evaluating-int-5-2-fx-dx-int-1-61-3fx-dx-int-6-53-2/3105093 | 1,566,125,638,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00469.warc.gz | 563,180,386 | 30,586 | # Evaluating $\int_{-5 }^{-2} f(x)\;dx + \int_{1/6}^{1/3}f(x)\,dx+\int_{6/5}^{3/2}f(x)\;dx$, where $f(x)=\left(\frac{x^2-x}{x^3-3x+1}\right)^2$
How to find the value of $$\int _ { - 5 } ^ { - 2 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx + \int _ { \frac 16 } ^ { \frac 13 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx+\int _ { \frac 65 } ^ {\frac 32 } \left( \frac { x ^ { 2 } - x} { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx$$
My attempt:
I tried to substitute in second and third integral, such that powers of integration become the same. But I could not find such a substitution. tried taking out powers of x from denominator such that differential coefficient gets formed in the numerator .
• Please use mathjax instead of including a picture of formulas – supinf Feb 8 at 10:06
• Hoping that you don't mind, I fixed the edit of the post. Cheers :-) – Claude Leibovici Feb 8 at 11:22
$$I= \int _ { - 5 } ^ { - 2 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx + \int _ { \frac 16 } ^ { \frac 13 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx+\int _ { \frac 65 } ^ {\frac 32 } \left( \frac { x ^ { 2 } - x} { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx$$
$$\displaystyle I_{1} = \int^{-2}_{-5}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2dx,$$ put $$\displaystyle x=1-\frac{1}{u}\rightarrow u=\frac{1}{1-x}$$ and $$\displaystyle dx = \frac{1}{u^2}du$$
$$\displaystyle I_{1}=\int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\frac{1}{x^2}dx$$
$$\displaystyle I_{3}=\int^{\frac{3}{2}}_{\frac{6}{5}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2dx,$$ put $$\displaystyle x=\frac{1}{1-v}\rightarrow v=1-\frac{1}{x}$$ and $$\displaystyle dx = \frac{1}{(1-v)^2}dv$$
$$I_{3} =\int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\frac{1}{(1-x)^2}dx$$
$$I = \int^{\frac{1}{3}}_{\frac{1}{3}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\bigg[\frac{1}{x^2}+1+\frac{1}{(1-x)^2}\bigg]dx$$
$$I = \int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\bigg[\frac{x^{4}-2x^{3}+3x^{2}-2x+1}{x^2(1-x)^2}\bigg]dx$$
$$I = \int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x-x^{2}}{x^{3}-3x+1}\bigg)'dx = \frac{x-x^{2}}{x^{3}-3x+1}\bigg|^{\frac{1}{3}}_{\frac{1}{6}}=$$ | 1,087 | 2,298 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-35 | latest | en | 0.419265 |
https://www.edrm.net/glossary/probability/ | 1,558,318,878,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255536.6/warc/CC-MAIN-20190520021654-20190520043654-00092.warc.gz | 772,073,868 | 10,701 | # Probability
## Definition(s)
The fraction (The fraction of a set of Documents having some particular property (typically Relevance).) of times that a particular outcome would occur, should the same action be repeated under the same conditions an infinite number of times. For example, if one were to flip a fair coin, the Probability of it landing “heads” is one-half, or 50%; as one repeats this action indefinitely, the fraction of times that the coin lands “heads” will become indistinguishable from 50%. If one were to flip two fair coins, the Probability of both landing “heads” is one-quarter, or 25%. 1
## Notes
1. Maura R. Grossman and Gordon V. Cormack, EDRM page & The Grossman-Cormack Glossary of Technology-Assisted Review, with Foreword by John M. Facciola, U.S. Magistrate Judge2013 Fed. Cts. L. Rev. 7 (January 2013). | 213 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-22 | longest | en | 0.915089 |
https://studydaddy.com/question/example-1-consider-another-example-where-a-pack-contains-4-blue-2-red-and-3-blac | 1,660,313,307,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00143.warc.gz | 470,579,703 | 8,306 | Answered You can buy a ready-made answer or pick a professional tutor to order an original one.
QUESTION
# Example 1: Consider another example where a pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probab
Example 1:
Consider another example where a pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?
Example 2:
What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement.
• @
• 9 orders completed
Solution 1;
Here, total number of pens = 9
Probability of drawing 1 blue pen = 4/9Probability of drawing another blue pen = 4/9Probability of drawing 1 black pen = 3/9Probability of drawing 2 blue pens and 1 black pen = 4/9 * 4/9 * 3/9 = 48/729 = 16/243
Solution 2;
Probability of drawing a king = 4/52 = 1/13
After drawing one card, the number of cards are 51.
Probability of drawing a queen = 4/51.
Now, the probability of drawing a king and queen consecutively is 1/13 * 4/51 = 4/663 | 333 | 1,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.92147 |
https://rbseguide.com/class-9-science-chapter-3/ | 1,716,129,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00884.warc.gz | 434,312,835 | 20,356 | # RBSE Solutions for Class 9 Science Chapter 3 Atomic Structure
RBSE Solutions for Class 9 Science Chapter 3 Atomic Structure are part of RBSE Solutions for Class 9 Science. Here we have given Rajasthan Board RBSE Class 9 Science Chapter 3 Atomic Structure.
Board RBSE Textbook SIERT, Rajasthan Class Class 9 Subject Science Chapter Chapter 3 Chapter Name Atomic Structure Number of Questions Solved 85 Category RBSE Solutions
## Rajasthan Board RBSE Class 9 Science Chapter 3 Atomic Structure
### Atomic Structure Textbook Questions Solved
Objective Type Questions
Question 1.
Plum pudding model of the atom was proposed by:
(A) Neils Bohr
(B) Thomson
(C) Rutherford
(D) Goldstein James
Question 2.
Neutron was discovered by:
(A) C.V. Raman
(B) Rutherford
(C) J.J. Thomson
Question 3.
The size of an atom is:
(A) 10-6cm
(B) 10-5cm
(C) 10-2 cm
(D) 10-8 cm
Question 4.
The number of neutrons in deuterium, an isotope of hydrogen is:
(A) one
(B) two
(C) three
(D) None
### Atomic Structure Very Short Answer Type Questions
Question 5.
Define isotopes.
Isotopes: Atoms of the same element having the same atomic number but different mass numbers are called isotopes.
For example, Chlorine has two isotopes
$$^{ 35 }{ Cl }\quad and\quad _{ 17 }^{ 37 }{ Cl }$$, in which atomic number is 17 for both, but mass numbers are 35 and 37 respectively.
Question 6.
Define isobars.
Isobars: Atoms of different elements having the same mass number, but different atomic numbers are known as isobars.
For example, Calcium and argon have the same mass number 40, but different atomic numbers 20 and 18, respectively.
Question 7.
What are the fundamental particles of an atom?
Electron, proton, and neutron are called fundamental particles of an atom.
Question 8.
Define atomic mass.
Atomic mass (or atomic weight) of an element is defined as the relative mass of the atom of the element as compared to the mass of an atom of C-12 isotope, i.e
$$Atomic mass (or Atomic Weight)=\cfrac { Mass\quad (or\quad Weight\quad of\quad the\quad element) }{ \cfrac { 1 }{ 12 } th\quad of\quad mass\quad of\quad C-12\quad isotope }$$
Question 9.
Define atomic number.
Atomic number: The number of unit positive charges (or protons) present in the nucleus of an atom of a particular element is called the atomic number of that element. Furthermore, the number of protons is exactly equal to the number of electrons present in a neutral atom. Atomic number is represented by Z. Thus Atomic Number (Z) = Number of Protons = Number of Electrons, in a neutral atom The atomic number is represented by putting ‘Z’ in subscript on the left-hand side of the symbol of the element, for e.g., 2He for helium as He has only two protons in its nucleus. Similarly, for the atomic number of potassium is 19, it means that the potassium atom has 19 protons and 19 electrons in its neutral atom and is depicted by K.
Question 10.
What is the charge on neutron?
Neutron carries no charge, i.e., the neutron is a neutral particle.
Question 11.
Write the value of Avogadro’s number.
The value of Avogadro’s number is 6.022 x 1023.
Question 12.
Who discovered proton?
Proton was discovered by Goldstein.
### Atomic Structure Short Answer Type Questions
Question 13.
What is an electric discharge tube? Explain with a diagram.
Electrical discharge tube:
Electrical discharge through gases is studied by using a specially designed glass tube commonly called a discharge tube. It consists of a cylindrical glass tube having a side tube, and two metal electrodes, one at each end. These electrodes can be connected to the respective terminals of a high tension power supply. The air inside the tube can be pumped out by connecting the side tube to a vacuum pump, and the desired pressure can be maintained inside the tube.
Question 14.
Explain the postulates of Thomson’s model of an atom.
Thomson suggested that an atom is positively charged sphere of radius 10-8 cm. Negatively charged electrons are embedded in the sphere like plums in the pudding, or like the seeds in the edible red interior of a watermelon.
Since the two types of charges are equal in amount, the atom on the whole electrically neutral. It is also known as the plum-pudding model.
Question 15.
Explain the mole concept.
Wilhelm Ostwald introduced the term mole (Latin ‘moles’ means heap), in 1896 which was accepted as a unit to represent a large number of atoms in 1967. The Mole is the amount of a substance which contains as many particles as in 12g of C-12. Thus, one mole of any species is that quantity in number which has a mass equal to its atomic or molecular mass in grams.
For example, 1 mole of carbon isotope (atomic mass = 12) is equal to 12g.
1 mole of oxygen (Oa, molecular mass = 2 x 16) is equal to 32 g.
1 mole of water (H2O, molecular mass = 2×1 + 1×16) is equal to 18 g.
Question 16.
Write the main postulates of Dalton’s atomic theory.
The main postulates of Dalton’s atomic theory are:
• Every matter is made up of very small particles, called the atoms.
• Atoms are indivisible particles which can neither be created nor destroyed, in a chemical reaction.
• Atoms of a given element are identical in mass as well as in chemical properties.
• Atoms of different elements have different masses and chemical properties.
• Atoms combine in the ratio of small whole numbers to form compounds.
