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3 years ago

A cube has a volume of 512 mm^3. If all sides of the cube are divided by 4, what will the resulting volume be?

1 answer:

3 0

$\bf \textit{volume of a cube}\\\\
V=s^3\qquad \boxed{s=\frac{s}{4}}\implies V=\left( \cfrac{s}{4} \right)^3\implies V=\cfrac{s^3}{4^3}\implies V=s^3\cdot \cfrac{1}{4^3}
\\\\\\
V=s^3\cdot \cfrac{1}{64}\impliedby \frac{1}{64}\textit{ of the original volume, if the original was 512}
\\\\\\
512\cdot \cfrac{1}{64}\implies \cfrac{512}{64}$

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Answer: 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

Step-by-step explanation:

Since we have given that

Number of alluvial wells = 349

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Number of alluvial wells that had concentrations above 0.1 = 182

Number of quaternary wells that had concentrations above 0.1 = 112

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Standard deviation = 0.4

Average of quaternary wells = 1.62

Standard deviation =1.70

So, 95% confidence interval gives

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$\dfrac{\alpha}{2}=2.5\%\\\\z_{\frac{\alpha}{2}}=1.96$

So, 95% confidence interval becomes,

$(1.62-0.27)\pm 1.96\sqrt{\dfrac{0.4^2}{349}+\dfrac{1.7^2}{143}}\\\\=1.35\pm 1.96\times 0.020\\\\=(1.35-0.040,1.35+0.040)\\\\=(1.31,1.39)$

Hence, 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

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3 years ago

How does the distributive property relate to the factoring expression ?

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3 years ago

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Step-by-step explanation:

$x^2 -7xy =3000y \: then \to \\ x^2 =3000y + 7xy \\ {(10 ^{3} )}^{2} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{5} ) \\ {10}^{6} = 300000 + 700000 \\ \boxed{{10}^{6} = 1000000} \\ hence \: the \: equation \to \: x^2 -7xy =3000y \\ \boxed{ is \: corret}$

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