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Answer:
There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.
Step-by-step explanation:
For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinatios of x objects from a set of n elements, given by the following formula.
And is the probability of X happening.
In this problem
15 drivers are randomly selected, so .
A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that .
What is the probability of getting 3 or more who were involved in a car accident last year?
This is .
Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:
In which
So
Finally
There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.
In the parallelogram ABCD, diagonals AC and BD intersect at point E.
According to the definition of parallelogram, opposite sides are equal and parallel to each other. That means, AB = DC
Now as AB and DC are parallel, so according to the property of Alternate Interior Angles, we will get:
∠EAB = ∠ECD and ∠EBA = ∠EDC
Thus , in two triangles ΔABE and ΔDCE, two angles and one side are equal. So, ΔABE and ΔDCE are congruent to each other.
That means, AE = CE and BE = DE
So, AE is congruent to CE and BE is congruent to DE
