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garri49 [273]
3 years ago
15

The quantity demanded each month of the walter serkin recording of beethoven's moonlight sonata, manufactured by phonola media,

is related to the price per compact disc. the equation p = −0.00054x + 9 (0 ≤ x ≤ 12,000) where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. the total monthly cost (in dollars) for pressing and packaging x copies of this classical recording is given by c(x) = 600 + 2x − 0.00002x2 (0 ≤ x ≤ 20,000)
Mathematics
1 answer:
Olenka [21]3 years ago
7 0
We can get the Profit function P(x) from the Hint. 

the Profit function is: P(x) = xp(x) - C(x) = -0.00041 x2 + 4x - 600
Attention: don't get confuse by the <span>big P of the profit with the small p of the price</span> To calculate the maximum profit, we need to find the derivative of P(x) then set it to 0 then find x: dP(x)/dx = -0.00082 x + 4 = 0 ,   so x = 4/0.00082 = 4,878 copies each month.
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SIZIF [17.4K]

Angle <QAB is =15° because the opposite angles of an isosceles triangle are equal.

The length of the straight line AB = 80cm

<h3>Calculation of angle of a triangle</h3>

The angle at a point = 360°

Angle AQB= 360 - 210° = 150

But the angle that makes up a triangle= 180°

180-150= 30°

But <QAB = <QBA because triangle AQB is an isosceles triangle.

30/2 = 15°

To calculate the length of the straight line the following is carried out using the sine laws.

a/ sina, = b sinb

a= 8cm, sin a { sin 15)

b= ? , sin B = 150

make b the subject formula;

8/sin15= b/sin 150

b= 8 × sin 150/sin 15

b= 80cm

Learn more about isosceles triangle here:

brainly.com/question/25812711

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7 0
2 years ago
A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

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3 years ago
4/7= 5/p. What is P? Pleas solve and show you r work!
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This is a proportion. First you take the numerator times the denominator that is already solved. In this case you will take 5 times 7 and get 35. You then take 35 divided by 4 and get your answer.

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I think 710 is the answer

266+266+89+89= 710

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