<img src="https://tex.z-dn.net/?f=3%5E%7B3%7D%20%2B%206%282%20%2B%20%5Cfrac%7B3%7D%7B6%7D%20%29" id="TexFormula1" title="3^{3} +

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saveliy_v [14]

3 years ago

6(2 + \frac{3}{6} )" alt="3^{3} + 6(2 + \frac{3}{6} )" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Drupady [299]3 years ago

8 0

Answer:

Step-by-step explanation:

9+6(2+3/6)

9+6(2+1/2)

2+1/2 = 2.5

9+6(2.5)

9+15

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Two sides of a triangle measure 8 cm and 15 cm. Whi[:h could be the length of the third side?

lutik1710 [3]

18 cm is correct .

Step-by-step explanation:

The sum of two smaller sides is greater that the largest side.

It is given that the two sides of a triangle measure 8 cm and 15 cm.

Case 1: Let 8 cm and 15 cm. are smaller side. So,

Third side < 8 + 15

Third side < 23

It means 3rd side must be less than 23

Case 2: Let 15 cm is the largest side.

15 < Third side + 8

15 - 8 < Third side

7 < Third side

It means 3rd side must be greater than 7.

Since only 18 is less than 23 and greater than 7, therefore the possible length of third sides is 18 cm and option 2 is correct.

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Step-by-step explanation:

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Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Alinara [238K]

Answer:

a) $P(X \leq 4) = 0.6289$

$P(X < 4) = 0.4335$

b) $P(4 \leq X \leq 8) = 0.5452$

c) $P(X \geq 8) = 0.0511$

d) 0.2605

Step-by-step explanation:

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

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

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

It is important to know that the variance has the same value as the mean in a Poisson distribution. The standard deviation is the square root of the variance.

In this problem, we have that:

$\mu = 4, \sigma = \sqrt{4} = 2$ .

To help our solution, i am going to find each of P(X = x) from x = 0 to 8[/tex]

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

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733$

$P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.1465$

$P(X = 3) = \frac{e^{-4}*4^{3}}{(3)!} = 0.1954$

$P(X = 4) = \frac{e^{-4}*4^{4}}{(4)!} = 0.1954$

$P(X = 5) = \frac{e^{-4}*4^{5}}{(5)!} = 0.1563$

$P(X = 6) = \frac{e^{-4}*4^{6}}{(6)!} = 0.1042$

$P(X = 7) = \frac{e^{-4}*4^{7}}{(7)!} = 0.0595$

$P(X = 8) = \frac{e^{-4}*4^{8}}{(8)!} = 0.0298$

(a) Compute both P(X ≤ 4) and P(X < 4)

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 = 0.6289$

---------

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335$

(b) Compute P(4 ≤ X ≤ 8).

$P(4 \leq X \leq 8) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1954 + 0.1563 + 0.1042 + 0.0595 + 0.0298 = 0.5452$

(c) Compute P(8 ≤ X)

That is $P(X \geq 8)$ .

The sum of the probabilities must be 1 in decimal. Either X is greater or equal to 8, or X is lesser than 8.

$P(X < 8) + P(X \geq 8) = 1$

$P(X \geq 8) = 1 - P(X < 8)$

From what we have in a) and b)

$P(X < 8) = P(X < 4) + P(4 \leq X < 8) = 0.4335 + 0.1954 + 0.1563 + 0.1042 + 0.0595 = 0.9489$

$P(X \geq 8) = 1 - P(X < 8) = 1 - 0.9489 = 0.0511$

(d) What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation

One standard deviation is 2, and the mean is 4. So, this is:

$P(4 < X \leq 6) = P(X = 5) + P(X = 6) = 0.1563 + 0.1042 = 0.2605$

6 0

3 years ago