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W*F?!

LoL you 

Hiya? lol

Hiya

What is the purpose of all the stuff u post?

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<3 <3 <3 <3 

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Turn the 3and3/4 into an improper fraction:  \(3\frac{3}{4} \rightarrow \frac{15}{4}\)

To divide by a fraction, turn the fraction upside down and multiply, so we have  \(6000\times \frac{4}{15} \)

I'll let you finish it from here.

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If z = a + ib  where i is the square root of -1 then the absolute value of z is given by:

\(|z| = \sqrt{a^2+b^2}\)

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1.7724538509055159

If you have a set of numbers, x_i where i runs from 1 to n, then the sample standard deviation is given by:

\(\sigma = \sqrt{\sum_{i=1}^n \frac{(x_i-\mu)^2}{n-1}}\) where  \(\mu = \frac{\sum_{i=1}^n x_i}{n}\)

The above holds when you have to find the mean value from the sample.  If you know the mean independently of the sample then you can replace the n-1 by n in the definition of the standard deviation.

Some calculators will automatically calculate the standard deviation for you, given a list of numbers.

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A square graphed on the coordinate plane has a diagonal with endpoints E(2,3) and F(0,-3). What are the coords of the endpoints of the other diagonal?

1. Diagonal: \(E (x_E=2,y_E=3)\) and \(F(x_F=0,y_F=-3)\).  

What are the coords of the endpoints of the other diagonal \(G(x_G,y_G)\) and \(H(x_H,y_H)\).

\(\begin{array}{rcll} \binom{x_G}{y_G} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} + \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} + \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} + \frac12 \cdot \binom{-6}{2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} + \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_G}{y_G} &=& \binom{0}{-3} +\binom{1}{3} + \binom{-3}{1} \\ \binom{x_G}{y_G} &=& \binom{0+1-3}{-3+3+1} \\ \binom{x_G}{y_G} &=& \binom{-2}{1} \\ \end{array}\)

\(\begin{array}{rcll} \binom{x_H}{y_H} &=& \binom{x_F}{y_F} +\frac12 \cdot \binom{x_E-x_F}{y_E-y_F} - \frac12 \cdot \binom{-(y_E-y_F)}{x_E-x_F} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2-0}{3-(-3)} - \frac12 \cdot \binom{-(3-(-3))}{2-0} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\frac12 \cdot \binom{2}{6} - \frac12 \cdot \binom{-6}{2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{\frac12\cdot 2}{\frac12\cdot 6} - \binom{\frac12\cdot (-6)}{\frac12\cdot 2} \\ \binom{x_H}{y_H} &=& \binom{0}{-3} +\binom{1}{3} - \binom{-3}{1} \\ \binom{x_H}{y_H} &=& \binom{0+1-(-3)}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{0+1+3}{-3+3-1} \\ \binom{x_H}{y_H} &=& \binom{4}{-1} \\ \end{array}\)

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