Answer:
26.31
Step-by-step explanation:
Use distance formula
d =
substitute in points and solve
d =
d =
d =
d =
d = 26.30589288
d = 26.31 rounded
On a car trip, Ben drove 65 miles more than half the number of miles Steve drove. Together they drove 500 miles. How many miles
1 answer:
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Answer:
(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.
(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.
(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.
Step-by-step explanation:
Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.
The arrival rate is, <em>λ</em>t = 8 per hour.
(a)
For <em>t</em> = 1 the average number of aircraft arrival is:
The probability distribution of a Poisson distribution is:
Compute the value of P (X = 6) as follows:
Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.
Compute the value of P (X ≥ 6) as follows:
Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.
Compute the value of P (X ≥ 10) as follows:
Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.
(b)
For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:
The expected value of the number of small aircraft that arrive during a 90-min period is 12.
The standard deviation is:
The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.
(c)
For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:
Compute the value of P (X ≥ 20) as follows:
Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.
Compute the value of P (X ≤ 10) as follows:
Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.