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elena-14-01-66 [18.8K]
3 years ago
5

Let C be the closed, piecewise smooth curve formed by traveling in straight lines between the points (-2,1), (-2,-3), (1,-1) , (

1,5), and back to (-2,1), in that order. Use Green's theorem to evaluate the integral:
Integral(C) (2xy) dx +(xy^2) dy
Mathematics
1 answer:
Greeley [361]3 years ago
6 0

The given points are the vertices of the quadrilateral

Q=\left\{(x,y)\mid-2\le x\le1,\dfrac{2x-5}3\le y\le\dfrac{4x+11}3\right\}

By Green's theorem, the line integral is

\displaystyle\int_C2xy\,\mathrm dx+xy^2\,\mathrm dy=\iint_Q\frac{\partial(xy^2)}{\partial x}-\frac{\partial(2xy)}{\partial y}\,\mathrm dA

=\displaystyle\int_{-2}^1\int_{(2x-5)/3}^{(4x+11)/3}y^2-2x\,\mathrm dy\,\mathrm dx=\boxed{61}

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I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
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The mean is 96.667

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b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

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The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

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P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

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