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const2013 [10]
3 years ago
5

this rectangular prism is cut by the plane shown. The plane is perpendicular to the base of the rectangular prism. What is the s

hape of the cross-section? A) rectangle. B) rhombus. C) square. D) trapezoid.
Mathematics
2 answers:
Julli [10]3 years ago
5 0

Answer: A i took the flvs test

Step-by-step explanation:

Musya8 [376]3 years ago
4 0

Answer: REACTANGLE !!!!!!!!!!!

Step-by-step explanation:

The shape of the cross-section is a rectangle. The plane is perpendicular to the base of the rectangular prism so the shape of the cross-section is the same as the top and bottom views of the rectangular prism.

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An office supply company produces yellow document envelopes. The envelopes come in
Ierofanga [76]

Answer:

6x+10

Step-by-step explanation:

2x×2(2x+5)

2x+4x+10

6x+10

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The sum of an acute angle and an obtuse angle is an acute angle always sometimes or never true ???​
xxTIMURxx [149]

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Never

Step-by-step explanation:

Adding together an obtuse angle (greater than 90°) and acute angle (less than 90°) should result in an obtuse angle every time

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120 to 52 what % decrease
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The formula is percent= part/whole so it would be 52/120 which is .43333333333<span> or 43%</span>
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There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl
insens350 [35]

We're told that

P(A\cap B)=0.15

P(A\cup B)^C=0.06\implies P(A\cup B)=0.94

P(B\mid A)=P(B^C\mid A)=0.5

where the last fact is due to the law of total probability:

P(A)=P(A\cap B)+P(A\cap B^C)

\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)

\implies 1=P(B\mid A)+P(B^C\mid A)

so that B\mid A and B^C\mid A are complementary.

By definition of conditional probability, we have

P(B\mid A)=P(B^C\mid A)

\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}

\implies P(A\cap B)=P(A\cap B^C)

We make use of the addition rule and complementary probabilities to rewrite this as

P(A\cap B)=P(A\cap B^C)

\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)

\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)

\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]

\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]

\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C

\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)

By the law of total probability,

P(B)=P(A\cap B)+P(A^C\cap B)

\implies P(A^C\cap B)=P(B)-P(A\cap B)

and substituting this into (*) gives

2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]

\implies P(B)=P(A\cup B)-P(A\cap B)

\implies P(B)=0.94-0.15=\boxed{0.79}

8 0
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Doing some homework right now :) Need help on this problem
pychu [463]

Answer:

both have 5

Step-by-step explanation:

10 divided by 2 equals 5

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3 years ago
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