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3 years ago

SECTION B

Mathematics

1 answer:

lapo4ka [179]3 years ago

3 0

Answer: The median value of data set B is -5.5, which is less than the median value of 3.1 in dataset A.

Step-by-step explanation:

Order the dataset from least to greatest:

-38 → -13 → -9 → -2 → 14 → 28

Then find the values that lies in the middle:

-38 → -13 → <u>-9 → -2</u> → 14 → 28

Since there are 2 values, find the average of those 2 values:

$\frac{-9+(-2)}{2} =\frac{-11}{2} =-5.5$

The median value = -5.5.

The median value of data set B is -5.5, which is less than the median value of 3.1 in dataset A.

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What expression is equivalent to ^3 square root of 2y^3 times 7 square root of 18y

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Answer:

$\sqrt[3]{2y^3} * 7\sqrt{18y} = 21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

Step-by-step explanation:

The question is poorly formatted.

Given

$\sqrt[3]{2y^3} * 7\sqrt{18y}$

Required

Derive an equivalent expression

$\sqrt[3]{2y^3} * 7\sqrt{18y}$

Express 18 as 9 * 2

$\sqrt[3]{2y^3} * 7\sqrt{9 * 2y}$

Split the expression as follows:

$\sqrt[3]{2y^3} * 7\sqrt{9} * \sqrt{2y}$

Take positive square root of 9

$\sqrt[3]{2y^3} * 7*3 * \sqrt{2y}$

$\sqrt[3]{2y^3} * 21 * \sqrt{2y}$

$21*\sqrt[3]{2y^3} * \sqrt{2y}$

The cube root can be rewritten to give:

$21*\sqrt[3]{2}*\sqrt[3]{y^3} * \sqrt{2y}$

$\sqrt[3]{y^3} = y^{3*\frac{1}{3}} = y$

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$21*\sqrt[3]{2} * y * \sqrt{2y}$

Rewrite as:

$21y *\sqrt[3]{2} * \sqrt{2y}$

Split $\sqrt{2y$

$21y *\sqrt[3]{2} * \sqrt{2} * \sqrt{y}$

Collect Like Terms

$21y*\sqrt{y} *\sqrt[3]{2} * \sqrt{2}$

Represent in index form

$21y*y^{\frac{1}{2}} *2^\frac{1}{3} *2^\frac{1}{2}$

Apply law of indices

$21*y^{1+\frac{1}{2}} *2^{\frac{1}{3} +\frac{1}{2} }$

$21*y^{\frac{2+1}{2}} *2^{\frac{2+3}{6}}$

$21*y^{\frac{3}{2}} *2^{\frac{5}{6}}$

$21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

Hence:

$\sqrt[3]{2y^3} * 7\sqrt{18y} = 21(y^{\frac{3}{2}})(2^{\frac{5}{6}})$

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