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3 years ago

Typically a point in a three dimensional Cartesian coordinate system is represented by which of the following

Mathematics

2 answers:

Anon25 [30]3 years ago

7 0

Answer:

B. (x,y,z).

Step-by-step explanation:

We represent the coordinate of a point in one dimension as x on the line. We represent the coordinate of a point in two dimension (plane) as (x,y). Similarly we represent the coordinate of a point which lie in the space (three dimension) as (x,y,z) .Here x is the x-coordinate of the point,

y is the y-coordinate of the point,

and z is the z-coordinate of the point. Hence (x,y,z) represent a point a point in a three dimensional Cartesian System.

tigry1 [53]3 years ago

3 0

Answer: B (x, y, z)

<u>Step-by-step explanation:</u>

In a two-dimensional plane, a coordinate is represented as (x, y).

In a three-dimensional plane, a coordinate is represented the same as the two-dimensional plane, except we need to add the third coordinate (z).

--> (x, y, z)

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3 years ago

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

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

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

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

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

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