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Valentin [98]
2 years ago
12

A battery company produces 12,000 batteries per day. Each day they randomly select 50 to test. If they find that 1 is dead, how

many dead batteries would they expect to find among the entire 12,000
Could someone help me solve this please:)
Mathematics
1 answer:
alexgriva [62]2 years ago
5 0

Using the binomial distribution, it is found that they would expect to find 240 dead batteries among the entire 12,000.

For each battery, there are only two possible outcomes, either it was dead or it was not. The probability of a battery being dead is independent of any other battery, hence the binomial distribution is used to solve this question.

<h3>What is the binomial probability distribution?</h3>

It is the <u>probability of exactly x successes on n repeated trials, with p probability</u> of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

In this problem, we have that:

  • One in 50 batteries are dead, hence p = 1/50 = 0.02.
  • There is a total of 12,000 batteries, hence n = 12000.

Then:

E(X) = np = 12000(0.2) = 240

They would expect to find 240 dead batteries among the entire 12,000.

You can learn more about the binomial distribution at brainly.com/question/24863377

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\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV

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\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2

Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}

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\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5

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\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k

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and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

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