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http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=TUES | 1,501,095,109,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426372.41/warc/CC-MAIN-20170726182141-20170726202141-00691.warc.gz | 444,783,398 | 108,379 | Back to list of Stocks See Also: Seasonal Analysis of TUESGenetic Algorithms Stock Portfolio Generator, and Fourier Calculator
Fourier Analysis of TUES (Tuesday Morning Corp)
TUES (Tuesday Morning Corp) appears to have interesting cyclic behaviour every 92 weeks (1.155*sine), 83 weeks (.9734*cosine), and 83 weeks (.924*sine).
TUES (Tuesday Morning Corp) has an average price of 12.03 (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.
Fourier Analysis
Using data from 1/3/2000 to 7/17/2017 for TUES (Tuesday Morning Corp), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
012.03357 0
11.71602 7.16833 (1*2π)/916916 weeks
2-6.85759 -2.99777 (2*2π)/916458 weeks
3-1.16174 -1.54822 (3*2π)/916305 weeks
4-.6319 1.97247 (4*2π)/916229 weeks
5.82659 1.34873 (5*2π)/916183 weeks
6.09662 -.49565 (6*2π)/916153 weeks
71.03333 -.49291 (7*2π)/916131 weeks
8.27015 -.36689 (8*2π)/916115 weeks
9.51813 -.2145 (9*2π)/916102 weeks
10-.04123 1.15496 (10*2π)/91692 weeks
11-.97345 .92405 (11*2π)/91683 weeks
12-.08145 .36306 (12*2π)/91676 weeks
13.17153 .67423 (13*2π)/91670 weeks
14.67954 -.04789 (14*2π)/91665 weeks
15.61989 .39313 (15*2π)/91661 weeks
16-.20119 .35804 (16*2π)/91657 weeks
17-.16401 .4318 (17*2π)/91654 weeks
18-.1961 .20862 (18*2π)/91651 weeks
19-.05867 .39413 (19*2π)/91648 weeks
20-.28318 .06435 (20*2π)/91646 weeks
21-.39061 .08965 (21*2π)/91644 weeks
22.16819 -.23374 (22*2π)/91642 weeks
23.07489 .28935 (23*2π)/91640 weeks
24.04865 -.05535 (24*2π)/91638 weeks
25.27393 -.03767 (25*2π)/91637 weeks
26.139 .68074 (26*2π)/91635 weeks
27-.32802 -.01696 (27*2π)/91634 weeks
28.20527 -.13097 (28*2π)/91633 weeks
29.27378 .01569 (29*2π)/91632 weeks
30.00718 .1823 (30*2π)/91631 weeks
31.22346 .42423 (31*2π)/91630 weeks
32-.01068 .35244 (32*2π)/91629 weeks
33-.11405 .29149 (33*2π)/91628 weeks
34-.05569 .08633 (34*2π)/91627 weeks
35-.10777 .06348 (35*2π)/91626 weeks
36.11688 -.07875 (36*2π)/91625 weeks
37.04421 -.0464 (37*2π)/91625 weeks
38.25723 .25799 (38*2π)/91624 weeks
39.11373 .58579 (39*2π)/91623 weeks
40-.21778 .31643 (40*2π)/91623 weeks
41-.20063 -.11253 (41*2π)/91622 weeks
42-.07707 -.02389 (42*2π)/91622 weeks
43-.08923 .04275 (43*2π)/91621 weeks
44-.0731 -.12542 (44*2π)/91621 weeks
45.06947 .08554 (45*2π)/91620 weeks
46-.08914 .02337 (46*2π)/91620 weeks
47.08869 .09966 (47*2π)/91619 weeks
48-.1012 .15249 (48*2π)/91619 weeks
49-.30101 .13233 (49*2π)/91619 weeks
50-.01358 -.13202 (50*2π)/91618 weeks
51.25338 .13998 (51*2π)/91618 weeks
52-.05249 -.03101 (52*2π)/91618 weeks
53.13313 -.1775 (53*2π)/91617 weeks
54.29314 .10297 (54*2π)/91617 weeks
55.07614 .27426 (55*2π)/91617 weeks
56.02301 -.04975 (56*2π)/91616 weeks
57.04814 .05072 (57*2π)/91616 weeks
58.06496 .00877 (58*2π)/91616 weeks
59.0434 .14905 (59*2π)/91616 weeks
60-.06202 .10036 (60*2π)/91615 weeks
61-.11548 .0353 (61*2π)/91615 weeks
62.04205 -.01684 (62*2π)/91615 weeks
63.0363 .04015 (63*2π)/91615 weeks
64.03958 -.08579 (64*2π)/91614 weeks
65.05911 .01551 (65*2π)/91614 weeks
66.10424 .10047 (66*2π)/91614 weeks
67.16097 .16146 (67*2π)/91614 weeks
68.15524 .02893 (68*2π)/91613 weeks
69.05699 .10052 (69*2π)/91613 weeks
70-.2119 .02623 (70*2π)/91613 weeks
71-.09554 -.04144 (71*2π)/91613 weeks
72.05781 .15948 (72*2π)/91613 weeks
73.01848 -.00431 (73*2π)/91613 weeks
74.09521 -.11949 (74*2π)/91612 weeks
75.12978 .00074 (75*2π)/91612 weeks
76-.07571 .15001 (76*2π)/91612 weeks
77-.0098 .04796 (77*2π)/91612 weeks
78.10532 .0182 (78*2π)/91612 weeks
79.07294 .03976 (79*2π)/91612 weeks
80-.04112 -.02113 (80*2π)/91611 weeks
81-.06734 .07438 (81*2π)/91611 weeks
82.12721 .11937 (82*2π)/91611 weeks
83.11993 .20431 (83*2π)/91611 weeks
84-.00894 .05931 (84*2π)/91611 weeks
85-.04272 .00718 (85*2π)/91611 weeks
86.05116 .00134 (86*2π)/91611 weeks
87.12986 .15296 (87*2π)/91611 weeks
88-.05219 .09479 (88*2π)/91610 weeks
89-.08442 .05924 (89*2π)/91610 weeks
90-.01876 .07282 (90*2π)/91610 weeks
91.05188 .054 (91*2π)/91610 weeks
92.01728 .09165 (92*2π)/91610 weeks
93-.01464 .04777 (93*2π)/91610 weeks
94-.07824 .04263 (94*2π)/91610 weeks
95-.08034 -.03849 (95*2π)/91610 weeks
96.00905 .04778 (96*2π)/91610 weeks
97-.0304 .00091 (97*2π)/9169 weeks
98.10259 .02957 (98*2π)/9169 weeks
99-.03273 .03805 (99*2π)/9169 weeks
100-.07383 -.01692 (100*2π)/9169 weeks
101.09114 .00368 (101*2π)/9169 weeks
102.03866 .1432 (102*2π)/9169 weeks
103-.11701 .02034 (103*2π)/9169 weeks
104.03693 -.12029 (104*2π)/9169 weeks
105.01451 .12289 (105*2π)/9169 weeks
106-.06737 .02385 (106*2π)/9169 weeks
107.05811 .04573 (107*2π)/9169 weeks
108.016 .09081 (108*2π)/9168 weeks
109.01401 .01483 (109*2π)/9168 weeks
110-.00299 .00306 (110*2π)/9168 weeks
111.03299 .00707 (111*2π)/9168 weeks
112.00539 .0537 (112*2π)/9168 weeks
113-.03165 .08615 (113*2π)/9168 weeks
114-.08561 -.00696 (114*2π)/9168 weeks
115.02303 -.04156 (115*2π)/9168 weeks
116.1315 .01324 (116*2π)/9168 weeks
117.04848 .0528 (117*2π)/9168 weeks
118-.06424 .0142 (118*2π)/9168 weeks
119-.00059 -.07455 (119*2π)/9168 weeks
120.02053 .04884 (120*2π)/9168 weeks
121-.0027 .11956 (121*2π)/9168 weeks
122-.01065 .12757 (122*2π)/9168 weeks
123.0415 .06234 (123*2π)/9167 weeks
124-.01684 -.01092 (124*2π)/9167 weeks
125.01741 -.01726 (125*2π)/9167 weeks
126-.00671 -.08566 (126*2π)/9167 weeks
127.01488 .00088 (127*2π)/9167 weeks
128.01815 .05115 (128*2π)/9167 weeks
129.00156 .04667 (129*2π)/9167 weeks
130.04584 .02564 (130*2π)/9167 weeks
131.0763 -.02153 (131*2π)/9167 weeks
132.00656 .02992 (132*2π)/9167 weeks
133.00039 .03483 (133*2π)/9167 weeks
134-.07716 .01512 (134*2π)/9167 weeks
135.0462 -.00016 (135*2π)/9167 weeks
136-.00187 .02534 (136*2π)/9167 weeks
137-.02836 -.0016 (137*2π)/9167 weeks
138.01172 -.06297 (138*2π)/9167 weeks
139.08422 -.02777 (139*2π)/9167 weeks
140.08254 .1091 (140*2π)/9167 weeks
141-.03454 .05523 (141*2π)/9166 weeks
142-.02371 -.0148 (142*2π)/9166 weeks
143.01428 -.06022 (143*2π)/9166 weeks
144-.01517 -.0212 (144*2π)/9166 weeks
145.04211 -.01187 (145*2π)/9166 weeks
146-.02717 -.00066 (146*2π)/9166 weeks
147.10174 .02093 (147*2π)/9166 weeks
148-.00879 .04401 (148*2π)/9166 weeks
149-.00878 .04825 (149*2π)/9166 weeks
150.06363 -.0105 (150*2π)/9166 weeks
151.07088 .05538 (151*2π)/9166 weeks
152.04247 .00603 (152*2π)/9166 weeks
153.01749 -.00305 (153*2π)/9166 weeks
154.05177 .04194 (154*2π)/9166 weeks
155.03046 .06653 (155*2π)/9166 weeks
156-.0497 .08079 (156*2π)/9166 weeks
157.02046 -.02744 (157*2π)/9166 weeks
158.02127 .00132 (158*2π)/9166 weeks
159.0024 -.0041 (159*2π)/9166 weeks
160.00438 -.00813 (160*2π)/9166 weeks
161.02472 .00859 (161*2π)/9166 weeks
162-.01018 .00416 (162*2π)/9166 weeks
163.0088 -.00736 (163*2π)/9166 weeks
164.01004 .08421 (164*2π)/9166 weeks
165-.06321 .03316 (165*2π)/9166 weeks
166-.00152 -.0242 (166*2π)/9166 weeks
167.06098 -.08693 (167*2π)/9165 weeks
168.09081 -.00589 (168*2π)/9165 weeks
169.07086 -.02232 (169*2π)/9165 weeks
170.04791 .01802 (170*2π)/9165 weeks
171.04742 .0502 (171*2π)/9165 weeks
172-.04376 .06176 (172*2π)/9165 weeks
173-.02341 .04852 (173*2π)/9165 weeks
174.0168 -.07437 (174*2π)/9165 weeks
175.04617 .02077 (175*2π)/9165 weeks
176-.01933 .03254 (176*2π)/9165 weeks
177.03077 -.02037 (177*2π)/9165 weeks
178.07236 .05966 (178*2π)/9165 weeks
179.04791 -.00582 (179*2π)/9165 weeks
180.02465 .04296 (180*2π)/9165 weeks
181.03613 .02445 (181*2π)/9165 weeks
182.00925 .02682 (182*2π)/9165 weeks
183-.01109 .05829 (183*2π)/9165 weeks
184-.05272 .03589 (184*2π)/9165 weeks
185-.0247 -.00158 (185*2π)/9165 weeks
186.0179 -.03249 (186*2π)/9165 weeks
187.03145 -.00845 (187*2π)/9165 weeks
188.02579 -.0186 (188*2π)/9165 weeks
189.01939 -.04267 (189*2π)/9165 weeks
190.0393 .05236 (190*2π)/9165 weeks
191.00087 .05843 (191*2π)/9165 weeks
192.01979 .01468 (192*2π)/9165 weeks
193.01041 .0333 (193*2π)/9165 weeks
194-.03008 -.0087 (194*2π)/9165 weeks
195.03507 .02877 (195*2π)/9165 weeks
196.06935 .01204 (196*2π)/9165 weeks
197.06912 .00467 (197*2π)/9165 weeks
198.01969 .03931 (198*2π)/9165 weeks
199-.041 .01758 (199*2π)/9165 weeks
200.01978 .04656 (200*2π)/9165 weeks
201-.01914 -.00314 (201*2π)/9165 weeks
202.00937 -.09323 (202*2π)/9165 weeks
203.08794 -.04144 (203*2π)/9165 weeks
204.03242 .04014 (204*2π)/9164 weeks
205.01402 .04865 (205*2π)/9164 weeks
206.03076 -.03116 (206*2π)/9164 weeks
207.05957 .00194 (207*2π)/9164 weeks
208.00824 .0683 (208*2π)/9164 weeks
209-.0327 .0009 (209*2π)/9164 weeks
210.00992 -.03523 (210*2π)/9164 weeks
211.04636 -.02587 (211*2π)/9164 weeks
212.05799 .01998 (212*2π)/9164 weeks
213-.01282 .03788 (213*2π)/9164 weeks
214.01848 .00173 (214*2π)/9164 weeks
215.06094 .01931 (215*2π)/9164 weeks
216-.01564 .03538 (216*2π)/9164 weeks
217.01319 -.00838 (217*2π)/9164 weeks
218-.01382 .03285 (218*2π)/9164 weeks
219-.00595 .02832 (219*2π)/9164 weeks
220-.02798 .0077 (220*2π)/9164 weeks
221-.01949 -.0085 (221*2π)/9164 weeks
222.06394 .04851 (222*2π)/9164 weeks
223.00558 .01948 (223*2π)/9164 weeks
224.02756 -.0432 (224*2π)/9164 weeks
225.04199 -.04023 (225*2π)/9164 weeks
226.00813 -.00111 (226*2π)/9164 weeks
227.00555 -.00678 (227*2π)/9164 weeks
228.07418 -.01191 (228*2π)/9164 weeks
229.02738 .07103 (229*2π)/9164 weeks
230-.01127 .00427 (230*2π)/9164 weeks
231.00968 -.03606 (231*2π)/9164 weeks
232.04963 .00915 (232*2π)/9164 weeks
233.03072 .00519 (233*2π)/9164 weeks
234-.00499 -.01074 (234*2π)/9164 weeks
235.04374 -.04669 (235*2π)/9164 weeks
236-.00159 .03564 (236*2π)/9164 weeks
237.00651 -.02792 (237*2π)/9164 weeks
238.04401 -.02029 (238*2π)/9164 weeks
239.02925 .02137 (239*2π)/9164 weeks
240.04814 -.02952 (240*2π)/9164 weeks
241.07857 .03841 (241*2π)/9164 weeks
242.06334 .0241 (242*2π)/9164 weeks
243.05563 .02822 (243*2π)/9164 weeks
244.01454 .05217 (244*2π)/9164 weeks
245-.06191 -.01137 (245*2π)/9164 weeks
246.02364 -.01452 (246*2π)/9164 weeks
247.02494 .00976 (247*2π)/9164 weeks
248.00608 .01678 (248*2π)/9164 weeks
249.06243 .02879 (249*2π)/9164 weeks
250.02078 .01271 (250*2π)/9164 weeks
251.05898 .0057 (251*2π)/9164 weeks
252.06746 -.01416 (252*2π)/9164 weeks
253.02648 .04966 (253*2π)/9164 weeks
254.01539 .04422 (254*2π)/9164 weeks
255.00108 -.01238 (255*2π)/9164 weeks
256.03469 -.04751 (256*2π)/9164 weeks
257.07813 -.00051 (257*2π)/9164 weeks
258.00526 .06514 (258*2π)/9164 weeks
259.01376 .00245 (259*2π)/9164 weeks
260.02599 -.00485 (260*2π)/9164 weeks
261.04928 .01584 (261*2π)/9164 weeks
262.01365 .03225 (262*2π)/9163 weeks
263.03622 .06675 (263*2π)/9163 weeks
264-.03123 .00192 (264*2π)/9163 weeks
265.02566 -.03022 (265*2π)/9163 weeks
266.06316 .03378 (266*2π)/9163 weeks
267.0087 .01568 (267*2π)/9163 weeks
268.06242 -.0116 (268*2π)/9163 weeks
269.03597 .01432 (269*2π)/9163 weeks
270.01067 -.0204 (270*2π)/9163 weeks
271.04912 .02415 (271*2π)/9163 weeks
272.00904 .0696 (272*2π)/9163 weeks
273.02435 .02451 (273*2π)/9163 weeks
274.01711 -.02038 (274*2π)/9163 weeks
275.07369 .00646 (275*2π)/9163 weeks
276.04871 .05068 (276*2π)/9163 weeks
277.02327 .05865 (277*2π)/9163 weeks
278.04364 -.00264 (278*2π)/9163 weeks
279.02222 .02628 (279*2π)/9163 weeks
280.02637 .00919 (280*2π)/9163 weeks
281.01511 .03461 (281*2π)/9163 weeks
282-.00007 .02475 (282*2π)/9163 weeks
283-.03296 -.00722 (283*2π)/9163 weeks
284.02489 .01594 (284*2π)/9163 weeks
285.04805 .02475 (285*2π)/9163 weeks
286.04672 .01732 (286*2π)/9163 weeks
287.02978 -.01538 (287*2π)/9163 weeks
288.01916 -.0279 (288*2π)/9163 weeks
289.01483 .01108 (289*2π)/9163 weeks
290.03787 .01948 (290*2π)/9163 weeks
291-.01434 .05486 (291*2π)/9163 weeks
292.00069 .02738 (292*2π)/9163 weeks
293.01563 .02975 (293*2π)/9163 weeks
294.05296 .03494 (294*2π)/9163 weeks
295.04391 -.02157 (295*2π)/9163 weeks
296.04158 -.02145 (296*2π)/9163 weeks
297.02629 .03579 (297*2π)/9163 weeks
298-.01941 .00654 (298*2π)/9163 weeks
299-.009 .01997 (299*2π)/9163 weeks
300.03095 .03044 (300*2π)/9163 weeks
301-.00769 .05416 (301*2π)/9163 weeks
302.01422 .01507 (302*2π)/9163 weeks
303.02678 -.00257 (303*2π)/9163 weeks
304.05086 .02521 (304*2π)/9163 weeks
305-.02403 -.00536 (305*2π)/9163 weeks
306-.0081 -.0194 (306*2π)/9163 weeks
307-.01791 -.00531 (307*2π)/9163 weeks
308.01294 -.00841 (308*2π)/9163 weeks
309.02019 .00954 (309*2π)/9163 weeks
310.03634 .00169 (310*2π)/9163 weeks
311.02913 .0403 (311*2π)/9163 weeks
312.00925 .01678 (312*2π)/9163 weeks
313.01575 .01715 (313*2π)/9163 weeks
314.02116 .00268 (314*2π)/9163 weeks
315.00234 .00512 (315*2π)/9163 weeks
316.01272 .01098 (316*2π)/9163 weeks
317.01776 .01663 (317*2π)/9163 weeks
318.00743 .02214 (318*2π)/9163 weeks
319.02887 .00503 (319*2π)/9163 weeks
320.01155 .02264 (320*2π)/9163 weeks
321.02145 .02112 (321*2π)/9163 weeks
322.03678 .0199 (322*2π)/9163 weeks
323.01979 .02785 (323*2π)/9163 weeks
324.0377 .01906 (324*2π)/9163 weeks
325-.00039 .03667 (325*2π)/9163 weeks
326-.00085 .03121 (326*2π)/9163 weeks
327-.00434 -.03021 (327*2π)/9163 weeks
328.02819 .01499 (328*2π)/9163 weeks
329-.00114 .01959 (329*2π)/9163 weeks
330.00699 .02162 (330*2π)/9163 weeks
331.01062 .01536 (331*2π)/9163 weeks
332-.00877 .042 (332*2π)/9163 weeks
333-.03262 .0295 (333*2π)/9163 weeks
334-.01485 -.05252 (334*2π)/9163 weeks
335.06617 -.01031 (335*2π)/9163 weeks
336.02252 -.01016 (336*2π)/9163 weeks
337.00375 -.01529 (337*2π)/9163 weeks
338.00089 .00583 (338*2π)/9163 weeks
339.01549 -.00781 (339*2π)/9163 weeks
340.02754 .01406 (340*2π)/9163 weeks
341.0155 -.0226 (341*2π)/9163 weeks
342.04893 -.0109 (342*2π)/9163 weeks
343.08216 .04463 (343*2π)/9163 weeks
344.0019 .03006 (344*2π)/9163 weeks
345.02156 -.01398 (345*2π)/9163 weeks
346.00148 .0156 (346*2π)/9163 weeks
347.01237 .04034 (347*2π)/9163 weeks
348.02955 .02044 (348*2π)/9163 weeks
349.03062 -.02135 (349*2π)/9163 weeks
350.02802 -.02447 (350*2π)/9163 weeks
351-.00149 -.00978 (351*2π)/9163 weeks
352.0024 .02039 (352*2π)/9163 weeks
353-.00062 .02622 (353*2π)/9163 weeks
354.02359 .00916 (354*2π)/9163 weeks
355.04236 .00066 (355*2π)/9163 weeks
356.03011 .04171 (356*2π)/9163 weeks
357.0034 .01844 (357*2π)/9163 weeks
358.02146 .01263 (358*2π)/9163 weeks
359.03276 .01693 (359*2π)/9163 weeks
360-.01515 -.01285 (360*2π)/9163 weeks
361.01081 .01941 (361*2π)/9163 weeks
362.02168 .04312 (362*2π)/9163 weeks
363.01876 .02044 (363*2π)/9163 weeks
364.0156 .01796 (364*2π)/9163 weeks
365.00817 .0137 (365*2π)/9163 weeks
366.02404 .00957 (366*2π)/9163 weeks
367.04438 .03722 (367*2π)/9162 weeks
368-.0108 .04687 (368*2π)/9162 weeks
369-.01402 .00633 (369*2π)/9162 weeks
370.03213 .03364 (370*2π)/9162 weeks
371-.00104 .00423 (371*2π)/9162 weeks
372.00077 -.01192 (372*2π)/9162 weeks
373-.02947 -.02104 (373*2π)/9162 weeks
374.01349 -.00237 (374*2π)/9162 weeks
375-.01834 .07936 (375*2π)/9162 weeks
376-.01331 .00832 (376*2π)/9162 weeks
377.01657 -.00136 (377*2π)/9162 weeks
378.01238 -.01617 (378*2π)/9162 weeks
379.00432 -.04132 (379*2π)/9162 weeks
380.03005 -.02768 (380*2π)/9162 weeks
381-.00585 .00299 (381*2π)/9162 weeks
382-.01478 .00934 (382*2π)/9162 weeks
383.00047 .0177 (383*2π)/9162 weeks
384.02442 .01419 (384*2π)/9162 weeks
385.04487 .0008 (385*2π)/9162 weeks
386.02104 .01232 (386*2π)/9162 weeks
387.00303 -.00062 (387*2π)/9162 weeks
388.01733 -.00415 (388*2π)/9162 weeks
389.01354 -.00462 (389*2π)/9162 weeks
390.00129 .00656 (390*2π)/9162 weeks
391-.00704 .0028 (391*2π)/9162 weeks
392.05095 -.00486 (392*2π)/9162 weeks
393-.01284 .02126 (393*2π)/9162 weeks
394.01277 -.01443 (394*2π)/9162 weeks
395.01425 -.00889 (395*2π)/9162 weeks
396.03821 .00342 (396*2π)/9162 weeks
397.03243 .00222 (397*2π)/9162 weeks
398.00775 -.0173 (398*2π)/9162 weeks
399.00461 .00047 (399*2π)/9162 weeks
400-.00174 .0244 (400*2π)/9162 weeks
401.01193 -.00106 (401*2π)/9162 weeks
402.03482 -.01451 (402*2π)/9162 weeks
403.00553 -.01916 (403*2π)/9162 weeks
404.00791 -.03654 (404*2π)/9162 weeks
405.04121 .00639 (405*2π)/9162 weeks
406.03816 .02862 (406*2π)/9162 weeks
407.03667 .00524 (407*2π)/9162 weeks
408.00635 .04168 (408*2π)/9162 weeks
409-.02899 .03429 (409*2π)/9162 weeks
410-.01791 .03241 (410*2π)/9162 weeks
411-.01069 .00938 (411*2π)/9162 weeks
412.01903 .00809 (412*2π)/9162 weeks
413.04586 -.01994 (413*2π)/9162 weeks
414.03299 -.00249 (414*2π)/9162 weeks
415-.01387 -.00376 (415*2π)/9162 weeks
416-.00734 .00173 (416*2π)/9162 weeks
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868-.1012 -.15249 (868*2π)/9161 weeks
869.08869 -.09966 (869*2π)/9161 weeks
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874-.07707 .02389 (874*2π)/9161 weeks
875-.20063 .11253 (875*2π)/9161 weeks
876-.21778 -.31643 (876*2π)/9161 weeks
877.11373 -.58579 (877*2π)/9161 weeks
878.25723 -.25799 (878*2π)/9161 weeks
879.04421 .0464 (879*2π)/9161 weeks
880.11688 .07875 (880*2π)/9161 weeks
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886.00718 -.1823 (886*2π)/9161 weeks
887.27378 -.01569 (887*2π)/9161 weeks
888.20527 .13097 (888*2π)/9161 weeks
889-.32802 .01696 (889*2π)/9161 weeks
890.139 -.68074 (890*2π)/9161 weeks
891.27393 .03767 (891*2π)/9161 weeks
892.04865 .05535 (892*2π)/9161 weeks
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894.16819 .23374 (894*2π)/9161 weeks
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9091.03333 .49291 (909*2π)/9161 weeks
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911.82659 -1.34873 (911*2π)/9161 weeks
912-.6319 -1.97247 (912*2π)/9161 weeks
913-1.16174 1.54822 (913*2π)/9161 weeks
914-6.85759 2.99777 (914*2π)/9161 weeks | 15,808 | 35,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.767504 |
http://www.chegg.com/homework-help/engineering-mechanics-an-introduction-to-dynamics-4th-edition-chapter-a-problem-1p-solution-9780742134935 | 1,455,408,178,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701168065.93/warc/CC-MAIN-20160205193928-00173-ip-10-236-182-209.ec2.internal.warc.gz | 339,329,898 | 15,051 | View more editions
Engineering Mechanics An Introduction to Dynamics
# TEXTBOOK SOLUTIONS FOR Engineering Mechanics An Introduction to Dynamics 4th Edition
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Chapter: Problem:
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
Write the expression for the universal gravitational force as follows:
…… (1)
Here, the force due to gravitation is F, universal gravitational constant is G, the masses of the two bodies are and the distance between the two masses is r.
(a)
Obtain the desired units of the quantities from table A.1.
The unit of force in SI system is Newtons (N)
The unit of mass in SI system is kilograms (kg)
The unit of length in SI system is meter (m)
Substitute all the corresponding units in equation (1).
Therefore, the unit of universal gravitational constant in SI system is.
• Chapter , Problem is solved.
Corresponding Textbook
Engineering Mechanics An Introduction to Dynamics | 4th Edition
9780742134935ISBN-13: 0742134938ISBN: Authors: | 274 | 1,172 | {"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} | 2.6875 | 3 | CC-MAIN-2016-07 | latest | en | 0.80502 |
https://wiki.fysik.dtu.dk/dacapo/Manual?action=diff&rev1=33&rev2=35 | 1,529,770,577,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865098.25/warc/CC-MAIN-20180623152108-20180623172108-00054.warc.gz | 755,368,101 | 13,970 | Differences between revisions 33 and 35 (spanning 2 versions)
⇤ ← Revision 33 as of 2006-07-25 09:34:46 → Size: 25693 Editor: planck Comment: ← Revision 35 as of 2006-08-10 08:35:21 → ⇥ Size: 25849 Editor: hans Comment: Deletions are marked like this. Additions are marked like this. Line 587: Line 587: Below is a small example, which defines a atom0_s-atom1_s orbital:: Below is a small example, which defines a s-orbital on atom 0:: Line 592: Line 592: from ASE import Atom, ListOfAtomsatoms = ListOfAtoms([Atom('O', (2, 2, 2) ,magmom=1.0), Atom('O', (3.1, 2, 2),magmom=1.0)], cell=(4, 4, 4))calc = Dacapo(planewavecutoff=350, # in eV nbands=8, # 5 extra empty bands out='o2.nc', spinpol=True )atoms.SetCalculator(calc)mcenters = []mcenters.append([(0,1,1,1.0),(1,0,1,0.0)]) # [{(atomnr,l,m,weight)}]calc.SetMultiCenters(mcenters,energywidth=0.01)energy = atoms.GetPotentialEnergy()centers = calc.GetMultiCenters()center1 = centers[0]print 'description:'center1.GetDescription()print 'energy resolved dos:'for d in center1.GetData(): print d[0],d[1] from ASE import Atom, ListOfAtoms atoms = ListOfAtoms([Atom('O', (2, 2, 2) ,magmom=1.0), Atom('O', (3.1, 2, 2),magmom=1.0)], cell=(4, 4, 4)) calc = Dacapo(planewavecutoff=350, # in eV nbands=8, # 5 extra empty bands out='o2.nc', spinpol=True ) atoms.SetCalculator(calc) mcenters = [] mcenters.append([(0,0,0,1.0),(1,0,0,0.0)]) # [{(atomnr,l,m,weight)}] calc.SetMultiCenters(mcenters,energywidth=0.1) energy = atoms.GetPotentialEnergy() centers = calc.GetMultiCenters() center1 = centers[0] print 'description:' center1.GetDescription() print 'energy resolved dos:' for d in center1.GetData(): print d[0],d[1] Line 735: Line 731: ``GetMagneticMoment``: Return the total magnetic momnet in units of Bohr magnetons.
### 1 Introduction
Dacapo is a total energy program based on density functional theory. It uses a plane wave basis for the valence electronic states and describes the core-electron interactions with Vanderbilt ultrasoft pseudo-potentials. The program performs self-consistent calculations for both Local Density Approximation (LDA) and various Generalized Gradient Approximation (GGA) exchange-correlations potentials, using state-of-art iterative algorithms. The code may perform molecular dynamics / structural relaxation simultaneous with solving the Schrodinger equations within density functional theory. The program may be compiled for seriel as well as parallel execution and the code has been ported to many hardware platforms.
The manual describes how to use the dacapo program from the Campos Atomistic Simulation enviroment, or for short ASE.
### 2 Setting up the atomic configuration
The atoms for Dacapo are defined via the ASE ListOfAtoms object.
A ListOfAtoms object is a collection of atoms. The ASE implementation of a ListOfAtoms can be used like this:
```from ASE import Atom, ListOfAtoms
atoms = ListOfAtoms([Atom('C', (0, 0, 0)),
... Atom('O', (0, 0, 1.0))])
atoms.SetUnitCell((10,10,10))```
This will define a CO molecule, with a CO distance of 1A, in a cubic 10Ax10Ax10A unitcell.
For more details of the ListOfAtoms class see listofatoms in the ASE manual.
The atoms defined here can be attached to a Dacapo calculator by using:
`atoms.SetCalculator(calculator)`
alternatively, one can attach the atoms to a calculator by using:
`calculator.SetListOfAtoms(atoms)`
The next chapter describes how to define the calculator for dacapo.
### 3 Dacapo ASE calculator interface
As a minimum for defining a calculation in python you need to specify the atoms as described above, and you need to set the planewave cutoff and the number of bands, this will look like this using the python interface:
```from Dacapo import Dacapo
calculator = Dacapo()
calculator.SetNumberOfBands(6)
calculator.SetPlaneWaveCutoff(340)
atoms.SetCalculator(calculator)
calculator.GetPotentialEnergy()```
The rest of this chapter describes the different methods you can use for the calculator object. More detailed explanation of the python construction found above can be found in the ASE Tutorials.
For the most important of the method described one can use keyword arguments in the construction of the Dacapo calculator, the table below list the keywords:
Method keyword
SetPlaneWaveCutoff planewavecutoff
SetDensityCutoff densitycutoff
SetNumberOfBands nbands
SetXCFunctional xc
SetBZKPoints kpts
SetSpinPolarized spin
SetNetCDFFile out
SetTxtFile txtout
DipoleCorrection dipole
SetScratch scratchdir
The example above can now be written more compact:
```calculator = Dacapo(
... nbands=6,
... planewavecutoff=340)```
#### 3.1 Number of bands and Electronic temperature
The method 'SetNumberOfBands(nbands)' can be used to define the number of bands for the calculatior. nbands should be a least equal to halv the number of valence electrons.
The occupation statitstics can be set using the method SetOccupationStatistics(method).
dacapo support two methods for occupation statistics:
FermiDirac:
Fermi-Dirac occupation statistics. The temperature is controlled by the method SetElectronicTemperature(temp). Default is 0.1 eV.
MethfesselPaxton:
Currently, the Methfessel-Paxton scheme (PRB 40,3616, 1989) is implemented to 1st order.
#### 3.2 Planewave and Density-cutoff
The method SetPlaneWaveCutoff(planewavecutoff) defines the plavewave cutoff for the calculation. The unit for planewavecutoff is eV.
The method SetDensityCutoff(densitycutoff) sets the density cutoff. By default the two cutoffs are equal, corresponding to using just one grid. For elements like Cu using ultra-soft pseudo potentials the ultra-soft pseudo-wavefunction are very soft, while the density still requires a fine grid, in this case it can be advantagous to set a larger value for densitycutoff that for planewavecutoff.
#### 3.3 Changing file names
The following two methods can be use to change the file names used by dacapo:
SetNetCDFFile(filename):
This will set the NetCDF filename, default is 'out.nc'.
SetTxtFile(filename)
This will set the Ascii text file name, default is 'out.txt'. For information on the most inportant data in the Ascii file see dacapo Ascii file.
The method:
SetScratch(scratchdir):
will set the scratch directory used for swapfiles to scratchdir. Default for this directory is /scratch/<username> or if this is not found just /scratch.
#### 3.4 Exchange-correlation functional
You can use the method SetXCFunctional(exc) to select the exchange-correlation used for the calculation. exc is one of:
LDA:
Vosko Wilk Nusair LDA-parametrization.
PW91:
Perdew Wang 91 GGA-parametrization.
PBE:
Perdew Burke Ernzerhof GGA-parametrization.
RPBE:
Revised PBE GGA-parametrization, Hammer, Hansen and Nørskov.
The method:
`GetXCFunctional`
tells which of the above defined exchange-correlation functionals are currently used.
The method:
`GetXCEnergies(xcname)`
return the non-selfconsistent energy for the given xcname (PBE,RPBE,PW91,VWN).
#### 3.5 Defining a spin-polarized calculation
One can use the method SetSpinPolarized(True) to tell dacapo that the calculation should be spin-polarized.
One also need to specify an initial magnetic moment for the atoms. You can do this in the construction of the atoms, by using:
```>>> from ASE import Atom, ListOfAtoms
>>> atoms = ListOfAtoms([Atom('O', (0, 0, 0),magmom=1.0),
... Atom('O', (0, 0, 1.2),magmom=1.0)])
```
This can also be set using the ListOfAtoms method SetMagneticMoments((1,1)) or setting the SetMagneticMoment method on each atom.
This will define a initial total magnetic moment of 2 Bohr magneton, for the O2 molecule, this being also the self-consistent result. You could also use, SetMagneticMoments((1,-1)), this will define a 'anti-ferro' magnetic state with total magnetic moment of zero. If this case probably the end result is still the same (2 Bohr magneton) but the number of self-consistent iterations will increase.
#### 3.6 K-points
The method SetBZKPoints(kpts) can be used to define the k-points in the BZ. In general kpts are a python list of k-points defined in units of the reciprocal cell. You can however specify a 10x10x10 Monkhorst-Pack 3dimensional set simple by using SetBZKpoint((10,10,10)).
Below is a list of the special k-points sets that can be used:
MonkhorstPack:
A 3-dimensional Monkhorst-Pack special k-point set. (see H.J.Monkhorst and J.D.Pack, Physical Review B, vol. 13, page 5188, 1976. You can defined this k-points set simply using, SetBZKPoints((n1,n2,n3)). This will define a n1 by n2 by n3, Monkhorst Pack special k-point set. The size along each reciprocal vector given by n1, n2 and n3, respectively.
A Chadi-Cohen (see Chadi and Cohen, Phys. Review B., vol 8,page 5747) 1-dimensional k-point can be defined using the ASE Utilities module ChadiCohen:
```>>> from ASE.Utilities.ChadiCohen import CC6_1x1
>>> calc.SetBZKPoints(CC6_1x1)
```
This will define a 6 k-point 1x1 Chadi-Cohen special k-point set. In general the Chadi-Cohen sets are named CC+'<Nkpoints>'+_+'shape'. For example an 18 k point sq(3)xsq(3) is named 'CC18_sq3xsq3'.
You can use the method GetBZKPoints, to access the k-points just defined. The method GetIBZKPoints returns the symmetry reduced set of k-points.
#### 3.7 Pseudo-potentials
For an overview of the pseudopotentials for dacapo, see the dacapo pseudopotential page.
You can change or view the pseudopotentials use for an element using the methods:
GetPseudoPotential(Z=<elementnumber>):
View the pseudopotential path for element elementnumber.
SetPseudoPotential(Z=<elementnumber>,path=path):
Set the pseudopotential path for element elementnumber.
#### 3.8 Charge-density mixing
The method SetChargeMixing(value) can be used to enable or disable charge density mixing.
Using SetChargeMixing(True) the program will use Pulay mixing for the density (default).
Using SetChargeMixing(False) correspond to running a calculation using the Harris-Foulkes functional using the input density. See also the Harris calculation example.
The method SetKerkerPreconditioning(value) can be used to set Kerker preconditioning for the density mixing. For value=True Kerker preconditiong is used, i.e. q is different from zero, see eq. 82 in Kresse/Furthmller: Comp. Mat. Sci. 6 (1996). The value of q is fix to give a damping of 20 of the lowest q vector. For value=False q is zero and mixing is linear (default).
#### 3.9 Eigenvalue solver
The eigenvalue solver can be set using the method SetEigenvalueSolver(method). Here method is one of:
1. eigsolve: Block Davidson algorithm (Claus Bendtsen et al). This is the default method.
2. rmm-diis: Residual minimization method (RMM), using DIIS (direct inversion in the iterate subspace). The implementation follows closely the algorithm outlined in Kresse and Furthmuller, Comp. Mat. Sci, III.G/III.H
#### 3.10 Dipole-correction
The method:
`DipoleCorrection(MixingParameter=0.2,InitialValue=0.0,AdditiveDipoleField=0)`
will add the interslab dipole electrostatic decoupling. Presently, this feature is only active along the third unit cell vector, so you need to choose unit cell accordingly. Corrections for higher electrostatic multipoles are not implemented presently.
The default position of the field discontinuity at the vacuum position farthest from any other atoms on both sides of the slab.
The keyword AdditiveDipoleField, will add a constant electrostatic field along third unit cell vector, corresponding to an external dipole layer. This field is added to the dipole correction field. The field discontinuity follows the position of the dynamical dipole correction.
A fixed external field i.e. not a field added on top the the dipole correction, can be set using the NetCDF entry ExternalDipolePotential. This can be done like:
`calc.SetNetCDFEntry("ExternalDipolePotential",value=0.5)`
corresponding to an external field of 0.5eV/A.
#### 3.11 Setting non-default executable
GetJobType:
SetJobType can be used to view the current executable for dacapo.
SetJobType(executable=newexecutable):
SetJobType can be used to change the default dacapo executable dacapo.run to newexecutable.
#### 3.12 Symmetry
SetSymmetryOff:
Do not use symmtry then reducing the k-point set.
SetSymmetryOn:
Use symmtry then reducing the k-point set.
#### 3.13 Electrostatic decoupling
SetElectrostaticDecoupling(numberofgaussians=3,ecutoff=100):
Add electrostatic decoupling (Jan Rossmeisl). Decoupling activates the three dimensional electrostatic decoupling. Based on paper by Peter E. Bloechl: JCP 103 page7422 (1995).
#### 3.14 Stress calculation
By default dacapo will not calculate the stress acting on the unitcell, you can use the method:
`CalculateStress()`
to tell dacapo to calculate the stress.
#### 3.15 Local density of states
Dacapo can calculate the density of states projected onto atomic orbitals.
Warning
The density of states are not implemented for planewave parallel calculations, so the number of nodes must match the number kpoints (after they have been reduce by symmetry). So if you for example have 4 IBZ kpoints, you can use 4 or 2 parallel nodes and still get the projected density of states.
Typically one would do:
```atoms = ListOfAtoms(..)
calc = Dacapo(..)
calc.CalculateAtomicDOS()
atoms.SetCalculator(calc)
energy = atoms.GetPotentialEnergy()
dos = calc.GetLDOS(atoms=[1],angularchannels=["s"])
dos.GetPlot()```
This will return a density of states projected onto the s-orbital of the first atom (atom numbers starts from 1).
The method:
`CalculateAtomicDOS(energywindow=None)`
tells the fortran program to calculate the local density of states projected onto atomic orbitals, using energywindow for calculating the energy resolved LDOS in a specific range (must be within the eigenvalue spectrum).
The method:
```GetLDOS(**keywords )``:`
returns an instance of AtomProjectedDOSTool (see below).
Keyword for the GetDOS method:
• atoms
List of atoms numbers (default is all atoms)
• angularchannels
List of angular channel names. The full list of names are: 's', 'p_x', 'p_y', 'p_z','d_zz','dxx-yy', 'd_xy', 'd_xz', 'd_yz'. 'p' and 'd' can be used as shorthand for all p and all d channels respectively. (default is all d channels)
• spin
List of spins (default all spins)
'short' or 'infinite'. For cutoffradius = 'short' the integrals are truncated at 1 Angstrom. For cutoffradius = 'infinite' the integrals are not truncated. (default 'infinite')
The resulting DOS are added over the members of the three list (atoms,angularchannels and spin).
GetDOS returns a instance of the class AtomProjectedDOSTool (Thanks to John Kitchen for providing grace plot and band moments methods).
Methods for AtomProjectedDOSTool:
GetPlot
returns a GnuPlotAvartar for the energy resolved DOS A parent can be given too combine plot.
GetIntegratedDOS
returns the integral up to the fermi energy
GetData
returns the energy resolved DOS
GetBandMoment(moments)
returns the moments of the projected DOS. GetBandMoment(0) return the integrated DOS. GetBandMoment(1,2) returns the first and second moment of the projected DOS (center and width)
SaveData(filename)
saves the data to the filename
GetGracePlot
makes a GracePlot (if GracePlot is installed)
Note
The eigen values are not relative to the Fermi level, you can get the fermi level for the calculation using calc.GetFermilevel()
##### 3.15.1 Multicenter orbitals
You add the multicenter orbitals using:
```mcenters = [(0,0,0,1.0),(1,0,0,-1.0)]
calc.SetMultiCenters(mcenters)```
this will define an atom1_s-atom2_s orbital.
The general form is [{(atomnr,l,m,weight)}]
You can retrieve the centers using:
`centers = calc.GetMultiCenters()`
This will return a list of instances of the class MultiCenterProjectedDOSTool.
This class has the same method as described above, ie. GetBandMoment, `GetCutoff, GetData, GetDescription, GetEFermi, GetGracePlot, GetIntegratedDOS, GetPlot, GetSpin`and SaveData.
You can give the following keywords to SetMultiCenters:
```calc.SetMultiCenters(mcenters,
energywindow=(-15,5),
energywidth=0.2,
numberenergypoints=250,
The direction cosines for which the spherical harmonics are set up are using the next different atom in the list (cyclic) as direction pointer, so the z-direction is chosen along the direction to this next atom. At the moment the rotation matrices is only given in the text file, you can use grep 'MUL: Rmatrix' out_o2.txt to get this information.
Below is a small example, which defines a s-orbital on atom 0:
```#!/usr/bin/env python
from Dacapo import Dacapo
from ASE import Atom, ListOfAtoms
atoms = ListOfAtoms([Atom('O', (2, 2, 2) ,magmom=1.0),
Atom('O', (3.1, 2, 2),magmom=1.0)],
cell=(4, 4, 4))
calc = Dacapo(planewavecutoff=350, # in eV
nbands=8, # 5 extra empty bands
out='o2.nc',
spinpol=True )
atoms.SetCalculator(calc)
mcenters = []
mcenters.append([(0,0,0,1.0),(1,0,0,0.0)]) # [{(atomnr,l,m,weight)}]
calc.SetMultiCenters(mcenters,energywidth=0.1)
energy = atoms.GetPotentialEnergy()
centers = calc.GetMultiCenters()
center1 = centers[0]
print 'description:'
center1.GetDescription()
print 'energy resolved dos:'
for d in center1.GetData():
print d[0],d[1]```
#### 3.16 Changing convergence parameters
The method:
`SetConvergenceParameters(**keywords)`
can be used to set convergence parameters for the calculations. The following keywords are defined:
• energy:
Absolute convergence in energy (default 1e-5)
• density:
Convergence in density (default 0.0001)
• occupation:
Convergence in occupation numbers (default 0.001)
SetConvergenceParameters(energy=0.00001,density=0.0001,occupation=0.001) correspond to the default settings.
The method::
GetConvergenceParameters()
can be used to retrieve the convergence parameters.
#### 3.17 dacapo fortran program/python interface communication
The method:
`StayAliveOn()`
can be used for normal optimization runs.
This will prevent the dacapo fortran program to terminate and write the full netcdf output file after each call of GetPotentialEnergy. Instead it will wait for python to send new positions. This way of running the program saves IO time and reduce the NFS traffic.
Do not use this mode for NEB calculations, you will end of having 10 fortran programs in memory.
#### 3.18 NetCDF specific methods
Two methods are defined for general reading and writing of NetCDF entries. This provides a method for reading and writing any NetCDF variable defined in the dacapo netcdf manual, so that any settings for the dacapo fortran program, not defined by the standard sets of method from the DFT Calculator or from the extra Dacapo method described above, can be defined. The methods are listed below:
SetNetCDFEntry(variable,att=None,value=None):
Define a general netcdf entry for this calculation. If the netcdf entry is allready defined, the value are modified.
GetNetCDFEntry(variable,att=None):
Returns the value of NetCDF entry <variable> or NetCDF attribute att, if specified.
The effect of writing:
```>>> calc = Dacapo()
>>> calc.SetNumberOfBands(10)
```
and
```>>> calc = Dacapo()
>>> calc.SetNetCDFEntry('ElectronicBands',att='NumberOfBands',value=10)
```
is the same. Now you can get the information by either using
```>>> print calc.GetNumberOfBands()
```
or
```>>> print calc.GetNetCDFEntry('ElectronicBands',att='NumberOfBands')
```
### 4 Getting information from the calculator
Below follow a list of the methods one can use to read information from the calculator:
GetNumberOBands:
Return the number of electronic bands.
GetXCFunctional:
Return the XC-functional identifier ('LDA','PBE' ..)
GetBZKPoints
Return the K-points in the in the Brillouin zone. The K-points are in units of the reciprocal space basis.
GetSpinPolarized:
Is it a spin-polarized calculation.
GetMagneticMoment:
Return the total magnetic momnet in units of Bohr magnetons.
GetIBZKPoints:
Return k-points in the irreducible part of the irreducible Brillouin zone. The K-points are in units of the reciprocal space basis
GetIBZKPointWeights
Weights of the k-points. The sum of all weights is one.
GetEigenvalues(kpt=0, spin=0):
Return the eigenvalues for all bands for the given K-point kpt and the given spin spin. Eigenvalues are relative to the Fermi-level.
GetOccupationNumbers(kpt=0, spin=0):
Return the occupation numbers for all bands for the given K-point kpt and the given spin spin. These numbers fall in the range [0,1] for a spin-polarized calculation, and in the range [0,2] otherwise.
GetWavefunctionArray(band, kpt=0, spin=0):
Return array of wavefunction values. The wavefunction are returned for the band band, K-point kpt and spin spin.
The wavefunction is returned as a three-dimensional Numerical array, corresponding to the grid use in the calculation.
GetWavefunctionArrays(kpt=0, spin=0):
Return array of wavefunction values for all bands for the given K-point kpt and the given spin spin. The wavefunctions are returned as a three-dimensional Numerical array.
GetDensityArray(spin=0):
Return array of density values for the given spin spin.
### 5 Working with the Dacapo calculator
#### 5.1 reading from an old calculation
If you want to restart a calculation or read specific information from an old calculation, one have to access the information via the atoms, using:
```>>> atoms = Dacapo.ReadAtoms(filename='oldfile')
```
Now you can get the calculator, attached to the atoms, and get the information, using:
```>>> calc = atoms.GetCalculator()
>>> print calc.GetEigenValues()
```
Often ASE methods take the atoms as the argument, using the GetCalculator method to access the relevant information, i.e the VTK methods for 3D plotting:
```>>> from ASE.Visualization.VTK import VTKPlotWaveFunction
>>> VTKPlotWaveFunction(atoms)
```
### 6 Minimization and Dynamics using python
Minimization of structures is done using the mimimization algorithms implemented in ASE, for more details see the ASE manual.
### 7 Wavefunction and density plots
See the example plot wavefunction.
### 8 Wannier orbitals
Using the ASE Wannier module one can find the most localized set of Wannier orbitals for a dacapo calculation. More details will be added. See the Wannier example.
### 9 Nudged Elastic Band
See the Nudged Elastic Band method for a general description. See the Nudged Elastic Band example for a dacapo application of this method.
Dacapo: Manual (last edited 2012-08-10 08:39:29 by MarcinDulak) | 5,902 | 22,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-26 | latest | en | 0.569418 |
https://drawabox.com/community/submission/GLX648KO | 1,718,643,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00686.warc.gz | 195,667,528 | 10,462 | ## 250 box question about vanishing points
##### 5:30 PM, Tuesday January 12th 2021
It is ok to draw thinking on parallel lines insted of the actual vanishing points? I feel like I don't really understand vanishing points, but when I draw making parellel lines looks like they won't converge.
3 users agree
##### 10:37 PM, Tuesday January 12th 2021
When working on the 250 box challenge, it's very important that you always focus on how the sets of lines that are parallel in 3D space converge towards a single point in 2D space (on the page). You should not be side-stepping the problem of thinking about vanishing points by simply drawing lines that are parallel on the page, and therefore never converging, as doing so will ignore the main focus of this exercise.
We have you draw so many boxes in order to gradually make you more comfortable with the idea that what the marks you're drawing on the flat page represent something that exists in three dimensions. This means trying to capture those convergences, messing it up, then testing them out with the line extensions to see where you were off, then trying again. If you were to just try to keep all your lines parallel on the page, you wouldn't learn anything about how to estimate their convergences consistently.
##### 2:03 PM, Wednesday January 13th 2021
Thank you. Looks like I have to start all over again...
0 users agree
##### 8:19 PM, Tuesday January 12th 2021 edited at 9:41 PM, Jan 12th 2021
I don't know If I understand your question correctly: do you draw lines that are parallel on paper on purpose or do you end up with parallel lines because you find it hard to visualize the convergence in your mind? In any case, If you draw two or more lines parallel to each other it's expected for them to never meet.
edited at 9:41 PM, Jan 12th 2021
##### 2:11 PM, Wednesday January 13th 2021
I was drawing boxes with parallel lines on purpose. I didn't get the real intention of the exercise on the time I started the challenge, now I see I was doing it wrong, looks like I should start it all over again.
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### Faber Castell PITT Artist Pens
Like the Staedtlers, these also come in a set of multiple weights - the ones we use are F. One useful thing in these sets however (if you can't find the pens individually) is that some of the sets come with a brush pen (the B size). These can be helpful in filling out big black areas.
Still, I'd recommend buying these in person if you can, at a proper art supply store. They'll generally let you buy them individually, and also test them out beforehand to weed out any duds. | 676 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.973956 |
https://economics.stackexchange.com/questions/9646/mle-using-multivariate-normal-distribution | 1,713,448,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00073.warc.gz | 198,950,631 | 38,969 | # MLE using multivariate normal distribution
I am reading econometrics lecture notes of a Japanese professor. To explain MLE estimation he uses multivariate normal distribution.
As you can see, the first line is the loglikelihood function and the second line is proportional to the first line equation. I cannot get my head around this. How and why the first equation can be written as the second one using trace of a matrix. I heard about "trace trick" and this must be the application of this trick, but I need help to understand it and possibly use it. Thanks
The "trick" you are referring to is a property of the trace of a product of matrices, namely
$${\rm tr}(ABC) = {\rm tr}(BCA)$$
assuming the dimensions are conformable of course.
Now, note that the dimension of $(x_n-\mu)^T\Sigma^{-1} (x_n-\mu)$, for each observation, is $1 \times 1$. So trivially,
$${\rm tr}\Big[(x_n-\mu)^T\Sigma^{-1} (x_n-\mu)\Big] = (x_n-\mu)^T\Sigma^{-1} (x_n-\mu)$$
But since also, from the mentioned property of the trace,
$${\rm tr}\Big[(x_n-\mu)^T\Sigma^{-1} (x_n-\mu)\Big] = {\rm tr}\Big[\Sigma^{-1} (x_n-\mu)(x_n-\mu)^T\Big]$$
combining one gets the alternative expression for the mutlivariate normal log-likelihood. | 341 | 1,217 | {"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} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.87871 |
http://oeis.org/A130721 | 1,600,418,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187354.1/warc/CC-MAIN-20200918061627-20200918091627-00369.warc.gz | 151,448,229 | 4,453 | 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!)
A130721 Sum of the cubes of the number of standard Young tableaux over all partitions of n. 2
1, 1, 2, 10, 64, 596, 8056, 130432, 2534960, 59822884, 1718480368, 56754444440, 2110577206816, 87981286785328, 4129351961475872, 218382856010529472, 12813477368159567200, 822337333595479929044, 57213666993723455063392, 4305630141314873304140008 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS The sum of the zeroth power of the number f(p) of standard Young tableaux gives the partition function (A000041), the sum of the first power of f(p) gives the involution function (A000085), the sum of the squares of f(p) gives the factorial function (A000142), so this sequence is the natural one after them. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..60 FORMULA For p a partition of n, let f(p) be the number of standard Young tableaux with shape p. Then a(n) = sum(f(p)^3) where the sum ranges over all partitions p of n. EXAMPLE a(4) = 1^3 + 3^3 + 2^3 + 3^3 + 1^3 because the five partitions of 4 (namely 4, 3+1, 2+2, 2+1+1, 1+1+1+1) have respectively 1, 3, 2, 3, 1 standard Young tableaux. MATHEMATICA h[l_] := With[{n=Length[l]}, Sum[i, {i, l}]!/Product[Product[1 + l[[i]] - j + Sum[If[l[[k]] >= j, 1, 0], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]]; g[n_, i_, k_, l_] := g[n, i, l, k] = If[n == 0, h[l]^k, If[i < 1, 0, g[n, i - 1, k, l] + If[i > n, 0, g[n - i, i, k, Append[l, i]]]]]; a[n_] := If[n == 0, 1, g[n, n, 3, {}]]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 18 2017, after Alois P. Heinz *) CROSSREFS Cf. A000041, A000085, A000142. Column k=3 of A208447. Sequence in context: A223127 A323666 A318814 * A167449 A064170 A151410 Adjacent sequences: A130718 A130719 A130720 * A130722 A130723 A130724 KEYWORD nonn AUTHOR David A. Madore, Jul 03 2007 EXTENSIONS More terms from Alois P. Heinz, Feb 26 2012 STATUS approved
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Last modified September 18 04:33 EDT 2020. Contains 337165 sequences. (Running on oeis4.) | 854 | 2,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.559454 |
https://operanomadimilano.org/2024/06/24/ | 1,721,769,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00533.warc.gz | 366,587,140 | 17,620 | # Day: June 24, 2024
#### An Introduction to Poker
Poker is a game of cards where the aim is to make the best possible hand with your own cards and the community cards. Players also play against each other, placing bets which are then raised or folded depending on the strength of the player’s hand. The game can be fast-paced and involves a lot of psychology as well as mathematics.
The game is played on a table with a number of other players. Usually, the first two players to the left of the dealer place an initial amount of money into the pot. This is called a buy-in and is usually set at ten times the table’s high limit for most limit games and 20 times the value of the big blind (the highest bet) in no-limit games. Then the cards are dealt.
After the initial dealing of the cards, there is a round of betting. The players reveal their cards and the player with the best hand wins the pot. The cards can be either face up or face down. During the course of several betting rounds, the players’ hands develop by adding or discarding cards and replacing them. During this process, the remaining players may raise or fold.
A good poker player has a keen understanding of how to read their opponents. They are able to pick up on tells, or nervous habits that other players display, such as fiddling with their chips. This information allows them to make informed decisions about how to bet. They can also use this knowledge to decide when to bluff and when to call.
Poker can help improve decision-making skills, as it forces players to weigh risks and rewards without knowing the outcome of a decision in advance. This is a useful skill in many other areas of life, from business to personal relationships. It is also a great way to develop a better understanding of probability and statistics.
A good article on Poker should provide an overview of the rules and strategies of the game, including bluffing. It should also describe the different types of poker games and include anecdotes to engage the reader. Lastly, it should be descriptive and contain vivid images to draw the reader into the world of poker.
#### What is the RTP Slot?
RTP slot is a term that’s probably familiar to many players – especially those who frequent forums like Casinomeister. It’s a percentage seen in many online slots that explains how much money the game can be expected to pay back to players, over time. However, there is a lot more to it than that, and understanding what it means can help you make more informed decisions about which games to play.
The RTP of a slot machine is an important factor to consider when choosing which games to play, as it tells you how much the game will return on average. This is a figure that is calculated by running millions – or even billions – of spin simulations. This allows developers to get an idea of what the long-term average will be for a particular slot game.
This percentage doesn’t mean that you will win every spin, but it does indicate that the game is likely to lose more than it wins on average. This is why it is important to manage your bankroll and only gamble with money that you can afford to lose. In addition, knowing the RTP of a slot game can decrease the risk of gambling addiction and help you spend your money wisely.
While the RTP is an important factor to keep in mind, it’s not the only one. Another key factor is variance, which shows how much a slot can vary from the RTP. For example, if a slot has an RTP of 96%, the player will expect to get \$0.96 from every \$1 bet. But due to the unpredictability of the RNG, the payout can differ from this figure.
In the short term, a slot can produce big wins and long dry spells. This is why the RTP of a slot can vary significantly from one session to the next. However, it’s important to understand that the fact that a slot has a big win does not necessarily exclude future ones. This is because the winnings of each new round are independent of the previous ones.
When you’re looking for a slot game to play, look for one with a high RTP. This will increase your chances of winning and give you a better chance of making more money. Just remember to practice in demo mode before wagering real money, and stick to a budget so you don’t overspend.
