- Stahuj zápisky z přednášek a ostatní studijní materiály
- Zapisuj si jen kvalitní vyučující (obsáhlá databáze referencí)
- Nastav si své předměty a buď stále v obraze
- Zapoj se svojí aktivitou do soutěže o ceny
- Založ si svůj profil, aby tě tví spolužáci mohli najít
- Najdi své přátele podle místa kde bydlíš nebo školy kterou studuješ
- Diskutuj ve skupinách o tématech, které tě zajímají
Studijní materiály
Zjednodušená ukázka:
Stáhnout celý tento materiál) =
integraldisplay 4
2,3
1
4 · (x− 1)dx
.= 0,914 (a112a176a101a115a110a165 731
800).
a65a49a49
a80a176a46 a49a46 a86 a108a97a98a111a114a97a116a111a114a110a237a109 a99a118a105a163a101a110a237 a115a112a111a108a117 a112a114a97a99a117a106a237 a118a101 a115a107a117a112a105a110a165 a80a101a112a97a44 a82a117a100a97 a97 a83a116a97a110a100a97a46 a80a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101
a80a101a112a97 a117a100a165a108a225 a112a176a105 a109a165a176a101a110a237 a99a104a121a98a117a44 a106a101 0,01a44 a117 a82a117a100a121 a106a101 a116a97a116a111 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116 0,03 a97 a117 a83a116a97a110a100a121 0,75a46
a80a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a98a117a100a101 a109a165a176a105a116 a80a101a112a97a44 a106a101 a115a116a101a106a110a225 a106a97a107a111 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a98a117a100a101 a109a165a176a105a116 a82a117a100a97a44
a97 a106a101 a100a101a115a101a116a107a114a225a116 a118a165a116a178a237 a110a101a186 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a98a117a100a101 a109a165a176a105a116 a83a116a97a110a100a97 a40a122 a112a111a99a104a111a112a105a116a101a108a110a253a99a104 a100a183a118a111a100a183 a104a111
a107 a116a111a109a117 a110a101a99a104a116a165a106a237 a112a117a115a116a105a116a41a46
a97a41 a74a97a107a225 a106a101 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a118 a110a225a104a111a100a110a165 a118a121a98a114a97a110a233a109 a99a118a105a163a101a110a237 a98a117a100a101 a109a165a176a101a110a237 a99a104a121a98a110a233a63
a98a41 a77a165a176a101a110a237 a106a101 a99a104a121a98a110a233a46 a74a97a107a225 a106a101 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a109a165a176a105a108 a83a116a97a110a100a97a63
a144a101a178a101a110a237a58 a79a122a110a97a163a101a110a237 a106a101a118a183a58 Aa46a46a46a109a165a176a101a110a237 a106a101 a99a104a121a98a110a233a59 P,R,Sa46a46a46a109a165a176a105a108 a80a101a112a97a44 a82a117a100a97a44 a83a116a97a110a100a97a46
a90a101 a122a97a100a225a110a237 a112a108a121a110a101a44 a186a101 P(P) = P(R) a97 P(P) = 10P(S)a46 a77a117a115a237 a112a108a97a116a105a116 P(P) + P(R) + P(S) = 1a44
a116a101a100a121 10P(S) + 10P(S) +P(S) = 1a46 a79a100a116a117a100 P(S) = 1/21a44 P(P) = 10/21a44 P(R) = 10/21a46
a97a41 P(A) = 1021 · 0,01 + 1021 · 0,03 + 121 · 0,75 .= 0,055 a40a112a176a101a115a110a165 a50a51a47a52a50a48a41
a98a41 P(S|A) = 121·0,75P(A) .= 0,652 a40a112a176a101a115a110a165 a49a53a47a50a51a41
a80a176a46 a50a46 a86a101 a115a107a117a112a105a110a165 a49a48 a111a115a111a98 a109a97a106a237 a52 a107a114a101a118a110a237 a115a107a117a112a105a110a117 a65a44 a51 a107a114a101a118a110a237 a115a107a117a112a105a110a117 a48a44 a50 a107a114a101a118a110a237 a115a107a117a112a105a110a117 a66 a97 a106a101a100a101a110
a107a114a101a118a110a237 a115a107a117a112a105a110a117 a65a66a46 a78a225a104a111a100a110a165 a118a121a98a101a114a101a109a101 a51 a108a105a100a105a46
