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Verizon [17]
3 years ago
6

A stadium has 50 comma 00050,000 seats. Seats sell for ​$3030 in Section​ A, ​$2424 in Section​ B, and ​$1818 in Section C. The

number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in ​$1 comma 288 comma 8001,288,800 from each​ sold-out event. How many seats does each section​ hold?
Mathematics
1 answer:
Mariulka [41]3 years ago
5 0

A= B+C

As given, A+B+C = 50000 seats

A+(B+C) = 50000

2A = 50000

A= 25000 seats

Now section A sells seats at $30 each

So, 25000\times30=750000

Now, B+C = 25000

B = 25000-C

B sells at $24 and C sells at $18

So, 24B+18C = 538800

Now as section A incurs an amount of $750000 from booking, so B and C will incur 1288800-750000 = 538800 where $1288800 is the total booking amount.

24B+18C = 538800

24(25000-C)+18C=538800

600000-24C+18C=538800

6C = 61200

C = 10200

Now, B = 25000-C

B = 25000 - 10200 = 14800

Hence, section A has 25000 seats

Section B has 14800 seats

Section C has 10200 seats




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Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\

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P(x\geq10)=1-P(x

P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\

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