In a group of 10 batteries, 3 are dead. you choose

Ответ:

riiahh22

07.01.2022

Good (0) | P(x) - .067 | (3/10)(2/9)
Good (1) | P(x) - .467 | (7/10)(3/9) + (3/10)(7/9)
Good (2) | P(x) - .467 | (7/10)(6/9)

The odds added up give 14/30
= 0.46666

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Ответ:

wbrandi118

06.08.2020

Mathematics

Refresh if you see weird display errors. $\sin 2x = -2 \sin\left(x - \frac{\pi}{2}\right) \cos\left(x - \frac{\pi}{2}\right)$ for the left hand side, sine double angle identity can be used: $\text{lhs} = \sin 2x = 2\sin x \cos x$ for the right hand side, we have $\text{rhs} = -2 \sin\left(x - \frac{\pi}{2}\right) \cos\left(x - \frac{\pi}{2}\right)$ i will deal with $\sin\left(x - \frac{\pi}{2}\right)$ separately. we can use this sine difference identity: $\begin{aligned} \sin(a-b) & = \sin a \cos b - \cos a \sin b\\ \sin\left(x - \tfrac{\pi}{2}\right) & = \sin x \cos\tfrac{\pi}{2}-\cos x \sin \tfrac{\pi}{2} \end{aligned}$ from unit circle, etc, we have $\cos\frac{\pi}{2} = 0,\ \sin\frac{\pi}{2} = 1$ , therefore $\begin{aligned}\sin\left(x - \tfrac{\pi}{2}\right) & = \sin x \cos\tfrac{\pi}{2}-\cos x \sin \tfrac{\pi}{2} \\ & = \sin x (0) - \cos x (1)\\ & = -\cos x \end{aligned}$ now i will deal with $\cos\left(x - \frac{\pi}{2}\right)$ separately. we can use this cosine difference identity: $\begin{aligned} \cos(a-b) & = \cos a\cos b + \sin a \sin b \\ \cos\left(x - \tfrac{\pi}{2}\right) & = \cos x \cos\tfrac{\pi}{2} + \sin x \sin \tfrac{\pi}{2} \\ & = \cos x (0) + \sin x (1) \\ & = \sin x \end{aligned}$ bringing these two simplifications together: $\begin{aligned} \text{rhs} & = -2 \sin\left(x - \tfrac{\pi}{2}\right) \cos\left(x - \tfrac{\pi}{2}\right) \\ & = -2 (-\cos x)(\sin x) \\ & = 2\cos x \sin x \\ & = 2\sin x \cos x \\ & = \text{lhs} \end{aligned}$

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In a group of 10 batteries, 3 are dead. you choose 2 batteries at random.
create a probability model for the number of good batteries.

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In a group of 10 batteries, 3 are dead. you choose 2 batteries at random. create a probability model for the number of good batteries.

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In a group of 10 batteries, 3 are dead. you choose 2 batteries at random.
create a probability model for the number of good batteries.