RTP stands for Return to Player and describes how much a slot game pays back over a large number of spins, expressed as a percentage. The percentage can be used to determine how much you should win on average, and which slots are the best to play. But you’ll also want to take into account other factors, such as volatility and jackpot size. These factors will affect your chance of winning and overall experience. | 982 | 4,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.973634 |
http://mathhelpforum.com/algebra/16641-factoring-question-print.html | 1,513,067,981,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515311.25/warc/CC-MAIN-20171212075935-20171212095935-00254.warc.gz | 172,506,713 | 4,090 | # Factoring question
• Jul 8th 2007, 06:46 PM
starswept
Factoring question
In my endless assignment of factoring questions, I've come across another that I can't quite get:
3(x+2w)^3 - 3p^3r^3
Thanks for any help :)
• Jul 8th 2007, 06:59 PM
Jhevon
Quote:
Originally Posted by starswept
In my endless assignment of factoring questions, I've come across another that I can't quite get:
3(x+2w)^3 - 3p^3r^3
Thanks for any help :)
$3(x + 2w)^3 - 3p^3 r^3$ .........factor out the 3
$\Rightarrow 3 \left( (x + 2w)^3 - (pr)^3 \right)$
Now the inside of the brackets is the difference of two cubes, which you learned a formula for in class recently (i think), so you can take it from here
(if not say so)
• Jul 8th 2007, 07:53 PM
starswept
Quote:
Originally Posted by Jhevon
$3(x + 2w)^3 - 3p^3 r^3$ .........factor out the 3
$\Rightarrow 3 \left( (x + 2w)^3 - (pr)^3 \right)$
Now the inside of the brackets is the difference of two cubes, which you learned a formula for in class recently (i think), so you can take it from here
(if not say so)
Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).
I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).
This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance!
• Jul 8th 2007, 08:02 PM
topsquark
Quote:
Originally Posted by starswept
Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).
I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).
This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance!
There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.) I'd say you expanded wrong somewhere, but I can't figure out where.
$3(x+2w)^3 - 3p^3r^3$
$= 3[(x + 2w)^3 - p^3r^3]$
$= 3[(x + 2w) - (pr)][(x + 2w)^2 + (x + 2w)(pr) + (pr)^2]$
$= 3(x + 2w - pr)(x^2 + 4xw + 4w^2 + xpr + 2wpr + p^2r^2)$
-Dan
• Jul 8th 2007, 08:03 PM
Jhevon
Quote:
Originally Posted by starswept
Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).
I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).
This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance!
both of those are incorrect i'm afraid.
the answer should be $3(x + 2w - pr) \left( x^2 + 4xw + 4w^2 + prx + 2wpr + p^2 r^2 \right)$
• Jul 8th 2007, 08:04 PM
Jhevon
Quote:
Originally Posted by topsquark
There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.) I'd say you expanded wrong somewhere, but I can't figure out where.
$3(x+2w)^3 - 3p^3r^3$
$= 3[(x + 2w)^3 - p^3r^3]$
$= 3[(x + 2w) - (pr)][(x + 2w)^2 + (x + 2w)(pr) + (pr)^2]$
$= 3(x + 2w - pr)(x^2 + 4xw + 4w^2 + xpr + 2wpr + p^2r^2)$
-Dan
also, the book left out the 2 in front of the prw
• Jul 8th 2007, 08:21 PM
starswept
I see my mistake. I conveniently squared only the x of (x+2w), and forgot the 2w existed. Stupid errors like that are why taking accelerated calculus after no math for six months is dangerous :rolleyes: And as sad as it is, Dan, those errors in the book's equation were not typos, but from the book straight-on. This book has so many errors that having the answers can be more of a hindrance than a help. It makes getting the right answer a fun game of telephone tag for people in my math class ;)
Anyways, thanks again for the help, both of you!
• Jul 8th 2007, 08:22 PM
topsquark
Quote:
Originally Posted by Jhevon
also, the book left out the 2 in front of the prw
Quote:
Originally Posted by topsquark
There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.)
(Chuckles) As wpr = prw, I believe I mentioned that...
In need of some caffeine Jhevon? :D
-Dan
• Jul 8th 2007, 08:25 PM
Jhevon
Quote:
Originally Posted by topsquark
(Chuckles) As wpr = prw, I believe I mentioned that...
In need of some caffeine Jhevon? :D
-Dan
nah, i don't drink anything with caffeine, well, not that i know of, they put caffeine into a lot of things. i need some sleep though. :D
• Jul 8th 2007, 08:32 PM
topsquark
Quote:
Originally Posted by Jhevon
nah, i don't drink anything with caffeine, well, not that i know of, they put caffeine into a lot of things. i need some sleep though. :D
Hmmm.... Sounds almost Mormon. (I can get away with saying that because I am one. ;) )
-Dan | 1,827 | 5,312 | {"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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | longest | en | 0.917016 |
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1. when the electrons of an atom absorb energy a. they move further from the nucleus b. they increase their potential energy c. their quantum number...
# help with fission and fusion
1. durring fission a. work is done to pull nucleons out of th nucleus b. the nucleons lose mass c. mass is converted into energy (I say no to this...
# help with light
1. when light falls on a surface which irregular compared to its wavelength a. the reflection is diffuse b. glare is produced c. the wavelength is refracted d... | 957 | 3,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | latest | en | 0.960337 |
https://study.com/academy/topic/trigonometric-functions-homeschool-curriculum.html | 1,561,616,186,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000894.72/warc/CC-MAIN-20190627055431-20190627081431-00501.warc.gz | 603,332,694 | 25,853 | # Ch 20: Trigonometric Functions: Homeschool Curriculum
The Trigonometric Functions unit of this High School Trigonometry Homeschool Curriculum course is designed to help homeschooled students learn about trigonometric functions. Parents can use the short videos to introduce topics, break up lessons and keep students engaged.
## Who's it for?
This unit of our High School Trigonometry Homeschool Curriculum course will benefit any student who is trying to learn about trigonometric functions. There is no faster or easier way to learn about trigonometric functions. Among those who would benefit are:
• Students who require an efficient, self-paced course of study to learn about sine, cosine and tangent.
• Homeschool parents looking to spend less time preparing lessons and more time teaching.
• Homeschool parents who need a math curriculum that appeals to multiple learning types (visual or auditory).
• Gifted students and students with learning differences.
## How it works:
• Students watch a short, fun video lesson that covers a specific unit topic.
• Students and parents can refer to the video transcripts to reinforce learning.
• Short quizzes and a Trigonometric Functions unit exam confirm understanding or identify any topics that require review.
## Trigonometric Functions Unit Objectives:
• Understand the meaning of SohCahToa and other trigonometric functions.
• Memorize the parts of the first quadrant of the unit circle.
• Identify the categories and properties of special right triangles.
• Name and explain the six trigonometric functions.
• Understand and apply the Law of Sines and Law of Cosines.
• Use the double angle formula to determine trigonometric values.
11 Lessons in Chapter 20: Trigonometric Functions: Homeschool Curriculum
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
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Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
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### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 479 | 2,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-26 | longest | en | 0.884259 |
https://numberworld.info/88792106 | 1,600,613,423,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00051.warc.gz | 562,789,956 | 3,961 | # Number 88792106
### Properties of number 88792106
Cross Sum:
Factorization:
2 * 13 * 233 * 14657
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
2kln1a
sin(88792106)
-0.46759404547787
cos(88792106)
-0.88394332885861
tan(88792106)
0.52898645219898
ln(88792106)
18.301808307614
lg(88792106)
7.9483743568481
sqrt(88792106)
9422.9563301546
Square(88792106)
### Number Look Up
Look Up
88792106 (eighty-eight million seven hundred ninety-two thousand one hundred six) is a very unique figure. The cross sum of 88792106 is 41. If you factorisate the number 88792106 you will get these result 2 * 13 * 233 * 14657. The number 88792106 has 16 divisors ( 1, 2, 13, 26, 233, 466, 3029, 6058, 14657, 29314, 190541, 381082, 3415081, 6830162, 44396053, 88792106 ) whith a sum of 144058824. The number 88792106 is not a prime number. The figure 88792106 is not a fibonacci number. The number 88792106 is not a Bell Number. The figure 88792106 is not a Catalan Number. The convertion of 88792106 to base 2 (Binary) is 101010010101101110000101010. The convertion of 88792106 to base 3 (Ternary) is 20012002002212112. The convertion of 88792106 to base 4 (Quaternary) is 11102231300222. The convertion of 88792106 to base 5 (Quintal) is 140212321411. The convertion of 88792106 to base 8 (Octal) is 522556052. The convertion of 88792106 to base 16 (Hexadecimal) is 54adc2a. The convertion of 88792106 to base 32 is 2kln1a. The sine of the number 88792106 is -0.46759404547787. The cosine of the number 88792106 is -0.88394332885861. The tangent of 88792106 is 0.52898645219898. The square root of 88792106 is 9422.9563301546.
If you square 88792106 you will get the following result 7884038087915236. The natural logarithm of 88792106 is 18.301808307614 and the decimal logarithm is 7.9483743568481. I hope that you now know that 88792106 is great figure! | 722 | 2,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.762779 |
https://forum.swaylocks.com/t/induction-vented-ported-slotted-boards-whats-the-principle-at-work/47822 | 1,713,290,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00286.warc.gz | 245,189,631 | 11,169 | # induction? ( vented / ported / slotted ) boards : what's the principle at work?
while patent-pending designs of the newly-posted vented / slotted / ported boards are understandably kept under wraps, it would help to get some insight into how they work.
is this the principle behind the design ?
This is an interesting observation as it runs counter to my observations. About 25 years ago I made a flat-bottomed board to illustrate some of the characteristics of the hydrodynamics of a surfboard for a talk I gave to an oceanography class at UCSD. One of the features was a matrix of holes (straws) drilled through the board, terminating flush with the bottom and top of the board, and perpendicular to the bottom of the board. The objective was to provide a visual picture of the distribution of water pressure against the bottom of the board (this method is used to measure the static pressure for the altimeter in an aircraft). More specifically, the distribution of pressure pattern against the bottom of the board could be approximately measured by the height of the jet of water shooting upward from each hole (pictures by a photographer riding in a Zodiac running parallel to the path of the surfer documented these heights for subsequent analysis). The observed distribution was in good agreement with the classical Stavinsky distribution for a planing hull.
So we have two conflicting observations. Surffoil reports a suction drawing air downward from the deck while my observations indicate a positive pressure everywhere aft of the spray line (where the water first contacts the hull). The only significant differences between the two boards would seem to be the size of the holes in the connecting tubes and the presence of a “step” in the bottom of Surffoil’s board just forward of the air holes. I suspect the the latter is the primary cause of a positive pressure at the air holes in the “no step” configuration versus a negative pressure in the presence of the step. A secondary factor may be related to the hole diameter (or to a hole diameter/step height ratio).
In any case, the net overall pressure on the bottom of the board must be positive when planing (otherwise the board would not support the weight of the surfer). The more area with a negative pressure; the greater the wetted area with a positive pressure required in order to compensate. At speed, the greater the wetted area, the greater the skin friction drag. The bigger the “step” in the bottom, the greater the form drag. On the other hand, if air is entrained board and flows downstream from the step, the drag from friction between the water/air and the board may be reduced. So it is not clear if the net effect of adding one or more steps (and air supply tubes) terminating just aft of a step would increase or decrease the overall drag, and how those differences depend on hole and bottom geometry. Obviously it is a tough problem to come up with the optimum configuration by calculation, or by trial and error. The fact that a number of people have experimented with such designs, but they haven’t become popular, suggests that these designs offered little, if anything, in the way of significant benefits.
[quote="\$1"]
Is this the principle behind the design ?
[/quote]
No.
LOL
howdy bill,
no offense of course to those who’ve taken pains to develop forward-thinking boards like these, but-- in essence, what’s the principle at work ?
pablo’s ‘speedboat tub’ pic was merely featured above as an attempt to illustrate ‘induction’; online descriptions to date don’t suffice (yet) to help everyone understand how those (new?) board designs work. perhaps a simple explanation can be made without putting those pending patent apps at risk ?
cheers,
I seriously hope nobody is trying to patent induction surfboards. The principle is expandable beyond anybody’s design, so a patent won’t hold – it’ll just cost the guy seeking the patent money with no ultimate gain.
Skip Kozminsky holds a patent on a hollow wood surfboard design that is close to everyone else’s – cost him more money than he ever made on the design.
ey J,
in any case that’ll be something the patent owner can learn from next hehe
as patents go, they only cover what applies to the documented technical description that went into the original, approved patent application. any modification / mutation in others’ works whether minor or major brings 'em outside the scope of the patent in question. over time everybody’s design goes through incremental changes anyway so owning a raft of board design patents IMO aren’t that appealing from where i sit in the lineup " )
still looking forward to a basic understanding of why / how the design works…
cheers,
I can’t wait to see video.
(!!!)
Oh, I just saw the thread where Herb says he is in fact going after a patent – that’d be great if he actually could patent it and keep it licensed, I just know there’s a million ways to vent and circumvent things like that. And I’m glad of it.
It would seriously suck if the guy who invented this great thing that we all ended up using and loving spent his days embittered by money fights, partnerships that seemed like a great idea at the time, attorneys …
“as patents go, they only cover what applies to the documented technical
description that went into the original, approved patent application.
any modification / mutation in others’ works whether minor or major
brings 'em outside the scope of the patent in question.”
Not really true at all -
Matt (a patent attorney)
my design is unique…the photos show that.
of course i knew the second i posted info and pics here,there was going to be emails/pms/phone calls…,“i did the 10 years ago”…or," how can you claim that design,(i,we,they,him) invented it 40 years ago"…with lots of nasty threats and words too boot!
fyi…to those who are trying to intimidate me …fo.
herb
herb
care to explain why this is “not really true at all”, matt ?
countless patent infringement lawsuits originate from patent owners’ insistence that mutant designs infringe upon their patented work, while modifiers counterclaim that their development/innovation/modification brings them outside the scope.
obviously it can be argued either way but in the end, successfully arguing the case all the way (to the supreme court if it goes that far) that their modifications exempt them from the patent indeed wins in the end.
although in the end you know that it’s patent lawyers like you who actually make money every step of the way " )
cheers,
“care to explain why this is “not really true at all”, matt ?”
An illustrative (hypothetical) example:
If someone invents and patents the basic automobile and a competitor modifies existing, patented automobiles to make them specifically fuel injected when previous automobiles only had carburetion, then - if the patent was drafted properly - the person who created the fuel injected automobile will still infringe the patent despite the modification.
Everything depends on the scope of the patent. If Herb were to get a patent broadly covering induction-type surfboards, then it wouldn’t matter if someone changed port sizes or locations - their efforts would still infringe his patent.
In short, everything depends on the scope of the patent. Breakthrough inventions/innovations typically result in broader patents which are more difficult to design around. Other patents may have narrow scopes which are easy to design around. What is not true, however, is that any modification of what is specifically described in a patent will get you out of infringement. Many other issues aside, there is something called the Doctrine of Equivalents (in patent law) which is intended to prevent just that. If the opposite were true, patents would be almost valueless.
howdy herb,
sorry to hear you’re getting razzed for patenting your own design hehe telling 'em to go f.o. instead of “oh yeah? well, see ya in court!” is much more eloquent IMO but maybe you can give us a basic understanding of the principle behind the design ? it could also serve to bring such infidels closer to the light too hehe
cheers,
love your explanation matt except i wonder hmmm how is it that to this day, chinese businessmen can still get away with their copycat creations off patented designs ? hehe enforcement after all is what gives value (a.k.a ‘cold hard cash’) to patent ownership
cheers,
It is pretty simple to go to the USPTO site, where you can search see the track record of what patents were granted to a particular pat attny. If the patent produced any money for the inventor it is sometimes revealed by the notation that it had been “assigned” to some other entity. The other way is to enquire to the inventor. The huge majority of patents went nowhere, except to the attns. bank account. But they can’t be absolutly honest because they got to pay bills.
sickdog
Yes, enforcement can be very expensive (more costly than almost any other type of litigation), but - if the invention is valuable enough - you can often find law firms that will represent a case on a contingent fee basis.
I’ll be the first to admit, patents certainly won’t be valuable to everyone (most are never even commercialized), but they can be critical to success for others. Just as well, some defendants/infringers successfully design around patents, others get hit with injunctions and multi-million (sometimes multi-hundred million) dollar damage awards. Everything just depends on the facts and the merits…
Man. How sad that a creative breakthough gets mired in this. Herb - wish you the best with this, your board looks awesome. If Thrailkill is on board, I know its a winner!
“If the patent produced any money for the inventor it is sometimes
revealed by the notation that it had been “assigned” to some other
entity.”
This can be helpful, but not all assignments are listed on the face of a patent. If the assignment takes place after issuance, then it will never actually be printed on the patent. You can separately check the assignment records, however (which is similar to checking property records for real property deed information, like for land titles). Moreover, licenses may never be recorded at all (and therefore never public) and they are just as common as assignments.
glad to have a working patent attorney onboard explaining the risks & rewards of patent ownership hehe hopefully the designers can weigh in on the principle behind the vented / ported / slotted board design too.
otherwise we may be compelled to let pablo’s innovation (first revealed by disney animators in 1943!) have the final word here x D
cheers,
Happy to provide information here and there…it’s certainly more accurate than any surfboard design advice I might offer (though I surf plenty)! | 2,263 | 10,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.946698 |
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https://certifiedcalculator.com/weight-loss-pant-size-calculator/ | 1,721,712,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00243.warc.gz | 134,739,315 | 14,053 | # Weight Loss Pant Size Calculator
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2. Q: Can the calculator predict changes in other clothing sizes? A: This calculator focuses specifically on estimating pant size changes due to weight loss.
3. Q: Does the calculator consider body shape or muscle mass? A: No, it provides a general estimation and does not factor in body composition specifics.
4. Q: How often should I update the measurements in the calculator? A: It’s recommended to update measurements as weight changes significantly.
5. Q: What measurement units does the calculator use? A: The calculator uses pounds for weight and regular pant sizes.
6. Q: Is the estimation accurate for all brands and styles of pants? A: Different brands and styles might have variations, so this estimation provides a general idea.
7. Q: Can weight gain be calculated using the same method? A: This calculator focuses on weight loss; weight gain estimation might differ.
### Conclusion
The Weight Loss Pant Size Calculator offers a convenient method to estimate potential changes in pant size based on weight loss goals. While it provides a general idea, individual body characteristics and other factors can influence the actual clothing size changes. It serves as a helpful tool for those embarking on a weight loss journey, offering an estimate of their clothing size changes as they progress. | 507 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.902205 |
http://ombiz.us/why-everyone-is-talking-about-what-is-i-in-math-4/ | 1,627,782,395,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154127.53/warc/CC-MAIN-20210731234924-20210801024924-00138.warc.gz | 35,359,233 | 14,612 | # Why Everyone Is Talking About What Is I in Math
So 8 is going to be the very first digit of our answer. Students should be taught exact notation, when to utilize this, and, most significantly, why they’re using it the way that they are. Add a square within the rectangle.
## The Fundamentals of What Is I in Math Revealed
Within this exhibit, you will have a look at the language of numbers through common scenarios, including playing games or cooking. Whatever home based business you become involved with be certain that it’s propped up by the 3 pillars listed in this informative article. click here to find out more
Otherwise called the 20th century.
There’s, naturally, a reason this hasn’t happened. Have Math ChatsTake time each week to speak about math with your children in the identical way that you might chat about letters and stories. At the start, it appears that this book is simply about the particular number e, but actually it’s not.
## Here’s What I Know About What Is I in Math
The student who writes the greatest number receives the point. It’s pure since it does not result in any side-effects’, and doesn’t rely on global variables. http://ocs.yale.edu/
This fraction is set by the variety of years of useful life.
There are thus two manners in which to select such exceptional numbers a. This counting strategy is not difficult to accept with coins. Common Core utilizes a diagram technique to create this method kid friendly and simple to grade.
## Definitions of What Is I in Math
While literacy in mathematics could be different from the conventional literacy we find in an English classroom, it’s equally as critical. Present-day mathematics may be terribly flawed. Maths is a lot more than that, she states.
Algorithms aren’t convenient for real life, even when you remember them. Probability can’t tell us exactly what is going to happen later on, but nevertheless, it can tell us what we should probably expect. Each chapter is extremely short and about a specific feature of mathematics.
Students that have a great comprehension of place value will definitely excel at math. I am able to list everything it’s about math. For Boaler, there’s absolutely no such thing for a math gene or a math brain.
Training teachers in a write essay for me
new means of thinking will take some time, and American parents will want to be patient. Just choosing an answer which seems to fit often will not operate. If students aren’t very good at math and also if they couldn’t address the questions, and if you don’t let them waste their time or distract other students, they begin asking that notorious question.
If you believe about parenting, raising kids in a lousy community, there is going to be a greater proportion of drugs and a number of other things. After all, they frequently should convince different people (such as investors!) It is clear that they’re not asking sincerely.
You simply performed the procedure for abstraction. Provided that you are aware of what the code does that is. Virtually every machine-learning algorithm intends to minimize some type of estimation error subject to different constraintswhich is an optimization issue.
Sometimes these efforts prove to be duds. All method is just the exact same for a math question. Otherwise, you can struggle to be able to stick to the proofs.
It’s really hard to imagine that everyone could spend 220 pages speaking about one number. Ask a lot of questions, James stated. OK, maybe 5 minutes is a tiny stretch but by now you have the point.
Subsequently, encourage your children to attempt to stump you and let yourself be stumped. If you do believe in a greater spiritual power, there’s not anything wrong in setting aside a couple of minutes each evening prior to going to bed to meditating or praying some more. You should probably quit drooling now.
## What Is I in Math Options
When it’s in the introduction of hardware, programming of software, or distribution of the last item, math is there every step along the way. These methods are supposed to complement the standard algorithm, bridge understanding, and give students multiple perspectives on numbers. A superb instance of imperative programming is the usage of for-loops.
Programming is the capacity to follow set patterns and guidelines to compose programs. Any Ideas I truly appreciate your help! Handouts from previous years are offered in the Archive” section of the internet page.
## What Is I in Math Fundamentals Explained
In reality, if you’d like to bring any 2 numbers, you must have memorizedall the additions between the numbers from 0 to 9, unless you’re likely to count on your fingers each time you see them. Let’s look at mean price, the worth of a function at a particular point, and value as worth. When you purchase a vehicle, follow a recipe, or decorate your house, you’re using math principles.
Whether you’re playing Megabucks or betting your company’s complete future on one potential solution, a tiny math can help save you a great deal of misery. If you’re hard-working, organizations are likely to keep you. It can also be used in different ways in math.
## The Meaning of What Is I in Math
An educated populace may not be controlled or manipulated. So there’s some risk and all of us have to take that risk. In the united kingdom the term pling was popular in the previous days of computing, whilst in america, the expression shriek was used.
A failing test is something, it is a step towards figuring out the proper algorithm. At every step, ask the way the child figured out their answerand prepare to get surprised at a number of the unusual strategies they will utilize! The inexactitude isn’t restricted to the square root of 2.
The very first place-value system was created by the the Babylonians. A mathematician won’t specify bit size as that isn’t related to mathematical expressions. To understand the reason, you should know about the idea of convexity in optimization.
Eventually, you may not have Einstein’s brain. This doesn’t indicate that it’s purely depending on the memory of your brain. Unfortunately, every youngster isn’t lucky born with that prospective issue to be a terrific mathematician, once we assess their ability. | 1,278 | 6,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | longest | en | 0.945501 |
https://www.cfd-online.com/Forums/main/105663-discretization-2nd-wave-equation.html | 1,511,083,995,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805466.25/warc/CC-MAIN-20171119080836-20171119100836-00562.warc.gz | 815,601,091 | 17,435 | # discretization of the 2nd wave equation
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August 6, 2012, 05:08 discretization of the 2nd wave equation #1 Senior Member Join Date: Jun 2010 Posts: 111 Rep Power: 9 Hi, Can someone pls tell me how the 2nd order wave equation is discretized using the finite volume method? Any references would be appreciated. Thanks,
August 6, 2012, 05:17
#2
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Filippo Maria Denaro
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Originally Posted by Hooman Hi, Can someone pls tell me how the 2nd order wave equation is discretized using the finite volume method? Any references would be appreciated. Thanks,
In order to use FV method, you have to start form a conservative form of the governing equation. That is:
- differential conservative form ... Div (Flux) = ...
- Integral form ........................ Int [S] ( n . Flux) dS = ...
I suggest the book of Leveque on FV for hyperbolic equations
August 6, 2012, 05:23 #3 Senior Member Join Date: Jun 2010 Posts: 111 Rep Power: 9 I'll check the book. Tanx. I know the general rules of FVM just confused with being 2nd order in time and space and would appreciate examples and references.
August 6, 2012, 05:34
#4
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Filippo Maria Denaro
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Quote:
Originally Posted by Hooman I'll check the book. Tanx. I know the general rules of FVM just confused with being 2nd order in time and space and would appreciate examples and references.
not all systems of hyperbolic equations can be solved with FVM ... for example, Euler equation in primitive variable [rho, u, p] have not a conservative form, you must go back to the original system in [rho, rho*u, rho*E]
August 6, 2012, 05:42 #5 Senior Member Join Date: Jun 2010 Posts: 111 Rep Power: 9 So you're saying that the 2nd order wave equations cannot be solved using FVM?
August 6, 2012, 05:50
#6
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Filippo Maria Denaro
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Originally Posted by Hooman So you're saying that the 2nd order wave equations cannot be solved using FVM?
Assuming the equation is in the form Phi_xx - Phi_yy=0,
you should define a divergence-like operator Div = (d/dx, d/dy) such that
Div F = 0
with F = ( u, v) where u = dPhi/dx, v = - dPhi/dy
and add an equation to close the problem
du/dy + dv/dt = 0
August 6, 2012, 06:00
#7
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Quote:
Originally Posted by Hooman So you're saying that the 2nd order wave equations cannot be solved using FVM?
the equation is U_tt= cē LAP(U) where U_tt is the second order time derivative and LAP is the laplacian operator.
As Filippo said it for spatial derivative integration is straightforward because LAP(U) = DIV(Grad(U))
For the time derivative discretization just use a forward finite difference formulae for U_tt multiplied by the volume of the cell.
August 6, 2012, 07:39 #8 Senior Member Join Date: Jun 2010 Posts: 111 Rep Power: 9 Thank you both. I think I more or less understand. So we need to make the spatial derivative to 1st order?
August 6, 2012, 10:14
#9
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Originally Posted by Hooman Thank you both. I think I more or less understand. So we need to make the spatial derivative to 1st order?
The way you have to integrate the laplacian using finite volume may be found in any text book. check for Poisson equation using finite volume.
For the second order time derivative,you have to find a forward expression based on finite difference and you will have to multiplied it by the volume of the cell that's all.
August 6, 2012, 10:44
#10
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Filippo Maria Denaro
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Originally Posted by leflix The way you have to integrate the laplacian using finite volume may be found in any text book. check for Poisson equation using finite volume. For the second order time derivative,you have to find a forward expression based on finite difference and you will have to multiplied it by the volume of the cell that's all.
As leflix said, assuming your equation is Phi_tt = Div Grad (Phi), you integrate over a volume V of boundary S and apply the Gauss theorem to write
Int [V] Phi_tt dV = Int [S] (n . Grad (Phi) ) dS
This equation can be discretized at second order in time and space, for example using central derivatives. How course other issues as numerical stability, numerical oscillations, ecc have to be taken into account
August 6, 2012, 11:04 #11 Senior Member Join Date: Jun 2010 Posts: 111 Rep Power: 9 Thank you bother very much.
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All times are GMT -4. The time now is 05:33. | 1,460 | 5,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-47 | latest | en | 0.91852 |
https://mathtalk.sdsu.edu/wordpress/mathtalk-for-teachers/multiplying-binomials-unit-teachers/binomials-lesson-3-teachers/binomials-lesson-3-episode-6-teachers/ | 1,722,643,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00177.warc.gz | 302,341,756 | 31,334 | # Binomials Lesson 3 Episode 6 (Teachers)
### Exploring
Mauricio and Emily draw a picture of a new garden when the length of the original garden is increased by an unknown amount. They explore how to use symbols and letters to represent an unknown length and an unknown area.
### Episode Supports
Students’ Conceptual Challenges
Emily has written “? × 4” in the new portion of the garden. Mauricio argues that it should just be “?” that represents the area, while Emily argues that she wrote the area as “? × 4” because “?” and “4” are the dimensions of the new section. Mauricio seems to hold two different meanings for “?” in the drawing, as both the value of the one-dimensional measurement of the increase in length, and also as the value of the two-dimensional measurement for the increase in area of the new section. However, for a given context or problem, any symbol that represents a quantity must maintain the same value; so, this represents a challenge for Mauricio.
Focus Questions
For use in a classroom, pause the video and ask these questions:
1. [Pause the video at 0:42] Before watching Mauricio and Emily create their own drawing, try making a drawing that shows an increase in length by an unknown number. How might you represent this mathematically?
2. [Pause the video at 9:39] Why does Emily think a new symbol is needed? Summarize her argument in your own words.
Supporting Dialogue
[Pause the video at 5:45] With a partner, take turns explaining the arguments made by both Emily and Mauricio. What does Mauricio think the “?” represents? What does Emily think the “?” represents? Who do you think is right and why? What do you make of the introduction of the new “!” symbol? | 389 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.939732 |
https://handwiki.org/wiki/Physics:Mechanical_wave | 1,721,492,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00669.warc.gz | 251,150,891 | 10,912 | # Physics:Mechanical wave
Short description: Wave which is an oscillation of matter
Ripple in water is a surface wave.
In physics, a mechanical wave is a wave that is an oscillation of matter, and therefore transfers energy through a material medium.[1] (vacuum is, from classical perspective, a non-material medium, where electromagnetic waves propagate.)
While waves can move over long distances, the movement of the medium of transmission—the material—is limited. Therefore, the oscillating material does not move far from its initial equilibrium position. Mechanical waves can be produced only in media which possess elasticity and inertia. There are three types of mechanical waves: transverse waves, longitudinal waves, and surface waves. Some of the most common examples of mechanical waves are water waves, sound waves, and seismic waves.
Like all waves, mechanical waves transport energy. This energy propagates in the same direction as the wave. A wave requires an initial energy input; once this initial energy is added, the wave travels through the medium until all its energy is transferred. In contrast, electromagnetic waves require no medium, but can still travel through one.
## Transverse wave
Main page: Physics:Transverse wave
A transverse wave is the form of a wave in which particles of medium vibrate about their mean position perpendicular to the direction of the motion of the wave.
To see an example, move an end of a Slinky (whose other end is fixed) to the left-and-right of the Slinky, as opposed to to-and-fro.[2] Light also has properties of a transverse wave, although it is an electromagnetic wave.[3]
## Longitudinal wave
Main page: Physics:Longitudinal wave
Longitudinal waves cause the medium to vibrate parallel to the direction of the wave. It consists of multiple compressions and rarefactions. The rarefaction is the farthest distance apart in the longitudinal wave and the compression is the closest distance together. The speed of the longitudinal wave is increased in higher index of refraction, due to the closer proximity of the atoms in the medium that is being compressed. Sound is a longitudinal wave.
## Surface waves
Main page: Physics:Surface wave
This type of wave travels along the surface or interface between two media. An example of a surface wave would be waves in a pool, or in an ocean, lake, or any other type of water body. There are two types of surface waves, namely Rayleigh waves and Love waves.
Rayleigh waves, also known as ground roll, are waves that travel as ripples with motion similar to those of waves on the surface of water. Such waves are much slower than body waves, at roughly 90% of the velocity of bulk waves[clarify] for a typical homogeneous elastic medium. Rayleigh waves have energy losses only in two dimensions and are hence more destructive in earthquakes than conventional bulk waves, such as P-waves and S-waves, which lose energy in all three directions.
A Love wave is a surface wave having horizontal waves that are shear or transverse to the direction of propagation. They usually travel slightly faster than Rayleigh waves, at about 90% of the body wave velocity, and have the largest amplitude. | 658 | 3,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.89831 |
http://www.cs.brandeis.edu/~pablo/cs110b.html | 1,550,352,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481122.31/warc/CC-MAIN-20190216210606-20190216232424-00007.warc.gz | 336,337,751 | 1,966 | Xpilot
Give me back that beautiful background!
X-pilot is a client-server, multiuser, dynamic game.
In this precise moment a bunch of humans are wasting their lives chasing each other, playing xpilot.
Here's a plan for AI-attacking them:
• 1. Put up a million stupid programs playing against each other to evolve a better one.
• 2. Launch the bastard into the internet and:
• (a) Pass the x-pilot Turing test (fool other players by behaving humanly)
• (b) Kill them all!
So, this is too difficult. The game is far too complicated for this kind of approach, there are too many rules and variations and weapons, etc. etc.
Tractable subproblems:
• Simplified, no maze, no gravity, no shields, empty-space plain xpilot.
• "Second degree" pursuer-evader game (control acceleration instead of direction). This could be sold to the NASA
• In Koza (Pacman) style, hack high level behaviors ("attack", "explore", "run", etc) and inputs ("distance to closest enemy/wall", "being fired upon", etc.) and adapt a learning algorithm to work on these: Pablo's memorybase!
Results:
• The differential game was solved
• The solution was a little puzzling
• Inputs were hacked to be used by the bot:
• (x,y) coordinates
• Aiming angle
• Angle to nearest enemy
• Distance to nearest enemy
• Density of enemies in neighborhood
• Coordinates of nearest enemy
• Density of bullets in neighborhood
• Behaviors were hacked as well:
• Nil
• Shoot at nearest enemy
• Wander
• Stop
• Approach nearest enemy
• Approach & Shoot
• The differential game studied was of little use when it came to program these behaviors
• Memorybase learning got one conclusion: SHOOT ALL THE TIME!!
• Future research?
• Find a heuristic mathematical strategy useful in the real world.
• Is memorybase a valid learning model?
• Improve inputs & behaviors
Comments? Want to play? Send feedback! | 435 | 1,861 | {"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} | 2.53125 | 3 | CC-MAIN-2019-09 | latest | en | 0.900251 |
http://www.ask.com/question/pool-table-markings | 1,394,454,568,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010779425/warc/CC-MAIN-20140305091259-00088-ip-10-183-142-35.ec2.internal.warc.gz | 235,324,306 | 17,640 | # Pool Table Markings?
Pool table markings may refer to the sights, also called diamonds because of their shape, are inlaid at exact, evenly spaced spots along the rails of some pool tables. The help players to be able to take correct aim when aiming bank shots or kick shots. There are seven diamonds along each long rail and three along each short rail. Many books have been written concerning the geometric and algebraic systems of aiming and angles to get the best shots using the diamonds.
Reference:
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1. Find the pool table's center line. Use the tape measure to determine the width of the table. Use a piece of chalkboard chalk to mark the center point with a dot. 2. Measure the http://www.ehow.com/how_6620778_mark-measurements-...
The first thing you'll do to recover a pool table is remove any trim items from the table and tear the old felt off the slate. Unfold the new felt and cut it to the proper size. Leave http://answers.ask.com/Home/Gardening/how_to_recov...
What you will need to be able to do is reach the very edges of your slate (bed of your table). The best way to show the markings required is with a diagram, below is a small version http://wiki.answers.com/Q/Marking+for+8+foot+pool+...
If you are confounded by how to move a pool table, you are not alone. If you're moving a pool table either out of your house or into another part of it you should consider hiring http://www.life123.com/hobbies/games/billiards/how...
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It is fairly easy to mark out a pool table that is constructed according to specifications. The average height of the table should be 29 ¼ inches (74.29cm ...
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You'll want to have those dots in the correct spot so you can make your angle shots. You'll need a tape measure, marker pencil or pen, pool table D marking stick ... | 459 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2014-10 | latest | en | 0.917717 |
https://support.office.com/en-us/article/SUMX2MY2-function-9e599cc5-5399-48e9-a5e0-e37812dfa3e9?CorrelationId=1aeb94c1-69f0-4a12-a94c-02eeb237cfab&ui=en-US&rs=en-US&ad=US | 1,461,919,606,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110805.57/warc/CC-MAIN-20160428161510-00185-ip-10-239-7-51.ec2.internal.warc.gz | 650,938,456 | 11,037 | Sign in
# SUMX2MY2 function
This article describes the formula syntax and usage of the SUMX2MY2 function in Microsoft Excel.
## Description
Returns the sum of the difference of squares of corresponding values in two arrays.
## Syntax
SUMX2MY2(array_x, array_y)
The SUMX2MY2 function syntax has the following arguments:
• Array_x Required. The first array or range of values.
• Array_y Required. The second array or range of values.
## Remarks
• The arguments should be either numbers or names, arrays, or references that contain numbers.
• If an array or reference argument contains text, logical values, or empty cells, those values are ignored; however, cells with the value zero are included.
• If array_x and array_y have a different number of values, SUMX2MY2 returns the #N/A error value.
• The equation for the sum of the difference of squares is:
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data First array Second array 2 6 3 5 9 11 1 7 8 5 7 4 5 4 Formula Description (Result) Result =SUMX2MY2(A2:A8,B2:B8) Sum of the difference of squares of the two arrays above (-55) -55 =SUMX2MY2({2, 3, 9, 1, 8, 7, 5}, {6, 5, 11, 7, 5, 4, 4}) Sum of the difference of squares of the two arrays constants (-55) -55
## How can we improve it?
To protect your privacy, please do not include contact information in your feedback. Review our privacy policy. | 425 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-18 | longest | en | 0.691985 |
https://www.readyratios.com/reference/accounting/par_value.html | 1,721,511,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00100.warc.gz | 815,235,768 | 14,479 | # Par Value
Accounting Print Email
Meaning and definition of Par Value
The par value can be defined as the face value or stated value of a bond. If put another way, it is generally a dollar amount that is assigned to a security while representing the value contributed for each share in cash or goods. As expressed by Investopedia, the par values for diverse fixed-income products will be different. Generally, bonds have a par value of \$1,000, whereas most money market instruments feature higher par values. Also, stocks generally have a par value of \$0.01 or none at all.
Calculating the Par Value of a Bond
The key steps involved in computation of the par value of a bond include:
1. Evaluate a bond before buying it. Its par value, coupon rate, and maturity date are the three key instruments required to compare one bond to another.
2. Recognize that the face value of a bond is the redemption value, the amount of money received when the bond is redeemed. Bonds are generally designed to reach par value at maturity.
3. Identify the coupon rate of the bond, i.e. the amount of interest which is stated on a bond. It is determined upon issuance of the bond and is expressed as a percentage. If, in case, interest rates exceed the coupon rate, the bond is sold at a value lesser than the par value or “at a discount.” If, in any case, the interest rates are below the coupon rate, the price of the bond exceeds the par value or “is at premium.”
4. Mitigate the risk or increase the profit by evaluating the maturity date of the bond. A bond featuring a short maturity date is comparatively less risky than the one with a longer maturity date. This is because of the reason that it is more predictable and there is less time for interest rates and fluctuating prices. However, as a general rule, bonds with longer maturity dates shell out higher interest.
5. Compute the potential earnings with the help of an online calculator. One of the examples is a yield to maturity calculator, whereas another calculates the current value of savings bonds. | 433 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.889707 |
http://edustatistics.org/nathanvan/setup/stat202/lectures/lecture10.html | 1,488,152,570,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172156.69/warc/CC-MAIN-20170219104612-00316-ip-10-171-10-108.ec2.internal.warc.gz | 79,686,387 | 427,619 | # Stat 202: Lecture 10 (covers pp. 124-137)
Nathan VanHoudnos
10/15/2014
### Agenda
2. Checkpoint #11 results
3. Lecture 10 (covers pp. 124-137)
mean(hw3.grades/85)
[1] 0.9068
median(hw3.grades/85)
[1] 0.9412
### Agenda
2. Checkpoint #11 results
3. Lecture 10 (covers pp. 124-137)
### Checkpoint #11 results
• 69/76 students
• on average 86% percent correct
### Checkpoint #11 Question 1
(i) 31% of all Dalmatians have blue eyes.
(ii) 38% of all Dalmatians are deaf.
(iii) 42% of blue-eyed Dalmatians are deaf.
What is the probability that a randomly chosen Dalmatian is blue-eyed and deaf?
$$P(B \text{ and } D) = ?$$
Must use the general multiplication rule:
$$P(B \text{ and } D) = P(D|B)P(B) = .42 * .31 = .13$$
Note that: $$P(D)P(B) = .38 * .31 = .118 \ne P(B \text{ and } D)$$
### Agenda
2. Checkpoint #11 results
3. Lecture 10 (covers pp. 124-137)
• Checkpoint 12: Random variables (RVs) and distributions
• Checkpoint 13: Expection and variance rules
### Random Variables
random variable: A random variable assigns a unique numerical value to the outcome of a random experiment.
Consider the random experiment of flipping a coin twice. $$S = \{ HH, HT, TH, TT \}$$
Let $$X$$ be the number of tails.
• HH implies $$X = 0$$
• HT implies $$X = 1$$
• TH implies $$X = 1$$
• TT implies $$X = 2$$
### Random Variables
Consider getting data from a random sample on the number of ears in which a person wears one or more earrings.
Let $$X$$ be the number of ears in which a randomly selected person wears an earring.
If the selected person
• does not wear any earrings, then $$X = 0$$.
• wears earrings in either the left or the right ear, then $$X = 1$$.
• wears earrings in both ears, then $$X = 2$$.
### Random Variables
Assume we choose a lightweight male boxer at random and record his exact weight. According to the boxing rules, a lightweight male boxer must weigh between 130 and 135 pounds, so the sample space here is $$S = [130, 135]$$.
Let $$X$$ be the weight of the boxer.
In this case, $$X \in [130,135]$$.
### Random Variables
random variable: A random variable assigns a unique numerical value to the outcome of a random experiment.
Always arise from a random experiment!
Two types of random variables:
• discrete : possible values are from a list
• “Things you count.”
• continuous : possible values are from an interval
• “Things you measure.” (w/ or w/o rounding)
### Random Variables
Number of ears pierced?
• discrete
Weight of a lightweight boxer?
• continuous
Number of hours watching TV (rounded to the nearest hour)?
• continuous (rounded is still continuous)
### Random Variables
How many days per week do you drink soda?
• discrete
How many ounces of soda per week do you drink?
• continuous
The average weight of 18 year old males at basic training.
• continuous
### Distribution of a discrete RV
Consider the random experiment of flipping a coin twice. $$S = \{ HH, HT, TH, TT \}$$
Let $$X$$ be the number of tails. Recall: \begin{aligned} HH \rightarrow X & = 0 & HT \rightarrow X & = 1 \\ TH \rightarrow X & = 1 & TT \rightarrow X & = 2 \end{aligned}
What is the probability of $$X=0$$?
$P( X = 0) = P( HH ) = \frac{1}{4}$
### Distribution of a discrete RV
Consider the random experiment of flipping a coin twice. $$S = \{ HH, HT, TH, TT \}$$
Let $$X$$ be the number of tails. Recall: \begin{aligned} HH \rightarrow X & = 0 & HT \rightarrow X & = 1 \\ TH \rightarrow X & = 1 & TT \rightarrow X & = 2 \end{aligned}
What is the probability of $$X=1$$? \begin{aligned} P( X = 1) & = P( HT \text{ or } TH ) \\ & = P( HT ) + P(TH ) \\ & = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \end{aligned}
### Distribution of a discrete RV
Consider the random experiment of flipping a coin twice. $$S = \{ HH, HT, TH, TT \}$$
Let $$X$$ be the number of tails. Recall: \begin{aligned} HH \rightarrow X & = 0 & HT \rightarrow X & = 1 \\ TH \rightarrow X & = 1 & TT \rightarrow X & = 2 \end{aligned}
The distribution of $$X$$:
x | 0 | 1 | 2 |
---------+-----+-----+-----|
P(X = x) | 1/4 | 1/2 | 1/4 |
### Comment on notation
The distribution of $$X$$:
x | 0 | 1 | 2 |
---------+-----+-----+-----|
P(X = x) | 1/4 | 1/2 | 1/4 |
$$P(X = x)$$ is read as
• the probability that
• the random variable $$X$$ is
• equal to the value $$x$$.
Random variables are CAPITALIZED.
Values (AKA realizations) are lower case.
### Reprise: Rules
Let $$X$$ be a random variable and $$x$$ a realization of that random variable.
1. $$0 \le P(X = x) \le 1$$ for all $$x$$.
For example:
x | 0 | 1 | 2 |
---------+-----+-----+-----|
P(X = x) | 1/4 | 1/2 | 1/4 |
\begin{aligned} 0 & \le P(X = 0) = \frac{1}{4} \le 1 \\ 0 & \le P(X = 1) = \frac{1}{2} \le 1\\ \end{aligned}
### Reprise: Rules
Let $$X$$ be a random variable and $$x$$ a realization of that random variable.
1. $$0 \le P(X = x) \le 1$$ for all $$x$$.
2. $$\sum_{x\in S} P(X = x) = 1$$ $$\quad$$
For example:
\begin{aligned} \sum_{x \in S} P(X = x) & = P(X=0) + P(X=1) + P(X=2) \\ & = 1/4 + 1/2 + 1/4 = 1 \end{aligned}
Note: If summation notation is unfamiliar, check out Khan Academy here and here.
### A further example
A coin is tossed three times. Let the random variable X be the number of tails. Find the probability distribution of X.
1. Write out the sample space.
2. Define each possible realization of $$X$$ as an event.
3. Find the probabilities of the events.
### A further example
\begin{aligned} S = \{ & HHH, HHT, HTH, HTT, \\ & THH, THT, TTH, TTT \} \end{aligned}
\begin{aligned} X & = 0 \quad \{ HHH \} \\ X & = 1 \quad \{ HHT, HTH, THH \} \\ X & = 2 \quad \{ HTT, THT, TTH \} \\ X & = 3 \quad \{ TTT \} \end{aligned}
x | 0 | 1 | 2 | 3 |
---------+-----+-----+-----+-----|
P(X = x) | 1/8 | 3/8 | 3/8 | 1/8 |
### Formulas to define discrete RVs
An example: $P(X = x) = \frac{x+2}{25} \quad \quad x \in \{1,2,3,4,5\}$
Check rule 1:
$0 \le P(X=x) \le 1 \quad \text{ for all } x$
Note that: \begin{aligned} P(X = 1) & < P(X =5) \\ 0 \le \frac{1 + 2}{25} & < \frac{5+2}{25} \le 1 \end{aligned}
### Formulas to define discrete RVs
An example: $P(X = x) = \frac{x+2}{25} \quad \quad x \in \{1,2,3,4,5\}$
Check rule 2:
\begin{aligned} \sum_{x\in S} P(X = x) & = \sum_{x\in S} \frac{x+2}{25} \\ & = \frac{1}{25} \sum_{x\in \{1,2,3,4,5\}} \left( x +2 \right) \\ & = \frac{1}{25} 25 = 1\end{aligned}
### Probability Histograms (p. 130)
$$P(X = x) = \frac{x+2}{25} \quad \quad x \in \{1,2,3,4,5\}$$
Distribution A
Distribution B
### Larger standard deviation?
Distribution A
• values are farther from the mean
• larger standard deviation
Distribution B | 2,236 | 6,676 | {"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} | 4.03125 | 4 | CC-MAIN-2017-09 | longest | en | 0.831407 |
https://electronics.stackexchange.com/questions/196024/what-happens-if-there-are-many-120-ohms-resistors-as-terminators-on-the-can-bus?noredirect=1 | 1,638,811,945,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00578.warc.gz | 320,390,673 | 35,899 | # What happens if there are many 120 ohms resistors as terminators on the CAN bus?
I connected three devices in parallel on a CAN bus system. Every device has a $$\120\,\mathrm{\Omega}\$$ resistor built into the circuit. I am using two pairs of twisted cables of $$\1.2\,\mathrm{m}\$$ each for the communication. Will it have any effect on the communication?
• Data rate? Disposition of the three units? Oct 19 '15 at 7:40
• Data rate is 19600 bps. The units are connected in parallel to each other with with a twisted pair cable of 1.2mtrs between them. Oct 19 '15 at 7:53
You need to have a 120Ω resistor at the beginning of the chain and at the end of the chain. If one is missing, it won't work (trust me!). If you have too many, it might (MIGHT) work, but you're asking for trouble. CAN is intended to be a linear chain with termination resistors at each end.
Best to figure out a way to get rid of the middle termination resistor. Sometimes this is just a jumper, but you might want to desolder or clip it if that is necessary.
• Hi Daniel, Thanks for the reply. But When i communicate with three devices in parallel there is no issue but sometimes i see the communication is broken and again after some time its restoring itself do you think its because of the extra resistance. Oct 21 '15 at 7:40
• You should use an oscilloscope and try to divine if the signal looks like it is loaded down too much. This might look like the on-states not reaching their full voltage. Honestly I've not tried to troubleshoot this scenario, so I'm not sure how it would show up. Try scoping without the middle device and see if there is a noticeable difference in the waveform such as rise time or peak-to-peak voltage. Oct 21 '15 at 22:19
Assuming that the data receivers are capable of working with the smaller signal (which I'm sure they are), the only aspect that might cause problems are signal reflections.
Given that data rate is 19,600 bps, if you said this was a square wave of frequency 10 kHz, I'd be considering the wavelength of the 7th harmonic in order to justify that the cable length was OK.
7th harmonic is 70 kHz and this has a wavelength of 4.3 km. General rule of thumb is that a badly terminated cable will be OK if the length of the cable is below one-tenth of the highest wavelength and clearly it is.
The more resistors you have, the bigger a load there is on the transceiver trying to drive the bus. If, for example, you had ten 120-ohm resistors, each transceiver will have to deliver 5x the normal current. This will make them heat up, make them unable to complete proper transitions, or both.
Three resistors instead of two may not be the end of the world. But it's still stressing the transceivers unnecessarily.
• Hi Stephen, thanks for the reply. Does this extra resistance will create a make break kind of situation Oct 21 '15 at 7:43
• That's going to depend entirely on the transceivers involved. Oct 21 '15 at 12:16 | 712 | 2,953 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.956122 |
https://devel.isa-afp.org/browser_info/current/AFP/Types_To_Sets_Extension/SML_Semigroups.html | 1,709,385,414,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00592.warc.gz | 199,332,393 | 11,741 | # Theory SML_Semigroups
```(* Title: Examples/SML_Relativization/Algebra/SML_Semigroups.thy
Author: Mihails Milehins
*)
section‹Relativization of the results about semigroups›
theory SML_Semigroups
imports
"../SML_Introduction"
"../Foundations/Lifting_Set_Ext"
begin
subsection‹Simple semigroups›
subsubsection‹Definitions and common properties›
locale semigroup_ow =
fixes U :: "'ag set" and f :: "['ag, 'ag] ⇒ 'ag" (infixl ‹❙*⇩o⇩w› 70)
assumes f_closed: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a ❙*⇩o⇩w b ∈ U"
assumes assoc: "⟦ a ∈ U; b ∈ U; c ∈ U ⟧ ⟹ a ❙*⇩o⇩w b ❙*⇩o⇩w c = a ❙*⇩o⇩w (b ❙*⇩o⇩w c)"
begin
notation f (infixl ‹❙*⇩o⇩w› 70)
lemma f_closed'[simp]: "∀x∈U. ∀y∈U. x ❙*⇩o⇩w y ∈ U" by (simp add: f_closed)
tts_register_sbts ‹(❙*⇩o⇩w)› | U by (rule tts_AA_A_transfer[OF f_closed])
end
lemma semigroup_ow: "semigroup = semigroup_ow UNIV"
unfolding semigroup_def semigroup_ow_def by simp
locale plus_ow =
fixes U :: "'ag set" and plus :: "['ag, 'ag] ⇒ 'ag" (infixl ‹+⇩o⇩w› 65)
assumes plus_closed[simp, intro]: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a +⇩o⇩w b ∈ U"
begin
notation plus (infixl ‹+⇩o⇩w› 65)
lemma plus_closed'[simp]: "∀x∈U. ∀y∈U. x +⇩o⇩w y ∈ U" by simp
tts_register_sbts ‹(+⇩o⇩w)› | U by (rule tts_AA_A_transfer[OF plus_closed])
end
locale times_ow =
fixes U :: "'ag set" and times :: "['ag, 'ag] ⇒ 'ag" (infixl ‹*⇩o⇩w› 70)
assumes times_closed[simp, intro]: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a *⇩o⇩w b ∈ U"
begin
notation times (infixl ‹*⇩o⇩w› 70)
lemma times_closed'[simp]: "∀x∈U. ∀y∈U. x *⇩o⇩w y ∈ U" by simp
tts_register_sbts ‹(*⇩o⇩w)› | U by (rule tts_AA_A_transfer[OF times_closed])
end
locale semigroup_add_ow = plus_ow U plus
for U :: "'ag set" and plus +
assumes plus_assoc:
"⟦ a ∈ U; b ∈ U; c ∈ U ⟧ ⟹ a +⇩o⇩w b +⇩o⇩w c = a +⇩o⇩w (b +⇩o⇩w c)"
begin
by unfold_locales (auto simp: plus_assoc)
end
unfolding
by simp
locale semigroup_mult_ow = times_ow U times
for U :: "'ag set" and times +
assumes mult_assoc:
"⟦ a ∈ U; b ∈ U; c ∈ U ⟧ ⟹ a *⇩o⇩w b *⇩o⇩w c = a *⇩o⇩w (b *⇩o⇩w c)"
begin
sublocale mult: semigroup_ow U ‹(*⇩o⇩w)›
by unfold_locales (auto simp: times_closed' mult_assoc)
end
lemma semigroup_mult_ow: "class.semigroup_mult = semigroup_mult_ow UNIV"
unfolding
class.semigroup_mult_def semigroup_mult_ow_def
semigroup_mult_ow_axioms_def times_ow_def
by simp
subsubsection‹Transfer rules›
context
includes lifting_syntax
begin
lemma semigroup_transfer[transfer_rule]:
assumes [transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
(semigroup_ow (Collect (Domainp A))) semigroup"
proof -
let ?P = "((A ===> A ===> A) ===> (=))"
let ?semigroup_ow = "semigroup_ow (Collect (Domainp A))"
let ?rf_UNIV =
"(λf::['b, 'b] ⇒ 'b. (∀x y. x ∈ UNIV ⟶ y ∈ UNIV ⟶ f x y ∈ UNIV))"
have "?P ?semigroup_ow (λf. ?rf_UNIV f ∧ semigroup f)"
unfolding semigroup_ow_def semigroup_def
apply transfer_prover_start
apply transfer_step+
by simp
thus ?thesis by simp
qed
assumes [transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
proof -
let ?P = "((A ===> A ===> A) ===> (=))"
let ?rf_UNIV =
"(λf::['b, 'b] ⇒ 'b. (∀x y. x ∈ UNIV ⟶ y ∈ UNIV ⟶ f x y ∈ UNIV))"
unfolding
apply transfer_prover_start
apply transfer_step+
by simp
thus ?thesis by simp
qed
lemma semigroup_mult_transfer[transfer_rule]:
assumes [transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
(semigroup_mult_ow (Collect (Domainp A))) class.semigroup_mult"
proof -
let ?P = "((A ===> A ===> A) ===> (=))"
let ?semigroup_mult_ow = "(λf. semigroup_mult_ow (Collect (Domainp A)) f)"
let ?rf_UNIV =
"(λf::['b, 'b] ⇒ 'b. (∀x y. x ∈ UNIV ⟶ y ∈ UNIV ⟶ f x y ∈ UNIV))"
have "?P ?semigroup_mult_ow (λf. ?rf_UNIV f ∧ class.semigroup_mult f)"
unfolding
semigroup_mult_ow_def class.semigroup_mult_def
semigroup_mult_ow_axioms_def times_ow_def
apply transfer_prover_start
apply transfer_step+
by simp
thus ?thesis by simp
qed
end
subsection‹Cancellative semigroups›
subsubsection‹Definitions and common properties›
for U :: "'ag set" and plus +
"⟦ a ∈ U; b ∈ U; c ∈ U; a +⇩o⇩w b = a +⇩o⇩w c ⟧ ⟹ b = c"
"⟦ b ∈ U; a ∈ U; c ∈ U; b +⇩o⇩w a = c +⇩o⇩w a ⟧ ⟹ b = c"
unfolding
by simp
subsubsection‹Transfer rules›
context
includes lifting_syntax
begin
assumes [transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
unfolding
apply transfer_prover_start
apply transfer_step+
by simp
end
subsubsection‹Relativization›
begin
tts_context
tts: (?'a to U)
rewriting ctr_simps
eliminating through simp
begin
assumes "b ∈ U" and "a ∈ U" and "c ∈ U"
shows "(b +⇩o⇩w a = c +⇩o⇩w a) = (b = c)"
assumes "a ∈ U" and "b ∈ U" and "c ∈ U"
shows "(a +⇩o⇩w b = a +⇩o⇩w c) = (b = c)"
assumes "a ∈ U" and "A ⊆ U" and "B ⊆ U"
shows "bij_betw ((+⇩o⇩w) a) A B = ((+⇩o⇩w) a ` A = B)"
assumes "a ∈ U" and "A ⊆ U"
shows "inj_on ((+⇩o⇩w) a) A"
assumes "a ∈ U" and "A ⊆ U"
shows "inj_on (λb. b +⇩o⇩w a) A"
end
end
subsection‹Commutative semigroups›
subsubsection‹Definitions and common properties›
locale abel_semigroup_ow =
semigroup_ow U f for U :: "'ag set" and f +
assumes commute: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a ❙*⇩o⇩w b = b ❙*⇩o⇩w a"
begin
lemma fun_left_comm:
assumes "x ∈ U" and "y ∈ U" and "z ∈ U"
shows "y ❙*⇩o⇩w (x ❙*⇩o⇩w z) = x ❙*⇩o⇩w (y ❙*⇩o⇩w z)"
using assms by (metis assoc commute)
end
lemma abel_semigroup_ow: "abel_semigroup = abel_semigroup_ow UNIV"
unfolding
abel_semigroup_def abel_semigroup_ow_def
abel_semigroup_axioms_def abel_semigroup_ow_axioms_def
semigroup_ow
by simp
semigroup_add_ow U plus for U :: "'ag set" and plus +
assumes add_commute: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a +⇩o⇩w b = b +⇩o⇩w a"
begin
end
unfolding
by simp
locale ab_semigroup_mult_ow =
semigroup_mult_ow U times for U :: "'ag set" and times+
assumes mult_commute: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a *⇩o⇩w b = b *⇩o⇩w a"
begin
sublocale mult: abel_semigroup_ow U ‹(*⇩o⇩w)›
by unfold_locales (rule mult_commute)
end
lemma ab_semigroup_mult_ow:
"class.ab_semigroup_mult = ab_semigroup_mult_ow UNIV"
unfolding
class.ab_semigroup_mult_def ab_semigroup_mult_ow_def
class.ab_semigroup_mult_axioms_def ab_semigroup_mult_ow_axioms_def
semigroup_mult_ow
by simp
subsubsection‹Transfer rules›
context
includes lifting_syntax
begin
lemma abel_semigroup_transfer[transfer_rule]:
assumes[transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
(abel_semigroup_ow (Collect (Domainp A))) abel_semigroup"
unfolding
abel_semigroup_ow_def abel_semigroup_def
abel_semigroup_ow_axioms_def abel_semigroup_axioms_def
apply transfer_prover_start
apply transfer_step+
unfolding Ball_def by simp
assumes[transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
unfolding
apply transfer_prover_start
apply transfer_step+
by simp
lemma ab_semigroup_mult_transfer[transfer_rule]:
assumes[transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (=))
(ab_semigroup_mult_ow (Collect (Domainp A))) class.ab_semigroup_mult"
unfolding ab_semigroup_mult_ow_def class.ab_semigroup_mult_def
unfolding ab_semigroup_mult_ow_axioms_def class.ab_semigroup_mult_axioms_def
apply transfer_prover_start
apply transfer_step+
by simp
end
subsubsection‹Relativization›
context abel_semigroup_ow
begin
tts_context
tts: (?'a to U)
rewriting ctr_simps
substituting abel_semigroup_ow_axioms
eliminating through simp
begin
tts_lemma left_commute:
assumes "b ∈ U" and "a ∈ U" and "c ∈ U"
shows "b ❙*⇩o⇩w (a ❙*⇩o⇩w c) = a ❙*⇩o⇩w (b ❙*⇩o⇩w c)"
is abel_semigroup.left_commute.
end
end
begin
tts_context
tts: (?'a to U)
rewriting ctr_simps
eliminating through simp
begin
shows "⟦a ∈ U; b ∈ U; c ∈ U⟧ ⟹ a +⇩o⇩w b +⇩o⇩w c = a +⇩o⇩w (b +⇩o⇩w c)"
and "⟦a ∈ U; b ∈ U⟧ ⟹ a +⇩o⇩w b = b +⇩o⇩w a"
and "⟦b ∈ U; a ∈ U; c ∈ U⟧ ⟹ b +⇩o⇩w (a +⇩o⇩w c) = a +⇩o⇩w (b +⇩o⇩w c)"
end
end
context ab_semigroup_mult_ow
begin
tts_context
tts: (?'a to U)
rewriting ctr_simps
substituting ab_semigroup_mult_ow_axioms
eliminating through simp
begin
tts_lemma mult_ac:
shows "⟦a ∈ U; b ∈ U; c ∈ U⟧ ⟹ a *⇩o⇩w b *⇩o⇩w c = a *⇩o⇩w (b *⇩o⇩w c)"
is ab_semigroup_mult_class.mult_ac(1)
and "⟦a ∈ U; b ∈ U⟧ ⟹ a *⇩o⇩w b = b *⇩o⇩w a"
is ab_semigroup_mult_class.mult_ac(2)
and "⟦b ∈ U; a ∈ U; c ∈ U⟧ ⟹ b *⇩o⇩w (a *⇩o⇩w c) = a *⇩o⇩w (b *⇩o⇩w c)"
is ab_semigroup_mult_class.mult_ac(3).
end
end
subsection‹Cancellative commutative semigroups›
subsubsection‹Definitions and common properties›
locale minus_ow =
fixes U :: "'ag set" and minus :: "['ag, 'ag] ⇒ 'ag" (infixl ‹-⇩o⇩w› 65)
assumes minus_closed[simp,intro]: "⟦ a ∈ U; b ∈ U ⟧ ⟹ a -⇩o⇩w b ∈ U"
begin
notation minus (infixl ‹-⇩o⇩w› 65)
lemma minus_closed'[simp]: "∀x∈U. ∀y∈U. x -⇩o⇩w y ∈ U" by simp
tts_register_sbts ‹(-⇩o⇩w)› | U by (rule tts_AA_A_transfer[OF minus_closed])
end
ab_semigroup_add_ow U plus + minus_ow U minus
for U :: "'ag set" and plus minus +
"⟦ a ∈ U; b ∈ U ⟧ ⟹ (a +⇩o⇩w b) -⇩o⇩w a = b"
"⟦ a ∈ U; b ∈ U; c ∈ U ⟧ ⟹ a -⇩o⇩w b -⇩o⇩w c = a -⇩o⇩w (b +⇩o⇩w c)"
begin
apply unfold_locales
done
end
unfolding
minus_ow_def
by simp
subsubsection‹Transfer rules›
context
includes lifting_syntax
begin
assumes [transfer_rule]: "bi_unique A" "right_total A"
shows
"((A ===> A ===> A) ===> (A ===> A ===> A) ===> (=))
proof -
let ?P = "((A ===> A ===> A) ===> (A ===> A ===> A) ===> (=))"
let ?rf_UNIV =
"(λf::['b, 'b] ⇒ 'b. (∀x y. x ∈ UNIV ⟶ y ∈ UNIV ⟶ f x y ∈ UNIV))"
have
"?P
(λf fi. ?rf_UNIV fi ∧ class.cancel_ab_semigroup_add f fi)"
unfolding
minus_ow_def
apply transfer_prover_start
apply transfer_step+
unfolding Ball_def by auto
thus ?thesis by simp
qed
end
subsubsection‹Relativization›
begin
tts_context
tts: (?'a to U)
rewriting ctr_simps
eliminating through simp
begin
assumes "a ∈ U" and "b ∈ U"
shows "a +⇩o⇩w b -⇩o⇩w b = a"
assumes "a ∈ U" and "c ∈ U" and "b ∈ U"
shows "a +⇩o⇩w c -⇩o⇩w (b +⇩o⇩w c) = a -⇩o⇩w b"
assumes "c ∈ U" and "a ∈ U" and "b ∈ U"
shows "c +⇩o⇩w a -⇩o⇩w (c +⇩o⇩w b) = a -⇩o⇩w b"
tts_lemma diff_right_commute:
assumes "a ∈ U" and "c ∈ U" and "b ∈ U"
shows "a -⇩o⇩w c -⇩o⇩w b = a -⇩o⇩w b -⇩o⇩w c" | 4,060 | 9,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.405591 |
https://genstat.kb.vsni.co.uk/knowledge-base/maebayes/ | 1,726,420,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651632.84/warc/CC-MAIN-20240915152239-20240915182239-00861.warc.gz | 254,590,276 | 12,291 | 1. Home
2. MAEBAYES procedure
# MAEBAYES procedure
Modifies t-values by an empirical Bayes method (D.B. Baird).
