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Anon25 [30]
3 years ago
11

A meteorologist is studying the monthly rainfall in a section of the Brazilian rainforest. She recorded the monthly rainfall, in

inches, for last year. They were:
10.4, 10.3, 11.7, 11.1, 8.0, 4.4, 2.6, 1.8, 2.5, 4.4, 7.3, 9.5

Calculate the following for the data set:
Mathematics
1 answer:
mario62 [17]3 years ago
7 0

Answer:

Mean = 7

Median = 7.65

Mode = 4.4

Range = 9.9

Step-by-step explanation:

The question is incomplete as it doesn't state what to find. Lets complete the question first. The complete question states that:

Calculate the following for the data set

Mean

Median

Mode

Range

<h3>A) Mean</h3>

Mean = sum of values/ total number of values

Mean = 84/12

Mean = 7

<h3>B) Median</h3>

Arrange the data in ascending order

1.8, 2.5, 2.6, 4.4, 4.4, 7.3, 8.0, 9.5, 10.3, 10.4, 11.1, 11.7

Median = Average of middle values

Median = (7.3+8.0) / 2

Median = 7.65

<h3>C) Mode</h3>

Mode = number which appears most = 4.4

<h3>D) Range</h3>

Range = Max value - Min value

Range = 11.7 - 1.8

Range = 9.9

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\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1

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|y|>7\text{, so }y=7 is not a valid argument

For the first argument: \text{if }x>7\text{, then }|x|>7

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For the second argument: |y|>7\text{, so }y=7

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|y|>7:=\left \{ {y>7\text{  if }y\ge0}\atop{-y>7\text{  if }y

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Learn more here: brainly.com/question/11897796

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