3 years ago

-3.25, The greater number is . The number farther from 0 is

Mathematics

1 answer:

astra-53 [7]3 years ago

3 0

25 and the farthest is also 24

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A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f

Vitek1552 [10]

Answer:

$P(X\cap Y\cap Z)=0.05$

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert $P(X)=0.28$

Percentage of audience in second concert $P(Y)=0.42$

Percentage of audience in third concert $P(Z)=0.30$

Audience Percentage at at-least one concert $P(X \cup Y \cup Z)=0.80$

Percentage of audience at first & second concert $P(X \cap Y)=0.10$

Percentage of audience in first & third concert $P(X \cap Z)=0.08$

Percentage of audience in second & third concert $P(Y\cap Z)=0.07$

Generally the equation for probability of attending all concerts $P(X\cap Y\cap Z)$ is mathematically given by

$P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)$

$P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)$

$P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70$

$P(X\cap Y\cap Z)=0.05$

Therefore the probability that a randomly selected audient attended all the concerts

$P(X\cap Y\cap Z)=0.05$

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3 years ago

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swat32

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