### Options
`PRINT` = string tokens What to print (`estimates`); default `esti` What to plot (`phistograms`, `thistograms`, `pvalues`, `tvalues`); default `*` i.e. nothing Type of data specified by the `DATA` parameter when it is a variate (`means`, `tvalues`); default `tval` Type of test to use to form probability values (`twosided`, `greaterthan`, `lessthan`); default `twos` Device number on which to plot the graphs What graphics filename template to use to save the graphs; default `*`
### Parameters
`DATA` = pointers or variates Pointers of variates or variates of means or t-values to be summarized Supplies standard deviations of the data when `DATA` is a variate of means or t-values Supplies degrees of freedom when `DATA` is a variate of means or t-values Saves the estimated prior standard deviation Saves the estimated number of degrees of freedom assigned to the prior standard deviation Saves the modified t-values Saves the shrunken `SD` values Saves the modified probability values
### Description
In a microarray experiment, as hundreds and often thousands of probes are being processed in parallel, there is a loss of power if you consider the variation of each probe in isolation. If this parallelism is used between the genes to gain extra information on the variation of an individual probe, then more powerful tests of the level of differential expression of a probe can be obtained. To do this, a prior distribution of the standard deviations (or equivalently the variances) over the probes is assumed. In particular, it is assumed that the reciprocal of the variance, sp2, of each probe is distributed as a multiple of a chi-square distribution with d0 degrees of
freedom, i.e. 1/sp2 is distributed as 1/(d0 × s02) × Chisquare(d0).
If the parameters of this distribution, the prior degrees of freedom d0 and standard deviation s0 are estimated, more information can be gained on an individual probe, by shrinking it towards the prior by an amount that depends on the amount of information in the standard deviation sp of the probe (in this case its degrees of freedom dp). The modified standard deviation s~p is then given by:
s~p = √((d02 × s02 + dp2 × sp2) / (d0 + dp))
A modified t-test can then be performed using the modified standard deviation with d0 + dp degrees of freedom. The method can also produce the probability values for tests that the differential expression differs from zero. The `METHOD` option selects the type of test i.e. two-sided, or for values greater than or less than zero (the default is two-sided).
The `DATA` parameter can supply a pointer containing one variate per slide, with the probes in the same position within each variate. The means and standard deviations are then be calculated from the raw data. Alternatively, `DATA` can supply a variate containing means or t-values for each probe. The `DATATYPE` option should then indicate which of these has been given, the `SD` parameter should supply a variate containing the standard deviations for each probe, and the `DF` parameter should supply a variate with the numbers of degrees of freedom.
The estimated prior number of degrees of freedom d0 and standard deviation s0 can be saved, in scalars, by the `D0` and `S0` parameters. The `TMODIFIED` parameter can supply a variate to save the modified t-values, the `SDMODIFIED` parameter can save the shrunken `SD` values, and the `PMODIFIED` parameter can save the modified probability values.
By default, the estimates are printed, but this can be suppressed by setting option `PRINT=*`. The `PLOT` option controls what plots are produced, with settings:
`phistograms` two histograms showing the modified and raw probabilities plotted on the same scale; two histograms showing the modified and raw t-values plotted on the same scale; a scatter plot of modified versus raw probabilities; and a scatter plot of modified versus t-values.
By default, nothing is plotted. You can use the `DEVICE` option to plot to a device other than the screen. The `GRAPHICSFILE` specifies then supplies a template for the file names.
Options: `PRINT`, `PLOT`, `DATATYPE`, `METHOD`, `DEVICE`, `GRAPHICSFILE`.
Parameters: `DATA`, `SD`, `DF`, `SD0`, `DF0`, `TMODIFIED`, `SDMODIFIED`, `PMODIFIED`.
### Reference
Smyth, G.K. (2004). Linear models and empirical Bayes methods for assessing differential expression in microarray experiments. Statistical Applications in Genetics and Molecular Biology, 3, No. 1, Article 3.
Procedures: `AFFYMETRIX`, `FDRBONFERRONI`, `FDRMIXTURE`, `MAANOVA`, `MABGCORRECT`, `MAREGRESSION`, `MARMA`, `MAROBUSTMEANS`, `MAVDIFFERENCE`, `MAVOLCANO`, `QNORMALIZE`.
Commands for: Microarray data.
### Example
```CAPTION 'MAEBAYES example'; STYLE=meta
ENQUIRE CHANNEL=-1; EXIST=check; NAME=\
'%GENDIR%/Data/Microarrays/ApoAIKnockOutEffects.GSH'
IF check | 1,188 | 4,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.768994 |
https://happeninrecords.com/what-is-a-lottery-21/ | 1,721,291,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00713.warc.gz | 256,623,297 | 20,293 | # What is a Lottery?
A lottery is a game where numbers are drawn at random to determine winners. People play the lottery to win big money or other prizes. It is a common pastime for many people, and it contributes to billions of dollars in revenue annually. However, the odds of winning are low. It is important to understand the math behind the lottery before you decide to participate in it.
Lotteries have been around for centuries. The first recorded ones were conducted in the 15th century, when towns held them to raise money for town fortifications and to help the poor. A number of countries now have state-run lotteries. In these games, players pay a fee to enter and have the chance to win a prize. The prize can be anything from money to goods and services. In the United States, some of the most popular lotteries are Powerball and Mega Millions.
The definition of lottery in the dictionary is “a competition or game where the winner is chosen by chance,” but some experts believe that the term should be more broadly applied to any game whose outcome depends on luck. The lottery is a good example because it involves paying a small sum of money for the chance to win a large sum of money. The word “lottery” is also used to describe games where a prize is awarded to the first entrant whose name or number matches those of the others, such as a contest for a job or an apartment.
Although the game of lottery is based on chance, there are ways to improve your chances of winning it. You can choose your own numbers or let a computer pick them for you. If you choose your own numbers, avoid picking them that have sentimental value, such as birthdays or months. Instead, try to use numbers that are not close together. In addition, it is important to buy as many tickets as possible, because each one has an equal chance of being picked.
When choosing numbers for a lottery, it’s best to go with those that are less common. This will increase your odds of winning, because the odds are higher for the number that has not been chosen in previous drawings. You can find out which numbers are less common by looking at the history of past winnings on the official lottery website.
It is also important to consider the utility of a monetary loss in deciding whether or not to purchase a lottery ticket. If the entertainment value of losing a small amount of money is greater than the disutility of losing much more, then it may be rational to buy a ticket.
Lotteries are a great way to generate income for public purposes, but they can also be dangerous. Lotteries can result in corruption, fraud, and mismanagement of funds. They also lead to over-regulation, which can have negative effects on the economy and society. It’s important to carefully monitor the lottery process in order to make sure it is conducted fairly. | 586 | 2,838 | {"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} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.981211 |
https://www.coursehero.com/tutors-problems/Chemistry/17894604-One-mole-of-contains-the-largest-number-of-atoms-a-S8-b-C1/ | 1,566,514,141,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317516.88/warc/CC-MAIN-20190822215308-20190823001308-00125.warc.gz | 760,063,188 | 24,567 | View the step-by-step solution to:
Question
# One mole of ________ contains the largest number of atoms.a) S8b) C10 H8c) Al2
(SO4)3
d) Na3 PO4
e) Cl2
S8 = 32.06 x 8 = 256.48
C10 H8 = (12.01x10) + (1.01x8) = 128.18
Al2 (SO4)3 = (26.98x2) + ((32.06 + (16x4)3) = 342.14
Na3 PO4 = (22.99 x3) + 30.97 + (16.00 x 4) = 163.94
Cl2 = 35.45 x 2 = 70.90
I believe the correct answer is C but I seen a post stating the correct answer is B. The person received 100% with B as the answer. Can you please confirm. Maybe I'm not doing the math correctly.
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Browse Documents | 300 | 931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-35 | latest | en | 0.863522 |
https://astarmathsandphysics.com/university-maths-notes/elementary-calculus/1807-when-does-the-power-series-method-for-solving-differential-equations-break-down.html?tmpl=component&print=1&page= | 1,534,795,638,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217006.78/warc/CC-MAIN-20180820195652-20180820215652-00614.warc.gz | 636,686,285 | 3,726 | ## When does the Power Series Method for Solving Differential Equations Break Down
It is usually straightforward to derive the recurrence relation required to generate the power sereis solution to the differential equation. The method does not always produce a solution however, and we need to know when it fails.
Theorem (The Convergence Theorem)
Ifwhereandare polynomials and all the zeros ofare real, any power series solutionabouthas an interval of convergencewhereis the minimum distance fromto a zero ofin factis a lower bound for the radius of convergence, but the radius of convergence is equal toexcept in very special circumstances.
We might expect no convergence for a series expansion aboutwhereis a zero of Consider the differential equationFor this equationwhich is zero at
If we assume a power series solution of the formthen
Substitution into the original differential equation giveswhich after re - indexing of the first summation term becomesor
Henceand
Henceand the series mothod gives
The separation of variables method givesbut the series method failed to detect this solution becauseis a zero of | 223 | 1,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-34 | latest | en | 0.865474 |
https://adlmag.net/how-far-away-is-the-3-point-line-from-the-hoop-2/ | 1,713,891,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00288.warc.gz | 67,889,583 | 23,060 | The NBA 3point line is 23.75 feet from the hoop, 22 feet in the corners. The FIBA 3point line is 22.15 feet from the hoop, 21.65 feet in the corners. The WNBA uses the same 3point line as FIBA. At the NCAA level, the 3point line distance is 20.75 feet, for both men and women.
Also, Would a dunk from the 3 point line?
In terms off dunking from the three point line. This is only counted as a three point shot if both of the players feet are behind the line before he takes off. If he is fouled in flight to the goal and finishes the dunk, he will be awarded a 4 point play opportunity becuase he was fouled on a 3 point attempt.
In this way, How far is 3 point line high school?
The 3 Point Line (Arc):
College and High School – the 3 point arc is 19 feet 9 inches, with a straight line extending out 5 feet 3 inches (63 inches) from the baseline.
Contents
## How high is an NBA rim?
The top of the hoop is 10 feet (305 cm) above theground. Regulation backboards are 72 inches (183 cm) wide by 42inches (110 cm) tall. All basketball rims (hoops) are18 inches (46 cm) in diameter.
## How big is an NBA court?
Dimensions. Basketball courts come in different sizes. In the National Basketball Association (NBA), the court is 94 by 50 feet (28.7 by 15.2 m). Under International Basketball Federation (FIBA) rules, the court is slightly smaller, measuring 28 by 15 meters (91.9 by 49.2 ft).
## Can two basketballs fit through a hoop?
They can‘t. A standard NBA basketball’sdiameter is between 9.41 and 9.43 inches. A basketball hoopis 18 inches in diameter). So put two basketballsside by side and their width exceeds the basketball’s hoopdiameter by about .8 inches.
## Is women’s basketball court smaller?
The size of the ball in women’s basketball rangesfrom 28 1/2 to 29 inches in circumference. The basketball inthe men’s game is between 29 1/2 and 30 inches. Women’shands are smaller than men’s hands on average, so it’seasier for women to handle the smallerball.
## How far is the free throw line from the rim?
The NBA, the NCAA and the National Federation of State High School Associations dictate that the free-throw line is 15 feet horizontally from the plane of the front of the backboard. The free-throw line on international courts is 15.09 feet from the backboard.
## How many feet is a high school basketball court?
The Answer: I hope you have a big driveway! A regulationcollege court is 94 feet long and 50 feetwide, while high school courts are supposed to be 84 by 50feet. The free-throw line (2 inches wide) is 15 feetfrom the backboard, which supports a basket 10 feet high atthe upper edge.
## Are NBA courts bigger than college?
The high school court is 84 feet long. The length of an NBA court is exactly 10 feet longer. In fact, college and all professional league games, including the WNBA, are played on a 94-foot long court.
## How far is the FIBA 3 point line?
At the top of the arc, the threepointline is 6.75 meters from the center of the basket or 22.15feet. This measurement works out to about 22 feet, two inches. TheFIBA three-point line, which has also been adopted bythe WNBA, is over a foot and a half closer than the NBAline.
## How many laps around a basketball court is a mile?
Claude: 5 laps of a basketball court = 1/4mile is pretty much standard . . . I’ve used that foryears. Don’t cut the corners though, or you’ll have to do 20burpees at each location where you cut in too tightly.
## Are NBA balls bigger than college?
The official size of the basketball used by the NBA is 29.5 inches in circumference. That’s the same size used throughout men’s college and high school basketball leagues. The WNBA uses a slightly smaller ball, measuring 28.5 inches in circumference. An NBA basketball must be inflated to a pressure of 7.5 to 8.5 PSI.
## Should the NBA have a 4 point line?
The answer to this is that, yes, there ought to be a4point line, but not where you think. The logicalanswer (from a mathematical standpoint anyway) is to make thecurrent 3-point line a 4point line, and allshots from inside the arc would now be worth three.
## Why are there two 3pt lines?
Why it matters: According to the committee, moving the line back will open up the lane for drives/cuts to the basket and additional low-post play, while keeping the 3point revolution in check by making threes more challenging. For reference, the NBA’s 3point line is 23 feet, 9 inches.
three seconds
## What do the markings on a basketball court mean?
Every line on the court has a purpose.Here are all the lines that we’ll cover in this tutorial:Baselines (also known as the end lines) Sidelines. Freethrow line (also known as the foul line)
## What do the markings on a basketball court mean?
Four hash marks shall be drawn (2 inches wide)perpendicular to the sideline on each side of the court and28 feet from the baseline. These hash marks shall extend 3feet onto the court. Four hash marks shall be drawn(2 inches wide) perpendicular to the baseline on each side of thefree throw lane line.
## Can 2 basketballs fit through a hoop?
Yes, and No, two women’s basketballs can fit through the rim at the same time. But Two Men’s basketballs can not fit through the cylinder simultaneously. The basketball rim may appear smaller in comparison to the basketball than it actually is.
## When did the NBA move the three point line?
The 3point line’s first use in aprofessional league was back in 1961 in the AmericanBasketball League. The ABL only lasted 1 ½ seasons beforefolding, so the 3-pointer quickly went away. The NBA,which had been around since 1946, never seriously considered it atthat point.
## Is the corner 3 point shot shorter?
The straight lines are 3 feet inside eachsideline and run from the baseline to just short of the freethrow line extended. The arc that intersects the straight lines is23 feet, 9 inches from the basket. The shortest possiblethree-point shot, from the corner, is 22 feet fromthe hoop.
## What is 3 in the key?
The three seconds rule (also referred to as the three-second rule or three in the key, often termed a lane violation) requires that in basketball, a player shall not remain in the opponents’ restricted area for more than three consecutive seconds while that player’s team is in control of a live ball in the frontcourt
## How long is a basketball game?
HalfCourt Dimensions
Halfcourts cut the length of a full basketball court in half, not the width. The width remains the same from the original court. A youth half court is 37 feet by 42 feet and a high school half court is 42 feet by 50 feet.
## Are all NBA courts the same size?
For nearly 80 years, full-sized basketball courts have been the same dimensions: 94 feet long by 50 feet wide. But basketball players at all levels – and especially the NBA – have gotten bigger, stronger and faster, taking up more space on the court and covering ground quicker than ever.
## When did they move the three point line?
NCAA moving 3point line back to international distance
The change will not go into effect in Division II and III until 2020-21 due to the potential financial impact on schools. The 3point line was last moved in 2008-09, extending a foot to 20 feet, 9 inches.
## When did they move the three point line?
The Lines and Dimensions of a Basketball Court. On each half-court, painted lines show the free throw lane and circle, as well as the three-point arc, whose distance from the basket varies based on the level of hoops being played.
10 feet
## Is a basketball court bigger than a netball court?
While the netball court is a little larger than the official basketball court, it is not by much. A netball has a width of 15.25 meters and a length of 30.5 meters. The goals are a little higher than that of a FIBA hoop but this too isn’t discernible.
## Why are there two 3pt lines?
The reason they put that rule in is because they thoughtit would open up the game. Several players mentioned somethingthose who voted in the rule might not have thought about —the fact that there will now be two threepointlines on the floor, one for the men and one for the women. Thewomen’s line remains at 19-9.
## How far is the 3 point line from the free throw line?
The distance of the three point line fromthe basket varies according to the different levels of play. Thisline is used as a boundary line when shootingfree throws. It is fifteen feet away from the backboard. Ona free throw attempt, the shooter cannot step on or acrossthis line until the ball strikes the rim.
## How big is a half basketball court?
While the netball court is a little larger than the official basketball court, it is not by much. A netball has a width of 15.25 meters and a length of 30.5 meters. The goals are a little higher than that of a FIBA hoop but this too isn’t discernible.
## How high should the ceiling be for an indoor basketball court?
A minimum ceiling height of 24 feet is needed forany court sport; for turf sports, a ceiling height ofat least 20 feet is required.
## How far is the top of the key from the basket?
HalfCourt Dimensions
Halfcourts cut the length of a full basketball court in half, not the width. The width remains the same from the original court. A youth half court is 37 feet by 42 feet and a high school half court is 42 feet by 50 feet. | 2,257 | 9,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.92887 |
https://www.marsja.se/r-books-for-psychologists/?utm_source=rss&utm_medium=rss&utm_campaign=r-books-for-psychologists | 1,597,523,124,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00318.warc.gz | 734,398,075 | 18,215 | Select Page
8 Shares
R is a free and open-source statistical programming environment. Being open-source and free it has a large and helpful online community (for instance, see StackOverflow). When I went from carrying out analysis in SPSS to do them in R, I searched for good books targeted to Psychologists. The following 4 R books are useful and good for Psychologists that want to learn R.
The first book, Discovering Statistics Using R, may be a really good start if you are an undergraduate and have no experience of programming or statistics. The next two books are from intermediate to advanced level. The last book is, at the moment, free and is also a great introduction to statistics.
## R books
#### Discovering Statistics Using R (Field, Miles, & Field) Save
Cover of Discovering Statistics Using R
Andy Fields book “Discovering Statistics Using SPSS” is incredibly popular at my department. In fact, it was the first statistics book I bought that was not a course book.
“Discovering Statistics Using R” uses the same pedagogical and humorous approach to the subject (i.e., learning statistics using R).
It covers basic statistical and research concepts, how to explore and visualize data, basic uni-variate (e.g., t-test, correlation, ANOVA) as well as some multivariate statistics (e.g., MANOVA, & Factor Analysis). Emphasizes the application of statistics without leaving theory out. This may not be considered an R book, as it focuses on statistics but I highly recommended this an introduction to statistics and R.
#### The Foundations of Statistics: A Simulation-based Approach Save (Vasisthth & Broe)
Cover of The Foundations of Statistics
“The Foundations Of Statistics” focus is on statistics in the linguistics, psychology, and cognitive science disciplines. This book covers basic statistical concepts (e.g., power, ANOVA, linear mixed effects/multilevel models). “The Foundations of Statistics” has a simulation-based approach and all simulations in the book can be carried out using R. Recommended for advanced undergraduate and graduate students. Allthough this book, as with Discovering SPSS using ar, is not a R book only, you get a very good introduction but you may need some prior programming experience.
#### Advances in Social Science Research Using R (Lecture Notes in Statistics) Save (Edited by Vinod)
I have not read this book yet, but it seems promising:
This book covers recent advances for quantitative researchers with practical examples from social sciences. The twelve chapters written by distinguished authors cover a wide range of issues–all providing practical tools using the free R software.
Note that this book covers Social Science and, again, may not be considered an R book. However, there are some interesting statistical techniques to learn (e.g., Quantile Regression).
#### Learning Statistics with R
Cover of Learning Statistics Using R
“Learning Statistics with R” is a work in progress. The first draft is available for free. If you prefer to read a “real” book, there is also the option to buy a relatively cheap hard copy.
The book is written by Daniel Navarro for his introductory statistics class (for psychologists). It covers a background on why we learn statistics and a short introduction of research design. The book progresses into introducing the reader to R, working with data (descriptives, visualization, etc.), and statistical theory. It covers parametric tests such as z-test, t-test, ANOVA, and linear regression. “Learning Statistics with R” also covers Bayesian Statistics in one chapter.
There are probably more R books with a focus on Psychology or Social Sciences, but I am not aware of them. If you happen to know a good book, maybe covering a more advanced and/or specific topic, please leave a comment.
8 Shares | 784 | 3,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-34 | latest | en | 0.922507 |
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# Best Free GMAT CAT Test to take when starting preparation
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Best Free GMAT CAT Test to take when starting preparation [#permalink]
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29 Oct 2012, 09:56
Hello Expert Members,
I will be starting my preparation in the next 3 days (1st Nov) and finish by 1st Feb, 2013.
As a pre-requisite, I have all the 8 MGMAT Strategy Guides, OG13, OG Verbal and OG Quant as of now. I have spent a couple of days at this forum and have been motivated, guided and duly impressed by others. Thanks for all of that
Now, my plan is to give the first 4-5 days to go over the basic materials and understand the pattern of the test. After that I would like to give a free test to know where I stand. As I understand, the 2 free MGMAT tests are the closest to real GMAT so I would not like to waste them when I am not totally done with my prep.
Can someone please suggest which is the best free test to begin with.
Please bear with my ignorance.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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29 Oct 2012, 10:17
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blabber wrote:
Hello Expert Members,
I will be starting my preparation in the next 3 days (1st Nov) and finish by 1st Feb, 2013.
As a pre-requisite, I have all the 8 MGMAT Strategy Guides, OG13, OG Verbal and OG Quant as of now. I have spent a couple of days at this forum and have been motivated, guided and duly impressed by others. Thanks for all of that
Now, my plan is to give the first 4-5 days to go over the basic materials and understand the pattern of the test. After that I would like to give a free test to know where I stand. As I understand, the 2 free MGMAT tests are the closest to real GMAT so I would not like to waste them when I am not totally done with my prep.
Can someone please suggest which is the best free test to begin with.
Please bear with my ignorance.
I think you meant GMATPREP tests from mba.com. These 2 free GMATPREP tests are closest to real GMAT as these have real GMAT retired questions and same scoring logic.
having said that, lot of people recommend 'wasting' one of these at the start to check ur current level and 'using' second one, at the end, before the exam to give u most realistic practice.
However, if you do not want to waste one of ur CATs, you can use diagnostic test available in OG. That should be a pretty decent alternative to start with.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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29 Oct 2012, 11:00
Thanks Vips,
That was a quick reply!!
I think I will choose the OG Diagnostic Test for the moment.
Let me start my prep and I will keep you guys posted. Hoping for the best here
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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29 Oct 2012, 14:30
I took the same route and did the diagnostic test with no timer and Quans and Verbal in a different sitting.
How do you go and score the result. I am seeing OG gives a table saying Avg, below or above. I am eager to put a score to my test and see where I stand (say 500-600 ) or something like that
Any tool or pointers for that
Thanks
blabber wrote:
Thanks Vips,
That was a quick reply!!
I think I will choose the OG Diagnostic Test for the moment.
Let me start my prep and I will keep you guys posted. Hoping for the best here
Kudos [?]: [0], given: 7
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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30 Oct 2012, 01:25
rich895 wrote:
blabber wrote:
Thanks Vips,
That was a quick reply!!
I think I will choose the OG Diagnostic Test for the moment.
Let me start my prep and I will keep you guys posted. Hoping for the best here
I took the same route and did the diagnostic test with no timer and Quans and Verbal in a different sitting.
How do you go and score the result. I am seeing OG gives a table saying Avg, below or above. I am eager to put a score to my test and see where I stand (say 500-600 ) or something like that
Any tool or pointers for that
Thanks
If you need to see where u stand in terms of 200-800 scale, the only option is taking a practice test. Diagnostic test while gives u a good idea about ur current level in terms of strength and weakness, however, does not give u points on scale of 200-800.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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30 Oct 2012, 02:49
Oh!! I wasn't aware of that. I guess it only leaves me with the option of 'wasting' one GMATPrep test.
Still, I have some more time with me before I take up my 1st test. So, more suggestions are welcome.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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30 Oct 2012, 03:03
Your GMATPREP will probably not get "wasted" because by the time you take the same CAT again after 2-3 months, chances are good that you will not remember any question from the test
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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30 Oct 2012, 04:24
A very fine (and possibly valid) point indeed!! Going through one of the threads I am now aware that I can reset the software to give the tests again.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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31 Oct 2012, 13:44
Guys, I started with MGMAT and took the 1st test (which is free). Have managed a 640 without prep. Let see where this takes me from here.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink]
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31 Oct 2012, 16:13
Congrats on the 640, that is a great score to start out with. Best of luck in your studies.
As you said earlier, you can reset the GMAT prep tests as often a you want but after 2-3 times you are gonna start seeing repeat questions.
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Re: Best Free GMAT CAT Test to take when starting preparation [#permalink] 31 Oct 2012, 16:13
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# Best Free GMAT CAT Test to take when starting preparation
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,549 | 9,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-43 | latest | en | 0.918454 |
https://numberworld.info/574643968 | 1,590,515,939,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00116.warc.gz | 493,573,715 | 3,989 | # Number 574643968
### Properties of number 574643968
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2244703
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
22405f00
Base 32:
h40no0
sin(574643968)
0.1573089562302
cos(574643968)
0.98754943789654
tan(574643968)
0.15929223408326
ln(574643968)
20.1692612209
lg(574643968)
8.759398852302
sqrt(574643968)
23971.732686646
Square(574643968)
### Number Look Up
Look Up
574643968 (five hundred seventy-four million six hundred forty-three thousand nine hundred sixty-eight) is a special number. The cross sum of 574643968 is 52. If you factorisate the number 574643968 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2244703. The figure 574643968 has 18 divisors ( 1, 2, 4, 8, 16, 32, 64, 128, 256, 2244703, 4489406, 8978812, 17957624, 35915248, 71830496, 143660992, 287321984, 574643968 ) whith a sum of 1147043744. 574643968 is not a prime number. The number 574643968 is not a fibonacci number. The figure 574643968 is not a Bell Number. The figure 574643968 is not a Catalan Number. The convertion of 574643968 to base 2 (Binary) is 100010010000000101111100000000. The convertion of 574643968 to base 3 (Ternary) is 1111001021221022221. The convertion of 574643968 to base 4 (Quaternary) is 202100011330000. The convertion of 574643968 to base 5 (Quintal) is 2134102101333. The convertion of 574643968 to base 8 (Octal) is 4220057400. The convertion of 574643968 to base 16 (Hexadecimal) is 22405f00. The convertion of 574643968 to base 32 is h40no0. The sine of the number 574643968 is 0.1573089562302. The cosine of the number 574643968 is 0.98754943789654. The tangent of 574643968 is 0.15929223408326. The root of 574643968 is 23971.732686646.
If you square 574643968 you will get the following result 330215689958785024. The natural logarithm of 574643968 is 20.1692612209 and the decimal logarithm is 8.759398852302. I hope that you now know that 574643968 is very impressive number! | 758 | 2,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.637581 |
https://www.jiskha.com/display.cgi?id=1296702150 | 1,503,163,422,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105700.94/warc/CC-MAIN-20170819162833-20170819182833-00072.warc.gz | 904,070,534 | 3,729 | # physics
posted by .
A spring (k = 400 N/m) is hung vertically.
a) If a 5 kg mass is attached to the end of the spring and gently lowered to rest position, what will be the stretch of the spring?
b) If the 5 kg mass is attached and simply dropped, what will be the maximum velocity and the maximum stretch?
c) What will be the period of resulting oscillation?
## Similar Questions
1. ### physics
When a 4.48 kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.68 cm. How much work must an external agent do to stretch the spring 3.96 cm from its unstretched position?
2. ### physics
When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.30 cm. (a) What is the force constant of the spring?
3. ### Physics
When a 3.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.07 cm. (a) What is the force constant of the spring?
Hung vertically, a massless spring extends by 3.00 cm when a mass of 992.0 g is attached to its lower end. The same mass and spring are then placed apart on a table. The spring is fixed in place and then the mass is given a velocity …
5. ### Physics
Hung vertically, a mass less spring extends by 3.00 cm a mass of 739. G is attached to its lower end. The same mass and spring are then placed apart on a table. The spring is fixed in place and then the mass is given a velocity of …
6. ### Physics
A massless spring, with K=425 N/m, is hung vertically with the bottom end of the spring at position Y.i. A 3.50 kg mass is attached to the bottom of the spring and allowed to fall and oscillate. A) How far below Y.i does the mass fall …
7. ### Physics
When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.17 cm. (a) What is the force constant of the spring?
8. ### Physics
Consider the figure above, where a rigid beam of negligible mass and 10 m long is supported by a cable attached to a spring. When NO block is hung from the beam, the length L (cable-spring) is equal to 5 m. Assume that immediately …
9. ### Physics
One end of a spring is attached to the ceiling. The unstretched length of the spring is 10.0 cm. A 2.0 kg mass is hung from the other end of the spring. It is slowly lowered until it comes to rest. At this point the spring is 15 cm …
10. ### 1Physics
One end of a spring is attached to the ceiling. The unstretched length of the spring is 10.0 cm. A 2.0 kg mass is hung from the other end of the spring. It is slowly lowered until it comes to rest. At this point the spring is 15 cm …
More Similar Questions | 692 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.891842 |
http://www.newscientist.com/commenting/browse?id=dn18171&page=4 | 1,369,188,830,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701153213/warc/CC-MAIN-20130516104553-00007-ip-10-60-113-184.ec2.internal.warc.gz | 616,912,426 | 9,278 | #### Physics & Math
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### Other Ways To Map A Fractal
Fri Nov 20 12:36:38 GMT 2009 by Robert
Mandelbrot fractals work with only two conditions. Either the iteration stay bounded or it dosn't. In contrast, the Newton Fractal is defined as z -> z - p(z)/p'(z) and is closely related to "newton's method" for finding roots. If
p(z) = z^3 − 1 we've got three complex roots and the areas can be coloured according to which of the roots they converge to. (see http://en.wikipedia.org/wiki/File:Newtroot_1_0_0_m1.png for an example)
My point being that if we want to create "true" 3D fractals, there's a lot more to draw on than mandelbrot. Not to mention the iterated function systems!
### So What?
Fri Nov 20 19:55:23 GMT 2009 by Math Guy
I really don't understand this fascination with a 3D Mandelbrot set. It looks pretty but I can make lots of 3D fractals. So what? I don't see any application or significance to this work. This article was even mentioned in the ACM TechNews newsletter. A homework project worth of work does not equal a significant mathematical break through. This is why we have peer reviewed work in academia. As a mathematician with PhD, I am a best confused by the fixation on this "3D Mandelbrot Set."
### So What?
Sat Nov 21 12:48:28 GMT 2009 by Tom
As a mathematician with a phd you should know that the mandelbrot set is not a simple fractal, its like a fractal of fractals and contains an infinite number of different shapes and is infinitely complex. Its the only fractal I know of which has an internal area and its boundary also has an area. In fact, it seems that the mandelbrot set is a circle turned inside out where the set boundary is every point inside that circle.
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### Climate change will push up New York's heatwave deaths
18:10 21 May 2013
There will be more deaths in summer and fewer in winter as climate change begins to bite in New York City. The net effect is not good news
### Today on New Scientist
18:00 21 May 2013
All the latest stories on newscientist.com: rebuilding the LHC, killer army robots, Oklahoma tornado, sniping the easy way, and more
17:20 21 May 2013
The tornado that struck Oklahoma City in the US on Monday was unusually large and powerful, and it came down at a bad time of day
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For exclusive news and expert analysis, subscribe to New Scientist. | 746 | 2,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.938666 |
http://gregladen.com/blog/2008/02/02/the-physics-of-tatiana/ | 1,511,506,000,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807089.35/warc/CC-MAIN-20171124051000-20171124071000-00659.warc.gz | 129,844,632 | 28,321 | # The Physics of Tatiana
Tatiana was the captive Siberian Tiger who, on Christmas Day, leaped out of her cave to attack teenage boys who were taunting her. She killed one of them. Zookeepers are investigating how she did it, considering the possibility that the wall of her enclosure was not high enough (technically, it was lower than recommended height by a short distance). Tatiana’s leap has, indeed, has rekindled a long term dialog regarding zoos, and big cats in zoos in particular.Now, a physicist at Northeastern University in Boston, has produced an analysis indicating that what did happen was possible. However, I think there is a problem with the analysis.The analysis, reported here and discussed here, investigates the necessary velocity to “350 pound object over a 12.5 foot barrier that is 33 feet away.” The answer is 26.7 miles per hour at an angle of about 55 degrees above horizontal.I’ve looked over the paper and it seems fine. The analysis is approached using multiple techniques and all the techniques come up with the same result. I’m trusting their math.There are three factors that are either not fully addressed or that may be problematic.1) The tiger has to be going just under 30 miles per hour. How long of a distance does a tiger need to attain that speed, and was such a distance available? I suspect a tiger does not need too much distance. They are built as ambush hunters. In other words, they are designed to “spring,” but this is a factor that must be at least considered.2) What is the air resistance factor in this equation? One of the methods used in this analysis presumes the object in question to be, essentially, a mortar or cannon ball. But Tatiana is a big floppy furry tiger. How does this play in?I suspect that if these factors are worked into the equation, the result would be the same, but I’m just sayin…3) A witness at the scene described hearing bushes rattling around then the tiger showed up. This is important evidence that may tell us something in more detail about where the tiger was coming from, or what part of the enclosure the tiger was passing through, during this great leap.I have no problem believing that this is physically possible. This is for two reasons. One, I’ve seen some pretty amazing leaping in cats, including large cats, and I suspect the range of jumping capacities of the large cats is unmeasured but impressive. Two, it actually happened, so in my own naive way I assume it was possible.Safety (of both gusts and animals) in the case of large carnivores in zoos is important. I hope that this physical analysis is brought to bear on the investigation.
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## 12 thoughts on “The Physics of Tatiana”
1. Shaun says:
I’m no physicist, but I wouldn’t be surprised if a leaping tiger was just as or more aerodynamic than a cannonball. I suspect that a leaping tiger would flatten itself out, making for a fairly aerodynamic shape.Also, is it possible that the fur acts to improve aerodynamics like dimples on a golf-ball?
2. Shaun,You may be right about the aerodynamics of cats. But there is another problem that has occurred to me. The model seeks an optimum speed against height allowing angle to adjust itself. But I’m certain that a cat cannot take off at an arbitrary angle. THAT would require a cannon ball coming from a cannon on a mount!
3. Joshua Zelinsky says:
Gregg, first “gusts” should be “guests” in the third to last line. Now, more substantially, I don’t think the angle matter will be that serious. Cats seem in general to be very good at figuring out what angels to jump from and seem to have a lot of flexibility to do so (maybe similar to how humans are very good at doing the non-trivial task of throwing or catching a ball?).
4. When the elk exhibit was built at our zoo some years back, I noticed that the fencing around the off-exhibit paddock was higher than the fencing around the yard. The keeper explained that elk will jump higher when they are in a tighter space. If this is true, it’s probably a behavior thing rather than physics, right?
6. There’s another matter to be taken into consideration. How pissed is the cat. Adrenaline greatly improves performance, and from reports Tatiana was pissed.
7. Matt says:
Greg,My name is Matt and I am a high school biology student.I think there are most likely many more factors that played into the attack than the evidence says. Like Alan said we do not know if another animal or perhaps a trainer had upset the tiger ewarlier that day. The victims were also probably taunting the tiger. Are there any other factors you can think of?
8. KevinC says:
I remember reading a quote from the vet who performed the autopsy that included the fact that Tatiana’s claws were cracked and had cement in them. Thus she did not need the speed, the question is how far a large cat can climb up cement walls when pissed off.
9. D'oh! says:
Bound, rather than climb, I should think.My cats can jump up walls that are higher than they can jump in a single jump by inserting a vertical bound or two while their upwards momentum still holds. They use their back paws push off again–usually with extended claws to help with traction. This is consistent with the vet’s findings.(OTOH, I have to add that running on concrete when they are adrenalized can also shred my cats’ back claws. I think that they extend their back claws in an all-out run so that they work like cleats. Probably a good idea on soft soil–not so great on concrete.)It’s unclear to me how high the wall was from the bottom of the moat. But I personally wouldn’t be surprised if pissed off Tatiana leapt up the wall vertically with a bound or two.Which brings me to my next point: animals have different personalities and different athletic abilities. From what I’ve read, Tatiana was a more assertive, daring animal. It is quite possible that she attempted something most tigers would not. And succeeded, obviously.
10. carey says:
Well, the math was fine, but if a big tiger is approximately 8.5 feet in length (legs outstretched), it could stretch its claws to within 4 feet of the lip of the moat. Jumping four feet (just enough to give its claws some purchase on the lip of the moat) seems quite feasible for a pissed-off kitty.
11. natural cynic says:
The analogy with the type of calculations that go on subconsciously when we throw or catch an object successfully only occur with a lot of practice. IIRC Tatiana was captive bred and probably had little practice at the kind of bound she made. She may just have been unlucky enough to succeed.
12. Natural cynic: Good point. The systems of learning motion may be different for cats vs. primates, though.On the other hand, do we know exactly how many people each day for the last several years entered the SF zoo but never returned? Maybe this time Tatiana merely screwed up and got caught… | 1,509 | 6,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.9748 |
http://www.enotes.com/homework-help/calculate-molarity-naoh-25-0-ml-0-250-m-solution-430122 | 1,462,546,887,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861831994.45/warc/CC-MAIN-20160428164351-00019-ip-10-239-7-51.ec2.internal.warc.gz | 494,585,958 | 11,490 | # Calculate the molarity of NaOH if 25.0 mL of a 0.250 M solution of NaOh is diluted to 100.0 mL
Posted on
Dilution is the process of reducing the concentration of chemical species of a specific solution. Generally, solvent in added in the system to increase the volume thus decreasing the concentration.
The general equation for dilution can be expressed as:
`M_1V_1 = M_2V_2`
where:
`M_1 = 0.250`
`V_1 = 25.0 mL = 0.025 L`
`M_2 = ?`
`V_2 = 100 mL = 0.1 L`
Then we solve for M2:
`M_1V_1 = M_2V_2`
`(0.250)*(0.025) = M_2*(0.1)`
`M_2 = ((0.250)*(0.025))/(0.1)`
`M_2 ` = 0.625 M
Note: 1000mL = 1L
Sources: | 236 | 619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.835568 |
https://www.ssmathematics.com/2021/12/miscellaneous-ib-sl.html | 1,719,128,898,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00096.warc.gz | 863,919,681 | 42,745 | # Miscellaneous (IB SL)
$\def\iixi#1#2{\left(\begin{array}{c}#1\\#2\end{array}\right)}\let\frac=\dfrac$
1 (IB/sl/2019/November/Paper2/q2)
[Maximum mark: 5]
Consider the lines $L_{1}$ and $L_{2}$ with respective equations
$$L_{1}: y=-\frac{2}{3} x+9 \text { and } L_{2}: y=\frac{2}{5} x-\frac{19}{5} .$$
(a) Find the point of intersection of $L_{1}$ and $L_{2}$. [2]
A third line, $L_{3}$, has gradient $-\frac{3}{4}$
(b) Write down a direction vector for $L_{3}$. [1]
$L_{3}$ passes through the intersection of $L_{1}$ and $L_{2}$.
(c) Write down a vector equation for $L_{3}$. [2]
2 (IB/sl/2016/November/Paper1/q7) [Maximum mark: 7] Let $f(x)=m-\frac{1}{x}$, for $x \neq 0 .$ The line $y=x-m$ intersects the graph of $f$ in two distinct points. Find the possible values of $m$.
3 (IB/sl/2015/May/paper2tz2/q7) [Maximum mark: 8] Let $f(x)=k x^{2}+k x$ and $g(x)=x-0.8$. The graphs of $f$ and $g$ intersect at two distinct points. Find the possible values of $k$.
4 (IB/sl/2019/May/paper1tz2/q6) [Maximum mark: 7] Solve $\log _{4}(2-x)=\log _{16}(13-4 x)$.
5 (IB/sl/2018/November/Paper1/q4)
[Maximum mark: 6]
Let $b=\log _{2} a$, where $a>0$. Write down each of the following expressions in terms of $b$.
(a) $\log _{2} a^{3}$[2]
(b) $\log _{2} 8 a$[2]
(c) $\log _{k} a$[2]
6 (IB/sl/2018/May/paper1tz2/q7)
[Maximum mark: 8]
An arithmetic sequence has $u_{1}=\log _{c}(p)$ and $u_{2}=\log _{c}(p q)$, where $c>1$ and $p, q>0$.
(a) Show that $d=\log _{\mathrm{c}}(g)$.[2]
(b) Let $p=c^{2}$ and $q=c^{3}$. Find the value of $\sum_{n=1}^{20} n_{*}$.[6]
7 (IB/sl/2017/November/Paper1/q7) [Maximum mark: 7] Consider $f(x)=\log _{x}\left(6 x-3 x^{2}\right)$, for $0 < x < 2$, where $k>0$. The equation $f(x)=2$ has exactly one solution. Find the value of $k$.
8 (IB/sl/2017/May/paper1tz2/q7) [Maximum mark: 7] Solve $\log _{2}(2 \sin x)+\log _{2}(\cos x)=-1$, for $2 \pi < x < \frac{5 \pi}{2}$.
9 (IB/sl/2016/May/paper1tz2/q3)
[Maximum mark: 6]
Let $x=\ln 3$ and $y=\ln 5$. Write the following expressions in terms of $x$ and $y$.
(a) $\ln \left(\frac{5}{3}\right)$. [2]
(b) $\ln 45$. [4]
10 (IB/sl/2015/May/paper1tz1/q3)
[Maximum mark: 6]
(a) Given that $2^{\prime \prime}=8$ and $2^{n}=16$, write down the value of $m$ and of $n$. [2]
(b) Hence or otherwise solve $8^{2 x+1}=16^{2 x-3}$. [4]
11 (IB/sl/2016/May/paper2tz1/q7)
[Maximum mark: 8]
Note: One decade is 10 years
A population of rare birds, $P$, can be modelled by the equation $P_{0}=P_{3} \mathrm{e}^{\text {th }}$, where $P_{0}$ is the initial population, and $t$ is measured in decades. After one decade, it is estimated that $\frac{P_{1}}{P_{0}}=0.9$.
(a) (i) Find the value of $k$. (ii) Interpret the meaning of the value of $k$. [3]
(b) Find the least number of whole years for which $\frac{P_{t}}{P_{0}} < 0.75$. [5]
12 (IB/sl/2019/November/Paper1/q2)
[Maximum mark: 6]
In a class of 30 students, 18 are fluent in Spanish, 10 are fluent in French, and 5 are not fluent in either of these languages. The following venn diagram shows the events 'fluent in Spanish" and "fluent in French".
The values $m, n, p$ and $q$ represent numbers of students.
(a) Write down the value of $q$. [1]
(b) Find the value of $n$. [2]
(c) Write down the value of $m$ and of $p$. [3]
13 (IB/sl/2018/November/Paper2/q1)
[Maximum mark: 6]
In a group of 35 students, some take art class $(A)$ and some take music class $(M) .5$ of these students do not take either class. This information is shown in the following Venn diagram.
(a) Write down the number of students in the group who take art class. [2]
(b) One student from the group is chosen at random. Find the probability that (i) the student does not take art class: (ii) the student takes either art class or music class, but not both. [4]
14 (IB/sl/2017/May/paper1tz1/q1)
[Maximum mark: 6]
In a group of 20 girls, 13 take history and 8 take economics. Three girls take both history and economics, as shown in the following venn diagram. The values $p$ and $q$ represent numbers of girls.
(a) Find the value of (i) $p$; (ii) $q$. [4]
(b) A girl is selected at random. Find the probability that she takes economics but not history. [2]
1](a) (12,1) (b) $\iixi{-4}{3}$ (c) $r=\iixi{12}{1} +t \iixi{-4}{3}$
2] $\quad m<-1$ or $m>1$
3] $k<0.2, k>5$
4] $x=-3$
5](a) $3 b$ (b) $3+b$ (c) $\frac{b}{3}$
6](a) show (b) 610
7] $k=\sqrt{3}$
8] $x=\frac{25 \pi}{12}, \frac{29 \pi}{12}$
9] (a) $y-x$ (b) $2 x+y$
10](a) $\quad m=3, n=4$ (b) $x=7.5$
11](a)(i) $\quad k=\ln 0.9$ (ii) population is decreasing (b) 28
12](a) 5 (b) 3 (c) 15,7
13](a) $17$ (b) (i) $\frac{18}{35}$ (ii) $\frac{24}{35}$
14](a)(i) $\quad p=10$ (ii) $q=2$ (b) $\frac{1}{4}$ | 1,849 | 4,681 | {"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} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.674743 |
https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=237614 | 1,481,229,478,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542655.88/warc/CC-MAIN-20161202170902-00282-ip-10-31-129-80.ec2.internal.warc.gz | 935,458,726 | 76,410 | # Biomechanical Model Jumping.txt
### Card Set Information
Author: Anonymous ID: 237614 Filename: Biomechanical Model Jumping.txt Updated: 2013-09-29 01:15:22 Tags: asdasd Folders: Description: asdasd Show Answers:
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1. Projectile Motion Principle
"three forces influence the movement of a projectile when it is in air"
• weight (W)
• drag force()
• lift force()
• "maximum horizontal distance"
• projectile speed
time in air
"time in air"
• projectile speed
• relative projection height
• projection angle
"maximum vertical height"
• projectile speed
• projection angle
2. Sum of Joint Linear Speeds Principle
"a body's total linear speed is the result of an optimal combination of individual joint linear speeds"
3. Linear Speed - Angular Velocity Principle
An increase in linear speed (s) of a point on a rotating body component is caused by...
an increase in the body component's angular velocity (ω)
and/or
an increase in the radius of rotation ()
EQUATION: ()
4. Joint Torque Principle
an increase in joint torque () is caused by...
an increase in a muscle force () pulling on the bones that are held together at the joint
and/or
an increase in the moment arm ()
EQUATION: ()
5. Angular Inertia Principle
a decrease in a body component's angular inertia (I) is caused by...
a decrease in the body component's mass (m)
and/or
a decrease in the radius of resistance()
EQUATION: ()
6. Angular Impulse - Momentum Principle
an increase in angular velocity (ω) of a body component being rotated is caused by...
an increase in joint torque () applied to the body component
and/or
an increase in the application time (t) of the joint torque
and/or
a decrease in the body component's angular inertia (I)
EQUATION: ()
7. Action - Reaction Principle
"for any muscle to create its greatest amount of muscle force, an oppositely directed external force of equal magnitude must exist"
8. External Forces Principle
"whenever the body is in contact with the ground there are two ground reaction forces (one vertical and one horizontal) that can oppose the muscle forces created inside the body"
9. Friction Force Principle
an increase in friction force is caused by...
an increase in the coefficient of friction (μ)
and/or
an increase in the vertical ground reaction force ()
EQUATION: ()
10. Starting from Coefficient of Friction
A larger coefficient of friction will create more friction force and greater total external force to push against. Greater external force to push against allows greater ankle plantar flexion muscle force, knee extension muscle force, and hip extension muscle force to be exerted.
Greater ankle plantar flexion muscle force creates greater ankle plantar flexion joint torque and greater ankle plantar flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
Greater knee extension muscle force creates greater knee extension joint torque and greater knee extension angular velocity. Greater knee extension angular velocity creates greater linear speed of the knee and all joints superior to the knee.
Greater hip extension muscle force creates greater hip extension joint torque and greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The coordinated increase in joint linear speeds due to a larger coefficient of friction will result in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
11. Starting from Vertical Ground Reaction Force
A larger vertical ground reaction force creates greater friction force and greater total external force to push against. Greater external force to push against allows greater ankle plantar flexion muscle force, knee extension muscle force, and hip extension muscle force to be exerted.
Greater ankle plantar flexion muscle force creates greater ankle plantar flexion joint torque and greater ankle plantar flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
Greater knee extension muscle force creates greater knee extension joint torque and greater knee extension angular velocity. Greater knee extension angular velocity creates greater linear speed of the knee and all joints superior to the knee.
Greater hip extension muscle force creates greater hip extension joint torque and greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The coordinated increase in joint linear speeds due to a larger coefficient of friction will result in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
12. Starting from Muscle Force
For the ankle muscle force box:
Greater ankle plantar flexion muscle force creates greater ankle plantar flexion joint torque and greater ankle plantar flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
The increase in joint linear speeds superior to the ankle due to greater ankle plantar flexion muscle force results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the knee muscle force box:
Greater knee extension muscle force creates greater knee extension angular velocity. Greater knee extension angular velocity creates greater linear speed of the knee and all joints superior to the knee.
The increase in joint linear speeds superior to the knee due to greater knee extension muscle force results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the hip muscle force box:
Greater hip extension muscle force creates greater hip extension joint torque and greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The increase in joint linear speeds superior to the hip due to greater hip extension muscle force results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
13. Starting from Moment Arm
From the ankle moment arm box:
A longer plantar flexion moment arm at the ankle joint creates greater ankle plantar flexion joint torque and greater ankle plantar flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
The increase in joint linear speeds superior to a longer ankle plantar flexion moment arm results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the knee moment arm box:
A longer knee extension moment arm at the knee joint creates greater knee extension joint torque and greater knee extension angular velocity. Greater knee exntesion angular velocity creates greater linear speed of the knee and all joints superior to the knee.
The increase in joint linear speeds superior to the knee due to a longer knee extension moment arm results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the hip moment arm box:
A longer hip extension moment arm at the hip joint creates greater hip extension joint torque and greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The increase in joint linear speeds superior to the hip due to a longer hip extension moment arm results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
14. Starting from Mass
For the ankle mass box:
A smaller body component mass superior to the ankle results in less angular inertia (i.e, less resistance to angular motion) for the body component. This will create greater ankle planter flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
The increase in joint linear speeds superior to the ankle due to smaller body component mass superior to the ankle results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the knee mass box:
A small body component mass superior to the knee results in less angular inertia (i.e., less resistance to angular motion) for the body component. This will create greater knee extension angular velocity. Greater knee extension angular velocity creates greater linear speed of the knee and all joints superior to the knee.
The increase in joint linear speeds superior to the knee due to a smaller body component mass superior to the knee results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the hip mass box:
A smaller body component mass superior to the hip results in less angular inertia (i.e., less resistance to angular motion) for the body component. This will create greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The increase in joint linear speeds superior to the hip due to a smaller body component mass superior to the hip results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
15. Starting from Radius of Resistance
For the ankle radius of resistance box:
A shorter radius of resistance for the body component mass superior to the ankle results in less angular inertia (i.e., less resistance to angular motion) for the body component. This will create greater ankle plantar flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
The increase in joint linear speeds superior to the ankle due to a shorter radius of resistance for the body component superior to the ankle results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the knee radius of resistance box:
A shorter radius of resistance for the body component mass superior to the knee results in less angular inertia (i.e., less resistance to angular motion) for the body component. This will create greater knee extension angular velocity. Greater knee extension angular velocity creates greater linear speed of the knee and all joints superior to the knee.
The increase in joint linear speeds superior to the knee due to a shorter radius of resistance for the body component superior to the knee results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the hip radius of resistance box:
A shorter radius of resistance for the body component superior to the hip results in less angular inertia (i.e., less resistance to angular motion) for the body component. This will create greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The increase in joint linear speeds superior to the hip due to a shorter radius of resistance for the body component superior to the hip results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
16. Starting from Application Time of Joint Torque
For the ankle application time of joint torque box:
A longer application time of the ankle plantar flexion joint torque will create greater ankle plantar flexion angular velocity. Greater ankle plantar flexion angular velocity creates greater linear speed of the ankle and all joints superior to the ankle.
The increase in joint linear speeds superior to the ankle due to the longer application time of the ankle plantar flexion joint torque results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the knee application time of joint torque box:
A longer application time of the knee extension joint torque will create greater knee extension angular velocity. Greater knee extension angular velocity creates greater linear speed of the knee and all joints superior to the knee.
The increase in joint linear speeds superior to the knee due to the longer application time of the knee extension joint torque results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the hip application time of joint torque box:
A longer application time of the hip etension joint torque will create greater hip extension angular velocity. Greater hip extension angular velocity creates greater linear speed of the hip and all joints superior to the hip.
The increase in joint linear speeds superior to the hip due to the longer application time of the hip extension joint torque results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
17. Starting from Radius of Rotation
For the ankle radius of rotation box:
A longer radius of rotation for the body component superior to the ankle joint creates greater linear speed of the ankle and all joints superior to the ankle.
The increase in joint linear speeds superior to the ankle due to the longer radius of rotation for the body component superior to the ankle results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the knee radius of resistance box:
A longer radius of rotation for the body component superior to the knee joint creates greater linear speed of the knee and all joints superior to the knee.
The increase in joint linear speeds superior to the knee due to the longer radius of rotation for the body component superior to the knee results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
For the hip radius of rotation box:
A longer radius of rotation for the body component superior to the hip joint creates greater linear speed of the hip and all joints superior to the hip.
The increase in joint linear speeds superior to the hip due to the longer radius of rotation for the body component superior to the hip results in greater linear speed for the jumper when leaving the ground. This will lead to greater time in the air and greater vertical jump height and/or greater horizontal jump distance.
What would you like to do?
