O’zgaruvchilar

u U
elementi;

b) ^A noravshan to’plam; ^U to’plam;
с) ^U ravshan to’plam; ^{u U} elementi;
e) ^A ravshan to’plam; ^{u U} elementi;

21. 
    Mansublik funksiyasini qurishning bevosita usuli ekspertlar tomonidan berilgan qanday baholashga asoslanadi?
    

^а)  _À
(U )  ^{n1 ; b)}
m
 ( A)  ^{n1 ; с)}
U _m
 (U )  ^{2n1 ; e)}
А _m
 ( A)  ^{n1  m .}
U _m

22. 
    A noravshan to’plamda berilgan
    

u U
elementning mansubligi to`g`risida

quyilgan savolga ekspertlardan
n₁ tasi to`g`ri va
ⁿ₂ tasi salbiy javob bergan.

Ekspertlarning umumiy soni jadvalda berilgan.
m  n₁  n₂
bo’lsin. Exspertlar javobi quyidagi

	U
m	1	2	3	4	5
n1	2	1	6	3	4
n2	4	5	0	3	2

Jadvaldagi ma’lumotlarga va
 _À (U )  ¹

n
m

formulaga asoslanib

A
 (1)  ?;
 (2)  ?;
 (3)  ?;
 (4)  ?;
 (5)  ?
hosil qiling va A ning

A

A

A

A
mansublik funksiyasini quring?

А
^а)  (U)   0.3/1, 0.2 / 2 ,1/ 3 , 0.5/ 4 , 0.7 / 5 ^;

^b) 
^с) 
(U)   0.3/1 ,  0.3/ 2 ,  1/ 3 ,  0.4 / 4 ,  0.7 / 5 ^;

А
(U)   0 /1 ,  0 / 2 ,  2 / 3 ,  1/ 4 ,  0.2 / 5 ^;

А

А
^e) 
(U)   0.2 /1 ,  0.3/ 2 ,  0 / 3 ,  0.5 / 4 ,  1/ 5 ^;

23. 
    Noravshan to’plamning normal to’plam bo’lishlik shartini ko’rsating?
    

_а) ^{hgt( A) } Sup ^ _A ^{(u)  1}_{; b)}
uU
^{hgt( A) } Sup ^{ A (u)  1}_;
uU

_с) ^{hgt( A) } Sup ^ _A ^{(u)  1}_{; e)}
uU
^{hgt( A) } Inf
uU
_A (u)  1.

24. 
    M fazo sifatida [0; 1] intervalda M=[0.4, 0.8] qiymatlarni olsak, u holda ܷ =
    

{ݔ_ଵ, ݔ_ଶ} noravshan to’plamlarning P(U) to’plamini hosil qiling?
а) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.4^)}, ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.8^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.8^)}};
b) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.4^)}, ^{(ݔ_ଵ^|0.3⁾, ⁽ݔ_ଶ^|0.6^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.8^)}};
с) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.5⁾, ⁽ݔ_ଶ^|0.5^)}, ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.8^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.6⁾, ⁽ݔ_ଶ^|0.6^)}}.
e) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.4^)}, ^{(ݔ_ଵ^|0.7⁾, ⁽ݔ_ଶ^|0.7^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.8^)}}.

25. 
    M fazo sifatida [0; 1] intervalda M=[0.4, 0.8] qiymatlarni olsak, u holda ܷ =
    

{ݔ_ଵ, ݔ_ଶ} noravshan to’plamlarning P(U) to’plamini hosil qiling?
а) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.4^)}, ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.8^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.8^)}};
b) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.4^)}, ^{(ݔ_ଵ^|0.3⁾, ⁽ݔ_ଶ^|0.6^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.8^)}};
с) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.5⁾, ⁽ݔ_ଶ^|0.5^)}, ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.8^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.6⁾, ⁽ݔ_ଶ^|0.6^)}}.
e) ܲ⁽ܷ⁾ = ^{(ݔ_ଵ^|0.4⁾, ⁽ݔ_ଶ^|0.4^)}, ^{(ݔ_ଵ^|0.7⁾, ⁽ݔ_ଶ^|0.7^)}, ^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.4^)},
^{(ݔ_ଵ^|0.8⁾, ⁽ݔ_ଶ^|0.8^)}}.
26. Bizga ܷ = ൛ݑ_ଵ, ݑ_ଶ,ݑ_ଷ, ݑ_ସൟ, ܯ = ^[0,1^],
ܣ = ^{(ݔ_ଵ^|0.3⁾, ⁽ݔ_ଶ^|0.2⁾, ⁽ݔ_ଷ^|0⁾, ⁽ݔ_ସ^|1^)},
ܤ = ^{(ݔ_ଵ^|0.7⁾, ⁽ݔ_ଶ^|0.4⁾, ⁽ݔ_ଷ^|1⁾, ⁽ݔ_ସ^|1^)}.
berilgan bo’lsin.
Yutilish qoidasidan foydalanib to’g’ri javobni ko’rsating?
а) A  B; b) B  A; с) B  A; e) ^{B  A.}
27. Bizga ܷ = ^{ݔ_ଵ, ݔ_ଶ, ݔ_ଷ, ݔ_ସ, ݔ_ହ, ݔ_଺^}, ܯ = ^[0,1^],
ܣ = ^{(ݔ_ଵ^|0.35⁾, ⁽ݔ_ଶ^|0.68⁾, ⁽ݔ_ଷ^|1⁾, ⁽ݔ_ସ^|0⁾, ⁽ݔ_ହ^|0.15^)}.
berilgan bo’lsin.
To’dirish amalidan foydalanib, ∼A=B ni toping?
а) ܤ = ^{(ݔ_ଵ^|0.65⁾, ⁽ݔ_ଶ^|0.32⁾, ⁽ݔ_ଷ^|0⁾, ⁽ݔ_ସ^|1⁾, ⁽ݔ_ହ^|0.85^)}.
b) ܤ = ^{(ݔ_ଵ^|0.65⁾, ⁽ݔ_ଶ^|0.032⁾, ⁽ݔ_ଷ^|1⁾, ⁽ݔ_ସ^|1⁾, ⁽ݔ_ହ^|0.85^)}.
с) ܤ = ^{(ݔ_ଵ^|0.065⁾, ⁽ݔ_ଶ^|0.320⁾, ⁽ݔ_ଷ^|0.99⁾, ⁽ݔ_ସ^|1⁾, ⁽ݔ_ହ^|0.85^)}.
e) ܤ = ^{(ݔ_ଵ^|0.650⁾, ⁽ݔ_ଶ^|0.32⁾, ⁽ݔ_ଷ^|0⁾, ⁽ݔ_ସ^|1⁾, ⁽ݔ_ହ^|0.85^)}.
28. Bizga ܷ = ^{ݔ_ଵ, ݔ_ଶ, ݔ_ଷ, ݔ_ସ, ݔ_ହ^}, ܯ = ^[0,1^],

