Does this graph show a function? Explain how you know.

Mathematics

1 answer:

uranmaximum [27]3 years ago

5 0

Answer:

it's C.

Step-by-step explanation:

hope this helps you out

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The following data set represents the age of

Julli [10]

$\huge\text{Hey there!}$

$\huge\textsf{Breaking the meaning down in simpler terms}$
$\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}$

$\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}$

$\huge\textsf{Formulas for each term}$

$\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }$

$\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}$

$\huge\textbf{SOLVING FOR THE QUESTION}$

$\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}$

$\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}$

$\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}$