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jarptica [38.1K]

1 year ago

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When the admission price for a baseball game was $4 per ticket, 50,000 tickets were sold. When the price was raised to $5, only

45,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ball park owners are $0.10 and $75,000 respectively.

1 answer:

anygoal [31]1 year ago

5 0

The profit function is P(x) = ( - x² / 5000 ) + 13.9x - 75000.

50,000 baseball game tickets were sold at $4 per ticket.

When the price is raised to $5, then 45,000 tickets were sold.

The variable and fixed costs for the ballpark owners are $0.10 and $75,000 respectively.

Let's say x is the number of tickets sold, and P is the profit.

Then,

P = ax + b

At P = 4,

4 = (50000)a + b ---------(1)

At P = 5,

5 = (45000)a + b --------(2)

Subtracting (2) from (1),

4 - 5 = (50000)a + b - (45000)a - b

- 1 = 5000(a)

a = ( - 1/5000)

So if a = ( - 1/5000),

Then,

4 = (50000)a + b

4 = (50000)( - 1 / 5000 ) + b

4 = -10 + b

b = 14

Therefore,

p(x) = ( - x /5000) + 14

Now, the profit function is:

Profit = Revenue - Costs

P(x) = R(x) - C(x)

Now, R(x) = xp(x)

R(x) = x[ ( - x/5000) + 14]

R(x) = ( - x² / 5000 ) + 14x

The fixed cost is F(x) = $75000

Hence, the costs will be:

C(x) = 75000 + (0.10)x

Therefore the profit function is:

P(x) = R(x) - C(x)

P(x) = ( - x² / 5000 ) + 14x - 75000 - (0.10)x

P(x) = ( - x² / 5000 ) + 13.9x - 75000

Learn more about profit function here:

brainly.com/question/21497949

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The complete question is mentioned below:

When the admission price for a baseball game was $4 per ticket, 50,000 tickets were sold. When the price was raised to $5, only 45,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ballpark owners are $0.10 and $75,000 respectively.

Find the profit P as a function of x, the number of tickets sold

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(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the true mean, </u> $\mu$ <u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

<em>Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 98% confidence interval for the true mean, </u> $\mu$ <u> is ;</u>

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

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= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

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4 years ago

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lara [203]

Answer:

x = -1 , - 11,

Step-by-step explanation:

| x | = x

|- x | = x

f(x) = 2 | x + 6 | - 4

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x = 1 , f( 1 ) = 2 | 1 + 6 | - 4 = 2 | 7 | - 4 = 14 - 4 = 10 , False

x = 11, f( 11 ) = 2 | 11 + 6 | - 4 = 2 | 17 | - 4 = 34 - 4 = 30 , False

x = - 11, f( - 11) = 2 | -11 + 6 | - 4 = 2 | - 5 | - 4 = 10 - 4 = 6 , True

x = 14 , f( 14 ) = 2 | 14 + 6 | - 4 = 2 | 20 | - 4 = 40 - 4 = 36 , False

x = -14 , f( - 14 ) = 2 | - 14 + 6 | - 4 = 2 | - 8 | - 4 = 16 - 4 = 12 , False

x = - 26 , f( - 26) = 2 | -26 + 6 | - 4 = 2 | -20 | - 4 = 40 - 4 = 36 , False

x = 26, f( 26 ) = 2 | 26 + 6 | - 4 = 2 | 32 | - 4 = 64 - 4 = 60 , False

OR

Given f ( x ) = 6

f( x ) = 2 | x + 6 | - 4

6 = 2 | x + 6 | - 4

6 + 4 = 2 | x + 6| - 4 + 4 [ adding 4 to both sides ]

10 = 2 | x + 6 | + 0

5 = | x + 6 | [ dividing both sides by 2 ]

$- 5 =\ ( x+ 6) \ = 5\\\\$ $[ \ | x | = a\ => \ -a \ = x \ = a \ ]$

$- 5 - 6 = \ x + 6 - 6 \ = \ 5 - 6\\$ $[ \ subtracting \ by \ 6 \ ]$

$-11 = \ x \ = - 1$

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3 years ago