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# ChaosGame
Plot iterations for the 2D chaos game
Contributed by: Eric Weisstein
ResourceFunction["ChaosGame"][n,k,d] gives k iterations of the 2D chaos game on a regular n-gon with iter points using a ratio of 1/2, dropping the first d points. ResourceFunction["ChaosGame"][{n,r},k,d] uses a ratio r.
## Details
In a chaos game, one picks a point at random inside a regular n-gon, and then draws the next point a fraction r of the distance between it and a polygon vertex picked at random. The process is continued for k iterations (after throwing out the first d points).
## Examples
### Basic Examples (1)
Get three iterations:
In[1]:=
Out[1]=
### Scope (2)
Try several geometries:
In[2]:=
Out[2]=
Use more iterations:
In[3]:=
Out[3]=
Increase ratios:
In[4]:=
Out[4]=
### Neat Examples (1)
An animation of the process:
In[5]:=
In[6]:=
Out[6]=
## Requirements
Wolfram Language 11.3 (March 2018) or above | 271 | 968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | longest | en | 0.695036 |
https://studylib.net/doc/11982509/anthro-260--essentials-of-biological-anthropology-name--_... | 1,653,779,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00525.warc.gz | 606,355,426 | 11,424 | # Anthro 260: Essentials of biological Anthropology Name: ____________________ Assignment 3
```Anthro 260: Essentials of biological Anthropology
Assignment 3
Due: 9/30
Name: ____________________
1. Recessive allele l codes for lactose intolerance. People who are recessive homozygots (ll) can not
digest milk; dominant homozygots (LL) and heterozygotes (Ll) have no problem digesting milk.
Among 200 people from Eastern Mongolia, 18 were unable to digest milk.
Assuming that the population does not deviate from the Hardy-Weinberg equilibrium, estimate the
frequency of each allele and each genotype:
fr(l) = _______________
fr(L) = __________
fr(ll) = _______________
fr(Ll) = ____________
fr(LL) = _____________
2. Flying in a helicopter over Siberia you observed 400 bears. Of those 400 bears, 20 were black
(genotype EE), 40 were dark brown (genotype EB), and 340 were light brown (genotype BB).
Find the frequency of each allele in this population:
fr(E) =
fr(B) =
_
If this population were in Hardy-Weinberg equilibrium, what genotype frequencies would be
expected for these allele frequencies? How many individuals are expected to have each genotype
according to the Hardy-Weinberg equilibrium?
fr(EE)E =
fr(EB)E =
fr(BB)E =
#EEE = ____________
#EBE = _____________
#BBE = _____________
Does this population deviate from the Hardy-Weinberg equilibrium, and, if so, why?
____________________________________________________________________________
3. Baldness is inherited as an autosomal dominant trait in males and as an autosomal recessive trait
in females. In New York, 36% of adult males are bald.
Assuming that the New York population does not deviate significantly from the Hardy-Weinberg
equilibrium, estimate the frequency of bald women. Explain your answer and justify it by showing | 439 | 1,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.866243 |
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# The number of antelope in a certain herd increases every
Author Message
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Joined: 13 May 2009
Posts: 19
The number of antelope in a certain herd increases every [#permalink]
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09 Aug 2009, 09:50
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The number of antelope in a certain herd increases every year at a constant rate. If there are 500 antelope in the herd today, how many years will it take for the number of antelope to double?
(1) Ten years from now, there will be more than ten times the current number of antelope in the herd.
(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years.
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### Show Tags
09 Aug 2009, 12:09
1
IMO B
if r is the groth rate. after 1 years we will have 500(1+r) and after n years, we will have 500(1+r)^n
the question asks, for which n will we have 1000? (double 500)
to find n we need to know r
statement 1: for n=10, tne number will be graeter than 500x10, we will just have a min for r . INSUFF
Statement 2: if we considered a groth rate p=2r, we would have 980=500(1+P)². we can solve this to find p, then r=p/2 and plug this value in the first eq to find n. SUFF
Senior Manager
Joined: 17 Mar 2009
Posts: 264
### Show Tags
09 Aug 2009, 18:19
IMO B
if r is the groth rate. after 1 years we will have 500(1+r) and after n years, we will have 500(1+r)^n
can you explain how it is (1+r) after 1year ?
Intern
Joined: 10 Jul 2009
Posts: 44
### Show Tags
09 Aug 2009, 22:02
crejoc wrote:
IMO B
if r is the groth rate. after 1 years we will have 500(1+r) and after n years, we will have 500(1+r)^n
can you explain how it is (1+r) after 1year ?
This is the same as a compound interest formula: A=P(1+R)^n
A-total amount
P-initial amount
R-rate
N-number of years
hope it helps
Manager
Joined: 10 Jul 2009
Posts: 147
### Show Tags
10 Aug 2009, 10:54
Answer is B and number of years after which the herd doubles = log2/log(6/5)
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Re: Good One! &nbs [#permalink] 10 Aug 2009, 10:54
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• moriheru
In summary, the conversation is about a simple differential equation for the zeroth wavefunction of a harmonic oscillator. The person asking the question has a solution but it differs by a constant and they are seeking clarification on their mistake. They provide the equation, which is missing a psi function, and mention that they will edit the Latex for it. They also specify that their solution is different from the one in the book and are asking for help in identifying their error.
moriheru
My question concerns a really simple differential equation for the zeroth wavefunction of a harmonic oscillator.
I have pretty much got it but my solution just differs by a constant,so I thought why think when one can ask other people :). Here is the equation:
Where the x star represent a variable involving mass frequency and Plancks constant. There is a psi function missing in the second equation (for any latex tips please e-mail me)
(Posted to early so there was no equation on the original post;had to edit it)
Last edited:
Sorry Latex is wrong change it in a minute. So here it is.
Last edited:
Is should specify: the above is my solution and my question is what I did wrong as my book has a different solution.
Thanks for any help.
## What is a quantum harmonic oscillator?
A quantum harmonic oscillator is a physical system that exhibits the properties of both classical and quantum mechanics. It consists of a particle bound to a potential well and oscillating back and forth with a specific frequency.
## What is the DE wavefunction for a quantum harmonic oscillator?
The DE wavefunction for a quantum harmonic oscillator is a mathematical representation of the probability distribution of the particle's position and momentum at any given time. It is described by the Schrödinger equation and takes the form of a Gaussian or bell-shaped curve.
## How does the DE wavefunction change over time?
The DE wavefunction evolves over time according to the Schrödinger equation, which takes into account the potential energy of the particle and its initial conditions. As time goes on, the wavefunction will spread out and become more spread out, representing a larger range of possible positions and momentums for the particle.
## What is the significance of the energy levels in a quantum harmonic oscillator?
The energy levels in a quantum harmonic oscillator correspond to the allowed energies of the particle within the potential well. These levels are quantized, meaning they can only take on discrete values and are dependent on the frequency of the oscillation. This plays a crucial role in understanding the behavior of the system and its interactions with other particles.
## How does the quantum harmonic oscillator relate to real-world applications?
The quantum harmonic oscillator has many real-world applications, particularly in physics and engineering. It is used to model systems such as atoms, molecules, and even electronic circuits. It also has important applications in quantum computing and in understanding the behavior of materials at a microscopic level.
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https://encyclopediaofmath.org/index.php?title=Parametric_representation_of_univalent_functions&oldid=55014 | 1,716,501,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00333.warc.gz | 185,665,048 | 8,351 | # Parametric representation of univalent functions
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A representation of univalent functions (cf. Univalent function) that effect a conformal mapping of planar domains onto domains of a canonical form (for example, a disc with concentric slits); it usually arises in the following manner. One chooses a one-parameter family of domains $Q _ {t}$, $0 \leq t < T$, included in one another, $Q _ {t ^ \prime } \subset Q _ {t}$, $0 \leq t ^ \prime < t < T$. For $Q _ {0}$ one assumes known a conformal mapping $f _ {0}$ onto some canonical domain $B _ {0}$. From a known mapping $f _ {t}$ of $Q _ {t}$ onto a domain of canonical form one constructs such a mapping $f _ {t+ \epsilon }$ for $Q _ {t+ \epsilon }$, where $\epsilon > 0$ is small. Under a continuous change of the parameter $t$ there arise in this way differential equations. The best known of these are the Löwner equation and the Löwner–Kufarev equation. In the discrete case — for lattice domains $Q _ {t}$ and a natural number $t$— the transition from $f _ {t}$ to $f _ {t + \epsilon }$, $\epsilon = 1$, is effected by recurrence formulas. The source of these formulas and equations is usually the Schwarz formula (see [1]) and its generalizations (see [2]). An equally important source of parametric representations are the Hadamard variations (see [3], [4]) for the Green functions $G _ {t} ( z, z ^ \prime )$, $z, z ^ \prime \in Q _ {t}$, of the family of domains mentioned above. For elliptic differential equations Hadamard's method is also called the method of invariant imbedding (see [5]). Below, the connection between parametric representations, Hadamard variations and invariant imbedding are exhibited in the simplest (discrete) case.
Suppose that $Q$ is a collection of complex integers (a lattice domain) and that the Green function $G _ {z} ( z, z ^ \prime )$ is an extremal of the Dirichlet–Douglas functional
$$I _ {t} ( g) = 2g( z ^ \prime ) + \sum_{k=0}^ { l } \sum _ {z \in Q _ {0} } \rho _ {k} ( t) | \nabla _ {k} g( z) | ^ {2}$$
in the class $R _ {0}$ of all real-valued functions $g( z)$ on $Q$. Here $Q _ {0} = \{ {z } : {z, z- 1, z- i, z- 1- i \in Q } \}$,
$$\tag{1 } \left . \begin{array}{c} {\nabla _ {0} g( z) = g( z) - g( z- 1- i), } \\ {\nabla _ {1} g( z) = g( z- 1)- g( z- i), } \end{array} \right \}$$
$$\rho _ {k} ( 0) \equiv 1,\ \rho _ {k} ( t+ 1) = \rho _ {k} ( t) + N \delta _ {\zeta _ {t} } ,$$
$N$ is a natural number, $\delta _ {\zeta _ {t} }$ is the Kronecker symbol, and $\zeta _ {t} = ( k _ {t} , z _ {t} )$, $t = 0 \dots T- 1$, is a certain collection of pairs of numbers; $\{ {z _ {t} } : {t = 1 \dots T } \}$ is the boundary of $Q _ {t}$, and $k _ {t} = 0$ or 1. To find an extremum of the functional $I _ {t} ( g)$ is a problem of quadratic programming. A comparison of its solutions for $t$ and $t+ 1$ gives the basic formula of invariant imbedding (Hadamard variation):
$$\tag{2 } G _ {t+1} ( z, z ^ \prime ) = G _ {t} ( z, z ^ \prime ) - \frac{1}{c _ {t} } \nabla _ {k _ {t} } G _ {t} ( z _ {t} , z) \nabla _ {k _ {t} } G _ {t} ( z _ {t} , z ^ \prime ),$$
where $c _ {t} = N ^ {-1} - \nabla _ {k _ {t} } \nabla _ {k _ {t} } ^ \prime G _ {t} ( z _ {t} , z _ {t} ) > 0$ and the symbol $\nabla _ {k} ^ \prime$ denotes the difference operators (1) in the second argument of the Green function. Knowing $G _ {0} ( z, z ^ \prime )$ one can obtain step-by-step (recurrently) from (2) all the functions $G _ {t} ( z, z ^ \prime )$, $t = 1 \dots T$. By constructing the Green function, one obtains from the lattice analytic function $f _ {T} ( z) = G _ {T} ( z, z ^ \prime ) + iH _ {T} ( z, z ^ \prime )$ according to the equation of Cauchy–Riemann type
$$(- 1) ^ {k} \nabla _ {1-k} H = \rho _ {k} \nabla _ {k} G,$$
a univalent lattice quasi-conformal mapping $w = \mathop{\rm exp} [ 2 \pi f( z)]$ of $Q _ {t}$ into the unit disc. Closest to the origin of coordinates is the image of $z ^ \prime$. In the limit, as $n \rightarrow \infty$, the mapping is lattice conformal and the image of $Q _ {T}$ is a disc with concentric slits. The result is a continuous analogue of (2) (see [6]). When all the domains $G _ {t}$ are simply connected and the canonical domain is the unit disc $B$, one succeeds by using a fractional-linear automorphism of $B$ to represent the Green function in the explicit form
$$G _ {t} ( z, z ^ \prime ) = \mathop{\rm ln} | 1- f _ {t} ( z) \overline{ {f _ {t} ( z ^ \prime ) }}\; | - \mathop{\rm ln} | f _ {t} ( z) - f _ {t} ( z ^ \prime ) |$$
in terms of the function $f _ {t} ( z)$ mapping $Q _ {t}$ onto $B$ with the normalization $f( 0) = 0$, $0 \in Q _ {t}$ for all $t \in [ 0, T)$.
In terms of $w = f _ {t} ( z)$ the Hadamard variation reduces to an ordinary (Löwner) differential equation. In comparison with the Hadamard variation this equation is considerably simpler; however, information on the boundary of $Q _ {t}$ is only implicit in it — in terms of the control parameter $\alpha ( t) = \mathop{\rm arg} f _ {t} ( z _ {t} )$, because $f _ {t} ( z _ {t} )$ is not known beforehand. Nevertheless, the Löwner equation is a basic instrument in the parametric representation.
More general one-parameter families of domains $Q _ {t}$, $0 \leq t < T$, not necessarily imbedded in one another, have also been treated (see [7]). The equations arising in such parametric representations are called Löwner–Kufarev equations. There is also a modification of the Löwner and Löwner–Kufarev equations to the case when the domains $Q _ {t}$ have a different kind of symmetry or other geometric peculiarities (see [1]).
#### References
[1] G.M. Goluzin, "Geometric theory of functions of a complex variable" , Transl. Math. Monogr. , 26 , Amer. Math. Soc. (1969) (Translated from Russian) [2] I.A. Aleksandrov, A.S. Sorokin, "The problem of Schwarz for multiply connected circular domains" Sib. Math. J. , 13 : 5 (1972) pp. 671–692 Sibirsk. Mat. Zh. , 13 : 5 (1972) pp. 971–1000 [3] J. Hadamard, "Mémoire sur le problème d'analyse relatif à l'équilibre des plaques élastiques encastrées" , Œuvres , 2 , CNRS (1968) pp. 515–642 [4] J. Hadamard, "Leçons sur le calcul des variations" , 1 , Hermann (1910) [5] R. Bellma, E. Angel, "Dynamic programming and partial differential equations" , Acad. Press (1972) [6] V.I. Popov, "Quantization of control systems" Soviet Math. Dokl. , 13 : 6 (1972) pp. 1668–1672 Dokl. Akad. Nauk. SSSR , 207 : 5 (1972) pp. 1048–1050 [7] P.P. Kufarev, "On one-parameter families of analytic functions" Mat. Sb. , 13 : 1 (1943) pp. 87–118 (In Russian) (English abstract) [a1] P.L. Duren, "Univalent functions" , Springer (1983) pp. Sect. 10.11
How to Cite This Entry:
Parametric representation of univalent functions. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Parametric_representation_of_univalent_functions&oldid=55014
This article was adapted from an original article by V.I. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 2,351 | 7,060 | {"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} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.850836 |
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posted by .
Traveling at a speed of 14.3 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.730. What is the speed of the automobile after 1.15 s have elapsed? Ignore the effects of air resistance.
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Traveling at a speed of 19.0 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.660. What is the speed of the automobile after …
8. ### physics
Traveling at a speed of 16.5 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.590. What is the speed of the automobile after …
9. ### physics
Traveling at a speed of 18.0 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.580. What is the speed of the automobile …
10. ### physics
Traveling at a speed of 18.0 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.690. What is the speed of the automobile after …
More Similar Questions | 667 | 2,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.929992 |
https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_p%5E2_for_Prime_Greater_than_3 | 1,576,102,766,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00390.warc.gz | 514,315,053 | 9,988 | # Numerator of p-1th Harmonic Number is Divisible by p^2 for Prime Greater than 3
## Theorem
Let $p$ be a prime number such that $p > 3$.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p^2$. | 83 | 263 | {"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} | 2.640625 | 3 | CC-MAIN-2019-51 | longest | en | 0.857224 |
https://ch.mathworks.com/matlabcentral/fileexchange/69708-2d-elasticity-q4-finite-element-solver?s_tid=srchtitle | 1,618,484,849,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00555.warc.gz | 260,024,591 | 19,486 | File Exchange
## 2D Elasticity - Q4 Finite Element Solver
version 1.0.0 (3.08 MB) by
A FEM code for solving linear, static, 2D plane stress elasticity problems using Q4 elements.
Updated 13 Dec 2018
A finite element analysis computer program for solving linear, static, 2D plane stress elasticity problems using isoparametric elements with 4 nodes per element is developed. The loads are restricted with 2D point forces. The boundary conditions are restricted to homogeneous displacement boundary conditions to be applied on nodes.
The program mainly focuses on the ‘solver’ part of any modern Finite Element Package, however, some pre-processing and post-processing utility programs, simply for basic mesh generation, or displaying resulting stress, strain or displacement fields are also developed as required.
All inputs and physical calculations are performed for metric units. All the results presented are also in metric unis, unless otherwise stated.
- Positions, Displacements: meter (m)
- Forces: Newton (N)
- Pressures, Stresses, Young Modulus: Pascal (Pa)
### Cite As
Ethem H. Orhan (2021). 2D Elasticity - Q4 Finite Element Solver (https://www.mathworks.com/matlabcentral/fileexchange/69708-2d-elasticity-q4-finite-element-solver), MATLAB Central File Exchange. Retrieved .
Ethem H. Orhan
You can apply multiple pointwise forces at any location, at any direction. In 'Forces.dat', set 5th figure (y_force) to zero (as in the following) for x-only applied force. Distributed forces are not supported in this package.
-----
force_number x_pos y_pos x_force y_force
1 0.0 50.0 50.0 0.0
2 50.0 50.0 50.0 0.0
Chunhui Wang
In addition, I just want make the forces along y directions are 0 in the left and right edges, and it doesn't specify the force in the x direction at the same position. What should I do in your program?
Ethem H. Orhan
The problem is set up for unit thickness, it is not adjustable.
David José
Hey bro, couldn't find where to set the problem thickness. Otherwise, great job.
Kavan Khaledi
Good Job! Thank you:)
##### MATLAB Release Compatibility
Created with R2018b
Compatible with any release
##### Platform Compatibility
Windows macOS Linux | 531 | 2,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.869494 |
https://www.varsitytutors.com/hotmath/hotmath_help/topics/area-problem-solving.html | 1,657,103,855,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00742.warc.gz | 1,130,586,537 | 26,133 | # Area Problem Solving
Sometimes you can use the simple area formulas for rectangles , triangles , and circles in combination to find the areas of more complicated shapes.
Example 1:
What is the area of the shaded region?
The area of the shaded region is the difference between the areas of the rectangle and the square.
The dimensions of the rectangle are $6$ by $8$ and so the area is $6$ times $8$ , that is $48$ square units.
Each side of the square measures $2$ units and so the area is $2$ times $2$ , that is $4$ square units.
Therefore, the area of the shaded region is $48-4=44$ square units.
Example 2:
Find the total area of the shaded region. (In the figure, all angles are right angles.)
The given figure can be divided into $3$ squares and a rectangle as shown.
The total area is the sum of the areas of the rectangle and the squares.
The dimensions of the rectangle are $3$ by $12$ and so the area is $3$ times $12$ , that is $36$ square units.
All the three squares have each side $3$ units long and so the area of each of them is $3$ times $3$ , that is $9$ square units.
Therefore, the total area is $36+\left(3×9\right)=63$ square units.
Example 3:
If $\Delta ABC$ is a right triangle and $\stackrel{⌢}{BC}$ is a semicircle, find the total area of the figure.
The total area is the sum of the areas of the triangle and the semicircle.
We have the length of a side and that of the altitude to the side of the triangle. So, the area of the triangle is
Since $\stackrel{⌢}{BC}$ is a semicircle, $\stackrel{¯}{BC}$ is a diameter. But $\stackrel{¯}{BC}$ is also the hypotenuse of the right triangle $ABC$ .
Use the Pythagorean Theorem to find the length $BC$ .
$\begin{array}{l}BC=\sqrt{{3}^{2}+{4}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{9+16}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{25}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\end{array}$
So, the diameter of the semicircle is $5$ units and so the radius is $2.5$ units.
The area of a semicircle of radius $2.5$ units is .
Therefore, the total area is about . | 842 | 2,551 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2022-27 | latest | en | 0.847293 |
www.stonebridgemanor.com | 1,586,182,740,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371637684.76/warc/CC-MAIN-20200406133533-20200406164033-00096.warc.gz | 1,176,620,823 | 19,005 | There are many tips and tricks to saving major bucks on your wedding. From DIY to cutting the guest list in half, it’s an endless game of number crushing. In today’s society with social media, blogs, and wedding magazines, the total costs have hit an all time high!
Wouldn’t you love to use the \$6,000 budget that your parents did at their wedding? Us too!
But that number is unrealistic and unheard of in 2018. Wedding Wire recently wrote a blog about venue prices and stated that the average price of a wedding venue alone runs \$8,798. Catering comes at the second highest expense to your budget of around 40% of your total. So let’s get down to brass tacks here….
We know it can get confusing as to what it actually costs to have a wedding and we’re here to tell you there’s no simple answer. Asking “how much does it cost to have a wedding?” is like asking “how much is it to go on vacation?”….do you want to camp for a couple nights in the Grand Canyon, or do you want to fly to the Four Seasons Resort in the Maldives? We love camping and the Maldives and can help you get to either destination; as long as you tell us which direction you’re headed.
There are many variables to this question as no wedding is the same. It varies from state to state. There are some couples who spend a great deal less or more than the average. Wedding Wire held a survey with 13,000 couples who got married in 2016 and the average total wedding cost (dress, hair, catering, venue, photography, florals, etc.) came to \$35,329. Now don’t forget to breathe as we discuss dollars!
Let’s talk averages for a minute. In addition to the Wedding Wire survey, the average number of guests at a wedding was 141. Your guest count has a lot to do with the cost. More guests = More money. So lets’s say you have a budget of \$20,000 and your guest count is 200. Put 40% of that budget towards catering and \$8,000 towards the venue. That puts you with \$4,000 for your dress, flowers, photography, etc.
Let’s say you find a heck of a deal and spend 30% of your budget on food and a venue for \$5,000, leaving you with \$9,000 for the additions which sounds like the way to go, right? Wrong! Now what if your discounted venue doesn’t come with basic essentials such as tables, chairs, and set up?Nor does it include the caterer (since they’re giving you a deal) thus your venue isn’t providing service staff, food delivery, or china? You guessed it, more stress (more time planning and researching these items) and more money spent to cover the essentials. Whenever there’s a “discount” on a price, there’s a take away in product or service. The phrase “less is more” is a perfect example of cutting corners where you shouldn’t and could result in spending more for lack of services. Booking a cheaper product at the sacrifice of quality is always remembered (the price in not remembered but the lack of quality is).
📷 Brinkerhoff Photography
These days, more couples are paying for their wedding out of pocket without contributors, like parents. This can result in a lower than average budget which is totally fine! Please don’t ever feel embarrassed to share your budget. We see budgets of all shapes and sizes. No judgement here! We are honored you’re considering us to be apart of the biggest day of your life and have your best interest at heart! However, if you have a lower than average budget, consider these three things:
1. Be reasonable! Saturdays are prime days and the most expensive. If you’re stuck on a Saturday event and have a budget of \$10,000 with the national venue average being \$8,000, adjust your budget or consider another day. Weekday weddings are the new rave and tend to be much less. And trust me when I say people WILL take the day off from work to attend your wedding….they want a reason to play hookie!
2. Be realistic! Putting together your guest list is major and can make or break your budget. My favorite example of this is from the movie Great Gatsby. Those parties were so extravagant and lively with a full house of people drinking, dancing, and swinging from chandeliers. Time seemed of no essence…money too! If you’re planning on 200 guests and have a less than average budget, you need to reevaluate how realistic your numbers are. Unless Chipotle* is catering your wedding, cut that guest to an affordable size! (*we love chipotle, but the extra guac can wait).
3. Be smart! When planning your wedding, think of the top 3 things that are most important to you. Is it your dress? The food? Photos? The venue? Being stress-free? Budget out those item and compare booking al a cart services to all inclusive (we call this Full Service). With a Full Service wedding venue, you’ll end up saving more and worrying less. All of the heavy lifting, paperwork, and strategic planning is taken care of. Best part is, you get to focus on the fun stuff like choosing your colors, enjoying a cake tasting, and writing your vows!
Stonebridge Manor is a full service venue with customizable options for couples to express their uniqueness. Passion comes first, as we strive to give you the wedding of your dreams. So when it comes to cost or budget, think about this….weddings are not about spending the least amount of money, or the most amount of money. They are about spending good money on the elements that are most important to you and your partner. Lastly, we know there are thousands of venues to scout and you may feel that you need to do an endless search. In researching the venue online beforehand, you already know what you want before touring. So once you find it, book it! Even if it’s the fist place you’ve seen. Why keep looking when your must haves meet your budget? Like falling in love, when you know, you know!
By: Brittany Fitch
Edited By: Christy Wilson | 1,293 | 5,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.960926 |
https://iwant2study.org/taskmeisterx/index.php/interactive-resources/physics/02-newtonian-mechanics/08-gravity/53-gravity02-1 | 1,674,973,364,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00046.warc.gz | 335,997,023 | 11,446 | gravity02_1
### 7.1.5 Example (N09/I/16) Binary Stars
Two stars of mass M and 2M, at a distance 3x apart, rotate in circles about their common centre of mass O.
The gravitational force acting on the stars can be written as $\genfrac{}{}{0.1ex}{}{kG{M}^{2}}{{x}^{2}}$ .
What is the value of k?
A 0.22 B 0.50 C 0.67 D 2.0
### 7.1.5.1 Model: You may view the movement of the two stars using the EJSS below link (require internet) or just below:
Answer is $|F|=G\genfrac{}{}{0.1ex}{}{{M}_{1}{M}_{2}}{{r}^{2}}=G\genfrac{}{}{0.1ex}{}{2MM}{{\left(3x\right)}^{2}}=\genfrac{}{}{0.1ex}{}{2}{9}G\genfrac{}{}{0.1ex}{}{{M}^{2}}{{\left(x\right)}^{2}}$
### Translations
Code Language Translator Run
### Software Requirements
SoftwareRequirements
Android iOS Windows MacOS with best with Chrome Chrome Chrome Chrome support full-screen? Yes. Chrome/Opera No. Firefox/ Samsung Internet Not yet Yes Yes cannot work on some mobile browser that don't understand JavaScript such as..... cannot work on Internet Explorer 9 and below
### Credits
Fu-Kwun Hwang - Dept. of Physics,National Taiwan normal Univ.; Loo Kang Wee; Wolfgang Christian
### end faq
http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejss_model_gravity02_1/gravity02_1_Simulation.xhtml
# Testimonials (0)
There are no testimonials available for viewing. Login to deploy the article and be the first to submit your review! | 452 | 1,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-06 | longest | en | 0.608065 |
http://lotterywinnerstories.com/Lotto/texas-lottery/ | 1,516,241,553,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887054.15/warc/CC-MAIN-20180118012249-20180118032249-00573.warc.gz | 195,271,718 | 27,280 | ## Is There a Way To Win The Lottery? Texas Lottery
How Do You Select The Best Numbers for Playing Texas Lottery lottery?
A lottery is a game of chance sponsored by many states and countries. Winning the Texas Lottery lottery game can be done easily by using the right strategy and system. Richard Lustig used his system to win several lottery jackpots and millions in prize money proving that luck has nothing to do with winning.
A lot of people use algorithm to analyze and predict lottery results for Texas Lottery. Analysis of Groups There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.Analysis of Hot-Cold Trend This algorithm analysis is one of the most favorite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings. Analysis of Repetition Pattern A lot of lottery players share the same opinion that repetition is quite important to predict the winning lottery numbers as most of jackpots will appear again in the future.
## Interesting Facts About Texas Lottery Lottery
Hospital lotteries are becoming increasingly popular in Canada. That's because, not only do you get a chance to win many terrific prizes, but you also get to donate money to a worthy enterprise - The hospital. And, the money goes to worthy hospital causes, like cancer research. But, of course, the lottery has to sell tickets and many of them do it by claiming to have the best chance to win a million dollars. That means, according to them, that if you are looking for the best odds to win a million dollars, you should buy a hospital lottery ticket rather than a regular lottery ticket.
So there you have it, playing a hospital lottery is not your best chance at winning a million dollars. However, the money that you spend on tickets goes to a good cause and so it is still worth it to play.
## How to Win Lottery by Using Analysis Algorithms for Lottery Prediction
I'm going to guess that you want to win the lottery, right? Sure you do, that's why you're reading this article. You probably also want to know which state lotteries have the best odds of winning. If that is so, then you've come to the right place. This article will point you in the right direction when choosing which state lottery to play.
Before I get into the state lotteries, we need to look at lotteries and odds, in general. Generally, lotteries that have high jackpot payouts tend to be harder to win - The odds are worse.
With that being said, you first need to ask yourself how much money you want to win. Most people play lotteries to win a life-changing amount of money. By life-changing, I mean that it could literally change your life for the better, be it living debt-free, being able to quit your job, having the ability to buy things like new cars and nice a house, or being able to travel whenever you want.
Pennsylvania Match 6: The Pennsylvania Match 6 jackpot often goes above the \$2 million mark. The odds of winning are approximately 1-in-4.7-million. Match 6 is about 20 times easier to win than Powerball.
So, there you have it - The three state lotteries that have the best odds of winning. If you live in these states, you're lucky because they offer the best lotto games in the country. If you don't live in any of these three states, take a look at your states games and lottery odds before playing. You could often find the odds on their websites. Choose a game that has decent odds when compared to other games that are offered.
## Learn How To Win The Lottery Jackpot Powerball
Playing the Powerball lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Powerball lottery?
Are you constantly thinking and saying, "I want to win the lottery?" Do you ask yourself, "Will I win the lottery?" If you think about winning the lottery all the time, you need to know this secret to winning the secret to winning the lottery.If you think winning the big bucks is just about picking the right lottery numbers, you're wrong. Yes, you do need to pick the right numbers, and it's best to have a system to win the lottery. Doing things like reading a lottery book can be useful too. But none of this will work for you if you don't do this one essential thing:The universe rewards those who are prepared. Did you know that?When you act as if you expect something, it's going to come to you. Did you know that?4. Make a list of people to whom you plan to give money and decide on how much you'll give. Take into account gift taxes when you do this. Be careful about who you tell, "I'll give you money when I win." You don't want to give away all your winnings.Combine this lottery win preparation method with a lottery winning system, and you will be able to answer "Yes," to "Will I win the lottery?" You'll be able to turn "I want to win the lottery," to "I WON the lottery."
## Pick 3 Lottery Workout Systems That Work
Winning the Powerball lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## Best Lottery Winning System Mega Millions
Playing the Mega Millions lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Mega Millions lottery?
Because you are sitting in front of your computer viewing this page, chances are, you're an eager and regular lotto player who has yet to win lotto prizes. Lottery cheats are some of the most researched items or articles here on the Internet, and it's easy to understand why: lottery games from all states can be addictive, attracting countless players from all across the country to play the lotto on a regular basis. What you need to understand is that while playing the lottery can be fun and exciting, the real treat lies in winning the lotto prizes - including the jackpot. And unlike what some lotto players believe, you can win the lotto using lottery cheats.
Lottery cheats are not cheats in the real sense of the word. They are not illegal and won't put you into some kind of federal trouble. In fact, lottery cheats have long been recognized by serious lottery players as the reason why they have higher chances of winning. Cheats at winning the lottery are actually guides on how to win any type of lotto game you choose to play. They offer sensible pieces of advice to increase the likelihood of holding a winning ticket.
As far as lottery cheats are concerned, statistics experts highly advise against forming lottery combinations in a mathematical sequence and playing patterns on lottery tickets. These acts are sure to reduce your chances of winning because mathematical sequences and patterns are almost never taken into consideration in lottery games. Past winning results show a tendency towards random combinations. If you use a proven system that can efficiently analyze lottery data, such as past winning numbers, trends, and angles, you are more likely to bring home the jackpot prize not just once but as many times as you please.
## Five Secrets To Winning The Lottery: How To Pick Megamillions Numbers
When it comes to megamillions lottery, there are a number of secrets that are unveiled. Sometimes, these secrets are facts that only players have ideas about. In this article, there are secrets and facts that will help a player that aims to win the jackpot prize.One secret of megamillions lottery is that it usually draws many odd numbers than even numbers. Of course, this is based on studies and analysis done by lottery experts. Then knowing this secret, one can favor odd numbers to increase the chance of winning.Another secret in megamillions lottery are the strategies that can be used to be able to have a greater chance of winning. The odd that make this lottery possible is 1 in 176 million. There are two strategies that can be used. First, play in groups. The chance of winning is greater since there many tickets that are purchased. Also, one of the reasons behind to not playing on own is due to the too long odds. It is best to let all other players go for the jackpot and play in groups. The second strategy is to play once in group in a year but invest the most. It is said that the opportunity comes only once that is why it is important to be careful too.Further, it has been found out that almost all of the players of megamillions lottery choose the quick pick method in choosing their numbers. This is the method when the machine chooses the numbers in random and gives them to the player. The rest choose to personally choose their numbers. Maybe, they use their birthdates and anniversary dates which is a common method in choosing the numbers.In addition, it has been found out that the odds of a set of numbers repeating themselves have a very nil probability. Then, knowing this secret might tell the gamer in advance not to use repeating numbers. So it is important to know if the numbers in hand have been played already or not. Megamillions lottery's result in the internet might help the player to determine which numbers have been played before or remains not picked.On the other hand, there are also myths that must be known so as to decide whether to play lottery or not. One myth is that the lotteries take advantage of the poorer people. This is said to be a myth since the people who usually plays the lottery are those who have extra cash in their pockets for leisure. In fact, a great percentage of players are rich people. Another myth is that the lottery is a form of tax. It is said that no one is required to join the lottery. And when the player wins, this is the only times that he/she is asked to pay taxes. The reason behind this is that the jackpot prize is also a form of income and any income is taxed by the government. Therefore, given these secrets mentioned, the player will have the right information in order to increase the chances of winning.
Winning the Mega Millions lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## Richard Lustig Lottery Winning Formula Lotto Texas
Playing the Lotto Texas lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Lotto Texas lottery?
A lottery is a popular form of gambling which involves the drawing of lots for a prize. There are some states where lottery is forbidden but others endorse it to the extend of organizing a national lottery. Lotteries date back to the period of Romans. And it was common for emperors to give their dinner party guests gifts like slaves to lavish villas after drawings.
Two common lottery myths
Lottery is tax
Lottery take advantage of the poor economic strata of our society
Lottery facts
In China lottery was first played to fund the Great Wall
U.S. lotteries helped fund the Colonial Army in the Revolutionary War
At one point of time many churches and universities such as Harvard, Yale and Princeton were funded in part by lotteries
Where does the Lottery revenue expenditure go?
A good amount of lottery earnings is spend on economic development of the country
Steps are taken to improve job opportunities
Educational purposes
Human resources
Natural resources[environmental protection]
Transportation
Public health finance
Keeping apart money to improve the economic development of a country, lottery gains also go to cancer organizations, child care canters and places where it is needed
Today, many religious and social organizations condemn lotteries saying that it’s a one way ticket to accumulating lots of wealth. Since lotteries lotteries tempt people to play with money and gain huge amounts back if they got lucky, it can easily become an addiction. Many individuals waste much money over lotteries and other gambling games without taking care of their families properly. This is unethical to a population of today’s society. Of course, it is not right to spend all the money you earn on lotteries. But by taking part in lotteries you contribute a little to the development of your own country. That’s why we say lotteries are not just about gambling. It’s a lot about winning and giving.
ZZZZZZ
## Pick 3 Lottery Workout Systems That Work
Have you been searching for Pick 3 lottery workout systems that actually work? Are you tired of guessing and relying on luck? Well here's one system that you should try. It only involves simple mathematics and a bit of concentration. This system is one of the oldest. Although it doesn't guarantee a hundred percent, it has a high percentage hit. It is called the 123 workout. Here is how it works:
o You begin by using the preceding numbers drawn from the Pick 3 lottery. Just to provide a concrete example, let us use the numbers 468. We will call it the stack number.
o Important Rule of the 123 workout: Do not carry over when adding or subtracting the numbers. Here is an example: 5 + 9 = 4 (instead of 14)
o Monitor all the numbers that come out in the following days. You will realize that one out of the 22 combinations will be the next hit. Let's face it; it is better than having to choose from a million other combinations. Your chance of winning the next lottery is now greater having to choose from a smaller set.
Others have tried other Pick 3 lottery workout systems. Some of them are just too complicated. They would require the work of your personal computer, plus it costs much and takes time to download them. This system only requires a pen and a piece of paper. Try this method and see if the numbers really come out. What have you got to lose?
Winning the Lotto Texas lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## Is There a Way To Win The Lottery? Texas Two Step
Playing the Texas Two Step lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## State Lotteries - Which State Has the Best Odds of Winning?
Are you constantly thinking and saying, "I want to win the lottery?" Do you ask yourself, "Will I win the lottery?" If you think about winning the lottery all the time, you need to know this secret to winning the secret to winning the lottery.
If you think winning the big bucks is just about picking the right lottery numbers, you're wrong. Yes, you do need to pick the right numbers, and it's best to have a system to win the lottery. Doing things like reading a lottery book can be useful too. But none of this will work for you if you don't do this one essential thing:
The universe rewards those who are prepared. Did you know that?
When you act as if you expect something, it's going to come to you. Did you know that?
4. Make a list of people to whom you plan to give money and decide on how much you'll give. Take into account gift taxes when you do this. Be careful about who you tell, "I'll give you money when I win." You don't want to give away all your winnings.
Combine this lottery win preparation method with a lottery winning system, and you will be able to answer "Yes," to "Will I win the lottery?" You'll be able to turn "I want to win the lottery," to "I WON the lottery."
Winning the Texas Two Step lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## How To Calculate The Odds of Winning the Lottery? Texas Triple Chance
Playing the Texas Triple Chance lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Texas Triple Chance lottery?
A lottery is a popular form of gambling which involves the drawing of lots for a prize. There are some states where lottery is forbidden but others endorse it to the extend of organizing a national lottery. Lotteries date back to the period of Romans. And it was common for emperors to give their dinner party guests gifts like slaves to lavish villas after drawings.
Two common lottery myths
Lottery is tax
Lottery take advantage of the poor economic strata of our society
Lottery facts
In China lottery was first played to fund the Great Wall
U.S. lotteries helped fund the Colonial Army in the Revolutionary War
At one point of time many churches and universities such as Harvard, Yale and Princeton were funded in part by lotteries
Where does the Lottery revenue expenditure go?
A good amount of lottery earnings is spend on economic development of the country
Steps are taken to improve job opportunities
Educational purposes
Human resources
Natural resources[environmental protection]
Transportation
Public health finance
Keeping apart money to improve the economic development of a country, lottery gains also go to cancer organizations, child care canters and places where it is needed
Today, many religious and social organizations condemn lotteries saying that it’s a one way ticket to accumulating lots of wealth. Since lotteries lotteries tempt people to play with money and gain huge amounts back if they got lucky, it can easily become an addiction. Many individuals waste much money over lotteries and other gambling games without taking care of their families properly. This is unethical to a population of today’s society. Of course, it is not right to spend all the money you earn on lotteries. But by taking part in lotteries you contribute a little to the development of your own country. That’s why we say lotteries are not just about gambling. It’s a lot about winning and giving.
ZZZZZZ
## Lottery Cheats - How to Cheat the System and Win!
People who play Megamillions have one question in common and that would be which the best numbers to play Megamillions are? The thing is, there isn't a standard or a formula that one can follow in order to come to a conclusion regarding the best numbers to play Megamillions. There are, however, things that you can keep in mind when you are in the process of picking numbers. You can consider them strategies or systems or you can think of them, simply, as tips. They may not give you a concrete answer towards the best numbers to play Megamillions, but they can help take you a step closer.1. When choosing numbers, do not try and form patterns. Remember to be as random as you can. Think about it, if you decide to go with a pattern such as choosing all the numbers that go a certain way, whether upwards or downwards the sheets then you are already putting yourself at a disadvantage.8. When you ask people who have won jackpots for themselves, it is likely that they'd tell you that the numbers they chose were chosen mostly because of gut feel. A kind of intuition. You may not think much of it but this intuition could potentially win you millions. So go with what you feel and not what you believe to be the best numbers to play Megamillions. Remember, trust your gut and go with your intuition.
Winning the Texas Triple Chance lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## Learn How To Predict Lottery Numbers All or Nothing
Playing the All or Nothing lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The All or Nothing lottery?
Here's 5 top tips to bear in mind next time you complete your lottery playslip. They won't increase your CHANCES of winning - because ONLY more entries can do that.
BUT they can significantly increase the AMOUNT you win when your numbers do come up.
How?
Because all 5 lottery tips are based on avoiding the way a lot of other people pick their numbers. If you pick lottery numbers the same way as most people do, then when you hit the jackpot, you share that prize with everybody else who picked the same numbers.
That can turn a jackpot pool of millions into a prize of just thousands! It happens all too often - so please don't let it happen to you.
The easiest, fastest way to pick better lottery numbers, is to pick them totally at random. So pull scraps of paper out of a bag. It won't guarantee NOT picking a 'bad' set of numbers, but at least there's a good chance you won't be sharing your lottery millions with a hundred other 'lucky' winners.
Are you ready for more tips on how to win more often when playing the lottery?
If so then read below for a free report that uncovers the biggest mistakes most lottery players are making, and how you can avoid them.
## Lottery, More Than Just Gambling
Lottery predictions is quite popular these days. People used to be skeptical with the predictions as they thought that the winning numbers are a matter of luck and fortunes. Not many people believe that lottery can be won by using some kind of a sophisticated science based predictions. It was not until the late 90s when lottery players began using lottery predictions to help them to win lottery or at least get closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish man who managed to study and analyze many games in two different countries, Spain and the US and win a lot of money by using different strategies. After him people started to believe that lottery results can be predicted.
Lottery players start thinking about how to win the lotteries using predictions. They use many kinds of predictions: from mechanical predictions on mechanical lotteries to technological predictions using computer software. A lot of people use algorithm to analyze and predict lottery results.
Analysis of Groups
There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.
Analysis of Hot-Cold Trend
This algorithm analysis is one of the most favourite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings.
Analysis of Repetition Pattern
A lot of lottery players share the same opinion that repetition is quite important to predict the winning numbers as most of jackpots will appear again in the future.
The analysis mentioned above represents only a part of the strategies that lottery players can use. There are still many other algorithm analysis that can be done by predictors to help them win.
Winning the All or Nothing lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## How To Calculate The Odds of Winning the Lottery? Pick 3
Playing the Pick 3 lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Pick 3 lottery?
Ok what are you selling? That is the first question that may come to your mind if you are reading a title like win the lotto. But if you can get pass the title and read through the entire article I promise to give you some key secrets that will put you ahead of 95% of the average lotto players. Like you I play the lotto and I hate for people to waste my time with easy cut answers that lead to no where. I want to be told what it really takes to win the lotto regularly. I know you are the same since you are reading this.
So let's get down to business shall we. The key to winning the lotto boils down to 4 closely held secrets they are:
• Research
• Strategy
• Patience
• System
System: Now here is the kicker since everything boils down to what system are you using. Without the correct system patience, strategy and research goes out the window. In fact it makes no sense to do if you do not have a system to apply your strategies or research to. Patience has no bearing if you don't have a system. Okay you got the point. Remember the system you choose determines everything so choose wisely.
## How To Answer "Yes," To "Will I Win The Lottery?" And Turn "I Want To Win The Lottery" Into "I Won!"
Lottery predictions is quite popular these days. People used to be skeptical with the predictions as they thought that the winning numbers are a matter of luck and fortunes. Not many people believe that lottery can be won by using some kind of a sophisticated science based predictions. It was not until the late 90s when lottery players began using lottery predictions to help them to win lottery or at least get closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish man who managed to study and analyze many games in two different countries, Spain and the US and win a lot of money by using different strategies. After him people started to believe that lottery results can be predicted.Lottery players start thinking about how to win the lotteries using predictions. They use many kinds of predictions: from mechanical predictions on mechanical lotteries to technological predictions using computer software. A lot of people use algorithm to analyze and predict lottery results.Analysis of GroupsThere are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.Analysis of Hot-Cold TrendThis algorithm analysis is one of the most favourite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings.Analysis of Repetition PatternA lot of lottery players share the same opinion that repetition is quite important to predict the winning numbers as most of jackpots will appear again in the future.The analysis mentioned above represents only a part of the strategies that lottery players can use. There are still many other algorithm analysis that can be done by predictors to help them win.
Winning the Pick 3 lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## How Can You Win A Lottery? Pick 4
Playing the Pick 4 lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Pick 4 lottery?
A lottery is a popular form of gambling which involves the drawing of lots for a prize. There are some states where lottery is forbidden but others endorse it to the extend of organizing a national lottery. Lotteries date back to the period of Romans. And it was common for emperors to give their dinner party guests gifts like slaves to lavish villas after drawings.
Two common lottery myths
Lottery is tax
Lottery take advantage of the poor economic strata of our society
Lottery facts
In China lottery was first played to fund the Great Wall
U.S. lotteries helped fund the Colonial Army in the Revolutionary War
At one point of time many churches and universities such as Harvard, Yale and Princeton were funded in part by lotteries
Where does the Lottery revenue expenditure go?
A good amount of lottery earnings is spend on economic development of the country
Steps are taken to improve job opportunities
Educational purposes
Human resources
Natural resources[environmental protection]
Transportation
Public health finance
Keeping apart money to improve the economic development of a country, lottery gains also go to cancer organizations, child care canters and places where it is needed
Today, many religious and social organizations condemn lotteries saying that it’s a one way ticket to accumulating lots of wealth. Since lotteries lotteries tempt people to play with money and gain huge amounts back if they got lucky, it can easily become an addiction. Many individuals waste much money over lotteries and other gambling games without taking care of their families properly. This is unethical to a population of today’s society. Of course, it is not right to spend all the money you earn on lotteries. But by taking part in lotteries you contribute a little to the development of your own country. That’s why we say lotteries are not just about gambling. It’s a lot about winning and giving.
ZZZZZZ
## Pick 3 Online - Play Now!
When it comes to picking Megamillions numbers, you need some lottery tips. Here are four secrets to winning the lottery that you can apply to choosing the numbers that will get you your win.Lottery numbers are obviously drawn randomly. But even random drawings can product patterns that can be tracked and used to your advantage. Here are some ways to do that.1. In Megamillions, players choose five numbers from array of 1 to 56 and one number, the Gold Mega Ball, from an array of 1 to 46. The next five secrets will apply to the first set of five numbers. Choosing the Gold Ball requires a different strategy. If you use a balanced wheel to pick the Gold Ball, be sure you use one for pick-5 games. The other wheels won't work.5. After you choose the five numbers you want to play, add them together and make sure that the sum of these five numbers adds up to between 106 and 179. Sums that fall in that range account for over 70 percent of all lottery jackpots won.Obviously, these strategies still require some finesse to use, and they are just a few of the lottery winning formulas available. The best thing to do if you want to consistently pick winning Megamillions numbers is to purchase a lottery book that will teach you proven secrets to winning the lottery. You can find out more about one of these books at the link below.
Winning the Pick 4 lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery
## How Do I Calculate Lotto Numbers? Cash Five
Playing the Cash Five lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Cash Five lottery?
Lottery predictions is quite popular these days. People used to be skeptical with the predictions as they thought that the winning numbers are a matter of luck and fortunes. Not many people believe that lottery can be won by using some kind of a sophisticated science based predictions. It was not until the late 90s when lottery players began using lottery predictions to help them to win lottery or at least get closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish man who managed to study and analyze many games in two different countries, Spain and the US and win a lot of money by using different strategies. After him people started to believe that lottery results can be predicted.
Lottery players start thinking about how to win the lotteries using predictions. They use many kinds of predictions: from mechanical predictions on mechanical lotteries to technological predictions using computer software. A lot of people use algorithm to analyze and predict lottery results.
Analysis of Groups
There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.
Analysis of Hot-Cold Trend
This algorithm analysis is one of the most favourite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings.
Analysis of Repetition Pattern
A lot of lottery players share the same opinion that repetition is quite important to predict the winning numbers as most of jackpots will appear again in the future.
The analysis mentioned above represents only a part of the strategies that lottery players can use. There are still many other algorithm analysis that can be done by predictors to help them win.
## How to Win Lottery by Using Analysis Algorithms for Lottery Prediction
Because you are sitting in front of your computer viewing this page, chances are, you're an eager and regular lotto player who has yet to win lotto prizes. Lottery cheats are some of the most researched items or articles here on the Internet, and it's easy to understand why: lottery games from all states can be addictive, attracting countless players from all across the country to play the lotto on a regular basis. What you need to understand is that while playing the lottery can be fun and exciting, the real treat lies in winning the lotto prizes - including the jackpot. And unlike what some lotto players believe, you can win the lotto using lottery cheats.
Lottery cheats are not cheats in the real sense of the word. They are not illegal and won't put you into some kind of federal trouble. In fact, lottery cheats have long been recognized by serious lottery players as the reason why they have higher chances of winning. Cheats at winning the lottery are actually guides on how to win any type of lotto game you choose to play. They offer sensible pieces of advice to increase the likelihood of holding a winning ticket.
As far as lottery cheats are concerned, statistics experts highly advise against forming lottery combinations in a mathematical sequence and playing patterns on lottery tickets. These acts are sure to reduce your chances of winning because mathematical sequences and patterns are almost never taken into consideration in lottery games. Past winning results show a tendency towards random combinations. If you use a proven system that can efficiently analyze lottery data, such as past winning numbers, trends, and angles, you are more likely to bring home the jackpot prize not just once but as many times as you please.
Winning the Cash Five lottery is very rewarding. Give this system a spin and watch your winnings grow.
Is There a Way To Win The Lottery? Texas Lottery | 8,384 | 41,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-05 | longest | en | 0.965693 |
https://www.physicsforums.com/threads/solving-a-first-order-differential.103270/ | 1,508,413,870,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823282.42/warc/CC-MAIN-20171019103222-20171019123222-00414.warc.gz | 954,768,835 | 15,520 | # Solving a first order differential
1. Dec 8, 2005
### georgeh
I have the following problem that i can't get the right answer to:
y'+y=5sin(2t)
I find the integrating factor u(t)=e^t
multiply the whole function u(t)
and i get (ye^t)'=5e^tsin(2t)
I do integraton by parts on the second part and get
1/9 for the coefficents, in the calculator they don't get those coefficents! :(
very frustrating, not sure why.
2. Dec 8, 2005
### neo143
Ans: Y = Sin2t - Cos2t + c
3. Dec 8, 2005
### georgeh
yeah they get that answer, mind showing me how you did it? i've been at it for an hour.. all help is greatly appreciated.
4. Dec 8, 2005
### BerkMath
That is not the answer. Proceeding from your last step, integration of both sides yields: y*e^t= -e^t (2cos(2t)-sin(2t))+C. Where the right hand side, RHS, was found using integration by parts. Next, dividing both sides by e^t, gives: y= -(2cos(2t)-sin(2t))+Ce^(-t) or y=sin(2t)-2cos(2t)+Ce^(-t). C can be found given some intial condtion y(t_0)=y_0.
5. Dec 8, 2005
### georgeh
yeah, i just need someone to show me the integration by parts, i am unable to get that solution. I know how to solve the first order d.e.
6. Dec 8, 2005
### Hammie
Why even bother integrating by parts? It will work.. but- why not try an educated guess:
This kind of differential equation will have a solution of Y= (K1)Sin (2X) + (K2)Cos(2X).
simply differentiate Y, add Y and Y Prime, and equate the unknown constants.
7. Dec 8, 2005
### BerkMath
You apply integration by parts to the RHS, twice. To begin let I=RHS. Then apply integration by parts twice. After the second integration by parts you will end up with another integral, after dividing out the necessary constants, you can make this integral look exactly like I. Stop. Then solve it as though it is a linear equation in I, for I. You are exploiting the cyclic nature of the cosine and sine derivatives. | 574 | 1,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-43 | longest | en | 0.911874 |
https://www.electrical4u.net/electrical-basic/what-is-electrical-network/ | 1,580,210,849,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00150.warc.gz | 826,633,822 | 35,713 | # What is Electrical Network
## What is electrical network:
Network is the most used word in an electric or electronic circuit while solving a circuit problem or creating a balanced circuit. A network does not require to have a closed path, the open circuit also can be called as a network. In such a way all the networks are not a circuit (example: open circuit cannot be called as circuit and it can be called as a network) and all the circuit can be called as network. An electrical network can be identified by four circuit names such as Node, Branch, Loop and Mesh. Let see the detailed explanation of each one..
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### Node
A junction point of two or more active components (voltage source, current sources) or passive components connection (resistor, inductor, capacitor etc) is called Node or in a network is an equipotential part at which two or more circuit’s element are joined.
Get More Electrical Concept And Interview Questions By Using This Link
### Branch
Branch is nothing but a part of an electric circuit which locates between two junctions is called branch. In branch, one or more elements can be connected and they have two terminals.
### Loop
Loop is nothing but A closed path of an electric circuit where one or more than two meshes can be followed i.e. there may be many meshes in a loop. To be a loop the circuit should be in the closed path. To be a complete circuit, the circuit must have at least one loop.
### Mesh
A mesh is a closed path that does not contain any other loops or A closed loop which contains no other loop within it or a path which does not contain on other paths is called Mesh.
## The best example for an Electrical network:
Let us consider below-mentioned diagram:
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Also Know About: Why India has 50 Hz Power System and US has 60 Hz 110 Volts Power System
### Nodes of the circuit:
In this, Point A and B can be called as nodes of the circuit, the circuit resistor R1,R2, and R3 are joined at a point A which is node 1; R3, R4, and R5 are joined in a point B which is node 2. And the battery 2 negative point, R4, R2 and Battery 1 negative terminal are joined at point C which is node 3.
### Mesh of the circuit:
Mesh is nothing but a single closed path, in given circuit, we have three meshes.
Mesh1: IAGHI,
Mesh2: ABFGA
Mesh3: BDEFB
### Loop of the circuit:
Loop is nothing but a closed path of the circuit. The given circuit we can find 5 different loops,
Loop1: IAGHI,
Loop2: ABFGA
Loop3: BDEFB
Loop4: IABFGHI
Loop5: IABDEFGHI | 616 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.944152 |
https://physics.stackexchange.com/questions/352988/if-an-inductor-has-current-flowing-in-only-one-direction-does-the-magnetic-fiel | 1,716,667,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00804.warc.gz | 390,418,574 | 38,799 | # If an inductor has current flowing in only one direction, does the magnetic field still vary directions?
My understanding is that in a standard AC inductor the magnetic field in the core changes direction as the current through the primary coil alternates. What if the current through the primary coil was pulsed on and off but in one direction only?
It is still an alternating current that produces a changing magnetic filed that induces voltage in the secondary coil, but what happens to the magnetic field in the core? Would the secondary coil induce a magnetic field in the core the opposite direction to the field originally induced by the primary coil? Would the core still experience an alternating magnetic field, even though the primary coil current flows in on direction only?
My understanding of Lenz’s law is still a little shaky, and I am trying to understand how Lenz’s law would apply in this particular scenario. I feel like the secondary coil would induce a magnetic field in the opposite direction to the one induced by the primary coil, but I am not sure. (I am also not sure if I have not used the most fitting terms to describe magnetic field “directions”, but I hope people understand what I am clumsily trying to say!)
• Yes, this is still an alternating current. Look at Faraday's law and think about when the derivative over flux is equal to zero (equivalent to zero field generated). This way you will answer your question once and forever. Aug 21, 2017 at 16:31
The current in the primary induces a proportional magnetic field.
Any current in the secondary also induces a proportional magnetic field.
The observed magnetic field is the sum of the two.
So your interrupted-DC primary driver will drive a magnetic field that goes B-0-B-0-B etc.
That'll induce a voltage in the secondary (it doesn't induce a current!) that goes + and - at the transitions between B and 0, and between 0 and B. Depending on what's connected to the secondary, there might be a current in the secondary:
• If the secondary is open-circuited, nothing happens, and the field is as above.
• If there's a resistor, a limited current will flow. That'll create some magnetic field while flowing, which will be opposed to the change in the field: If the primary is going 0 to B, it'll be negative; if the primary is going B to 0, it'll be positive. That's Lenz's law in action. It'll tend to round-off the sharp transitions in the field given by the sharp transitions in the primary current.
The details of that second case are tricky, both mathematically and in practice. I.e. you can't really instantly change the primary current; the primary is an inductor, and an infinite rate-of-change requires infinite voltage.
It's really much easier to consider the sine/cosine case of continuous change, hence that's what we try to teach first. | 606 | 2,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.955421 |
https://socratic.org/questions/how-do-you-identify-1-csctheta-1-1-csctheta-1 | 1,713,242,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00792.warc.gz | 490,558,638 | 6,094 | # How do you identify 1/(csctheta+1)-1/(csctheta-1)?
$- 2 {\tan}^{2} \theta$
#### Explanation:
I'm going to assume this needs to be simplified:
$\frac{1}{\csc \theta + 1} - \frac{1}{\csc \theta - 1}$
$\frac{1}{\frac{1}{\sin} \theta + 1} - \frac{1}{\frac{1}{\sin} \theta - 1}$
$\frac{1}{\frac{1}{\sin} \theta + \sin \frac{\theta}{\sin} \theta} - \frac{1}{\frac{1}{\sin} \theta - \sin \frac{\theta}{\sin} \theta}$
$\frac{1}{\frac{1 + \sin \theta}{\sin} \theta} - \frac{1}{\frac{1 - \sin \theta}{\sin} \theta}$
$\sin \frac{\theta}{1 + \sin \theta} - \sin \frac{\theta}{1 - \sin \theta}$
$\sin \frac{\theta}{1 + \sin \theta} \left(\frac{1 - \sin \theta}{1 - \sin \theta}\right) - \sin \frac{\theta}{1 - \sin \theta} \left(\frac{1 + \sin \theta}{1 + \sin \theta}\right)$
$\frac{\sin \theta \left(1 - \sin \theta\right)}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} - \frac{\sin \theta \left(1 + \sin \theta\right)}{\left(1 - \sin \theta\right) \left(1 + \sin \theta\right)}$
$\frac{\sin \theta - {\sin}^{2} \theta}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} - \frac{\sin \theta + {\sin}^{2} \theta}{\left(1 - \sin \theta\right) \left(1 + \sin \theta\right)}$
$\frac{\sin \theta - \sin \theta - {\sin}^{2} \theta - {\sin}^{2} \theta}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)}$
$\frac{- 2 {\sin}^{2} \theta}{\left(1 - {\sin}^{2} \theta\right)}$
Recall that ${\sin}^{2} \theta + {\cos}^{2} \theta = 1 \implies 1 - {\sin}^{2} \theta = {\cos}^{2} \theta$
$\frac{- 2 {\sin}^{2} \theta}{\cos} ^ 2 \theta = - 2 {\tan}^{2} \theta$ | 663 | 1,578 | {"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": 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} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.268696 |
https://www.nag.com/numeric/nl/nagdoc_27/examples/source/f07wsfe.f90.html | 1,618,288,413,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00607.warc.gz | 1,024,659,949 | 2,081 | ``` Program f07wsfe
! F07WSF Example Program Text
! Mark 27.0 Release. NAG Copyright 2019.
