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3 years ago

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suppose that there are two types of tickets to a show: advance and same-day. the combined cost of one advance ticket and one sam

e-day ticket is $65. for one performance, 25 advance tickets and 40 same-day tickets were sold. the total amount paid for the tickets was $2150. what was the price of each kind of ticket?

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

The Answer is: Price of Advanced Tickets: $30. Price of Same Day tickets: $35.

Step-by-step explanation:

Amount of Advanced + Amount of Same = $65

a + s = $65

a = 65 - s

25 tickets times the adult price plus 40 tickets at the same day price equals $2,150. Setup the equation, substitute, and solve for s:

25a + 40s = $2150

25(65-s) + 40s = 2150

1625 - 25s + 40s = 2150

15s = 2150 - 1625

15s = 525

s = 525 / 15 = $35.

a = 65 - 35 = $30.

Proof:

25(30) + 40(35) =

$750 + $1,400 = $2,150

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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$Z = \frac{X - \mu}{\sigma}$

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When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.71$

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here $n = 16$ , we want $P(X \geq 13)$ . So

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591$

$P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835$

$P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273$

$P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042$

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741$

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now $n = 160$ . So

$\mu = E(X) = np = 160*0.71 = 113.6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74$

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$Z = \frac{109.5 - 113.6}{5.74}$

$Z = -0.71$

$Z = -0.71$ has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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