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Deffense [45]
3 years ago
11

Calculate the mean, median, mode, variance and standard deviation for the following set of data. Show all calculations.

Mathematics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

Mean (Average) 16.4125

Median 18.3  

Mode 10.1, appeared 2 times

Step-by-step explanation:

Mean:10.1, 20.2, 16.7, 9.9, 10.1, 24.4, 19.9, 20.0 / 8 = 16.4125

Median: This is the number in the middle. So if we where to put all of the numbers in number order from least to greatest, 18.3 would be in the middle.

Here are the numbers in number order,

9.9, 10.1, 10.1, 16.7, 19.9, 20.0, 20.2, 24.4

now we would just find the difference between 16.7 and 19.9 and that gets us 18.3.

Mode: the mode is the value in a data set that has the highest number of recurrences. Because 10.1 appeared twice, that's why its the mode.  

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Answer:

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Step-by-step explanation:

Given \displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1, start by squaring both sides to work towards isolating x:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2

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\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}

Square both sides:

\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2

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\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36

Move everything to one side to get a quadratic:

\displaystyle-\frac{1}{4}x^2+7x-24=0

Solving using the quadratic formula:

A quadratic in ax^2+bx+c has real solutions \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. In \displaystyle-\frac{1}{4}x^2+7x-24, assign values:

\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24

Solving yields:

\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}

Only x=4 works when plugged in the original equation. Therefore, x=24 is extraneous and the only solution is \boxed{x=4}

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