{"id":2550,"date":"2023-02-17T20:00:00","date_gmt":"2023-02-17T19:00:00","guid":{"rendered":"http:\/\/matma.com.pl\/?p=2550"},"modified":"2023-02-19T15:19:47","modified_gmt":"2023-02-19T14:19:47","slug":"zadanie-64-0-1-zbior-podstawa","status":"publish","type":"post","link":"http:\/\/matma.com.pl\/?p=2550","title":{"rendered":"zadanie 64 (0-1) zbi\u00f3r podstawa"},"content":{"rendered":"<p>Zr\u00f3bmy rysunek pomocniczy<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 42.7273%;\"><img fetchpriority=\"high\" decoding=\"async\" class=\"wp-image-2551 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/02\/Zrzut-ekranu-2023-02-17-195232-300x265.jpg\" alt=\"\" width=\"368\" height=\"325\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/02\/Zrzut-ekranu-2023-02-17-195232-300x265.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/02\/Zrzut-ekranu-2023-02-17-195232.jpg 363w\" sizes=\"(max-width: 368px) 100vw, 368px\" \/><\/td>\n<td style=\"width: 57.2727%;\">Widzimy, \u017ce z ka\u017cdego wierzcho\u0142ka mo\u017cemy poprowadzi\u0107 odcinek \u0142\u0105cz\u0105cy dwa wierzcho\u0142ki, kt\u00f3ry b\u0119dzie przek\u0105tn\u0105 lub bokiem naszego pi\u0119ciok\u0105ta. Zatem wszystkich mo\u017cliwo\u015bci mamy<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;\\Omega&amp;space;\\right&amp;space;|=5\\cdot&amp;space;4=20\" alt=\"\\left | \\Omega \\right |=5\\cdot 4=20\" align=\"absmiddle\" \/><\/p>\n<p>A- zdarzenie polegaj\u0105ce na tym, \u017ce nasz odcinek b\u0119dzie przek\u0105tn\u0105 pi\u0119ciok\u0105ta<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;|&amp;space;A&amp;space;\\right&amp;space;|=5\\cdot&amp;space;2\" alt=\"\\left | A \\right |=5\\cdot 2\" align=\"absmiddle\" \/><\/p>\n<p>bo z ka\u017cdego wierzcho\u0142ka wychodz\u0105 dwie przek\u0105tne.<\/p>\n<p>Prawdopodobie\u0144stwo zdarzenia <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?A\" alt=\"A\" align=\"absmiddle\" \/> jest r\u00f3wne:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?P\\left&amp;space;(&amp;space;A&amp;space;\\right&amp;space;)=\\frac{10}{20}=\\frac{1}{2}\" alt=\"P\\left ( A \\right )=\\frac{10}{20}=\\frac{1}{2}\" align=\"absmiddle\" \/><\/p>\n<p>Prawid\u0142owa odpowied\u017a to\u00a0<strong>D.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>Zr\u00f3bmy rysunek pomocniczy Widzimy, \u017ce z ka\u017cdego wierzcho\u0142ka mo\u017cemy poprowadzi\u0107 odcinek \u0142\u0105cz\u0105cy dwa wierzcho\u0142ki, kt\u00f3ry b\u0119dzie przek\u0105tn\u0105 lub bokiem naszego pi\u0119ciok\u0105ta. Zatem wszystkich mo\u017cliwo\u015bci mamy A- zdarzenie polegaj\u0105ce na tym, \u017ce nasz odcinek b\u0119dzie przek\u0105tn\u0105 pi\u0119ciok\u0105ta bo z ka\u017cdego wierzcho\u0142ka wychodz\u0105 dwie przek\u0105tne. Prawdopodobie\u0144stwo zdarzenia jest r\u00f3wne: Prawid\u0142owa odpowied\u017a to\u00a0D.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[44],"tags":[],"class_list":["post-2550","post","type-post","status-publish","format-standard","hentry","category-matura"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.1 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>zadanie 64 (0-1) zbi\u00f3r podstawa - Matma.com.pl<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/matma.com.pl\/?p=2550\" \/>\n<meta property=\"og:locale\" content=\"pl_PL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"zadanie 64 (0-1) zbi\u00f3r podstawa - Matma.com.pl\" \/>\n<meta property=\"og:description\" content=\"Zr\u00f3bmy rysunek pomocniczy Widzimy, \u017ce z ka\u017cdego wierzcho\u0142ka mo\u017cemy poprowadzi\u0107 odcinek \u0142\u0105cz\u0105cy dwa wierzcho\u0142ki, kt\u00f3ry b\u0119dzie przek\u0105tn\u0105 lub bokiem naszego pi\u0119ciok\u0105ta. Zatem wszystkich mo\u017cliwo\u015bci mamy A- zdarzenie polegaj\u0105ce na tym, \u017ce nasz odcinek b\u0119dzie przek\u0105tn\u0105 pi\u0119ciok\u0105ta bo z ka\u017cdego wierzcho\u0142ka wychodz\u0105 dwie przek\u0105tne. 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