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3 years ago

3. Multiply: (x + 8)(x - 12) Please show work

Mathematics

2 answers:

alexira [117]3 years ago

5 0

Answer:

${x}^{2} + 20x + 96$

Step-by-step explanation:

$(x + 8) \times (x - 12)$

➡️ $x \times (x + 12) + 8(x - 12)$

➡️ ${x}^{2} + 12x + 8x + 96$

➡️ ${x}^{2} + 20x + 96$

Sedaia [141]3 years ago

4 0

Answer:

x^2-4x-96

Step-by-step explanation:

x^2-12x+8x-96

=x^2-4x-96

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Step-by-step explanation:

Given:

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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.

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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:

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

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

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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

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3 years ago