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Z.Sobirov - Uch karrali integral (amaliy mashg'ulot 1 - qism)
#uch_karrali #karrali
#integral
@universal_library
#uch_karrali #karrali
#integral
@universal_library
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🔥Diqqat!
E'lonimiz ba'zilarga tegishli, hammaga ham emas, oldindan uzrli:
Azizlar, ongingiz ishlamasligini hammaga isbotlashingiz shart emas - Makro avval ishchilarini ta'minlab olsin, vaqtida oyligini bersin. Nahotki shunchalik ong ishlamaydi?!
Nima emish, Makro sovg'a tarqatmoqda ekan... .
// Post tugadi😄👍
tushungan tushundi.
@Universal_Library - haqiqatga yaqin yuraylik🕊
E'lonimiz ba'zilarga tegishli, hammaga ham emas, oldindan uzrli:
Azizlar, ongingiz ishlamasligini hammaga isbotlashingiz shart emas - Makro avval ishchilarini ta'minlab olsin, vaqtida oyligini bersin. Nahotki shunchalik ong ishlamaydi?!
Nima emish, Makro sovg'a tarqatmoqda ekan... .
// Post tugadi😄👍
tushungan tushundi.
@Universal_Library - haqiqatga yaqin yuraylik🕊
👍22😁9🔥2
Ehtimollar Nazariyasi
Uchta merganning o'qni nishonga tekkizish ehtimollari mos ravishda 0.4, 0.7 va 0.8 ga teng. Demak, P(a)=0.4, P(b)=0.7 va P(c)=0.8.
Bu merganlar baravariga o'q uzganda:
1. Hech birining nishonga tekkiza olmaslik ehtimolini toping.
P(d)=[1-P(a)]•[1-P(b)]•[1-P(c)]=0.6•0.3•0.2=0.036.
2. Uchalasining ham o'qni nishonga tekkizish ehtimolini toping.
P(f)=P(a)•P(b)•P(c)=0.4•0.7•0.8=0.224
3. Birinchi mergan nishonga tekkizib, ikkinchi va uchinchi merganlar nishonga tekkiza olmaslik ehtimolini toping.
P(g)=P(a)•[1-P(b)]•[1-P(c)]=0.4•0.3•0.2=0.024.
4. Ikkinchi merganniki tegib birinchi va uchinchi merganniki tegmaslik ehtimoli:
P(h)=[1-P(a)]•P(b)•[1-P(c)].
5. Uchinchi merganniki tegib, birinchi va ikkinchi merganniki tegmaslik ehtimoli:
P(j)=[1-P(a)]•[1-P(b)]•P(c).
6. Birinchi va ikkinchi merganniki tegib, uchinchi merganniki tegmaslik ehtimoli:
P(k)=P(a)•P(b)•[1-P(c)].
7. Birinchi va uchinchi merganniki tegib, ikkinchi merganniki tegmaslik ehtimoli:
P(l)=P(a)•[1-P(b)]•P(c).
8. Ikkinchi va uchinchi merganniki tegib, birinchi merganniki tegmaslik ehtimoli:
P(z)=[1-P(a)]•P(b)•P(c).
9. Aniq (faqat) bittasiniki nishonga tegish ehtimoli:
P(x)=P(g)+P(h)+P(j).
10. Hech bo'lmaganda (kamida) bittasining nishonga tekkizish ehtimoli:
P(v)=1-P(d).
11. Aniq (faqat) 2 tasining nishonga tekkizish ehtimoli:
P(n)=P(k)+P(l)+P(z).
12. Hech bo'lmaganda (kamida) ikkitasining nishonga tekkizish ehtimoli:
P(m)=P(n)+P(f).
@Universal_Library - siz izlagan manzil🕊
Uchta merganning o'qni nishonga tekkizish ehtimollari mos ravishda 0.4, 0.7 va 0.8 ga teng. Demak, P(a)=0.4, P(b)=0.7 va P(c)=0.8.
