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STatiana [176]
3 years ago
11

A company makes and sells charm bracelets. The cost of producing x bracelets is represented by the function

Mathematics
2 answers:
Serggg [28]3 years ago
8 0

P(x) = 12x – 180

Given:

A company makes and sells charm bracelets.

The cost of producing x bracelets is represented by the function C(x) = 180 +

8x

The revenue earned from selling x bracelets is represented by the function

R(x) = 20x.


Explanation of terms:

For x bracelets,

Cost of production is C(x); C(x) = 180 + 8x

Revenue earned is R(x); R(x) = 20x.Profit made is P(x); P(x) is unknown

Profit made = Revenue earned – Cost of production

∴ P(x) = R(x) – C(x)

P(x) = 20x – (180 + 8x)

P(x) = 20x – 180 – 8x

P(x) = 12x – 180

The profit made from selling x bracelets is represented by the function

<em> P(x) = 12x – 180</em>



Klio2033 [76]3 years ago
5 0

15 bracelets are required to break even.

180+ 8x=20x

180=12x

15=x

Hope this helps!

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Answer:

False

Step-by-step explanation:

3.17t - 2.431 = 5.06 \\ 3.17t = 5.06 + 2.431 \\ 3.17t = 7.491 \\  \frac{3.17t}{3.17}  =  \frac{7.491}{3.17}  \\ t = 2.363

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Nata [24]

Answer:

400,000km is represented as 8cm on the map!!!

Step-by-step explanation:

1cm=50000km

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5 0
2 years ago
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

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3 years ago
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Free_Kalibri [48]

Answer:

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T(2, -7) = (4, -10)

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2 years ago
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