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3 years ago

4y-12x+8=0; Solve for y

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

4 0

$4y-12x+8=0$

$4y = 12x-8$

$y = 3x - 2$

Luba_88 [7]3 years ago

3 0

$4y-12x+8=0\ \ \ \ /+12x\\\\4y+8=12x\ \ \ \ /-8\\\\4y=12x-8\ \ \ \ /:4\\\\y=\frac{12x}{4}-\frac{8}{4}\\\\y=3x-2$

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a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you

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Answer:

0.64 = 64% probability that the student passes both subjects.

0.86 = 86% probability that the student passes at least one of the two subjects

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.

In this question:

Event A: Passing subject A

Event B: Passing subject B

The probability of passing subject A is 0.8.

This means that $P(A) = 0.8$

If you have passed subject A, the probability of passing subject B is 0.8.

This means that $P(B|A) = 0.8$

Find the probability that the student passes both subjects?

This is $P(A \cap B)$ . So

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

$P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64$

0.64 = 64% probability that the student passes both subjects.

Find the probability that the student passes at least one of the two subjects

This is:

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

Considering $P(B) = 0.7$ , we have that:

$p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86$

0.86 = 86% probability that the student passes at least one of the two subjects

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