• The relative numbers, as well as the kind of atoms, are constant in a given compound.
Question 17.
Write the characteristics of Cathode rays.
The characteristics of Cathode rays are:
• When these rays strike the glass walls of the tube, a greenish glow is produced.
• When an electric field is applied through metal plates, placed over the discharge tube, cathode rays are deflected towards the positive terminal, showing that cathode rays consist of negatively charged particles.
• When a strong magnetic field is applied, cathode rays are deflected which shows cathode rays consist of moving charged particles.
• If a solid is placed in their path, its shadow is observed on a screen, showing that cathode rays move in a straight line.
• When a wheel is placed in their path, it starts moving, showing that cathode rays consist of material particles, possessing kinetic energy.
Question 18.
Explain covalent radius of an atom with an example.
Covalent radius is the half of the distance between the nuclei of two covalently bonded atoms of the same element in a molecule. Example, Covalent radius of chlorine molecules is 99Å.
### Atomic Structure Long Answer Type Questions
Question 19.
Describe Rutherford’s nuclear model of an atom with the help of the gold foil experiment and well-labeled diagram.
Rutherford designed an experiment in which fast-moving alpha-particles were made to fall on a thin gold foil. On the basis of the experiment, Rutherford proposed a model of an atom as follows:
(a) There is a positively charged region located at the center in an atom called the nucleus. Nearly, all the mass of an atom resides in the nucleus.
(b)The electrons revolve around the nucleus in well-defined orbits.
(c) The size of the nucleus is very small as compared to the size of the atom.
Question 20.
Describe the main postulates of the Neils Bohr model of the atom. Draw and explain the atomic structure of Na and K based on it.
Bohr’s model of an atom:
1. An atom is made up of charged particles, electrons, and protons. Electrons are negatively charged, whereas protons are positively charged.
2. The electrons revolve around the nucleus in fixed orbits called energy levels or shells which are represented by numbers 1, 2, 3, — or letters K, L, M, N, — counted from the nucleus outwards.
3. Protons are present within the nucleus. Due to the presence of protons, the nucleus is positively charged.
4. Each energy level has a fixed amount of energy. The energy of orbits increases with distance from the nucleus.
5. The number of electrons and protons are equal so that an atom, on the whole, is electrically neutral.
6. There is no change in the energy of electrons as long as they revolve in the same orbit and remain stable. However, if an electron gains energy, it jumps to a higher energy level and if it loses energy, it falls back to the lower energy level.
Structure of potassium: The atomic number of potassium is 19 and its mass the number is 39. It is represented by the symbol $${ 19 }^{ 39 }{ K }$$
Being a neutral atom, it has: Number of electrons = 19 = Number of protons & Number of neutrons = 39 – 19 = 20 The nucleus of potassium contains 19 protons and 20 neutrons i.e. 39 nucleons.
According to rule 1, K – shell (n = 1) can not have more than 2 electrons. L-shell according to rule 1
(n = 2) can not have more than 8 electrons. The new shell i.e. the M-shell begins to fill and may contain a maximum of 18 electrons. However, being the outermost shell, it can not have more than 8 electrons (Rule 2). Thus, no. of electrons in N-shell = 8.
Now, according to rule 4, the filling of N-shell starts and since only one electron is left after filling K, L, M shell it occupies N-shell.
Thus, the electronic configuration of potassium would be 2, 8, 8, 1. or K – shell = 2 electrons, L – shell = 8 electrons M – shell = 8 electrons, N – shell = 1 electron Structure of Sodium: Same rule applies for sodium element also. Thus, the electronic configuration of sodium would be 2, 8, 1 K shell = 2 electron L shell = 8 electron M shell = 1 electron.
Question 21.
What are positive rays? How are they produced? Write their properties also.
While working with cathode rays, Sir J.J. Thomson had noticed a red glow around the cathode, on the opposite side of the anode. To know the reason for this glow, he designed a discharge tube with a perforated cathode fixed in the middle of the tube, as shown in the figure.
On applying an electrical potential across the two electrodes, green fluorescence was seen in the glass tube at one end, while on the other end, red fluorescence was seen. The green fluorescence was due to the bombardment of glass by the cathode rays. Obviously, the red fluorescence was due to some other type of rays. J.J. Thomson later showed that these rays causing red fluorescence consisted of positively charged particles and were thus, named as positive or anode rays. These rays were also called canal rays.
Production of positive rays: Cathode rays consist of negatively charged particles called electrons. These electrons move away from the cathode with very high speeds. These fast-moving electrons split the molecule into atoms, and remove one or more electrons from the atoms. Thus, the atoms get converted into the positive ions due to the loss of electrons. These positive ions pass through the holes in the cathode plate to produce a glow on the glass wall of the discharge tube. A stream of these positively charged particles is called positive rays or anode rays.
Properties of positive rays:
(a) Positive rays consist of positively charged particles.
(b) The nature of these rays depends on the gas used, in the discharge tube.
(c) These rays travel in straight lines.
(d) These rays get deflected by an electrical field and bend towards the negative plate. Thus, the deflection of the positive rays is in a direction opposite to that shown by the cathode rays.
(e) These rays are also deflected by the magnetic fields, in a direction opposite to that of the cathode rays.
(f) These rays can produce mechanical as well as chemical effects.
(g)The ratio of charge (e) to mass (m), i.e., (e/m) for the particles in the positive rays is not the same for all gases.
(h)The ratio e/m for the positive rays is very low as compared to the e/m value for cathode rays. This shows that the particles in the positive rays are heavier than the particles in the cathode rays.
Numerical Questions Solved
Question 1.
An isotope of an element having a number of neutrons 9 and its mass number is 17. Write the name and an atomic number of the element.
Mass number = Z + n
17 = Z + 9
Z = 17 – 9
Z = 8
Atomic number = 8 The name of the element is oxygen.
Question 2.
Calculate the weight of nitrogen (in gms) at 22.4 liters at NTP.
At NTP, one mole of a gas occupies a volume of 22.4 liters. Since, N, molecules have
molecular weight = 28 grams.
Thus, 22.4 liters of N0 at NTP has weight = 28 grams.
Question 3.
How many numbers of atoms of C is present in 1.5 moles of carbon?
1 mole of C = 6.022 x 1023 atoms
1.5 mole of C = 6.022 x 10a x 1.5 = 9.033 x 1023 atoms of carbon.
Question 4.
How many water molecules are present in 9 gm of water?
We know that 1 mole of any compound contains 6.022>< 1023 molecules of that compound.
Here, mass of water = 9 gm
Molar mass of water (H2O) = (2 x 1 + 1 x 16)g/mol = 18g/mol
∵ 18 g (= 1 mol of H2O contains 6.02 x 1023 molecules.
∴ 9 gm of H2O contains $$= [latex]\cfrac { 6.022\times { 10 }^{ 23 }\times 9g }{ 18g } =3.011\times { 10 }^{ 23 } molecules$$
### Atomic Structure Additional Questions Solved
I. Multiple Choice Questions (MCQs)
Question 1.
The accepted unit of atomic and molecular masses is:
(A) kilogram
(B) gram
(C) pound
(D) atomic mass unit
Question 2.
An element whose gram-atomic mass and gram-molecular mass are the same is:
(A) hydrogen
(B) oxygen
(C) nitrogen
(D) helium
Question 3.
Mass of Avogadro’s number of oxygen (O) atoms is equal to:
(A) 16 u
(B) 16 g
(C) 32 g
(D) 6 kg
Question 4.
The mass percentage of hydrogen (H) in water (H2O) is:
(A) 1.11
(B) 88.9
(C) 8.89
(D) 11.11
Question 5.
The number of molecules in 8 g of oxygen gas is:
(A) 6.02 x 1023
(B) 3.01 x 1023
(C) 1.5 x 1023
(D) 3.01 x 1012
Question 6.
Which of the following is used as the ‘reference standard’ used at present, for describing the atomic and molecular masses?
(A) Carbon-13
(B) Chlorine-35
(C) Oxygen-8
(D) Carbon-12
Question 7.
The gas which has a molecular mass twice that of oxygen gas is:
(A) CO2
(B) CO
(C) SO2
(D) H2S
Question 8.
The element which can form gases having different atomicity is:
(A) chlorine
(B) oxygen
(C) hydrogen
(D) nitrogen
Question 9.
A number of elements present in ammonium phosphate are:
(A) 2
(B) 3
(C) 4
(D) 5
Question 10.
If atomic mass of nitrogen is 14 u and atomic mass of hydrogen is 1 u, what is the molar mass of ammonia?
(A) 17 u
(B) 15 u
(C) 34 u
(D) 30 u
Question 11.
Molar mass of HNO3 is 63 u, and an atomic mass of hydrogen and nitrogen are 1 u and 14 u respectively. Atomic mass of oxygen will be:
(A) 3 u
(B) 16 u
(C) 48 u
(D) 24 u
Question 12.
One mole of BaCf ionises completely, total moles of barium and chloride ions will be:
(A) 1 mole
(B) 2 moles
(C) 3 moles
(D) 02 x 1023 moles
Question 13.
A neutral atom contains an equal number of………
(A) Protons and Electrons
(B) Protons and Neutrons
(C) Neutrons and Electrons
(D) Protons, Electrons, and Neutrons
Question 14.
Particles present in the nucleus are:
(A) Proton and Electron
(B) Proton and Neutron
(C) Neutron and Electron
(D) Only Neutron
Question 15.
The presence of the nucleus at the centre of the atom was suggested by:
(A) J.J. Thomson
(B) Rutherford
(C) Goldstein
(D) Bohr
Question 16.
A maximum number of electrons which can be accommodated in the nth energy level is given by:
(A) n2
(B) 2n2
(C) 2n
(D) n
Question 17.
The fundamental particles with proper electric charge in an atom are:
(A) Electron (- 1), Proton (+ 1) and Neutron (1)
(B) Electron (- 1), Proton (0) and Neutron (0)
(C) Electron (+ 1), Proton (0) and Neutron (- 1)
(D) Electron (- 1), Proton (+ 1) and Neutron (0)
Question 18.
The correct electronic configuration of sodium atom is:
(A) 2, 8
(B) 2, 8, 1
(C) 8, 2, 1
(D) 1, 2, 8
Question 19.
