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3 years ago

12

2 sobre 5 por 7 sobre 8 por 10 sobre 21 PARA MAÑANA PLISS

1 answer:

expeople1 [14]3 years ago

3 0

Answer:I don’t speak Spanish .

Step-by-step explanation:

You might be interested in

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

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

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

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

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

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

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

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

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

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

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

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

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

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

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

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

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

c. Compute P(8≤ X).

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

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

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

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

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

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

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

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

How metric units and place value units are alike and unlike

dem82 [27]

Well metric unit is for measurement and place value units is for numbers

8 0

2 years ago

Read 2 more answers

Solve for x<br> angle given: 32<br> answer choices<br> -64<br> -32<br> -90<br> -58

marishachu [46]

Answer:

58

Step-by-step explanation:

Angle NPO is a right angle, then the angle 32 is equal to angle PNO, so x = 90 - 32 = 58.

3 0

2 years ago

Please help!! Thank you so much! (I hate school) :,)

Romashka-Z-Leto [24]

The first and second one

(7x2)=14+5=19
(28/2)=14+5=19

Have a nice day!:)

5 0

2 years ago

Please help quickly

Airida [17]

Answer:

m∠PQR = 133

Step-by-step explanation:

∠PQR and ∠RQS are supplementary angles, meaning they have a sum of 180°.

3x - 5 + x + 1 = 180

4x - 4 = 180

4x = 184

x = 46

Now we simply substitute 46 in for x:

3(46) - 5 = 133

5 0

3 years ago