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Veseljchak [2.6K]
3 years ago
12

At the beginning of the day the stock market goes up 30 1/2 points. At the end of the day, the stock market goes down 100 3/4 po

ints. what is the change from the high to the end of the day?
Mathematics
1 answer:
Akimi4 [234]3 years ago
5 0
Well, is namely their difference, so, let's first convert the mixed fractions to "improper", and subtract.

\bf \stackrel{mixed}{30\frac{1}{2}}\implies \cfrac{30\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{61}{2}}
\\\\\\
\stackrel{mixed}{100\frac{3}{4}}\implies \cfrac{100\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{403}{4}}\\\\
-------------------------------\\\\
\cfrac{61}{2}-\cfrac{403}{4}\impliedby \textit{so our \underline{LCD is 4}}\implies \cfrac{(2\cdot 61)~-~(1\cdot 403)}{4}
\\\\\\
\cfrac{122-403}{4}\implies \cfrac{-281}{4}\implies -70\frac{1}{4}
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(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

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The formula for margin of error is:

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MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, <em>MOE</em> = 0.02.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

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Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

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