Which of the following is correct about the energy of the shells in an atom?
(A) M > L > K
(B) M < L < K
(C) K > M > L
(D) M > K > L
Question 20.
An atom of an element with atomic number 18 and mass number 40 has the following articles:
(A) 18 protons, 18 electrons, 22 neutrons
(B) 18 protons, 22 electrons, 18 neutrons
(C) 22 protons, 18 electrons, 18 neutrons
(D) 22 protons, 22 electrons, 18 neutrons
Question 21.
Rutherford’s scattering experiment of resulted in the discovery of:
(A) electron
(B) proton
(C) neutron
(D) nucleus in an atom
Question 22.
A mass number of an atom is equal to the:
(A) number of protons in the nucleus of the atom
(B) number of electrons and protons in the atom
(C) number of neutrons and protons in the nucleus of an atom
(D) number of neutrons in the nucleus of an atom
Question 23.
A number of electrons in the valence shell of aluminum atom (at. no. 13) is:
(A) 1
(B) 2
(C) 3
(D) 0
Question 24.
Which of the following pairs represents a pair of isotopes?
(A) 8p + 8n ; 8p + 9n
(B) 9p + 9n ; 8p + 9n
(C) 18p + 22n ;16p + 24n
(D) 9p + 9n ; 8p + 10n
Question 25.
The number of electrons in an element X is 15 and the number of neutrons is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element?
(A) $$_{ 15 }^{ 31 }{ X }$$
(B) $$_{ 16 }^{ 31 }{ X }$$
(C) $$_{ 16 }^{ 15 }{ X }$$
(D) $$_{ 15 }^{ 16 }{ X }$$
Question 26.
The ion of an element has 3 unit positive charge. A mass number of its atom is 27 and number of neutrons in it is 14. What is the number of electrons in the ion?
(A) 14
(B) 13
(C) 27
(D) 10
Question 27.
The mass number of an element is 7 and number of neutrons in its atom is 4. What is the number of electrons in the atom?
(A) 3
(B) 4
(C) 7
(D) 11
Question 28.
Which of the following configuration is not possible as per the Bohr’s model of an atom?
(A) 2, 5
(B) 2, 8, 8
(C) 2, 7, 3
(D) 2, 8, 2
Question 29.
Which of the following pairs represents a pair of isobars?
(A) 8p + 8n ; 8p + 9n
(B) 18p + 22n ; 20p + 20n
(C) 18p + 22n ; 16p + 22n
(D) 8p + 9n ; 8p + 8n
### Atomic Structure Very Short Answer Type Questions
Question 1.
Define the term atom.
An atom is the smallest particle of an element that can take part in a chemical reaction. It may or may not exist independently.
Question 2.
Define law of conservation of mass for a chemical reaction.
Law of conservation of mass states that “mass can neither be created nor destroyed in a chemical reaction”.
Question 3.
State the law of constant proportion.
Law of Constant Proportion: In a chemical substance, the elements are always present in a definite proportion by mass.
Question 4.
‘Atoms of most elements are not able to exist independently. ’Name two atoms which exist as independent atoms.
Noble gases such as argon (Ar), Helium (He) exist as independent atoms.
Question 5.
Define atomicity.
The number of atoms present in one molecule of an element or a compound is known as its atomicity.
Question 6.
Give one relevant reason why scientists, initially choose 1/16 of the mass of an atom of naturally occurring oxygen as the atomic mass unit.
Initially, 1/16th of the mass of naturally occurring oxygen was taken as the atomic mass unit because this unit gave masses of most of the elements as whole numbers.
Question 7.
What is molar mass? What are its units?
The mass of one mole of a substance is called its molar mass. Its unit is gram per mole (g/mol.)
Question 8.
How many moles of CO2 are there is 11g of carbon dioxide? Molar mass of CO2 is 44 g/mol.
Question 9.
Name the scientists who proposed the law of conservation of mass and law of constant proportions.
Ans.
Lavoisier proposed the law of conservation of mass and Proust proposed the law of constant proportions.
Question 10.
Which two fundamental particles are present inside the nucleus?
Protons and neutrons.
Question 11.
Which particles were bombarded by Rutherford on a very thin gold foil?
α-particles.
Question 12.
Chemical properties of all the isotopes of an element are similar. Give one reason.
Isotopes of an element have similar electronic configuration and the same number of valence electrons.
Question 13.
What is Orbit?
According to Bohr’s model, the electron in an atom moves around the nucleus in a fixed circular path. This path is called orbit.
Question 14.
An element has mass number 28 and atomic number 14. Find the number of protons, neutrons, and electrons in it. Write the complete symbol of the element.
Number of protons =14
Number of neutrons =14
Number of electrons =14
Symbol of the element is $$_{ 14 }^{ 28 }{ Si }$$
Question 15.
If the atomic number of an element is 11 and its mass number is 23, then what is the number of neutrons in its nucleus?
Atomic mass of sodium atom is 23 and its atomic number is 11.
Hence a number of neutrons (N) will be 23 – 11 = 12.
### Atomic Structure Short Answer Type Questions
Question 1.
Why do the canal rays obtained by using different gases have different e/m values?
When the electrons emitted from the cathode collide with the neutral atoms of the gas present in the tube, they remove one or more electrons present in them. This leaves behind positively charged particles, which travel towards the cathode. As the atoms of different gases have a different number of protons present in them. Therefore, different gases are found to have different e/m values.
Question 2.
What happens when the cathode rays are passed through an electric field between two parallel plates? Can one determine the nature of the charge of the particles constituting the cathode rays from this experiment? If Yes, how?
When the cathode rays are passed through an electric field, they are deflected towards the positively charged plate. This deflection of cathode rays towards the positively charged electrode indicated that the particles in the cathode rays are negatively charged particles.
Question 3.
The electronic configuration of an element X is 2, 8, 7.
1. What is the atomic number of the element X?
2. Is X metal or non-metal?
3. What is the valency of X?
1. Z = 17
2. Non-metal
3. 1.
Question 4.
If an atom carries one proton and one electron, will it carry ant) next charge or not?
This atom will not carry any next charge, because the positive charge of a proton will be balanced by the negative charge of the electron.
Question 5.
What is the difference between isobar: and isotopes ? Give one example for each
Isobars Isotopes 1. They have a same mass number but different atomic numbers. 2. They have different numbers of protons. 3. Example: $$_{ 20 }^{ 40 }{ A }\quad and\quad _{ 18 }^{ 40 }B\quad$$ 1. They have different mass numbers but the same atomic number. 2. They have the same number of protons. 3. Example: $$_{ 92 }^{ 238 }{ A }\quad and\quad _{ 92 }^{ 235 }B\quad$$
Question 6.
State the law of conservation of mass. Is this law applicable to the chemical reactions? Elaborate your answer with the help of an example.
(1) Law of conservation of mass states that mass can neither be created nor destroyed.
(2) Yes, this law is applicable to the chemical reactions. In all chemical reactions, there is an only exchange of reactants taking place when products are formed. There is no loss or gain of mass.
For example: In the following reaction, the total mass of the reactants is equal to the total mass of the products formed.
Question 7.
(1) What is the difference between molecular mass and molar mass?
(2) What relation exists between the atomic mass of an element and number of atoms? Explain with an example.
(1) The average mass of a molecule of a substance in the atomic mass unit is called molecular mass, whereas, the mass of 1 mole or 6.02 x 1023 molecules of the substance (in grams) is called its molar mass. Both these terms are numerically the same but differ in units. Molecular mass is expressed in the atomic mass unit (u), whereas, molar mass is expressed in grams.
(2) The mass of 6.02 x 1023 atoms, in grams is equal to its atomic mass. In other words, the atomic mass of every element contains 6.02 x 1023 atoms of that element.
For example, Atomic mass of magnesium is 24. So 24-gram magnesium has 6.02 x 1023 atoms of Mg, and a mass of 6.02 x 1023 atoms of magnesium is 24 gram.
Question 8.
What is Avogadro’s number and how is it related to the molecular mass of compounds? Explain with an example?
The number 6.02 x 1023 is called Avogadro’s number.
The mass, in grams of 6.02 x 1023 molecules of any substance is equal to its molar mass. In other words, molecules mass of any substance has 6.02 x 1023 molecules.
For example, molecular mass of water H2O = 2 + 16 = 18
.’. Molar mass of H2O = 18 g
.’. In 18 g of H2O molecules,
6.022 x 1023 molecules are there.
Question 9.
(a) What is the maximum number of electrons that can be present in the outermost shell of an atom?
(b) Draw a sketch of Bohr’s model of an atom with three shells.
(a) Eight.
Question 10.
(a) How are canal rays different from electrons in terms of charge and mass?
(b) What are canal rays? Who discovered them? What is the charge and mass of canal ray?
(a) Canal rays consist of positively charged particles, protons, having appreciable mass, while electrons are negatively charged with negligible mass.
(b) New radiations in a gas discharge tube which are positively charged are known as canal rays. They were discovered by E. Goldstein. Charge on canal rays is positive and its mass is one unit.
Question 11.
What information do you get from the diagrams given below about the atomic number, mass number, and valency of the atoms X, Y, and Z? Give your answer in a tabular form
Atom Atomic Number Mass Valency X 5 11 3 Y 8 18 2 Z 15 31 3, 5
Question 12.
State the major drawback in Rutherford’s model of an atom. Mention two features of Bohr’s Model which helped compensate this drawback.
The major drawback of Rutherford’s model of an atom is that it does not explain the stability of an atom. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus.
Two features of Bohr’s model which helped compensate this drawback are as follows:
1. Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.
2. While revolving in these discrete orbits, the electrons do not radiate energy.
### Atomic Structure Long Answer Type Questions
Question 1.
Find the % composition of the elements in the following compounds:
(1) Water
(2) Sodium Sulphate
Question 2.
(1) Define one mole, illustrate its relationship with Avogadro constant,
(2) Calculate the number of moles in:
• 12.044 x 1023 atoms of carbon
• 64g of oxygen atoms
• 66g of carbon dioxide molecules.
