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Dominik [7]
3 years ago
8

A football team plays in a large stadium. With a ticket price of ​$19​, the average attendance at recent games has been 50 comma

000. A market survey indicates that for each​ $1 increase in the ticket​ price, attendance decreases by 300. a. Express the number of spectators at a football​ game, N, as a function of the ticket​ price, x. b. Express the revenue from a football​ game, R, as a function of the ticket​ price, x.
Mathematics
1 answer:
tino4ka555 [31]3 years ago
8 0

Answer:

Part 1: N(x) = 50,000 - 300(x-19)

Part 2: R(x) =-300x² + 55700x

Step-by-step explanation:

Given,

The original price of each ticket = $ 19,

The original attendance = 50,000

Part 1 :

∵ For the each​ $1 increase in the ticket​ price, attendance decreases by 300.

Let x represents the price of each ticket after increment,

Thus, if price increment = (x-19) dollars,

New attendance, N(x) = 50,000 - 300(x-19)

Part 2 :

Since, revenue = price of each ticket × attendance

Thus, the revenue from the football​ game,

R(x) = x(50,000 - 300(x-19))

R(x) = 50000x - 300x²+ 5700x

⇒ R(x) =-300x² + 55700x

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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
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(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

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Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

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The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

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                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

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(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

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