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Olasılık sorusu
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y.nurdan
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Apr 1, 2009, 2:16:02 PM
4/1/09
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to TMOZ - Öğretmen & Öğrenci
N={x: 20/x-3=y x ve y tamsayı} kümesinin elemanlarından biri
rastgele
seçiliyor.Seçilen bu elemanın doğal sayı olma olasılığı kaçtır?
Şahin Danişman
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Apr 1, 2009, 4:29:43 PM
4/1/09
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y nin tamsayı olması için, x-3 ün 20 nin tam boleni olması gerekir.
20=2^2.5
3.2=6 tane pozitif bolen, 12 tane tam bolen olur.
pozitif bolenlerin hepsinde x doğal sayıdır. -1 -2 -3 için x yine doğal sayı olur.
istenen koşulda 9 eleman var.
9/12=3/4
01 Nisan 2009 Çarşamba 21:16 tarihinde y.nurdan
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OSMAN TIKNAZOĞLU
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Apr 1, 2009, 4:38:25 PM
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şshin hocam güzel ama 8/12 olmazmı çünkü payda sadece -1 ve -2 iken x pozitif olur
01 Nisan 2009 Çarşamba 21:16 tarihinde y.nurdan
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Şahin Danişman
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Apr 1, 2009, 4:51:07 PM
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doğal sayı deniyor osman hocam.
01 Nisan 2009 Çarşamba 23:38 tarihinde OSMAN TIKNAZOĞLU
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