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fenix001 [56]
3 years ago
5

A sales team estimates that the number of new phones they will sell is a function of the price that they set. They estimate that

is they set the price at x dollars, they will sell f(x)=4080-10x. Therefore, the company's revenue is x(4080-10x).
Mathematics
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

$204

Step-by-step explanation:

The question is at what price x will the company maximize revenue.

The revenue function is:

R(x) = 4,080x-10x^2

The price for which the derivate of the revenue function is zero is the price the maximizes revenue:

R(x) = 4,080x-10x^2\\\frac{dR(x)}{dx}=0=4,080-20x\\x=\frac{4,080}{20}\\ x=\$204

The company will maximize its revenue when the price is $204.

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Answer:

Answer is 360

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What is 0.8 as a fraction
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Plot the foci of the hyperbola represented by the equation x^2/625 − y^2/3600 =1 100pts
Pavel [41]

The attached image represents the foci of the hyperbola

<h3>How to determine the foci?</h3>

The equation of the hyperbola is given as:

\frac{x^2}{625} - \frac{y^2}{3600} = 1

Rewrite as:

\frac{x^2}{25^2} - \frac{y^2}{60^2} = 1

A hyperbola is represented as:

\frac{(x - h)^2}{b^2} - \frac{(y - k)^2}{a^2} = 1

This means that:

h = 0

k = 0

b = 25

a = 60

Next, calculate c the distance from the center to the focus using:

c = \sqrt{a^2 -b^2}

This gives

c = \sqrt{60^2 -25^2}

Evaluate

c = \pm \sqrt{2975}

This means that:

Foci = (0, -√2975) and (0, √2975)

See attachment for the hyperbola and the foci

Read more about hyperbola at:

brainly.com/question/16735067

#SPJ1

3 0
2 years ago
Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard
ziro4ka [17]

Answer:

a) P(X

b)P(\bar X

And using a calculator, excel ir the normal standard table we have that:

P(Z

c)  Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the female pulse rates of a population, and for this case we know the distribution for X is given by:

X \sim N(72,12.5)  

Where \mu=72 and \sigma=12.5

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

Part b

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

The new z score is given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

P(\bar X

And using a calculator, excel ir the normal standard table we have that:

P(Z

Part c

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

4 0
3 years ago
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