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professor190 [17]
3 years ago
6

22% of the customers visiting the suit department of a certain store will purchase a suit, 30% will purchase a shirt and 28% wil

l purchase a tie. 11% of customers purchase a shirt and a suit, 14% both a suit and a tie and 10% buy a shirt and a tie. Only 6% of customers buy all three items. Let A be the event that the customer purchases a suit, B the event that the customer purchases a shirt and C the event that the customer purchases a tie. If there are 1000 customers in the store one week, how many will purchase exactly one of these items
Mathematics
1 answer:
Ulleksa [173]3 years ago
6 0

Answer:

If there are 1000 customers in the store one week, how many will purchase exactly one of these items

1000 CUSTOMERS*28%=280

Step-by-step explanation:

A The event that a persons buys a suit  

B The event that a person buys a shirt  

C The event that a person buys a tie  

 

P(A)=  22%

P(B)=  30%

P(C)=  28%

P(AB)= 11%

P(AC)= 14%

P(BC)= 10%

P(ABC)= 6%

A u B u C Is the event that any item is bougth

AC u AC u BC Is the event that any two events occured

So the wanted probability is

P[(A u B u C )(AB u AC u BC)^c

P[(A u B u C )=P(AB)+ P(BC)+P(BC)

P[(A u B u C ) =0.22+0.30+0.28-0.11-0.14.-0.10+0.06

                      =0,51

0,51=+0,23+P[(A u B u C )(AB u AC u BC)^c

        =0,28

1000 CUSTOMERS*28%=280

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Answer:

Step-by-step explanation:

The standard form of an arithmetic sequence is

aₙ = a₁ + d(n - 1)

where aₙ is the number of the term in the sequence (in order from first term where n = 1, to second term where n = 2, to third term where n = 3, etc) a₁ is the the first term in the sequence, and d is the arithmetic difference or means.  This is what we are looking to solve for.  

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Because we have the fifth term, we can write our standard form to fit our needs:

a₅ = a₁ + d(n-1).  Therefore,

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93 = -3 + d(4) so

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a₅ = -3 + 24(4) so

a₅ = 96 - 3 so

a₅ = 93 (Phew!)  ; )

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