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vichka [17]
3 years ago
8

When two men get married to each other, do they both go to the same bachelor party?

Mathematics
2 answers:
Verizon [17]3 years ago
7 0

Answer:

yeah?

Step-by-step explanation:

Elis [28]3 years ago
5 0

Answer:

yes they do

Step-by-step explanation:

i guess

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Find the missing number 9:4::63: ?
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Answer: 28

Step-by-step explanation:

28 is thhe missing number

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How to find the volume of a room
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multiply the height x the length x the width to find the volume of a room (or another 3d object)

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sweet T has 2 orange picks for every 5 green.If there are 21 picks in all,How many picks are orange?
kipiarov [429]
6 of the picks are orange.
8 0
3 years ago
Read 2 more answers
Write the equation of the graph
olchik [2.2K]

Answer:

y = 2.cos(x) - 2

Step-by-step explanation:

1. Shifting cos(x) right or left, top or down, as the graph in the question has the same illustration of  cos(x). Consider

y = a.cos(x) + b

2. find a and b by using two sample points from the graph: (0, 0) and (\pi, -4)

0 = a.cos(0) + b  => a + b = 0\\ -4 = a.cos(\pi) + b => b-a=-4\\\left \{ {{2b=-4} \atop {2a=4}} \right.\\a = 2, b=-2\\

5 0
3 years ago
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
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