a78a225a104a111a100a110a225 a118a101a108a105a163a105a110a97 X a117a100a225a118a225a44 a107a111a108a105a107 a122 a110a105a99a104 a109a225 a115a107a117a112a105a110a117 a65a46
a86a121a112a111a163a116a165a116a101 a115a116a176a101a100a110a237 a104a111a100a110a111a116a117 a110a225a104a111a100a110a233 a118a101a108a105a163a105a110a121 X a97 a104a111a100a110a111a116a117 a100a105a115a116a114a105a98a117a163a110a237 a102a117a110a107a99a101 a118 a98a111a100a165 x = 2a46
a144a101a178a101a110a237a58 a77a183a186a101a109a101 a115a105 a118a178a105a109a110a111a117a116a44 a186a101 X a109a225 a104a121a112a101a114a103a101a111a109a101a116a114a105a99a107a233 a114a111a122a100a165a108a101a110a237 a115 a112a97a114a97a109a101a116a114a121 N = 10a44
M = 4a44 n = 3a46 a83a116a176a101a100a110a237 a104a111a100a110a111a116a97 a106a101 a112a114a111a116a111
EX = n· MN = 3 · 410 = 1,2
a74a105a110a225 a109a111a186a110a111a115a116 a106a101 a118a121a112a111a163a237a116a97a116 a104a111a100a110a111a116a121 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a110a237 a102a117a110a107a99a101 a21 a117a118a225a100a237a109a101 a100a118a165 a109a111a186a110a111a115a116a105 a118a253a112a111a163a116a117a58
p(0) =
parenleftbig6
3
parenrightbig
parenleftbig10
3
parenrightbig = 610 · 59 · 48 = 16, p(1) = 4 ·
parenleftbig6
2
parenrightbig
parenleftbig10
3
parenrightbig = 3 · 410 · 69 · 58 = 12,
p(2) =
parenleftbig4
2
parenrightbig· 6
parenleftbig10
3
parenrightbig = 3 · 410 · 39 · 68 = 310, p(3) =
parenleftbig4
3
parenrightbig
parenleftbig10
3
parenrightbig = 410 · 39 · 28 = 130.
EX = 0 ·p(0) + 1 ·p(1) + 2 ·p(2) + 3 ·p(3) = 1,2.
a72a111a100a110a111a116a97 a100a105a115a116a114a105a98a117a163a110a237 a102a117a110a107a99a101 a118 a98a111a100a165 a50a58
F(2) = P(X < 2) = p(0) +p(1) = 23 .= 0,667.
a80a176a46 a51a46 a78a225a104a111a100a110a225 a118a101a108a105a163a105a110a97 X a109a225 a115a112a111a106a105a116a233 a114a111a122a100a165a108a101a110a237 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a105 a115 a104a117a115a116a111a116a111a117
f(x) = a(1 +x2) pro x ∈ 〈0,1〉; f(x) = 0 jinak.
a97a41 a78a97a106a100a165a116a101 a104a111a100a110a111a116a117 a107a111a110a115a116a97a110a116a121 a a116a97a107a44 a97a98a121 f a115a107a117a116a101a163a110a165 a98a121a108a97 a104a117a115a116a111a116a97 a110a165a106a97a107a233 a110a225a104a111a100a110a233 a118a101a108a105a163a105a110a121a46
a98a41 a86a121a112a111a163a116a165a116a101 P(X = 0,2)a44 P(X ∈ 〈0,3; 0,8〉) a97 F(0,6)a46
a144a101a178a101a110a237a58 a77a117a115a237 a112a108a97a116a105a116 integraltext∞−∞f(x)dx = 1a58
integraldisplay 1
0
a(1 +x2)dx = a
bracketleftbigg
x+ x
3
3
bracketrightbigg1
0
= a· 43; 43 ·a = 1 ⇒ a = 34
P(X = 0,2) = 0
P(X ∈ 〈0,3; 0,8〉 =
integraldisplay 0,8
0,3
3
4(1 +x
2)dx = 3
4
bracketleftbigg
x+ x
3
3
bracketrightbigg0,8
0,3
= 0,49625 .= 0,496
F(0,6) = P(X < 0,6) =
integraldisplay 0,6
0
3
4(1 +x
2)dx = 0,504
a66a49a49
a80a176a46 a49a46 a86 a108a97a98a111a114a97a116a111a114a110a237a109 a99a118a105a163a101a110a237 a115a112a111a108a117 a112a114a97a99a117a106a237 a118a101 a115a107a117a112a105a110a165 a80a101a112a97a44 a82a117a100a97 a97 a83a116a97a110a100a97a46 a80a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101
