TheXSquaredFactor

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UsernameTheXSquaredFactor
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 #2
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A perpendicular bisector is a line that intersects a segment at a right angle and at that segment's midpoint. In order to generate a unique equation of any line, you need to know at least one point and the slope; without these, it is impossible to generate a specific equation.

 

The definition I provided should give a clue as to how to find one point of the perpendicular bisector. Since we know that perpendicular bisectors intersect a segment at its midpoint, we can find the midpoint of \(\overline{JK}\) :

 

\(m_{\overline{JK}}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\) This is the midpoint formula. The midpoint formula just finds the average of the two coordinates.
\(m_{\overline{JK}}=\left(\frac{-4+2}{2},\frac{2+2}{2}\right)\) The only thing left to do is to simplify.
\(m_{\overline{JK}}=(-1,2)\) Yes! We have already found one part of the perpendicular bisector, a point.
   


Finding the slope requires a slightly indirect approach. We can determine the slope of the given segment and adjust with the knowledge of perpendicular lines. I need not use the slope formula here because the y-coordinates of both points are the same, which means that the segment that connects J and K is horizontal. All horizontal lines have a slope of 0. The line of the perpendicular bisector, therefore, is vertical because a vertical line is perpendicular to a horizontal line. A vertical line has an undefined slope. Look at that! We have determined the slope of the perpendicular bisector. 

 

Vertical lines are written in the form of x=c, where c is a constant. We know, from the information gathered from solving for the midpoint, that \(x=-1\) is the equation of the perpendicular bisector.

TheXSquaredFactor Sep 28, 2018
 #1
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+1

I may not be a lexicographer, but I do know what affixes are. An affix is a word element attached to either the beginning or ending of the root of a word with the purpose of modifying the original word's meaning. Here are a few examples of an affix. 

 

Look at how many words, for example, build off of the root -clos-. This root, at its core, means to "shut." Here are a few I could think of:
 

Closet
Closely
Closest
Disclose
Foreclosure
Closeness
Closemouthed
 

 

Notice how combinations of letters added to the word change the meaning of the word while still preserving the original meaning. Let's try an easier root, -spec-, which means "to see." This one has so many examples that I will not even attempt to list them all. I tried listing ones with unique affixes and lengths:

 

Retrospectively
Specifications
Disrespectful
Spectrograph
Spectacular
Inspectors
Suspected
Spectrum
Aspects
Speck

 

All of these words (and many others I omitted) are some variation of "to see." 

 

I do not think this extreme diversity of words would be possible without affixes. Imagine if there was a completely different word for every variation of -spec- or -clos-? These are only two examples. You can attempt to investigate -aqua- or -jur- or -chrono-. 

 

Despite this analysis, I am not sure whether or not it would change the number of words in a language. Maybe there is some other way of organizing symbols to create words that we are not implementing in the modern day; you have to start from somewhere. Maybe other people can chime in and provide their thoughts. I suspect that this is a toss-up question and that no definite answer exists. 

TheXSquaredFactor Sep 27, 2018
 #1
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+2
TheXSquaredFactor Sep 27, 2018
 #5
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+2

1) Variance measures how spread out a particular data set is. It does this by measuring the average distance each data point is from the true mean. Variance has a formula. It is the following:
 

\(\sigma^2=\frac{\sum_{i=1}^{N}(X_{i}-\mu)^2}{N}\)

 

Yes, this formula looks very complicated, but it is first important to define a few variables.

 

\(N = \text{sample size}\\ \mu = \text{mean of population}\\ X_i=\text{term of data}\\ \sigma^2=\text{variance}\)

 

What is this formula really trying to tell you to do? Well, here is an elementary breakdown:

 

1. Find the mean of the data. 

 

2. Subtract each individual element from the mean 

 

3. Square all results from the previous step and find the sum of the squared difference. 

 

4. Divide by the sample size.

 

Let's get to it!
 

1) Find the mean of the data. 

 

In order to find the mean, we must first find the sum of all the data points. 

 

  \(x_i\)
\(x_1\) 182
\(x_2\) 274
\(x_3\) 207
\(x_4\) 216
\(x_5\) 285
\(x_6\) 190
\(x_7\) 138
\(x_8\) 240
\(x_9\) 291
\(x_{10}\) 288
\(x_{11}\) 176
\(x_{12}\) 186
\(x_{13}\) 190
\(x_{14}\) 213
\(x_{15}\) 236
\(x_{16}\) 242
\(x_{17}\) 186
\(x_{18}\) 259
\(x_{19}\) 164
\(x_{20}\) 172
\(x_{21}\) 233
\(x_{22}\) 268
\(x_{23}\) 152
\(x_{24}\) 274
\(\Sigma\) 5262

 

If the sum of all the data points is 5262, then the mean is this number divided by the number of data points, 24. 

