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Alex73 [517]
3 years ago
7

............................

Mathematics
1 answer:
Aleks [24]3 years ago
5 0
I'm pretty sure you're supposed to use Pythagorean Theorem, so here goes:
a^2+b^2=c^2
=8^2+b^2=10^2
=64+b^2=100
=b^2=100-64
=b^2=36
=b=6
So, AD=6
Then, 6*2=12
AC=12cm
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The following data summarizes results from 939 pedestrian deaths that were caused by accidents . If one of the pedestrian deaths
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Hello!

The contingency table is attached.

The total of pedestrians is 909, not 939, there are 30 accidents less in the given data.

I've calculated the asked probabilities using a total of 909.

1. The probability that the pedestrian was intoxicated or the driver was intoxicated.

Two events are mutually exclusive when the occurrence of one prevents the occurrence of the other in one single performance of the experiment. (i.e there is no intersection between the events P(A∩B)=0)

When two events are mutually exclusive, the probability of the union of both elements is equal to the sum of their individual probabilities P(A∪B)= P(A) + P(B).

When the events aren't mutually exclusive, the probability of their union is equal to the summary of their probabilities minus the probability of the intersection between these two events P(A∪B)= P(A) + P(B) - P(A∩B)

Where:

∪ is the union of both events, in colloquial language represents "or"

∩ is the union of both events, or intersection of the events, in colloquial language represents "and"

In this case P(P₁ ∪ D₁) = P(P₁) + P(D₁) - P(P₁ ∩ D₁) = \frac{306}{909} + \frac{132}{909} - \frac{81}{909} = 0.393

P₁ and D₁ are not mutually exclusive

2. The probability of the pedestrian was intoxicated or the driver was not intoxicated.

These two events aren't mutually exclusive.

P(P₁ ∪ D₂)= P(P₁) + P(D₂) - P(P₁ ∩ D₂)= \frac{306}{909} +\frac{777}{909} *-\frac{225}{909}= 0.944

I hope it helps!

3 0
3 years ago
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