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Semenov [28]
3 years ago
12

Solve for x and y. *

Mathematics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

beeebooobeeeeeboooooobeeeeebooooooo

Step-by-step explanation:

option a

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the data represents the heights of fourteen basketball players, in inches. 69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 8
Daniel [21]
If you would like to know the interquartile range of the new set and the interquartile range of the original set, you can do this using the following steps:

<span>The interquartile range is the difference between the third and the first quartiles.

The original set: </span>69, 70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77, 82
Lower quartile: 72
Upper quartile: 76.25
Interquartile range: upper quartile - lower quartile = 76.25 - 72 = <span>4.25
</span>
The new set: <span>70, 72, 72, 74, 74, 74, 75, 76, 76, 76, 77, 77
</span>Lower quartile: 72.5
Upper quartile: 76
Interquartile range: upper quartile - lower quartile = 76 - 72.5 = 3.5

The correct result would be: T<span>he interquartile range of the new set would be 3.5. The interquartile range of the original set would be more than the new set.</span>
6 0
3 years ago
Read 2 more answers
PLZ HELP WILL MARK BRANLIEST (No dum answers or I will delete your question - seriously don't be annoying)
Vsevolod [243]

Answer:

A. tan(2π/3) = -\sqrt{3}

4 0
3 years ago
Choose the number sentence that illustrates the distributive property of multiplication over addition.
marin [14]

Answer:

3*(7 + 2) = 3*7 + 3*2

Option 3

Step-by-step explanation:

Distributive property of multiplication over addition:

a*(b+ c) = (a*b ) + (a * c)

6 0
3 years ago
Read 2 more answers
Can you find the sum of (-6y-2)+5(3+2.5y) ?
hjlf
(-6y-2)+5(3+2.5y) expend -6y-2+15+12.5 6.5y+13
6 0
3 years ago
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Rewrite the following integral in spherical coordinates.​
lora16 [44]

In cylindrical coordinates, we have r^2=x^2+y^2, so that

z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}

correspond to the upper and lower halves of a sphere with radius \sqrt2. In spherical coordinates, this sphere is \rho=\sqrt2.

1 \le r \le \sqrt2 means our region is between two cylinders with radius 1 and \sqrt2. In spherical coordinates, the inner cylinder has equation

x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)

This cylinder meets the sphere when

x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)

which occurs at

\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi

where n\in\Bbb Z. Then \frac\pi4\le\phi\le\frac{3\pi}4.

The volume element transforms to

dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi

Putting everything together, we have

\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3

4 0
2 years ago
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