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zheka24 [161]
2 years ago
12

Help me ASAP! Will give brainliest to whoever answers these correctly.​

Mathematics
1 answer:
hammer [34]2 years ago
5 0
#9. There are 180 degrees in a triangle. We already know two of the sides which added together make 112. Then we do, 180-112=68

#10. We do the same process. 90 + 60 = 150
150-180= 30

#11. To find whether or not any three numbers can make a triangle, two of the numbers must equal more than the remaining number. But this has to be the case for any combination. 6+7=13. 13 is larger than 11. So yes it can.

#12. 8+2=10 which may be bigger than 6 but, 6+2=8. Which is equal to 8. So no, it cannot make a triangle.

I’m still working out the first question
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5 \times 5 \times 10 = 250
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Step-by-step explanation:

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Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e
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Answer:

0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, then:

P(A \cap B) = P(A)*P(B)

Conditional probability:

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: At least one tag is lost

Event B: Exactly one tag is lost.

Each tag has a 40% = 0.4 probability of being lost.

Probability of at least one tag is lost:

Either no tags are lost, or at least one is. The sum of the probabilities of these events is 1. Then

p + P(A) = 1

p is the probability none are lost. Each one has a 60% = 0.6 probability of not being lost, and they are independent. So

p = 0.6*0.6 = 0.36

Then

P(A) = 1 - p = 1 - 0.36 = 0.64

Intersection:

The intersection between at least one lost(A) and exactly one lost(B) is exactly one lost.

Then

Probability at least one lost:

First lost(0.4 probability) and second not lost(0.6 probability)

Or

First not lost(0.6 probability) and second lost(0.4 probability)

So

P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48

Find the probability that exactly one tag is lost, given that at least one tag is lost (write it up to second decimal place).

P(B|A) = \frac{0.48}{0.64} = 0.75

0.75 = 75% probability that exactly one tag is lost, given that at least one tag is lost

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