(Atomic masses O = 16, C = 12)
(1) One mole of any species (atoms, molecules, ions or particles) is that quantity in number which has a mass equal to its atomic or molecular mass (in grams). The number of particles (atoms), molecules or ions) present in 1 mole of any substance is fixed, with a value of 6.022 x 1023. This number is called Avogadro constant or Avogadro number.
Question 3.
(a) Define:
(1) Molecular mass
(b) Calculate the number of molecules in 50g of CaCO3
(Atomic mass of Ca = 40 u, C = 12u and O=16u)
(c) If one mole of sodium atom weighs 23g, what is the mass (in g) of one atom of sodium?
(a)
(1) Molecular mass:
It is equal to the sum of the masses of atoms present in one molecule of a substance.
The number of atoms present in 12 gms of C-12-isotope.
$$\cfrac { Gram\quad atom\quad mass\quad of\quad carbon }{ Mass\quad of\quad one\quad carbon\quad atom }$$
Question 4.
A number of electrons, protons, and neutrons in chemical species A, B, C, and D is given below.
Element Electrons Protons Neutrons A 2 3 4 B 10 9 8 C 8 8 8 D 8 8 10
1. What is the mass number of A and B?
2. What is the atomic number of B?
3. Which two elements represent a pair of isotopes and why?
4. What is the valency of element C? Also, justify your answers.
(1) Mass number of A = 3 + 4 = 7
Mass number of B = 9 + 8= 17
(2) The atomic number of B
= Number of protons = 9
(3) Elements C and D represent a pair of isotopes because their atomic numbers are the same, but mass numbers are different.
(4) Electronic configuration of C(8)=2, 6
So, its valency is 2.
Question 5.
Give reasons for the following:
(a) The nucleus of an atom is heavy.
(b) The nucleus of an atom is positively charged.
(c) An atom is electrically neutral. | 7,462 | 28,152 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | longest | en | 0.822538 |
https://ccssanswers.com/spectrum-math-grade-4-chapter-1-pretest-answer-key/ | 1,723,018,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00896.warc.gz | 136,185,448 | 49,158 | This handy Spectrum Math Grade 4 Answer Key Chapter 1 Pretest provides detailed answers for the workbook questions.
Question 1.
a.
Explanation:
Adding the ones we get 5 + 3 = 8
Adding the tens we get 3 + 0 = 3
So, the sum of both addends 35 and 3 is 38
35 + 3 = 38
b.
Explanation:
Adding the ones we get 5 + 3 = 8
Adding the tens we get 2 + 1 = 3
So, the sum of both addends 25 and 13 is 38
25 + 13 = 38
c.
Explanation:
Adding the ones we get 5 + 4 = 9
Adding the tens we get 7 + 2 = 9
So, the sum of both addends 75 and 24 is 99
75 + 24 = 99
d.
Explanation:
Adding the ones we get 3 + 2 = 5
Adding the tens we get 1+ 1 = 2
So, the sum of both addends 13 and 12 is 25
13 + 12 = 25
e.
Explanation:
Adding the ones we get 2 + 7 = 9
Adding the tens we get 4+ 0 = 4
So, the sum of both addends 42 and 7 is 49
42 + 7 = 49
f.
Explanation:
Adding the ones we get 4 + 3 = 7
Adding the tens we get 5+ 3 = 8
So, the sum of both addends 54 and 33 is 87
54 + 33 = 87
Question 2.
a.
Explanation:
Adding the ones we get 3 + 4 = 7
Adding the tens we get 4+ 2 = 6
So, the sum of both addends 43 and 24 is 67
43 + 24 = 67
b.
Explanation:
Adding the ones we get 4 + 5 = 9
Adding the tens we get 5+ 0 = 5
So, the sum of both addends 54 and 5 is 59
54 + 5 = 59
c.
Explanation:
Adding the ones we get 3 + 1 = 4
Adding the tens we get 6+ 3 = 9
So, the sum of both addends 63 and 31 is 94
63 + 31 = 94
d.
Explanation:
Adding the ones we get 2 + 6 = 8
Adding the tens we get 8+ 1 = 9
So, the sum of both addends 82 and 16 is 98
82 + 16 = 98
e.
Explanation:
Adding the ones we get 2 + 3 = 5
Adding the tens we get 3+ 2 = 5
So, the sum of both addends 32 and 23 is 55
32 + 23 = 55
f.
Explanation:
Adding the ones we get 4 + 5 = 9
Adding the tens we get 7+ 1 = 8
So, the sum of both addends 74 and 15 is 89
74 + 15 = 89
Question 3.
a.
Explanation:
Adding the ones we get 0 + 3 = 3
Adding the tens we get 5+ 3 = 8
So, the sum of both addends 50 and 33 is 83
50 + 33 = 83
b.
Explanation:
Adding the ones we get 5 + 2 = 7
Adding the tens we get 9+ 0 = 9
So, the sum of both addends 95 and 2 is 97
95 + 2 = 97
c.
Explanation:
Adding the ones we get 2 + 5 = 7
Adding the tens we get 3+ 2 = 5
So, the sum of both addends 32 and 25 is 57
32 + 25 = 57
d.
Explanation:
Adding the ones we get 3 + 5 = 8
Adding the tens we get 7+ 2 = 9
So, the sum of both addends 73 and 25 is 98
73 + 25 = 98
e.
Explanation:
Adding the ones we get 6 + 3 = 9
Adding the tens we get 5+ 1 = 6
So, the sum of both addends 56 and 13 is 69
56 + 13 = 69
f.
Explanation:
Adding the ones we get 7 + 2 = 9
Adding the tens we get 4+ 3 = 7
So, the sum of both addends 47 and 32 is 79
47 + 32 = 79
Question 4.
a.
Explanation:
Adding the ones we get 2 + 7 = 9
Adding the tens we get 1+ 0 = 1
So, the sum of both addends 12 and 7 is 19
12 + 7 = 19
b.
Explanation:
Adding the ones we get 6 + 2 = 8
Adding the tens we get 3+ 1 = 4
So, the sum of both addends 36 and 12 is 48
36 + 12 = 48
c.
Explanation:
Adding the ones we get 5 + 3 = 8
Adding the tens we get 5+ 2 = 7
So, the sum of both addends 55 and 23 is 78
55 + 23 = 78
d.
Explanation:
Adding the ones we get 0 + 9 = 9
Adding the tens we get 7+ 1 = 8
So, the sum of both addends 70 and 19 is 89
70 + 19 = 89
e.
Explanation:
Adding the ones we get 2 + 4 = 6
Adding the tens we get 9+ 0 = 9
So, the sum of both addends 92 and 4 is 96
92 + 4 = 96
f.
Explanation:
Adding the ones we get 4 + 3 = 7
Adding the tens we get 5+ 2 = 7
So, the sum of both addends 54 and 23 is 77
54 + 23 = 77
Question 5.
a.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 5 – 4 = 1
Subtract the tens we get 4 – 0 = 4
So, the difference of 45 and 4 is 41
45 – 4 = 41
b.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 5 – 3 = 2
Subtract the tens we get 7 – 2 = 5
So, the difference of 75 and 23 is 52
75 – 23 = 52
c.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 6 – 4 = 2
Subtract the tens we get 6 – 1 = 5
So, the difference of 66 and 14 is 52
66 – 14 = 52
d.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 5 – 1 = 4
Subtract the tens we get 9 – 3 = 6
So, the difference of 95 and 31 is 64
95 – 31 = 64
e.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 4 – 2 = 2
Subtract the tens we get 8 – 2 = 6
So, the difference of 84 and 22 is 62
84 – 22 = 62
f.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 5 – 2 = 3
Subtract the tens we get 2 – 1 = 1
So, the difference of 25 and 12 is 13
25 – 12 = 13
Question 6.
a.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 7 = 2
Subtract the tens we get 4 – 2 = 2
So, the difference of 49 and 27 is 22
49 – 27 = 22
b.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 7 – 6 = 1
Subtract the tens we get 5 – 4 = 1
So, the difference of 57 and 46 is 11
57 – 46 = 11
c.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 8 = 1
Subtract the tens we get 3 – 1 = 2
So, the difference of 39 and 18 is 21
39 – 18 = 21
d.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 7 = 2
Subtract the tens we get 7 – 2 = 5
So, the difference of 79 and 27 is 52
79 – 27 = 52
e.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 7 – 6 = 1
Subtract the tens we get 2 – 0 = 2
So, the difference of 27 and 6 is 21
27 – 6 = 21
f.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 8 – 6 = 2
Subtract the tens we get 8 – 5 = 3
So, the difference of 88 and 56 is 32
88 – 56 = 32
Question 7.
a.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 5 – 5 = 0
Subtract the tens we get 6 – 5 = 1
So, the difference of 65 and 55 is 10
65 – 55 = 10
b.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 8 – 3 = 5
Subtract the tens we get 7 – 3 = 4
So, the difference of 78 and 33 is 45
78 – 33 = 45
c.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 4 – 2 = 2
Subtract the tens we get 5 – 4 = 1
So, the difference of 54 and 42 is 12
54 – 42 = 12
d.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 7 – 6 = 1
Subtract the tens we get 9 – 2 = 7
So, the difference of 97 and 26 is 71
97 – 26 = 71
e.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 5 = 4
Subtract the tens we get 2 – 1 = 1
So, the difference of 29 and 15 is 14
29 – 15 = 14
f.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 8 = 1
Subtract the tens we get 5 – 4 = 1
So, the difference of 59 and 48 is 11
59 – 48 = 11
Question 8.
a.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 4 – 3 = 1
Subtract the tens we get 5 – 2 = 3
So, the difference of 54 and 23 is 31
54 – 23 = 31
b.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 8 = 1
Subtract the tens we get 2 – 1 = 1
So, the difference of 29 and 18 is 11
29 – 18 = 11
c.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 7 – 7 = 0
Subtract the tens we get 4 – 3 = 1
So, the difference of 47 and 37 is 10
47 – 37 = 10
d.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 6 = 3
Subtract the tens we get 9 – 6 = 3
So, the difference of 99 and 66 is 33
99 – 66 = 33
e.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 9 – 7 = 2
Subtract the tens we get 8 – 2 = 6
So, the difference of 89 and 27 is 62
89 – 27 = 62
f.
