#89 will give brainliest for best answer!

Mathematics

1 answer:

Lady_Fox [76]2 years ago

6 0

Answer:

Median: 11

Range: 12

25th Percentile: 9

75th Percentile: 14

Interquartile Range: 5

Step-by-step explanation:

The median is the line in the 'middle' of the box. The range is the largest number minus the smallest number. 25th Percentile is the first line of the box, 75th is 3rd line. IQR is the 75th Percentile - 25th Percentile.

Hope this helps! :)

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Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

Alex787 [66]

Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$