a80a101a112a97 a117a100a165a108a225 a112a176a105 a109a165a176a101a110a237 a99a104a121a98a117a44 a106a101 0,1a44 a117 a82a117a100a121 a106a101 a116a97a116a111 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116 0,05 a97 a117 a83a116a97a110a100a121 0,07a46
a80a114a97a118a100a165a112a111a100a111a98a110a111a115a116a105a44 a186a101 a98a117a100a101 a109a165a176a105a116 a80a101a112a97 a97 a82a117a100a97a44 a106a115a111a117 a115a116a101a106a110a233a46 a80a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a98a117a100a101 a109a165a176a105a116
a83a116a97a110a100a97a44 a106a101 a178a101a115a116a107a114a225a116 a118a165a116a178a237 a110a101a186 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a98a117a100a101 a109a165a176a105a116 a80a101a112a97 a40a106a101 a112a114a183a98a111a106a110a165a106a178a237 a97 a110a101a99a104a99a101
a107 a116a111a109a117 a111a115a116a97a116a110a237 a112a117a115a116a105a116a41a46
a97a41 a74a97a107a225 a106a101 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a118 a110a225a104a111a100a110a165 a118a121a98a114a97a110a233a109 a99a118a105a163a101a110a237 a98a117a100a101 a109a165a176a101a110a237 a115a112a114a225a118a110a233a63
a98a41 a77a165a176a101a110a237 a106a101 a115a112a114a225a118a110a233a46 a74a97a107a225 a106a101 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a44 a186a101 a109a165a176a105a108 a82a117a100a97a63
a144a101a178a101a110a237a58 a79a122a110a97a163a101a110a237 a106a101a118a183a58 Aa46a46a46a109a165a176a101a110a237 a106a101 a115a112a114a225a118a110a233a59 P,R,Sa46a46a46a109a165a176a105a108 a80a101a112a97a44 a82a117a100a97a44 a83a116a97a110a100a97a46
a90a101 a122a97a100a225a110a237 a112a108a121a110a101a44 a186a101 P(P) = P(R) a97 P(S) = 6P(P)a46 a77a117a115a237 a112a108a97a116a105a116 P(P) + P(R) + P(S) = 1a44
a116a101a100a121 P(P) +P(P) + 6P(P) = 1a46 a79a100a116a117a100 P(P) = P(R) = 1/8a44 P(S) = 3/4a46
a97a41 P(A) = 18 · (1 − 0,1) + 18 · (1 − 0,05) + 34 · (1 − 0,93) = 0,92875 .= 0,929 a40a122a108a111a109a107a101a109 a55a52a51a47a56a48a48a41
a98a41 P(R|A) = 18·0,95P(A) .= 0,128 a40a112a176a101a115a110a165 a57a53a47a55a52a51a41
a80a176a46 a50a46 a86a101 a115a107a117a112a105a110a165 a49a50 a111a115a111a98 a109a225 a53 a108a105a100a237 a107a114a101a118a110a237 a115a107a117a112a105a110a117 a65a44 a51 a107a114a101a118a110a237 a115a107a117a112a105a110a117 a48a44 a51 a107a114a101a118a110a237 a115a107a117a112a105a110a117 a66 a97 a106a101a100a101a110
a107a114a101a118a110a237 a115a107a117a112a105a110a117 a65a66a46 a78a225a104a111a100a110a165 a118a121a98a101a114a101a109a101 a52 a108a105a100a105a46
a78a225a104a111a100a110a225 a118a101a108a105a163a105a110a97 X a117a100a225a118a225a44 a107a111a108a105a107 a122 a110a105a99a104 a109a225 a115a107a117a112a105a110a117 a48a46
a86a121a112a111a163a116a165a116a101 a115a116a176a101a100a110a237 a104a111a100a110a111a116a117 a110a225a104a111a100a110a233 a118a101a108a105a163a105a110a121 X a97 a104a111a100a110a111a116a117 a100a105a115a116a114a105a98a117a163a110a237 a102a117a110a107a99a101 a118 a98a111a100a165 x = 2a46
a144a101a178a101a110a237a58 a77a183a186a101a109a101 a115a105 a118a178a105a109a110a111a117a116a44 a186a101 X a109a225 a104a121a112a101a114a103a101a111a109a101a116a114a105a99a107a233 a114a111a122a100a165a108a101a110a237 a115 a112a97a114a97a109a101a116a114a121 N = 12a44
M = 3a44 n = 4a46 a83a116a176a101a100a110a237 a104a111a100a110a111a116a97 a106a101 a112a114a111a116a111
EX = n· MN = 4 · 312 = 1.