1
A  (x | 0.5), (x | 0.9), (x | 0), (x |1), (x | 0),
2 3 4 5
B  (x | 0.7), (x | 0.6), (x |1), (x | 1), (x | 0).

1 2
berilgan bo’lsin.
Kesishish amalidan foydalanib,
3 4 5


A  B ni toping?

1
^а) A  B  (x
| 0.5), (x₂ | 0.6),(x₃
| 0),(x₄ |1),(x₅
| 0);

1
^b) A  B  (x
| 0.12), (x₂ | 0.15),(x₃ |1),(x₄
| 2),(x₅ | 0);

1
^с) A  B  (x
| 0.5), (x₂ | 0.9),(x₃ | 0),(x₄
|1),(x₅ | 0);

1
^e) A  B  (x
| 0.35), (x₂ | 0.54),(x₃
| 0),(x₄ |1),(x₅ | 0).

29. Bizga ܷ = ^{ݔ_ଵ, ݔ_ଶ, ݔ_ଷ, ݔ_ସ, ݔ_ହ^}, ܯ = ^[0,1^],

1
A  (x | 0.5), (x | 0.9), (x | 0), (x
2 3 4
B  (x | 0.7), (x | 0.6), (x |1), (x
|1), (x₅
| 1), (x
| 0),
| 0).

1 2
berilgan bo’lsin.
Birlashtirirish amalidan foydalanib,
3 4 5


A  B ni toping?

1
^а) A  B  (x
| 0.7), (x₂ | 0.9),(x₃
|1),(x₄ |1),(x₅
| 0);

1
^b) A  B  (x
| 0.12), (x₂ | 0.15),(x₃ |1),(x₄
| 2),(x₅ | 0);

1
^с) A  B  (x
| 0.5), (x₂ | 0.9),(x₃ | 0),(x₄
|1),(x₅ | 0);

1
^e) A  B  (x
| 0.35), (x₂ | 0.54),(x₃
| 0),(x₄ |1),(x₅ | 0).

30.
Х = {12, 18, 28} va Y = {3 ,7}
ko`rinishdagi asosiy to`plamlar berilgan. Ushbu

to`plamlarda noravshan top’lamostilari
A<sub>1</sub>  {< 1/12 >, < 0.8/18 >, < 0.5/28 >} <sup>va</sup> A<sub>2</sub> {< 0.6/3>, < 0.4/7 >}
belgilangan. Ular ustida dekart ko`paytmasi amali bajarish natijasida
toping?
A₁  A₂ ⁿⁱ

_а) A₁  A₂ = {< 0.6/(12,3) >, < 0.6/(18,3)
>, < 0.5/(28,3) >

, < 0.4/(12,7) >, < 0.4/(18,7)
>, < 0.4/(28,7) >};

_b) A₁  A₂ = {< 1/(12,3) >, < 0.8/(18,3)
>, < 0.5/(28,3) >

, < 0.4/(12,7) >, < 0.4/(18,7)
>, < 0.4/(28,7) >};

_с) A₁  A₂ = {< 1.6/(12,3) >, < 1.4/(18,3)
>, < 0.9/(28,3) >

, < 0.4/(12,7) >, < 0.4/(18,7)
>, < 0.4/(28,7) >};

_e) A₁  A₂ = {< 0.6/(12,3) >, < 0.6/(18,3)
>, < 0.5/(28,3) >

, < 1.4/(12,7) >, < 1.2/(18,7)
>, < 0.9/(28,7) >};

31. Bizga ܷ = ^{ݔ_ଵ, ݔ_ଶ, ݔ_ଷ, ݔ_ସ^}, ܯ = ^[0,1^],
ܣ = ^{(ݔ_ଵ^|0.35⁾, ⁽ݔ_ଶ^|0.65⁾, ⁽ݔ_ଷ^|1⁾, ⁽ݔ_ସ^|0^)},
ܤ = ^{(ݔ_ଵ^|0.63⁾, ⁽ݔ_ଶ^|0.55⁾, ⁽ݔ_ଷ^|0⁾, ⁽ݔ_ସ^|1^)}.
berilgan bo’lsin.

Ayirma amalidan foydalanib