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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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Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

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P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42

b) We have, 1-(1-p)^n=P, Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.

1-0.99^n\geq 0.95\\0.99^n\leq 1-0.95\\0.99^n\leq 0.05\\n\geq 297

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