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2 years ago

2x+1 divided by 8x^3+18x^2+11x+2

Mathematics

1 answer:

yuradex [85]2 years ago

8 0

Answer:

1/4x^2+7x+2

Step-by-step explanation:

hope this helps

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3 years ago

Alana has 100 mL of water in her water bottle. She needs to share it with 3 other students so that all 4 of them have the same a

elena-14-01-66 [18.8K]

Answer:

each person (including Alana) gets 25mL of water

Step-by-step explanation:

since there is 4 ppl, and 100mL of water, you divide 100/4. Therefore, each person gets 25mL of water:)

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3 years ago

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gayaneshka [121]

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3 years ago

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What is the slope of this line

Illusion [34]

Answer:

-8/5 is the slope.

Step-by-step explanation:

To find slope, it's the change in y over the change in x. Or rise over run.

That would be 8/5.

And this line would be negative because it is going from left to right.

So your final answer would be -8/5 as your slope.

3 0

3 years ago

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

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

5 0

3 years ago