PLEASE HELP HURRY!!!

Since you mentioned the fundamental counting principle, I assume you are looking for the total number of possibilities.

If you are picking a month, there are 12 options.
If you are selecting a day of the week, there are 7 options.

If you are doing both, you multiply the two numbers together.
12 x 7 = 84

8 0

3 years ago

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o

timurjin [86]

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 X = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, λt = 8 per hour.

(a)

For t = 1 the average number of aircraft arrival is:

$\lambda t=8\times 1=8$

The probability distribution of a Poisson distribution is:

$P(X=x)=\frac{e^{-8}(8)^{x}}{x!}$

Compute the value of P (X = 6) as follows:

$P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214$

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:

$P(X\geq 6)=1-P(X$

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:

$P(X\geq 10)=1-P(X$

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For t = 90 minutes = 1.5 hour, the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 1.5=12$

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

$SD=\sqrt{\lambda t}=\sqrt{12}=3.464$

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For t = 2.5 the value of λ, the average number of aircraft arrival is:

$\lambda t=8\times 2.5=20$

Compute the value of P (X ≥ 20) as follows:

$P(X\geq 20)=1-P(X$

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:

$P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108$

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0

3 years ago

-4zsquared +zsquared

TEA [102]

-4

Step-by-step explanation:

-4z² + z²

let's assume you have z² and you're owing 4z², you'll pay the z² you have and you'll still owe 4

so that is -4

5 0

3 years ago