 

\(\mu=\frac{5262}{24}=219.25\)

 

2) Subtract each individual element from the mean and square that result. I will replicate the table above. 

 

  \(x_i\) \(\mu\) \(x_i-\mu\)
\(x_1\) 182   -37.25
\(x_2\) 274   54.75
\(x_3\) 207   -12.25
\(x_4\) 216   -3.25
\(x_5\) 285   65.75
\(x_6\) 190   -29.25
\(x_7\) 138   -81.25
\(x_8\) 240   20.75
\(x_9\) 291   71.75
\(x_{10}\) 288   68.75
\(x_{11}\) 176   -43.25
\(x_{12}\) 186   -33.25
\(x_{13}\) 190   -29.25
\(x_{14}\) 213   -6.25
\(x_{15}\) 236   16.75
\(x_{16}\) 242   22.75
\(x_{17}\) 186   -33.25
\(x_{18}\) 259   39.75
\(x_{19}\) 164   -55.25
\(x_{20}\) 172   -47.25
\(x_{21}\) 233   13.75
\(x_{22}\) 268   48.75
\(x_{23}\) 152   -67.25
\(x_{24}\) 274   54.75
\(\Sigma\) 5262 219.25  

 

3) Square all results from the previous step and find the sum of the squared difference. 

 

  \(x_i\) \(\mu\) \(x_i-\mu\) \((x_i-\mu)^2\)
\(x_1\) 182   -37.25 1387.5625
\(x_2\) 274   54.75 2997.5625
\(x_3\) 207   -12.25 150.0625
\(x_4\) 216   -3.25

10.5625

\(x_5\) 285   65.75 4323.0625
\(x_6\) 190   -29.25 855.5625
\(x_7\) 138   -81.25 6601.5625
\(x_8\) 240   20.75 430.5625
\(x_9\) 291   71.75 5148.0625
\(x_{10}\) 288   68.75 4726.5625
\(x_{11}\) 176   -43.25 1870.5625
\(x_{12}\) 186   -33.25 1105.5625
\(x_{13}\) 190   -29.25 855.5625
\(x_{14}\) 213   -6.25 39.0625
\(x_{15}\) 236   16.75 280.5625
\(x_{16}\) 242   22.75 517.5625
\(x_{17}\) 186   -33.25 1105.5625
\(x_{18}\) 259   39.75 1580.0625
\(x_{19}\) 164   -55.25 3052.5625
\(x_{20}\) 172   -47.25 2232.5625
\(x_{21}\) 233   13.75 189.0625
\(x_{22}\) 268   48.75 2376.5625
\(x_{23}\) 152   -67.25 4522.5625
\(x_{24}\) 274   54.75 2997.5625
\(\Sigma\) 5262 219.25   49356.5

 

4) Divide by the sample size

 

  \(x_i\) \(\mu\) \(x_i-\mu\) \((x_i-\mu)^2\) \(\frac{(x_i-\mu)^2}{N}\)
\(x_1\) 182   -37.25 1387.5625  
\(x_2\) 274   54.75 2997.5625  
\(x_3\) 207   -12.25 150.0625  
\(x_4\) 216   -3.25

10.5625

 
\(x_5\) 285   65.75 4323.0625  
\(x_6\) 190   -29.25 855.5625  
\(x_7\) 138   -81.25 6601.5625  
\(x_8\) 240   20.75 430.5625  
\(x_9\) 291   71.75 5148.0625  
\(x_{10}\) 288   68.75 4726.5625  
\(x_{11}\) 176   -43.25 1870.5625  
\(x_{12}\) 186   -33.25 1105.5625  
\(x_{13}\) 190   -29.25 855.5625  
\(x_{14}\) 213   -6.25 39.0625  
\(x_{15}\) 236   16.75 280.5625  
\(x_{16}\) 242   22.75 517.5625  
\(x_{17}\) 186   -33.25 1105.5625  
\(x_{18}\) 259   39.75 1580.0625  
\(x_{19}\) 164   -55.25 3052.5625  
\(x_{20}\) 172   -47.25 2232.5625  
\(x_{21}\) 233   13.75 189.0625  
\(x_{22}\) 268   48.75 2376.5625  
\(x_{23}\) 152   -67.25 4522.5625  
\(x_{24}\) 274   54.75 2997.5625  
\(\Sigma\) 5262 219.25 0 49356.5 \(2056.5208\overline{3}\)

 

I realize that this answer differs from Cphill's. I have reviewed my work many times, and I do not believe that I have done anything incorrect. 

 

2) The standard deviation, thankfully, does not require this much computation. Take the square root of the previous result, \(2056.5208\overline{3}\)

 

\(\sigma=\sqrt{2056.5208\overline{3}}\approx45.3489\)

 

I realize that this is a lot of information to take in at once. If you have any questions, then do not hesitate to ask.

TheXSquaredFactor Sep 26, 2018