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Vinvika [58]
3 years ago
5

Please help with all 3

Mathematics
2 answers:
DerKrebs [107]3 years ago
8 0

Answer:

1, volume=23×9×5

=1035

2, volume=(1/2×5×4)×20

... =200

3.

Vlad [161]3 years ago
6 0
<h2>Ah, volume! </h2>

1) Use the formula of rectangular prism: width x height x length

2) Use the formula of triangular prism: (Base x height x length) ÷ 2

3) Use the formula of cylinder: π x radius^{2} x height

--------------------------------------------------------------------------------------------------

1)

  • width = 5 cm
  • height = 9 cm
  • length = 23 cm

width x height x length

= 5 x 9 x 23

= 1035

<h2>answer: 1035 cubic centimeters</h2>

------------------------------------------------------------------------------------------------

2)

  • Base = 5 in
  • height = 4 in
  • length = 20 in

(Base x height x length) ÷ 2

= (5 x 4 x 20) ÷ 2

= 400 ÷ 2

= 200

<h2>answer: 200 cubic inches</h2>

------------------------------------------------------------------------------------------------

3)

  • radius = 5/2 = 2.5 m
  • height = 12 m

π x radius^{2} x height

= π x 2.5^{2} x 12

= π x 6.25 x 12

= 75π

= 236 (rounded to neareast whole number)

<h2>answer: 236 cubic meters</h2>
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Answer:

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Step-by-step explanation:

Let's start defining the random variables for this exercise :

X_{1}: '' The proportion of the particulates that are removed by the first pass ''

X_{2}: '' The proportion of what remains after the first pass that is removed by the second pass ''

Y: '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

Y=(1-X_{1})(1-X_{2})

We also assume that X_{1} and X_{2} are independent random variables with common pdf.

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The first step to solve this exercise is to find the expected value for X_{1} and X_{2}.

Because the variables have the same pdf we write :

E(X_{1})= E(X_{2})=E(X)

Using the pdf to calculate the expected value we write :

E(X)=\int\limits^a_b {xf(x)} \, dx

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Now we need to use some expected value properties in the expression of Y ⇒

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Y=1-X_{2}-X_{1}+X_{1}X_{2}

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Using that X_{1} and X_{2} have the same expected value E(X) and given that X_{1} and X_{2} are independent random variables we can write E(X_{1}X_{2})=E(X_{1})E(X_{2})   ⇒

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3 years ago
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