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dem82 [27]
3 years ago
11

Evaluate

" align="absmiddle" class="latex-formula"> for a= -6 and b= -2
A) 3

B) 12

C) \frac{4}{3}

D) –12


Please explain your answers.
Thanks! : )
Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
8 0
Hey there! :D

Place the value of a and b in the fraction. 

\frac{a}{b}

a= -6 and b=-2

\frac{-6}{-2}

Divide both by -2. 

\frac{-6/-2}{-2/-2}

\frac{3}{1}

\frac{3}{1} = 3

"A" is the answer.

I hope this helps!
~kaikers
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Answer:

6.56% probability that a real emergency situation exists.

Step-by-step explanation:

We have these following probabilities:

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The problem can be formulated as the following question:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

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Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

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What is the probability of a real emergency situation existing, given that the alarm has sounded.

P(B) is the probability of there being a real emergency situation. So P(B) = 0.004.

P(A/B) is the probability of the alarm sounding when there is a real emergency situation. So P(A/B) = 0.95.

P(A) is the probability of the alarm sounding. This is 95% of 0.4%(real emergency situation) and 2% of 99.6%(no real emergency situation). So

P(A) = 0.95*0.04 + 0.02*0.996 = 0.05792

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P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.004*0.95}{0.05792} = 0.0656

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