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Juli2301 [7.4K]
3 years ago
13

There are 6 women and 5 men interviewing for 4 cashier positions at Walmart. In how many ways can the positions be filled if 2 w

omen and 2 men are hired?
Mathematics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

The positions can be filled in 150 ways.

Step-by-step explanation:

The order in which the people are chosen is not important. For example, A, B, C, D is the same outcome as B, A, C, D. So we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In how many ways can the positions be filled if 2 women and 2 men are hired?

2 women from a set of 6.

2 men from a set of 5.

So

T = C_{6,2} \times C_{5,2} = \frac{6!}{2!(6-2)!} \times \frac{5!}{2!(5-2)!} = 120

The positions can be filled in 150 ways.

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Five times the sum of a number and 6 is 48
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5(n+6)=48

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4 0
3 years ago
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Elis [28]

Answer:

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4 0
2 years ago
A store selling newspapers orders only n = 4 of a certain newspaper because the manager does not get many calls for that publica
umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

  • PY(k) = PX(k) = ε^(-3)*(3^k)/k! for k in {0,1,2,3}
  • PY(4) = 1 -PX(0) - PX(1) - PX(2) - PX (3)

we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

8 0
3 years ago
(5,2) (8,2), (8,5) and (?,?)
Luba_88 [7]
<h2>Answer:</h2><h2>(5,5)</h2><h2 /><h2>Hope this helps!!</h2><h2></h2>

6 0
2 years ago
Can some one do this and help me out
alexira [117]

Answer:

<h2><em><u>4</u></em></h2>

Step-by-step explanation:

Numbers less than 10 are = 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Not multiple of 2 = 3, 5, 7, and 9

Composite number = 4

I eliminated the numbers to get the answer.

Hope this helped!

5 0
3 years ago
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