a74a105a110a225 a109a111a186a110a111a115a116 a106a101 a118a121a112a111a163a237a116a97a116 a104a111a100a110a111a116a121 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a110a237 a102a117a110a107a99a101 a21 a117a118a225a100a237a109a101 a100a118a165 a109a111a186a110a111a115a116a105 a118a253a112a111a163a116a117a58
p(0) =
parenleftbig9
4
parenrightbig
parenleftbig12
4
parenrightbig = 912 · 811 · 710 · 69 = 1455, p(1) = 3 ·
parenleftbig9
3
parenrightbig
parenleftbig12
4
parenrightbig = 4 · 312 · 911 · 810 · 79 = 2855,
p(2) =
parenleftbig3
2
parenrightbig·parenleftbig9
2
parenrightbig
parenleftbig12
4
parenrightbig = 6 · 312 · 211 · 910 · 89 = 1255, p(3) = 9parenleftbig12
4
parenrightbig = 4 · 312 · 211 · 110 · 99 = 155.
EX = 0 ·p(0) + 1 ·p(1) + 2 ·p(2) + 3 ·p(3) = 1.
a72a111a100a110a111a116a97 a100a105a115a116a114a105a98a117a163a110a237 a102a117a110a107a99a101 a118 a98a111a100a165 a50a58
F(2) = P(X < 2) = p(0) +p(1) = 4255 .= 0,764.
a80a176a46 a51a46 a78a225a104a111a100a110a225 a118a101a108a105a163a105a110a97 X a109a225 a115a112a111a106a105a116a233 a114a111a122a100a165a108a101a110a237 a112a114a97a118a100a165a112a111a100a111a98a110a111a115a116a105 a115 a104a117a115a116a111a116a111a117
f(x) = 1 +ax
2
2 pro x ∈ 〈0,1〉; f(x) = 0 jinak.
a97a41 a78a97a106a100a165a116a101 a104a111a100a110a111a116a117 a107a111a110a115a116a97a110a116a121 a a116a97a107a44 a97a98a121 f a115a107a117a116a101a163a110a165 a98a121a108a97 a104a117a115a116a111a116a97 a110a165a106a97a107a233 a110a225a104a111a100a110a233 a118a101a108a105a163a105a110a121a46
a98a41 a86a121a112a111a163a116a165a116a101 P(X = 0,7)a44 P(X ∈ 〈0,1; 0,6〉) a97 F(0,3)a46
a144a101a178a101a110a237a58 a77a117a115a237 a112a108a97a116a105a116 integraltext∞−∞f(x)dx = 1a58
integraldisplay 1
0
1 +ax2
2 dx =
1
2
bracketleftbigg
x+ax
3
3
bracketrightbigg1
0
= 12(1 + a3); 12(1 + a3) = 1 ⇒ a = 3
P(X = 0,7) = 0
P(X ∈ 〈0,1; 0,6〉 = 12
integraldisplay 0,6
0,1
(1 + 3x2)dx = 12 bracketleftbigx+x3bracketrightbig0,60,1 = 0,3575 .= 0,358
F(0,3) = P(X < 0,3) = 12
integraldisplay 0,3
0
(1 + 3x2)dx = 0,1635 .= 0,164.
Vloženo: 28.05.2009
Velikost: 127,89 kB
Komentáře
Tento materiál neobsahuje žádné komentáře.
