<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx-2%7D%7Bx%5E2%2B4x-12%7D" id="TexFormula1" title="\frac{x-2}{x^2+4x-12}" alt="\frac

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Liula [17]

3 years ago

{x-2}{x^2+4x-12}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

arsen [322]3 years ago

3 0

Answer:

1/(x+6)

Step-by-step explanation:

I'm assuming you want me to factorize

x^2+4x-12 would factorize into (x-2) and (x+6)

Therefore, the numerator (x-2) would cancel with the part of the numerator (x-2) leaving 1/(x+6)

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3 years ago

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Rainbow [258]

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4 0

1 year ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

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$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)

= 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0

3 years ago