Subtract ones Subtract tens
Explanation:
Subtract the ones we get 6 – 5 = 1
Subtract the tens we get 3 – 1 = 2
So, the difference of 36 and 15 is 21
36 – 15 = 21
Solve each problem.
Question 9.
Mr. Dimas has 15 new students in his fourth-grade class. He already has 21 students in the class. How many students are in Mr. Dimas’s class?
There are ___________ students in his class.
Question 10.
There are 35 pages in Kendrick science book. Last night, Kendrick read 14 pages. How many more pages does Kendrick have left to read?
There are ____ pages left to read.
21
Explanation:
There are 35 pages in Kendrick science book
So, there are more 21 pages left to read.
Total pages – read pages = left pages
35 – 14 = 21
Question 11.
Kono’s father gave him 75 apples so he could pass them out to his friends. If Kono gave 43 away, how many apples does he have left?
There are ____ apples left.
32
Explanation:
Kono’s father gave him 75 apples so he could pass them out to his friends.
If Kono gave 43 away, then
There will be 32 apples left.
75 – 43 = 32
Question 12.
Monica and Tania want to throw a surprise party for Rosa. They plan to send out 45 invitations. If Tania writes 24, how many invitations does Monica need to write?
Monica needs to write ___________ invitations.
21
Explanation:
Monica and Tania want to throw a surprise party for Rosa
They plan to send out 45 invitations
If Tania writes 24, then
Monica needs to write 21 invitations.
24 + 21 = 45
Question 13.
Seki’s soccer team is in the State Cup Tournament. There were 23 goals made in the entire tournament. Seki’s team made 12 of them. How many goals were made by the other teams?
The other teams scored ____ goals. | 3,429 | 9,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-33 | latest | en | 0.844294 |
https://web2.0calc.com/questions/algebra-2-help_8 | 1,560,684,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00323.warc.gz | 638,816,105 | 6,444 | +0
# Algebra 2 help
0
203
4
+299
Q1)
Q2)
Q3)
Q4)
Oct 14, 2018
#1
+101084
+2
1. v^2 -8v - 59 = -2 add 59 to both sides
v^2 - 8v = 57 take 1/2 of 8 = 4...square it = 16....add to both sides
v^2 - 8x + 16 = 57 + 16 factor the left side, simplify the right
(v - 4)^2 = 73 take both roots
x - 4 = sqrt73 x - 4 = -sqrt73
x = 4 + sqr73 or x = 4 - sqrt73
Oct 14, 2018
#2
+101084
+2
16x^2 + 9y^2
[ 4x + 3iy ] [ 4x - 3iy ]
Oct 14, 2018
#3
+101084
+2
x^2 + 4x + 13 = 0 subtract 13 from both sides
x^2 + 4x = -13 take 1/2 of 4 = 2...square it = 4 add to both sides
x^2 + 4x + 4 = -13 + 4 factor the left...simplify the right
(x + 2)^2 = -9 take both roots
x + 2 = sqrt (-9) or x + 2 = -sqrt(-9)
x+ 2 = 3i or x + 2 = -3i subtract 2 from both sides
x = -2 + 3i or x = -2 - 3i
Oct 14, 2018
#4
+101084
+2
x^2 + 27 = 0
x^2 = -27 take both roots
x = sqrt (-27) or x = -sqrt (-27)
x = i *sqrt (27) or x = -i * sqrt (27)
x = i * sqrt ( 9 * 3) or x = -i * sqrt (9 * 3)
x = 3isqrt(3) or x = -3isqrt(3)
Oct 14, 2018 | 621 | 1,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.087124 |
https://www.rhayden.us/regression-models/info-owf.html | 1,576,330,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541157498.50/warc/CC-MAIN-20191214122253-20191214150253-00342.warc.gz | 833,731,399 | 6,312 | ## Info
FIGURE 15.4 Probit model: (a) given I,, read P, from the ordinate; (b) given P,, read I, from the abscissa.
610 PARTTHREE: TOPICS IN ECONOMETRICS
Probit Estimation with Grouped Data: gprobit
We will use the same data that we used for glogit, which is given in Table 15.4. Since we already have Pi, the relative frequency (the empirical measure of probability) of owning a house at various income levels as shown in Table 15.5, we can use it to obtain Ii from the normal CDF as shown in Table 15.10, or from Figure 15.5.
Once we have the estimated Ii, estimating and 02 is relatively straightforward, as we show shortly. In passing, note that in the language of probit analysis the unobservable utility index Ii is known as the normal equivalent deviate (n.e.d.) or simply normit. Since the n.e.d. or Ii will be negative whenever Pi < 0 . 5, in practice the number 5 is added to the n.e.d. and the result is called a probit.
TABLE 15.10 ESTIMATING THE INDEX i FROM THE STANDARD NORMAL CDF
TABLE 15.10 ESTIMATING THE INDEX i FROM THE STANDARD NORMAL CDF | 293 | 1,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.873312 |
http://www.physicsforums.com/showthread.php?p=3887465 | 1,369,151,345,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700168711/warc/CC-MAIN-20130516102928-00059-ip-10-60-113-184.ec2.internal.warc.gz | 550,118,953 | 8,841 | Fortran: help with arguments to a pre-built software
There are two functions that I need help with:
http://www.maths.uq.edu.au/expokit/fortran/dgchbv.f
The problem I'm having is that the arguments on the line DGPADM(....) don't match identically with the explanation. For example, H, ldh and m vs. m and H(ldh,m) not sure why there is a parenthesis. If someone could give me a simple example of acceptable parameters that I could try out,that would be greatly appreciated.
I understand what to put for ideg, m, t, but not so sure about wsp, ipiv , H or the others. For example, H is a matrix, so I guess it needs a matrix as an argument..... regardless could someone help me out here?! Thanks!
In the first case:
subroutine DGPADM( ideg,m,t,H,ldh,wsp,lwsp,ipiv,iexph,ns,iflag )
* ideg : (input) the degre of the diagonal Pade to be used.
* a value of 6 is generally satisfactory.
*
* m : (input) order of H.
*
* H(ldh,m) : (input) argument matrix.
*
* t : (input) time-scale (can be < 0).
*
* wsp(lwsp) : (workspace/output) lwsp .ge. 4*m*m+ideg+1.
*
* ipiv(m) : (workspace)
*
*>>>> iexph : (output) number such that wsp(iexph) points to exp(tH)
* i.e., exp(tH) is located at wsp(iexph ... iexph+m*m-1)
* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
* NOTE: if the routine was called with wsp(iptr),
* then exp(tH) will start at wsp(iptr+iexph-1).
*
* ns : (output) number of scaling-squaring used.
*
* iflag : (output) exit flag.
* 0 - no problem
* <0 - problem
Second case:
subroutine DGCHBV( m, t, H,ldh, y, wsp, iwsp, iflag )
* m : (input) order of the matrix H
*
* t : (input) time-scaling factor (can be < 0).
*
* H(ldh,m): (input) argument matrix.
*
* y(m) : (input/output) on input the operand vector,
* on output the resulting vector exp(t*H)*y.
*
* iwsp(m) : (workspace)
*
* wsp : (workspace). Observe that a double precision vector of
* length 2*m*(m+2) can be used as well when calling this
* routine (thus avoiding an idle complex array elsewhere)
PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
Mentor Mod note: moved to Prog & CS section
Mentor
Quote by brydustin There are two functions that I need help with: http://www.maths.uq.edu.au/expokit/fortran/dgpadm.f http://www.maths.uq.edu.au/expokit/fortran/dgchbv.f The problem I'm having is that the arguments on the line DGPADM(....) don't match identically with the explanation. For example, H, ldh and m vs. m and H(ldh,m) not sure why there is a parenthesis. If someone could give me a simple example of acceptable parameters that I could try out,that would be greatly appreciated. I understand what to put for ideg, m, t, but not so sure about wsp, ipiv , H or the others. For example, H is a matrix, so I guess it needs a matrix as an argument..... regardless could someone help me out here?! Thanks!
I'll take a stab at the first one.
H - the name of a two-dimensional matrix - the matrix contains double precision numbers.
ldh - # of rows in matrix H - an integer.
m - # of cols in matrix H - an integer.
t - timescale. This is the constant t in etH -- double precision.
wsp - workspace - an output array with lwsp elements, each of type double precision.
lwsp - # of elements of the wsp array - an integer. It needs to be > 4m2 + ideg + 1.
ipiv - an output array with m integer elements.
iexph - an index into the wsp array - an integer.
The computed elements of etH are located in the wsp array. They start at wsp(iexph) and end at wsp(iexph + m2 - 1). This section of the array contains all of the elements of etH.
ns - no clue - integer.
iflag - a flag that indicates whether the calculation was successful or not. Negative value indicates some kind of error.
I would start with a very simple matrix, like
$$H = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$
and a value of 1.0 for t.
You should end up with
$$e^{tH} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
The variables that are identified as being output don't need to be initialized before calling this routine. The other variables need to be initialized before the call.
Quote by brydustin In the first case: subroutine DGPADM( ideg,m,t,H,ldh,wsp,lwsp,ipiv,iexph,ns,iflag ) * ideg : (input) the degre of the diagonal Pade to be used. * a value of 6 is generally satisfactory. * * m : (input) order of H. * * H(ldh,m) : (input) argument matrix. * * t : (input) time-scale (can be < 0). * * wsp(lwsp) : (workspace/output) lwsp .ge. 4*m*m+ideg+1. * * ipiv(m) : (workspace) * *>>>> iexph : (output) number such that wsp(iexph) points to exp(tH) * i.e., exp(tH) is located at wsp(iexph ... iexph+m*m-1) * ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ * NOTE: if the routine was called with wsp(iptr), * then exp(tH) will start at wsp(iptr+iexph-1). * * ns : (output) number of scaling-squaring used. * * iflag : (output) exit flag. * 0 - no problem * <0 - problem Second case: subroutine DGCHBV( m, t, H,ldh, y, wsp, iwsp, iflag ) * m : (input) order of the matrix H * * t : (input) time-scaling factor (can be < 0). * * H(ldh,m): (input) argument matrix. * * y(m) : (input/output) on input the operand vector, * on output the resulting vector exp(t*H)*y. * * iwsp(m) : (workspace) * * wsp : (workspace). Observe that a double precision vector of * length 2*m*(m+2) can be used as well when calling this * routine (thus avoiding an idle complex array elsewhere)
Tags fortran software | 1,634 | 5,520 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2013-20 | latest | en | 0.71461 |
https://socratic.org/questions/58223f8fb72cff56e22bc133 | 1,571,755,358,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822098.86/warc/CC-MAIN-20191022132135-20191022155635-00249.warc.gz | 679,229,532 | 6,579 | # Question #bc133
Jun 30, 2017
Let the maximum uniform acceleration with which the man can ascend $5 m$ to reach the balcony with the help of the hanging rope (the other end of which is tied with a chandelier of mass $124 k g$ be $a m \text{/} {s}^{2}$
The accelerated motion of the man will create extra tension on the rope more than the weight of the man. This tension should not be greater than than weight of the chandelier of mass $124 k g$.