Mohlo by tě zajímat:
Skupina předmětu BMA3 - Matematika 3
Reference vyučujících předmětu BMA3 - Matematika 3
Podobné materiály
- BEL1 - Elektrotechnika 1 - Vzorové řešení1
- BEL1 - Elektrotechnika 1 - Vzorové řešení2
- BEL1 - Elektrotechnika 1 - Vzorové řešení3
- BEL1 - Elektrotechnika 1 - Vzorové řešení4
- BEL1 - Elektrotechnika 1 - Vzorové řešení5
- BEL1 - Elektrotechnika 1 - Vzorové řešení6
- BEL1 - Elektrotechnika 1 - Vzorové řešení7
- BEL1 - Elektrotechnika 1 - Vzorové řešení8
- BMA3 - Matematika 3 - Vzorové příklady
- BVPA - Vybrané partie z matematiky - Semestrálka vzorové řešení 2 květen 07
- BVPA - Vybrané partie z matematiky - Semestrálka vzorové řešení květen 07
- BMA3 - Matematika 3 - Vzorové řešení
- BEL2 - Elektrotechnika 2 - Zkoušky BEL2 2007 až 2009 + vzorové zadání 2010
- BMA3 - Matematika 3 - 1. písemka 2012 vzorové řešení
- BEVA - Elektromagnetické vlny, antény a vedení - Vzorové zadání zkoušky 2013
- BAEO - Analogové elektronické obvody - OSZ1_19.1.07_reseni
- BAEO - Analogové elektronické obvody - OSZ2_26.1.07_reseni
- BAEO - Analogové elektronické obvody - sem_zk_20.1.2006_reseni
- BAEO - Analogové elektronické obvody - sem_zk_27.1.2006_reseni
- BAEO - Analogové elektronické obvody - SZ_11.1.07_reseni
- BAEO - Analogové elektronické obvody - zkouska_11.1.2006_reseni
- BFY2 - Fyzika 2 - Semestrální zkouška A řešení
- BFY2 - Fyzika 2 - Semestrální zkouška B řešení
- BFY2 - Fyzika 2 - Řešení 2 AB
- BFY2 - Fyzika 2 - Řešení 3
- BFY2 - Fyzika 2 - Řešení náhradní AB
- BFY2 - Fyzika 2 - Řešení radny AB
- BKEZ - Konstrukce elektronických zařízení - zkouška řešení 1
- BKEZ - Konstrukce elektronických zařízení - zkouška řešení 2
- BKEZ - Konstrukce elektronických zařízení - zkouška řešení 3
- BKEZ - Konstrukce elektronických zařízení - zkouška řešení 4
- BKEZ - Konstrukce elektronických zařízení - zkouška řešení 5
- BKEZ - Konstrukce elektronických zařízení - zkouška řešení 6
- BMA3 - Matematika 3 - Zkouška řešení 02A
- BMA3 - Matematika 3 - Zkouška řešení 02B
- BMA3 - Matematika 3 - Zkouška řešení 02C
- BMA3 - Matematika 3 - Zkouška řešení 02D
- MMUT - Multitaktní systémy - pisemka-reseni 1
- MMUT - Multitaktní systémy - pisemka-reseni 10
- MMUT - Multitaktní systémy - pisemka-reseni 11
- MMUT - Multitaktní systémy - pisemka-reseni 12
- MMUT - Multitaktní systémy - pisemka-reseni 2
- MMUT - Multitaktní systémy - pisemka-reseni 3
- MMUT - Multitaktní systémy - pisemka-reseni 4
- MMUT - Multitaktní systémy - pisemka-reseni 5
- MMUT - Multitaktní systémy - pisemka-reseni 6
- MMUT - Multitaktní systémy - pisemka-reseni 7
- MMUT - Multitaktní systémy - pisemka-reseni 8
- MMUT - Multitaktní systémy - pisemka-reseni 9
- BCZA - Číslicové zpracování a analýza signálů - CZA_3_reseni
- BASS - Analýza signálů a soustav - Písemka 1 řešení
- BEL1 - Elektrotechnika 1 - Test s řešením
- BEL1 - Elektrotechnika 1 - Řešení semestrálky 1
- BEL1 - Elektrotechnika 1 - Řešení semestrálky 2
- BEL1 - Elektrotechnika 1 - Řešení semestrálky 3
- BFY2 - Fyzika 2 - Test-semestrálka a řešení
- BMVE - Měření v elektrotechnice - Semestrálky řešení
- BVPA - Vybrané partie z matematiky - Semestrálka řešení x
- BVPA - Vybrané partie z matematiky - Semestrálka řešení y
- BVPA - Vybrané partie z matematiky - Řešení semstrálek
- BFY2 - Fyzika 2 - Řešení 2AB
- BMOD - Modelování a simulace - řešení 1 termín
- BMOD - Modelování a simulace - řešení 2 termín
- BSIS - Signály a soustavy - 2004 BSIS zkouška a řešení
- BMVA - Měření v elektrotechnice - zkouška BMVA 3-1-2011 řešení, řádný termín
- BMA1 - Matematika 1 - BMA1 zkouška 3-1-2011 řešení
- BVMT - Vysokofrekvenční a mikrovlnná technika - Malá písemka BVMT říjen 2011 + řešení
- BVMT - Vysokofrekvenční a mikrovlnná technika - Malá písemka BVMT říjen 2011 + řešení
- BFY1 - Fyzika 1 - Řešení Semestrální práce – domácího úkolu části Mechanika
- BFY1 - Fyzika 1 - Řešení Semestrální práce – domácího úkolu části El. pole
- BFY1 - Fyzika 1 - Řešení Semestrální práce – domácího úkolu části Magnetizmus
- MTEO - Teorie elektronických obvodů - MTEO-PC-cviceni07-reseni-Mathcad-11-2013
Copyright 2025 unium.cz


