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Rina8888 [55]
2 years ago
12

Which of the pairs of events below is mutually exclusive?

Mathematics
1 answer:
Nady [450]2 years ago
3 0

Answer:

Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards.

Step-by-step explanation:

Given four pair of events:

1. Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards

2. Drawing a 2 and drawing a 4 with replacement from a standard deck of cards

3. Drawing a heart and then drawing a spade without replacement from a standard deck of cards

4. Drawing a jack and then drawing a 7 without replacement from a standard deck of cards

To find:

Which of the pairs is mutually exclusive?

Solution:

First of all, let us have a look at the definition of mutually exclusive events.

Mutually exclusive events do not have any case in common.

In other words, two events are known as mutually exclusive when both the events can not occur at the same time.

Now, let us have a look at the given options one by one:

1. Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards.

We know that there is only one ace of spades in a standard deck.

Therefore, the two events can not occur one after the other, so these are mutually exclusive events.

2. Drawing a 2 and drawing a 4 with replacement from a standard deck of cards.

The two events can occur one after the other, so these are not mutually exclusive.

3. Drawing a heart and then drawing a spade without replacement from a standard deck of cards.

Separate cards are drawn one after the other, so not mutually exclusive.

4. Drawing a jack and then drawing a 7 without replacement from a standard deck of cards.

Separate cards are drawn one after the other, so not mutually exclusive.

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Find the values of x and y
zhuklara [117]

You can use a tangent:

tangent=\dfrac{opposite}{adjacent}

We have opposite = 17 and adjacent = x.

\tan30^o=\dfrac{\sqrt3}{3}

substitute:

\dfrac{17}{x}=\dfrac{\sqrt3}{3}       cross multiply

x\sqrt3=(3)(17)     multiply both sides by √3

x(\sqrt3)(\sqrt3)=51\sqrt3

3x=51\sqrt3      divide both sides by 3

x=17\sqrt3

Use the Pythagorean theorem:

y^2=(17\sqrt3)^2+17^2\\\\y^2=289(\sqrt3)^2+289\\\\y^2=289\cdot3+289\\\\y^2=867+289\\\\y^2=1156\to y=\sqrt{1156}\\\\y=34

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

Other method.

30^o-60^o-90^o triangle.

The sides are in the ratio 1:2:\sqrt3\to17:y:x

Therefore

17:(2\cdot17):(17\sqrt3)\to17:34:17\sqrt3\to x=34,\ y=17\sqrt3

 


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