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3 years ago

Plz help me.

Mathematics

1 answer:

Vlad [161]3 years ago

4 0

Assuming you want to find x?
You know that the angles at C and B have to be the same (6x) because it’s an isosceles triangle (as indicated by the lines going through AC and AB), and angles in a triangle add to 180*

So 3x + 6x + 6x = 180*
15x = 180*
x = 180*/15
x = 12*

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leva [86]

Answer:

a) 0.34 = 35% probability that at least one member of a married couple will vote.

b) 0.7143 = 71.43% probability that a wife will vote

c) 0.0968 = 9.68% probability that a husband will vote

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

This is used for itens B and C. For item a, we treat the probabilities as Venn sets.

A) What is the probability that at least one member of a married couple will vote?

I am going to say that:

Event A: Husband votes.

Event B: Wife votes.

The probability that the husband will vote on a bond referendum is 0.21

This means that $P(A) = 0.21$

The probability that the wife will vote on the referendum is 0.28

This means that $P(B) = 0.28$

The probability that both the husband and the wife will vote is 0.15.

This means that $P(A \cap B) = 0.15$

At least one votes:

This is $P(A \cup B)$ , which is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cup B) = 0.21 + 0.28 - 0.15$

$P(A \cup B) = 0.34$

0.34 = 35% probability that at least one member of a married couple will vote.

B) What is the probability that a wife will vote, given that her husband will vote?

Here, we use conditional probability:

Event A: Husband votes:

Event B: Wife votes

The probability that the husband will vote on a bond referendum is 0.21

This means that $P(A) = 0.21$

Intersection of events A and B:

Intersection between husband voting and wife voting is both voting, which means that $P(A \cap B) = 0.15$

The desired probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.21} = 0.7143$

0.7143 = 71.43% probability that a wife will vote.

C) What is the probability that a husband will vote, given that his wife does not vote?

Event A: Wife does not vote.

Event B: Husband votes.

The probability that the wife will vote on the referendum is 0.28

So 1 - 0.28 = 0.62 probability that she does not vote, which means that $P(A) = 0.62$

Probability of husband voting and wife not voting:

0.21 probability husband votes, 0.15 probability wife votes, so 0.21 - 0.15 = 0.06 probability husband votes and wife does not, so $P(A \cap B) = 0.06$

The desired probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.62} = 0.0968$

0.0968 = 9.68% probability that a husband will vote

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3 years ago