Malia found the margin of error for the percent of lengths of 100 willow leaves greater than 5 cm. If she increases her sample t
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Assume x <span>> 0
</span>√(2x⁵/18)
= √(2/18) * √(x⁵)
= √(1/9) * √(x⁴ · x)
= 1/3 * √x⁴ * √x
= 1/3 * x² * √x
= 1/3 * x²√x
=
</span>√(2x⁵/18)
= √(2/18) * √(x⁵)
= √(1/9) * √(x⁴ · x)
= 1/3 * √x⁴ * √x
= 1/3 * x² * √x
= 1/3 * x²√x
=