PLEASE HELPPPPP!!!!!!!!! (show work if you can)

Mathematics

2 answers:

Hunter-Best [27]3 years ago

7 0

Answer:

$\mathrm{Factor\:}18x^4y^3z= 2\cdot \:3^2\cdot \:x^4\cdot \:y^3\cdot \:z\\\\\mathrm{Factor\:}30xy^2z^2= 2\cdot \:3\cdot \:5\cdot \:x\cdot \:y^2\cdot \:z^2\\\\\mathrm{Factor\:}12x^3y^2= 2^2\cdot \:3\cdot \:x^3\cdot \:y\\\\\mathrm{Multiply\:each\:factor\:with\:the\:highest\:power}= 2^2\cdot \:3^2\cdot \:5\cdot \:x^4\cdot \:y^3\cdot \:z^2\\\\180x^4y^3z^2$

Kryger [21]3 years ago

3 0

Answer:

$180x^4y^3z^2$

Step-by-step explanation:

Start by finding the LCM of the coefficients of each polynomial:

$LCM(12,18,30)=180$

Next, to find the least common multiple of each of the following terms, we need to take the absolute minimum we can of each term ( $x$ , $y$ , and $z$ ). The largest term of $x$ is $x^4$ in the first polynomial, so we'll take exactly that for our $x$ term in the LCM, absolutely nothing more. Similarly, the largest term of $y$ in any of the three polynomials is $y^3$ (also in the first polynomial) and the largest $z$ term in any of the three polynomials is $z^2$ in the second polynomial. Thus, the LCM of all our polynomials is:

$180\cdot x^4\cdot y^3\cdot z^2=\boxed{180x^4y^3z^2}$

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The total scores on the Medical College Admission Test (MCAT) in 2013 follow a Normal distribution with mean 25.3 and standard d

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In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of $Q_1$ such that

$\mathbb P(X$

You have

$\mathbb P(X$

For the standard normal distribution, the first quartile is about $z\approx-0.6745$ , and by symmetry the third quartile would be $z\approx0.6745$ . In terms of the MCAT score distribution, these values are

$\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9$
$\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7$

The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.

The central 80% of the scores have z-scores $\pm z$ such that

$\mathbb P(-z$

That leaves 10% on either side of this range, which means

$\underbrace{\mathbb P(-z$

You have

$\mathbb P(Z$

Converting to MCAT scores,

$-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0$
$1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6$

So the interval that contains the central 80% is $(17.0,33.6)$ (give or take).