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kiruha [24]
3 years ago
10

What is the order of rotation symmetry of these shapes?????

Mathematics
2 answers:
iren [92.7K]3 years ago
8 0

Hope it helps...

<em>Benjemin</em>

Zigmanuir [339]3 years ago
5 0

Answer:

i think the first 1 is 4 and the second is 2

Step-by-step explanation:

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OLEGan [10]
Well it have to be D because 32 and 20 can not go into 64 to 40 because 32 times 64 is 2048 and 64 times 20 is 1280 and if it was 40 times 32 and 20 it will still be higher. I HAVE FAITH IN YOU.
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The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
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I hope this helps you

8 0
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13 m is the volume of the prism
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