So we can write $70 \times \left(a + g\right) = 124 \times g$
where $g =$ acceleration due to gravity $= 9.8 m \text{/} {s}^{2}$
So the above equation becomes
$70 \times \left(a + 9.8\right) = 124 \times 9.8$
$\implies \left(a + 9.8\right) = 124 \times \frac{9.8}{70}$
$\implies a = 124 \times \frac{9.8}{70} - 9.8 = 7.56 m \text{/} {s}^{2}$
So if time taken by the man to reach the balcony with this maximum acceleration staring from rest be $t$ sec then by equation of kinematics we can write
$0 \times t - \frac{1}{2} a {t}^{2} = 5$
$= t = \sqrt{\frac{10}{a}} = \sqrt{\frac{10}{7.56}} \approx 1.15 s$ | 356 | 1,061 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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-2019-43 | longest | en | 0.796607 |
https://codereview.stackexchange.com/questions/166028/suffixing-numeric-values | 1,718,567,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00192.warc.gz | 150,333,518 | 39,830 | Suffixing numeric values
This is a part of my Factorio mod Visual Signals.
This code takes a numeric value as input and returns a string for the value, for example:
-12 --> -12
1 --> 1
50 --> 50
999 --> 999
1500 --> 1.5k
24999 --> 24.9k
123456 --> 123k
1234567 --> 1.2M
12345678 --> 12M
123456789 --> 123M
1234567890 --> 1.2G
and so on...
The idea is to replicate the same way Factorio show signals in the circuit network.
The code works by determining a prefix, which simply is a potential minus sign. It then determines the middle part (the numbers and potential comma separator to use) and the suffix (the letter at the end).
The code seems to work perfectly fine and now I simply wonder: Can I improve this code somehow?
local suffixChars = { "", "k", "M", "G", "T", "P", "E" }
function CountString(count)
local absValue = math.abs(count)
local prefix = ""
if count < 0 then
prefix = "-"
end
local suffix = 1
while absValue >= 1000 do
absValue = absValue / 1000
suffix = suffix + 1
end
local str = tostring(absValue)
if absValue < 10 then
return prefix .. string.sub(str, 1, 3) .. suffixChars[suffix]
end
if absValue < 100 then
return prefix .. string.sub(str, 1, 2) .. suffixChars[suffix]
end
return prefix .. string.sub(str, 1, 3) .. suffixChars[suffix]
end
First off, your naming convention. You have a function name CountString, which to me states that you are either counting a string (?) or something along those lines. A better name would be FriendlyNumber or HumaneNumber.
count in a similar manner, means tracking an iterator. It could become input, rawValue etc.
Now, to the actual function. You can make use of a log function to compute your suffix (base 10), and the same for referring to your index in suffix list, as well as getting the absValue.
Instead of concatenating different parts of string, use a string.format, which would also remove your call to tostring.
Lua has a ternary operator, which can be done using and..or statements. Based on your version of lua engine (4.x, 5.{1, 2, 3}), you might have different implementation of the log function. A sample code would be:
-- For Lua 5.1
local suffixChars = { "", "k", "M", "G", "T", "P", "E" }
local l, f, a = math.log10, math.floor, math.abs -- Lua 5.2+ have math.log(x, base)
function FriendlyNumber(rawValue)
local absValue = a(rawValue)
local prefix = rawValue < 0 and "-" or ""
local log = f(l(absValue) / 3)
-- I was calculating log first, but I needed floor(1/3rd) of it,
-- and later had to multiply by 3 again, so computing the 1/3rd here only
local suffix = suffixChars[1 + log]
absValue = absValue / (10 ^ (log * 3))
local str = ("%%.%df"):format((absValue < 1000 and absValue > 10) and 0 or 1):format(absValue)
return ("%s%s%s"):format(prefix, str, suffix)
end
-12 -12
1.0 1
50 50
999 999
1.5k 1500
25k 24999
123k 123456
1.2M 1234567
12M 12345678
123M 123456789
1.2G 1234567890
Use built-in tools
The idea is to replicate the same way Factorio show signals in the circuit network.
Using the built-in sprite-button element and the number property it allows you to show signals just like Factorio shows them in a circuit network. | 908 | 3,156 | {"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} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.744239 |
https://bathmash.github.io/HELM/6_5_modelling_exercises-web/6_5_modelling_exercises-webse3.html | 1,695,897,822,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00362.warc.gz | 136,526,914 | 3,485 | 3 Exponential decrease
Consider the value, $£D$ , of a car subject to depreciation, in terms of the age $A$ years of the car. The car was bought for $£10500$ . The function
$\phantom{\rule{2em}{0ex}}D=10500{e}^{-0.25A}\phantom{\rule{2em}{0ex}}\left(0\le A\le 6\right)$
could be considered appropriate on the ground that
1. $D$ had a fixed value of $£10500$ when
$A$ = 0,
2. $D$ decreases as $A$ increases and
3. $D$ decreases faster when $A$ is small than when $A$ is large. A plot of this function is shown in Figure 8.
Figure 8 :
Produce the linearised model of $D=10500{e}^{-0.25A}$ .
$lnD=ln10500+ln\left({e}^{-0.25A}\right)$
so $lnD=ln10500-0.25A$ | 242 | 662 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.847911 |
http://flaguide.org/extra/download/tools/math/convincing/convincing2.txt | 1,723,783,111,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00187.warc.gz | 9,660,655 | 1,577 | Always, Sometimes or Never True - Set #2 Malcolm Swan Mathematics Education University of Nottingham Malcolm.Swan@nottingham.ac.uk Jim Ridgway School of Education University of Durham Jim.Ridgway@durham.ac.uk Introduction: You will be given a number of statements. You must decide if each statement is · always true, or · sometimes true, or · never true You must provide full and convincing reasons for your decision. If you think that a statement is sometimes true, you must fully explain when it is true and when it is not true. Here is an example of what we mean: Example: Weaker response: This statement is sometimes true. It is true when both numbers are 0 and when both numbers are 2. It is not true when one number is 2 and one number is 3. Stronger response: This statement is sometimes true. Suppose one number is x and one number is y. The statement says that: x+y = xy This simplifies to the condition that y = x/(x-1) A few pairs of numbers when it works are therefore: (0, 0); (2, 2); (3, 3/2); (4, 4/3); (5, 5/4) ...... There are also other pairs which work! ____________________________________________________ The aim of this assessment is to provide the opportunity for you to: · test statements to see how far they are true; · provide examples or counterexamples to support your conclusions · provide convincing arguments or proofs to support your conclusions For each statement, say whether it is always, sometimes or never true. You must provide several examples or counterexamples to support your decision. Try also to provide convincing reasons for your decision. You may even be able to provide a proof in some cases. 1. The center of a circle that circumscribes a triangle is inside the triangle. Is this always, sometimes or never true? ...................................................... Reasons or examples: 2. An altitude subdivides a triangle into two similar triangles. Is this always, sometimes or never true? ...................................................... Reasons or examples: 3. (a + b)2 = a2 + b2 Is this always, sometimes or never true? ...................................................... Reasons or examples: 4. 3x2 = (3x)2 Is this always, sometimes or never true? ...................................................... Reasons or examples: 5. A shape with a finite area has a finite perimeter. Is this always, sometimes or never true? ...................................................... Reasons or examples: 6. A shape with a finite perimeter has a finite area. Is this always, sometimes or never true? ...................................................... Reasons or examples: | 553 | 2,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.871268 |
http://nrich.maths.org/public/leg.php?code=-334&cl=4&cldcmpid=5741 | 1,484,629,065,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279468.17/warc/CC-MAIN-20170116095119-00107-ip-10-171-10-70.ec2.internal.warc.gz | 211,081,307 | 9,542 | # Search by Topic
#### Resources tagged with Games similar to Developing a Framework for Mathematical Enrichment:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to Developing a Framework for Mathematical Enrichment
Games. Mathematical Thinking. Rich Tasks. Enrichment. Problem solving. Pedagogy. Mathematical modelling. Questioning. Learning mathematics. Number - generally.
### There are 62 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Games
### Using Games in the Classroom
##### Stage: 2, 3 and 4
Gillian Hatch analyses what goes on when mathematical games are used as a pedagogic device.
### Behind the Rules of Go
##### Stage: 4 and 5
This article explains the use of the idea of connectedness in networks, in two different ways, to bring into focus the basics of the game of Go, namely capture and territory.
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Nim
##### Stage: 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Fraction and Percentage Card Game
##### Stage: 3 and 4 Challenge Level:
Match the cards of the same value.
### Tower Rescue
##### Stage: 2, 3, 4 and 5 Challenge Level:
Help the bee to build a stack of blocks far enough to save his friend trapped in the tower.
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Nim-like Games
##### Stage: 2, 3 and 4 Challenge Level:
A collection of games on the NIM theme
### One, Three, Five, Seven
##### Stage: 3 and 4 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### The Unmultiply Game
##### Stage: 2, 3 and 4 Challenge Level:
Unmultiply is a game of quick estimation. You need to find two numbers that multiply together to something close to the given target - fast! 10 levels with a high scores table.
