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kirill [66]
3 years ago
15

5/3 × 20 must show work

Mathematics
1 answer:
Romashka [77]3 years ago
5 0

Answer:

5/3*20

5/3*20/1

100/3

33.3333

Step-by-step explanation:

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Find the average value of the function f(x, y, z) = 5x2z 5y2z over the region enclosed by the paraboloid z = 9 − x2 − y2 and the
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The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin. 

<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>

<span>If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV </span>

<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>

<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>


<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>

<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>

<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>

<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>

<span>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>
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Step-by-step explanation:

I attached a file.

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Adding or subtracting a constant from each value of a distribution changes the mean, but not the standard deviation.


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You just hung a picture twelve inches above the wall trim.
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Yes I sure did hang the picture what about it
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2 years ago
Help me with this please I will give you extra points
Ierofanga [76]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Total outcomes in which Billy can get 5 or more : 2 <u>outcomes</u>

Total outcomes = 6

So, the probability of getting 5 or higher is ~

\qquad \sf  \dashrightarrow \: \dfrac{favorable \: outcomes}{total \: outcomes}

\qquad \sf  \dashrightarrow \: \dfrac{2}{6}

\qquad \sf  \dashrightarrow \: \dfrac{1}{3}

Therefore, the correct choice is C

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

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