Bu merganlar baravariga o'q uzganda:
1. Hech birining nishonga tekkiza olmaslik ehtimolini toping.
P(d)=[1-P(a)]•[1-P(b)]•[1-P(c)]=0.6•0.3•0.2=0.036.
2. Uchalasining ham o'qni nishonga tekkizish ehtimolini toping.
P(f)=P(a)•P(b)•P(c)=0.4•0.7•0.8=0.224
3. Birinchi mergan nishonga tekkizib, ikkinchi va uchinchi merganlar nishonga tekkiza olmaslik ehtimolini toping.
P(g)=P(a)•[1-P(b)]•[1-P(c)]=0.4•0.3•0.2=0.024.
4. Ikkinchi merganniki tegib birinchi va uchinchi merganniki tegmaslik ehtimoli:
P(h)=[1-P(a)]•P(b)•[1-P(c)].
5. Uchinchi merganniki tegib, birinchi va ikkinchi merganniki tegmaslik ehtimoli:
P(j)=[1-P(a)]•[1-P(b)]•P(c).
6. Birinchi va ikkinchi merganniki tegib, uchinchi merganniki tegmaslik ehtimoli:
P(k)=P(a)•P(b)•[1-P(c)].
7. Birinchi va uchinchi merganniki tegib, ikkinchi merganniki tegmaslik ehtimoli:
P(l)=P(a)•[1-P(b)]•P(c).
8. Ikkinchi va uchinchi merganniki tegib, birinchi merganniki tegmaslik ehtimoli:
P(z)=[1-P(a)]•P(b)•P(c).
9. Aniq (faqat) bittasiniki nishonga tegish ehtimoli:
P(x)=P(g)+P(h)+P(j).
10. Hech bo'lmaganda (kamida) bittasining nishonga tekkizish ehtimoli:
P(v)=1-P(d).
11. Aniq (faqat) 2 tasining nishonga tekkizish ehtimoli:
P(n)=P(k)+P(l)+P(z).
12. Hech bo'lmaganda (kamida) ikkitasining nishonga tekkizish ehtimoli:
P(m)=P(n)+P(f).
@Universal_Library - siz izlagan manzil🕊
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Biror A Sonning barcha qismlari.
✅ Avval A sonni tub ko'paytuvchilarga ajratib olamiz: A=a^m•b^n•c^k....
❗️ A sonning tub ko'paytuvchilarga ajratilish yoyilmasi A sonning kanonik yoyilmasi deyiladi.
❕a - juft, b, c - toq shunday bo'lishi mumkin degan maqasadda shunday deb yozdim.
1. NBS(A) - natural bo'luvchilar soni(umumiy bo'luvchilar soni):
NBS(A)=(m+1)×(n+1)×(k+1)....
2. TBS - tub bo'luvchilar soni:
TBS(A)= A sonning kanonik yoyilmasidagi tub sonlar soni.
3. TBBS - tub bo'lmagan bo'luvchilar son:
TBBS(A)=NBS(A)-TBS(A)
4. MBS - murakkab bo'luvchilar soni:
MBS(A)=NBS(A)-TBS(A)-1=TBBS(A)-1
5. BBS - butun bo'luvchilar soni:
BBS(A)=2NBS(A)
6. TBS - toq bo'luvchilar soni. men boshida b, c - toq deb o'tdim, toq bo'luvchilar sonini topish uchun b va c darajalariga bir qo'shib ikkalasini ko'paytirib qo'ysak tub bo'luvchilar soni kelib chiqadi:
TBS(A)=(n+1)×(k+1)
7. JBS - juft bo'luvchilar soni:
JBS(A)=NBS(A)-TBS(A)
8. NBY - natural no'luvchilar yig'indisi:
NBY(A)=(a^(m+1)-1)/(a-1)×(b^(n+1)-1)/(b-1)×(c^(k+1)-1)/(c-1)
9. BBY - butun bo'luvchilar yig'indisi:
BBY(A)=0
10. TBY - tub bo'luvchilari yig'indisi:
TBY(A)=a+b+c
11. TBBY - tub bo'lmagan bo'luvchilar yig'indisi:
TBBY(A)=NBY(A)-TBY(A)
12. MBY - murakkab bo'luvchilar yig'indisi:
MBY(A)=NBY(A)-TBY(A)-1=TBBY(A)-1
13. TBY - toq bo'luvchilar yig'indisi
b, c - toq edi. Kanonik yoyilmadagi toq sonlarni olamiz:
