In analyzing hits by certain bombs in a war, an area was partitioned into 553 regions, each with an area of 0.95 km2. A total of 535 bombs hit the combined area of 553 regions. Assume that we want to find the probability that a randomly selected region had exactly two hits. In applying the Poisson probability distribution formula, P(x)equalsStartFraction mu Superscript x Baseline times e Superscript negative mu Over x exclamation mark EndFraction , identify the values of mu, x, and e. Also, briefly describe what each of those symbols represents. Identify the values of mu, x, and e.

Probability of having two hits in the same region = 0.178

mu: average number of hits per region

x: number of hits

e: mathematical constant approximately equal to 2.71828.

Step-by-step explanation:

We can describe the probability of k events with the Poisson distribution, expressed as:

Being μ the expected rate of events.

If 535 bombs hit 553 regions, the expected rate of bombs per region (the events for this question) is:

For a region to being hit by two bombs, it has a probability of:

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

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