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3 years ago

Hii please help i’ll give brainliest

Mathematics

2 answers:

Scrat [10]3 years ago

6 0

Answer:

-19

Step-by-step explanation:

lys-0071 [83]3 years ago

5 0

Answer:

-19

hope it helps. Plz mark brainliest!

explanation:

To find its opposite, change the negative sign to a positive sign. The answer is that the opposite of 19 is -19.

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Jonathan and Byron both have comic book collection the number of books in the collection form a ratio 4:9 of Jonathan's to Byron

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3 years ago

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Step-by-step explanation:

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3 years ago

Verify the identity cos (pi/2-x)= tanx

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3 years ago

In a large school, it was found that 77% of students are taking a math class, 74% of student are taking an English class, and 70

Iteru [2.4K]

Answer:

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

Step-by-step explanation:

We solve this question working with the probabilities as Venn sets.

I am going to say that:

Event A: Taking a math class.

Event B: Taking an English class.

77% of students are taking a math class

This means that $P(A) = 0.77$

74% of student are taking an English class

This means that $P(B) = 0.74$

70% of students are taking both

This means that $P(A \cap B) = 0.7$

Find the probability that a randomly selected student is taking a math class or an English class.

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

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

$P(A \cup B) = 0.77 + 0.74 - 0.7 = 0.81$

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

Find the probability that a randomly selected student is taking neither a math class nor an English class.

This is

$1 - P(A \cup B) = 1 - 0.81 = 0.19$

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

6 0

3 years ago