TBY(A)=(b^(n+1)-1)/(b-1)×(c^(k+1)-1)/(c-1)
14. JBY - juft bo'luvchilar yig'indisi:
JBY(A)=NBY(A)-TBY(A)
15. NBK - natural bo'luvchilar ko'paytmasi:
NBK(A)=A^(NBS(A)/2)
16. BBK - butun bo'luvchilar ko'paytmasi:
BBK(A)=A^NBS(A)
17. TBK - tub bo'luvchilar ko'paytmasi:
TBK(A)=a×b×c
18. TBBK - tub bo'lmagan bo'luvchilar ko'paytmasi:
TBBK(A)=NBK(A)/TBK(A)
19. MBK - murakkab bo'luvchilar ko'paytmasi:
MBK(A)=NBK(A)/TBK(A)=TBBK(A)
20. TBK - toq bo'luvchilar ko'paytmasi:
TBK(A)=(b^n×c^k)^(TBS(A)/2)
21. JBK - juft bo'luvchilar ko'paytmasi:
JBK(A)=NBK(A)/TBK(A)
@Universal_Library - siz izlagan manzil🕊
✅ Avval A sonni tub ko'paytuvchilarga ajratib olamiz: A=a^m•b^n•c^k....
❗️ A sonning tub ko'paytuvchilarga ajratilish yoyilmasi A sonning kanonik yoyilmasi deyiladi.
❕a - juft, b, c - toq shunday bo'lishi mumkin degan maqasadda shunday deb yozdim.
1. NBS(A) - natural bo'luvchilar soni(umumiy bo'luvchilar soni):
NBS(A)=(m+1)×(n+1)×(k+1)....
2. TBS - tub bo'luvchilar soni:
TBS(A)= A sonning kanonik yoyilmasidagi tub sonlar soni.
3. TBBS - tub bo'lmagan bo'luvchilar son:
TBBS(A)=NBS(A)-TBS(A)
4. MBS - murakkab bo'luvchilar soni:
MBS(A)=NBS(A)-TBS(A)-1=TBBS(A)-1
5. BBS - butun bo'luvchilar soni:
BBS(A)=2NBS(A)
6. TBS - toq bo'luvchilar soni. men boshida b, c - toq deb o'tdim, toq bo'luvchilar sonini topish uchun b va c darajalariga bir qo'shib ikkalasini ko'paytirib qo'ysak tub bo'luvchilar soni kelib chiqadi:
TBS(A)=(n+1)×(k+1)
7. JBS - juft bo'luvchilar soni:
JBS(A)=NBS(A)-TBS(A)
8. NBY - natural no'luvchilar yig'indisi:
NBY(A)=(a^(m+1)-1)/(a-1)×(b^(n+1)-1)/(b-1)×(c^(k+1)-1)/(c-1)
9. BBY - butun bo'luvchilar yig'indisi:
BBY(A)=0
10. TBY - tub bo'luvchilari yig'indisi:
TBY(A)=a+b+c
11. TBBY - tub bo'lmagan bo'luvchilar yig'indisi:
TBBY(A)=NBY(A)-TBY(A)
12. MBY - murakkab bo'luvchilar yig'indisi:
MBY(A)=NBY(A)-TBY(A)-1=TBBY(A)-1
13. TBY - toq bo'luvchilar yig'indisi
b, c - toq edi. Kanonik yoyilmadagi toq sonlarni olamiz:
TBY(A)=(b^(n+1)-1)/(b-1)×(c^(k+1)-1)/(c-1)
14. JBY - juft bo'luvchilar yig'indisi:
JBY(A)=NBY(A)-TBY(A)
15. NBK - natural bo'luvchilar ko'paytmasi:
NBK(A)=A^(NBS(A)/2)
16. BBK - butun bo'luvchilar ko'paytmasi:
BBK(A)=A^NBS(A)
17. TBK - tub bo'luvchilar ko'paytmasi:
TBK(A)=a×b×c
18. TBBK - tub bo'lmagan bo'luvchilar ko'paytmasi:
TBBK(A)=NBK(A)/TBK(A)
19. MBK - murakkab bo'luvchilar ko'paytmasi:
MBK(A)=NBK(A)/TBK(A)=TBBK(A)
20. TBK - toq bo'luvchilar ko'paytmasi:
TBK(A)=(b^n×c^k)^(TBS(A)/2)
21. JBK - juft bo'luvchilar ko'paytmasi:
JBK(A)=NBK(A)/TBK(A)
@Universal_Library - siz izlagan manzil🕊
👍9🔥2❤1😱1🎉1
: :
Yangi nazariyani o'zlashtirishda hammaning o'z prinsipi bor. Masalan, shaxsan men matematikani video darslardan o'rgana olish qobiliyatimni past darajada deb bilaman.
