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;
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
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.
|
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 ;
Noravshan to’plamning normal to’plam bo’lishlik shartini ko’rsating?
а) hgt( A) Sup A (u) 1; b)
uU
hgt( A) Sup A (u) 1;
uU
с) hgt( A) Sup A (u) 1; e)
uU
hgt( A) Inf
uU
A (u) 1.
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 )}}.
M fazo sifatida [0; 1] intervalda M=[0.4, 0.8] qiymatlarni olsak, u holda ܷ =
{ݔଵ, ݔଶ} noravshan to’plamlarning P(U) to’plamini hosil qiling?
а) ܲ(ܷ) = {(ݔଵ|0.4), (ݔଶ|0.4)}, {(ݔଵ|0.4), (ݔଶ|0.8)}, {(ݔଵ|0.8), (ݔଶ|0.4)},
{(ݔଵ|0.8), (ݔଶ|0.8)}};
b) ܲ(ܷ) = {(ݔଵ|0.4), (ݔଶ|0.4)}, {(ݔଵ|0.3), (ݔଶ|0.6)}, {(ݔଵ|0.8), (ݔଶ|0.4)},
{(ݔଵ|0.8), (ݔଶ|0.8)}};
с) ܲ(ܷ) = {(ݔଵ|0.5), (ݔଶ|0.5)}, {(ݔଵ|0.4), (ݔଶ|0.8)}, {(ݔଵ|0.8), (ݔଶ|0.4)},
{(ݔଵ|0.6), (ݔଶ|0.6)}}.
e) ܲ(ܷ) = {(ݔଵ|0.4), (ݔଶ|0.4)}, {(ݔଵ|0.7), (ݔଶ|0.7)}, {(ݔଵ|0.8), (ݔଶ|0.4)},
{(ݔଵ|0.8), (ݔଶ|0.8)}}.
26. Bizga ܷ = ൛ݑଵ, ݑଶ,ݑଷ, ݑସൟ, ܯ = [0,1],
ܣ = {(ݔଵ|0.3), (ݔଶ|0.2), (ݔଷ|0), (ݔସ|1)},
ܤ = {(ݔଵ|0.7), (ݔଶ|0.4), (ݔଷ|1), (ݔସ|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 ܷ = {ݔଵ, ݔଶ, ݔଷ, ݔସ, ݔହ, ݔ}, ܯ = [0,1],
ܣ = {(ݔଵ|0.35), (ݔଶ|0.68), (ݔଷ|1), (ݔସ|0), (ݔହ|0.15)}.
berilgan bo’lsin.
To’dirish amalidan foydalanib, ∼A=B ni toping?
а) ܤ = {(ݔଵ|0.65), (ݔଶ|0.32), (ݔଷ|0), (ݔସ|1), (ݔହ|0.85)}.
b) ܤ = {(ݔଵ|0.65), (ݔଶ|0.032), (ݔଷ|1), (ݔସ|1), (ݔହ|0.85)}.
с) ܤ = {(ݔଵ|0.065), (ݔଶ|0.320), (ݔଷ|0.99), (ݔସ|1), (ݔହ|0.85)}.
e) ܤ = {(ݔଵ|0.650), (ݔଶ|0.32), (ݔଷ|0), (ݔସ|1), (ݔହ|0.85)}.
28. Bizga ܷ = {ݔଵ, ݔଶ, ݔଷ, ݔସ, ݔହ}, 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 ܷ = {ݔଵ, ݔଶ, ݔଷ, ݔସ, ݔହ}, 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
A1 {< 1/12 >, < 0.8/18 >, < 0.5/28 >} va A2 {< 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 ܷ = {ݔଵ, ݔଶ, ݔଷ, ݔସ}, ܯ = [0,1],
ܣ = {(ݔଵ|0.35), (ݔଶ|0.65), (ݔଷ|1), (ݔସ|0)},
ܤ = {(ݔଵ|0.63), (ݔଶ|0.55), (ݔଷ|0), (ݔସ|1)}.
berilgan bo’lsin.
Ayirma amalidan foydalanib
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