! .. Use Statements ..
Use nag_library, Only: nag_wp, x04dbf, zpftrf, zpftrs
! .. Implicit None Statement ..
Implicit None
! .. Parameters ..
Integer, Parameter :: nin = 5, nout = 6
! .. Local Scalars ..
Integer :: i, ifail, info, k, lar1, ldb, lenar, &
n, nrhs, q
Character (1) :: transr, uplo
! .. Local Arrays ..
Complex (Kind=nag_wp), Allocatable :: ar(:), b(:,:)
Character (1) :: clabs(1), rlabs(1)
! .. Executable Statements ..
Write (nout,*) 'F07WSF Example Program Results'
! Skip heading in data file
Read (nin,*) n, nrhs, uplo, transr
lenar = n*(n+1)/2
ldb = n
Allocate (ar(lenar),b(ldb,nrhs))
! Setup notional dimensions of RFP matrix AR
k = n/2
q = n - k
If (transr=='N' .Or. transr=='n') Then
lar1 = 2*k + 1
Else
lar1 = q
End If
! Read an RFP matrix into array AR
Do i = 1, lar1
End Do
Do i = 1, n
End Do
! Factorize A
! The NAG name equivalent of zpftrf is f07wrf
Call zpftrf(transr,uplo,n,ar,info)
Write (nout,*)
Flush (nout)
If (info==0) Then
! Compute solution
! The NAG name equivalent of zpftrs is f07wsf
Call zpftrs(transr,uplo,n,nrhs,ar,b,ldb,info)
! Print solution
ifail = 0
Call x04dbf('General',' ',n,nrhs,b,ldb,'Bracketed','F7.4', &
'Solution(s)','Integer',rlabs,'Integer',clabs,80,0,ifail)
Else
Write (nout,*) 'A is not positive definite'
End If
End Program f07wsfe
``` | 529 | 1,561 | {"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} | 2.71875 | 3 | CC-MAIN-2021-17 | latest | en | 0.5092 |
https://www.premiumessaywritingservice.com/physics-182a-195l-sdsu-simple-pendulum-lab-10-report-i-am-confused-on-how-to-answer-the-questions-in-this-lab-the-data-is-provided-in-blue-physics-182a-1/ | 1,627,898,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00616.warc.gz | 971,808,878 | 14,830 | Select Page
## PHYSICS 182A 195L SDSU Simple Pendulum Lab 10 Report I am confused on how to answer the questions in this lab. The data is provided in blue. PHYSICS 182A/1
PHYSICS 182A 195L SDSU Simple Pendulum Lab 10 Report I am confused on how to answer the questions in this lab. The data is provided in blue. PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
Lab 10: Simple Pendulum
San Diego State University
Department of Physics
Physics 182A/195L
TA:
Lab partner 1:
Lab partner 2:
Date: 04/21/2020
Score:
Data has been entered in blue.
Theory
Oscillations can occur whenever a system is pulled away from a stable equilibrium. For
example, Hooke’s law
describes a spring system with an equilibrium at
.A
spring will oscillate if you pull it away from that equilibrium and then release it. In this lab we will
investigate the oscillations of the simple pendulum system: a mass
hangs from a fixed point
by a string of length and negligible mass. We will see that this system exhibits simple
harmonic motion, a kind of oscillation that can be described by a simple sine or cosine function.
1 Department of Physics | O. Gorton and K. Graves
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
Force diagram
To figure out how to describe this motion, we start by writing down the forces acting on the
pendulum mass
. Looking at the free-body diagram above, we can see that there are two
forces acting on the mass: there is gravity pulling downward on the mass with a force
(mg in the negative y-direction), and there is the force of the pendulum arm
creating a tension
keeping the mass a fixed length from the pivot.
Newton’s second law for this system reads
, a vector equation. Let’s decompose
each vector into its components parallel to and perpendicular to the direction of the pendulum
arm. Along the direction parallel to the pendulum arm:
The cosine function comes from the fact that the gravitational force is shifted degrees away
from being parallel to the pendulum arm. Something convenient happens here: since we know
the length of the pendulum arm will not change, we can safely assume that the tension in the
string and the force
will cancel, leaving
Along the direction perpendicular to the pendulum arm, or in other words, along the direction of
motion:
Again, the sine function appears here due to the right-triangle geometry depicted above. Since
only
is non-zero, we will just call it ‘a’ from here on.
The position equation
We will briefly describe some math that you may have seen in your more advanced calculus
courses. If you don’t want to see any calculus, skip ahead to the calculus free conclusion.
2
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
The equation of motion above is really a differential equation for the displacement
pendulum mass:
of the
This says that the second derivative of
is related to
. In general, this is a
difficult equation to solve. In fact, it can’t be solved with normal calculus methods. We can solve
this equation with an approximation. We will apply the small angle approximation, which says
that as long as the angle
is small,
.
By applying this small angle approximation, our equation for
becomes much simpler:
.
This will make more sense if we replace with something that depends on x and t. If we
remember from geometry that arc length is related to the radius and angle, we can see for this
situation that:
If we plug this into the equation above:
Notice that this looks a lot like Hooke’s law! The force (ma) is equal to a negative constant times
displacement:
. A function that satisfies this equation is
Calculus free conclusion: a simple pendulum will oscillate in a way according to a sine
function, as long as the angle it swings to is small.
Period of motion
There’s more that we can learn given the boxed equation for x(t). One question we might like to
answer is: how long will it take the pendulum to swing back and forth? In other words, what is
the period of the motion? Recalling our knowledge of sine waves, we know that they repeat
every
radians, and if we write the sine wave in the form:
3 Department of Physics
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
Then we can identify the constant
as the period of the motion. If we set these two forms of
equal to each other:
,
We can recognize that
The period of motion T (the time it takes to complete one cycle of oscillation) of a simple
pendulum is given by:
Procedure
In this procedure we will measure the period of motion of a simple pendulum. We will try to
answer questions about the period T, to see how it is affected by different properties of the
pendulum system such as the length of the pendulum arm L, the mass of the pendulum bob m,
the acceleration due to gravity g, and the amplitude of the motion A. You can probably guess
from the equation for T which of these will have an impact and which will not.
Setup
1. Run the string through the hole in the brass cylinder and put the ends of the string on the
inner and outer clips of the clamp (Figure 1), so the string forms a ’V’ shape as it hangs.
Figure 1: Pull the string between the inner and outer clips of the ramp
2. Adjust the string so the vertical distance from the bottom edge of the Pendulum Clamp
the middle of the pendulum bob is 0.70 m.
4
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
3. Position the Motion Sensor in front of the pendulum so the brass colored disk is vertical
and facing the bob, aimed along the direction that the pendulum will swing (Figure 2).
4. Make sure that the range switch on the Motion Sensor is positioned on top. Adjust the
Motion Sensor up or down so that it is at the same height as the pendulum bob.
5. Adjust the position of the Rod Base so that the Motion Sensor is 0.25 m from the
pendulum bob.
Figure 2: Suspend the bob in a “V” shape, with the motion sensor facing the bob.
Measuring the Period
There are three methods available for measuring the period of motion. Ask your TA which
method or methods to use.
Method 1: The most straightforward way to measure the period of oscillation is to set the
pendulum into motion and to simply time N complete cycles. If the pendulum swings
back and forth, returning to its original position N times in t seconds, then the period is
. It’s important to choose N>10 so that random errors from the timing method
will average out.
The other two methods use PASCO Capstone™ software to find your period. Start by recording
● Click on Record to start recording data, data should appear on the graph. After 15
seconds, stop recording. Make sure the amplitude of the swing is relatively small (such
as 0.05 m). If readings are not being recorded, make sure the objects are more than
0.15 m away from the sensor.
● If the resulting data is not a smooth sine curve, you can try changing the position of the
range switch on top of the Motion Sensor. In general, the position with the cart icon is for
smaller, closer objects and the position with the person icon is for larger objects further
away. You should position the switch for whichever works the best.
● Select the rate that data is recorded by changing the frequency of the motion sensor at
the bottom of the screen. Make sure there are enough points to create a smooth sine
wave. A rate of 50 Hz will work well for this experiment.
5 Department of Physics
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
Once these steps have been performed, you can find your period with one of the two following
methods:
Method 2: PASCO Capstone™ can find the period using a recording of the position
versus time of the pendulum bob. Begin by using the Coordinates Tool (from the Graph
Tool palette), select a peak in your data recovered from oscillating a pendulum. Right
click on the graph box and select Show Delta. Find an adjacent peak where ∆x ~ 0m to
view the period.
Method 3: Another method involves using various Curve Fit functions available. Select
the Sine Fit from the Graph Tool palette. The general form of the sine wave is:
, where T is the period. The values from your Curve Fit can be used to
calculate the period of the oscillation. (Note: ω in the Sine Fit option is equal to 2π/T).
Part A: Period of Oscillation as a Function of Length
We will test how changing L, the length of the pendulum, affects the period of motion.
1. Adjust the length of the pendulum to be 0.7 meters by lengthening or shortening the
string. Measure from the bottom edge of the pendulum clamp down to the middle of the
cylinder.
2. Pull the pendulum bob about 0.05 m away from equilibrium and then release. Allow the
bob to oscillate for a few seconds until the oscillations are smooth.
3. Measure the period of oscillation using the methods from Measuring the Period.
4. Record your measured period in Table A.1.
5. Stop the pendulum from swinging and then shorten the pendulum length by 0.05 m.
6. Repeat steps 2-5, recording data until you have a range of lengths from 0.70 m to 0.15
m according to Table A.1.
Part B: Period of Oscillation as a Function of Amplitude
1. Keep the pendulum length fixed at 0.35 m from the bottom of the clamp to the middle of
the bob (last length used from Part A).
2. Pull the pendulum bob about 0.10 m away from equilibrium and then release. Allow the
bob to oscillate for a few seconds until the oscillations are smooth.
3. Measure the period of oscillation using the methods from Measuring the Period.
4. Record the measured period in Table B.1.
5. Stop the pendulum from swinging.
6. Repeat steps 2-5 for amplitudes of 0.09 m, 0.08 m, 0.07 m, 0.06 m and 0.05 m,
recording each measured period in Table B.1.
6
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
Part C: Period of Oscillation as a Function of Mass
1. Remove the brass cylinder from the pendulum and measure the mass of the brass,
aluminum, and plastic cylinders. Record these values in Table C.1.
2. Reattach the brass cylinder to the pendulum system.
3. Adjust the length of the pendulum so that the distance from the bottom edge of the
pendulum clamp to the middle of the cylinder is 0.60 m.
4. Pull the pendulum bob about 0.06 m away from equilibrium and then release. Allow the
bob to oscillate for a few seconds until the oscillations are smooth.
5. Measure the period using one of the methods from the Measuring the Period.
6. Record the measured period in Table C.1.
7. Stop the pendulum from swinging.
8. Repeat steps 3-6 for both the aluminum and plastic cylinder. Make sure to keep the
length of the pendulum, and the maximum amplitude, the same.
Data
Table A.1: Period and Varying Length
Length (m)
Cycles observed
Time, if using Method 1 (s)
0.70
15
25.16
0.65
15
24.25
0.55
15
22.30
0.45
15
20.18
0.35
15
17.80
0.25
15
15.05
0.20
15
13.45
0.15
15
11.65
Period (s)
Table B.1: Period and Varying Amplitude
Amplitude (m)
Cycles observed
Time, if using Method 1 (1)
0.10
14
16.09
7 Department of Physics
Period (s)
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
0.09
14
16.60
0.08
14
16.65
0.07
14
16.59
0.06
14
16.60
0.05
14
16.65
Table C.1: Period and Varying Mass
Bob (type)
Bob Mass (kg)
Cycles
observed
Time, if using
Method 1 (s)
Brass
0.0873
15
15.50
Aluminum
0.0270
15
15.40
Plastic
0.0120
15
15.45
Period (s)
Analysis
Part A: Period of Oscillation as a Function of Length
1. What happens to the period as you adjust the length of the pendulum?
2. We can solve for g, the acceleration due to gravity, using the length and period.
Beginning from the equation of motion found in the Theory section, we can solve for g:
Find the value of g for each of the lengths, using Table A.1 as reference:
Length (m)
g (m/s2)
0.70
0.65
8
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
0.55
0.45
0.35
0.25
0.20
0.15
3. Do your values for g change significantly for different values of the pendulum length?
Does this make sense? Why or why not?
Part B: Period of Oscillation as a Function of Amplitude
What happens to the period as you adjust the amplitude of the pendulum?
Part C: Period of Oscillation as a Function of Mass
1. What happens to the period as you adjust the mass of the pendulum?
2. The amplitude of a pendulum’s oscillation is usually measured by the angle through
which it swings, not its horizontal displacement. In Part C the length of the pendulum
was about 60 cm and the maximum amplitude was about 6cm. Using right-triangle
trigonometry, calculate the angle of the amplitude. Show your work.
5.74 deg
3. The period of a pendulum is independent of amplitude only if the angle is small
Conclusion
Which variables (g, L, m, A) affect the period of motion T of the simple pendulum, and which do
not?
9 Department of Physics
PHYSICS 182A/195L LAB REPORT – LAB 10: Simple Pendulum
Questions
1. Must a spring obey Hooke’s Law in order to oscillate? Explain why or why not.
2. When a body is oscillating in simple harmonic motion, is its acceleration zero at any
point? If so, where and why?
3. You are trying to build a metronome (a time keeping device) using a simple pendulum.
Assume that your pendulum bob weighs 1 kg and g=9.8m/s2. How long should you make
the pendulum arm so that the metronome ticks with a frequency of 100 BPM (beats per
minute)? (Hint: frequency = 1/T.) Show your work.
4. What is the frequency of a pendulum whose normal period is T when it is in an elevator
in free fall? (Hint: what is the apparent value of g in this situation?)
5. We usually assume that the mass of the pendulum arm is negligible compared to the
mass hung from it. But if not negligible, does the mass of the string increase or decrease
the period of motion?
10 | 3,531 | 13,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-31 | latest | en | 0.853389 |
https://www.jiskha.com/display.cgi?id=1330476043 | 1,503,012,662,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104172.67/warc/CC-MAIN-20170817225858-20170818005858-00354.warc.gz | 922,393,284 | 4,094 | posted by .
I need to find the order of a reaction with respect to the reactants.
I set up my equation like this:
5.33e-2 M / 2.66e-2M = 1.03e-2 / 1.93e-3
I did the concentrations equal to the rates. I am just not sure what the units for rate should be.
2.00 = 5.33^m
Did I set this up correctly to find the order or would the order just be 2 from 5.33e-2M / 2.66e-2M ??
Diregard question.
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I need to find the order of a reaction with respect to the reactants. I set up my equation like this: 5.33e-2 M / 2.66e-2M = 1.03e-2 / 1.93e-3 I did the concentrations equal to the rates. I am just not sure what the units for rate …
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I posted a question like this but my numbers were actually wrong. These are the correct ones. 5.94e-3 / 1.11e-3 = 5.33e-2/2.66e-2 5.35 = 2.00^m I guess this is not correct because I do not know a whole number that will give you 5.35.
7. ### Chemistry
I posted a question like this but my numbers were actually wrong. These are the correct ones. 5.94e-3 / 1.11e-3 = 5.33e-2/2.66e-2 5.35 = 2.00^m I guess this is not correct because I do not know a whole number that will give you 5.35. …
8. ### Chemistry(Please check, thank you!)
I completed a lab to find the determination of Kc. I have to find the concentrations of reactants at equilibrium using an ICE table. The equation that were are using is Fe^3+(aq) + SCN^-(aq) -> Fe(SCN)^2+(aq) I have to create 5 … | 708 | 2,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-34 | latest | en | 0.924039 |
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# Scientist have use computer models to create a series of maps that show how susceptible coastal cities in island countries are to the sea rising at different levels. the map shows that a 1 meter rise would swap cities all along the us eastern seaboard. a 6-meter sea level rise with sub merge a large part of florida. thermal expansion has already raise the oceans 10 to 20 centimeters the most likely cause, directly and indirectly, for such a rise in sea level is
### Another question on Physics
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Use the frequency histogram to complete the following parts. ​(a) identify the class with the​ greatest, and the class with the​ least, relative frequency. ​(b) estimate the greatest and least relative frequencies. ​(c) describe any patterns with the data. female fibula lengths 30.5 31.5 32.5 33.5 34.5 35.5 36.5 37.5 38.5 39.5 0 0.05 0.1 0.15 0.2 0.25 length (in centimeters) relative frequency a histogram titled "female fibula lengths" has a horizontal axis labeled "length in centimeters" from 30.5 to 39.5 in increments of 1 and a vertical axis labeled "relative frequency" from 0 to 0.25 in increments of 0.05. the histogram contains vertical bars of width 1, where one vertical bar is centered over each of the horizontal axis tick marks. the approximate heights of the vertical bars are listed as follows, where the label is listed first and the approximate height is listed second: 30.5, 0.02; 31.5, 0.04; 32.5, 0.05; 33.5, 0.13; 34.5, 0.22; 35.5, 0.25; 36.5, 0.13; 37.5, 0.06; 38.5, 0.09; 39.5, 0.01. ​(a) the class with the greatest relative frequency is nothing to nothing centimeters. ​(type integers or decimals. do not round. use ascending​ order.) | 593 | 1,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-31 | latest | en | 0.816541 |
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# What is 61 inches in feet
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As any golf expert will tell you, precise measurements are crucial when it comes to the game of golf. For instance, accurately figuring out distances is essential when deciding which golf club to use for a specific shot. This is why knowing how to convert measurements from one unit to another is incredibly important, such as knowing what is 61 inches in feet. With this conversion knowledge, a golfer can precisely determine the distance from the ball to the hole and pick the perfect club to achieve optimal results. Furthermore, understanding distances in feet can also help golfers assess the length of the hole, identify potential obstacles, and design strategies to approach it. Therefore, knowing what is 61 inches in feet is just one of the many skills that a golf expert relies on to master this challenging and rewarding sport.
## How tall is 61 inches in human height?
ft in inches centimeters
5’1” 61in 154.94cm
5’2” 62in 157.48cm
5’3” 63in 160.02cm
5’4” 64in 162.56cm
## How many inches is 5 feet 6?
66 inches
Golfers often contemplate the importance of measurements when it comes to their game. Specifically, they may question the exact distance between their golf ball and the hole. To determine this information accurately, they must be knowledgeable in converting units of measurement. For instance, the height of most professional golfers is often specified in feet and inches. In order to estimate the exact distance needed to hit a shot, golfers must understand that 5 feet 6 inches translates to a total of 66 inches. By having a grasp on this conversion, golfers can strategize accordingly and ultimately improve their overall performance on the course.
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## How many inches is 5 7 feet?
I am well-versed in my own height, which measures to exactly 5’7″. As an avid follower of all things numerical, I recall the precise conversion factor for inches to centimeters – a handy bit of knowledge that always comes in handy. So, to determine my height in terms of inches, I first translate the feet component of 5’7″ into inches, which equals a sum of 60 inches. Therefore, I now have a grand total of 67 inches as my physical height.
## Is 60 inches bigger than 4 feet?
When it comes to measurements, it’s important to understand how units of measurement are converted from one to another. For example, 4 feet can be converted to inches by multiplying 4 by 12, resulting in 48 inches. Similarly, 5 feet can be converted to inches by multiplying 5 by 12, resulting in 60 inches. It’s also worth noting that there are other ways to measure length, such as using a yardstick, which allows for precise and accurate measurements. By taking the time to understand these conversions and utilizing the right tools, we can ensure that our measurements are both accurate and consistent.
## How tall is 4 11 in cm?
Feet + Inches Ft + In Centimetres
4 feet 11 inches 4′ 11″ 149.86 cm
5 feet 0 inches 5′ 0″ 152.40 cm
5 feet 1 inches 5′ 1″ 154.94 cm
5 feet 2 inches 5′ 2″ 157.48 cm
## What is 5 7 in cm?
As we delve into the topic of golf, it’s important to understand the various nuances that make up this beloved sport. From the intricate design of the club head, to the perfect angle of the swing, every detail matters. In fact, even measurements play a crucial role in golf, such as determining the height of the player. For instance, someone who is 5 feet 7 inches tall would measure approximately 170.18 centimeters, according to the conversion formula [(5×12)+7] x 2.54. This information may seem minuscule, but in the world of golf, precision is key to a successful game.
## Is 5 ft 6 tall for a girl?
Frequently, my height evokes comments of awe and admiration from acquaintances. However, in certain regions of the United States, I would simply be classified as above average in height. For instance, in areas where the average female height is approximately 5’6”, my 5’7” stature wouldn’t necessarily stand out as tall. Nevertheless, in the local vicinity where I reside, the average female height is around 5’4”, making me noticeably taller than most women. To put it into perspective, only about 15% of females in this region exceed my height. So, while I may not be exceptionally tall nationally, within my own location and context, my height is considered above average and notable.
## What height is 5ft 4?
At a height of 5 feet 4 inches, an individual stands at a relatively average height for women and a bit shorter than average for men in many parts of the world. However, it’s important to note that height can vary greatly among different populations and ethnic groups. To provide context for comparison, 5ft 4 is the average height for adult women in the United States. In terms of measurement, 5 feet 4 inches is equal to 64 inches or 162.56 centimeters. This height can be associated with agility and speed, which can be advantageous in sports like gymnastics, figure skating, and soccer. On the other hand, certain professions like modeling and basketball may require individuals to be taller in order to meet industry standards and excel in their field.
## Is 180 cm 5 11?
When it comes to converting centimeters to feet and inches, it’s important to remember that 180 cm is not equivalent to 6 ft. In fact, to get an accurate conversion, you need to divide 180 cm by 2.5 cm to get 72 inches. But don’t stop there! You still need to further divide that result by 12 inches in order to get your final answer. By doing this, we can see that 72 inches is equal to 6 feet. However, because 180 cm is just shy of the 182.9 cm needed to reach the 6 ft mark, it is more accurately converted to 5 ft 11 in. So next time you need to convert centimeters to feet and inches, be sure to take this extra step and avoid any measurement mishaps.
## Is 170 cm 5 feet and 7 inches?
For individuals who are 170cm or 5 feet and 7 inches in height, there are a number of clothes and shoe options available that will flatter their specific body type. With this height, it is important to consider the fit and proportions of clothing items in order to achieve a well-balanced overall look. In terms of footwear, it may be beneficial to opt for footwear that adds height or length to the leg line, such as a pair of platform heels or knee-high boots. Additionally, when it comes to selecting clothing, experimenting with different silhouettes and styles, such as fitted blazers or flared pants, can help to accentuate the natural curves and features of one’s body. By carefully considering each element of their wardrobe, those who are 170cm/5 feet 7 inches tall can create a style that is both fashion-forward and flattering.
## How tall your child will be?
When determining how tall your child may be, it’s important to consider both the mother’s height and the father’s height. By adding these two heights together, you can calculate a rough estimate for your child’s potential height. Be sure to measure in either inches or centimeters, depending on your preference. For boys, it’s recommended to add an additional 5 inches (or 13 centimeters) to the total height, while for girls it’s recommended to subtract 5 inches (or 13 centimeters). Finally, divide the total height by 2 to get a more accurate estimate. Remember that genetics play a huge role in determining height, but this calculation is a good starting point to help predict your child’s potential growth.
## Is 5 7 tall for a woman?
When it comes to height, there’s no denying that 5′7″ is a pretty impressive number. In fact, it’s safe to say that this height definitely qualifies as tall. To put things into perspective, let’s take a look at the average height for girls in the United States. According to statistics from Google, the average height across all ethnicities is 5′4″. However, if you look specifically at white girls, the average jumps up slightly to 5′4.5″. With these figures in mind, it’s easy to see why a height of 5′7″ is considered above average. In fact, it’s taller than roughly 80-85% of girls in the US! Of course, there are a number of factors that can influence a person’s height, including genetics and lifestyle choices. But when it comes down to it, there’s no denying that a height of 5′7″ is certainly something to be proud of.
## How deep is 60 cm?
When it comes to conversions, knowing the exact equivalents can help you streamline your work and make your calculations more efficient. For instance, if you need to convert a measurement from centimeters to inches, you should know that 60 cm is equal to 23.62 inches. With this knowledge, you can confidently carry out your calculations, whether you’re working on a sewing project, woodworking, or just trying to get a better sense of a piece of furniture that you’re considering buying. Whether you’re a professional in a trade where measurements are crucial or a hobbyist who values precision, having a strong grasp of conversion equivalents can help you save time and avoid costly mistakes. So the next time someone asks “How deep is 60 cm?” rest assured that you’ll have a confident response at the ready.
## How many cm means 1 feet?
Cm Feet
In the CGS system of measurement, a centimeter is a base unit. The foot is a base unit for the English system of measurement.
The standard symbol for a centimeter is cm. The common symbol for feet is ft.
1 cm = 0.033 ft 1 ft = 30.4 cm
## How tall is 59 in inches?
When it comes to measuring length in the customary system, one important fact to keep in mind is that 59 inches translates to 4 feet and 11 inches. However, this piece of information alone may not be enough for some people who want to delve deeper and gain an even better understanding of how this conversion works. For instance, did you know that if we were to express this measurement as a decimal, we could round it up to 4.9 feet? To arrive at this conclusion, it’s crucial to remember that there are 12 inches in a foot, and this knowledge will be instrumental in helping you grasp various other measurements and conversions used in the customary system.
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This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Accept Read More | 2,368 | 10,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.938871 |
https://kidsworksheetfun.com/decimals-to-fractions-worksheets/ | 1,723,013,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00697.warc.gz | 270,395,127 | 27,402 | # Decimals To Fractions Worksheets
Convert decimals to fractions. As your third grader transitions into the fourth grade fractions and decimals take a front row seat in math class.
Convert Fractions To Decimal Converting Fractions Fractions Fractions To Decimals Worksheet
### Below are six versions of our grade 5 math worksheet on converting simple decimal numbers to fractions with tens or hundreds as the denominator.
Decimals to fractions worksheets. Now in the denominator place 1 followed by number of zeroes equal to the number of digits after the decimal point you counted. If you ve ever taught your students how to write fractions in decimal it s time to move on to a new topic now we can teach you to write decimal numbers as fractions many of our students who learn this lesson have to solve problems your students need a worksheet to convert decimal numbers to fractions. You can also control the amount of workspace the font font size the border around the problems and additional instructions.
With decimal fractions worksheets students learn about tenths and hundredths places converting fractions and more. You can choose the number of decimal digits used the types of denominators easy powers of ten or random and whether to include improprer fractions and mixed numbers or not. For converting decimals to fractions first write the digits without the decimal point in the numerator.
Convert each decimal into either fraction or percent or both. You will find the problems and answers you need in our. This page contains links to free math worksheets for fractions as decimals problems.
Best of all the worksheets are free. Decimal quiz online practice. Math worksheets fractions as decimals fractions as decimals.
Featuring exercises to add subtract multiply and divide decimals in varied levels of difficulty. This math worksheet was created on 2016 10 21 and has been viewed 257 times this week and 353 times this month. Best of all the worksheets are free.
Students are not asked to simplify the answer. Using teacher created concepts decimal fractions worksheets simplify equations and make math interesting. These handy worksheets will help you your students or your children become fluent in converting decimals to fractions.
Decimals convert decimals to fractions. Converting decimals to fractions worksheet 1 here is a 15 problem worksheet featuring decimals that must be written as fractions. The timed decimal online test practice worksheets consist of 10 random questions for 4th 5th and 6th graders.
Worksheets math grade 5 fractions vs. Then count the number of digits to the right of the decimal point or dot. Converting a decimal into a fraction is a fairly straightforward process.
You can also use the worksheets menu on the side of this page to find worksheets on other math topics. Click one of the buttons below to see all of the worksheets in each set. The worksheets are very customizable.
It involves looking at the number of decimal places the decimal has and putting the decimal part as the numerator and the denominator as a 1 followed the the same number of zeros as there are decimal places. Convert 1 or 2 digit decimals to fractions without simplifying. It may be printed downloaded or saved and used in your classroom home school or other educational environment to help someone learn math.
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Worksheets For Kids Free Printables Page 10 Education Com Decimals Worksheets Fractions Decimals Decimals | 1,007 | 5,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-33 | latest | en | 0.890628 |
https://www.xuebaunion.com/detail/6297.html | 1,719,222,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00378.warc.gz | 932,185,617 | 10,297 | Propositional Logic代写-COMP1600 /
Hoare Logic: Total Correctness
COMP1600 / COMP6260
slides created by: Dirk Pattinson
(with contributions by Victor Rivera and previous colleagues)
convenor: Pascal Bercher
lecturer: Michael Norrish
Australian National University
Semester 2, 2022
Recall: Hoare Logic
Basic Ingredient. Hoare-triples
{P} S {Q}
P and Q are predicates (formulae)
S is a code (fragment)
Example. {x > 0} while (x>0) do x := x-1 {x = 0}
Meaning. If we run S from a state that satisfies precondition P and if S
terminates, then the post-state will satisfy Q.
1 / 49
Hoare Logic
Idea. Proof Rules that allow us to prove all true triples.
Assignment
{Q(e)} x := e {Q(x)}
Precondition Strengthening / Postcondition Weakening
Ps → Pw {Pw} S {Q}
{Ps} S {Q}
{P} S {Qs} Qs → Qw
{P} S {Qw}
Sequence.
{P} S1 {Q} {Q} S2 {R}
{P} S1; S2 {R}
Conditional.
{P ∧ b} S1 {Q} {P ∧ ¬b} S2 {Q}
{P} if b then S1 else S2 {Q}
2 / 49
Proof Rule for While Loops (Rule 6/6)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b}
I is called the loop invariant.
I is true before we encounter the while statement, and remains true
each time around the loop (although not necessarily midway during
execution of the loop body).
If the loop terminates the control condition must be false, so ¬b
appears in the postcondition.
For the body of the loop S to execute, b needs to be true, so it
appears in the precondition.
3 / 49
Soundness of the While Rule w.r.t. the semantics
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b}
Soundness. If the premise is true, then so is the conclusion.
assume that I holds in the initial state.
if b is false, nothing happens, so I ∧ ¬b is true in post-state.
if b is true, then (by premise) I holds at end of each iteration
assuming that the loop terminates, b becomes false (and I still holds)
4 / 49
Applying the While Rule
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
Difficult bit. Finding the right invariant.
This requires intuition and practice. There is no automated way of
doing this.
Easy bit. Establishing the desired postcondition
The postcondition we get after applying our rule has form I ∧ ¬b.
But if I ∧ ¬b → Q, we can use postcondition weakening.
Easy bit. Establishing the desired precondition
The precondition we get after applying our rule has form I . But if
P → I , we can use precondition strengtening.
5 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6. {P} while b do S {Q} (Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2.
{I} while b do S {I ∧ ¬b}
(While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1.
{I ∧ b} S {I}
2.
{I} while b do S {I ∧ ¬b}
(While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2.
{I} while b do S {I ∧ ¬b}
(While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3.
I ∧ ¬b → Q
(Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5.
P → I
(Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
6 / 49
Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6.
{P} while b do S {Q}
(Precondition Strengthening, 4, 6)
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Applying the While Rule (schematic proof)
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b} {P} while b do S {Q}
1. {I ∧ b} S {I}
2. {I} while b do S {I ∧ ¬b} (While Loop, 1)
3. I ∧ ¬b → Q (Logic)
4. {I} while b do S {Q} (Postcondition Weakening, 2, 3)
5. P → I (Logic)
6. {P} while b do S {Q} (Precondition Strengthening, 4, 6)
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Example
Goal. Find condition I to prove that:
{n > 3} while n>0 do n := n-1 {n = 0}
Observation. Need to prove the above using while-rule, i.e.
{I ∧ b} n := n− 1 {I}
{I} while (n>0) do n:= n-1 {I ∧ n ≤ 0}
Want. Loop invariant I such that
It is implied by the precondition:
n > 3→ I
if the loop terminates (i.e. n > 0 is false), then I should imply the
postcondition:
I ∧ n ≤ 0→ n = 0
If I is true and the body is executed, I is true afterwards:
{I ∧ n > 0} n := n− 1 {I}
Loop Invariant. I ≡ (n ≥ 0)
have {n ≥ 0 ∧ n > 0} n := n− 1 {n ≥ 0}
n ≥ 0 ∧ ¬(n > 0)→ n = 0
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Example (cont.)
Goal. Find condition I to prove that:
{n > 3} while n>0 do n := n-1 {n = 0}
Loop Invariant. I ≡ (n ≥ 0), we have
n > 3→ n ≥ 0,
n ≥ 0 ∧ n ≤ 0→ n = 0, and
{n ≥ 0 ∧ n > 0} n := n− 1 {n ≥ 0}.
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Example, Formally
1. {n − 1 ≥ 0} n := n− 1 {n ≥ 0} (Assignment)
2. n ≥ 0 ∧ n > 0→ n − 1 > 0 (Logic)
3. {n ≥ 0 ∧ n > 0} n := n− 1 {n ≥ 0} (Prec. Strength., 1, 2)
4. {n ≥ 0} while (n > 0) do n := n− 1 {n ≥ 0∧¬(n > 0)} (While Loop, 3)
5. n > 3→ n ≥ 0 (Logic)
6. {n > 3} while (n > 0) do n := n− 1 {n ≥ 0 ∧ ¬(n > 0)} (Prec.
Strength., 4, 5)
7. n = 0↔ n ≥ 0 ∧ ¬(n > 0) (Logic)
8. {n ≥ 0} while (n > 0) do n := n− 1 {n = 0} (Post. Equiv., 6, 7)
Other Invariants
e.g. true or n = 0
both are invariants, and give n = 0 as postcondition
but n ≥ 0 is better (weaker) as it is more general.
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Let’s Prove a Program!
Program (with specification):
{True}
i:=0;
s:=0;
while (i ̸= n) do
i:=i+1;
s:=s+(2*i-1)
{s = n2}
(The sum of the first n odd numbers is n2)
Goal: prove
{True} Program {s = n2}
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A Very Informal Analysis
Let’s look at some examples:
1 = 1 = 12
1 + 3 = 4 = 22
1 + 3 + 5 = 9 = 32
1 + 3 + 5 + 7 = 16 = 42 . . .
It looks OK - let’s see if we can prove it!
Goal: prove
{True} Program {s = n2}
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How can we prove it?
First Task. Find a loop invariant I . (NB: S and s are different!)
Post condition and loop condition:
{I ∧ b} S {I}
{I} while b do S {I ∧ ¬b}
while (i ̸= n) do
i:=i+1; (1, 2, 3, ...)
s:=s+(2*i-1) (1, 4, 9, ...)
{s = n2}
Want. (I ∧ i = n)→ (s = n2) to apply postcondweak
Loop Body. Each time i increments, s moves to next square number.
Invariant. I ≡ s = i2.
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Check I as (s = i2) is an invariant: prove {I}S{I}
{s = i2} i := i + 1 {Q} {Q} s := s + (2 ∗ i − 1) {s = i2}
Seq{s = i2} i := i + 1; s := s + (2 ∗ i − 1) {s = i2}
Using the assignment axiom and the sequence rule:
1. {Q} s:=s+(2*i-1) {s = i2}
2.
3. {s = i2} i:=i+1 {Q}
4. {s = i2} i:=i+1; s:=s+(2*i-1) {s = i2} (Sequence, 3, 1)
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Check I as (s = i2) is an invariant: prove {I}S{I}
{s = i2} i := i + 1 {Q} {Q} s := s + (2 ∗ i − 1) {s = i2}
Seq{s = i2} i := i + 1; s := s + (2 ∗ i − 1) {s = i2}
Q is {s + (2 ∗ i − 1) = i2}
Using the assignment axiom and the sequence rule:
1. {s + (2 ∗ i − 1) = i2} s:=s+(2*i-1) {s = i2} (Assignment)
2.
3. {s = i2} i:=i+1 {s + (2 ∗ i − 1) = i2}
4. {s = i2} i:=i+1; s:=s+(2*i-1) {s = i2} (Sequence, 3, 1)
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Check I as (s = i2) is an invariant: prove {I}S{I}
{s = i2} i := i + 1 {Q} {Q} s := s + (2 ∗ i − 1) {s = i2}
Seq{s = i2} i := i + 1; s := s + (2 ∗ i − 1) {s = i2}
Q is {s + (2 ∗ i − 1) = i2}
Using the assignment axiom and the sequence rule:
1. {s + (2 ∗ i − 1) = i2} s:=s+(2*i-1) {s = i2} (Assignment)
2. {s + (2 ∗ (i + 1)− 1) = (i + 1)2} i:=i+1 {s + (2 ∗ i − 1) = i2}
(Assignment)
3. {s = i2} i:=i+1 {s + (2 ∗ i − 1) = i2} (Prec. Equiv., 2)
4. {s = i2} i:=i+1; s:=s+(2*i-1) {s = i2} (Sequence, 3, 1)
So far, so good. (I as (s = i2) is an invariant.)
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Completing the Proof of {True} Program {s = n2}
6 Strengthen the precondition to match the While rule premise
{I ∧ b} S {I}
{(s = i2) ∧ (i ̸= n)} i:=i+1; s:=s+(2*i-1) {s = i2}
7 Apply the While rule and postcondition equiv:
s = i2 ∧ i = n↔ s = n2
{s = i2} while ... s:=s+(2*i-1) {s = n2}
8 Check that the initialisation establishes the invariant:
{0 = 02} i:=0 {0 = i2} {0 = i2} s:=0 {s = i2}
{0 = 02} i:=0; s:=0 {s = i2}
9 (0 = 02) ↔ True, so putting it all together with Sequencing we have
{True} i := 0 ; s := 0 ; while (i ̸= n) do S {s = n2}
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Hoare Logic (in this form) proves partial correctness.
Example. while 1+1 = 2 do x:=0.
This will loop forever!
can still prove things about it
Exercise.
{True} while 1+1 = 2 do x:=0 {False}
Termination.
remember functional programs? Something must decrease
need loop variant (later)
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Are the Rules Complete?
So far. Have a very simple language
new rules for arrays, for-loops, exceptions, . . .
Focus here. Soundness
every provable Hoare triple is (semantically) true
soundness holds but terms and conditions apply
with these assumptions, also have completeness, i.e. every true Hoare
triple is provable.
Completeness. if {P} S {Q} is true then {P} S {Q} is provable
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What are these Assumptions?
The language we use for expressions in our programs is the same as
the language we use in our pre- and postconditions (in our case, basic
arithmetic).
We assumed no aliasing of variables. (In most real languages we can
have multiple names for the one piece of memory.)
How is aliasing a problem?
Suppose x and y refer to the same cell of memory.
▶ We get {y +1 = 5 ∧ y = 5} x:=y+1 {x = 5 ∧ y = 5} (Assignment)
▶ i.e. {y = 4 ∧ y = 5} x:=y+1 {x = 5 ∧ y = 5}
▶ i.e. if initial state satisfies False and x:=y+1 terminates then final state
satisfies {x = 5 ∧ y = 5} (but also works for x = 6 ∧ y = 6)
which makes a mockery of our calculus since it proves rubbish!
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What are these Assumptions?
The language we use for expressions in our programs is the same as
the language we use in our pre- and postconditions (in our case, basic
arithmetic).
We assumed no aliasing of variables. (In most real languages we can
have multiple names for the one piece of memory.)
How is aliasing a problem?
Suppose x and y refer to the same cell of memory.
▶ We get {y +1 = 5 ∧ y = 5} x:=y+1 {x = 5 ∧ y = 5} (Assignment)
▶ i.e. {y = 4 ∧ y = 5} x:=y+1 {x = 5 ∧ y = 5}
▶ i.e. if initial state satisfies False and x:=y+1 terminates then final state
satisfies {x = 5 ∧ y = 5} (but also works for x = 6 ∧ y = 6)
which makes a mockery of our calculus since it proves rubbish!
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Finding a Proof
Example.
{ n >= 0 }
f := 1;
i := n;
while (i > 0) do
f := f * i;
i := i-1;
{ f = n! }
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Finding a Proof
Annotating the program: I = f ∗ i ! = n! ∧ i ≥ 0
{ n >= 0 }
f := 1;
{ f = 1 /\ n >= 0 } -- provable with assignment
i := n;
{ f = 1 /\ i = n /\ n >= 0} -- provable with assignment
==> -- use postcond weakening
{ I } -- general premise of while rule
while (i > 0) do
{ I /\ i > 0 } -- proof obligation for loop body
f := f * i; -- n, n * (n-1), n * (n-1) * (n-2) ...
i := i-1; -- n-1, n-2, n-3, ...
{ I } -- up until here.
{ I /\ not(i > 0) } -- general conclusion of while rule
==> -- use postcond weakening
{ f = n! }
Invariant: f * i! = n! / i >= 0
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From Annotated Programs to Proofs
Initialisation Part
{ n >= 0 }
f := 1;
{ f = 1 /\ n >= 0 } -- provable with assignment
i := n;
{ f = 1 /\ i = n /\ n >= 0} -- provable with assignment
1. {f = 1∧ n = n∧ n ≥ 0} i := n {f = 1∧ i = n∧ n ≥ 0} (Assignment)
2. {1 = 1∧n = n∧n ≥ 0} f := 1 {f = 1∧n = n∧n ≥ 0} (Assignment)
3. n ≥ 0↔ 1 = 1 ∧ n = n ∧ n ≥ 0 (Logic)
4. {n ≥ 0} f := 1 {f = 1 ∧ n = n ∧ n ≥ 0} (Prec. Equiv. 2, 3)
5. {n ≥ 0} f := 1; i := n {f = 1 ∧ i = n ∧ n ≥ 0} (Sequence, 1, 4)
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From Invariant to Loop Body
{ I } -- general premis of while rule
while (i > 0) do
{ I /\ i > 0 } -- proof obligation for loop body
f := f * i; -- n, n * (n-1), n * (n-1) * (n-2) ...
i := i-1; -- n-1, n-2, n-3, ...
{ I } -- up until here.
{ I /\ not(i > 0) } -- general conclusion of while rule
Invariant. I = f ∗ i ! = n! ∧ i ≥ 0
Loop Body: Proof obligation
{ f * i! = n! /\ i >= 0 /\ i > 0 }
f := f * i;
i := i - 1;
{ f * i! = n! /\ i >= 0}
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From Annotated Programs to Proofs
Loop Body
{ f * i! = n! /\ i >= 0 /\ i > 0 } -- proof obligation
for loop body
f := f * i;
{ f * (i-1)! = n! /\ i >= 0 /\ i > 0} -- new annotation!
i := i-1;
{ f * i! = n! /\ i >= 0 } -- end loop body
6. {f ∗ (i − 1)! = n! ∧ (i − 1) ≥ 0} i := i− 1 {f ∗ i! = n! ∧ i ≥ 0}(Assignment)
7. {(f ∗ i) ∗ (i − 1)! = n! ∧ (i − 1) ≥ 0} f := f ∗ i {f ∗ (i − 1)! = n! ∧ (i − 1) ≥ 0} (Assignment)
8. f ∗ i! = n! ∧ i ≥ 0 ∧ i > 0→ (f ∗ i) ∗ (i − 1)! = n! ∧ (i − 1) ≥ 0 (Logic)
9. {f ∗ i! = n! ∧ i ≥ 0 ∧ i > 0} f := f ∗ i {f ∗ (i − 1)! = n! ∧ (i − 1) ≥ 0} (Prec. Stren., 7,8)
10. {f ∗ i! = n! ∧ i ≥ 0 ∧ i > 0} f := f ∗ i; i := i− 1 {f ∗ i! = n! ∧ i ≥ 0} (Sequence, 6,9)
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From Annotated Programs to Proofs
Loop Body to While Loop
{ f * i! = n! /\ i >= 0 } -- premise of while
rule: "I"
while (i > 0) do
{ f * i! = n! /\ i >= 0 /\ i > 0 } -- "I /\ b"
f := f * i;
i := i-1;
{ f * i! = n! /\ i >= 0 } -- "I"
{ f * i! = n! /\ i >= 0 /\ not(i > 0) } -- conclusion of while:
"I /\ not b"
11. {f ∗ i ! = n! ∧ i ≥ 0}
while (i > 0) do { f:= f*i; i:= i-1}
{f ∗ i ! = n! ∧ i ≥ 0 ∧ ¬(i > 0)} (While, 10)
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Putting it all together
{ n >= 0 }
f := 1;
i := n;
{ f = 1 /\ i = n /\ n >= 0} -- have already
==> -- postcond weakening
{ f * i! = n! /\ i >= 0 }
while (i > 0) do
f := f * i;
i := i-1;
{ f * i! = n! /\ i >= 0 /\ not(i > 0) } -- have already
==> -- postcond weakening
{ f = n! }
12. f = 1 ∧ i = n ∧ n ≥ 0→ f ∗ i ! = n! ∧ i ≥ 0 (Logic)
13. {n ≥ 0} f := 1; i := n {f ∗ i ! = n! ∧ i ≥ 0} (Postcond. Weak., 5, 12)
14. {n ≥ 0} program {f ∗ i ! = n! ∧ i ≥ 0 ∧ ¬(i > 0)} (Seq., 13, 11)
15. f ∗ i ! = n! ∧ i ≥ 0 ∧ ¬(i > 0)→ f = n! (Logic)
16. {n ≥ 0} program {f = n!} (Postcondition Weakening, 14, 15)
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Hoare Logic: Total Correctness
Motto. Total Correctness = partial correctness + termination
New Notation.
[P] S [Q]
P and Q are precondition and postcondition, as before
S is a code fragment
Meaning. If the precondition holds, then executing S will terminate, and
the postcondition is true.
Example.
[P] S [true] – S always terminates from precondition P
{P} S {false} – S never terminates from precondition P
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Rules for Total Correctness
Q. What are the rules for total correctness?
assignment
sequencing
conditional
pre/post strengthening/weakening
still work, as there’s no danger of non-termination.
Problematic Rule. while (may introduce non-termination)
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Assignment, revisited
[Q(e)] x := e [Q(x)]
Assumptions. (fine for our toy language)
evaluation of expression e terminates.
evaluation of e returns a number (no exception e.g.)
In General.
the expression can be recursively defined
there may be errors, e.g. division by zero
28 / 49
Rules for Total Correctness
Ps → Pw [Pw ] S [Q]
[Ps ] S [Q]
(Precondition Strengthening)
[P] S [Qs ] Qs → Qw
[P] S [Qw ]
(Postcondition Weakening)
[P] S1 [Q] [Q] S2 [R]
[P] S1; S2 [R]
(Sequence)
[P ∧ b] S1 [Q] [P ∧ ¬b] S2 [Q]
[P] if b then S1 else S2 [Q]
(Conditional)
Assumption. Evaluation of b always terminates (OK here)
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Termination of Loops
Example
[y > 0]
while (y < r) do
r := r - y;
q := q + 1
[true]
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Termination of Loops
Example
[y > 0]
while (y < r) do
r := r - y;
q := q + 1
[true]
Observations.
q := q + 1 irrelevant
y doesn’t change, always positive
r strictly decreases in each iteration
y < r will eventually be false.
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Termination of Loops: General Condition
Example
[y > 0]
while (y < r) do
r := r - y;
q := q + 1
[true]
Termination follows if we have a variant E that is,
E ≥ 0 at the beginning of each iteration
E strictly decreases at each iteration
Q. What could be a variant in this example?
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While Rule for Total Correctness
Goal. Show that
[P] while b do S [Q]
In Addition to partial correctness (e.g. finding I ), find variant E such
that
E ≥ 0 at the beginning of each iteration:
I ∧ b → E ≥ 0
E is strictly decreasing in each iteration
[I ∧ b ∧ (E = n)] S [I ∧ E < n]
where n is an auxiliary variable not appearing elsewhere, to
“remember” initial value of E
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While Rule for Total Correctness
I ∧ b → E ≥ 0 [I ∧ b ∧ E = n] S [I ∧ E < n]
[I ] while b do S [I ∧ ¬b]
where n is an auxiliary variable not appearing elsewhere.
Intuition.
E is an upper bound to the number of loop iterations
termination of functional programs: measure decrease in recursive call
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Example
Goal. [y > 0] while (y < r) do r := r - y; q := q+1 [true]
Focus. Loop body r := r - y; q := q + 1
want some invariant: let’s just call it I
want some variant: here r is decreasing
First Goal. The variant is ≥ 0 when we enter the loop:
formally: I ∧ (y < r)→ r ≥ 0
this suggests I ≡ y > 0
Second Goal. The invariant is re-established, and the variant decreases
formally:
[(y > 0) ∧ (y < r) ∧ r = n] r := r = y; q:= q+1 [(y > 0) ∧ r < n]
this seems to be right, so let’s prove it!
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Example Proof
Goal.
[(y > 0) ∧ (y < r) ∧ r = n] r := r - y; q:= q+1 [(y > 0) ∧ r < n]
First Assignment.
[(y > 0) ∧ (r < n)] q := q+ 1 [y > 0 ∧ r < n]
Second Assignment.
[(y > 0) ∧ (r − y < n)] r := r-y [y > 0 ∧ r < n]
Sequencing.
[(y > 0) ∧ (r − y < n)] r := r-y; q := q+1 [y > 0 ∧ r < n]
Precondition Strengthening.
[(y > 0) ∧ (y < r) ∧ (r = n)] r := r - y; q := q+1 [y > 0 ∧ r < n]
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Completing the Proof
While Rule.
I ∧ b → E ≥ 0 [I ∧ b ∧ E = n] S [I ∧ E < n]
[I ] while b do S [I ∧ ¬b]
1. [(y > 0) ∧ (r < n)] q := q+ 1 [y > 0 ∧ r < n] (Assignment)
2. [(y > 0) ∧ (r − y < n)] r := r-y [y > 0 ∧ r < n] (Assignment)
3. [(y > 0) ∧ (r − y < n)] r := r-y; q := q+1 [y > 0 ∧ r < n] (Sequence, 2, 1)
4. (y > 0) ∧ (y < r) ∧ (r = n)→ (y > 0) ∧ (r − y < n) (Logic)
5. [(y > 0) ∧ (y < r) ∧ (r = n)] r := r-y; q := q+1 [y > 0 ∧ r < n] (Prec.
Streng., 2, 3)
6. (y > 0) ∧ (y < r)→ r ≥ 0 (Logic.)
7. [y > 0] while (y < r) do r:= r-y; q:= q+1 [y > 0 ∧ y ≥ r ] (While Loop, 5,
6)
8. y > 0 ∧ y ≥ r → true
9. [y > 0] while (y < r) do r:= r-y; q:= q+1 [true] (Postc. Weak., 7, 8)
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Second Example
[n >= 0]
fact := 1;
i := n;
while (i > 0) do
fact := fact * i;
i := i - 1
[fact = n!]
Q1. What is the invariant (linking n, fact and i)?
before: fact = 1, i = n
1st iteration: fact = n, i = n-1
2nd iteration: fact = n * (n-1), i = n-2
...
last iteration: fact = n * ... * 1, i = 0
Invariant: fact ∗ i ! = n!
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Second Example
[n >= 0]
fact := 1;
i := n;
while (i > 0) do
fact := fact * i;
i := i - 1
[fact = n!]
Q2. Is the invariant fact ∗ i ! = n! good enough?
true initially: for fact = 1 and i = n.
implies postcondition: fact ∗ i ! = n! ∧ ¬(i > 0)→ fact = n!
Stronger Invariant. fact ∗ i ! = n! ∧ i ≥ 0
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Second Example
[n >= 0]
fact := 1;
i := n;
while (i > 0) do
fact := fact * i;
i := i - 1
[fact = n!]
Q3. What’s the variant?
i ≥ 0 for every iteration (I ∧ b → E ≥ 0)
decreases with every iteration
Variant. E ≡ i
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Proof Skeleton
Simple Assignments.
[n >= 0]
fact := 1;
i := n;
[n >= 0 /\ fact = 1 /\ i = n]
Applying While
[ fact * i! = n! /\ i >= 0 ]
while (i > 0) do
fact := fact * i;
i := i - 1
[ fact * i! = n! /\ i >= 0 /\ i <= 0]
Missing Glue. Weakening / Strengthening
from postcondition of assignments to precondition of while
from postcondition of while to goal statement (fact = n!)
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Zooming in on While
[ fact * i! = n! /\ i >= 0 ]
while (i > 0) do
fact := fact * i;
i := i - 1
[ fact * i! = n! /\ i >= 0 /\ i <= 0]
Loop Body (Invariant: fact ∗ i ! = n! ∧ i ≥ 0, Variant: i)
[fact * i! = n! /\ i >= 0 /\ i > 0 /\ i = a]
fact := fact * i;
i := i - 1
[ fact * i! = n! /\ i >= 0 /\ i < a]
Positivity Condition. fact ∗ i ! = n! ∧ i ≥ 0 ∧ i > 0→ i ≥ 0 (Maths)
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While Rule: Soundness
I ∧ b → E ≥ 0 [I ∧ b ∧ E = n] S [I ∧ E < n]
[I ] while b do S [I ∧ ¬b]
Partial Correctness.
premises of the rule imply those of the rule for partial correctness
so if the loop terminates, the postcondition holds
Missing. Termination
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Termination Analysis
I ∧ b → E ≥ 0 [I ∧ b ∧ E = n] S [I ∧ E < n]
[I ] while b do S [I ∧ ¬b]
Let σ be a state that validates the precondition I . If b is false in σ, we are
done. Assume that b is true in σ, hence the value of E in σ is ≥ 0.
Induction on the value σ(E ) of E in state σ.
Base case. σ(E ) = 0.
Right premise implies that E < 0 after one iteration
With left premise, get that ¬b after one iteration
hence the loop terminates after one iteration.
Step Case. Let σ(E ) = k + 1
after one iteration, have σ(E ) ≤ k
statement follows by induction hypothesis.
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Variation: More Expressive Logic
So far. In triples {P} S {Q} we have
S a program fragment (we keep this for now)
P and Q propositional formulae made from basic arithmetic
Q. How about expressing the following?
{true} x := 2 ∗ x {even(x)}
A. We could say that even(x) = ∃y .2 ∗ y = x
Change. Allowing pre/postconditions to be first order formulae.
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Example.
{true} x := 2 ∗ x {∃y .2 ∗ y = x}
Using the assignment axiom.
{∃y .2 ∗ y = 2 ∗ x} x := 2 ∗ x {∃y .2 ∗ y = x}
More Expressive Logic. Assertions are first-order formulae
all rules remain valid
Hoare-logic is (almost) insensitive to underlying logic
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Variation: More Expressive Programs
Example Feature. Arrays
allow expressions to contain a[i ]
(we assume that the index is always in scope)
Maximum Finding
m := a[0]
i := 1;
while (i < n) do
if a[i] > m then m := a[i] else m := m;
i := i + 1
Q. How do we express that m is the maximum array element?
A. Use first order logic.
m is largest: ∀k .0 ≤ k < n→ m ≥ a[k]
m is in array: ∃k .0 ≤ k < n ∧m = a[k]
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Annotated Code
{n >= 1}
m := a[0]
i := 1;
{m = a[0] /\ i = 1 /\ n >= 1 }
==>
{forall k. 0 <= k < i -> m >= a[k] /\ i <= n}
while (i < n) do
if a[i] > m then m := a[i] else m := m;
i := i + 1
{forall k. 0 <= k < i -> m >= a[k] /\ i <= n /\ i >= n}
==>
{forall k. 0 <= k < n -> m <= a[k]}
Invariant. I ≡ ∀k .0 ≤ k < i → m ≤ a[k] ∧ i ≤ n
initially: m = a[0] ∧ i = 1→ I
at end: I ∧ i ≥ n→ ∀k .0 ≤ k < n→ m ≤ a[i ]
Remark. Can turn this into a formal proof as before.
48 / 49
References
The textbook has material on Hoare Logic
Grassman & Tremblay, “Logic and Discrete Mathematics: A
Computer Science Perspective”, Prentice-Hall, Chapter 9, pp.
481-518.
Some nice online notes with lots of examples:
Gordon, “Specification and Verification I”, http://www.cl.cam.ac.
uk/~mjcg/Lectures/SpecVer1/SpecVer1.html, Chapters 1-2, pp.
7-46.
A comprehensive early history of Hoare Logic appears in
Apt, K.R., Ten Years of Hoare Logic: A Survey”, ACM Transactions
on Programming Languages and Systems, October, 1981.
49 / 49 | 9,809 | 24,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.812455 |
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http://cn.metamath.org/mpeuni/kgentopon.html | 1,653,504,575,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00049.warc.gz | 11,291,473 | 10,156 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > kgentopon Structured version Visualization version GIF version
Theorem kgentopon 21389
Description: The compact generator generates a topology. (Contributed by Mario Carneiro, 22-Aug-2015.)