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Ratio Pairs 3
##### Stage: 3 and 4 Challenge Level:
Match pairs of cards so that they have equivalent ratios.
### Sufficient but Not Necessary: Two Eyes and Seki in Go
##### Stage: 4 and 5
The game of go has a simple mechanism. This discussion of the principle of two eyes in go has shown that the game does not depend on equally clear-cut concepts.
### Stop or Dare
##### Stage: 2, 3 and 4 Challenge Level:
All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### FEMTO: Follow Up
##### Stage: 4 Challenge Level:
Follow-up to the February Game Rules of FEMTO.
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Crossword 1
##### Stage: 5 Challenge Level:
A mathematically themed crossword.
### Turnablock
##### Stage: 4 Challenge Level:
A simple game for 2 players invented by John Conway. It is played on a 3x3 square board with 9 counters that are black on one side and white on the other.
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Charlie's Delightful Machine
##### Stage: 3 and 4 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Twin Corresponding Sudoku III
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Advent Calendar 2010
##### Stage: 1, 2, 3 and 4 Challenge Level:
Advent Calendar 2010 - a mathematical game for every day during the run-up to Christmas.
### 9 Hole Light Golf
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We think this 3x3 version of the game is often harder than the 5x5 version. Do you agree? If so, why do you think that might be?
### 18 Hole Light Golf
##### Stage: 1, 2, 3 and 4 Challenge Level:
The computer starts with all the lights off, but then clicks 3, 4 or 5 times at random, leaving some lights on. Can you switch them off again?
### Quadruple Clue Sudoku
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Twin Corresponding Sudokus II
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Ratio Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
### Square it for Two
##### Stage: 1, 2, 3 and 4 Challenge Level:
Square It game for an adult and child. Can you come up with a way of always winning this game?
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Domino Magic Rectangle
##### Stage: 2, 3 and 4 Challenge Level:
An ordinary set of dominoes can be laid out as a 7 by 4 magic rectangle in which all the spots in all the columns add to 24, while those in the rows add to 42. Try it! Now try the magic square...
### Low Go
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players. Take turns to place a counter so that it occupies one of the lowest possible positions in the grid. The first player to complete a line of 4 wins.
### Square It
##### Stage: 1, 2, 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Diagonal Sums Sudoku
##### Stage: 2, 3 and 4 Challenge Level:
Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells.
### Mathematical Anagrams
##### Stage: 5 Challenge Level:
1. LATE GRIN (2 solutions)
### Slippery Snail
##### Stage: 2, 3 and 4 Challenge Level:
A game for two people, who take turns to move the counters. The player to remove the last counter from the board wins.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### Game of PIG - Sixes
##### Stage: 2, 3, 4 and 5 Challenge Level:
Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one?
### Corresponding Sudokus
##### Stage: 3, 4 and 5
This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.
### Sprouts
##### Stage: 2, 3, 4 and 5 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Everthing you have always wanted to do with dominoes! Some of these games are good for practising your mental calculation skills, and some are good for your reasoning skills.
### Football World Cup Simulation
##### Stage: 2, 3 and 4 Challenge Level:
A maths-based Football World Cup simulation for teachers and students to use.
### Intersection Sums Sudoku
##### Stage: 2, 3 and 4 Challenge Level:
A Sudoku with clues given as sums of entries.
### Factors and Multiples Game
##### Stage: 2, 3 and 4 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### Pole Star Sudoku
##### Stage: 4 and 5 Challenge Level:
A Sudoku based on clues that give the differences between adjacent cells.
### Intersection Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Going First
##### Stage: 4 and 5
This article shows how abstract thinking and a little number theory throw light on the scoring in the game Go.
### Ratio Sudoku 3
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios or fractions.
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning. | 2,235 | 9,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-04 | longest | en | 0.847648 |
https://www.dsprelated.com/freebooks/pasp/First_Order_Allpass_Interpolation.html | 1,717,081,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00291.warc.gz | 627,047,646 | 11,031 | ### First-Order Allpass Interpolation
A delay line interpolated by a first-order allpass filter is drawn in Fig.4.3.
Intuitively, ramping the coefficients of the allpass gradually grows'' or hides'' one sample of delay. This tells us how to handle resets when crossing sample boundaries.
The difference equation is
Thus, like linear interpolation, first-order allpass interpolation requires only one multiply and two adds per sample of output.
The transfer function is
(5.2)
At low frequencies (), the delay becomes
(5.3)
Figure 4.4 shows the phase delay of the first-order digital allpass filter for a variety of desired delays at dc. Since the amplitude response of any allpass is 1 at all frequencies, there is no need to plot it.
The first-order allpass interpolator is generally controlled by setting its dc delay to the desired delay. Thus, for a given desired delay , the allpass coefficient is (from Eq.(4.3))
From Eq.(4.2), we see that the allpass filter's pole is at , and its zero is at . A pole-zero diagram for is given in Fig.4.5. Thus, zero delay is provided by means of a pole-zero cancellation! Due to inevitable round-off errors, pole-zero cancellations are to be avoided in practice. For this reason and others (discussed below), allpass interpolation is best used to provide a delay range lying wholly above zero, e.g.,
Note that, unlike linear interpolation, allpass interpolation is not suitable for random access'' interpolation in which interpolated values may be requested at any arbitrary time in isolation. This is because the allpass is recursive so that it must run for enough samples to reach steady state. However, when the impulse response is reasonably short, as it is for delays near one sample, it can in fact be used in random access mode'' by giving it enough samples with which to work.
The STK class implementing allpass-interpolated delay is DelayA.
#### Minimizing First-Order Allpass Transient Response
In addition to approaching a pole-zero cancellation at , another undesirable artifact appears as : The transient response also becomes long when the pole at gets close to the unit circle.
A plot of the impulse response for is shown in Fig.4.6. We see a lot of ringing'' near half the sampling rate. We actually should expect this from the nonlinear-phase distortion which is clearly evident near half the sampling rate in Fig.4.4. We can interpret this phenomenon as the signal components near half the sampling rate being delayed by different amounts than other frequencies, therefore sliding out of alignment'' with them.
For audio applications, we would like to keep the impulse-response duration short enough to sound instantaneous.'' That is, we do not wish to have audible ringing'' in the time domain near . For high quality sampling rates, such as larger than kHz, there is no issue of direct audibility, since the ringing is above the range of human hearing. However, it is often convenient, especially for research prototyping, to work at lower sampling rates where is audible. Also, many commercial products use such sampling rates to save costs.
Since the time constant of decay, in samples, of the impulse response of a pole of radius is approximately
and since a 60-dB decay occurs in about 7 time constants ('') [451, p. 38], we can limit the pole of the allpass filter to achieve any prescribed specification on maximum impulse-response duration.
For example, suppose 100 ms is chosen as the maximum allowed at a sampling rate of . Then applying the above constraints yields , corresponding to the allowed delay range .
Next Section:
Linear Interpolation as Resampling
Previous Section:
Linear Interpolation | 770 | 3,691 | {"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} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.920104 |
https://share.cocalc.com/share/5d54f9d642cd3ef1affd88397ab0db616c17e5e0/www/papers/thesis-old/symbols.tex?viewer=share | 1,601,520,284,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00452.warc.gz | 558,414,651 | 3,861 | CoCalc Public Fileswww / papers / thesis-old / symbols.tex
Author: William A. Stein
Compute Environment: Ubuntu 18.04 (Deprecated)
1\mbox{}
2\vspace{7ex}
3\section*{\Huge List of Symbols}
5
6\begin{tabular}{llr}
7{\bf \large Symbol} \hspace{4em} & {\bf \large Definition}
8& {\bf \large Page}\\
9&\vspace{-2ex}\\
10$\Adual$ & dual of $A$ & \pageref{pg:dual}\\
11$\sB_k(N,\eps)$ & boundary modular symbols & \pageref{def:boundarysymbols}\\
12$c_A$ & Manin index of~$A$ & \pageref{defn:maninconstant}\\
13$\delta_A$ & modular degree & \pageref{defn:modulardegree}\\
14$\e$ & winding element $\e=\{0,\infty\}$ &\pageref{defn:windingelement}\\
15$\sM_k(N,\eps;K)$& modular symbols & \pageref{defn:modsym}\\
16$\esM_k(N,\eps;K)$& extended modular symbols & \pageref{defn:extendedmodsyms}\\
17$M[I]$ & $\intersect_{a\in I} \ker(a)$ & \\
18$P(X,Y)\{\alp,\beta\}$ & higher weight modular symbol & \pageref{pg:higherweightmodsym}\\
19$[P(X,Y),(u,v)]$ & higher weight Manin symbol & \pageref{defn:maninsymbols}\\
20%$R[\eps]$ & $R(\{\eps(a) : a \in \Z/N\Z\})$ & \pageref{defn:keps}\\
21$\Sha(A/K)^{\circ}$ & visible part of~$\Sha$ & \pageref{defn:visiblepart}\\
22$\sS_k(N,\eps;K)$& cuspidal modular symbols & \pageref{defn:cuspidalmodularsymbols}\\
23$T_n$ & $n$th Hecke operator & \pageref{subsec:heckeonmanin}\\
24$V_k$ & homogeneous polys. of degree $k$ in $\Z[X,Y]$ & \pageref{defn:vk}\\
25$W_n$ & $n$th Atkin-Lehner involution & \pageref{sec:atkin-lehner}\\
26$\cX_p(M)$ & character group of torus of $J_0(pM)$ at~$p$ & \pageref{defn:chargroup}\\
27$\alp_t$, $\beta_t$ & degeneracy maps & \pageref{pg:degeneracymaps}\\
28$\Theta_f$ & rational period mapping & \pageref{sec:ratperiod}\\
29$\sigma$, $\tau$ & $\sigma=\abcd{0}{-1}{1}{\hfill 0}$,
30 $\tau=\abcd{0}{-1}{1}{-1}$ & \pageref{defn:sigmatau}\\
31$\Phi_f$ & analytic period mapping & \pageref{defn:periodmapping}\\
32$\Phi_{A,p}$ & component group of~$A$ at~$p$ & \pageref{defn:componentgroup}\\
33$\Omega_A$ & real volume & \pageref{defn:omega}\\
34$\langle \,\, , \, \rangle$ & integration pairing & \pageref{thm:perfectpairing}\\
35$*$ & conjugation involution & \pageref{sec:starinvolution}\\
36\end{tabular}
37
38
39
40
41 | 849 | 2,227 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.221599 |
https://www.slideserve.com/wilson/module-b-4-processing-ict-survey-data-powerpoint-ppt-presentation | 1,725,942,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00581.warc.gz | 945,194,213 | 21,671 | 1 / 14
# Module B-4: Processing ICT survey data
TRAINING COURSE ON THE PRODUCTION OF STATISTICS ON THE INFORMATION ECONOMY. TRAINING COURSE ON THE PRODUCTION OF STATISTICS ON THE INFORMATION ECONOMY. Module B-4 Processing ICT Survey data. Module B-4: Processing ICT survey data. Unctad Manual Chapter 7. Objectives.