Aynan, 😉menga o'xshaganlar uchun yaxshi bir dars topdim:
1, 2 va 3 o'zgaruvchili limitlar bo'yicha bugun 20'00 da bizni kuting.
Albatta, ingliz tilida👍🔥
: :
@Universal_Library - siz izlagan manzil🕊
Yangi nazariyani o'zlashtirishda hammaning o'z prinsipi bor. Masalan, shaxsan men matematikani video darslardan o'rgana olish qobiliyatimni past darajada deb bilaman.
Aynan, 😉menga o'xshaganlar uchun yaxshi bir dars topdim:
1, 2 va 3 o'zgaruvchili limitlar bo'yicha bugun 20'00 da bizni kuting.
Albatta, ingliz tilida👍🔥
: :
@Universal_Library - siz izlagan manzil🕊
👍6
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Limits of Multivariable Functions.
YouTube channel
Link for this lesson
: :
Join to our channel:
@Universal_Library🕊
Limits of Multivariable Functions.
YouTube channel
Link for this lesson
: :
Join to our channel:
@Universal_Library🕊
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Forwarded from Universal Library || New Moon (𝑺𝒉𝒂𝒉𝒛𝒐𝒅𝒃𝒆𝒌 𝑮'𝒖𝒍𝒐𝒎𝒋𝒐𝒏𝒐𝒗)
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1-tartibli bir jinsli va chiziqli differensial tenglama
#bir_jinsli_va_chiziqli_dif_tenglamalar
#bir_jinsli_tenglamalar
#chiziqli_differensial_tenglamlar
#differensial
#dif
@universal_library
#bir_jinsli_va_chiziqli_dif_tenglamalar
#bir_jinsli_tenglamalar
#chiziqli_differensial_tenglamlar
#differensial
#dif
@universal_library
Forwarded from Universal Library || New Moon (Shahzodbek G'ulomjonov)
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dif.ur.dan ma'ruzalar 15 ta.docx
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Oddiy differensial tenglamalarni taqribiy yechishning Eyler usuli
#differensial_tenglamalar
#differensial
#Eyler_usuli
#Eyler
#diff
#sòragan_edingiz
@universal_library
#differensial_tenglamalar
#differensial
#Eyler_usuli
#Eyler
#diff
#sòragan_edingiz
@universal_library
Forwarded from Universal Library || New Moon (Shahzodbek G'ulomjonov)
@universal_library dif2.pdf
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#darsxonam
Chiziqli, Bernulli, To'liq differensialli birinchi tartibli differensial tenglamalar.
#dif #chiziqli #bernulli #differensial
Kanalimiz:
@universal_library
Jamlovchi:
@TUIT_Shahzodbek
Chiziqli, Bernulli, To'liq differensialli birinchi tartibli differensial tenglamalar.
#dif #chiziqli #bernulli #differensial
Kanalimiz:
@universal_library
Jamlovchi:
@TUIT_Shahzodbek
😢1