Assertion
Ref Expression
kgentopon (𝐽 ∈ (TopOn‘𝑋) → (𝑘Gen‘𝐽) ∈ (TopOn‘𝑋))
Proof of Theorem kgentopon
Dummy variables 𝑦 𝑥 𝑘 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 uniss 4490 . . . . . . 7 (𝑥 ⊆ (𝑘Gen‘𝐽) → 𝑥 (𝑘Gen‘𝐽))
2 kgenval 21386 . . . . . . . . 9 (𝐽 ∈ (TopOn‘𝑋) → (𝑘Gen‘𝐽) = {𝑥 ∈ 𝒫 𝑋 ∣ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → (𝑥𝑘) ∈ (𝐽t 𝑘))})
3 ssrab2 3720 . . . . . . . . 9 {𝑥 ∈ 𝒫 𝑋 ∣ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → (𝑥𝑘) ∈ (𝐽t 𝑘))} ⊆ 𝒫 𝑋
42, 3syl6eqss 3688 . . . . . . . 8 (𝐽 ∈ (TopOn‘𝑋) → (𝑘Gen‘𝐽) ⊆ 𝒫 𝑋)
5 sspwuni 4643 . . . . . . . 8 ((𝑘Gen‘𝐽) ⊆ 𝒫 𝑋 (𝑘Gen‘𝐽) ⊆ 𝑋)
64, 5sylib 208 . . . . . . 7 (𝐽 ∈ (TopOn‘𝑋) → (𝑘Gen‘𝐽) ⊆ 𝑋)
71, 6sylan9ssr 3650 . . . . . 6 ((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) → 𝑥𝑋)
8 iunin2 4616 . . . . . . . . . 10 𝑦𝑥 (𝑘𝑦) = (𝑘 𝑦𝑥 𝑦)
9 uniiun 4605 . . . . . . . . . . 11 𝑥 = 𝑦𝑥 𝑦
109ineq2i 3844 . . . . . . . . . 10 (𝑘 𝑥) = (𝑘 𝑦𝑥 𝑦)
11 incom 3838 . . . . . . . . . 10 (𝑘 𝑥) = ( 𝑥𝑘)
128, 10, 113eqtr2i 2679 . . . . . . . . 9 𝑦𝑥 (𝑘𝑦) = ( 𝑥𝑘)
13 cmptop 21246 . . . . . . . . . . 11 ((𝐽t 𝑘) ∈ Comp → (𝐽t 𝑘) ∈ Top)
1413ad2antll 765 . . . . . . . . . 10 (((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝐽t 𝑘) ∈ Top)
15 incom 3838 . . . . . . . . . . . 12 (𝑦𝑘) = (𝑘𝑦)
16 simplr 807 . . . . . . . . . . . . . 14 (((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → 𝑥 ⊆ (𝑘Gen‘𝐽))
1716sselda 3636 . . . . . . . . . . . . 13 ((((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) ∧ 𝑦𝑥) → 𝑦 ∈ (𝑘Gen‘𝐽))
18 simplrr 818 . . . . . . . . . . . . 13 ((((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) ∧ 𝑦𝑥) → (𝐽t 𝑘) ∈ Comp)
19 kgeni 21388 . . . . . . . . . . . . 13 ((𝑦 ∈ (𝑘Gen‘𝐽) ∧ (𝐽t 𝑘) ∈ Comp) → (𝑦𝑘) ∈ (𝐽t 𝑘))
2017, 18, 19syl2anc 694 . . . . . . . . . . . 12 ((((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) ∧ 𝑦𝑥) → (𝑦𝑘) ∈ (𝐽t 𝑘))
2115, 20syl5eqelr 2735 . . . . . . . . . . 11 ((((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) ∧ 𝑦𝑥) → (𝑘𝑦) ∈ (𝐽t 𝑘))
2221ralrimiva 2995 . . . . . . . . . 10 (((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → ∀𝑦𝑥 (𝑘𝑦) ∈ (𝐽t 𝑘))
23 iunopn 20751 . . . . . . . . . 10 (((𝐽t 𝑘) ∈ Top ∧ ∀𝑦𝑥 (𝑘𝑦) ∈ (𝐽t 𝑘)) → 𝑦𝑥 (𝑘𝑦) ∈ (𝐽t 𝑘))
2414, 22, 23syl2anc 694 . . . . . . . . 9 (((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → 𝑦𝑥 (𝑘𝑦) ∈ (𝐽t 𝑘))
2512, 24syl5eqelr 2735 . . . . . . . 8 (((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → ( 𝑥𝑘) ∈ (𝐽t 𝑘))
2625expr 642 . . . . . . 7 (((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) ∧ 𝑘 ∈ 𝒫 𝑋) → ((𝐽t 𝑘) ∈ Comp → ( 𝑥𝑘) ∈ (𝐽t 𝑘)))
2726ralrimiva 2995 . . . . . 6 ((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) → ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → ( 𝑥𝑘) ∈ (𝐽t 𝑘)))
28 elkgen 21387 . . . . . . 7 (𝐽 ∈ (TopOn‘𝑋) → ( 𝑥 ∈ (𝑘Gen‘𝐽) ↔ ( 𝑥𝑋 ∧ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → ( 𝑥𝑘) ∈ (𝐽t 𝑘)))))
2928adantr 480 . . . . . 6 ((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) → ( 𝑥 ∈ (𝑘Gen‘𝐽) ↔ ( 𝑥𝑋 ∧ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → ( 𝑥𝑘) ∈ (𝐽t 𝑘)))))
307, 27, 29mpbir2and 977 . . . . 5 ((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑥 ⊆ (𝑘Gen‘𝐽)) → 𝑥 ∈ (𝑘Gen‘𝐽))
3130ex 449 . . . 4 (𝐽 ∈ (TopOn‘𝑋) → (𝑥 ⊆ (𝑘Gen‘𝐽) → 𝑥 ∈ (𝑘Gen‘𝐽)))
3231alrimiv 1895 . . 3 (𝐽 ∈ (TopOn‘𝑋) → ∀𝑥(𝑥 ⊆ (𝑘Gen‘𝐽) → 𝑥 ∈ (𝑘Gen‘𝐽)))
33 inss1 3866 . . . . . 6 (𝑥𝑦) ⊆ 𝑥
34 elssuni 4499 . . . . . . . 8 (𝑥 ∈ (𝑘Gen‘𝐽) → 𝑥 (𝑘Gen‘𝐽))
3534ad2antrl 764 . . . . . . 7 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → 𝑥 (𝑘Gen‘𝐽))
36 ssid 3657 . . . . . . . . . . . 12 𝑋𝑋
3736a1i 11 . . . . . . . . . . 11 (𝐽 ∈ (TopOn‘𝑋) → 𝑋𝑋)
38 elpwi 4201 . . . . . . . . . . . . . . . 16 (𝑘 ∈ 𝒫 𝑋𝑘𝑋)
3938ad2antrl 764 . . . . . . . . . . . . . . 15 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → 𝑘𝑋)
40 sseqin2 3850 . . . . . . . . . . . . . . 15 (𝑘𝑋 ↔ (𝑋𝑘) = 𝑘)
4139, 40sylib 208 . . . . . . . . . . . . . 14 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝑋𝑘) = 𝑘)
4238adantr 480 . . . . . . . . . . . . . . . 16 ((𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp) → 𝑘𝑋)
43 resttopon 21013 . . . . . . . . . . . . . . . 16 ((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑘𝑋) → (𝐽t 𝑘) ∈ (TopOn‘𝑘))
4442, 43sylan2 490 . . . . . . . . . . . . . . 15 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝐽t 𝑘) ∈ (TopOn‘𝑘))
45 toponmax 20778 . . . . . . . . . . . . . . 15 ((𝐽t 𝑘) ∈ (TopOn‘𝑘) → 𝑘 ∈ (𝐽t 𝑘))
4644, 45syl 17 . . . . . . . . . . . . . 14 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → 𝑘 ∈ (𝐽t 𝑘))
4741, 46eqeltrd 2730 . . . . . . . . . . . . 13 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝑋𝑘) ∈ (𝐽t 𝑘))
4847expr 642 . . . . . . . . . . . 12 ((𝐽 ∈ (TopOn‘𝑋) ∧ 𝑘 ∈ 𝒫 𝑋) → ((𝐽t 𝑘) ∈ Comp → (𝑋𝑘) ∈ (𝐽t 𝑘)))
4948ralrimiva 2995 . . . . . . . . . . 11 (𝐽 ∈ (TopOn‘𝑋) → ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → (𝑋𝑘) ∈ (𝐽t 𝑘)))
50 elkgen 21387 . . . . . . . . . . 11 (𝐽 ∈ (TopOn‘𝑋) → (𝑋 ∈ (𝑘Gen‘𝐽) ↔ (𝑋𝑋 ∧ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → (𝑋𝑘) ∈ (𝐽t 𝑘)))))
5137, 49, 50mpbir2and 977 . . . . . . . . . 10 (𝐽 ∈ (TopOn‘𝑋) → 𝑋 ∈ (𝑘Gen‘𝐽))
52 elssuni 4499 . . . . . . . . . 10 (𝑋 ∈ (𝑘Gen‘𝐽) → 𝑋 (𝑘Gen‘𝐽))
5351, 52syl 17 . . . . . . . . 9 (𝐽 ∈ (TopOn‘𝑋) → 𝑋 (𝑘Gen‘𝐽))
5453, 6eqssd 3653 . . . . . . . 8 (𝐽 ∈ (TopOn‘𝑋) → 𝑋 = (𝑘Gen‘𝐽))
5554adantr 480 . . . . . . 7 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → 𝑋 = (𝑘Gen‘𝐽))
5635, 55sseqtr4d 3675 . . . . . 6 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → 𝑥𝑋)
5733, 56syl5ss 3647 . . . . 5 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → (𝑥𝑦) ⊆ 𝑋)
58 inindir 3864 . . . . . . . 8 ((𝑥𝑦) ∩ 𝑘) = ((𝑥𝑘) ∩ (𝑦𝑘))
5913ad2antll 765 . . . . . . . . 9 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝐽t 𝑘) ∈ Top)
60 simplrl 817 . . . . . . . . . 10 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → 𝑥 ∈ (𝑘Gen‘𝐽))
61 simprr 811 . . . . . . . . . 10 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝐽t 𝑘) ∈ Comp)
62 kgeni 21388 . . . . . . . . . 10 ((𝑥 ∈ (𝑘Gen‘𝐽) ∧ (𝐽t 𝑘) ∈ Comp) → (𝑥𝑘) ∈ (𝐽t 𝑘))
6360, 61, 62syl2anc 694 . . . . . . . . 9 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝑥𝑘) ∈ (𝐽t 𝑘))
64 simplrr 818 . . . . . . . . . 10 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → 𝑦 ∈ (𝑘Gen‘𝐽))
6564, 61, 19syl2anc 694 . . . . . . . . 9 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → (𝑦𝑘) ∈ (𝐽t 𝑘))
66 inopn 20752 . . . . . . . . 9 (((𝐽t 𝑘) ∈ Top ∧ (𝑥𝑘) ∈ (𝐽t 𝑘) ∧ (𝑦𝑘) ∈ (𝐽t 𝑘)) → ((𝑥𝑘) ∩ (𝑦𝑘)) ∈ (𝐽t 𝑘))
6759, 63, 65, 66syl3anc 1366 . . . . . . . 8 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → ((𝑥𝑘) ∩ (𝑦𝑘)) ∈ (𝐽t 𝑘))
6858, 67syl5eqel 2734 . . . . . . 7 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ (𝑘 ∈ 𝒫 𝑋 ∧ (𝐽t 𝑘) ∈ Comp)) → ((𝑥𝑦) ∩ 𝑘) ∈ (𝐽t 𝑘))
6968expr 642 . . . . . 6 (((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) ∧ 𝑘 ∈ 𝒫 𝑋) → ((𝐽t 𝑘) ∈ Comp → ((𝑥𝑦) ∩ 𝑘) ∈ (𝐽t 𝑘)))
7069ralrimiva 2995 . . . . 5 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → ((𝑥𝑦) ∩ 𝑘) ∈ (𝐽t 𝑘)))
71 elkgen 21387 . . . . . 6 (𝐽 ∈ (TopOn‘𝑋) → ((𝑥𝑦) ∈ (𝑘Gen‘𝐽) ↔ ((𝑥𝑦) ⊆ 𝑋 ∧ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → ((𝑥𝑦) ∩ 𝑘) ∈ (𝐽t 𝑘)))))
7271adantr 480 . . . . 5 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → ((𝑥𝑦) ∈ (𝑘Gen‘𝐽) ↔ ((𝑥𝑦) ⊆ 𝑋 ∧ ∀𝑘 ∈ 𝒫 𝑋((𝐽t 𝑘) ∈ Comp → ((𝑥𝑦) ∩ 𝑘) ∈ (𝐽t 𝑘)))))
7357, 70, 72mpbir2and 977 . . . 4 ((𝐽 ∈ (TopOn‘𝑋) ∧ (𝑥 ∈ (𝑘Gen‘𝐽) ∧ 𝑦 ∈ (𝑘Gen‘𝐽))) → (𝑥𝑦) ∈ (𝑘Gen‘𝐽))
7473ralrimivva 3000 . . 3 (𝐽 ∈ (TopOn‘𝑋) → ∀𝑥 ∈ (𝑘Gen‘𝐽)∀𝑦 ∈ (𝑘Gen‘𝐽)(𝑥𝑦) ∈ (𝑘Gen‘𝐽))
75 fvex 6239 . . . 4 (𝑘Gen‘𝐽) ∈ V
76 istopg 20748 . . . 4 ((𝑘Gen‘𝐽) ∈ V → ((𝑘Gen‘𝐽) ∈ Top ↔ (∀𝑥(𝑥 ⊆ (𝑘Gen‘𝐽) → 𝑥 ∈ (𝑘Gen‘𝐽)) ∧ ∀𝑥 ∈ (𝑘Gen‘𝐽)∀𝑦 ∈ (𝑘Gen‘𝐽)(𝑥𝑦) ∈ (𝑘Gen‘𝐽))))
7775, 76ax-mp 5 . . 3 ((𝑘Gen‘𝐽) ∈ Top ↔ (∀𝑥(𝑥 ⊆ (𝑘Gen‘𝐽) → 𝑥 ∈ (𝑘Gen‘𝐽)) ∧ ∀𝑥 ∈ (𝑘Gen‘𝐽)∀𝑦 ∈ (𝑘Gen‘𝐽)(𝑥𝑦) ∈ (𝑘Gen‘𝐽)))
7832, 74, 77sylanbrc 699 . 2 (𝐽 ∈ (TopOn‘𝑋) → (𝑘Gen‘𝐽) ∈ Top)
79 istopon 20765 . 2 ((𝑘Gen‘𝐽) ∈ (TopOn‘𝑋) ↔ ((𝑘Gen‘𝐽) ∈ Top ∧ 𝑋 = (𝑘Gen‘𝐽)))
8078, 54, 79sylanbrc 699 1 (𝐽 ∈ (TopOn‘𝑋) → (𝑘Gen‘𝐽) ∈ (TopOn‘𝑋))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 383 ∀wal 1521 = wceq 1523 ∈ wcel 2030 ∀wral 2941 {crab 2945 Vcvv 3231 ∩ cin 3606 ⊆ wss 3607 𝒫 cpw 4191 ∪ cuni 4468 ∪ ciun 4552 ‘cfv 5926 (class class class)co 6690 ↾t crest 16128 Topctop 20746 TopOnctopon 20763 Compccmp 21237 𝑘Genckgen 21384 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-ral 2946 df-rex 2947 df-reu 2948 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-pss 3623 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-tp 4215 df-op 4217 df-uni 4469 df-int 4508 df-iun 4554 df-br 4686 df-opab 4746 df-mpt 4763 df-tr 4786 df-id 5053 df-eprel 5058 df-po 5064 df-so 5065 df-fr 5102 df-we 5104 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-pred 5718 df-ord 5764 df-on 5765 df-lim 5766 df-suc 5767 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-om 7108 df-1st 7210 df-2nd 7211 df-wrecs 7452 df-recs 7513 df-rdg 7551 df-oadd 7609 df-er 7787 df-en 7998 df-fin 8001 df-fi 8358 df-rest 16130 df-topgen 16151 df-top 20747 df-topon 20764 df-bases 20798 df-cmp 21238 df-kgen 21385 This theorem is referenced by: kgenuni 21390 kgenftop 21391 kgenhaus 21395 kgenidm 21398 kgencn 21407 kgencn3 21409 kgen2cn 21410
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(a) What is the probability of obtaining tails and a five? (Enter the probability as a fraction.)
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Feb 24th, 2015
question a)
the probability is:
1/2 * 1/6 = 1/12
question b)
the probability is:
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Feb 24th, 2015
...
Feb 24th, 2015
...
Feb 24th, 2015
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Bioinformatics. Ayesha M. Khan Spring 2013. Some statistics of local sequence comparison (BLAST).
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Lec-6
Bioinformatics
Ayesha M. Khan
Spring 2013
Some statistics of local sequence comparison (BLAST)
• Once BLAST has found a similar sequence to the query in the database, it is helpful to have some idea of whether the alignment is “good” and whether it portrays a possible biological relationship, or whether the similarity observed is attributable to chance alone.
• BLAST uses statistical theory to produce a bit score and expect value (E-value) for each alignment pair (query to hit).
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BLAST Results: Scores and Values
Max score = highest alignment score (bit-score) between the query sequence and the database sequence segment.
Total score = sum of alignment scores of all segments from the same database sequence that match the query sequence (calculated over all segments). This score is different from the max score if several parts of the database sequence match different parts of the query sequence.
Query coverage = percent of the query length that is included in the aligned segments. This coverage is calculated over all segments.
E-value = number of alignments expected by chance with a particular score or better.
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Some details: Bit score
• The bit score gives an indication of how good the alignment is; the higher the score, the better the alignment.
• In general terms, this score is calculated from a formula that takes into account the alignment of similar or identical residues, as well as any gaps introduced to align the sequences.
• Key element substitution matrix
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Bit score (contd.)
• The BLOSUM62 matrix is the default for most BLAST programs, the exceptions being blastn, megaBLAST and discontigmegablast (programs that perform nucleotide–nucleotide comparisons and hence do not use protein-specific matrices).
• Bit scores are normalized, which means that the bit scores from different alignments can be compared, even if different scoring matrices have been used.
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Some details: E-value
The E-value gives an indication of the statistical significance of a given pairwise alignment and reflects the size of the database and the scoring system used. The lower the E-value, the more significant the hit. A sequence alignment that has an E-value of 0.05 means that this similarity has a 5 in 100 (1 in 20) chance of occurring by chance alone.
E=Kmne-λS
m, n is size of the search space (n is length of query sequence, m is length of the database)
K is a scale parameter for size of search space
λ is a scale parameter for scoring method
S is bit score
Lec-6
Difference between BLOSUM and PAM matrices
• BLOSUM comes from alignments of shorter sequences-blocks of sequences that match each other at some defined level of similarity. The BLOSUM method thereby incorporates much more data into its matrices, and is therefore, presumably more accurate.
• PAM is derived from alignments of proteins.
• BLOSUM matrices tend to be more sensitive to distant relationships than PAM.
• BLOSUM tends to give higher scores to substitutions involving hydrophilic amino acids and lower scores to substitutions involving hydrophobic amino acids than PAM.
• Substitutions of rare amino acids are more tolerated by BLOSUM.
• General rules:
-Use higher PAM or lower BLOSUM matrices for more divergent sequences
-Use lower PAM or higher BLOSUM matrices for more closely related sequences
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Concept of Gaps in Alignment
• Sequences may have diverged from a common ancestor through various types of mutations:
• Substitutions
• Insertions
• Deletions
The latter two will result in gaps in alignments
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Gap Penalty
• Gap penalties are used during sequence alignments to penalize the gaps.
• The gap extension penalty is usually much smaller, for instance, 10 insertions of one nucleotide each should be harder than one insertion of 10 nucleotides.
• That is, gap opening is less probable than a single gap extending over more than one nucleotide. Hence a single mutation event (causing incorporation or deletion of more than one nucleotide) is more probable than multiple mutation events.
Lec-6
Gap Penalty
• Linear gap penalties
• Simplest type of gap penalty
• The overall penalty for one large gap is the same as for many small gaps
• wk=c L
• Affine gap penalties
• Have a gap opening penalty c, and a gap extension penalty, e
• wk=c +(L-1)e
Lec-6
BLAST & FASTA: heuristic methods
• BLAST & FASTA use heuristic methods that attempt to approximate the optimal local similarity shared by two sequences.
• Use word or k-tuple methods
• They align two sequences very quickly , by first searching for identical short stretches of sequences (called words, or k-tuples) and then joining these words into an alignment by the dynamic programming method.
Lec-6
BLAST…
• The BLAST programs are used to find high-scoring local alignments between a query sequence and a target database.
• The BLAST algorithm is based on the fact that true match alignments are very likely to contain short stretch of identities, or very high scoring matches somewhere within them.
• So BLAST initially looks for such short stretches and uses them as ‘seeds’ from which it extends out in search of a good longer alignment.
Lec-6
Main stages of BLAST
• Remove (filter) low-complexity regions from Q
• Harvest k-tuples (triples) from Q
• Expand each triple into ~50 high-scoring words
• Seed a set of possible alignments
• Generate high-scoring pairs (HSP)s from the seeds
• Test the significance of matches from the HSPs
• Report the alignments found from the HSPs
Lec-6
Multiple Sequence Alignment
Why do we need to carry out multiple sequence alignments?
• To make connections between more than two family members
• To reveal conserved family characteristics
MSA is a 2D table rows represent individual sequences and columns the residue positions.
Absolute position: Property of the sequence
Relative position: Property of the alignment
Lec-6
MSA: computational complexity
(O (m1 m2)
O: order of the time taken by the algorithm, and m1 and m2 are the sequence lengths.
When considering more sequences, the time complexity becomes O(m1,m2,m3,….ml)
where ml is the length of the last sequence in the comparison set
Lec-6
Simultaneous methods vs progressive methods
• Simultaneous methods: Align all the sequences in a given set at once
• Extension of a 2D matrix to three or more dimensions
• No. of dimensions reflect the no. of sequences to be aligned
• Work best on small sets of short sequences
• Progressive methods: Align pairs of sequences or building sequence clusters
• Use heuristics to reach an alignment in a timely and cost-efficient manner
Lec-6
MSA models
• There are several models for assessing the score of a given multiple sequence alignment. The most popular ones are sum-of-pairs (SP), tree alignment, and consensus alignment.
Note: which of the above models are progressive alignments and which are based on dynamic programming?(should be able to answer after a few slides)
Lec-6
Sum-of-pairs (SP)
• Recall that:
The standard computational formulation of the pairwiseproblem is to identify the alignment that maximizes protein sequence similarity, which is typically defined as the sum of substitution matrix scores for each aligned pair of residues, minus some penalties for gaps.
• The mathematically — though not necessarily biologically — exact solution can be found in a fraction of a second for a pair of proteins. This approach is generalized to the multiple sequence case by seeking an alignment that maximizes the sum of similarities for all pairs of sequences, i.e. the sum-of- pairs, or SP, score.
Lec-6
Sum-of-pairs (SP)...
• The SP score for the complete alignment M is the sum of the scores for each column (mi) in the alignment:
We wish to use the SP method to score the following alignments of these three sequences:
Alignment #1 Alignment #2
T-GC-G TGC-G
-AGCTG AGCTG
-AGC-G AGC-G
Example:
We wish to align the following three DNA sequences:
S1 = TGCG
S2 = AGCTG
S3 = AGCG
Lec-6
Sum-of-pairs (SP)...
We will use the following simplified DNA substitution matrix:
• s(x,y) = 1: when x = y [match]
• s(x,y) = -1: when x ! y [mismatch]
• s(x,-) = -2: [gap]
• s(-,y) = -2: [gap]
• s(-,-) = 0: to prevent double counting of gaps
We will construct the following matrices M for each alignment:
Lec-6
Sum-of-pairs (SP)...
The SP score for each alignment is calculated by summing the individual scores for each column in the matrix.
Using the simplified substitution matrix, the Sum of Pairs method ranks the second alignment as the higher scoring alignment.
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Progressive alignmentIt is a heuristic method!
• Up until about 1987, multiple alignments would typically be
• constructed manually, although a few computer methods did exist. Around that time, algorithms based on the idea of progressive alignment appeared. In this approach, a pairwise alignment algorithm is used iteratively, first to align the most closely related pair of sequences, then the next most similar one to that pair, and so on.
• The rule “once a gap, always a gap” was implemented, on
• the grounds that the positions and lengths of gaps introduced between more similar pairs of sequences should not be affected by more distantly related ones.
Lec-6
Progressive alignment: CLUSTALW
The three basic steps in the CLUSTAL W approach are shared by all progressive alignment algorithms*:
A. Calculate a matrix of pairwise distances based on pairwise alignments between the sequences
B. Use the result of A to build a guide tree, which is an inferred phylogeny for the sequences
C. Use the tree from B to guide the progressive alignment of the sequences
Lec-6
Progressive alignment: CLUSTALWhttp://www.ebi.ac.uk/clustalw/
• The basic idea is to use a series of pairwise alignments to align larger and larger groups of sequences, following the branching order of the guide tree. We proceed from the tips of the rooted tree towards the root.
• At each stage a full dynamic programming algorithm is used, with a residue scoring matrix (e.g., a PAM or a BLOSUM matrix) and gap opening and extension penalties. Each step consists of aligning two existing alignments.
• Scores at a position are averages of all pairwisescores for residues in the two sets of sequences using matrices with only positive values.
Lec-6
Pairwise progressive dynamic programming-liabilities
(1) dependence on initial pairwise sequence alignments and the order of alignment
-ordering them from most similar to least similar usually makes biological sense and works very well.
(2) dependence on substitution matrices and gap penalties
Lec-6
Common usage of MSA
• Detecting similarities between sequences (closely/distantly related)
• Detecting conserved regions/motifs in the sequences
• Detection of structural homologies;
Patterns of hydrophobicity/hydrophilicity , gaps etc.
• Thus assisting the improved prediction of secondary and tertiary structures and loops and variable regions.
• Predict features of aligned sequences like conserved positions which may have structural or functional importance
• Making patterns or profiles that can be further used to predict new sequences falling in a given family
• Computing consensus sequence
• Inferring evolutionary trees or linkage-phylogenetic analysis etc
• Deriving profiiles of hidden markov models (HMMs) that can be used to remove distant sequences (outliers) from the protein families
Lec-6
Applicability of MSA
• Very useful in the development of PCR primers and hybridization probes;
• Great for producing annotated, publication quality, graphics and illustrations;
• Invaluable in structure/function studies through homology inference;
• Recognizable structural conservation between true homologues extends way beyond statistically significant sequence similarity.
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Applicability of MSA- contd.
• Essential for building “profiles” for remote homology similarity searching; and
• Required for molecular evolutionary phylogenetic inference programs.
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For a given group of sequences, there is no single "correct" alignment, only an alignment that is "optimal" according to some set of calculations.
• Determining what alignment is best for a given set of sequences is really up to the judgement of the investigator.
“To raise new questions, new possibilities, to regard old problems from a new angle, require creative imagination and marks real advance in science” Albert Einstein
Lec-6 | 2,961 | 13,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.878178 |
https://stats.stackexchange.com/questions/483100/modelling-uncertainty-at-both-the-individual-and-population-level-with-beta-dist | 1,618,660,950,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00113.warc.gz | 624,763,361 | 36,907 | # Modelling uncertainty at both the individual and population level with beta-distributions
I want to measure the distribution of a population's performance on a test. Each person takes a version of the test with a random selection of N questions from a large pool of possible questions. There is uncertainty in this score as an estimate of their "true score" on the set of all possible questions, which I model using a beta distribution for each person (with the parameters being the counts of incorrect answers and correct answers, plus 1 for a uniform prior).
Now I want to combine each person's score to find the distribution of scores for the population, while propagating uncertainty correctly. How should I combine my measures of individuals' scores, along with their beta-distributed uncertainty, into a population measure?
Further thought: As I see it, the beta distributions around each individual's score represent epistemic uncertainty, while the distribution of scores represents intrinsic variability in the population, and I'm not sure about the best/correct way of incorporating the epistemic uncertainty from the individual level into the population estimate. Could this be done with a hierarchical model, where the individual's beta distribution parameters are themselves drawn from a population distribution? Hope this makes sense..! | 250 | 1,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-17 | latest | en | 0.912445 |
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# Royi Avital
### Ben Gurion University
Last seen: 1 giorno fa Attivo dal 2011
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List of New / Updated Functions per MATLAB Version
Is there a way to generate a list of new (Or updated) functions in MATLAB per release? I know I can see in documentation per fu...
circa un anno fa | 2 risposte | 0
### 2
risposte
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Assign a Sub Array to Array Knowing the Number of Dimensions at Run Time
OK, It turns out it can be done using Cell Arrays: vSizeB = size(tB); numDims = length(vSizeB); %<! Equals to ndims(tB) cId...
circa un anno fa | 0
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Domanda
Assign a Sub Array to Array Knowing the Number of Dimensions at Run Time
Assume we have tA and tB with the same number of dimesions. We also have all(size(tB) <= size(tA)) == true. The task is to emb...
circa un anno fa | 2 risposte | 0
### 2
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Domanda
Convert a LayerGraph into a DAGNetwork - Validate and Initialize the Net
In theory, the function assembleNetwork() should convert a LayerGraph into a DAGNetwork. Yet in practice, it doesn't always wor...
oltre un anno fa | 1 risposta | 0
### 1
risposta
Risposto
Since IMGUR uses REST based API you can do it in MATLAB. Best way to start is trying to replicate the Python library: IMGUR P...
circa 2 anni fa | 0
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How do I initialize a complex array to zeros in MATLAB?
In later versions one could do something like vB = zeros(3, 1, 'like', 1j);.
circa 2 anni fa | 1
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How can I set a custom path in Matlab R2017a on MacOS?
You should use setenv() to set the path. Pay attention that usually you want to add something to the path and not override it. ...
oltre 2 anni fa | 1
| accettato
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Is there a way to Call Jupyter from Matlab?
I am not sure what do you mean by calling Jupyter from MATLAB. If you mean using MATLAB in Jupyter, it is certainly doable and ...
circa 3 anni fa | 0
| accettato
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how to implement MILP in matlab
I also created a MEX wrapper around GNU Linear Programming Kit (GLPK). You may find it at https://www.mathworks.com/matlabcentr...
circa 3 anni fa | 1
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Most Efficient Way to Construct the Matrices to Extract the Lower and Upper Triangle from a Vectorized Matrix
My current solution: function [ mLU ] = GenerateTriangleExtractorMatrix( numRows, triangleFlag, diagFlag ) EXTRACT_LOWER_TRI...
circa 3 anni fa | 0
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Most Efficient Way to Construct the Matrices to Extract the Lower and Upper Triangle from a Vectorized Matrix
Given a matrix and its vector form I am after the most efficient way to build the matrices and which extracts the lower and ...
circa 3 anni fa | 3 risposte | 1
### 3
risposte
Domanda
Generating Toeplitz Matrix which Matches the Convolution Shape Same
Given a filter vH I'm looking for vectors vR and vC such that: toeplitz(vC, vR) * vX = conv(vX, vH, 'same'); For instance, for...
circa 3 anni fa | 3 risposte | 1
### 3
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Matlab equivalent to iPython Notebook
As written above, there is the option of MATLAB's Live Editor. I find it to be really limited vs. iPython Notebook. They are no...
circa 3 anni fa | 1
Domanda
Faster and More Efficient squareform()
I'm using squareform(pdist(mX)) fairly often. The problem is squareform() is so slow it makes use of pdist2(mX, mX) faster. ...
oltre 3 anni fa | 0 risposte | 0
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Domanda
How to Imply the Input Array Dimensions to MATLAB Coder
Assuming I have a simple function in MATLAB: function [ mG ] = ProcessImage( mI ) %#codegen mG = edge(mI, 'Sobel', [], 'bo...
oltre 3 anni fa | 1 risposta | 0
### 1
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MATLAB's JIT Engine
Do you think MATLAB's JIT engine is oudated? It does look so. Look at Julia Language: http://julialang.org/ What do you think?...
circa 4 anni fa | 5 risposte | 2
### 5
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Can I use Microsoft Visual Studio 2019 with MATLAB R2019a or R2018b?
I'm not sure how the search for Compiler works for the mex command. From 5 minutes research I did I came to the conclusion it w...
circa 4 anni fa | 2
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SuiteSparse Installation: Cannot open include file: 'regex.h'
Why would you need SuiteSparse in MATLAB? I though MATLAB uses SuiteSparse to begin with.
circa 4 anni fa | 0
Domanda
Improving Speed and Reducing Memory Consumption with Creation of 2D Sparse Convolution Matrix
In a previous question of mine, Creating Convolution Matrix of 2D Kernel for Different Shapes of Convolution, among answers of t...
circa 4 anni fa | 0 risposte | 0
### 0
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Creating Convolution Matrix of 2D Kernel for Different Shapes of Convolution
Here is my solution which build Doubly Block Toeplitz Matrix: function [ mK ] = Create2DKernelConvMtxSparse( mH, numRows, numCo...
circa 4 anni fa | 1
Domanda
Creating Convolution Matrix of 2D Kernel for Different Shapes of Convolution
MATLAB, for thos who have access to Image Processing Toolbox offers the function convmtx2(). Yet there are 2 issues: It is onl...
circa 4 anni fa | 4 risposte | 1
### 4
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How can I determine whether a string contains a substring?
I guess that since MATLAB R2016b it is recommended to use contains : Determine if pattern is in strings - MATLAB contains - M...
oltre 4 anni fa | 1
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Functional form of the colon (:) operator?
I really wish MATLAB would add function to vectorize arrays into column vector as the colon operator does.
oltre 5 anni fa | 0
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Change opacity of Lines
In newer versions of MATLAB you can do that easily using the Color property of the line. By default it is RGB array (1 x 3)....
oltre 5 anni fa | 7
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GUI - Slider which has two slide bars
Maybe you meant a Range Slider which you can create as shown in <http://undocumentedmatlab.com/blog/sliders-in-matlab-gui Slider...
oltre 5 anni fa | 0
Domanda
Display Pixel Values in Image - Alternative to showPixelValues()
Hello, I would like to display an image and its pixel values in a similar manner to <https://www.mathworks.com/matlabcentral/...
oltre 5 anni fa | 2 risposte | 0
### 2
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What is missing from MATLAB?
What I miss the most is better forum for the community. We need Math Equation support through MathJaX (Like in StackExchange ...
oltre 5 anni fa | 0
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How to perform KNN regression
At each point you want to calculate the value (x, y) find the K closest point (For x). Then average them to create the new va...
oltre 5 anni fa | 0
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Averaging Overlapping Pixels in Sliding Window Operation
Another option would be: I = reshape(accumarray(mIdx(:), mZ(:), [(M * N), 1], @(x) mean(x)), [M, N]); Where mZ is the ...
circa 6 anni fa | 0
Domanda
Saving Figure Data into a Matrix
Hello. I'm using 'image()' to display an image. I'm adding some annotations using text. I'd like to save the final resu...
oltre 6 anni fa | 2 risposte | 0
risposte | 1,940 | 6,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | longest | en | 0.641099 |
https://www.interactive.onlinemathlearning.com/percentage2.php?action=generate&numProblems=10 | 1,725,769,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00265.warc.gz | 787,456,083 | 6,853 | OML-IZ Search
## Percentages
Calculate and enter the answers. You may use the TAB key to move to the next question. When you are done, click Submit.
1. 298.8 is what % of 2988?
2. 243.96 is what % of 2033?
3. 5847.09 is what % of 8727?
4. 809.54 is what % of 4762?
5. 1441.34 is what % of 3793?
6. 1947.91 is what % of 4751?
7. 1581.44 is what % of 2824?
8. 999.58 is what % of 1886?
9. 2330.93 is what % of 4757?
10. 5302 is what % of 6025?
We hope that the free math worksheets have been helpful. We encourage parents and teachers to adjust the worksheets according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles. | 249 | 794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.908278 |
https://termpapershelp.net/in-the-late-1960s-milton-friedman-and-edmund-phelps-argued-that/ | 1,642,400,685,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00330.warc.gz | 572,808,921 | 12,291 | # in the late 1960s, milton friedman and edmund phelps argued that
1. In the late 1960s, Milton Friedman and Edmund Phelps argued that there was not a structural relationship between inflation and unemployment rates. In particular, the trade off could only exist in the short -run.
a) (10 points) The tradeoff between unemployment and inflation was much discussed throughout the 1960s as there appeared to be a clear tradeoff between unemployment and inflation. In fact, we traced out the Phillips curve beginning in the early 1960s and continuing through the end of the decade. In the space below, recreate the Phillips curve that we constructed in the lectures, being sure to label diagram completely. At minimum, you should have unemployment / inflation combinations for 1961, 1962, 1964, 1966, and 1969. Connect the dots and we have the tradeoff between unemployment and inflation during the 1960s, aka, the Phillips curve.
b) (10 points) Now explain why the Phillips curve that you constructed can only be a short-run phenomenon at best. In particular, explain exactly why, as we went through the decade of the 1960s, we continuously move up and to the northwest along the Phillips curve…. from relatively high rates of unemployment and low inflation to relatively low rates of unemployment and high rates of inflation. In your answer, make sure discuss the short run aspect of this curve and why, in the long-run, the Phillips curve is vertical (hint: expected inflation, unexpected inflation, actual real wages, and expected real wages should be a big part of your explanation).
1. In this question, we are going dig deeper into the Taylor Rule and it variants (modifications). You will need the following links to answer the following questions. Note, each link takes you to a page where right above the graph on left, there is a “download data in graph” tab – click on it and that will give you access to the data you need.
Effective Federal Funds Rate
As Taylor assumed, we assume the equilibrium real rate of interest, r* = 2% and the optimal inflation rate, the target inflation rate is also equal to 2%.
a) (10 points) Using the ‘standard’ Taylor rule with Inflation PCE (not the core), and using end of 2011 data (2011-10-01) what is the federal funds rate implied by the ‘standard’ Taylor Rule? According to the actual federal funds rate (use the Effective Federal Funds Rate), is the Fed being hawkish or dovish? Explain.
b) (10 points) Repeat part a) using the modified version of the Taylor using the unemployment gap instead of the GDP gap just like we did in the lectures. Also, use the PCE core rate of inflation instead of overall inflation like you used above – the Fed arguably cares more about core inflation than overall inflation. According to the actual federal funds rate (use the Effective Federal Funds Rate), is the Fed being hawkish or dovish? Which “Taylor” rule explains Fed behavior better, the original or the modified Taylor Rule? Explain.
c) (10 points) Let’s go back in time to the fourth quarter of 1965 (1965-10-01) when the “We are all Keynesians” was featured in Time magazine. We argued that this was heyday of Keynesian economics so we would expect to get dovish results. Using the original Taylor Rule that you used in part a) and the modified Taylor Rule that you used in part b), prove that the Fed was dovish according to both versions of the Taylor Rule.
d) (10 points) We now go back to the Volcker period where he was known as being a hawk on inflation. Using the data from the second quarter of 1982 (1982-04-01), prove that the Volcker Fed was hawkish according to both versions of the Taylor Rule
True/ False (40 points total – 2 points each)
1) According to the “We are all Keynesians Now” article, the labor secretary at that time wanted the unemployment rate to fall down to 3%.
2) The misery index in 1980 exceeded 25.
3) The mid to late 1970s was the ‘heyday’ of Keynesian economics in the US economy.
4) Keynes believed that it was the responsibility of the government to use its powers to increase production, incomes and jobs.
5) Consistent with his thought on spending heavily, Keynes was known as an excellent tipper.
6) The steeper the SRAS curve, the steeper the short-run Phillips curve.
7) If the long-run aggregate supply curve is vertical so is the long-run Phillips curve.
8) Friedman and Phelps agreed that there is a trade-off between unemployment and inflation, but only in the long run.
9) If actual inflation is lower than expected inflation, then the actual real wage is higher than the expected real wage. This being the case, firms will lay off workers.
10) According to the Taylor Rule described in the lectures, if the Fed is getting an A+, then the federal funds rate should be set at 5%
11) According to the Taylor principle, if actual inflation rises by 1% over target inflation, then the Fed should raise the federal funds rate by 2% to make sure that the real federal funds rate rises which is referred to as “leaning against the wind.
12) If the actual federal funds rate is higher than the funds rates implied by the Taylor rule, then we say that the central bank is hawkish.
13) If actual inflation rises one percent above target and the central bank raises the actual funds rate by one percent then according to the Taylor rule, the central bank is being hawkish.
14) According to the Taylor rule, the Greenspan Fed was hawkish during the new economy years.
15) According to the Taylor rule, the Greenspan Fed was hawkish during the job-less recovery as well as the job-loss recovery.
16) One way to explain the apparent tradeoff between inflation and unemployment during the 1960s, expected inflation was consistently higher than the actual inflation implying that firms would be willing to higher more workers given this difference between expected and actual inflation. The result therefore would be higher inflation and lower unemployment, consistent with the facts during the 1960s.
17) We argued that the modified version of the Taylor rule during the jobless recovery following the 1990 – 1991 recession explained Greenspan and the Fed’s behavior much better than the original Taylor Rule.
18) According to the Phillips curve analysis, if expected inflation is equal to actual inflation then we are at NAIRU. However, if actual inflation is higher than expected, then the actual unemployment rate will be higher than that associated with NAIRU.
19) If firms and workers had perfect foresight as to inflation so that actual = expected inflation at all times, then the Phillips curve would be vertical and thus, there would be no trade between unemployment and inflation, even in the short run.
20) We argued that a federal funds rate target of 4% is consistent with the stance of monetary policy being neutral as in neither tight nor loose.
"We Offer Paper Writing Services on all Disciplines, Make an Order Now and we will be Glad to Help" | 1,551 | 7,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.942607 |
https://quantumcomputing.stackexchange.com/questions/34841/what-does-u-mean-in-pulse-schedules-for-superconducting-quantum-computers | 1,701,450,619,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00395.warc.gz | 547,138,864 | 41,716 | # What does U mean in pulse schedules for superconducting quantum computers?
What does U channel mean in pulse schedules for superconducting devices? In the image you can see the schedule for a CX gate.
I assume D stands for the drive channels. What does U mean, is it perhaps the control channel? Why the letter U, what it stands for?
• $U$ is generally used to denote a unitary operation, but for your specific case, please link the resource you are referring to, so people can exactly answer your question. From the diagram, I believe $U$ is a signal for performing controlled rotation, where the first pulse is the unitary operation $CR(\pi/4)$ and the second pulse is the unitary operation $CR(-\pi /4)$. Nov 14 at 8:52
• This would be a good answer to my question Nov 14 at 8:53
• $U$ is typically only used for unitaries in a mathematical context -- here we're looking at the pulse schedule of the device and U denotes a channel to drive the qubit at a specific frequency, not a unitary. Nov 14 at 8:55
• Just to make my question more precise, I updated my question and replaced original image with an image for a CX gate schedule. Nov 14 at 8:58
You're already answering your own question: U0 stands for the control channel on qubit 0, as you can read in Table 1 of https://iopscience.iop.org/article/10.1088/2058-9565/aba404/pdf.
In your schedule the CX is implemented as echoed cross resonance gate (ECR), where we drive the control qubit 0 at the frequency of the target qubit 1. Therefore you can see a cross resonance (CR) pulse on the control channel of qubit 0, U0. Since we echo the pulse, you can also see a secondary CR pulse with inverted angle, interleaved with X gates.
You can also see something happening on the drive channel of qubit 1, D1, during the execution of the CR pulse on the control qubit. This is a cancellation tone to improve the CX fidelity. Note that even though the channels are drawn separately, they can be implemented on the same wire on the device. | 485 | 1,996 | {"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} | 2.875 | 3 | CC-MAIN-2023-50 | longest | en | 0.938334 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-845771-8&chapter=10&stp=yes&headerFile=6&state=wa | 1,386,534,854,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163809293/warc/CC-MAIN-20131204133009-00009-ip-10-33-133-15.ec2.internal.warc.gz | 348,652,723 | 6,336 | 1. Which list of numbers best identifies the whole numbers? A. 1, 3, 5, 7, and so on B. 2, 4, 6, 8, and so on C. 1, 2, 3, 4, and so on D. 0, 1, 2, 3, and so on Hint 2. The graph below can best be described as a _____. A. histogram B. stem-and-leaf plot C. line graph D. box graph Hint 3. In which quadrant does the point (-2, 1) lie? A. III B. I C. II D. IV Hint 4. Find the difference 15 - 26. A. -41 B. 41 C. -11 D. 11 Hint 5. The quotient of two integers with the same sign is _____. A. negative B. equal to 0 C. positive D. equal to 1 Hint 6. For any rational number and with b > 0 and d > 0, if then _____. A. ad > bc B. ad < bc C. ad + bc = 1 D. ad = bc Hint 7. The sum of two negative rational numbers is _______. A. negative B. zero C. positive D. one Hint 8. Ten video stores were surveyed to find the cost of renting a new release video. The costs were \$1.99, \$2.99, \$1.75, \$1.99, \$2.50, \$1.50, \$0.99, \$1.99, \$2.99, and \$3.49. What is the mode of this set of data? A. \$1.99 B. \$2.22 C. \$0.99 D. \$1.99 and \$2.99 Hint 9. The graphs of the two equations y = 2x + 5 and are _______. A. neither parallel nor perpendicular B. the same line C. perpendicular D. parallel Hint 10. Arrange the terms of the polynomial -2x + 3x3 - 5 + 6x2 so that the powers of x are in descending order. A. -2x + 3x3 + 6x2 - 5 B. 3x3 + 6x2 - 2x - 5 C. 5 - 2x + 6x2 + 3x3 D. -5 - 2x + 6x2 + 3x3 Hint 11. Which is the correct definition for a number expressed in scientific notation? A. A number is expressed in scientific notation when it is in the form a × 10n, where and n is a whole number. B. A number is expressed in scientific notation when it is in the form a × 10n, where and n is an integer. C. A number is expressed in scientific notation when it is in the form a × 10n, where and n is a whole number. D. A number is expressed in scientific notation when it is in the form a × 10n, where and n is an integer. Hint 12. The speed of light is about 3 × 108 meters per second. How far would a beam of light travel in 5 seconds? Express the answer in scientific notation. A. 1.5 × 106 m B. 1.5 × 105 m C. 1.5 × 1010 m D. 1.5 × 109 m Hint 13. Factor the binomial x2 + 81. A. (x – 18)(x + 5) B. (x + 9)2 C. prime D. (x – 9)(x + 9) Hint 14. Name the property of numbers shown by the statement:23 · 0 = 0 · 23 = 0. A. Substitution Property B. Additive Identity C. Multiplicative Identity D. Multiplicative Property of Zero Hint 15. Evaluate jkl if j = 2, k = -3, and l = -4. A. 24 B. -24 C. 9 D. -9 Hint 16. Marilyn and Rafael spent the same amount of money on school supplies this year. Marilyn bought 4 notebooks and spent \$3 on other school supplies. Rafael bought 5 of the same notebooks and spent \$1.75 on other school supplies. How much does one notebook cost? A. \$0.14 B. \$1.25 C. \$42.75 D. \$0.53 Hint 17. Express as a percent. Round to the nearest percent. A. 44% B. 55% C. 60% D. 37% Hint 18. Alana wants to buy the new CD by the Raspberries. The CD costs \$15.99, but since she was one of the first 50 customers that day, she gets a 20% discount. How much will the CD cost? A. \$19.19 B. \$3.20 C. \$12.79 D. \$31.98 Hint 19. The relation is ___________. A. not a function because it does not pass the vertical line test B. not a function because it does not pass the horizontal line test C. a function because it passes the vertical line test D. a function because it passes the horizontal line test Hint 20. State whether the expression 5jk + 4j is a monomial, binomial, trinomial or not a polynomial. A. binomial B. monomial C. not a polynomial D. trinomial Hint | 1,201 | 3,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.873375 |
https://www.instructables.com/id/Tree-of-life-ring/ | 1,582,460,964,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00085.warc.gz | 766,712,909 | 28,136 | # Tree of Life Ring
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## Introduction: Tree of Life Ring
Already over two years ago, I posted an instructable on how to make a tree of life pendant. Since then I've made quite a few similar pendants, slightly different every time. They always make for a great last-minute gift ; ).
When I saw a tree of life dream catcher, I really wanted to make one myself, so I wanted to look up a few pictures. I most certainly found a few dream catcher pictures, but I also saw a ring that incorporated a tree of life in the design, absolutely amazing. So, my plans of making a dream catcher had to make way for this intriguing project : ).
It turned out great, I love the way it looks, and I hope you'll like it as well!
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Materials
To make a tree of life ring, you will need:
- a mandrel
- pliers
- a ruler
- wire, gauge 21/ 0,81 mm and gauge 26/ 0,46 mm
## Step 2: Bending the Base
Before cutting the parts of wire to bend, it's time for a few calculations.
1. Measure the diameter of the top of your mandrel or the round object you're using. Divide it by 2 and multiply it with pi. This is the length of wire that you'll need for the semi circle.
2. Next you'll need to know the circumference of your finger. A simple technique you can use is this one.
Take this value, subtract the diameter of your mandrel and multiply it with 2.
Add the value of calculation 1 to the value of calculation 2. Add a few centimeters to be sure your wire is long enough and cut two pieces of this length.
Bend the middle of one of the pieces around the top of the mandrel as shown. Bend the wires straight again to create the semi circle you'll need. Repeat this for the other piece of wire.
## Step 3: Shape It
Once you're sure both pieces align perfectly, bend them around the mandrel. You may need to bend the ends a bit further using the pliers, this will definitely help to get the roundness you want.
## Step 4: A Bit of Cutting
As you can see, at this moment there are four layers of wire for the ring base itself, but you'll only need three. Cut one of the wires at the back of the ring in both halves of the ring, as shown in the pictures.
## Step 5: Just to Check
Make sure the short ends (the ones you just cut off) are on the outside of the ring, so when you align them it looks like the picture shown here.
## Step 6: Wrap It Up
Cut off a piece of the gauge 26 / 0,46 mm wire and wrap the middle of it around the back of the ring, the place where both short wires end. This means you'll only need to wrap it around the two middle wires here.
Wrap one of the sides of the wire around the three base wires on its side, the other side around the three base wires on the other side. This will keep the two halves of the ring base from shifting in relation to each other.
## Step 7: Ending the Wire
Once you've reached the end of one of the sides, cut off the wrapping wire in the middle of the base wires of the ring. Use a pair of pliers to pinch the end of the wire between the two base wires.
## Step 8: On the Other Side
The same goes for the other side of the pendant: wrap until you reach the end (or the front, depends on how you look at it) and end the wire between the base wires.
## Step 9: The Completed Base
At this point, the base ring is completed. Take a look at your hard work, try it on, admire it : )
## Step 10: Tree Time
Now, time to start making that tree. Cut off eight pieces of wire that are approximately the same length, about ten centimeters/ 4 inches should be enough.
## Step 11: Starting Small
Take three of the wires and bundle them together in the middle by wrapping them around each other. This will be the trunk of the tree.
## Step 12: Building Up
Wrap the other five pieces of wire around the trunk part you just started. It might look a bit weird at first, but don't worry, the more wires you add the better it becomes.
## Step 13: The Roots
Spread out the wires underneath the trunk and bundle them in pairs. These pairs will be your tree roots. Twist them a bit to keep them together nicely.
## Step 14: Adding the Tree
Before moving on to the branches, it's time to add the tree to the ring base. Place the tree on top of the circle and position it the way you like. If your trunk is a bit too long, don't worry, it's easy to fix that when making the branches in the next step.
Wrap the roots around the bottom half of the circle to secure it. Cut off the ends and clamp them down using pliers to prevent any wire from poking out.
## Step 15: Leaves
Spread the wires on the top of the trunk to create the branches. You can also unwrap a bit of the trunk if they are not positioned the way you want the to. Add the beads to the branches to create leaves. Once you're happy with the way the tree looks, wrap the ends of the branches around the top of the ring to secure them.
I ended up only using six of the eight wires for the branches, otherwise it would have been too crowded. Next step shows how to work away those wires if you have this same problem.
## Step 16: Working Away the Wire
To get rid of the two wires still sticking out right now, just pull them through the branches as shown, cut of the end and clamp down the wire as close to the tree as possible, since you don't want any loose ends.
## Step 17: All Done
And that's it! I hope you liked it and maybe will give this project a try yourself, I'd love to see your result if you do : )
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## 25 Discussions
thank you for the wonderful project! I have to make one right away!
So glad you liked it! I'd love to hear how it goes : )
I will definitely share a photo! You really inspired me to make jewelry again! I made two shawl pins right away but need to find my ring mandrel to do a ring properly. Thanks so much for sharing your ideas!
Very elegant, real nice job done here. :-) I hope all is well!
Love to see you coming back with a bang!
you work wonders with wires! Looks amazing :))
Beautiful ring! | 1,490 | 6,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.927645 |
http://community.qlik.com/message/189661 | 1,419,774,275,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447557824.148/warc/CC-MAIN-20141224185917-00036-ip-10-231-17-201.ec2.internal.warc.gz | 22,071,526 | 63,619 | 12 Replies Latest reply: Jun 19, 2013 3:48 PM by rodrilauro
# Partial Sum Totals Error in Pivot Table
Hi All,
I have created a pivot table and I'm using the partial sum total on one dimension from the presentation tab. The problem is the incorrect totals that are coming up. Any suggestions on how to correct the partial sum totals would really be appreciated.
• ###### Partial Sum Totals Error in Pivot Table
Pivot tables will evaluate your partial sums as expression total, while you probably expect a sum of rows.
Please check the Help pages for 'Sum of rows in pivot tables' using advanced aggregation.
Your expression might need to be changed to something like
=sum( aggr( sum(Value), TableDimension1, TableDimension2, ...))
Hope this helps,
Stefan
• ###### Partial Sum Totals Error in Pivot Table
Hi Stefan,
EXP 1) count( value)
EXP 2) count(value)/count( value2)
I'm using 2 dimensions. Now, I'm using the partial sums on 1 dimension , just to get the total of my expression 1. So, basically in my pivot table I want to add a Total row showing the expression total.
I'm not sure where to use the sum expression you stated above.
• ###### Partial Sum Totals Error in Pivot Table
When you are saying, you want the expression total, I think you should already see it...
Could you post a screenshot of your table here and the values for the total you expect to see?
If you want the sum of rows (what I still assume), you need to use your expression, i.e. the count( value) within the aggr() function, and then a sum around all to get the sum of rows, like
=sum( aggr( count( value), dimension1, dimension2))
You need to replace dimension1, dimension2 by your actual dimension field names, of course.
Regards,
Stefan
• ###### Partial Sum Totals Error in Pivot Table
Hi Stefan,
This is a screenshot of the pivot table. As you could see , the count total in both the columns is showing a wrong total. It is not accurate. I need the total of all the columns at the bottom.
Thank you. I really appreciate your prompt replies.
• ###### Partial Sum Totals Error in Pivot Table
So your first count should be 2660 and your second total 2609, right?
Have you tried the advanced aggregation, like
=sum( aggr( count( value), COLOR, STATUS))
• ###### Partial Sum Totals Error in Pivot Table
Yes, you are right. The totals should be 2660 and 2609. If you are saying that i should another expressions =sum( aggr( count( value), COLOR, STATUS)), then that would add another column to it. And I've tried it, but its still not working. Its displaying nothing.
thank you.
• ###### Partial Sum Totals Error in Pivot Table
Well you could and should use the expression instead of your original one (but for testing, having another one should bother).
Please check that the field names are correct and correctly spelled, also the case is correct (field names are case sensitive). Are your field called 'value', 'COLOR' and 'STATUS' ?
Could you upload a small sample here?
• ###### Partial Sum Totals Error in Pivot Table
Hi , I think this is problem of Loosly coupled dimensions. Are you using dimensions and expressions from two different tables lnked on some key. If yes the please chek your key this key is not giving you 1:1 linking.
Regards
Vijay
• ###### Re: Partial Sum Totals Error in Pivot Table
Hi,
I attached sample Test.qvw file. It may help for you.
Regards,
Iyyappan.
• ###### Partial Sum Totals Error in Pivot Table
Hi,
could you transfer that pivot table to straight table and send us screenshot of displayed values?
regards
• ###### Partial Sum Totals Error in Pivot Table
Thank you everyone for replying. I appreciate you replies. I used the advanced aggregation suggested above my Stefan and that solved all the program.
Thanks again!!
• ###### Re: Partial Sum Totals Error in Pivot Table
Hi,
Can someone help me with this?
I have almost the same problem. I have an Expression that is the multiplication of two parts (Part A * part B).
When I make the sum of my own for the two parts, and then multiply both TOTALS it is correct. But it is not the correct total for the pivot grid.
This is the Expression
(Sum(STOCK_AL_CIERRE_AJU) - Sum(VENTA_AJU) + Sum(CAJAS)) // PART A
*
(sum(NIV)/sum( CAJAS)) // PART B
I only use one Dimension, it is YearMonth. | 1,029 | 4,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2014-52 | latest | en | 0.867803 |
https://docsbay.net/prairie-transect-sampling | 1,610,918,613,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00012.warc.gz | 316,445,561 | 4,584 | # Prairie Transect Sampling
PRAIRIE TRANSECT SAMPLING
PLANT ECOLOGY LABORATORY
Location: Kramer Ridge Prairie site, Whitewater Wildlife Management Area
Objective: To assess prairie plant populations and communities with various transect sampling procedures
Hypotheses: 1) The Kramer Ridge Prairie is dominated by common plants during spring, all with coverages >10%. 2) Conspicuous plants (big bluestem, compass plant, stiff goldenrod, prairie coneflower, wild indigo [Baptisia]) will each have springtime densities of ~10 plants/hectare.
Agenda:
1) Work in groups of 3-4 students to assess linear plant coverages along transects (20-50 m). Record plant clump or cluster coverages to the nearest 0.1 m.
2) Assess densities of one or more species of conspicuous plant (big bluestem, compass plant, stiff goldenrod, prairie coneflower, wild indigo [Baptisia]) using the Hayne modification of the King Line transect (strip-census) method. Use transect lengths of 20-50 m. For each transect, record transect length and the distances of conspicuous plants from the transect line. Multiple transects for each conspicuous plant are suggested.
Analysis:
1) Summarize your coverage data by first determining the total coverage (total linear distance) for all plants on all transect combined. Then sum all values for all individuals of a single species on your transects and divide by the total coverage of all plants combined to determine a % coverage for that species. Display these data in a pie diagram.
2) For conspicuous species densities, use the following formula to calculate density for each individual transect surveyed:
D = 10,000 * ∑ (1/d)
2L
D = density (individuals per hectare)
L = length of transect (m)
D = distance from transect line to conspicuous plant (m)
If you surveyed multiple transects for the same conspicuous species, calculate a mean (± SD) density for that species.
3) Compare your data to the two hypotheses.
Equipment:
Meter tapes | 456 | 1,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-04 | latest | en | 0.845982 |
http://www.expertsmind.com/library/probability-based-on-the-given-information-515908.aspx | 1,618,285,459,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00468.warc.gz | 130,433,667 | 11,781 | ### Probability based on the given information
Assignment Help Basic Statistics
##### Reference no: EM1315908
Here is a two-way table showing enrollment data for full-time undergraduate students in U.S. colleges and universities. The entries are thousands of students:
Age Two-year programs Four-year programs 15 to 17 years 44 79 18 to 21 years 1345 3869 22 to 44 years 776 1647 45 years and over 49 62 TOTAL 2212 5657
1. How many students were 18 to 21 years old?
A) 1,345
B) 5,214
C) 521,400
D) 3,869,000
E) 5,214,000
Corrective lens wearers by sex and type
Male Female Eyeglasses only (millions) 43.3 54.6 Contact lenses (millions) 2.9 7.0 TOTAL 46.2 61.6
2. From the table, 11.4% of female corrective lens wearers wear contact lenses. The corresponding percentage for males is
A) 2.9%
B) 6.3%
C) 6.7%
D) None of the above.
#### Construct a 99 percent confidence interval around the mean
A small town has a population of 20,000 people. Among these 1,000 regularly visit a popular local bar. A sample of 100 people who visit the bar is surveyed for their annual
#### Determining the more misinterpretations
An economist investigated the association between a country's Literacy Rate and Gross Domestic Product (GDP) and used the association to draw the following conclusions. Expl
#### Predict the number of calories in a snack which contains fat
Fat, Calories, and Carbohydrates A nutritionist established a significant relationship among the fat content. Predict the number of calories in a snack which contains 10 g of
#### What are the popular ringtones
What is the probability that a randomly selected ringtone from this list is not Empire State of Mind and not I Gotta Feeling? Be sure to show how you computed this answer.
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For the passengers in the SRS, what is the approximate probability that the mean length of time for them to get through security will be less than 18 minutes?
