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Andreas93 [3]
2 years ago
9

2(r+7) ———— < r+3 -6

Mathematics
1 answer:
zhuklara [117]2 years ago
5 0

Answer:

3 +7> 9-^2 = 47

Step-by-step explanation: <em>i did this</em>

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Answer:

The lowest common denominator or least common denominator (abbreviated LCD) is the least common multiple of the denominators of a set of fractions. It is the smallest positive integer that is a multiple of each denominator in the set. Fractions write with fraction bar / like 3/4 .

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I need help plzzz this is algebra
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Which question is biased?
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A certain shop repairs both audio and video components. Let A denote the event that the next
alexdok [17]

Answer:

0.12 is the required probability.

Step-by-step explanation:

We are given the following in the question:

A: next  component brought in for repair is an audio component

B: event that the next component  is a compact disc player

B \in A

P(A)= 0.6\\P(B) =0.05

We have to evaluate:

P(B|A) = \dfrac{P(B\cap A)}{P(A)}\\P(B|A) = \dfrac{P(B)}{P(A)}\\\\P(B|A) = \dfrac{0.05}{0.6} = 0.12

0.12 is the required probability.

3 0
3 years ago
At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a
laila [671]

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

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According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

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