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2 years ago

What is the least common denominator of the rational expressions below?

Mathematics

1 answer:

Lyrx [107]2 years ago

4 0

Answer:

B.(x-6)

Step-by-step explanation:

12/(x^2 - 11x+30) - 43/(x^2-6x)

12/((x - 5) (x - 6)) - 43/x(x-6) [Factoring the denominator]

Both have x-6 common in the denominator,

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How do you solve what is 30% of $240

DerKrebs [107]

The Answer would be x=72. Why? Because $\frac{x}{240} = \frac{30}{100}. Fraction \frac{30}{100}$ is your %. You Cross multiply 240 * 30 = 100x
240 time 30 = 7200. you want x to be alone so you do 7200 divided by 100 equaling 72. So you get x=72.

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3 years ago

Solve for x.<br> .<br> Simplify your answer as much as possible.

Nady [450]

2x = 15 - 3x -- add 3x to both sides

5x = 15 -- divide both sides by 5

x = 3 -- simplest form

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2 years ago

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Can someone help me with this? Thanks a bunch! ;)

dexar [7]

Answer:

idk, can I?

Step-by-step explanation:

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2 years ago

What is -10=10(x+9) and could you tell me how to do it cause I don't understand.

timofeeve [1]

-10=10x+90
80=10x
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2 years ago

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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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2 years ago