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Aleonysh [2.5K]
3 years ago
10

Ron asked 18 classmates whether they prefer granola bars over muffins. he used a calculator to compare the number of classmates

who said yes to the total number he surveyed. the calculator showed the result as 0.66666667. how many students prefer granola bars over muffins?
Mathematics
1 answer:
muminat3 years ago
7 0
<h3>Answer: 12</h3>

Explanation:

Multiply 0.66666667 with 18 to get

0.66666667*18 = 12.00000006

The 6 at the very end is due to rounding error. In reality the 6's in 0.66666667 should go on forever. That 7 is from rounding up.

Anyways, the result 12.00000006 should be 12 exactly.

Note how

12/18 = 0.66666667

when using a calculator.

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(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

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\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}

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\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

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