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s344n2d4d5 [400]
3 years ago
7

2. The average distance from the sun to the planet Mercury is about 58,000,000 km. The diameter of a human hair is about 0.0025

cm.
(a) What is the distance from the sun to Mercury written in scientific notation?
5.8 x 10^7
(b) What is the diameter of a human hair written in scientific notation?
2.5 x 10^-3
(c) In comparing the measurements in Parts (a) and (b), what else must be done before a comparison is made?
Mathematics
1 answer:
kodGreya [7K]3 years ago
3 0
Distance of Sun and Mercury = 58,000,000 km.In Scientific notation, it would be: 5.8 × 10⁷ Km
Diameter of a human hair = 0.0025 cmIn scientific notation, it would be: 2.5 × 10⁻³ cm
Before Comparison, we need to bring cms to Kms or vice-versa, so we can easily calculate then, here, we will go from cm to Km.2.5 × 10⁻³ × 10⁻⁵ Km = 2.5 × 10⁻⁸
Now, we can compare the values, 5.8 × 10⁷ Km > 2.5 × 10⁻⁸
[ reason. - power of a smaller number is in negative, it means, it is denominator, which would be very small as compared to other ]
Hope this helps!
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damaskus [11]

Answer:

z=\frac{24-40}{8}=-2

z=\frac{40-40}{8}=0

And then the percentage between 24 and 40 would be \frac{95}{2}= 47.5 \%

Step-by-step explanation:

For this problem we have the following parameters given:

\mu = 40, \sigma =8

And for this case we want to find the percentage of lightbulb replacement requests numbering between 24 and 40.

From the empirical rule we know that we have 68% of the values within one deviation from the mean, 95% of the values within 2 deviations and 99.7% within 3 deviations.

We can find the number of deviations from themean for the limits with the z score formula we got:

z=\frac{X-\mu}{\sigma}

And replacing we got:

z=\frac{24-40}{8}=-2

z=\frac{40-40}{8}=0

And then the percentage between 24 and 40 would be \frac{95}{2}= 47.5 \%

5 0
3 years ago
Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = xyi + 5zj + 7yk
REY [17]

By Stokes' theorem,

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Parameterize S by

\vec\sigma(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(8-u\cos v)\,\vec k

with 0\le u\le9 and 0\le v\le2\pi. Then take the normal vector to S to be

\vec\sigma_u\times\vec\sigma_v=u\,\vec\imath+u\,\vec k

Then the line integral is equal to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9(2\,\vec\imath-u\cos v\,\vec k)\cdot(u\,\vec\imath+u\,\vec k)\,\mathrm du\,\mathrm dv

\displaystyle=\int_0^{2\pi}\int_0^9(2u-u^2\cos v)\,\mathrm du\,\mathrm dv=\boxed{162\pi}

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Hope this helped :)
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