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Maksim231197 [3]
3 years ago
8

In a class of 78 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both

courses. How many students are not enrolled in either course?
a. 6
b. 15
c. 24
d. 33
e. 54
Mathematics
2 answers:
galina1969 [7]3 years ago
7 0
78-41-22+9=24 the students are mutually exclusive. The answer is C
Nezavi [6.7K]3 years ago
7 0
The answer is A (6) because 41+22+9=72, and there are 72 students enrolled in either of the courses. When you subtract 72 from 78 you get 6, which is the ending amount of students  hat are not enrolled in French class, or German class.
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Given P = x^0.3 y^0.7 is the chicken lay eggs production function, where P is the number of eggs lay, x is the number of workers
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Answer:

Part A)

\displaystyle \frac{dy}{dx}=-\frac{3}{7}P^\frac{10}{7}x^{-\frac{10}{7}}

Part B)

The daily operating cost decreases by about $143 per extra worker.

Step-by-step explanation:

We are given the equation:

\displaystyle P=x^{\frac{3}{10}}y^{\frac{7}{10}}

Where <em>P</em> is the number of eggs laid, <em>x</em> is the number of workers, and <em>y</em> is the daily operating budget (assuming in US dollars $).

A)

We want to find dy/dx.

So, let’s find our equation in terms of <em>x</em>. We can raise both sides to 10/7. Hence:

\displaystyle P^\frac{10}{7}=\Big(x^\frac{3}{10}y^\frac{7}{10}\Big)^\frac{10}{7}

Simplify:

\displaystyle P^\frac{10}{7}=x^\frac{3}{7}y

Divide both sides by<em> </em>the <em>x</em> term to acquire:

\displaystyle y=P^\frac{10}{7}x^{-\frac{3}{7}}

Take the derivative of both sides with respect to <em>x: </em>

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\Big[P^\frac{10}{7}x^{-\frac{3}{7}}\Big]

Apply power rule. Note that P is simply a constant. Hence:

\displaystyle \frac{dy}{dx}=P^\frac{10}{7}(-\frac{3}{7})(x^{-\frac{10}{7}})

Simplify. Hence, our derivative is:

\displaystyle \frac{dy}{dx}=-\frac{3}{7}P^\frac{10}{7}x^{-\frac{10}{7}}

Part B)

We want to evaluate the derivative when <em>x</em> is 30 and when <em>y</em> is $10,000.

First, we will need to find <em>P</em>. Our original equations tells us that:

P=x^{0.3}y^{0.7}

Hence, at <em>x</em> = 30 and at <em>y</em> = 10,000, <em>P </em>is:

P=(30)^{0.3}(10000)^{0.7}

Therefore, for our derivative, we will have:

\displaystyle \frac{dy}{dx}=-\frac{3}{7}\Big(30^{0.3}(10000^{0.7})\Big)^\frac{10}{7}\Big(30^{-\frac{10}{7}}\Big)

Use a calculator. So:

\displaystyle \frac{dy}{dx}=-\frac{1000}{7}=-142.857142...\approx-143

Our derivative is given by dy/dx. So, it represents the change in the daily operating cost over the change in the number of workers.

So, when there are 30 workers with a daily operating cost of $10,000 producing a total of about 1750 eggs, the daily operating cost decreases by about $143 per extra worker.

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