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Semmy [17]
3 years ago
15

Help me please I don’t know what the answer is

Mathematics
1 answer:
Anon25 [30]3 years ago
5 0
I believe that B) m<3 would need to double. Since m<6 and m<3 are corresponding angles, they are always equal. So if 6 doubles, 3 needs to as well in order to be equal.
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1 question~Which inequality is true? Use the number line to help.
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What is the area of a rectangle with a length of 7.5 feet and a width of 8 feet?
Pepsi [2]

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Step-by-step explanation:

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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.}


\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>
~\frak{Amphitrite1040:)}</h3>
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Vikentia [17]
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