Find the median of the set of data 8.4, 4.9, 1.9, 1.3, 8.7, 9.4, 4.9, 5.5

Mathematics

2 answers:

Andrews [41]2 years ago

8 0

Answer:

Median is 5.2

Step-by-step explanation:

Median is the middle number when everything is in order

8.4, 4.9, 1.9, 1.3, 8.7, 9.4, 4.9, 5.5

--->

1.3, 1.9, 4.9, 4.9, 5.5, 8.4, 8.7, 9.4

Since there are two middle numbers we need the middle of those two

(4.9 + 5.5) / 2 = 5.2

sergejj [24]2 years ago

7 0

Answer:

5.2

Step-by-step explanation:

1.3, 1.9, 4.9, 4.9, 5.5, 8.4, 8.7, 9.4

the median is the middle number, so it would be in between 4.9 and 5.5

the answer is 5.2

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Rationalize the numerator. <img src="https://tex.z-dn.net/?f=%20%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B144x%20%7D%20%7D%7B%20%5Csq

Simora [160]

rationalizing the numerator, or namely, "getting rid of that pesky radical at the top".

we simply multiply top and bottom by a value that will take out the radicand in the numerator.

$\bf \cfrac{\sqrt[3]{144x}}{\sqrt[3]{y}}~~
\begin{cases}
144=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\\
\qquad 2^3\cdot 18
\end{cases}\implies \cfrac{\sqrt[3]{2^3\cdot 18x}}{\sqrt[3]{y}}\implies \cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}}
\\\\\\
\cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}}\cdot \cfrac{\sqrt[3]{(18x)^2}}{\sqrt[3]{(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)(18x)^2}}{\sqrt[3]{(y)(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)^3}}{\sqrt[3]{18^2x^2y}}$

$\bf \cfrac{2(18x)}{\sqrt[3]{324x^2y}}~~
\begin{cases}
324=2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3\\
\qquad 12\cdot 3^3
\end{cases}\implies \cfrac{36x}{\sqrt[3]{12\cdot 3^3x^2y}}
\\\\\\
\cfrac{36x}{3\sqrt[3]{12x^2y}}\implies \cfrac{12x}{\sqrt[3]{12x^2y}}$