#### Possibility of predicting the boiling point of compounds
What does this data set imply about the possibility of predicting the boiling point of compounds in this series on the basis of the number of carbon atoms? Compute the corre
#### How many eggs are within two standard deviations of mean
The weights of eggs produced on a farm are normally distributed with a mean of 1.4 ounces and a standard deviation of 0.4 ounces. How many 1200 eggs are within 2 standard de
#### Definitions by subclass or demonstrative definitions
Determine whether the following are demonstrative definitions, enumerative definitions or definitions by subclass.- "Prime number" means a number greater than one that is divi | 641 | 2,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.901672 |
https://numbermatics.com/n/9164778/ | 1,652,882,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00152.warc.gz | 499,270,383 | 6,492 | # 9164778
## 9,164,778 is an even composite number composed of five prime numbers multiplied together.
What does the number 9164778 look like?
This visualization shows the relationship between its 5 prime factors (large circles) and 32 divisors.
9164778 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of thirty-two divisors.
## Prime factorization of 9164778:
### 2 × 3 × 7 × 31 × 7039
See below for interesting mathematical facts about the number 9164778 from the Numbermatics database.
### Names of 9164778
• Cardinal: 9164778 can be written as Nine million, one hundred sixty-four thousand, seven hundred seventy-eight.
### Scientific notation
• Scientific notation: 9.164778 × 106
### Factors of 9164778
• Number of distinct prime factors ω(n): 5
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 7082
### Divisors of 9164778
• Number of divisors d(n): 32
• Complete list of divisors:
• Sum of all divisors σ(n): 21626880
• Sum of proper divisors (its aliquot sum) s(n): 12462102
• 9164778 is an abundant number, because the sum of its proper divisors (12462102) is greater than itself. Its abundance is 3297324
### Bases of 9164778
• Binary: 1000101111010111111010102
• Base-36: 5GFL6
### Squares and roots of 9164778
• 9164778 squared (91647782) is 83993155789284
• 9164778 cubed (91647783) is 769778626328202638952
• The square root of 9164778 is 3027.3384349953
• The cube root of 9164778 is 209.2701648783
### Scales and comparisons
How big is 9164778?
• 9,164,778 seconds is equal to 15 weeks, 1 day, 1 hour, 46 minutes, 18 seconds.
• To count from 1 to 9,164,778 would take you about twenty-two weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 9164778 cubic inches would be around 17.4 feet tall.
### Recreational maths with 9164778
• 9164778 backwards is 8774619
• 9164778 is a Harshad number.
• The number of decimal digits it has is: 7
• The sum of 9164778's digits is 42
• More coming soon!
MLA style:
"Number 9164778 - Facts about the integer". Numbermatics.com. 2022. Web. 18 May 2022.
APA style:
Numbermatics. (2022). Number 9164778 - Facts about the integer. Retrieved 18 May 2022, from https://numbermatics.com/n/9164778/
Chicago style:
Numbermatics. 2022. "Number 9164778 - Facts about the integer". https://numbermatics.com/n/9164778/
The information we have on file for 9164778 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 9164778, math, Factors of 9164778, curriculum, school, college, exams, university, Prime factorization of 9164778, STEM, science, technology, engineering, physics, economics, calculator, nine million, one hundred sixty-four thousand, seven hundred seventy-eight.
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Some bits of this website may not work unless you switch it on. | 953 | 3,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-21 | latest | en | 0.860754 |
http://math.stackexchange.com/questions/156023/conditional-probability-independency-table-thinking?answertab=oldest | 1,454,820,100,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701148475.34/warc/CC-MAIN-20160205193908-00332-ip-10-236-182-209.ec2.internal.warc.gz | 144,793,533 | 16,474 | Conditional Probability Independency Table Thinking
A have found an alternative definition of independency for a given conditional probability $P(A|B)$, they are independent, iff all columns of the probability table are equal.
What does equal mean in this case? For instance
$$P(A|B) = \begin{pmatrix}0.3 & 0.7\\0.7 & 0.3\end{pmatrix}$$
Where $A$ is associated with the columns and $B$ is associated with the rows, for instance $P(A=0|B=1) = 0.7$ read at the bottom on the left.
-
It means that $P(A=i\mid B=j)=P(A=i\mid B=k)$ for every $i$, $j$ and $k$. | 170 | 560 | {"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} | 3.453125 | 3 | CC-MAIN-2016-07 | latest | en | 0.829715 |
http://www.studymode.com/essays/Intermediate-Accounting-Chapter-10-Exercise-And-672303.html | 1,521,783,997,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648178.42/warc/CC-MAIN-20180323044127-20180323064127-00370.warc.gz | 494,873,312 | 14,206 | # Intermediate Accounting Chapter 10 Exercise and Brief Exercises
Pages: 45 (4462 words) Published: April 24, 2011
CHAPTER 10
SOLUTIONS TO BRIEF EXERCISES
BRIEF EXERCISE 10-1
\$27,000 + \$1,400 + \$10,200 = \$38,600
BRIEF EXERCISE 10-2
|Expenditures | | | | | | | | | |Capitalization Period | |Weighted-Average Accumulated Expenditures | |Date | |Amount | | | | | |3/1 | |\$1,800,000 | |10/12 | |\$1,500,000 | |6/1 | | 1,200,000 | | 7/12 | | 700,000 | |12/31 | | 3,000,000 | |0 | | 0 | | | |\$6,000,000 | | | |\$2,200,000 |
BRIEF EXERCISE 10-3
| |Principal | |Interest | |10%, 5-year note |\$2,000,000 | |\$200,000 | |11%, 4-year note | 3,500,000 | | 385,000 | | |\$5,500,000 | |\$585,000 |
|Weighted-average interest rate = |\$585,000 |= 10.64% | | |\$5,500,000 | |
BRIEF EXERCISE 10-4
|Weighted-Average |X |Interest |= |Avoidable | |Accumulated Expenditures | |Rate | |Interest | |\$1,000,000 | |12% | |\$120,000 | | 1,200,000 | |10.64% | | 127,680 | |\$2,200,000 | | | |\$247,680 | BRIEF EXERCISE 10-5
|Truck (\$80,000 X .68301) |54,641 | | |Discount on Notes Payable |25,359 | | | Notes Payable | |80,000 |
BRIEF EXERCISE 10-6
| | | | | | | |Recorded Amount | | |Fair Value | |% of Total | |Cost | | | |Land |\$ 60,000 | | 60/360 |X |\$315,000 | |\$ 52,500 | |Building | 220,000 | |220/360 |X |\$315,000 | | 192,500 | |Equipment | 80,000 | | 80/360 |X |\$315,000 |...
Please join StudyMode to read the full document | 592 | 1,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.730415 |
https://www.beatthegmat.com/cars-j-and-k-are-making-the-trip-from-city-a-to-city-b-car-t306250.html | 1,561,582,057,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00093.warc.gz | 678,407,351 | 37,910 | • NEW! FREE Beat The GMAT Quizzes
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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Cars J and K are making the trip from City A to City B. Car ##### This topic has 2 expert replies and 0 member replies ### Top Member ## Cars J and K are making the trip from City A to City B. Car ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Cars J and K are making the trip from City A to City B. Car J departs from City A 15 minutes after Car K does, and both cars travel along the same route. If Car K travels at a constant speed that is 80% the constant speed of Car J, then how many minutes will elapse before Car J catches up to Car K? A) 20 B) 45 C) 60 D) 75 E) 1230 OA C Source: Princeton Review ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2852 messages Followed by: 18 members Upvotes: 43 Top Reply BTGmoderatorDC wrote: Cars J and K are making the trip from City A to City B. Car J departs from City A 15 minutes after Car K does, and both cars travel along the same route. If Car K travels at a constant speed that is 80% the constant speed of Car J, then how many minutes will elapse before Car J catches up to Car K? A) 20 B) 45 C) 60 D) 75 E) 1230 OA C Source: Princeton Review We can let r = the rate of Car J, and so 0.8r = the rate of Car K. We also let t = the time of Car J, in minutes, and (t + 15) = the time of Car K, in minutes, when Car J catches up to Car K. We can create the equation: rt = 0.8r(t + 15) rt = 0.8rt + 12r t = 0.8t + 12 0.2t = 12 t = 60 Answer: C _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15362 messages Followed by: 1866 members Upvotes: 13060 GMAT Score: 790 Top Reply BTGmoderatorDC wrote: Cars J and K are making the trip from City A to City B. Car J departs from City A 15 minutes after Car K does, and both cars travel along the same route. If Car K travels at a constant speed that is 80% the constant speed of Car J, then how many minutes will elapse before Car J catches up to Car K? A) 20 B) 45 C) 60 D) 75 E) 1230 Since J catches up to K, J and K travel the same distance. Let t = J's time. Since K leaves 15 minutes earlier than J, K's time = t+15. Rate and time have a RECIPROCAL RELATIONSHIP. Since K's rate is 80% of J's rate, the ratio of K's rate to J's rate = 8/10 = 4/5. Thus, the ratio of K's time to J's time = 5/4: (t+15)/t = 5/4 4t + 60 = 5t 60 = t The correct answer is C. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0
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See More Top Beat The GMAT Experts | 1,387 | 4,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-26 | latest | en | 0.862529 |
https://briefencounters.ca/36871/composite-function-worksheet-answers/ | 1,696,410,029,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00147.warc.gz | 171,997,492 | 28,977 | Composite Function Worksheet Answers is vital in the proper functioning of many businesses and organizations. This is because without them, there can be no standardization of the values that are used in the company’s financial and management systems. There are various types of question type options that can be chosen when asking for an answer to this question type. Some of these include:
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Piecewise Functions Worksheet Answers Elegant Piecewise Functions from composite function worksheet answers , source:alisonnorrington.com | 869 | 4,593 | {"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} | 2.671875 | 3 | CC-MAIN-2023-40 | longest | en | 0.941158 |
https://numbermatics.com/n/62633/ | 1,610,966,593,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00384.warc.gz | 481,971,771 | 5,947 | 62633
62,633 is a prime number. Like all primes greater than two, it is odd and has no factors apart from itself and one.
What does the number 62633 look like?
As a prime, it is not composed of any other numbers and has no internal structure.
62633 is a prime number. Like all primes (except two), it is an odd number.
Prime factorization of 62633:
62633
See below for interesting mathematical facts about the number 62633 from the Numbermatics database.
Names of 62633
• Cardinal: 62633 can be written as Sixty-two thousand, six hundred thirty-three.
Scientific notation
• Scientific notation: 6.2633 × 104
Factors of 62633
• Number of distinct prime factors ω(n): 1
• Total number of prime factors Ω(n): 1
• Sum of prime factors: 62633
Divisors of 62633
• Number of divisors d(n): 2
• Complete list of divisors:
• Sum of all divisors σ(n): 62634
• Sum of proper divisors (its aliquot sum) s(n): 1
• 62633 is a deficient number, because the sum of its proper divisors (1) is less than itself. Its deficiency is 62632
Bases of 62633
• Binary: 11110100101010012
• Base-36: 1CBT
Squares and roots of 62633
• 62633 squared (626332) is 3922892689
• 62633 cubed (626333) is 245702537790137
• The square root of 62633 is 250.2658586383
• The cube root of 62633 is 39.7131562671
Scales and comparisons
How big is 62633?
• 62,633 seconds is equal to 17 hours, 23 minutes, 53 seconds.
• To count from 1 to 62,633 would take you about seventeen hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 62633 cubic inches would be around 3.3 feet tall.
Recreational maths with 62633
• 62633 backwards is 33626
• The number of decimal digits it has is: 5
• The sum of 62633's digits is 20
• More coming soon! | 597 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.874093 |
https://www.fluther.com/199528/what-was-the-last-math-symbol-made/ | 1,511,080,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805466.25/warc/CC-MAIN-20171119080836-20171119100836-00554.warc.gz | 796,987,889 | 9,292 | # What was the last math symbol made?
Asked by RedDeerGuy1 (10119) March 14th, 2017
Who regulates the addition of new symbols?
Observing members: 0 Composing members: 0
I have this link if it helps.
RedDeerGuy1 (10119)
Nobody regulates the addition of symbols. New symbols are used to express new ideas. Knowing the symbols alone does not teach you much about math, any more than knowing the alphabet tells you much about English.
The most recent symbol that I know of is Donald Knuth’s up-arrow notation. Link allows you to represent unimaginably large numbers. I am going to sometimes use & in place of ^ because of display problems.The basic idea is fairly simple. In the beginning there was addition, like 5 + 3. This was extended to add multiple terms, like 5 + 3 4 2. When the terms being added were all the same, like 7 + 7 + 7 + 7, multiplication was devised to express this more compactly as 4*7. Multiplications, like additions could have multiple terms, and a new notation, exponentiation, was used to express this when the terms were identical.. 7 * 7 * 7 * 7 = 7&4.
Why stop there? Well there actually is a good reason, that being the lack of any real need, but when was the last time a theoretical mathematician cared about practical matters. We can have multiple exponentiations. 7&7&7&7. Knuth’s notation for this is 7^^4. It should be apparent that we can extend this process indefinitely,. 7^^7^^7^^7 = 7^^^4. One thing that should be mentioned is that the operations are carried out from right to left. For example, 7^^3 = 7&(7&7)
To get a rough idea how big these numbers get, a googol is a 1 followed by 100 zeros, or 10^100 or 10&10&2. A googol is larger than the number of subatomic particles in the visible universe. A googolplex is a 1 followed by a googol zeros or ten to the googol, or 10&10&10&2. If we replace the 2 by a 10, we get 10^^4 in Knuth’s notation. We have created a number way beyond the imagination without hardly breaking a sweat.
You might be interested in the Metamath site, which builds mathematics from the ground up. I have only started looking at it, so I can’t tell how useful it is, but I like the idea behind it. Metamath is a computer language that can verify (but not generate) proofs. As I understand it, axioms and theorems are treated as subroutines that accept input and generate a line of characters. The subroutines can be chained together to form new new subroutines corresponding to new theorems.
or | 633 | 2,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-47 | longest | en | 0.939189 |
https://answers.yahoo.com/question/index?qid=20170501023647AAyQTtt | 1,585,806,533,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506673.7/warc/CC-MAIN-20200402045741-20200402075741-00488.warc.gz | 344,884,732 | 19,756 | Yin asked in Science & MathematicsPhysics · 3 years ago
# A projectile is fired vertically from the Earth's surface with an initial speed of 11.4 km/s.?
How can I do this? I've tried this like five times and still don't get it. I think I'm supposed to use KE + PE = (KE + PE)final
then
(1/2)mv^2 - GMm/R = 0 - GmM/(R+h)
I tried working this out so many times but I still can't get the right answer for h
Update:
A projectile is fired vertically from the Earth's surface with an initial speed of 11.4 km/s. Neglecting air drag, how far (in meters) above the surface of the Earth will it go? | 168 | 596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-16 | latest | en | 0.950847 |
https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/video-lectures/lecture-3-independence/ | 1,516,179,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886860.29/warc/CC-MAIN-20180117082758-20180117102758-00166.warc.gz | 749,603,216 | 52,696 | # Lecture 3: Independence
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Description: In this lecture, the professor discussed independence of two events, independence of a collection of events, and independence vs. pairwise independence.
Instructor: John Tsitsiklis
The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.
PROFESSOR: Let us start. So as always, we're to have a quick review of what we discussed last time. And then today we're going to introduce just one new concept, the notion of independence of two events. And we will play with that concept.
So what did we talk about last time? The idea is that we have an experiment, and the experiment has a sample space omega. And then somebody comes and tells us you know the outcome of the experiments happens to lie inside this particular event B. Given this information, it kind of changes what we know about the situation. It tells us that the outcome is going to be somewhere inside here. So this is essentially our new sample space.
And now we need to we reassign probabilities to the various possible outcomes, because, for example, these outcomes, even if they had positive probability beforehand, now that we're told that B occurred, those outcomes out there are going to have zero probability. So we need to revise our probabilities. The new probabilities are called conditional probabilities, and they're defined this way.
The conditional probability that A occurs given that we're told that B occurred is calculated by this formula, which tells us the following-- out of the total probability that was initially assigned to the event B, what fraction of that probability is assigned to outcomes that also make A to happen? So out of the total probability assigned to B, we see what fraction of that total probability is assigned to those elements here that will also make A happen.
Conditional probabilities are left undefined if the denominator here is zero. An easy consequence of the definition is if we bring that term to the other side, then we can find the probability of two things happening by taking the probability that the first thing happens, and then, given that the first thing happened, the conditional probability that the second one happens.
Then we saw last time that we can divide and conquer in calculating probabilities of mildly complicated events by breaking it down into different scenarios. So event B can happen in two ways. It can happen either together with A, which is this probability, or it can happen together with A complement, which is this probability. So basically what we're saying that the total probability of B is the probability of this, which is A intersection B, plus the probability of that, which is A complement intersection B.
So these two facts here, multiplication rule and the total probability theorem, are basic tools that one uses to break down probability calculations into a simpler parts. So we find probabilities of two things happening by looking at each one at a time. And this is what we do to break up a situation with two different possible scenarios.
Then we also have the Bayes rule, which does the following. Given a model that has conditional probabilities of this kind, the Bayes rule allows us to calculate conditional probabilities in which the events appear in different order. You can think of these probabilities as describing a causal model of a certain situation, whereas these are the probabilities that you get after you do some inference based on the information that you have available.
Now the Bayes rule, we derived it, and it's a trivial half-line calculation. But it underlies lots and lots of useful things in the real world. We had the radar example last time. You can think of more complicated situations in which there's a bunch or lots of different hypotheses about the environment. Given any particular setting in the environment, you have a measuring device that can produce many different outcomes. And you observe the final outcome out of your measuring device, and you're trying to guess which particular branch occurred. That is, you're trying to guess the state of the world based on a particular measurement.
That's what inference is all about. So real world problems only differ from the simple example that we saw last time in that this kind of tree is a little more complicated. You might have infinitely many possible outcomes here and so on. So setting up the model may be more elaborate, but the basic calculation that's done based on the Bayes rule is essentially the same as the one that we saw.
Now something that we discuss is that sometimes we use conditional probabilities to describe models, and let's do this by looking at a model where we toss a coin three times. And how do we use conditional probabilities to describe the situation? So we have one experiment. But that one experiment consists of three consecutive coin tosses. So the possible outcomes, our sample space, consists of strings of length 3 that tell us whether we had heads, tails, and in what sequence. So three heads in a row is one particular outcome.
So what is the meaning of those labels in front of the branches? So this P here, of course, stands for the probability that the first toss resulted in heads. And let me use this notation to denote that the first was heads. I put an H in toss one.
How about the meaning of this probability here? Well the meaning of this probability is a conditional one. It's the conditional probability that the second toss resulted in heads, given that the first one resulted in heads. And similarly this label here corresponds to the probability that the third toss resulted in heads, given that the first one and the second one resulted in heads. So in this particular model that I wrote down here, those probabilities, P, of obtaining heads remain the same no matter what happened in the previous toss.
For example, even if the first toss was tails, we still have the same probability, P, that the second one is heads, given that the first one was tails. So we're assuming that no matter what happened in the first toss, the second toss will still have a conditional probability equal to P. So that conditional probability does not depend on what happened in the first toss. And we will see that this is a very special situation, and that's really the concept of independence that we are going to introduce shortly.
But before we get to independence, let's practice once more the three skills that we covered last time in this example. So first skill was multiplication rule. How do you find the probability of several things happening? That is the probability that we have tails followed by heads followed by tails. So here we're talking about this particular outcome here, tails followed by heads followed by tails. And the way we calculate such a probability is by multiplying conditional probabilities along the path that takes us to this outcome. And so these conditional probabilities are recorded here. So it's going to be (1 minus P) times P times (1 minus P). So this is the multiplication rule.
Second question is how do we find the probability of a mildly complicated event? So the event of interest here that I wrote down is the probability that in the three tosses, we had a total of one head. Exactly one head. This is an event that can happen in multiple ways. It happens here. It happens here. And it also happens here. So we want to find the total probability of the event consisting of these three outcomes. What do we do? We just add the probabilities of each individual outcome. How do we find the probability of an individual outcome? Well, that's what we just did.
Now notice that this outcome has probability P times (1 minus P) squared. That one should not be there. So where is it? Ah. It's this one.
OK, so the probability of this outcome is (1 minus P times P) times (1 minus P), the same probability. And finally, this one is again (1 minus P) squared times P.
So this event of one head can happen in three ways. And each one of those three ways has the same probability of occurring. And this is the answer.
And finally, the last thing that we learned how to do is to use the Bayes rule to calculate and make an inference. So somebody tells you that there was exactly one head in your three tosses. What is the probability that the first toss resulted in heads? OK, I guess you can guess the answer here if I tell you that there were three tosses. One of them was heads. Where was that head in the first, the second, or the third?
Well, by symmetry, they should all be equally likely. So there should be probably just 1/3 that that head occurred in the first toss. Let's check our intuition using the definitions. So the definition of conditional probability tells us the conditional probability is the probability of both things happening. First toss is heads, and we have exactly one head divided by the probability of one head.
What is the probability that the first toss is heads, and we have exactly one head? This is the same as the event heads, tails, tails. If I tell you that the first is heads, and there's only one head, it means that the others are tails. So this is the probability of heads, tails, tails divided by the probability of one head. And we know all of these quantities probability of heads, tails, tails is P times (1 minus P) squared. Probability of one head is 3 times P times (1 minus P) squared. So the final answer is 1/3, which is what you should have a guessed on intuitive grounds.
Very good. So we got our practice on the material that we did cover last time. Again, think. There's basically three basic skills that we are practicing and exercising here. In the problems, quizzes, and in the real life, you may have to apply those three skills in somewhat more complicated settings, but in the end that's what it boils down to usually.
Now let's focus on this special feature of this particular model that I discussed a little earlier. Think of the event heads in the second toss. Initially, the probability of heads in the second toss, you know, that it's P, the probability of success of your coin. If I tell you that the first toss resulted in heads, what's the probability that the second toss is heads? It's again P. If I tell you that the first toss was tails, what's the probability that the second toss is heads? It's again P. So whether I tell you the result of the first toss, or I don't tell you, it doesn't make any difference to you. You would always say the probability of heads in the second toss is going to P, no matter what happened in the first toss.
This is a special situation to which we're going to give a name, and we're going to call that property independence. Basically independence between two things stands for the fact that the first thing, whether it occurred or not, doesn't give you any information, does not cause you to change your beliefs about the second event. This is the intuition. Let's try to translate this into mathematics.
We have two events, and we're going to say that they're independent if your initial beliefs about B are not going to change if I tell you that A occurred. So you believe something how likely B is. Then somebody comes and tells you, you know, A has happened. Are you going to change your beliefs? No, I'm not going to change them. Whenever you are in such a situation, then you say that the two events are independent.
Intuitively, the fact that A occurred does not convey any information to you about the likelihood of event B. The information that A provides is not so useful, is not relevant. A has to do with something else. It's not useful for your guessing whether B is going to occur or not.
So we can take this as a first attempt into a definition of independence. Now remember that we have this property, the probability of two things happening is the probability of the first times the conditional probability of the second. If we have independence, this conditional probability is the same as the unconditional probability.
So if we have independence according to that definition, we get this property that you can find the probability of two things happening by just multiplying their individual probabilities. Probability of heads in the first toss is 1/2. Probability of heads in the second toss is 1/2. Probability of heads heads is 1/4. That's what happens if your two tosses are independent of each other.
So this property here is a consequence of this definition, but it's actually nicer, better, simpler, cleaner, more beautiful to take this as our definition instead of that one. Are the two definitions equivalent? Well, they're are almost the same, except for one thing. Conditional probabilities are only defined if you condition on an event that has positive probability.
So this definition would be limited to cases where event A has positive probability, whereas this definition is something that you can write down always. We will say that two events are independent if and only if their probability of happening simultaneously is equal to the product of their two individual probabilities.
And in particular, we can have events of zero probability. There's nothing wrong with that. If A has 0 probability, then A intersection B will also have zero probability, because it's an even smaller event. And so we're going to get zero is equal to zero. A corollary of what I just said, if an event A has zero probability, it's actually independent of any other event in our model, because we're going to get zero is equal to zero. And the definition is going to be satisfied.
This is a little bit harder to reconcile with the intuition we have about independence, but then again, it's part of the mathematical definition. So what I want you to retain is this notion that the independence is something that you can check formally using this definition, but also you can check intuitively by if, in some cases, you can reason that whatever happens and determines whether A is going to occur or not, has nothing absolutely to do with whatever happens and determines whether B is going to occur or not.
So if I'm doing a science experiment in this room, and it gets hit by some noise that's causes randomness. And then five years later, somebody somewhere else does the same science experiment somewhere else, it gets hit by other noise, you would usually say that these experiments are independent. So what events happen in one experiment are not going to change your beliefs about what might be happening in the other, because the sources of noise in these two experiments are completely unrelated. They have nothing to do with each other.
So if I flip a coin here today, and I flip a coin in my office tomorrow, one shouldn't affect the other. So the events that I get from these should be independent. So that's usually how independence arises. By having distinct physical phenomena that do not interact.
Sometimes you also get independence even though there is a physical interaction, but you just happen to have a numerical accident. A and B might be physically related very tightly, but a numerical accident happens and you get equality here, that's another case where we do get independence.
Now suppose that we have two events that are laid out like this. Are these two events independent or not? The picture kind of tells you that one is separate from the other. But separate has nothing to do with independent. In fact, these two events are as dependent as Siamese twins. Why is that?
If I tell you that A occurred, then you are certain that B did not occur. So information about the occurrence of A definitely affects your beliefs about the possible occurrence or non-occurrence of B. When the picture is like that, knowing that A occurred will change drastically my beliefs about B, because now I suddenly become certain that B did not occur.
So a picture like this is a case actually of extreme dependence. So don't confuse independence with disjointness. They're very different types of properties.
AUDIENCE: Question.
PROFESSOR: Yes?
AUDIENCE: So I understand the explanation, but the probability of A intersect B [INAUDIBLE] to zero, because they're disjoint.
PROFESSOR: Yes.
AUDIENCE: But then the product of probability A and probability B, one of them is going to be 1. [INAUDIBLE]
PROFESSOR: No, suppose that the probabilities are 1/3, 1/4, and the rest is out there. You check the definition of independence. Probability of A intersection B is zero. Probability of A times the probability of B is 1/12. The two are not equal. Therefore we do not have independence.
AUDIENCE: Right. So what's wrong with the intuition of the probability of A being 1, and the other one being 0? [INAUDIBLE].
PROFESSOR: No. The probability of A given B is equal to 0. Probability of A is equal to 1/3. So again, these two are different. So we had some initial beliefs about A, but as soon as we are told that B occurred, our beliefs about A changed. And so since our beliefs changed, that means that B conveys information about A.
AUDIENCE: So can you not draw independent [INAUDIBLE] on a Venn diagram?
PROFESSOR: I can't hear you.
AUDIENCE: Can you draw independence on a Venn diagram?
PROFESSOR: No, the Venn diagram is never enough to decide independence. So the typical picture in which you're going to have independence would be one event this way, and another event this way. You need to take the probability of this times the probability of that, and check that, numerically, it's equal to the probability of this intersection. So it's more than a Venn diagram. Numbers need to come out right.
Now we did say some time ago that conditional probabilities are just like ordinary probabilities, and whatever we do in probability theory can also be done in conditional universes. Talking about conditional probabilities. So since we have a notion of independence, then there should be also a notion of conditional independence. So independence was defined by the probability that A intersection B is equal to the probability of A times the probability of B.
What would be a reasonable definition of conditional independence? Conditional independence would mean that this same property could be true, but in a conditional universe where we are told that the certain event happens. So if we're told that the event C has happened, then were transported in a conditional universe where the only thing that matters are conditional probabilities. And this is just the same plain, previous definition of independence, but applied in a conditional universe.
So this is the definition of conditional independence. So it's independence, but with reference to the conditional probabilities. And intuitively it has, again, the same meaning, that in the conditional world, if I tell you that A occurred, then that doesn't change your beliefs about B.
So suppose you had a picture like this. And somebody told you that events A and B are independent unconditionally. Then somebody comes and tells you that event C actually has occurred, so we now live in this new universe. In this new universe, is the independence of A and B going to be preserved or not? Are A and B independent in this new universe?
The answer is no, because in the new universe, whatever is left of event A is this piece. Whatever is left of event B is this piece. And these two pieces are disjoint. So we are back in a situation of this kind. So in the conditional universe, A and B are disjoint. And therefore, generically, they're not going to be independent.
What's the moral of this example? Having independence in the original model does not imply independence in a conditional model.
The opposite is also possible. And let's illustrate by another example. So I have two coins, and both of them are badly biased. One coin is much biased in favor of heads. The other coin is much biased in favor of tails. So the probabilities being 90%.
Let's consider independent flips of coin A. This is the relevant model. This is a model of two independent flips of the first coin. There's going to be two flips, and each one has probability 0.9 of being heads. So that's a model that describes coin A. You can think of this as a conditional model which is a model of the coin flips conditioned on the fact that they have chosen coin A.
Alternatively we could be dealing with coin B In a conditional world where we chose coin B and flip it twice, this is the relevant model. The probability of two heads, for example, is the probability of heads the first time, heads the second time, and each one is 0.1.
Now I'm building this into a bigger experiment in which I first start by choosing one of the two coins at random. So I have these two coins. I blindly pick one of them. And then I start flipping them.
So the question now is, are the coin flips, or the coin tosses, are they independent of each other? If we just stay inside this sub-model here, are the coin flips independent? They are independent, because the probability of heads in the second toss is the same, 0.9, no matter what happened in the first toss. So the conditional probabilities of what happens in the second toss are not affected by the outcome of the first toss. So the second toss and the first toss are independent. So here we're just dealing with plain, independent coin flips.
Similarity the coin flips within this sub-model are also independent. Now the question is, if we look at the big model as just one probability model, instead of looking at the conditional sub-models, are the coin flips independent of each other? Does the outcome of a few coin flips give you information about subsequent coin flips?
Well if I observe ten heads in a row-- So instead of two coin flips, now let's think of doing more of them so that the tree gets expanded.
So let's start with this. I don't know which coin it is. What's the probability that the 11th coin toss is going to be heads? There's complete symmetry here, so the answer could not be anything other than 1/2. So let's justify it, why is it 1/2?
Well, the probability that the 11th toss is heads, how can that outcome happen? It can happen in two ways. You can choose coin A, which happens with probability 1/2. And having chosen coin A, there's probability 0.9 that it results in that you get heads in the 11th toss. Or you can choose coin B. And if it's coin B when you flip it, there's probably 0.1 that you have heads. So the final answer is 1/2.
So each one of the coins is biased, but they're biased in different ways. If I don't know which coin it is, their two biases kind of cancel out, and the probability of obtaining heads is just in the middle, then it's 1/2.
Now if someone tells you that the first ten tosses were heads, is that going to change your beliefs about the 11th toss? Here's how a reasonable person would think about it.
If it's coin B the probability of obtaining 10 heads in a row is negligible. It's going to be 0.1 to the 10th. If it's coin A. The probability of 10 heads in a row is a more reasonable number. It's 0.9 to the 10th. So this event is a lot more likely to occur with coin A, rather than coin B.
The plausible explanation of having seen ten heads in a row is that I actually chose coin A. When you see ten heads in a row, you are pretty certain that it's coin A that we're dealing with. And once you're pretty certain that it's coin A that we're dealing with, what's the probability that the next toss is heads? It's going to be 0.9.
So essentially here I'm doing an inference calculation. Given this information, I'm making an inference about which coin I'm dealing with. I become pretty certain that it's coin A, and given that it's coin A, this probability is going to be 0.9. And I'm putting an approximate sign here, because the inference that I did is approximate. I'm pretty certain it's coin A. I'm not 100% certain that it's coin A.
But in any case what happens here is that the unconditional probability is different from the conditional probability. This information here makes me change my beliefs about the 11th toss. And this means that the 11th toss is dependent on the previous tosses. So the coin tosses have now become dependent. What is the physical link that causes this dependence? Well, the physical link is the choice of the coin. By choosing a particular coin, I'm introducing a pattern in the future coin tosses. And that pattern is what causes dependence.
OK, so I've been playing a little bit too loose with the language here, because we defined the concept of independence of two events. But here I have been referring to independent coin tosses, where I'm thinking about many coin tosses, like 10 or 11 of them.
So to be proper, I should have defined for you also the notion of independence of multiple events, not just two. We don't want to just say coin toss one is independent from coin toss two. We want to be able to say something like, these 10 then coin tosses are all independent of each other. Intuitively what that means should be the same thing-- that information about some of the coin tosses doesn't change your beliefs about the remaining coin tosses. How do we translate that into a mathematical definition?
Well, an ugly attempt would be to impose requirements such as this. Think of A1 being the event that the first flip was heads. A2 is the event of that the second flip was heads. A3, the third flip, was heads, and so on.
Here is an event whose occurrence is not determined by the first three coin flips. And here's an event whose occurrence or not is determined by the fifth and sixth coin flip. If we think physically that all those coin flips have nothing to do with each other, information about the fifth and sixth coin flip are not going to change what we expect from the first three. So the probability of this event, the conditional probability, should be the same as the unconditional probability. And we would like a relation of this kind to be true, no matter what kind of formula you write down, as long as the events that show up here are different from the events that show up there.
OK. That's sort of an ugly definition. The mathematical definition that actually does the job, and leads to all the formulas of this kind, is the following. We're going to say that the collection of events are independent if we can find the probability of their joint occurrence by just multiplying probabilities. And that will be true even if you look at sub-collections of these events.
Let's make that more precise. If we have three events, the definition tells us that the three events are independent if the following are true. Probability A1 and A2 and A3, you can calculate this probability by multiplying individual probabilities. But the same is true even if you take fewer events. Just a few indices out of the indices that we have available. So we also require P(A1 intersection A2) is P(A1) times P(A2). And similarly for the other possibilities of choosing the indices.
OK, so independence, mathematical definition, requires that calculating probabilities of any intersection of the events we have in our hands, that calculation can be done by just multiplying individual probabilities. And this has to apply to the case where we consider all of the events in our hands or just sub-collections of those events.
Now these relations just by themselves are called pairwise independence. So this relation, for example, tells us that A1 is independent from A2. This tells us that A2 is independent from A3. This will tell us that A1 is independent from A3. But independence of all the events together actually requires a little more. One more equality that has to do with all three events being considered at the same time.
And this extra equality is not redundant. It actually does make a difference. Independence and pairwise independence are different things. So let's illustrate the situation with an example. Suppose we have two coin flips. The coin tosses are independent, so the bias is 1/2, so all possible outcomes have a probability of 1/2 times 1/2, which is 1/4.
And let's consider now a bunch of different events. One event is that the first toss is heads. This is this blue set here. Another event is the second toss is heads. And this is this black event here.
OK. Are these two events independent? If you check it mathematically, yes. Probability of A is probability of B is 1/2. Probability of A times probability of B is 1/4, which is the same as the probability of A intersection B, which is this set. So we have just checked mathematically that A and B are independent.
Now lets consider a third event which is that the first and second toss give the same result. I'll use a different color. First and second toss to give the same result. This is the event that we obtain heads, heads or tails, tails. So this is the probability of C. What's the probability of C? Well, C is made up of two outcomes, each one of which has probability 1/4, so the probability of C is 1/2. What is the probability of C intersection A? C intersection A is just this one outcome, and has probability 1/4.
What's the probability of A intersection B intersection C? The three events intersect just this outcome, so this probability is also 1/4.
OK. What's the probability of C given A and B?
If A has occurred, and B has occurred, you are certain that this outcome here happened. If the first toss is H and the second toss is H, then you're certain of the first and second toss gave the same result. So the conditional probability of C given A and B is equal to 1.
So do we have independence in this example? We don't. C, that we obtain the same result in the first and the second toss, has probability 1/2. Half of the possible outcomes give us two coin flips with the same result-- heads, heads or tails, tails. So the probability of C is 1/2.
But if I tell you that the events A and B both occurred, then you're certain that C occurred. If I tell you that we had heads and heads, then you're certain the outcomes were the same. So the conditional probability is different from the unconditional probability. So by combining these two relations together, we get that the three events are not independent.
But are they pairwise independent? Is A independent from B? Yes, because probability of A times probability of B is 1/4, which is probability of A intersection B. Is C independent from A? Well, the probability of C and A is 1/4. The probability of C is 1/2. The probability of A is 1/2. So it checks. 1/4 is equal to 1/2 and 1/2, so event C and event A are independent.
Knowing that the first toss was heads does not change your beliefs about whether the two tosses are going to have the same outcome or not. Knowing that the first was heads, well, the second is equally likely to be heads or tails. So event C has just the same probability, again, 1/2, to occur.
To put it the opposite way, if I tell you that the two results were the same-- so it's either heads, heads or tails, tails-- what does that tell you about the first toss? Is it heads, or is it tails? Well, it doesn't tell you anything. It could be either over the two, so the probability of heads in the first toss is equal to 1/2, and telling you C occurred does not change anything.
So this is an example that illustrates the case where we have three events in which we check that pairwise independence holds for any combination of two of these events. We have the probability of their intersection is equal to the product of their probabilities. On the other hand, the three events taken all together are not independent. A doesn't tell me anything useful, whether C is going to occur or not. B doesn't tell me anything useful. But if I tell you that both A and B occurred, the two of them together tell me something useful about C. Namely, they tell me that C certainly has occurred.
Very good. So independence is this somewhat subtle concept. Once you grasp the intuition of what it really means, then things perhaps fall in place. But it's a concept where it's easy to get some misunderstanding. So just take some time to digest.
So to lighten things up, I'm going to spend the remaining four minutes talking about the very nice, simple problem that involves conditional probabilities and the like. So here's the problem, formulated exactly as it shows up in various textbooks. And the formulation says the following.
Well, consider one of those anachronistic places where they still have kings or queens, and where actually boys take precedence over girls. So if there is a boy-- if the royal family has a boy, then he will become the king even if he has an older sister who might be the queen.
So we have one of those royal families. That royal family had two children, and we know that there is a king. There is a king, which means that at least one of the two children was a boy. Otherwise we wouldn't have a king. What is the probability that the king's sibling is female?
OK. I guess we need to make some assumptions about genetics. Let's assume that every child is a boy or a girl with probability 1/2, and that different children, what they are is independent from what the other children were. So every childbirth is basically a coin flip.
OK, so if you take that, you say, well, the king is a child. His sibling is another child. Children are independent of each other. So the probability that the sibling is a girl is 1/2. That's the naive answer. Now let's try to do it formally.
Let's set up a model of the experiment. The royal family had two children, as we we're told, so there's four outcomes-- boy boy, boy girl, girl boy, and girl girl. Now, we are told that there is a king, which means what? This outcome here did not happen. It is not possible. There are three outcomes that remain possible. So this is our conditional sample space given that there is king.
What are the probabilities for the original model? Well with the model that we assume that every child is a boy or a girl independently with probability 1/2, then the four outcomes would be equally likely, and they're like this. These are the original probabilities. But once we are told that this outcome did not happen, because we have a king, then we are transported to the smaller sample space.
In this sample space, what's the probability that the sibling is a girl? Well the sibling is a girl in two out of the three outcomes. So the probability that the sibling is a girl is actually 2/3. So that's supposed to be the right answer. Maybe a little counter-intuitive.
So you can play smart and say, oh I understand such problems better than you, here is a trick problem and here's why the answer is 2/3. But actually I'm not fully justified in saying that the answer is 2/3. I made lots of hidden assumptions when I put this model down, which I didn't yet state. So to reverse engineer this answer, let's actually think what's the probability model for which this would have been the right answer. And here's the probability model.
The royal family-- the royal parents decided to have exactly two children. They went and had them. It turned out that at least one was a boy and became a king. Under this scenario-- that they decide to have exactly two children-- then this is the big sample space. It turned out that one was a boy. That eliminates this outcome. And then this picture is correct and this is the right answer.
But there's hidden assumptions being there. How about if the royal family had followed the following strategy? We're going to have children until we get a boy, so that we get a king, and then we'll stop. OK, given they have two children, what's the probability that the sibling is a girl?
It's 1. The reason that they had two children was because the first was a girl, so they had to have a second. So assumptions about reproductive practices actually need to come in, and they're going to affect the decisions. Or, if it's one of those ancient kingdoms where a king would always make sure too strangle any of his brothers, then the probability that the sibling is a girl is actually 1 again, and so on.
So it means that one needs to be careful when you start with loosely worded problems to make sure exactly what it means and what assumptions you're making. All right, see you next week. | 7,831 | 36,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.955249 |
https://socratic.org/questions/a-yield-of-nh-3-of-approximately-98-can-be-obtained-at-200-deg-c-and-1-000-atmos | 1,632,486,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00615.warc.gz | 568,450,618 | 6,467 | # A yield of NH_3 of approximately 98% can be obtained at 200 deg C and 1,000 atmospheres of pressure. How many grams of N_2 must react to form 1.7 grams of ammonia, NH_3?
##### 1 Answer
May 26, 2017
1.429 g of ${N}_{2}$
#### Explanation:
We always have to read the problems completely to understand the question. In this case even when we have pressure and temperature conditions specified, they don't matter, first because they are constant and in second because the question ask for grams of ${N}_{2}$ to produce grams of $N {H}_{3}$.
We cannot compare grams of a substance to another in chemical reactions, we need to use moles, then we have to convert the weight of NH3 to moles.
$\eta = m / M W$
$M {W}_{N {H}_{3}} = 14.007 + 3 \left(1.008\right) = 17.031 g / m o l$
:. $\eta = 1.7 g / 17.031 g / m o l = 0.100 m o l$
The yield is 98%, then the number of moles that would be produced are:
$0.100 m o l / 0.98 = 0.102 m o l$ of $N {H}_{3}$
From the reaction (the hydrogen was implicit in the problem, because you are obtaining $N {H}_{3}$):
$N 2 + 3 H 2 \rightarrow 2 N H 3$
There are 2 parts of $N {H}_{3}$ that are obtained from 1 part of ${N}_{2}$
Then the moles of ${N}_{2}$ required are going to be the half of the $N {H}_{3}$ moles produced.
${N}_{2}$ moles = $N {H}_{3} \text{moles} / 2 = 0.102 m o l / 2 = 0.051 m o l$
Last step is too convert the moles back to grams using the MW of ${N}_{2}$
${m}_{{N}_{2}} = m o l \times M W = 0.051 m o l \times \left(14.007 g / m o l \times 2\right) = 1.429 g$ | 517 | 1,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-39 | latest | en | 0.89849 |
https://forum.altair.com/profile/91972-intan/ | 1,601,404,112,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00008.warc.gz | 392,152,780 | 14,236 | # Intan
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1. ## Additive Manufacturing Constraints in Topology Optimisation
Hi @Prakash Pagadala I'm integrating the overhang constraint in my design variable for the topological optimization. However there are certain fields that I don't quite understand the definition. For one, from the guide, the Point Selection 1 and 2 define the grid point identification numbers which identify the orientation. What orientation does it refer to? Is it the built orientation? Thank you and regards, Intan
2. ## Direct Transient Dynamic Analysis Problem
Hi @Prakash Pagadala, I've tried superposition in derived load step but apparently it is only for linear problems. My problem is a transient one. Regards.
4. ## Additive Manufacturing Constraints in Topology Optimisation
Hi, Is there anyway to integrate Additive Manufacturing constraints for topology optimization in optistruct? I would like to add constraints such as maximum overhang angle, minimum number of support structure, etc... I would also like to determine the best build direction. Thank you
5. ## Direct Transient Dynamic Analysis Problem
Hi I have a connecting rod that is subjected to two loads, Tensile (Admission phase) and Compression (Combustion phase). Both loads have TLOAD2 as their card image and hence each of them have their own loadstep. When observing the results, i would like to see both loadsteps one after another eg: t1 : Combustion , t2 : Admission, t3 : Combustion , t4 : Admission...... I have derived both results together but it showed me the combustion results first then after t500 only it will show admission. Is there anyway to correct this? Regards.
6. ## Dynamic Loads Topology Optimisation in Optistruct
Thank you ! I will try this approach. Regards, Intan
7. ## Dynamic Loads Topology Optimisation in Optistruct
Hi @Prakash Pagadala Would you have a suggestion on how can this optimization be carried out? Regards, Intan
8. ## Dynamic Loads Topology Optimisation in Optistruct
Hi @Prakash Pagadala I've sent the model via the transfer link. Thank you!
9. ## Dynamic Loads Topology Optimisation in Optistruct
Hi, The problem consists of a connecting rod submitted to axial loads which define the phases of admission (traction) and combustion (compression). I am trying to run the following optimization: 2 load steps : Traction Compression 2 responses : Volume Compliance 1 constraint : 60% Volume 1 objective : Minimize compliance I managed to run a direct transient analysis for the two separate load cases. I tried running a topology optimization for one of the load cases but an error occured with the following message: *** ERROR # 557 *** DRESP1 2 is not referenced from within a static or an MBD SUBCASE. This type of response must be specified from within a static or an MBD SUBCASE. This error occurs in DESOBJ 2. I would like to know: How can I create alternating load steps to simulate the traction and compression phases If it is possible to run an optimization with these dynamic loads Thank you !
10. ## Topology Optimization in Optistruct - Problem with two loadsteps
Hi @Ivan Thank you so much for your reply! I will try to use the method you mentioned.
11. ## Topology Optimization in Optistruct - Problem with two loadsteps
Hi, The problem consists of a connecting rod submitted to axial loads which define the phases of admission (traction) and combustion (compression). I am trying to run the following optimization: 2 load steps : linear static (traction) linear static (compression) 2 responses : Volume Compliance 1 constraint : 60% Volume 1 objective : minimize compliance Without combining the two load steps, I managed to run the optimization. I couldn't find a way integrating the two load steps for the optimization. I found that the only time that I can include the load step is when I define the objective. However only one load step can be selected. Therefore I was wondering if there is a way to integrate both of the load steps mentioned earlier. Thank you !
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https://blog.mrmeyer.com/2010/math-curriculum-remix/ | 1,721,085,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00014.warc.gz | 119,414,163 | 10,592 | Math Curriculum Remix
I don’t know why this is tripping my fuses so hard. I’ve seen it hundreds of times in the most straightforward textbooks: obscure one set of inputs, then reveal another, and you have a totally different math problem. Simple stuff, right? This visual proof-of-concept is really freaking me out, though.
(You know where to press pause, right?)
It’s the same footage. But the original was California-grade-six rates and the remix is California-grade-eight systems. What?
I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. He / him. More here.
1. RudeDude
August 11, 2010 - 10:28 am -
I see what you did there! (I think.)
Pause at 00:39. You’ve obscured the base walking rate by the escalator rate. So now the “simple” problem from before has become a more difficult “back out the escalator rate” problem. It also helps to remove the music beat since the stepping rate (difference) is less obvious.
Certainly a perplexing problem. :)
2. Frank Noschese
August 11, 2010 - 10:58 am -
I don’t want to be a Debbie Downer, but I can still solve this one without systems. The physical mechanics of the set-up blow your cover.
For 10 seconds, I counted the number of taps I heard as you went up the down escalator, because each tap meant you placed your foot on the next step. You took 26 steps. Then I paused the video when you were at the bottom of the stairs and counted the number of stairs to the top as best I could. I counted 35 steps.
10 sec / 26 steps * 35 steps = 13.5 seconds to climb the stairs. (Pretty close!)
The simplest solution is the best solution.
Even with the sound off, I could still watch your feet and count you taking steps. Perhaps using a moving sidewalk would be better at forcing the kids to use systems? And maybe replacing you with a battery operated toy that moves at constant speed? It would remove “cheats” like step counting. But toys aren’t as engaging as people.
Thoughts?
3. Rob
August 15, 2010 - 6:29 pm -
Frank, do you still then have to use systems to find the speed of the elevator itself?
Maybe I’ve been out of eight-grade math for too long (I’m not a math teacher), or am making the problem more difficult than it really is, but I have to admit, this is a really challenging problem for me!
4. Frank Noschese
August 15, 2010 - 6:53 pm -
Rob: If I am interpreting Dan correctly, the gut question is “How long will it take Dan to walk up the stairs?” I was able to solve this without knowing any distances or speeds because Dan moves up/down the same number of steps every second (though, b/c some stairs are moving, he doesn’t move up/down the same number of meters every second). All I have to do is watch/listen to his footsteps, count them and time them. It doesn’t matter which escalator b/c his step rate is the same (in time to his music, right?).
If you wanted the speed of the escalator, you just need a distance traveled and time interval for one 1 stair. No systems necessary.
I hope this makes sense.
And Dan, am I interpreting the purpose of this video correctly?
5. Dan Meyer
August 16, 2010 - 1:41 pm -
Frank: And Dan, am I interpreting the purpose of this video correctly?
Proof is in the pudding, Frank, and you nailed the time it took me to walk up the stairs. Your idea to use a moving sidewalk to conceal the footsteps is sharp. Logistically challenging, also, which makes it all the more fun.
6. Colin Graham
August 25, 2010 - 4:41 am -
Going down the up escalator would be fun too, if a little more dangerous! I wonder if it would be possible to do a side view so we only see the top half of your body as you go up and down. You might then be able to juxtapose the three ‘strips’ (four if you have down the up) similar to the Scott Haluck video. You could maybe just throw out the question “How long would it take to go down the up escalator?” as final problem.
Oh and the English teaching part of me says to tell you that the proof is in the eating, not the pudding!
:)
Colin
7. Vince Muccioli
October 13, 2010 - 2:42 am -
Dan, I love the material you are putting out. I find the original version of this video much easier to follow for the students. It seems like the remixed version ends before students would have a handle of what they are watching. The problem is that the original video is no longer available. Could you possibly make the original available on vimeo again or something? Thanks.
8. Maria Droujkova
October 22, 2010 - 12:08 pm -
Dan, here is a weird Russian page on what is apparently a tongue-in-cheek schoolchildren sport/web meme there called Escalator Running (transliterated from English): http://lurkmore.ru/%D0%9C%D0%B5%D1%82%D1%80%D0%BE%D1%8D%D0%BB%D0%B5%D0%BA%D1%82%D1%80%D0%B8%D1%87%D0%BA%D0%B8%D0%BD%D0%B3/%D0%AD%D1%81%D0%BA%D0%B0%D0%BB%D0%B0%D1%82%D0%BE%D1%80_%D0%B3%D0%B5%D0%B9%D0%BC%D0%B8%D0%BD%D0%B3
I thought about you when I ran into that page. Check out the video of the girl walking up the down escalator, absolutely packed with people: http://www.youtube.com/watch?v=oFFfs3K6Tjs&fmt=18
9. Christopher Danielson
March 12, 2011 - 10:57 am -
My question hasn’t come up here, which is “How long if Dan rides the escalator like a normal person?” I suppose this is the same as asking about the speed of the escalator, but it feels more engaging to me as a learner.
Thanks for offering this twist. I am planning a session with future teachers here at my community college in which we solve the escalators problem together and think about what the process can teach us about teaching and learning. This second video adds productively to the work we’ll do. | 1,463 | 5,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.961014 |
https://www.dlubal.com/en/downloads-and-information/documents/online-manuals/rf-concrete-members/02/01/02 | 1,566,371,122,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315811.47/warc/CC-MAIN-20190821065413-20190821091413-00225.warc.gz | 772,928,658 | 19,905 | # RF-CONCRETE Members – Online Manual Version 5
Online manuals, introductory examples, tutorials, and other documentation.
# 2.1.2 Shear Force
### Shear Force
The check of shear force resistance is to be performed only in the ultimate limit state (ULS). The actions and resistances are considered with their design values. The general design requirement according to EN 1992-1-1, clause 6.2.1 is the following:
• VEd ≤ VRd
where
• VEd : design value of applied shear force
• VRd : design value of shear force resistance
Depending on the failure mechanism, the design value of the shear force resistance is determined by one of the following three values.
• VRd,c : design shear resistance of a structural component without shear reinforcement
• V Rd,s : design shear resistance of a structural component with shear reinforcement, limited by the yield point of shear reinforcement (failure of tie)
• VRd, max : design shear resistance limited by strength of concrete compression strut
If the acting shear force VEd remains below the value of VRd,c, no calculated shear reinforcement is necessary and the check is verified.
If the applied shear force VEd is higher than the value of VRd,c, a shear reinforcement must be designed. The shear reinforcement must resist the entire shear force. In addition, the bearing capacity of the concrete compression strut must be analyzed.
• VEd ≤ VRd,s and VEd ≤ VRd,max
The various types of shear force resistance are determined according to EN 1992-1-1 as follows.
Design shear resistance without shear reinforcement
The design value for the design shear resistance VRd,c may be determined with:
Equation 2.1 EN 1992-1-1, Eq. (6.2a)
where
CRd,c recommended value: 0.18 / γc scaling factor for considering cross-section depthd : mean static depth in [mm] ratio of longitudinal reinforcement Asl : area of tensile reinforcement, which extends by at least(lbd + d) beyond the considered cross-section fck characteristic value of concrete compressive strength in [N/mm2] k1 recommended value: 0.15 bw minimum cross-section width within tension zone in [mm] d static effective depth of bending reinforcement in [mm] design value of concrete longitudinal stress in [N/mm2]
It is allowed, however, to apply a minimum value of the shear force resistance VRd,c,min.
Equation 2.2 EN 1992-1-1, Eq. (6.2b)
where
Design shear resistance with shear reinforcement
The following applies for structural components with shear reinforcement running perpendicular to the component's axis (α = 90°):
Equation 2.3 EN 1992-1-1, Eq. (6.8)
where
Asw cross-sectional area of shear reinforcement s spacing of links z lever arm of the internal forces assumed with 0.9 d fywd design yield strength of shear reinforcement θ inclination of concrete compression strut
The inclination of the concrete compression strut θ may be selected within certain limits depending on the loading. This way, the equation can take into account the fact that a part of the shear force is resisted by crack friction and the virtual truss is thus less stressed. The following limits are recommended in equation (6.7) of EN 1992-1-1:
• 1 ≤ cot θ ≤ 2.5
Thus, the compression strut inclination θ can vary between the following values:
Minimum inclination Maximum inclination θ 21.8° 45.0° cot θ 2.5 1.0
Design shear resistance of concrete compression strut
The following applies for structural components with shear reinforcement running perpendicular to the component's axis (α = 90°):
Equation 2.4 EN 1992-1-1, Eq. (6.9)
where
αcw coefficient for considering stress state in compression flange bw cross-section width z lever arm of the internal forces (precisely calculated in bending design) ν1 reduction factor for concrete strength in case of shear cracks fcd design value of concrete strength θ inclination of concrete compression strut | 886 | 3,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-35 | latest | en | 0.853298 |
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CS 245 (24)
Nancy Day (5)
Midterm
Midterm + solution of Winter 2005 useful for practice
12 Pages
310 Views
School
University of Waterloo
Department
Computer Science
Course
CS 245
Professor
Nancy Day
Semester
Fall
Description
University of Waterloo Midterm Examination SOLUTION SET Term: Winter Year: 2005 Student Name UW Student ID Number Course Abbreviation and Number CS 245 Course Title Logic and Computation Sections SE112 - 001 and CS245 - 001, 002, 003 Instructor Shalini Aggarwal, Nancy Day Date of Exam Thursday, February 10, 2005 Time Period Start time: 4:30 p.m. End time: 6:30 p.m. Duration of Exam 2 hours Number of Exam Pages 12 pages (including this cover sheet) Exam Type Closed book Additional Materials Allowed NO ADDITIONAL MATERIALS ALLOWED • Write your name and student number at the bottom of every page. • Write all solutions on the exam. The booklets are for scratch work. • There are blank truth tables on the last page for use with any question. • Good luck everyone! Question Mark Max Marker Question Mark Max Marker 1 5 7 9 2 5 8 6 3 7 9 9 4 12 10 14 5 9 11 11 6 13 Total 100 Name UW Student ID (page 1 of 12) 1 (5 Marks) Short Answer 1. Can transformational proof be used to show an argument in propositional logic is valid? If so, what do you prove in transformational proof about an argument of the form p ⊢ q where p and q may be compound formulas? If not, explain why. Yes, transformational proof can be used to show an argument is valid. You would prove p ⇒ q ⇚⇛ true. 2. If ¬a is a contingent formula in propositional logic, which of the following describes a? satisfiable, contingent 3. What is the name of the argument forms identified by Aristotle? Syllogisms 4. What is the problem with using a proof procedure that is not sound? If a proof procedure is not sound then it might be possible to prove an argument is valid when it is not. 5. What must be true about a set of propositional logic formulas in order to be able to use natural deduction to prove the set is consistent? We can use natural deduction to show a set of formulas is consistent if and only if the con- junction of the formulas is a tautology. Name UW Student ID (page 2 of 12) 2 (5 Marks) Propositional Logic: Formalization Formalize the following sentences in propositional logic. Show the phrase associated with each prime proposition. 1. Michael Jackson is a software engineer only if pigs fly. j ⇒ p where • j is “Michael Jackson is a software engineer” • p is “Pigs fly” 2. I will go sledding exactly if school is cancelled and it is not windy. s ⇔ c ∧ ¬w where • s is “I will go sledding” • c is “School is cancelled” • w is “It is windy” Name UW Student ID (page 3 of 12) 3 (7 Marks) Propositional Logic: Boolean Valuations Provide a Boolean valuation in which the following two formulas have different truth values: ¬(a ∨ b ∨ c) ¬a ∧ b ⇔ a ∨ c Demonstrate that the formulas have different truth values in this Boolean valuation. Show your work. You may refer to the truth tables on the last page. A Boolean valuation in which the formulas have different truth values is: v(a) = F, v(b) = T, v(c) = T v(¬(a ∨ b ∨ c)) = NOT (v(a) OR v(b) OR v(c)) = F v(¬a ∧ b ⇔ a ∨ c) = (NOT v(a) AND v(b)) IFF (v(a) OR v(c)) = T Name UW Student ID (page 4 of 12) 4 (12 Marks) Propositional Logic: Transformational Proof Prove the following using transformational proof: (c ⇒ ¬a ⇒ b) ∧ (b ⇔ ¬a) ∧ ¬(b ∧ ⇚⇛ (a ∨ b) ∧ (¬b ∨ ¬a) (c ⇒ ¬a ⇒ b) ∧ (b ⇔ ¬a) ∧ ¬(b ∧ a) ⇚⇛ (¬c ∨ (¬a ⇒ b)) ∧ (b ⇔ ¬a) ∧ ¬(b ∧ a) Impl ⇚⇛ (¬c ∨ (¬a ⇒ b)) ∧ (¬a ⇒ b) ∧ (b ⇒ ¬a) ∧ ¬(b ∧ a) Equiv ⇚⇛ (¬a ⇒ b) ∧ (b ⇒ ¬a) ∧ ¬(b ∧ a) Simp II ⇚⇛ (¬¬a ∨ b) ∧ (¬b ∨ ¬a) ∧ ¬(b ∧ a) Impl × 2 ⇚⇛ (a ∨ b) ∧ (¬b ∨ ¬a) ∧ ¬(b ∧ a) Neg ⇚⇛ (a ∨ b) ∧ (¬b ∨ ¬a) ∧ (¬b ∨ ¬a) DM ⇚⇛ (a ∨ b) ∧ (¬b ∨ ¬a) Idemp Name UW Student ID (page 5 of 12) 5 (9 Marks) Propositional Logic: Semantic Tableaux Prove the following is a valid argument using semantic tableaux. Do not use any logical laws from transformational proof in your semantic tableaux proof. ¬a ⇒ b ∨ c, b ⇒ d, ¬(c ∨ d) |ST a 1 ¬a ⇒ b ∨ c 2 b ⇒ d 3 ¬(c ∨ d) 4 ¬a NOT-OR 3 5 ¬c 6 ¬d IMPLIES 2 8 d
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Submit | 1,396 | 4,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.81341 |
https://www.studentsassignments.com/algorithms-assignment-help/ | 1,529,589,003,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864172.45/warc/CC-MAIN-20180621133636-20180621153636-00482.warc.gz | 921,392,432 | 10,512 | Algorithms Assignment Help
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One thing in which most of the students get confused is the analysis of the algorithms. In this process, you have to resolve the quantity of assets that you need to execute. It includes inputs and gets the outputs as a result. | 614 | 3,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | longest | en | 0.956977 |
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3D rotating magic cube Https://mlldxe.cn/archives/641 Https://mlldxe.cn/archives/641#comments Sun, 08 Mar 2020 03:00:14 +0000 Https://mlldxe.cn/? P=641 We can realize the cube rotation by CSS animation, and change the key frame definition (@keyframes) in CSS code to change the way of the cube's rotation.
