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## More answers

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- anonymous

help me!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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help me!

- schrodinger

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- anonymous

- anonymous

is this one stems or flower..?

- anonymous

flower

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- anonymous

and this one would be stem?
Buds are found on the (1 point)flowers.
leaves.
stems.
stems and roots.

- anonymous

The police department is keeping track of distracted drivers and accidents. They have found that if a driver is distracted, the driver has a 30% chance of being in an accident. If the driver is not distracted, the driver has a 2% chance of being in an accident. The probability of a driver being distracted is 10%. If needed, create a tree diagram on a separate piece of paper. Then use the diagram to answer the questions.
a. What is the probability a driver will be in an accident? Explain.
b. What is the probability that a driver who was in an accident was distracted? Explain.

- anonymous

- jim_thompson5910

Draw out a tree to get this
|dw:1364249812316:dw|

- jim_thompson5910

Now find the probability of each branch
so for instance the first uppermost branch would be 0.1*0.3 = 0.03
|dw:1364249966115:dw|

- jim_thompson5910

To find the probability of being in an accident, add up the accident probabilities to get
0.03+0.018 = 0.048

- jim_thompson5910

What is the probability that a driver who was in an accident was distracted?
P(Distracted|Accident) = P(Distracted and Accident)/P(Accident)
P(Distracted|Accident) = ???

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