## Module B-4: Processing ICT survey data
E N D
### Presentation Transcript
1. TRAINING COURSE ON THE PRODUCTION OF STATISTICS ON THE INFORMATION ECONOMY TRAINING COURSE ON THE PRODUCTION OF STATISTICS ON THE INFORMATION ECONOMY Module B-4 Processing ICT Survey data Module B-4: Processing ICT survey data Unctad Manual Chapter 7
2. Objectives After completing this module you will know how to do: • Data processing • Data weighting (grossing-up) • Data editing • Data analysis Contents of this module 4. Data processing and analysis 4.1 Data editing 4.2 Data weighting 4.3 Estimating ICT indicators
3. Page 82 B4.1. Data editing Data editing Editing! What is editing? • Statistical information provided by businesses can contain errors such as • Wrong or missing data, • Incorrect classifications • Inconsistent or illogical responses. • Solutions to minimize such errors • Ex ante optimize the effectiveness of • data capture instruments • collection procedures. • Ex post application of robust data editing techniques
4. B4.1. Data editing Phases of data processing Editing! Raw data Quality controls during data collection and entry Treatment of internal errors and inconsistencies Micro-editing(input) Estimation of missing data Data editing Outlier analysis Macro-editing(output) Re-weighting procedures Editing of aggregates Clean data file
5. Page 82 B4.3. Estimating ICT indicators Internal inconsistencies and errors • Validity control of an individual data item requires: • To define a valid set of responses (in general, gender should be = 0 or 1, age should not be 110 years, etc; in ICT use of Internet by business should be 0 or 1) • To check questions against valid responses - Definition of rules based in relationships between questions (see Box 15 of the Manual: some logical tests) • Arithmetic checks during data entry or batch mode (totals, subtotals, frequencies)
6. Page 84 B4.3. Estimating ICT indicators Treatment of missing data • Final non-response (missing data) should be treated to avoid biased estimates. • Unit non-response treatment: Corrective weighting. • Sample-based methods (the original weights are modified with sample information) • Population-based method (the weights are modified with population information, the classical post stratification procedure)
7. B4.3. Estimating ICT indicators Page 151 Annexe 5 Treatment of missing data (cont.) • Final non-response (missing data) should be treated in order to avoid biased estimates. • Item non-response treatment: Imputation. • Deterministic imputation (a law). • Hot deck imputation (let’s do it now). • Cold deck imputation (using other information, models, econometrics…). • Mean or modal value imputation ( it is clear). • Historical imputation (long series).
8. B4.3. Estimating ICT indicators Page 86 Misclassified units • Two cases of misclassification • Non-eligibility unit erroneously included • This will reduce the effective sample size unless a reserve list is prepared • Eligible unit included in the wrong stratum or omitted from the frame altogether • The technical solution consists of recalculating sample weights (see Box 17)
9. B4.2. Data weighting Some simple weighting methods • The sample average in stratum h is defined as • The estimate for the total for stratum h can be obtained by multiplying the stratum average by the total number of businesses in the stratum (Nh)
10. B4.3. Estimating ICT indicators Some simple weighting methods (cont.) • The estimate for the total in the population is just or See boxes 18 and 19 pag 89
11. B4.3. Estimating ICT indicators Estimating proportions and ratios • A proportion: • Four different types of estimates are very usual • Simple random sampling of a non-stratified population • Stratified random sampling • With one or several strata exhaustively investigated • Ratio estimates with simple random sampling • Ratio estimates with stratified random sampling • ICT indicators are mainly proportions and ratios. A ratio :
12. B4.3. Estimating ICT indicators CASE 1: Simple random sampling of a non-stratified population • The indicator can be expressed as the sample proportion: • The standard error (SE) of the sample proportion is estimated by: • SE expression valid with a sampling fraction of 10% or less
13. B4.3. Estimating ICT indicators CASE 2: Stratified random sampling • An unbiased estimate ofp is: Where, L: the number of strata Nh : the population in stratum h (h=1, 2, ... L) nh : the sample size in stratum h (h=1, 2, ... L) • The estimate of the SE of: • See Annex 4 of the Manual for more details
14. B4.3. Estimating ICT indicators CASE 3: Ratio estimates with simple random sampling • The indicator to estimate is: • The natural estimate of ratio p is: • Finally, one approximation of the SE is: where is the sample average of nX observations, This is a reference outside the scope of our course
More Related | 1,170 | 5,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.666809 |
http://mathcenter.oxford.emory.edu/math117/rAndNormalDistributions/ | 1,506,255,512,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690016.68/warc/CC-MAIN-20170924115333-20170924135333-00031.warc.gz | 213,967,537 | 2,133 | # R Functions Related to Normal Distributions
Recall that a normal distribution with mean $\mu$ and standard deviation $\sigma$ is one characterized by the normal density function: $$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}}$$ The value of this function is given by the R function
dnorm
Usage
dnorm(x, mean=μ, sd=σ)
Example
The height at $x=13$ of the normal curve with mean 10 and standard deviation 7 is given by
> dnorm(13,10,7)
[1] 0.05199096
To find the probability that a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$ results in a value less than $x$ (i.e. the area under the normal curve to the left of $x$.) is given by the following R function:
pnorm
Usage
pnorm(x, mean=μ, sd=σ)
Examples
Suppose the manufacturer of a certain type of snack knows that the total weight of the snack packet they sell is normally distributed with a mean of $80.2$ grams and a standard deviation of $1.1$ grams. What is the probability that a selected packet is less than $78$ grams?
> pnorm(78,80.2,1.1)
[1] 0.02275013
Under the same assumptions, what is the probability that a selected packet has a weight within 2 standard deviations of the mean weight?
> pnorm(82.4,80.2,1.1) - pnorm(78,80.2,1.1)
[1] 0.9544997
Notice, in the last example, we find the area under the normal curve between $x=a$ and $x=b$ by finding a difference of two left-tailed areas.
R also provides the following function that gives the $x$ value for which a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$ will produce an outcome less than $x$ with some given probability of $p$. Said another way, this function gives the $x$ value where there is an area of $p$ to the left of $x$ and under the related normal curve.
qnorm
Usage
qnorm(p,mean=μ,sd=σ)
Example
A math contest author knows that the length of time taken by students to complete a certain section of math questions is normally distributed with a mean of 17 minutes and standard deviation of 4.5 minutes. $90\%$ of students should have completed the section of questions in how many minutes?
> qnorm(0.90,17,4.5)
[1] 22.76698
Under the same assumptions, how long should the middle $50\%$ of students take to complete this section?
> qnorm(0.75,17,4.5)
[1] 20.0352
> qnorm(0.25,17,4.5)
[1] 13.9648
# ...so roughly between 14 and 20 minutes
Lastly, R also has a function that lets you simulate random variables that follow a Normal distribution:
rnorm
Usage
rnorm(n, mean=μ, sd=σ)
Note: $n$ represents the number of realizations of your random variable you wish to produce.
Examples
Suppose the error in a particular model follows a normal distribution with mean $\mu = 5$ and standard deviation $\sigma = 1$. Simulate the errors seen in 15 applications of the model in question.
> rnorm(15, mean=5, sd=1)
[1] 5.625824 6.438588 6.642417 3.537867 4.853430
[6] 4.405501 5.684504 6.041753 4.434506 5.550997
[11] 5.095519 5.188711 5.077366 5.102386 4.247265 | 881 | 3,010 | {"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} | 4.5 | 4 | CC-MAIN-2017-39 | latest | en | 0.875671 |
https://www.physicsforums.com/threads/local-density-and-speed-of-an-aircraft.603976/ | 1,539,806,714,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00197.warc.gz | 1,021,215,202 | 12,130 | # Homework Help: Local Density And Speed Of An Aircraft
1. May 7, 2012
### imamul
1. The problem statement, all variables and given/known data
A Pitot is used to measure the air speed of a light aircraft. The pressure difference recorded by the tube in flight was 3300 N/m^2. The local air pressure was 950 kN/m^2 and the temperature was 7 Degrees (280K). Calculate the local air density and the speed of the aircraft.
2. Relevant equations
Not sure if relevent but aattempted to use this formula:
p+1/2pV^2+pgh= constant
3. The attempt at a solution
Attempting as of right now any help appreciated.
2. May 7, 2012
### vela
Staff Emeritus
You need to show details of your attempt before you'll get any help. | 193 | 715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.918887 |
https://www.easycalculation.com/physics/classical-physics/physical-pendulum.php | 1,576,400,897,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307797.77/warc/CC-MAIN-20191215070636-20191215094636-00334.warc.gz | 687,966,458 | 7,465 | English
# Physical Pendulum Calculator English Español
A pendulum is a mass that is attached to a pivot, from which it can swing freely. Pendulum consisting of an actual object allowed to rotate freely around a horizontal axis. This is called Physical Pendulum.
## Physical Pendulum Calculation
kg-m2
kg
m/s2
m
s
A pendulum is a mass that is attached to a pivot, from which it can swing freely. Pendulum consisting of an actual object allowed to rotate freely around a horizontal axis. This is called Physical Pendulum.
#### Distance from Center of Mass to Pivot:
where, T = Period, I = Center of Mass or Moment of Inertia, M = Mass, g = Acceleration of Gravity, D = Distance from Center of Mass to Pivot. | 165 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.876737 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.