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Novay_Z [31]
3 years ago
6

5) help plz I need help I don't know if I am right so if I am not can you plz help me

Physics
2 answers:
seraphim [82]3 years ago
7 0

the answer is waning Gibbous

The Waning Gibbous is an intermediary Moon phase. It starts right after the Full Moon, and it lasts until the Third Quarter.

inessss [21]3 years ago
7 0

Answer:

the answer is waning gibbous

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Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v
vladimir1956 [14]

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P(X\ge 1) = 0.9502

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

\lambda = density

\lambda = \frac{3}{10}

\lambda = 0.3

T = the light years

T = 10

Calculating P(X \ge 1)

In probability:

P(X \ge 1) = 1 - P(X = 0)

Calculating P(X=0)

Substitute 0 for k and the values for \lambda and T in

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}

P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}

P(X=0) = (3)^0 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-3}

P(X=0) = e^{-3}

P(X=0) = 0.04979

Substitute 0.04979 for P(X=0) in P(X \ge 1) = 1 - P(X = 0)

P(X\ge 1) = 1 - 0.04979

P(X\ge 1) = 0.95021

P(X\ge 1) = 0.9502 ---  approximated

<em>Hence, the required probability is 0.9502</em>

3 0
3 years ago
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