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Eduardwww [97]
3 years ago
9

A blue die and a red die are thrown. B is the event that the blue comes up with a 6. E is the event that both dice come up even.

Write the sizes of the sets |E ∩ B| and |B|a. |E ∩ B| = ___b. |B| = ____
Mathematics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

Size of |E n B| = 2

Size of |B| = 1

Step-by-step explanation:

<em>I'll assume both die are 6 sides</em>

Given

Blue die and Red Die

Required

Sizes of sets

- |E\ n\ B|

- |B|

The question stated the following;

B = Event that blue die comes up with 6

E = Event that both dice come even

So first; we'll list out the sample space of both events

B = \{6\}

E = \{2,4,6\}

Calculating the size of |E n B|

|E n B| = \{2,4,6\}\ n\ \{6\}

|E n B| = \{2,4,6\}

<em>The size = 3 because it contains 3 possible outcomes</em>

Calculating the size of |B|

B = \{6\}

<em>The size = 1 because it contains 1 possible outcome</em>

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I kind forgot how to do this
joja [24]

Answer:

The green one

Step-by-step explanation:

In an obtuse triangle, one of the angles is greater than 90 degrees.

4 0
3 years ago
2. Lab groups of three are to be randomly formed (without replacement) from a class that contains five engineers and four non-en
Anna11 [10]

Answer:

The number of different lab groups possible is 84.

Step-by-step explanation:

<u>Given</u>:

A class consists of 5 engineers and 4 non-engineers.

A lab groups of 3 are to be formed of these 9 students.

The problem can be solved using combinations.

Combinations is the number of ways to select <em>k</em> items from a group of <em>n</em> items without replacement. The order of the arrangement does not matter in combinations.

The combination of <em>k</em> items from <em>n</em> items is: {n\choose k}=\frac{n!}{k!(n-k)!}

Compute the number of different lab groups possible as follows:

The number of ways of selecting 3 students from 9 is = {n\choose k}={9\choose 3}

                                                                                         =\frac{9!}{3!(9 - 3)!}\\=\frac{9!}{3!\times 6!}\\=\frac{362880}{6\times720}\\ =84

Thus, the number of different lab groups possible is 84.

8 0
3 years ago
I don't understand how we do this PLEASE help me!!!
sashaice [31]

28 = (45 + KL)/2

56 = 45+ KL

KL = 56-45

KL = 11

8 0
3 years ago
a bag contains 5 red marbles 4 blue marbles and 6 yellow marbles if 3 marbles are selected in succession what is the probability
Alex

Answer:

5/243

Step-by-step explanation:

4/18 × 6/18 × 5/18 = 5/243

I think

6 0
4 years ago
Convert the decimal expansion 0.291666... to a fraction
Alexus [3.1K]
7/24
You could try rounding up to 0.292 first and try guessing it it gets to 73/250 which you could guess into 70/240.
3 0
3 years ago
Read 2 more answers
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