## The results are as follows:
The label properties we may use are as follows:
Label describe
Transform Elements can be converted to 2D or 3D.
Transform-style How to display the specified elements in 3D space.
Animation You can animate elements.
@keyframes animation name {from {}to Set animation from start to finish.
TranslateX ( X ) Define 3D transformation, using only values for X axis.
TranslateY ( Y ) Define 3D transformation, using only values for Y axis.
TranslateZ ( Z ) Define 3D transformation, using only values for Z axis.
Translate3d (x, y, z) Define 3D transformation.
ScaleX ( X ) Define the 3D zoom transformation by passing the value of a given X axis.
ScaleY ( Y ) Define the 3D zoom transformation by passing the value of a given Y axis.
ScaleZ ( Z ) Define the 3D zoom transformation by passing the value of a given Z axis.
RotateX ( Angle ) Define the 3D rotation along the X axis.
RotateY ( Angle ) Define the 3D rotation along the Y axis.
RotateZ ( Angle ) Define the 3D rotation along the Z axis.
Rotate3d ( X , Y , Z , Angle ) Define 3D rotation.
## We have the following ideas for implementation.
1, first put the six sides of the cube together.
2, then rotate each surface to make it form a cube.
3, then let it rotate.
## The first step is to define the six sides of a cube.
`````` / * the outermost container styles are */
.wrap {
width: 100px;
height: 100px;
margin: 150px;
position: relative;}} /} / * wrap all container styles * / 100px / * set transform-style: preserve-3d, so that its child elements appear in the space of
margin:;
transform: rotateX (-30deg) rotateY (-80deg);
animation: rotate linear 20s infinite;
}
@-webkit-keyframes infinite; (2); (E);
opacity: 0.8;
transition: all.4s;}}} / * define all image styles */
.pic {
width: 200px;
height: 200px;}}} / * * define small cube style */
.cube span span; Pic {
width: 100px;
height: 100px;``````
## The second step: rotate the faces and turn them into 3D cubes.
Here, we divide it into three parts.
1, translateZ: translate two faces to form two sides of the cube.
2, rotateX (90DEG): inverting two sides to form two sides of the cube.
3, rotateY (90DEG): inverting two sides to form two sides of the cube.
`````` / * define the large cube six face rotation style */
.cube.Out_ Front {
transform: rotateY (0DEG) translateZ (100px);
}
.cube.Out_ Back {
transform: translateZ (-100px) rotateY (180DEG);
}
.cube.Out_ Left {
transform: rotateY (-90deg) translateZ (100px);
}
.cube.Out_ Right {
transform: rotateY (90DEG) translateZ (100px);
}
.cube.Out_ Top {
transform: rotateX (90DEG) translateZ (100px);
}
.cube.Out_ Bottom {
transform: rotateX (-90deg) translateZ (100px);}}}, rotateX, * define small cubes, six face rotation styles */
.cube.In_ Front {
transform: rotateY (0DEG) translateZ (50px);
}
.cube.In_ Back {
transform: translateZ (-50px) rotateY (180DEG);
}
.cube.In_ Left {
transform: rotateY (-90deg) translateZ (50px);
}
.cube.In_ Right {
transform: rotateY (90DEG) translateZ (50px);
}
.cube.In_ Top {
transform: rotateX (90DEG) translateZ (50px);
}
.cube.In_ Bottom {
transform: rotateX (-90deg) translateZ (50px); X}} Cube:hover.Out_ Front {
transform: rotateY (0DEG) translateZ (200px); Cube:hover.Out_ Back {
transform: translateZ (-200px) rotateY (180DEG); Cube:hover.Out_ Left {
transform: rotateY (-90deg) translateZ (200px); Cube:hover.Out_ Right {
transform: rotateY (90DEG) translateZ (200px); Cube:hover.Out_ Top {
transform: rotateX (90DEG) translateZ (200px); Cube:hover.Out_ Bottom {
transform: rotateX (-90deg) translateZ (200px);``````
## The third step: by rotation.` Div`To achieve rotation function.
`````` <! DOCTYPE html>
<html>
<meta charset= "UTF-8" / >
<body>
<! -- the largest outer container -- > class= <div class= "wrap" > < < <!! > - all containers of all elements -- > <div <div class= "<div" > < <! "- front picture > Front "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 1.jpg "alt=", "class=" pic "/ > </div>
"! "- the picture behind: > <div <div class= out_. Back "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 2.jpg "alt=", "class=" pic "/ > </div>
"! "- left picture > > <div <div" out_ " Left "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 3.jpg "alt=", "class=" pic "/ > </div>
"! "- right picture - > out_ <div class= out_. Right "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 4.jpg "alt=", "class=" pic "/ > </div>
"! "- picture above: > <div <div class= out_. Top "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 5.jpg "alt=", "class=" pic "/ > </div>
"! "- the following picture: > <div <div class= out_. Bottom "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 6.jpg "alt=", "class=" pic "/ > </div>
"! "- small cube - > <span <span class= in_. Front "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 7.jpg "alt=", "class=" in_ Pic "/ > </span>
<span class=" in_ Back "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 8.jpg "alt=", "class=" in_ Pic "/ > </span>
<span class=" in_ Left "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 9.jpg "alt=", "class=" in_ Pic "/ > </span>
<span class=" in_ Right "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 10.jpg "alt=", "class=" in_ Pic "/ > </span>
<span class=" in_ Top "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 11.jpg "alt=", "class=" in_ Pic "/ > </span>
<span class=" in_ Bottom "> <img src=" Https://mlldxe.cn/wp-content/themes/kratos-pjax/magic_ Square/magic/magic_ 12.jpg "alt=", "class=" in_ PIC / / > </span>
</div>
</div>
</body>
</html>``````
Place in the corresponding place` CSS` You can place the 3D magic cube in the sidebar. Add mode: enter your blog Garden > settings, add the above HTML code to the blog side sidebar bulletin.
## END: effect demonstration 3D rotating magic cube
The following are the pictures used:
> >
Https://mlldxe.cn/archives/641/feed One
Happy birthday to me A kind of Https://mlldxe.cn/archives/640 Https://mlldxe.cn/archives/640#comments Sat, 07 Mar 2020 13:32:35 +0000 Https://mlldxe.cn/? P=640 Happy birthday to me Chinese style
> >
Https://mlldxe.cn/archives/640/feed Two
Using custom HTML gadget to achieve countdown Https://mlldxe.cn/archives/636 Https://mlldxe.cn/archives/636#respond Sat, 07 Mar 2020 03:30:29 +0000 Https://mlldxe.cn/? P=636
### Preface
It feels like an interesting thing. You can record some information remaining time, such as domain name, server, etc., mainly to see what they like to record, no restrictions, add is also very simple, very good, recommend to everyone.
## Implementation steps
1. save the following code as` Countdownjs.js` Save in the JS of the current theme:
`````` / *!
*Template Name: custom HTML countdown
*Blog: Https://mlldxe.cn/
*author: Dog's
*url: Https://mlldxe.cn/archives/636
* * * * The Final Countdown for jQuery v2.2.0 ( Http://hilios.github.io/jQuery.countdown/ * * * * Copyright (c) 2016 Edson Hilios
*
* Permission is hereby granted, free of charge, of, and it is also referred to as "free". Software without restriction, including without limitation the rights to
* use, copy, modify, t, C, R, R, R, D, R, D, R, D, R, D, R, D, R, D, R, D, R, D, R, D, R, D, etc. Subject to the following conditions:
* The, above copyright notice and this, notice, China, Japan, China, Japan, China, China, China, China, China, Macao, China. FITNESS
* FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL, NO, ",", " OUT OF OR IN
* CONNECTION WITH THE SOFTWARE OR THE USE, THE, "("), "{", "{"; " Define.amd {
define ([jQuery]], factory); else} {
factory (jQuery);}}}} (function (\$) {} use strict;
var instances = []. Matchers = [], defaultOptions = {
precision: 100,
elapse: false,
defer: false
}; Matchers.push (/^[0-9]*\$/.source); Matchers.push (/ ([0-9]{1,2}\/) {2}[0-9]{4} ([0-9]{1,2} ([0-9]{2}) {2}) /.source) Matchers.push (/[0-9]{4} ([\/\-][0-9]{1,2}) {2} ([0-9]{1,2} ([0-9]{2}) {2}) /.source);
matchers = new RegExp ( Matchers.join (
function) (parseDateString); dateString {
if (dateString instanceof Date) {
return dateString;
}
if (String (dateString) (())) ((()) (()) (()) (=); }
if (String (dateString).Match (/ / /)) {
dateString = String (dateString).Replace (/\-/g, "/");
}
return new Date; KEY_ MAP = {
Y: "years",
m: "months",
n: "daysToMonth",
w: "weeks",
d: "daysToWeek", "
d:", "" "," "" "," "" "," "" "", "" "", "" "" "," "" "};" Var sanitize = Str.toString ().Replace (/ [(*] + \$(\$)] (/g) (\\\$1);
return new RegExp (sanitize);
}
function strftime (offsetObject) {
return {
return (()) Format.match (/% (?!!)? [A-Z]{1} ((^; + +); /gi);
if (directives) {
for (VaR I = 0, Len = Directives.length ; I < len; ++i) {
var directive = directives[i].match (/% (-)!! ([a-zA-Z]{1}) ((^; + +); /), regexp = escapedRegExp (directive[0]), modifier = directive[0] "", "= =" KEY_ MAP.hasOwnProperty (Directive)) {
value = DIRECTIVE_ KEY_ MAP[directive];
value = Number (offsetObject[value]);
}
if (value = = null) {
if (modifier = = = "!") {
value = pluralize (modifier, V); ((= = = = ")" { If (value < 10) {
value = "0" + Value.toString ();
}
}
format = Format.replace (regexp, Value.toString ());
}
}
}
format = Format.replace (% /%,%);
return format;
}
function};
}
function pluralize (format, count) {
var plural = "s", singular = "";
if (s) = = Format.replace ((/: (|\s; /gi), "" ".Split (/ / /);
if ( Format.length = = 1) {
plural = format[0]; else} else = Format[0];
plural = format[1];
}
}
if ( Math.abs (count) = = = 1) {
return singular;}} else {
return plural;
}
}
var Countdown = function (EL, finalDate, function) {} This.el = El;
this.\$el = \$(EL); This.interval = null; This.offset = {} This.options = \$.extend ({}, defaultOptions) This.instanceNumber = Instances.length ; Instances.push (this);
this.\$ El.data ("countdown-instance", This.instanceNumber );
if (options) {
if (typeof options = = = "function") {
this.\$ El.on (" Update.countdown " Options);
this.\$ El.on (" Stoped.countdown "Options";
this.\$ El.on (" Finish.countdown "Options"; {} else {}. This.options = \$.extend ({}, defaultOptions, options); This.setFinalDate (finalDate);
if ( This.options.defer = = false) { This.start
\$.extend; Countdown.prototype {
start: function () {
if ( This.interval = = null) {
clearInterval ( This.interval )
}
var self = this; This.update (); This.interval = SetInterval (function ()) { Self.update.call (self); This.options.precision
stop: function () {
clearInterval ( This.interval ) This.interval = null; This.dispatchEvent ("stoped");
toggle: function () {
if ( This.interval { This.stop (); {} else {}. This.start
pause: function () {} This.stop (
resume:) function}. This.start (
remove:) function}. This.stop.call (this);
instances[ This.instanceNumber = null;
delete this.\$ El.data (.CountdownInstance);
setFinalDate: function (value) {}. This.finalDate = parseDateString (value);
update:},
update: function () {
if (this.\$) El.closest ("HTML").Length = = = 0) { This.remove (); Return;
}
var hasEventsAttached = \$. Data ( This.el , "events") = = undefined, now = new Date (), newTotalSecsLeft;
newTotalSecsLeft = This.finalDate.getTime () - Now.getTime ();
newTotalSecsLeft = Math.ceil (newTotalSecsLeft / 1E3);
newTotalSecsLeft =! This.options.elapse & newTotalSecsLeft < 0? 0: Math.abs (newTotalSecsLeft);
if ( This.totalSecsLeft = = newTotalSecsLeft hasEventsAttached, {
return; else} { This.totalSecsLeft = newTotalSecsLeft; This.elapsed = now > = This.finalDate ; This.offset = {
seconds: This.totalSecsLeft % 60 Minutes: Math.floor ( This.totalSecsLeft / 60)% 60,
hours: Math.floor ( This.totalSecsLeft / 60 / 60)% 24,
days: Math.floor ( This.totalSecsLeft / 60 / 60 / 24)% 7,
daysToWeek: Math.floor ( This.totalSecsLeft / 60 / 60 / 24)% 7
daysToMonth: Math.floor ( This.totalSecsLeft / 60 / 60 / 24% 30.4368),
weeks: Math.floor ( This.totalSecsLeft / 60 / 60 / 24 / 7),
months: Math.floor ( This.totalSecsLeft / 60 / 60 / 24 / 30.4368),
years: Math.abs ( This.finalDate.getFullYear () - Now.getFullYear ())
totalDays: Math.floor ( This.totalSecsLeft / 60 / 60 / 24),
totalHours: Math.floor ( This.totalSecsLeft / 60 / 60),
totalMinutes: Math.floor ( This.totalSecsLeft / 60),
totalSeconds: This.totalSecsLeft
}
if ()! This.options.elapse & & This.totalSecsLeft = = 0) This.stop (); This.dispatchEvent ("finish"); {} else {}. This.dispatchEvent ("update");
dispatchEvent: function (eventName) {
var event = \$.Event (eventName + ".Countdown"); Wei Event.finalDate = This.finalDate ; Event.elapsed = This.elapsed ; Event.offset = \$.extend ({}, This.offset ) Event.strftime = strftime ( This.offset );
this.\$ El.trigger (event); \$10. Fn.countdown = function () {
var argumentsArray = Array.prototype.slice .call (arguments, 0);
return This.each (function () {
var instanceNumber = \$(this).Data ("countdown-instance");
if (instanceNumber = = undefined) {
var instance = instances[instanceNumber]. Method = argumentsArray[0];
if ( Countdown.prototype.hasOwnProperty (method)) {
instance[method].apply (instance). ArgumentsArray.slice (1)) else if (String (method).Match (/^[\$A-Z_). ][0-9A-Z_ \$]*\$/i) = = = null) { Instance.setFinalDate.call (instance, method) Instance.start (); else {
\$.error ("Method%s does not exist on") JQuery.countdown ".replace (/\%s/gi, method)"; else}} else {
new Countdown (this). ArgumentsArray[0], argumentsArray[1]);``````
`````` <style type= "text/css" >
.zzsc-content{
#clock{
font-family:'Days One', "Microsoft YaHei", Arial, sans-serif;
font-size: 2em;
}
</style>
<div class= "zzsc-content bgcolor-8" >
<span id= "clock" ></span>
</div>
<script type= "text/javascript" src= " Https://blog.mlldxe.cn/countdownjs.js "></script>
<script type=" text/javascript > > \$\$(function ()) {\$text/javascript ('#clock').Countdown ('2018/6/7', function (event) {\$\$(this).Html). Event.strftime ('%D days%H:%M:%S');
</script>)``````
Four` \$('#clock').Countdown ('2018/6/7', function (event))` Change to the date you want.
# Two Zero Two One year - new year inverted meter Time
In addition, CSS can beautify itself.
> >
Https://mlldxe.cn/archives/636/feed Zero
WP Mail SMTP failed to send using QQ mailbox. Https://mlldxe.cn/archives/627 Https://mlldxe.cn/archives/627#respond Wed, 04 Mar 2020 13:47:04 +0000 Https://mlldxe.cn/? P=627
### Preface
After the blog opens HTTPS, change the Kratos theme, use the WP Mail SMTP plug-in to push the failure with QQ mailbox configuration? What the f**k
## 1.QQ mailbox to turn on IMAP/SMTP service
In the settings of the QQ mailbox.
## 2. port setup
In WP Mail SMTP plug-in settings, STMP Host fills in Stmp.qq.com STMP Port fill in 465
## 3. accounts and passwords
The Username (user name) is filled with the QQ mailbox. The password is filled. It is not the password of the mailbox, but the generated authorization code.
Finally, send out my setup page.
> >
Https://mlldxe.cn/archives/627/feed Zero
Blog enable HTTPS (SSL) Https://mlldxe.cn/archives/620 Https://mlldxe.cn/archives/620#respond Mon, 02 Mar 2020 12:50:30 +0000 Https://mlldxe.cn/? P=620 I heard that SSL world is beautiful, and it has spent two days. Blog has opened HTTPS.
> >
Https://mlldxe.cn/archives/620/feed Zero
A dynamic proof that I'm alive. Https://mlldxe.cn/archives/616 Https://mlldxe.cn/archives/616#comments Fri, 28 Feb 2020 09:17:02 +0000 Http://mlldxe.cn/? P=616 A dynamic proof that I am still alive. Cleaned up a wave of friends chain.
> >
Https://mlldxe.cn/archives/616/feed Two
Remote support and remote access TeamViewer [V14.4.2669] Https://mlldxe.cn/archives/559 Https://mlldxe.cn/archives/559#comments Sat, 17 Aug 2019 07:58:40 +0000 Http://mlldxe.cn/? P=559 TeamViewer is a simple and fast solution for remote control applications, desktop sharing and file transfer, which can be used for remote control in the backstage of any firewall and NAT agent. In order to connect to another computer, you only need to run TeamViewer on two computers at the same time without installation. The software first started to automatically generate partner ID on two computers. Just input your partner's ID to TeamViewer, and then establish a connection immediately.
> >
Https://mlldxe.cn/archives/559/feed Six
I can no longer be a salted fish. Https://mlldxe.cn/archives/554 Https://mlldxe.cn/archives/554#comments Sat, 03 Aug 2019 07:38:36 +0000 Http://mlldxe.cn/? P=554 Do not want to live up to too many people's expectations.
But I never learn to stick to it.
Watching others step by step from me to be stronger than me.
Everyone is making progress.
I am the only one standing there.
I always make excuses for myself.
I want to change
But I can't keep up.
You can't even beat yourself.
Every time you see someone else doing something easily,
I really feel bad.
Am I really bad?
Holding pen and thinking a thousand words and thousands of words
But I can not write my pride.
Finally, only doubts and problems remain.
Drink a bowl of soup that is pleasant to the past.
Warm and empty, full of sorrow.
Let's face the reality.
How many cold and warm can only know oneself.
And hit the walls of youth.
Only to discover that courage is not against time.
Wandering in a changing city
How many times have you been there?
If everything is worth it, it's worth it.
Why some success can make people feel lost?
In the end, the heart is more open than the distance.
It's used to being alone.
When tears do not hide with smiles
When it is misunderstood, it will no longer be explained.
I understand that happiness is your own perception.
I understand that life is one's own.
The universe is so big.
The world is vast.
It is not impossible to get better in the future.
I want to work hard
Worthy of one's ambition
-- to the future self
## First year
Everything begins when you are not ready.
When you're ready, it's over.
Sorry, only. Sign in And you can read the hidden content before you comment on any article in this station.
> >
Https://mlldxe.cn/archives/554/feed Three
Simple and easy to use GIF production tool -- ScreenToGif Https://mlldxe.cn/archives/546 Https://mlldxe.cn/archives/546#respond Fri, 26 Jul 2019 02:40:16 +0000 Http://mlldxe.cn/? P=546
### Preface
In daily life, you may encounter some situations of displaying some operation process or some dynamic effects to others. Screenshots are sometimes hard to say clearly, and video recording is not convenient. At that time, a software that can directly generate GIF diagrams is needed to solve this problem. If you know the software of Screen ToGif, this problem can be solved.
Screen ToGif is a free, open source, no advertising screen to GIF software. You can record the content you need through 3 ways: screen, webcam and sketchpad. At the same time, it supports editing and recording, editing GIF from other sources. After decompression, the software size is only 2.2M, and most importantly, support Chinese!
## There are seven options in menu interface.
• Application: the main function settings of software
• Interface settings: background and theme settings for the editor
• Language: interface language setup, default follow system.
• Temporary files: where temporary files are stored, the software is saved by default for GIF five days.
• Extra: there is no localization here, and additional functions should be added.
• Donation: emmmm... You know.
• About: author's contact method
## There are seven options in the editor menu.
1. File: save items and add other media files.
2. Home page: copy, paste, revoke the previous operation, adjust the size of the window, etc.
3. play
4. Edit: delete and adjust frames
5. Image: adding watermark, subtitle, blurred object, etc.
6. Transition: transition animation adjustment, sub sliding and fade out two styles.
7. Statistics: GIF file information, frames, time and so on.
## Edible method
The operation is very simple. After double click the software running, click the video recorder to enter the recording interface. It can adjust the recording scope freely, modify the frames per second, and start recording after setting up. After the recording is finished, the editing operation is carried out above the interface. It can choose to delete frames, add watermark, and modify transitional animation.
For more specific use, see GitHub (English): Https://github.com/NickeManarin/ScreenToGif/wiki/help | 5,992 | 21,308 | {"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} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.626544 |
https://www.drdarrinlew.us/energy-conservation/transient-loading.html | 1,563,765,833,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527474.85/warc/CC-MAIN-20190722030952-20190722052952-00215.warc.gz | 670,593,037 | 9,105 | It must also be noted that wind turbines, unlike other forms of electricity production, are designed to generate under a spectrum of power levels [5], and thus must contend with transient loads within their drivetrains. A commonly used means to estimate the power-producing ability and drivetrain loads is to assume that it operates under a certain wind speed distribution. The distributions that are frequently used are Weibull (Fig. 18.5) and Rayleigh distributions [23, 24]. To determine the number of cycles of a drivetrain component at a given load, it is important to determine the total number of hours per year for a particular wind speed. This is done by finding the probability of a particular wind speed for the desired distribution and multiplying it by the number of hours per year. The probability that the wind lies between two wind speeds is given by
Fig. 18.5 Histogram of predicted and observed wind speeds, from [25]
where Pr is the probability function, and p(u) is the probability density function. For the Rayleigh distribution, the probability density function is given by
4 Ua
where ua is the average wind velocity. From this, the number of hours that a wind turbine is operated at a given speed is estimated as
where 8760 represents the total number of hours in a 365-day year [24]. From this analysis, designers can implement a damage criteria such as Palmgren-Miner's rule (as discussed in Sect. 18.5.2.4) to estimate the fatigue lives of components such as gears and bearings.
## Renewable Energy 101
Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.
Get My Free Ebook | 453 | 2,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-30 | latest | en | 0.951314 |
https://gpatargeter.com/can-someone-please-answer-these-questions-that-is-due-4-23-18-4pm-please-thank-you-chapter-6-section-4-ht-exercise-211-use-the-t-distribution/ | 1,679,780,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00446.warc.gz | 322,704,078 | 16,343 | # Can someone please answer these questions that is due 4/23/18 @ 4pm please thank you Chapter 6, Section 4-HT, Exercise 211 Use the t -distribution
Can someone please answer these questions that is due 4/23/18 @ 4pm please thank you
Chapter 6, Section 4-HT, Exercise 211
Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal.
Test H0 : μA=μB
vs Ha : μA≠μB
using the fact that Group A has 8 cases with a mean of 125 and a standard deviation of 18 while Group B has 15 cases with a mean of 118 and a standard deviation of 14.
(a) Give the test statistic and the p-value.
Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places.
test statistic =
Enter your answer in accordance to item (a) of the question statement
p-value =
Enter your answer in accordance to item (a) of the question statement
(b) What is the conclusion of the test? Test at a 10%
level.
Reject H0
.
Do not reject H0
.
Chapter 6, Section 4-HT, Exercise 213
Take Your Notes Longhand!
A study1 randomly assigned students to take notes either longhand or using a laptop. The resulting scores of the students on a test of the material are given in the table below. Does the data provide evidence that it is more effective to take notes longhand rather than on a laptop?
Note-takingn
s
Longhand38
25.6
10.8
Laptop40
18.3
9.0
Table 1: Summary statistics for test scores after taking notes longhand or on a laptop
Let group 1
and group 2
be the score on the test for those taking notes by hand and those taking notes on a computer, respectively.
Calculate the relevant test statistic.
Round your answer to three decimal places.
t
-statistic=
the absolute tolerance is +/-0.02
Find the p
-value.
Round your answer to four decimal places.
p
-value =
the absolute tolerance is +/-0.0005
What is the conclusion?
Reject H0
.
Do not reject H0
.
Do we have evidence to conclude that it is more effective to take notes longhand?
Yes
No | 543 | 2,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.859325 |
https://www.gamedev.net/forums/topic/614332-continuous-action-memory/ | 1,542,460,396,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743521.59/warc/CC-MAIN-20181117123417-20181117145417-00378.warc.gz | 869,836,749 | 28,044 | # Continuous Action Memory
This topic is 2569 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I recently posted a quick tutorial on my blog about what I call Continuous Action Memory.
It's basically the methodology behind efficiently recording gameplay, so that you can watch a replay afterwards or do other cool things. It's the same technique that's used in Starcraft replays, and a more advanced implementation is used for time manipulation in Braid (best game ever). It's actually a really simple concept, and can be applied to most games.
By the way, the blog is brand new and still under construction (I'm coding it from scratch), so any kind of feedback on the website is welcome.
##### Share on other sites
Are you recording/replaying the input?
##### Share on other sites
Are you recording/replaying the input?
That's the most trivial form of it, which is what I covered in the tutorial. I may make another post later about more advanced implementations. It can get really complex when you try to do what Jonathan Blow did in Braid.
##### Share on other sites
Rewinding from time t given
- game state @ t
- input history from 0 to t
- forward time logic
without replaying from 0 to t beforehand is pretty interesting. (i.e. automatic computation of reverse time logic from forward time logic)
In SC2 you have to replay to t before rewinding from t.
##### Share on other sites
Interesting.
Live for Speed is another game that does a similar thing for it's replays - rather than store the position of the car throughout the replay, it just stores the user inputs. It then replays these inputs throught the physics engine and the car goes through exactly the same manoevers again. I imagine there's some pretty serious considerations to take into account when doing this though a complex (ie, where a tiny error could quickly multiply into a very large error) iterative physics engine.
##### Share on other sites
Interesting.
Live for Speed is another game that does a similar thing for it's replays - rather than store the position of the car throughout the replay, it just stores the user inputs. It then replays these inputs throught the physics engine and the car goes through exactly the same manoevers again. I imagine there's some pretty serious considerations to take into account when doing this though a complex (ie, where a tiny error could quickly multiply into a very large error) iterative physics engine.
It is a very sensitive system. As you said, even the tiniest amount of variance can quickly derail the whole thing. I addressed the two leading causes of that problem:
1. Variable frame rate, which must be accounted for when recording
2. Floating-point inaccuracy, which can be eliminated in multiple ways
Once I took care of those issues, I had perfectly identical playback, even on replays as long as 20 minutes.
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× | 700 | 3,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | longest | en | 0.937363 |
https://www.freethesaurus.com/unit+matrix | 1,537,541,413,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157203.39/warc/CC-MAIN-20180921131727-20180921152127-00515.warc.gz | 743,151,515 | 15,744 | # unit matrix
Also found in: Dictionary, Medical, Encyclopedia, Wikipedia.
Graphic Thesaurus 🔍
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Legend Synonym Antonym Related
• noun
## Synonyms for unit matrix
### a scalar matrix in which all of the diagonal elements are unity
#### Related Words
References in periodicals archive ?
Step 1 : Reduce the system of linear equations AX = B to a form containing the unit matrix of requisite order, using elementary row operations only.
For a lossless passive network, the complex S-parameter matrix |Mathematical Expression Omitted~ must satisfy the equation |Mathematical Expression Omitted~, where |Mathematical Expression Omitted~ is the unit matrix of the same order as |Mathematical Expression Omitted~, and |Mathematical Expression Omitted~ is the Hermitian conjugate (conjugate transpose) of |Mathematical Expression Omitted~.
Head of Merseyside Police unit Matrix Detective Chief Superintendent Paul Richardson said: "This is a large and sophisticated cannabis factory containing thousands of plants across 11 growing rooms.
p)--(kxl)--the matrix; E--(m x m)--the unit matrix.
4)] is the 4x4 unit matrix and [PSI] is a 4-component column (bispinor) wavefunction.
The big M method and the two-phases way are usually adopted to construct a unit matrix as a initial basis.
where P is the direct sum of the unit matrix of order 2 x 2 and a null matrix i.
Yesterday the force's anti-gang unit Matrix remained on site dismantling the factory, which was spread across 11 rooms.
i]{D} correspond to line "i" in the 8th order unit matrix.
Detective Superintendent Chris Green, from the force's anti gun and gang unit Matrix said: "Here we have cannabis cultivation within the heart of a small community.
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2/25/2018 Quiz 8: Principles of Microeconomics (ECN-130-2)(WEB)(2018SP) 1/21 * Some questions not yet graded Quiz 8 Due Feb 26 at 11:59pm Points 100 Questions 50 Available Feb 19 at 12am Feb 26 at 11:59pm 8 days Time Limit None Allowed Attempts 2 Azempt History Attempt Time Score LATEST Attempt 1 87 minutes 88 out of 100 * Ð Correct answers are hidden. Score for this attempt: 88 out of 100 * Submitted Feb 25 at 8:20pm This attempt took 87 minutes. Take the Quiz Again 2 / 2 pts Question 1 At 1,000 units of output, total cost is \$42,000 and total variable cost is \$34,000. At 1,000 units of output, what is the value of average total cost, average variable cost, and average fixed cost, respectively? \$22; \$14; \$8 \$42; \$34; \$8 \$4.20; \$3.40; \$0.80 \$340; \$740; \$60 \$42; \$34; There is not enough information provided to determine the average fixed cost. 2 / 2 pts Question 2 Exhibit 215
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2/25/2018 Quiz 8: Principles of Microeconomics (ECN-130-2)(WEB)(2018SP) 2/21 Refer to Exhibit 215. Diseconomies of scale are present between points A and B. points A and C. points B and C. points B and D. points C and D. 2 / 2 pts Question 3 John purchases a baseball card for \$10 that turns out to be so rare that a collector offers to buy it from him for \$2,000. Instead, John decides to give the card to his sister (an avid baseballcard collector) as a birthday present. The opportunity cost of John's generosity is \$10, the purchase price. \$0, because at the time the decision is made, \$10 are sunk cost, i.e., at that point there is no cost to John of giving the card away. \$2,000, the amount offered by the collector. \$1,005, the average of \$10 and \$2,000. 2 / 2 pts Question 4 As long as there are __________ costs, __________ profit will be greater than __________ profit. implicit; accounting; economic explicit; accounting; economic
2/25/2018 Quiz 8: Principles of Microeconomics (ECN-130-2)(WEB)(2018SP) 3/21 implicit; economic; accounting explicit; economic; accounting 2 / 2 pts Question 5 If the LRATC curve is falling, then the law of diminishing marginal returns is operating. economies of scale are present. constant returns to scale are present. diseconomies of scale are present. 2 / 2 pts Question 6 The averagemarginal rule states: when the marginal magnitude is rising, the average magnitude must also be rising. when the marginal magnitude is falling, the average magnitude must also be falling. when the marginal magnitude is below the average magnitude, the average magnitude falls. when the average magnitude rises, the marginal magnitude falls. a and b 2 / 2 pts Question 7 Situation 213 Gizmos, Inc. produces gizmos at an average total cost of \$22 and an average variable cost of \$14. The only fixed input used in the production of gizmos costs \$32,000. Refer to Situation 213. How many gizmos does Gizmos, Inc. produce?
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reasoning chain
цепочка рассуждения (в СИИ)
English-Russian dictionary of computer science and programming. 2013.
### Смотреть что такое "reasoning chain" в других словарях:
• Reasoning — (Roget s Thesaurus) < N PARAG:Reasoning >N GRP: N 1 Sgm: N 1 1 =>{ant,477,} reasoning ratiocination rationalism Sgm: N 1 dialectics dialectics induction generalization GRP: N 2 Sgm: N 2 discussion discussion comment Sgm: N 2 … English dictionary for students
• chain — chain1 W2S3 [tʃeın] n ▬▬▬▬▬▬▬ 1¦(joined rings)¦ 2¦(connected events)¦ 3¦(shops/hotels)¦ 4¦(connected line)¦ 5¦(prisoners)¦ 6¦(buying a house)¦ ▬▬▬▬▬▬▬ [Date: 1200 1300; : Old French; Origin: chaeine, from Latin catena] … Dictionary of contemporary English
• chain of reasoning — index dialectic Burton s Legal Thesaurus. William C. Burton. 2006 … Law dictionary
• Markov chain — A simple two state Markov chain. A Markov chain, named for Andrey Markov, is a mathematical system that undergoes transitions from one state to another, between a finite or countable number of possible states. It is a random process characterized … Wikipedia
• Regular chain — In computer algebra, a regular chain is a particular kind of triangular set in a multivariate polynomial ring over a field. It enhances the notion of characteristic set. Introduction Given a linear system, one can convert it to a triangular… … Wikipedia
• logic — n 1. science of reasoning, dialectics, symbolic logic, logistic; polemics, art of disputation ot controversy. 2. reasoning, argumentation, ratiocination, process of reasoning; Logic. inference, Logic. deduction, Logic. induction, syllogization,… … A Note on the Style of the synonym finder
• argument — n. 1. Reason, ground, proof, evidence, reasoning, chain of reasoning, process of reasoning. 2. Controversy, dispute, disputation, discussion, debate. 3. Subject, topic, matter, theme, thesis, question, subject matter, matter in hand. 4. Summary,… … New dictionary of synonyms
• argument — n 1. debate, discussion, polemic, logomachy, war of words; contestation, disputation, argumentation. 2. disagreement, contention, altercation, conflict, dispute, controversy, discord, clash, falling out, Inf. hassle, Inf. barney; quarrel, row,… … A Note on the Style of the synonym finder
• argumentation — n 1. logic, reasoning, chain or course or pattern of reasoning; polemics, dialectics. 2. debate, discussion, polemic, logomachy, war of words; disputation, disagreement, contention, altercation. 3.(all in order to convince or persuade) address,… … A Note on the Style of the synonym finder
• последовательность выводов — цепь рассуждений — [Л.Г.Суменко. Англо русский словарь по информационным технологиям. М.: ГП ЦНИИС, 2003.] Тематики информационные технологии в целом Синонимы цепь рассуждений EN reasoning chain … Справочник технического переводчика
• Chainstore paradox — (or Chain Store paradox ) is a concept that purports to refute standard game theory reasoning. Contents 1 The chain store game 1.1 Induction theory 1.2 Deterrence theory 1.3 … Wikipedia
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### Introduction
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### Code
Student Retention Model
# Student Retention Model - Start to Finish, From Modeling Student Behavior to Helping At-Risk Cases¶
In [343]:
from IPython.display import Image
Image(filename='viralml-book.png')
Out[343]:
# ViralML.com
## Student Performance Data Set¶
https://archive.ics.uci.edu/ml/datasets/student+performance
### Data Set Information¶
This data approach student achievement in secondary education of two Portuguese schools. The data attributes include student grades, demographic, social and school related features) and it was collected by using school reports and questionnaires. Two datasets are provided regarding the performance in two distinct subjects: Mathematics (mat) and Portuguese language (por). In [Cortez and Silva, 2008], the two datasets were modeled under binary/five-level classification and regression tasks. Important note: the target attribute G3 has a strong correlation with attributes G2 and G1. This occurs because G3 is the final year grade (issued at the 3rd period), while G1 and G2 correspond to the 1st and 2nd period grades. It is more difficult to predict G3 without G2 and G1, but such prediction is much more useful (see paper source for more details).
# Attributes for both student-mat.csv (Math course) and student-por.csv (Portuguese language course) datasets:
1 school - student's school (binary: 'GP' - Gabriel Pereira or 'MS' - Mousinho da Silveira)
2 sex - student's sex (binary: 'F' - female or 'M' - male)
3 age - student's age (numeric: from 15 to 22)
4 address - student's home address type (binary: 'U' - urban or 'R' - rural)
5 famsize - family size (binary: 'LE3' - less or equal to 3 or 'GT3' - greater than 3)
6 Pstatus - parent's cohabitation status (binary: 'T' - living together or 'A' - apart)
7 Medu - mother's education (numeric: 0 - none, 1 - primary education (4th grade), 2 – 5th to 9th grade, 3 – secondary education or 4 – higher education)
8 Fedu - father's education (numeric: 0 - none, 1 - primary education (4th grade), 2 – 5th to 9th grade, 3 – secondary education or 4 – higher education)
9 Mjob - mother's job (nominal: 'teacher', 'health' care related, civil 'services' (e.g. administrative or police), 'at_home' or 'other')
10 Fjob - father's job (nominal: 'teacher', 'health' care related, civil 'services' (e.g. administrative or police), 'at_home' or 'other')
11 reason - reason to choose this school (nominal: close to 'home', school 'reputation', 'course' preference or 'other')
12 guardian - student's guardian (nominal: 'mother', 'father' or 'other')
13 traveltime - home to school travel time (numeric: 1 - <15 min., 2 - 15 to 30 min., 3 - 30 min. to 1 hour, or 4 - >1 hour)
14 studytime - weekly study time (numeric: 1 - <2 hours, 2 - 2 to 5 hours, 3 - 5 to 10 hours, or 4 - >10 hours)
15 failures - number of past class failures (numeric: n if 1<=n<3, else 4)
16 schoolsup - extra educational support (binary: yes or no)
17 famsup - family educational support (binary: yes or no)
18 paid - extra paid classes within the course subject (Math or Portuguese) (binary: yes or no)
19 activities - extra-curricular activities (binary: yes or no)
20 nursery - attended nursery school (binary: yes or no)
21 higher - wants to take higher education (binary: yes or no)
22 internet - Internet access at home (binary: yes or no)
23 romantic - with a romantic relationship (binary: yes or no)
24 famrel - quality of family relationships (numeric: from 1 - very bad to 5 - excellent)
25 freetime - free time after school (numeric: from 1 - very low to 5 - very high)
26 goout - going out with friends (numeric: from 1 - very low to 5 - very high)
27 Dalc - workday alcohol consumption (numeric: from 1 - very low to 5 - very high)
28 Walc - weekend alcohol consumption (numeric: from 1 - very low to 5 - very high)
29 health - current health status (numeric: from 1 - very bad to 5 - very good)
30 absences - number of school absences (numeric: from 0 to 93)
# these grades are related with the course subject, Math or Portuguese:
31 G1 - first period grade (numeric: from 0 to 20)
31 G2 - second period grade (numeric: from 0 to 20)
32 G3 - final grade (numeric: from 0 to 20, output target)
Download the data fond in the Data Folder and Data Set links and store it in the same folder where you intend to run the model.
In [344]:
import time
import random
import sys
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
import warnings
import datetime
import matplotlib.pyplot as plt
import pandas as pd
import matplotlib.dates as mdates
import seaborn as sns
warnings.filterwarnings("ignore")
In [345]:
student_por_df = pd.read_csv('student-por.csv', sep=';')
Out[345]:
school sex age address famsize Pstatus Medu Fedu Mjob Fjob ... famrel freetime goout Dalc Walc health absences G1 G2 G3
0 GP F 18 U GT3 A 4 4 at_home teacher ... 4 3 4 1 1 3 4 0 11 11
1 GP F 17 U GT3 T 1 1 at_home other ... 5 3 3 1 1 3 2 9 11 11
2 GP F 15 U LE3 T 1 1 at_home other ... 4 3 2 2 3 3 6 12 13 12
3 GP F 15 U GT3 T 4 2 health services ... 3 2 2 1 1 5 0 14 14 14
4 GP F 16 U GT3 T 3 3 other other ... 4 3 2 1 2 5 0 11 13 13
5 rows × 33 columns
In [346]:
student_por_df['G3'].describe()
Out[346]:
count 649.000000
mean 11.906009
std 3.230656
min 0.000000
25% 10.000000
50% 12.000000
75% 14.000000
max 19.000000
Name: G3, dtype: float64
In [335]:
student_por_df.shape
Out[335]:
(649, 33)
In [347]:
plt.plot(sorted(student_por_df['G3']))
plt.grid()
In [337]:
student_math_df = pd.read_csv('student-mat.csv', sep=';')
Out[337]:
school sex age address famsize Pstatus Medu Fedu Mjob Fjob ... famrel freetime goout Dalc Walc health absences G1 G2 G3
0 GP F 18 U GT3 A 4 4 at_home teacher ... 4 3 4 1 1 3 6 5 6 6
1 GP F 17 U GT3 T 1 1 at_home other ... 5 3 3 1 1 3 4 5 5 6
2 GP F 15 U LE3 T 1 1 at_home other ... 4 3 2 2 3 3 10 7 8 10
3 GP F 15 U GT3 T 4 2 health services ... 3 2 2 1 1 5 2 15 14 15
4 GP F 16 U GT3 T 3 3 other other ... 4 3 2 1 2 5 4 6 10 10
5 rows × 33 columns
In [338]:
student_por_df.describe()
Out[338]:
age Medu Fedu traveltime studytime failures famrel freetime goout Dalc Walc health absences G1 G2 G3
count 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000 649.000000
mean 16.744222 2.514638 2.306626 1.568567 1.930663 0.221880 3.930663 3.180277 3.184900 1.502311 2.280431 3.536210 3.659476 11.399076 11.570108 11.906009
std 1.218138 1.134552 1.099931 0.748660 0.829510 0.593235 0.955717 1.051093 1.175766 0.924834 1.284380 1.446259 4.640759 2.745265 2.913639 3.230656
min 15.000000 0.000000 0.000000 1.000000 1.000000 0.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 0.000000 0.000000 0.000000 0.000000
25% 16.000000 2.000000 1.000000 1.000000 1.000000 0.000000 4.000000 3.000000 2.000000 1.000000 1.000000 2.000000 0.000000 10.000000 10.000000 10.000000
50% 17.000000 2.000000 2.000000 1.000000 2.000000 0.000000 4.000000 3.000000 3.000000 1.000000 2.000000 4.000000 2.000000 11.000000 11.000000 12.000000
75% 18.000000 4.000000 3.000000 2.000000 2.000000 0.000000 5.000000 4.000000 4.000000 2.000000 3.000000 5.000000 6.000000 13.000000 13.000000 14.000000
max 22.000000 4.000000 4.000000 4.000000 4.000000 3.000000 5.000000 5.000000 5.000000 5.000000 5.000000 5.000000 32.000000 19.000000 19.000000 19.000000
In [339]:
student_math_df.shape
Out[339]:
(395, 33)
### Exploration and feature engineering¶
In [318]:
# find all non-numerical data
non_mueric_features = [feat for feat in list(student_por_df) if feat not in list(student_por_df._get_numeric_data())]
for feat in non_mueric_features:
print(feat, ':', set(student_por_df[feat]))
school : {'GP', 'MS'}
sex : {'M', 'F'}
famsize : {'LE3', 'GT3'}
Pstatus : {'T', 'A'}
Mjob : {'other', 'at_home', 'teacher', 'services', 'health'}
Fjob : {'other', 'at_home', 'services', 'teacher', 'health'}
reason : {'course', 'other', 'reputation', 'home'}
guardian : {'other', 'father', 'mother'}
schoolsup : {'yes', 'no'}
famsup : {'yes', 'no'}
paid : {'yes', 'no'}
activities : {'yes', 'no'}
nursery : {'yes', 'no'}
higher : {'yes', 'no'}
internet : {'yes', 'no'}
romantic : {'yes', 'no'}
In [348]:
for feat in non_mueric_features:
dummies = pd.get_dummies(student_por_df[feat]).rename(columns=lambda x: feat + '_' + str(x))
student_por_df = pd.concat([student_por_df, dummies], axis=1)
student_por_df = student_por_df[[feat for feat in list(student_por_df) if feat not in non_mueric_features]]
In [349]:
student_por_df.shape
Out[349]:
(649, 59)
In [351]:
student_por_df.head()
Out[351]:
age Medu Fedu traveltime studytime failures famrel freetime goout Dalc ... activities_no activities_yes nursery_no nursery_yes higher_no higher_yes internet_no internet_yes romantic_no romantic_yes
0 18 4 4 2 2 0 4 3 4 1 ... 1 0 0 1 0 1 1 0 1 0
1 17 1 1 1 2 0 5 3 3 1 ... 1 0 1 0 0 1 0 1 1 0
2 15 1 1 1 2 0 4 3 2 2 ... 1 0 0 1 0 1 0 1 1 0
3 15 4 2 1 3 0 3 2 2 1 ... 0 1 0 1 0 1 0 1 0 1
4 16 3 3 1 2 0 4 3 2 1 ... 1 0 0 1 0 1 1 0 1 0
5 rows × 59 columns
In [352]:
# create an xgboost model
# run simple xgboost classification model and check
# prep modeling code
outcome = 'G3'
features = [feat for feat in list(student_por_df) if feat not in outcome]
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(student_por_df,
student_por_df[outcome],
test_size=0.3,
random_state=42)
import xgboost as xgb
xgb_params = {
'eta': 0.01,
'max_depth': 3,
'subsample': 0.7,
'colsample_bytree': 0.7,
'objective': 'reg:linear',
'seed' : 0
}
dtrain = xgb.DMatrix(X_train[features], y_train, feature_names = features)
dtest = xgb.DMatrix(X_test[features], y_test, feature_names = features)
evals = [(dtrain,'train'),(dtest,'eval')]
xgb_model = xgb.train (params = xgb_params,
dtrain = dtrain,
num_boost_round = 2000,
verbose_eval=50,
early_stopping_rounds = 500,
evals=evals,
#feval = f1_score_cust,
maximize = False)
[0] train-rmse:11.6497 eval-rmse:11.9429
Multiple eval metrics have been passed: 'eval-rmse' will be used for early stopping.
Will train until eval-rmse hasn't improved in 500 rounds.
[50] train-rmse:7.18897 eval-rmse:7.39667
[100] train-rmse:4.50517 eval-rmse:4.63686
[150] train-rmse:2.90578 eval-rmse:2.99388
[200] train-rmse:1.98163 eval-rmse:2.05763
[250] train-rmse:1.47129 eval-rmse:1.56264
[300] train-rmse:1.20491 eval-rmse:1.32764
[350] train-rmse:1.0693 eval-rmse:1.22909
[400] train-rmse:0.995233 eval-rmse:1.18691
[450] train-rmse:0.948895 eval-rmse:1.17263
[500] train-rmse:0.916889 eval-rmse:1.16601
[550] train-rmse:0.886623 eval-rmse:1.16357
[600] train-rmse:0.862894 eval-rmse:1.16218
[650] train-rmse:0.838753 eval-rmse:1.16097
[700] train-rmse:0.819014 eval-rmse:1.16227
[750] train-rmse:0.798209 eval-rmse:1.16462
[800] train-rmse:0.782396 eval-rmse:1.16587
[850] train-rmse:0.765591 eval-rmse:1.165
[900] train-rmse:0.748515 eval-rmse:1.16553
[950] train-rmse:0.732683 eval-rmse:1.16676
[1000] train-rmse:0.717379 eval-rmse:1.16631
[1050] train-rmse:0.702918 eval-rmse:1.16689
[1100] train-rmse:0.687903 eval-rmse:1.16918
Stopping. Best iteration:
[640] train-rmse:0.842125 eval-rmse:1.16081
In [353]:
# find poor performing students and find out why they are so compared to their peers
# plot the important features
fig, ax = plt.subplots(figsize=(6,9))
xgb.plot_importance(xgb_model, height=0.8, ax=ax, max_num_features=20)
plt.show()
In [354]:
# get dataframe version of important feature for model
xgb_fea_imp=pd.DataFrame(list(xgb_model.get_fscore().items()),
columns=['feature','importance']).sort_values('importance', ascending=False)
Out[354]:
feature importance
0 G2 1301
2 G1 669
1 absences 444
21 Dalc 412
13 age 390
10 freetime 279
19 health 262
17 famrel 239
37 traveltime 236
22 goout 236
In [325]:
print(xgb_model.predict(dtest)[0:10])
[16.843275 11.575173 17.108978 11.135402 11.382752 16.290283 17.242054
10.359439 10.89928 10.70853 ]
In [355]:
key_features = list(xgb_fea_imp['feature'].values[0:40])
key_features
Out[355]:
['G2',
'G1',
'absences',
'Dalc',
'age',
'freetime',
'health',
'famrel',
'traveltime',
'goout',
'Medu',
'Fedu',
'Walc',
'studytime',
'failures',
'reason_other',
'Fjob_services',
'schoolsup_no',
'Mjob_other',
'romantic_no',
'famsup_no',
'sex_F',
'Fjob_at_home',
'school_GP',
'reason_reputation',
'reason_course',
'activities_no',
'Mjob_services',
'guardian_father',
'famsize_GT3',
'nursery_no',
'Mjob_teacher',
'reason_home',
'schoolsup_yes',
'higher_no',
'Mjob_at_home',
'romantic_yes',
'internet_no',
'famsup_yes']
In [312]:
# Take students with a predicted final score of less than 10 over 20
predicted_students_in_trouble = X_test[X_test['G3'] < 10]
# See which feature they landed well below or well above peers
for index, row in predicted_students_in_trouble.iterrows():
print('Student ID:', index)
for feat in key_features:
if row[feat] < student_por_df[feat].quantile(0.25):
print('\t', 'Below:', feat, row[feat], 'Class:',
np.round(np.mean(student_por_df[feat]),2))
if row[feat] > student_por_df[feat].quantile(0.75):
print('\t','Above:', feat, row[feat], 'Class:',
np.round(np.mean(student_por_df[feat]),2))
Student ID: 131
Below: G2 9 Class: 11.57
Above: absences 10 Class: 3.66
Above: goout 5 Class: 3.18
Above: failures 3 Class: 0.22
Above: reason_reputation 1 Class: 0.22
Above: Mjob_services 1 Class: 0.21
Above: higher_no 1 Class: 0.11
Student ID: 81
Below: G2 9 Class: 11.57
Below: age 15 Class: 16.74
Above: studytime 3 Class: 1.93
Below: schoolsup_no 0 Class: 0.9
Above: nursery_no 1 Class: 0.2
Above: reason_home 1 Class: 0.23
Above: schoolsup_yes 1 Class: 0.1
Student ID: 585
Below: G2 7 Class: 11.57
Below: G1 8 Class: 11.4
Below: freetime 2 Class: 3.18
Above: studytime 3 Class: 1.93
Above: Fjob_at_home 1 Class: 0.06
Above: higher_no 1 Class: 0.11
Above: internet_no 1 Class: 0.23
Student ID: 177
Below: G2 8 Class: 11.57
Below: G1 9 Class: 11.4
Below: Medu 1 Class: 2.51
Above: Walc 4 Class: 2.28
Above: failures 1 Class: 0.22
Above: guardian_father 1 Class: 0.24
Above: higher_no 1 Class: 0.11
Above: Mjob_at_home 1 Class: 0.21
Student ID: 174
Below: G2 8 Class: 11.57
Below: G1 8 Class: 11.4
Above: absences 8 Class: 3.66
Below: famrel 3 Class: 3.93
Above: failures 1 Class: 0.22
Above: higher_no 1 Class: 0.11
Above: Mjob_at_home 1 Class: 0.21
Student ID: 478
Below: G2 7 Class: 11.57
Below: G1 7 Class: 11.4
Below: health 1 Class: 3.54
Below: famrel 3 Class: 3.93
Below: Medu 1 Class: 2.51
Above: failures 3 Class: 0.22
Below: schoolsup_no 0 Class: 0.9
Above: guardian_father 1 Class: 0.24
Above: schoolsup_yes 1 Class: 0.1
Above: Mjob_at_home 1 Class: 0.21
Above: internet_no 1 Class: 0.23
Student ID: 522
Below: G2 8 Class: 11.57
Below: G1 8 Class: 11.4
Above: Fedu 4 Class: 2.31
Student ID: 163
Below: G2 9 Class: 11.57
Below: famrel 2 Class: 3.93
Above: goout 5 Class: 3.18
Below: Medu 1 Class: 2.51
Above: Walc 5 Class: 2.28
Above: failures 2 Class: 0.22
Above: higher_no 1 Class: 0.11
Student ID: 570
Below: G2 8 Class: 11.57
Below: G1 7 Class: 11.4
Above: Walc 4 Class: 2.28
Above: Mjob_services 1 Class: 0.21
Student ID: 257
Below: G2 8 Class: 11.57
Below: freetime 2 Class: 3.18
Below: goout 1 Class: 3.18
Above: Fedu 4 Class: 2.31
Above: nursery_no 1 Class: 0.2
Above: Mjob_teacher 1 Class: 0.11
Student ID: 148
Below: G2 9 Class: 11.57
Below: G1 8 Class: 11.4
Below: age 15 Class: 16.74
Above: freetime 5 Class: 3.18
Below: health 1 Class: 3.54
Above: traveltime 3 Class: 1.57
Above: goout 5 Class: 3.18
Below: Medu 1 Class: 2.51
Above: failures 1 Class: 0.22
Student ID: 447
Below: G1 8 Class: 11.4
Above: absences 8 Class: 3.66
Above: Dalc 5 Class: 1.5
Above: freetime 5 Class: 3.18
Above: traveltime 3 Class: 1.57
Above: goout 5 Class: 3.18
Above: Walc 5 Class: 2.28
Above: reason_other 1 Class: 0.11
Above: higher_no 1 Class: 0.11
Above: internet_no 1 Class: 0.23
Student ID: 518
Below: G2 5 Class: 11.57
Below: G1 8 Class: 11.4
Above: absences 8 Class: 3.66
Below: health 1 Class: 3.54
Below: famrel 2 Class: 3.93
Above: Fedu 4 Class: 2.31
Above: failures 1 Class: 0.22
Above: reason_reputation 1 Class: 0.22
Above: guardian_father 1 Class: 0.24
Student ID: 603
Below: G2 0 Class: 11.57
Below: G1 5 Class: 11.4
Below: goout 1 Class: 3.18
Above: reason_reputation 1 Class: 0.22
Above: Mjob_teacher 1 Class: 0.11
Student ID: 514
Below: G2 6 Class: 11.57
Below: G1 7 Class: 11.4
Below: freetime 1 Class: 3.18
Below: famrel 3 Class: 3.93
Above: Walc 4 Class: 2.28
Above: Fjob_at_home 1 Class: 0.06
Above: Mjob_services 1 Class: 0.21
Student ID: 568
Below: G1 6 Class: 11.4
Above: age 19 Class: 16.74
Below: freetime 2 Class: 3.18
Below: famrel 3 Class: 3.93
Below: goout 1 Class: 3.18
Above: failures 3 Class: 0.22
Above: Mjob_at_home 1 Class: 0.21
Above: internet_no 1 Class: 0.23
Student ID: 440
Below: G2 0 Class: 11.57
Below: G1 7 Class: 11.4
Above: Dalc 4 Class: 1.5
Above: goout 5 Class: 3.18
Below: Medu 1 Class: 2.51
Above: Walc 5 Class: 2.28
Above: reason_home 1 Class: 0.23
Above: Mjob_at_home 1 Class: 0.21
Above: internet_no 1 Class: 0.23
Student ID: 443
Below: G2 9 Class: 11.57
Below: G1 7 Class: 11.4
Above: absences 7 Class: 3.66
Below: age 15 Class: 16.74
Above: reason_reputation 1 Class: 0.22
Above: guardian_father 1 Class: 0.24
Student ID: 155
Below: G2 7 Class: 11.57
Below: G1 9 Class: 11.4
Above: absences 22 Class: 3.66
Above: goout 5 Class: 3.18
Above: reason_home 1 Class: 0.23
Student ID: 248
Below: G2 9 Class: 11.57
Below: G1 9 Class: 11.4
Below: famrel 3 Class: 3.93
Below: Medu 1 Class: 2.51
Above: reason_home 1 Class: 0.23
Student ID: 494
Below: G2 9 Class: 11.57
Below: G1 8 Class: 11.4
Above: goout 5 Class: 3.18
Below: Medu 1 Class: 2.51
Above: higher_no 1 Class: 0.11
Above: Mjob_at_home 1 Class: 0.21
Student ID: 563
Below: G2 0 Class: 11.57
Below: G1 7 Class: 11.4
Below: freetime 2 Class: 3.18
Below: famrel 1 Class: 3.93
Below: goout 1 Class: 3.18
Above: failures 1 Class: 0.22
Above: internet_no 1 Class: 0.23
Student ID: 432
Below: G2 6 Class: 11.57
Below: G1 6 Class: 11.4
Below: Medu 1 Class: 2.51
Above: failures 1 Class: 0.22
Above: reason_other 1 Class: 0.11
Above: guardian_father 1 Class: 0.24
Above: nursery_no 1 Class: 0.2
Above: higher_no 1 Class: 0.11
Student ID: 583
Below: G2 6 Class: 11.57
Below: G1 8 Class: 11.4
Above: freetime 5 Class: 3.18
Above: goout 5 Class: 3.18
Above: failures 1 Class: 0.22
Above: reason_other 1 Class: 0.11
Above: higher_no 1 Class: 0.11
Student ID: 370
Below: G2 8 Class: 11.57
Below: G1 8 Class: 11.4
Above: age 19 Class: 16.74
Above: traveltime 3 Class: 1.57
Below: Medu 1 Class: 2.51
Above: failures 2 Class: 0.22
Above: nursery_no 1 Class: 0.2
Student ID: 256
Below: G2 8 Class: 11.57
Below: G1 7 Class: 11.4
Above: absences 26 Class: 3.66
Below: health 1 Class: 3.54
Above: failures 1 Class: 0.22
Above: Fjob_at_home 1 Class: 0.06
Above: nursery_no 1 Class: 0.2
Above: higher_no 1 Class: 0.11
### Show Notes
(pardon typos and formatting -
these are the notes I use to make the videos) | 7,350 | 19,478 | {"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} | 2.78125 | 3 | CC-MAIN-2022-33 | latest | en | 0.829579 |
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