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Archy [21]
4 years ago
9

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 3, 5, 7, 9}, B = {2, 4, 6, 8, 10}, and C = {1, 2, 4, 5, 8, 9}. List the element

s of each set. (Enter your answers using roster notation. Enter EMPTY or ∅ for the empty set.)
(a) Ac ∩ (B ∩ C c)

(b) (A ∪ Bc) ∪ (B ∩ C c)

(c) (A ∪ B)c ∩ C c
Mathematics
1 answer:
podryga [215]4 years ago
5 0

Answer:

(a) {6,10}

(b) {1, 3, 5, 6, 7, 9, 10}

(c) ∅

Step-by-step explanation:

Given,

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},

A = {1, 3, 5, 7, 9},

B = {2, 4, 6, 8, 10},

C = {1, 2, 4, 5, 8, 9},

(a) Since,

A^c =U -A= \{ 2, 4, 6, 8, 10\} ,

C^c = U - C= \{ 3, 6, 7, 10\},

B\cap C^c = \{6, 10\},

\implies A^c\cap (B\cap C^c)=\{6,10\}

(b)  

B^c= U-B=\{1, 3, 5, 7, 9\}

A\cup B^c= \{1, 3, 5, 7, 9\}

\implies (A\cup B^c)\cup (B\cap C^c)=\{1,3,5,6,7,9, 10\}

(c)

A\cup B=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

(A\cup B)^c=U-(A\cup B) = \phi

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4 0
4 years ago
Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

5 0
4 years ago
Simplify <br> (rs) (25) -2 (47)4
Katen [24]

Answer:

there should be multiplication sign...so value will be (-351)

5 0
3 years ago
A box contains 10 red marbles, 13 blue marbles, 6 green marbles and 7 white
Zarrin [17]

Answer:

Total number of marbles = 10+20+30 = 60 out of which 20 are blue.

Probability of first marble being blue = 20/60

Probability of second marble being blue = 19/59

Probability of third marble being blue = 18/58

Probability of fourth marble being blue = 17/57

Probability of fifth marble being blue = 16/56

So probability of all marble being blue when five marbles are drawn from the box = (20/60)*(19/59)*(18/58)*(17/57)*(16/56) = 34/11977 = 0.00283877432

Step-by-step explanation:

7 0
3 years ago
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62 marbles is the answet
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3 years ago
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