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2 years ago

1/2 (6x-10) - x = What's the other side of the equation?

Mathematics

1 answer:

LiRa [457]2 years ago

4 0

Answer:

Step-by-step explanation:

1/2 (6x-10) - x

Opening bracket

= 6x/2 - 10/2 - x

= 3x - 5 - x

= 2x - 5

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Marine scientists categorize signature whistles of bottlenose dolphins by typelong dash—type a, type b, type c, etc. In one s

r-ruslan [8.4K]

Answer:

The 95% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 185, \pi = \frac{100}{185} = 0.5405$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5405 - 1.96\sqrt{\frac{0.5405*0.4595}{185}} = 0.4687$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5405 + 1.96\sqrt{\frac{0.5405*0.4595}{185}} = 0.6123$

The 95% confidence interval for the true proportion of bottlenose dolphin signature whistles that are type a whistles is (0.4687, 0.6123).

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Read 2 more answers

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

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3 years ago